12908714 Heat and Mass Transfer R K Rajput (2)

350 Heat and Mass Transfer

Graetz number, G = Pe - r? Grashoff number, Gr = P ~ P g At L3

p2 3. The following similarities for testing of 'heat transfer equipment' must be ensured between

the model and the prototype: (i) Geometric similarity (ii) Kinematic similarity (iii) Dynamic similarity (iv) Similarity of fluid entry conditions (v) Siniilarity of boundary temperature field.

THEORETICAL QUESTIONS 1. What is dimensional analysis? 2. What are the uses of dimensional analysis? 3. Explain the term dimensional homogeneity. 4. Describe the Rayleigh's method for dimensional analysis. 5. Describe Buckingham's method or x-theorem to formulate a dimensionally homogeneous equation

between the various physical quantities effecting a certain phenomenon. 6. What are dimensionless numbers? 7. Discuss the physical significance of the following dimensionless number Re, Nu, Pr, St, Gr. 8. Show by dimensional analysis for forced convection, Nu = (I (Re, Pr) 9. Show by dimensional analysis for free convection, Nu = (I (Pr, Gr)

10. What are the advantages and limitations of 'Dimensional analysis' ? 11. What do you mean by 'Characteristic length or Equivalent diameter' ? 12. For testing of 'heat transfer equipment' which of the similarities must be ensured between the model and

the prototype ?

'4

Forced Convection

7.1. Laminarflow over aflat plate: Introduction to boundary layer - Boundary layer definitions and characteristics - Momentum equation for hydrodynamic boundary layer over a flat plate - Blasius solution for laminar boundary layer flows - Von Karman integral momentum equation (Approximate hydrodynamic boundary layer analysis) - Thermal boundary layer - Energy equation of thermal boundary layer over a flat plate - Integral energy equation (Approximate solution of energy equation). 7.2. Laminar tube flow: Development of boundary layer - Velocity distribution - Temperature distribution, 7.3. Turbulent flow over aflat plate; Turbulent boundary layer - Total drag due to laminar and turbulent layers - Reynolds analogy. 7.4 Turbulent tubeflow. 7.5 Empirical correlations - Typical Examples - Highlights -Theoretical Questions - Unsolved Examples.

A. LAMINAR FLOW 7.1. LAMINAR FLOW OVER A FLAT PLATE

7.1.1. Introduction to Boundary Layer The concept of boundary layer was first introduced by L. Prandtl in 1904 and since then it

has been applied to several fluid flow problems. When a real fluid (viscous fluid) flows past a stationary solid boundary, a layer of fluid '

which comes in contact with the boundary surface, adheres to it (on account of viscosity) and condition of no slip occurs (The no-slip condition implies that the velocity of fluid at a solid boundary must be same as that of boundary itself). Thus the layer of fluid which cannot slip away from the boundary surface undergoes retardation; this retarded layer further causes retardation for the adjacetit layers of the fluid, thereby developing a small region in the immediate vicinity of the boundary surface in which the velocity of the flowing fluid increases rapidly from zero at the boundary surface and approaches the velocity of main stream. The layer adjacent to the boundary is known as boundary layer. Boundary layer is formed whenever there is relative motion

between the boundary and the fluid. Since zo = , the fluid exerts a shear stress on the

boundary and boundary exerts an equal and opposite force on fluid known as the shear resistance. According to boundary layer theory the extensive fluid medium around bodies moving in

fluids can be divided into following two regions: ( i ) A thin layer adjoining the boundary called the boundary layer where the viscous shear

takes place. (ii) A region outside the boundary layer where the flow behaviour is quite like that of an ideal

fluid and the potentialflow theory is applicable.

7.1.1.1 Boundary Layer Definitions and Characteristics Consider the boundary layer formed on a flat plate kept parallel to flow of fluid of velocity

U (Fig. 7.1) (Though the growth of a boundary layer depends upon the body shape, flow over a flat plate aligned in the direction of flow is considered, since most of the flow surface can be approximated to a Jlat plate and for simplicity).


- The edge facing the direction of flow is called leading edge. - The rear edge is called the trailing edge. - Near the leading edge of a flat plate, the boundary layer is wholly laminar. For a laminar

- boundary layer the velocity distribution is parabolic. - The thickness of the boundary layer (6) increases with distance from the leading edge x,

as more and mare fluid is slowed down by the viscous boundary, becomes unstable and breaks into turbulent boundary layer over a transition region.

Laminar --4 Tr?nsi- Turbulentboundary layer -1 boundary layer tlon Y I

1 Fig. 7.1. Boundary layer on a flat plate.

For a turbulent boundary layer, if the boundary is smooth, the roughness projections are covered by a very thin layer which remains laminar, called laminar sublayer. The velocity distribution in the turbulent boundary layer is given by Log law of Prandtl's one-seventh power law.

The characteristics of a boundary layer may be summarised as follows: ( i ) 6 (thickness of boundary layer) increases as distance from leading edge x increases.

(ii) 6 decreases as U increases. (iii) 6 increases as kinematic viscosity (v ) increases.

1 (iv)\,q, = p [f) hence ro decreases as x increases. However, when boundary layer becomes I

turbulent, it shows a sudden increase and then decreases with increasing x. ( v ) When U decreases in the downward direction, boundary layer growth is reduced.

(vi) When U decreases in the downward direction. flow near the boundary is further retarded. boundary layer growth is faster and is susceptible to separation.

(vii) The various characteristics of the boundary layer on flat plate (e.g.. variation of 6, $ or force F) are governed by inertial and viscous forces: hence they are functions of either ux UL - or -. v v

I (viii) 1f UI < 5 x lo5 ... boundary layer is laminar (velocity distribution is parabolic). v ux If - > 5 x 10' ... boundary layer is turbulent on that portion (velocity distribution follows v

Log law or a power law).

(ir) Critical value of at which boundary layer changes from laminar to turbulent depends v on : - turbulence in ambient flow,

Forced Convection 353

- surface roughness, - pressure gradient, - plate curvature, and - temperature difference between fluid and bounary.

(x) Though the velocity distribution would be a parabolic curve in the laminar sub-layer zone, but in view of the very small thickness we can reasonably assume that velocity distribution is linear and so the velocity gradient can be considered constant.

Boundary layer thickness (6) : The velocity within the boundary layer increases from zero at the boundary surface to the

velocity of the main stream asymptotically. Therefore, the thickness of the boundary layer is arbitrarily defined as that distance from the boundary in which the velocity reaches 99 per cent of the velocity of the free stream (u = 0.99U). It is denoted by the symbol 6 This definition. however. nives an approximate value of the boundary layer thickness and hence 6 is generally , " m *

termed as nominal thickness of the boundary layer. The boundary layer thicknpss for greater accuracy is defined in terms of certain mathematical

expressions which are the metsure of the boundary layer on the flow. The commonly adopted definitions of the boundary layer thickness are:

1. Displacement thickness (6*) 2. Momentum thickness (0) 3. Energy thickness (6,).

Displacement thickness (a*) : The displacement thickness can be defined as follws: "It is the distance, measured perpendicular to the boundby, by whi

is displaced on account of formation of boundary layer." Or

"It is an additional "wall thickness" that would have to be added to compensate for the reduction in jlow rate on accounibf boundary layer formation. "

The displacement thickness is denoted by 6*. Let fluid of density p flow past a stationary plate with velocity U as shown in Fig. 7.2.

Consider an elementary strip of thickness dy at a distance y from the plate. Assuming unit width, the mass flow per second through the elementary strip

= oudv . . . ( i ) . . Mass flow per second through the elementary strip (unit width) if the plate were not there

= p U.dy . ...( ii)

Boundary layer ___,

___, Stationary plate 2

Fig. 7.2. Displacement thickness.

Forced Convection Heat and Mass Transfer

Energy thickness is defined as the distance, measured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in K,E. o f the flowing fluid on account of boundary layer formation. It is denoted by 6,

Reduction of mass flow rate through the elementary strip

= p ( U - u ) d y [The difference (U - u) is called velocity of defect] Total reduction of mass flow rate due to introduction of plate

S

= J P ( u - U ) d~ . . . (iii) 0

(if the fluid is incompressible) Let the plate is displaced by a distance S* and ,velocity of flow for the distance 6* is equal

to the maidfree stream velocity (i.e., U). Then, loss of the mass of the fluidlsec. flowing through the distance S*

J ., - 7

Refer Fig. 7.2. Mass of flow per second through the elementaIy strip = pudy K.E. of this fluid inside the boundary layer

1 1 = - 2 rn u2 - - (pudy) u2 - 2

K.E. of the same mass of fluid before 1 entering the boundary Iayer

= 7 (pudy) tJ2

Loss of K.E. through elementary strip 1 1 2 d .,.(i) = - 2 (pudy) u2 - 5(pudy)u2 = ?pu (U -u ) y

= pU6* . . .(iv) Equating Eqns. (iii) and (iv), we get

6

pU6* = j p ( U - u ) d y 0 A

6 1

;. Total loss of K.E. of fluid = I -; pu (u' - u2) dy , - A 0

Let 8, = distance by which the plate is displaced to cornpia te for the reduction in K.E. Then loss of K.E. through 6, of fluid flowing with velocity 0

1 = I (pU6.) u2 ...( ii)

Equating Eqns. (i) and (ii), we have s

IcpusA 2 u2 = J 0 + p u ( u 2 - u 2 ) d y

Momentum thickness (0) : Momentum thickness is defined as the distance through which the total loss of momentum

per second be equal to if it were passing a stationary plate. It is denoted by 8. It may also be defined as the distance, measured perpendicular to the boundary of the solid

body, by which the bounhry should be displaced to compensate for reduction in momentum of the flowing fluid on account of boundary layer formation.

Refer Fig. 7.2. Mass of flow per second through the elementary strip = p u dy Momentudsec of this fluid inside the boundary layer = p u dy x u = pu2 dy Momentum/sec of the same mass of fluid before entering boundary layer = pu Udy Loss of momentudsec = puUdy - pu2dy = pu (U - u ) dy :. Total loss of momentudsec.

S

= J ~ " ( 0 - u ) d ~ . ..(i) 0

u Example 7.1. The velocity distribution in the boundoly layer is given by: 3 = i, where u

is the velocity at a distance y from the plate and u = U at y = 6, 6 being boundary layer thickness. Find:

(i) The displacement thickness, (ii) The momentum r* thickness,

Let 8 = distance by which plate is displaced when the fluid is flowing with a constant velocity U .

Then loss of momentudsec of fluid flowing through distance 8 with a velocity U

= p 8 U 2 . . .(ii) Equating Eqns. ( i ) and (ii), we have (iii) The energy thickness, and (iv) The value of 8 '

u Y Solution. Velocity distribution : - = 11 6 ( i ) The displacement thickness, 6*: II

...( Given)

...I Eqn. (7.111

The momentum thickness is useful.in kinetics.

Heat and Mass Transfer

(ii) The momentum thickness, €I : S

€I = j: (l - :Idy ...[ Eqn. (7.2)] 0

(iii) The energy thickness, 6, :

F (iv) The value of - . €I ' - --

or - - - ---- ' ' I 2 - 3.0 6 - 120 €I - 6/6

Example 7.3. I f velocity distribution in laminar boundary layer over a flat plate is assumed U Example 7.2. The velocity distribution in the boundary layer is given by = to be given by second order polynomial u = a + by + c y , determine its form using the necessary

boundary conditions. -- - -- I y2 6 being boundary layer thickness. Solution. Velocity distribution: u = a + by + cyL 2 6 2 lj2'

The following boundary conditions must be satisfied: Calculate the following: (i) At Y = 0, u = 0

(i) The ratio of displacement thickness to boundary layer thickness . . u = a + by + cy2 O = a + O + O ; . a = O

(ii) The ratio of momentum thickness to boundary layer thickness (ii) ~t y = 6, u = U U = b6 + c6' ...( i) . .

u 3 y I y 2 Solution. Velocity distribution : - = -- - -- ...( Given) du U 2 6 2 6 2 (iii) At y = 6, - = o dy

( i) 6*/6: . . d = - (a + by + cy2) = b + 2cy = b + 2c6 = 0 ... (dl) 6 6

6 . ~ 1 l - - d y = [ l - ? ~ + l d ( :) [ 2 6 2 @ ) d y i

o Substituting the value of b (= - 2 ~ 6 ) from(ii) in (i), we get

U = (- 2c6)6 + cs2 = - 2cs2 + c6' - c6' I

* I

358 Heat and Mass Transfer .

. , Hence form of the velocity distribution is

7.1.2. Momentum Equation for Hydrodynamic Boundary Layer Over a Flat Plate Consider a fluid flowing over a stationary flat plate and the development of hydrodynamic

boundary layer as shown in Fig. 7.3. (a) . In order to derive a differential equation for the boundary layer, let us consider an elemental, two-dimensional control volume (dx x dy x unit depth) within the boundary layer region [enlarged view shown in Fig. 7.3 (b)]. Following assumptions are made:

1. The flow is steady and the fluid is incompressible. 2. The viscosity of the fluid is constant. 3. The pressure variations in the direction perpendicular to the plate are negligible. 4. Viscous-shear forces in the Y-direction are negligible. - - 5. Fluid is continuous both in space and time.

Refer Figure 7.3. (b). Let u = velocity of fluid flow at left hand face AB.

Velocity distribution

_+

Stationary plate

--b

Elemental control volume

U

4 r ' d x B .t b c

(b) '5

Fig. 7.3. Equation of motion for boundary layer.

Forced Convection

au Then, u + - . dx = velocity of fluid flow at the right hand face CD (since the flow velocity ax au

changes in the X-direction at the rate of change given by - and the ax

change in velocity during distance dr will be - . dr . G )I Similarly, let, v = fluid velocity at the bottom face BC,

av then, v + y dy = fluid velocity at the top-face AD.

OY The mass flow rate along X-direction, m, = pu (dy ,X 1) = pu dy ...( 7.4)

The change of momentum of the mass mx along X-direction is given by dM, = m, x change in vebcity in X-direction

The mass flow rate along Y-direction, m, = pv (dr x 1 ) = pvdx ...( 7.6)

The change of momentum of the mass my along Y-direction is given by -

au = pv -. h.dy . ... (7.7)

JY Total viscous force along the X-direction is given by

F, = [(z + 6t) - z] x area au a

= [Ips& + 3-(.-$).dy) - p$] ( h x 1)

a 2 ~ = p ---i- a~ . dr.dy . . .(7.8)

Assuming the gravitational forces are balanced by buoyancy forces for equilibrium of the element, we have

Inertia forces = viscous forces au au azu

:. p u - . h . d y + p v - . h . d y = p 5 . d x . d y ax a~ a~

or au au p azu u - + v - = - S T

ax a~ P ay

or au au a 2 ~ u- + v - = ax ay 'v ...( 7.9)

(substituting v = ' P 1

Equation (7.9) is known as equation of motion or momentum equation for hydrodynamic boundary layer.

7.1.3. Blasius Exact Solution for Laminar Boundary Layer Flows The velocity distribution in the boundary layer can be obtained by solving the equation of

Heat and Mass Transfer Forced Convection

motion for hydrodynamic boundary layer [Eqn. (7.9)]. The following boundary conditions should be satisfied.

(i) At y = 0, u = 0 (ii) At y = 0, v = 0 (iii) At y = =, u = U The Blasius technique for an exact solution of the hydrodynamic boundary layer lies in

the conversion of the following differential equations into a single differential equation.

I au au a 2 ~ The hydrodynamic equation for boundary layer: u - aX"5"p ...( Eqn. 7.9)

Continuity equation: du dv - + - = o ax ay Here f is abbreviated as f (q) - Prandtl suggested that the solution of Eqn. (7.9) can be obtained by reducing the number

of variables with the help of magnitude analysis of the boundary layer thickness and transforming the partial differential equation into ordinary differentials.

The inertia forces represented by the left terms, in the Eqn. (7.9), must be balanced by the viscous forces represented by the right terms.

As u 2 v, therefore, we may write as Now, - a ' - . - ( - a & a & ? ? I . u 8f ) ( ) a ay a y m h h a ~

Similarly, @u fl d"f ...( 7.18)

Again,

au u Also as u = U and - = - along A plate length L, therefore, we have ax L'

1

I I From experiments it has been o b s e r v e W velocity profiles at different locations along the

plate are geometrically similar, i.e., they differ only by a stretching factor in the Y-direction.

i u This implies that the dimensionless velocity - can be expressed at any location x as a function U

1 of the dimensionless distance from the wall 1 6'

Substituting the value of 6 from Eqn. (7.1 1) in Eqn. (7.12), we obtain, r 1

where, q = y g d e n o f e s the stretching factor.

In order to account for the fact that the vertical camponent of velocity occurs in the boundary layer equation of motion (7.9)' it is essential to define a stream function \y such that, - -. au au azu

Inserting the values of u, - - - and v from Eqns. (7.15). (7.16), (7.17), (7.18) and a x 9 ay' a2 (7.19) in Eqn. (7.9), we get

- V 8f vx ' 4 3 or J7.14 (a)]

The continuous stream function is the mathematical postulation such that its partial differential with respect to x gives the velocity in the Y-direction (generally taken as negative) and, its partial differential with respect to y gives the velocity in the X-direction:


or 2f"'+ ff" = 0 ...( 7.20) which is an ordinary (but non-linear) differential equation for f. The number of primes on f denotes the number of successive derivatives off (q) with respect toy. The physical and transformed boundary conditions are:

Physical boundary conditions Transformed boundary conditions

(ii) At y = 0, v = O At ?l = 0. f = O

(iii) At y = w u = U A t q = w , g= f J = I 4

Fig. 7.4. Velocity distribution in boundary layer on flat plate by Howarth

The numerical solution of Eqn. (7.20) with the corresponding values of u and v are plotted in Fig. 7.4, and the results are listed in table 7.1.

The following results are of particular interest:

v 1. The single curve I1 shows the variation of normal velocity - It is to be noted that at the u '

outer edge of the boundary layer where q + -, this does not go to zero but approaches the value


Table 7.1. Laminar boundary layer solution for a flat plate

2. The graphlcurve I (i.e., the velocity distribution parallel to the surface) enables us to calculate the parameters : (i) Boundary layer thickness, 6 and (ii) skin friction coefficient,

c,. (i) Boundary layer thickness, 6:

The boundary layer thickness 6 is taken to. be the distance from the plate surface to a point U

at which the velocity is within 1% of the asymptotic limit, i.e., - U = 0.99; it occurs at q = 5.0

(Fig. 7.4). Therefore, the value of q at the edge of boundary layer O, = 6) is given by - 7

where Ux

Re, = - v is the local Reynolds number based on distance x from the leading

edge of the plate. (ii) Skin friction coefficient; Cf: The skin friction coefficient (Cf) is defined as the ratio of shear stress zo at the plate to

I the dyMmic head 5 p ~ ' caused by free stream velocity. Thus the local skin friction coefficient

Cfi at any value of x is

Heat and Mass Transfer Forced Convection 365

To Cfx = - = 1 1 ...( 7.23) u 2 p2

From Fig. 7.4, the gradient at 11 = 0 is

Let ABCD be a small element of a boundary layer of the boundary layer).

(the edge DC represents the outer edge

layer + flat plate

Fig. 7.5. Momentum equation for boundary layer by Von Karman.

Mass rate of fluid entering through AD

0

Mass rate of fluid leading through BC

= ipudy o 2 l P u d y J h

:. Mass rate of fluid entering the control \ -1ume through the surface DC The average value of the skin friction coefficient Zf can be determined by integrating the

local skin friction coefficient Cfx from x = 0 to x = L (where L is the plate length) and then dividing the integrated result by the plate length = mass rate of fluid through BC - mass rate of fluid through AD

The fluid is entering through DC with a uniform velocity U. where ReL is the Reynolds number based upon total length L of the plate.

7.1.4. Von Karman Integral Momentum equation (Approximate Hydrodynamic Boundary Layer Analysis)

Momentum rate of fluid entering the control volume in x-direction through AD 6

= I pu2 dy

Since it is difficult to obtain the exact solution of hydrodynamic boundary layer [Eqn. (7.9)] even for as simple geometry as flat plate (moreover the proper similarity variable is not known or does not exist. for many practical shapes), therefore, a substitute procedure entailing adequate accuracy has been developed which is known as "Approximate lntegral Method" and this is based upon a boundary-layer momentum equation derived by Von Karman.

- - --

Momentum rate of fluid leaving the control volume in x-direction through BC 8 r 6 1

Momentum rate of fluid entering the control volume through DC in x-direction r 6 1

von Karman suggested a method based on the momentum equation by the use of which the1 growth of a boundry layer along a flat plate, the wall shear stress and the drag force could be determined (when the velocity distribution in the boundary layer is known). Starting from the beginning of the plate, the method can be used for both laminar and turbulent boundnry layers

Figure 7.5, shows a fluid flowing over a thin plate (placed at zero incidence) with a free stream velocity equal to U. Consider a small length dr of the plate at a distance x from the leading edge as shown in Fig. 7.5 (a); the enlarged view of the small length of the plate is shown in Fig. 7.5 (b) . Consider unit width of plate perpendicular to the direction of flow.

(': velocity = I / )

? L"

:. Rate of change of momentum of control volume


= momentum rate of fluid through BC - momentum rate of fluid through AD - momentum rate of fluid through DC.

d ( p is constant for incompressible fluid)

0

As per momentum principle the rate of change of momentum on the control volume ABCD must be equal to the total force on the control volume in the same direction. The only external force acting on the control volume is the shear force acting on the side AB in the direction B to A (Fig. 7.56). The value of this force (drqg force) is given by

A F D = z 0 x d r Thus the total external force in the direction of rate of change of momentum

= - , r 0 x d x ...( 7.27) Equating the Eqns. (7.26) and (7.27), we have

But 6 ! ( 1 - $) dy = momentum thickness (8)

...(7. 28)

...[ 7.28 (a)]

Euation (7.28) is known as Von Karman momentum equation for boundary layerflow, and is used to find out the frictional drag on smooth flat plate for both laminor and turbulent boundary layers. Evidently, this integral equation (7.28) u expresses the wall shear stress,% as a function of

the non-dimensional velocity distribution -a u is a point velocity at the boundary layer and U U' is the velocity at the outer edge of boundary layer.

The following boundary conditions must be satisfied for any assumed velocity distribution du

(i) At the surface of the plate: y = 0, u = 0, - dY = finite value

(ii) At the outer edge of boundary layer: y = 6, u = U

The shear stress zo for a given velocity profile in laminar, transition or turbulent zone is obtained from Eqn. [7.28 (a)] or [7.28 (b)]. Then drag force on a small distance dr of a plate is given by

A FD = shear stress x area = zo x (B x dx) = zo x B x dr (where B = width of the plate)

1 :. Total drag on the plate of L length L one side.

F D = j ~ ~ D = ] b ~ ~ ~ & o ..(7.29)

1 - The ratio of the shear stress zo to the quantity - pu2 is known as the "Local coefficient 2

of drag" (or coeflcient of skin friction) and is denoted by Cfr

1 - The ratio of the total drag force to the quantity - ~ A U ~ is called 'Average coefficient of 2

drag' and is denote by CD. - - FD

1. e., Cf = 7 ...( 7.31) - PA u2 2

where p = mass density of fluid, A = area of surfacelplate, and U = free stream velocity.

It has been observed through experiments that for laminar boundary layer, the velocity distribution is parabolic and the velocity profiles at different locations u along the plate are

geometrically similar. This means that the dimensionless velocity - can be expressed at any u locatjon x as a function of the dimensionless distance from the wall, 6'

...[ 7.28 (b)] \ , . ,

, The constants can be evaluated by using the following boundary conditions:

Forced Convection Heat and Mass Transfer

a 2 ~ (i) At y = 0 (wall surface), u = U and - = 0

a9 au

(ii) At y = 6 (outer edge of the boundary layer), u = U and - = 0 ay

By applyiug boundary conditions the constants are evaluated which gives the velocity profile as:

, , . , In order to determine the boundary layer thickness and average skin-friction coefficient for

laminar flow over a flat plate, let us now use this above velocity profile in the Von Karman U integral equation. Now putting the value of - in Eqn. (7.28), we get u

or 39 d6 To = - pu2 - ...( 7.34) 280 du

Newton's law of viscosity, at solid surface, gives

= ($Izo 3

d 3 Y = 4% p ( 6 ) - 3$) llv=o

or 70 = 3CLLI 26 ...( 7.35)

From equations (7.34) and (7.35), we have 39 6 3pU % pu2 dx - 26

or = @.%fx 13 pu

Since, 6 is a function of x only, integration yields

B B+c - - - 2 1 3 p U

By using the boundary condition 6 = 0 at x = 0, we obtain the integration constant C = 0.

@-B or 62= 7 - 140x 2 x~ . . 2 1 3 p U 13 PU

This can be expressed in the non-dimensional form as

...( 7.36) X

x U where Re, = is the Reynolds number based on distance x from the leading edge of the plate.

U r-

Further, in order to estimate the value of zo, substituting the value of 6 Eqn. (7.36) in Eqn. (7.39, we get

20 = 3CLI/ + 4A4 on simplification, we get 6 K'

p@ 0.646 =o- 2 7iq'

Therefore, the local skin friction coefficient.

Average value of skin friction coefficient, 1 0.646 dx - L j L cfx*=- J - - E f - 0 L o d $ z i f i

where ReL = k@! is Reynolds number based on total length L of the flate. u

J

Mass pow through boundary layer The mass flow rate per unit width through the boundary layer is calculated assuming parabolic

velocity distribution in the boundary layer as 3 1

-- 370 Heat and Mass Transfer , Forced Convection

Example 7.4. The velocity profile for laminar boundary is in the form given below:

Find the thickness of boundary layer at the endof the plate 1.5 m long and 1 m wide when placed in water flowing with a velocity of 0.12 d s . Calculate the value of coefJicienr of drag also.

Take p for water = 0.001 Ws/rn2

U Solution. Velocity distribution - - u - 2 6 ) - br

The length of the plate, L = 1.5 m The width of the plate, B = 1 m Free stream velocity, U = 0.12 m/s p for water = 0.001 Ns/m2

Thickness of the boundary layer, 6: Reynolds number at the end of the plate (i.e., at a distance o

is given by

...( Given)

f 1.5 m from the leading edge)

Since ReL < 5 x lo5, therefore, this is case of laminar boundary layer. Thickness of boundary layer at a distance of 1.5 m is given by

Coefficient of drag Cf :

Example 7.5. Air is flowing over a smooth flat plate with a velocity of 12 m/s. The velocity profile is in the form:

The length of the plate is 1.1 m and width 0.9mJf laminar boundary layer exists upto a value of Re = 2 x 16 and kinematic viscosity of air is 0.15 stokes, find:

(i) The maximum distance from the leading edge upto which laminar boundary layer exists, and

(ii) The maximum thickness of boundary layer.

U Solution. Velocity distribution:

Velocity of air, U = 12 m/s Length of plate, L = l.lm Width of plate, B = 0.9m Reynolds number upto which laminar boundary exists, Re = 2 x lo5 Kinematic viscosity of air, v = 0.15 stokes = 0.15 x lo4 m2/s

( i ) The maximum distance from the leading edge upto which laminar boundary layer exists, X:

Ux or 2 x lo5 = 12 x x

Re, = - v 0.15 x lo4

(ii) The maximum thickness of boundary layer, 6 : For the given velocity profile, the maximum thickness of boundary layer is given by

5.48 x =

Example 7.6. A plate of length 750 mm and width 250 mm has been placed longitudinally in a stream of crude oil which flows with a velocity of 5 m/s. I f the oil has a specific gravity of 0.8 and kinematic viscosity of 1 stoke, calculate:

( i ) Boundary layer thickness at the middle of plate, (ii) Shear stress at the middle of plate, and

(iii) Friction drag on one side of the plate. Solution. Length of the plate, L = 750 mm = 0.75 m

Width of the plate, B = 250 mm = 0.25 m

Velocity of oil, U = 5 mls

Specific gravity of oil = 0.8 Kinematic viscosity of oil, v = 1 stoke = 1 x lod4 m2/s

( i ) Boundary layer thickness at middle of the plate 6:

Reynold number,

( '.' At the middle of plate, x = 0.792 = 0.375 m)

Since Re, < 5 x lo5, therefore, boundary layer is of laminar character and Blasius solution gives

(ii) Shear stress at the middle of plate, T,: According to Blasius, the local coefficient of drag (C,J is given by

By definition,

3x2 Heat and Mass Transfer

(iii) Friction drag on one side of the plate, FD:

As the boundary layer is laminar even at the trailing edge, therefore, the average drag (friction) coefficient,

:. Firction drag, 1 FD = x - U2 x area of plate on one side

-' 2 P I

1 = 6.858 x x - x (0.8 x 1000) x 52 x 0.75 x 0.25 2

= 12.86 N (Note : If velocity profile is not given in the problem, but boundary layer is laminar then

Blasius's solution is used.). Example 7.7. Atnzospheric air at 20°C is flowing parallel to a flat plate at a velocity of

2.8 4 s . Assuming cubic velocity profile and using exact Blasius solution, estimate the boundary layer thickness and the local coeficient of drag (or skin friction) at x = 1.2 m from the leading edge of the plate. Also find the deviation of the approximate solution from the exact solution.

Take the kinenmtic viscosity of air at 20°C = 15.4 x lo4 m2/s Solution. Velocity of air, U = 2.8 m/s

Distance from the leading edge of the plate, x = 1.2 m

Reynolds number

Blasius solution:

Boundary layer thickness, 6 = -& = = 0.01285 m = 12.85 mm 2 18 x 16

Local coefficient of drag, 0.664 = 0.001422 cfx = K = xlOI

Approximate solution (with assumption of cubic velocity profile):

The approximate solution deviates from the exact solution by

Derivation for 6: 12.85 - 11.92 x 100 = 7.34% 12.85

Drviation for

Example 7.8. Air is flowing over a flat plate 5 m long and 2.5 m wide with a v&ocity of 4 m/s at 15°C. I f p = 1.208 kg/m3 alzd v = 1.47 x 1 ~ ' m2/s, wlculate:

( i ) Length of plate over which the boundary layer is laminar, and thickness of thk'boundary layer (laminar). i

(ii) Shear stress at the location where boundary layer ceases to be laminar, andt,,.. : (iii) Total drag force on the both sides on that portion of plate where boundary layer is

laminar.

Solution. Length of the plate, L = 5 m

Width of the plate, B = 2.5 m . . Velocity of air, U = 4 mls

Density of air, p = 1.208 kg/m3

Kinematic viscosity of air, v = 1.47 x 1w5 m2/s

( i ) Length of plate over which the boundary layer is laminar:

Reynolds number uL 4 x 5 = 1 . 3 6 1 ~ 1 @ - Re, = - - V 1 .47x10-~

Hence on the front portion, boundary layer Ux is laminar and on the rear, it is turbulent.

Re, = - = 5 x 1 d v

Hence the boundary layer is laminar on 1.837 m length of the plate. Thickness of the boundary layer (laminar), 6

(ii) Shear stress at the location where boundary layer ceases to be laminar, g :

Local coefficient of drag, = O.CM0939 Cf, = K i F

1 1 b = CfxXgp u ~ = o . w ~ ~ ~ ~ x ~ x I . ~ o ~ ~4~

= 0.00907 ~ / m ~

(iii) Total drag force on both sides of plate, FD 1 :

F ~ = Z X C X - ~ A U ~ f 2

- 1.328 where. Cf = average coefficient of drag (or skin friction) = ,- = 1.878 x l(r3

5 x 105

and A = area of the plate = 1.837 x 2.5 = 4.59 m2 1

. . Fo = 2 x 1.878 x x 5 x 1.208 x 4.59 x 42 = 0.167 N

Example 7.9. Airflows over a plate 0.5 m long and 0.6 wide with a velocity of 4 d s . The

u - = sin (: i) u - If p = 1 . Z l k g / d and v = 0.1 5 x 1 m2/s, calculate:

( i ) Boundary layer thickness at the end of the plate, (ii) Shear stress at 250 mm from the leading edge, and (iii) Drag force on one side of theptate. Solution. Length of plate, L = 0.5 m

Width of plate, B = 0.6 m

Velocity of air, U = .4 mls


Density of air, p = 1.24 kg/m3 Kinematic viscosity of air, v = 0.15 x m2/s

Velocity profile, , u = sin (f 3 ( i ) Boundary layer thickness at the end of the plate, 6:

Reynolds number, ux ' 4 x 0e5 = 1 . 3 3 ~ 10s Rex = - =

V 0.15 x lo4 Since Rex c 5 x loS, therefore, the boundary layer is laminar over the entire length of the

plate.

We know,

= 0.00657 m = 6.57 mm (ii) Shear stress at 250 mm from the leading edge, 5:

...[ Eqn. (7.30)J

But, Cf.= -=-- 0.654 0.654 0.654 = 0.002533

0.15 x

(iii) Drag force on one side of the plate, F, :

where 1.31 CD = = - 0.003592

1.33 x loS - and A = area of the plate = L x B = 0.5 x 0.6 = 0.3 m2

. . 1 FD = 0.003592 x 7 x 1.24 x 0.3 x 4' = 0.01069 N

7.1.5. Thermal Boundary Layer Whenever a flow of fluid takes place past a heated or cold surface, a temperature field is

set up in the field next to the surface. If the surface of the plate is hotter than fluid, the temperature

Limit of thermal boundary layer

Free stream

Y

t r Plate surface (I,)

t --* I+ Laminar Turbulent

Fig. 7.6. Thermal boundary layer formed during flow of cool fluid over a warm plate.


distribution will be as shown in the Fig. 7.6. The zone or this layer wherein the temperature fiefd exists is called the thermal boundary layer. Due to the exchange of heat between the plate and the fluid, temperature gradient occurs/results.

The thermal boundary layer thickness (6,,,) is arbitrarily defined as the distance y from the plate surface at which

It c-4 - " 7 f I I f I ! Temperature profile

Fig. 7.7. Thermal boundary layer formed during flow of warm fluid over a cool plate.

Figure 7.7, shows the shape of the thermal boundary layer when the free stream temperature t, is above the plate surface temperature t,.

The thermal boundary layer concept is analogous to hydrodynamic boundary layer; the parameters which affect their growth are however different. Whereas the velocity profile of the hydrodynamic boundary layer depends mainly on the fluid viscosity. the temperature profile of the thermal boundary layer depends upon the viscosity, velocity of flow, specific heat and thermal conducitivity of the fluid. The relative magnitude of 6 and 6th are affected by the thermo-physical properties of the fluid; the governing parameter, however, is h e non-dimensional Prandtel number,

pr = 7. ( i ) 6th = 6 ........ when Pr = 1; (ii) 4, < 6 ........ P r > 1 (iii) 6 , > 6 ........ when Pr < 1.

7.1.6. Energy Equation of Thermal Boundary Layer Over a Flat Plate Figure 7.8 (a) shows a hot fluid flowing over a cool flat plate, and development of the

thermal boundary layer. In order to derive an energy equation, consider control volume (dr x dy x unit depth) in the boundary layer so that end effects are neglected. The enlarged view of this control volume is shown in Fig. 7.8 (b) in which the quantities of energy entering and leaving have been indicated.

Involving principle of conservation of energy for the steady state condition, we have: Heat energy convected (E,,,,) through the control volume in X and Y directions + heat

energy conducted (Econd,) through the control volume in X and Y directions + heat generated due to fluid friction (viscous heat generation) in the conuol volume = 0.

But as the rate of temperature change in the X-directidn is small and can be neglected the conservation of energy becomes:

Heat energy convected in X and Y directions + heat energy conducted in Y-direction + viscous S t heat generation = 0.

or d(E,,), + d(EcOnv.), + d(ECod), + viscous heat generation = 0 ...( 7.41)

(i) (ii) (iii) (iv)


boundary layer

Plate surface (r,>

Viscous heat (Ec~nv.!v +dy generation . t

(iii) The heat conduction in tbe y-direction:

d(Ecod), = (E~md.)~ - (~cotuf.h+dy at

. ..(7.44)

the element). = [Shear stress (r) x area upon which it acts)]

Fig. 7.8. Energies entering and leaving the control volume.

(i) The Energy convected in X-direction: (Econ,.)x = mass x specific heat x temperature

= [pu(dy x l)] c,t = (pu dy) c,,t

- ... neglecting the product of small quantities . an

(ii) The energy convected in Y-direction: + . The net energy cor+vected in Y-direction, " L

. - x distance travelled

= [,$ (dr x l)] x ($ dy)

Substituting the values in Eqn. (7.411, we get

. . .(7.46)

Form the continuity equation for two-dimensional flow, we have au av -+-=0: thus the Eqn. (7.46) reduces to ---

% ,, at at n a2t p (?= u- + v- = - ... (7.47) ' s + P . , ay ax a~ PC, ay

Equation (7.47) is the requid di#erential energy qUUt0n f0rflDw PaR a fLY p*1te. If heat genemhon is neglected [when the value of U is relatively 10" and difference of temperature bewen the swam and the plate is s d l (of the order of 40°c)1, fhe energ Cqudlon reduces

at k a2t 2 t at +,-=--=a u - &= ...( 7.48)

ax ay PC, a 3 fi may noted, the energy equation is similar to the momentum equation. Further the

dimensions of kinematic viscosity v and thermal diffusivity a are the same. The auation (7.48) has been derived with the following assumptiom:

1. Steady incompressible flow. t, - ts - - are constant.

2. m e properties of the fluids evaluated at the film temperature $ - 2


3. The body forces, viscous heating and conduction in the flow direction are negligible.

Pohlhausen solution for the 'Energy equation':

at at By using the following variables the energy equation u - + v - = a- can be recast ax JY ay2

into an ordinary differential equation as follows: "1 .

(Stretching factor) = Y 45, yf (StreaIn function) = f (ll), and

Also, the values of the &locity components u and v already calculated earlier are:

u = u & 4 ...[ Eqn. (7.15)l .

... [Eqn. (7.19)J 4

Further, from temperature parameter 0 (non-dimensional) defined above. we have

and

or

Also,

t = t,# + (t, - t,) e at - - ae

ax ax - (t- - tJ - = (tm - t#) -

atl ax

at ax J7.50)

at - - ae ay

" , h - 0,. - ts) - = (t- - t,) - JY atl ay

at

or - - 8 t tJ&e $2 - (1.. - 4) -

vxdq2 bsening the above values in the energy equation, we get:

at at a2t u - + v - = a - ax a~ ay2

U d28 = a (t, - t,) - -Z vx 4 ...( 7.53)

After simplification and arrangement of the above equation, we obtaL

Forced Convection

.' Pr (Randtl number) =

Thus the partial differential equation (7.48) has been converted into ordinary differential equation. The boundary conditions to the satisfied are:

At t = t,, y - 0 At t = t,, y = m

At q = 0, e(q) = ) values in terms of new variable ...( 7.55) ~t q = - em) = 1 The solution obtained by Pohlhausen for energy equation is given by:

de The factor fr) represents the dimensionless dope of the temperature profile at the

\Wh=o surface where q = 0; its value can be obtained by applying the boundary condition at r\ = w,

801) = 1. Thus,

. ,,

Evidently the dimensionless slope is a function of Prandtl number and the calculations made by Prandtl gave the following result :

For 0.6 < Pr 15, [ = 0.332 (PI)'" =o

Figure 7.9 shows the values of 0 (dimensionless temperature distribution) plotted for various values of Pr (Prandtl number).

- The curve for Pr = 0.7 is typical for air and several other gases. - The curve for Pr = 1 is the same as that of curve I in Fig. 7.4. - These curves also enable us to determine the thickness of thermal boundary layer 6,h and,

local average heat transfer coefficients h. Thermal

Fdrced Convection Heat and Mass Trapfpr

Thickness of thermal boundary layer, 6, : Case I: When Pr = 1. - -

in Eqn. (7.63). we obtain Substituting for - : \'I'.

k 1i2 pr)in = hx i t s - tJ = 0.332 ; (t, - tm)(Rex)

A k

hx = 0.332 ; ( ~ e ~ ) l " (pr)'" ...( 7.64

or

Since y = 6, at the outer edge of thermal boundary layer, therefore, 7

h2-X 112 pr)ln NU, = - = 0.332 (Re,) ( ...( 7.65)

or k ... (In non-dimensional form) This equation shows that for Pr = 1, the thickness of thermal boundary layer, 6, is equal

[where hx = local convective heat transfer coefficient;

Nu. = local value of Nusselt number (... at a distance x from the leading edge

to hydrodynamic boundary layer, 6. Case 11. When Pr < 1.

,- -7 A

of the plate].

The average heat transfer coefficient is given by

This equation shows that for Pr < 1, 6, > 6 Case 111. When Pr > 1.

\ /

If we compare the Eqns. (7.64) and (7.66), we find that - h = 2hx

This equation shows that for Pr > 1, ath < 6 Pohlhausen has suggested that the following relation is general may be between ~ 1 1 the results in Eqns. (7.64), (7.65) and (7.68) valid for Pr > 0e5.

\the thermal and hydrodynamic boundary layers: 7.1.7. lote& Energy aquation (Approximate ~olution of enera equation) 6 6, = - ... (7.62) Consider a volume shown in Fig. 7.10. Assume that p, cp and (thermo-plastic

(PrV3 Hydrodynamic 7 boundary layer . ,

The local and average heat transfer coefficients: At the surface of the plate, since there is no fluid motion and the heat transfer can occur

u, t , only through conduction, the heat flux may be written as

Qh Thermal boundary I

B I f I--- #"#-- ' I

I - I I

Control I .mlnrne I

4 From the relation 7.63, we may develop (i.e., surface temperature gradient) as

- 4

~-xoL .__ ) l Q c k e n - 4

Thin flat plate

I+----x -. _ rr 9 n 1 , + - ~ - ~ 1 ener~V eauation - control volume.

382 Heat and Mass Transfer. ,

propertids) of fluid remain constant within the operating range of the temperature, and the heating of the plate commences at a distance xo from the leading edge of the plate (so that the boundary layer initiates at x = xo and develops and grows beyond that). For unit width of the plate we have:

H Mass of fluid entering through face AB = jo pu dy ...( 7.69)

H a " Mass of fluid leaving through face CD = 4 pu dy + [Io pu d y ] d r ...( 7.70)

:. Mass of fluid entering the control volume through face BC

= [jOH Pu dy + $ {joH pu dy) dx] - joH pu dy = $ [joH pu dy] dx -47.71;

Heat influx through the face AB, Q, = mass x specific heat x temperature

Heat efflux through the face CD,

Heat (energy) influx through the face BC (which is outside thermal boundary layer and there the temperature is constant at t,),

Heat conducted into the control volume through face AD,

The energy balance for the element is given by

After simplification and rearrangement, we have

Equation (7.76) is the integral equation for the boundary layer for constant properties and constant free stream temperature t,.

If the viscous work done within the element is considered, then Eqn. (7.76) becomes

[when d r dy = viscous work done within the element ... Eqp. (7.8)] P C , 0 ay2

Usually the viscous dissipation term is very small and is neglected (and may be considered only when velocity of flow field becomes very large).

Exp-ion for the convective heat transfer coefficient for laminar flow over a flat blab: In order to derive an expression for convective heat transfer coefficient for laminar flow

Forced Convection

over a flat plate (that has an unheated starting length xO), let US use cubic velocity and temperature distributions in the integral boundary layer energy equation as follows:

he cubic velocity profile within the boundary layer is of the fom; ...[ Eqn. (7.33)l

U

The conditions which are satisfied by the temperature distribution within the boundary layer are:

(i) At y = 0, t = ts

- , r

W e temperature distribution takes the following fom:

-=-- ... (7.79)

B,, putting the proper values of velocity distribution and temperature distribution into the

integrind would be zero). 6th

After putting a = and earlying out the integration, Eqn. (7.80) gets reduced ... (7.81)

\-,, -

Neglecting the term involving r4 (because c 6, r < 11, we have

3 d = - 20 U (t, - t,) 5 (62)

Fu~her , from Eqn. (7.77). we kave

386

Heat and Mass Transfer . Forced Convection , (v) Thickness of the boundary layer,

(vi) Local convective heat transfer coeflcient, (vii) Average convective heat transfer coeflcient,

(viiil Rate of heat transfer by convection, (k) Total drag force on the plate, and (x) Total mass flow rate through the boundary.

Solution. Given: U = 3mls. x = 280 mm = 0.28 m, p = 1.1374 kg/m3, k = 0.02732 W/m°C, cP = 1.005 W g K , v = 16.768 x 10d m2/s.

- (vii) Average convective heat transfer eoeff~cient, h:

/

(viii) Rate of heat transfer by convection, Qco,,: -

Let us first amxtain the type of the flow, whether laminar or turbulent.

ux - Re, = - - = 5.0 x 104 V 16.768 x lo4

Since Re, < 5 ' ~ lo5, hence flow in laminar.

(ix) Total drag force on the plate, FD: FD = q, x area of plate on one side upto 0.28 m

= 0.01519 x 0.28 x 0.28 = 0.00119 N (Ans.)

(x) Total mass flow 5 rate through the boundary, m :

m = - pU (6, - 61) 8

(0 Boundary layer thickness at x = 0.28 m, 6:

or 5 x 0.28 6 = 7 = 0.00626 m or 6.26 mm 5 x lo4

(2) Local friction coefficient, C'.: x 1.1374 x 3(0.00626 - 0) = 0.01335 kgls (Ans.)

= 8 Example 7.11. Air at atmospheric pressure and 200°C flows over a plate with a velocity

of 5 d s . The plate is 15 mm wide and is maintained at a temperature of 120°C. Calculate the thicknesses of hydrodynamic and thermal boundaiy layers and the local heat transfer coefficient at a distance of 0.5 m from the leading edge. Assume that /low is on one side of the plate. or 0.664 4 = - = 0.002969

5 x lo4 (iii) Average friction coefficient, C'

p = 0.815 kg/rn3; p = 24.5 x 1 r 6 iVs/m2, Pr = 0.7. k = 0.0364 W/m K . (AMIE Summer, 1997)

Pr = 0.7; k = 0.0364 W / m K.

Let us first ascertain the type of flow, whether laminar or tubulent. or - = 0.005939

c f = x = z F (iv) Shearing stress due to friction, To :

Since Re, < 5 x Id, hence flow is laminar.

Boundary layer thickness at x = 0.5 m, 6 : = 0.002969 x 32 = 0.01519 N / ~ ~ 2 (v) Thickness of thermal boundary layer, &,, :

5x 5 x 0.5 6 = - = - = 8.669 x lo-) m or 8.669 mm (Ans.) f ie" 4iZE

Thickness of thermal boundary layer, at x = 0.5 m, St.: 6 6, = - (Pr) I r n ...[ Eqn. (7.62) s t h = - - = 8.669 9.763 mm (Ans.)

( P ~ ) ' / ~ - (0.7)"~ - 0.00626 - - = 0.00705 m or 7.05 mm (0.7)'"

(vi) Local convective heat transfer coefficient, 4 : Local heat transfer coefficient, h, : . . . [Eqn. (7.64)l

k hx = a332 - (Re,)'" (pr)'" X = 0.332 x x (83l63)ln x (0.7)'" = 6.189 w/m2 K (Ans.) ...[ Eqn. (7.64)]

-.- Example 7.12. Air at atmospheric pressure and 40°C flows with a velociry of (I= 5 m/s

over a 2 m long f k ~ plde whose surface is kept at a uniform temperature of 120°C. Determine

Heat and Mass'Transfer

the average heat transfer coeflcient over the 2 m length of the plate. Also find out the rate of heat transfer between the plate and the air per I m width of the plate. [Air at 1 arm. and 80°C,

v = 2.107 x 10- m2/s, k = 0.03025 W/m K, Pr = 0.69651 (AMIE Winter, 1998) Solution. Given : t, = 40°C; U = 5 m/s; L = 2 m, t, = 120°C; B = 1 m

The properties of air at mean bulk temperature of

v = 2.107 x 10- m2/s; k = 0.03025 W/m K; Pr = 0.6965. Average heat transfer coefficient, h :

Assuming Re,,= 5 x lo5, the flow is laminar.

Using exact solution, the average Nusselt number is given by

= 0.664 (~e,)'" ( ~ r ) - ...[ Eqn. (7.68)]

or hL = 0.664 (4.746 x 1@)ln (0.6965)'" = 405.48 k

Rate of heat transfer, Q :

Q = ZA, ( t , - t,) = 6.133 x (2 x 1 ) (120 - 40) = 981.28 W (Ans.)

Example 7.13. Air at 27°C and I barflows over a plate at a speed of 2 m/s. ( i ) Calculate the boundary layer thickness at 400 mm from the leading edge of the plate.

Find the massflow rate per unit width of the plate.

For air p = 19.8 x 10- kg/ms at 27°C. (ii) If the plate is maintained at 60°C, calculate the heat transferred per hour.

The properties of air at mean temperature of (27 + 6 0 ) d = 43.5' C are given below :

v = 17.36 x 10- m2/s; k = 0.02749 W/m°C

c,=lOOBJ/kgK; R=287Nm/kgmK; Pr=0.7. (M.U.)

Solution. Given : t = 27°C; p = 1 bar, U = 2 m/s; x = 400 rnm = 0.4 m

(i) Boundary layer thickness, 6 :

4.64 x Boundary layer thickness, 6 = - * 4.64 x 0.4

6 = ,-J===- = 0.00857 m or 8.57 mm (Ans.) 46869 The mass flow rate per metre width is given by,

... assumed Now,

5 = z x 1.16 x 2 x 0.00857 = 0.01242 kgh (Ans.) -.

8

Note, ~f h e mass added in the boundary is to be calculated when h e fluid moves from X , 10 + along the main flow direction then it is given by

S Am =: PU (8, - 8,)

where 6 , and l& are the boundary layer thicknesses at X l and q. - (ii) Heat transferred per hour, Q : -

- 15'76 3600 = 416.74 kl/h (Ans.) - 1000

Example 7.14. Air at 1 bar and at a temperature of 30°C (Y. = 0.06717 kghrn) flows at a speed of I.2ds over a fiat plate. Determine the boundary layer thickness at distance of So mm and 500 mm from the leading edge of the plate. Also, calculate the mass entrainment bemeen

these m o sections. Assume the parabolic velocity distribution as: - = -

Solution. Given : t.. = 30°c, )I = 0.06717 kglhm, U = 1.2 mls Boundary layer thicknesses:

E-= 1 x lo5 The density of air,

= 1.15 kg/m3 = RT 287 x (30 + 273)


4 . 6 4 ~ Boundary layer thickness, 6 , = - ... [Eqn. (7.36)]

6 , = 4.64 x 0.25 = 0.00853 m or 8.53 mm (Ans.) .I18490

Re, = 1.15 x 1.2 x 0.5 x 3600 0.067 17 = 36981

.. Boundary layer thickness, 62 = 4.64 x 0.5 = 0.01206 m = 12.06 mm (Ans.) Gzi-

Mass entrainment: The mass flow rate at any position in the boundary layer is given by

:. The mass entrainment between the two sections be., at x = 250 mm and x = 500 mm.

= 3.045 x kg/s = 10.96 kg/h (Ans.) Example 7.15. Air at 20°C is flowing over a flat plate which is 200 mm wide and 500 mm

long. The plate is maintained at 100°C. Find the heat loss per hour from the plate if the air is flowing parallel to 500 mm side with 2m4 velocity. What will be the efect on heat transfer if the flow is parallel to-200 mm side.

The properties of dir at (100 + 20)D = 60°C are: v = 18.97 x 1@ m2/s, k = 0.025 W/m°C and Pr = 0.7.

(M.U.) . --, Solution. U = 2 m/s, v = 18.97 x lo4 m2/s, k = 0.025 W/m°C and Pr = 0.7. Heat loss per hour from the plate, Q: Case I. When the flow is parallel to 500 mm side:

- hL N u = - - - 0.664 (Re,)'" (pr)'" k ...[ Eqn. (7.68)]

- where, Re, = - - uL OS5 = 127 x lo4

V 18.97 x 10"

I I . . Q = ~ s ( t s - t , ) = 6 . 7 6 7 ~ ( 0 . 5 ~ 0 . 2 ) ( l W - 2 0 ) = 5 4 . 1 4 W (Am.) Case 11. When the flow is parallel to 200 mm side:

Re, = 2 x 0.2 = 2.11 x lo4

18.97 x lo6

. . h = - -

''O2' 0.2 x 0.664 x (2.11 x 1 0 ~ ) " ~ (0.7)"~ = 10.7 w/m2~c (Ans.)

. . Q = 3; xA, ( t , " -t,) = 10.7 x (0.2 x 0.5) x (100- 20) = 85.6W (Ans.)

Example 7.16. In a certain glass making process, a square plate of glass 1 m2 area and 3 mm thick heated uniformly to 90°C is cooled by air at 2U'Cflowing over both sides parallel to the plate at 2 m/s. Calculate the initial rate of cooling the plate.

Neglect temperature gradient in the glass plate and consider only forced connection. Take for glass : p = 2500 kg/m3 and cp = 0.67 W / k g K Take the following properties of air : - .

p = 1.076 kg/mJ; cp = 1008 JAcg K, k = 0,0286 W/mO C and p = 19.8 x l(T6 N-$/mL. (N.U., 1997)

Solution. Given : A = 1 m2; t, = 90°C; t, = 2Q°C, U = 2 d s . 7 Plate at 90°C -

Fig. 7.11

Initial rate o f cooling : The average heat transfer co-efficient for the air flow parallel to the plate is given by - - hL

Nu = - K = 0.664 ( ~ e , ) " ~ (pr)'l3 .[Eqn. 7.68)]

(valid for Pr > 0.5)

Substituting the values in the abovgeqn, we get - hL= k 0.664 x ( 1 .087x 10")"~ x (0.698)]/~ = 194.19

The heat flow (Q) from both sides of the platet is given by :

~ = 2 h A ( t , - t J = 2 ~ 5 . 5 5 ~ 1 x ( 9 0 - 2 0 ) = 7 7 7 ~

The heat lost by the plate instantaneously is given by :

Q=mc,,At=777 E

where m = (area x thicknes$ x p

/

Example 7.17. A flat plate, l m wide and 1.5 m Long is to be maintained at 90°C in air with a free stream temperature of 10°C. Determine the velocity with which air must flow over

392 Heat and Mass Transfer - f

flat plate along 1.5 m side so that the rate of energy dissipation from the plate is 3.75 kW. Take the following properties of air at 50°C:

p = 1.09 kg/m3, k = 0.028W/m°C, cp = 1.007 kJ/kg°C, p = 2.03 x I @ kg/m-s Pr = 0.7.

(M.U.) Solution, Given : L = 1.5 m, B = lm, t, = 90°C, t, = 10°C, Q = 3.75 kW

Properties of air at (90 + 10)/2 = 50°C : p = 1.09 kg/m3, k = 0.028 W/m°C, cp = 1.007 kT/kg°C, p = 2.03 x lo-' kg/m-s

Free stream velocity, U: The heat flow from the plate to air is given by . s ' <

Q = j; A, (t, - t,) k where h = - x 0.664 (~e,)'" (pr)ln

L ...[ Eqn. (7.68)]

or U = 100 m/s (Ans.) Example 7.18. Air at 20°C and at atmospheric pressure jlows over aflat plate at a velocity

of 1.8 4 s . If the length of the plate is 2.2 m and is maintained at 100°C, calculate the heat transfer rate per unit width using (i) exact and (ii) approximate methods.

The properties of air at mean bulk temperature of (100 + 20)D = 60°C are:

p = 1.06 kg/m3, cp = 1 .OO5 kJ& OC, k = 0.02894 W/m°C, Pr = 0.696

v = 18.97 x 106 m2/s

Solution. Given : t, = 20°C, t, = 100°C, U = 1.8 mls, L = 2.2 m, B = 1 m Heat transfer rab per unit width:

Reynolds number,

Since Reynolds number is less than 5 x 10' hence flow is laminar. (i) Using exact solution: The average Nusselt number is given by -

Nu = 0.664 (Re, ) I n ( ~ r ) ' " . . . [Eqn. (7.68)J

;. Heat transfer rate from the plate,

Q = A, (t, - t,) = 3.536 x (2.2 x \l) (100 - 20) = 62234 W (Am.)

(ii) Using approximate solution: - = && = 0.646 (Re, )'" (pr)lD

- k

or = 0.646 (2.087 x 1 0 ' ) ' ~ (0.696)'" = 261.53 k - 261.53 k 261.53 X 0.02894 = 3.44 Wlm2 OC

or h=-= L 2.2

r- -

(i) The average skin friction, CI : UL 1.8 x 0.75 = 84375

Reynolds number, - Re, = --$ -

1.6 x lo4

laminar in nature.

(ii) The average shear stress, T,: 1 -

rw = -pu2 X Cf 2

= & x 1.165 x 1.g2 x 0.004572 = 0.008629 ~ l m ' (Ans.) L

(iii) The ratio of average shear stress to the shear stress at the trailing edge: The skin friction coefficient at the trailing edge (x = L),

0.664 Cfx =

... [Eqn. (7.24)J

;. Shear stress at the trailing edge, 1

r , = 5 p d Cgi

= x 1.165 x 1.8~ x 0.002286 = 0.004314 ~ 1 n - i ~ 2

Heat and Mass TrAhsfer

Example 7.20. Air at 3U°CJlows with a velocity of 2.8 m/s over a plate 1000 mm (length) x 600 mm (width) x 25 mm (thickness). The top sugace of the plate is maintained at 90°C. I f the thermal conductivity of the plate material is 25W/m°C, calculate :

( i ) Heat lost by the plate; (ii) Bottom temperature of the plate for the steady state condition.

The thermo-physical properties of air at mean film temperature (90 + 30)Q = 60°C are:

p = 1.06 kg/m3, c,, = 1.005 kJAg K, k = 0.02894 W/m°C, v = 18.97 x I@ m2/s, Pr = 0.696.

Solution. Given : t, = 30°C, ts = 90°C, U = 2.8 d s , kplate = 25 W/m°C, L = 1000 mm = lm, B = 600 mm = 0.6 m, 6 = 25 mm = 0.025m.

( i) Heat lost by the plate: Reynolds number at the trailing edge,

Since Reynolds number is less than 5 x lo5, hence flow is laminar throughout the length,

h~ Nu = 0.664 (ReL )'I2 (pr)'I3 = - k ...[ Eqn. (7.68)]

Nu x k 0.664 ( ~ e , ) " ~ (pr)'l3 x k or Ti (average heat transfer coefficient) = - - L - L

= 6.542 w/m2 "C :. Heat lost by the plate, Q = h A, (t, - t,) or Q = 6.542 x (1.0 x 0.6) (90 - 30) = 235.5 W (Ans.) (ii) Bottom temperature of the plate, t,: Heat lost by the plate Q (calculated above) must be conducted through the plate, hence

exchange from top to bottom surface is

tb = 90 + 0.025 x 235.5 25 (1.0 x 0.6)

= 90.39"C (Ans.)

Example 7.21. Air at 30°C and at atmospheric pressure flows at a velocity, of 2.2 m/s over a plate maintained at 90°C. The length and the width of the plate &re 900 mm and 450 mm respectively. Using exact solution, calculate the heat transfer rate from,

( i ) first half of the plate, (ii) full plate, and (iii) next half of the plate.

The properties of air at mean bulk temperature (90 + 30)Q = 60°C are:

p = 1.06 kg/m3, y = 7.211 kghm, v = 18.97 x I@ m2/s, Pr = 0.696, k = 0.02894 W/m°C.

Solution. Given : t, = 30°Q, U = 2.2 mls, t, = 90°C, L = 900mm = 0.9m, B = 450 mm = 0.45 m.

Forced Convection

( i ) Heat transfer rate from first half of the plate:

For first half of the plate,

Since Re < 5 x lo5 hence the flow is laminar. The local Nusselt number is given by,

Nu, = 0.332 (~e,) '" ( ~ r ) " ' ...I Eqn. 7.651

But,

Average heat transfer coefficient, = 2h, = 2 x 4.322 = 8.644 wlm' "C

:. Heat transfer rate from first half of the plate, Q = h A, (t, - 2,)

= 8.644 x (0.45 x 0.45) (90 - 30) = 105 W (Ans.)

(ii) Heat transfer rate from full plate: For full plate, x = L = 0.9 m

The heat transfer rate - from entire plate, Q, = h A, (t, - t,) = 6.1 13 x (0.9 x 0.45) x (90 - 30) = 148.54 W (Ans.)

(iii) Heat transfer rate from next half of the plate: Heat transfer rate from the next half of the plate

= Q, - Q, = 148.54 - 105 = 43.54 W (Ans.) Example 7.22. Castor oil at 25°C flows at a velocity of 0.1 m/s past a flat plate, in a

certain process. I f the plate is 4.5 m long and is maintained at a un@rm temperature of 9j°C, calculate the following using exact solution:

( i ) The hydrodynamic and thermal boundary layer thicknesses on one side of the plate, (ii) The total drag force per unit width on one side of the plate, (iii) The local heat transfer coeflcient at the trailing edge, and (iv) The heat transfer rate.

The thermo-physical properties of oil at mean film temperature of (95 + 25)R = 60°C are:

p = 956.8 kg/m3; a = 7.2 x I@ m2/s; k = 0.213 W/m°C; v = 0.65 x I @ m2/s.

Solution. Given : t, = 25"C, t, = 95"C, L = 4.5m, U = 0.1 mls.

&g Heat and Mass Transfer Forced Convection

(iii) Local and average convective heat transfer coeflcients; (iv) Heat transfer rate from both sides for unit width of the plate, Iv) Mass entrainment in the boundary layer,

( i ) The hydrodynamic and thermal boundary layer thicknesses, 6, tith: Reynolds number at the end of the plate,

\ ,

(vi) The skin friction coefficient. Assume cubic velocity profile and approximate method.

The thermo-physical properties of air at mean film temperature (60 + 2Oyr' = 40°C are: p = 1.128 kg/rn3, v = 16.96 x 1@ m2/s, k = 0.02755 W/m°C, PI = 0.699.

Solution. Given: t, = 20°C, t, = 60°C, U = 4.5 mls. Ux-

Since Reynolds number is less than 5 x lo5, hence the flow is laminar in nature. The hydrodynamic boundary layer thickness,

= 0.2704 m or 270.4 mm (Ans.) ...[ Eqn. (7.22)] The thermal boundary layer thidkness, according to Pohlhausen, is given by:

Re, x v 5 x id x 16.96 x lo4 = or x , = = U 4.5

(where x, = distance from the leading edge at which the flow in the boundary layer changes from where Pr (Prandtl number) = (0'65 lob4 = 902.77

a 7.2 x lo4 laminar to turbulent).

( i ) Thickness of hydrodynamic layer, 6: m e thickness of hydrodynamic layer for cubic velocity profile is given by

i 4.64 x,

6 = 7 ...[ Eqn. (7.36)]

0'2704 = 0.02798 m or 27.98 mm (Ans.)

6th = (902.77)'" (ii) The - total drag force per unit width on one side of the plate FD: The average skin friction coefficient is given by,

...[ Eqn. (7.25)1

(ii) Thickness of thermal boundary layer, The thermal boundary layer is given by

0.975 6 ...[ Eqn. (7.8911 1 The drag force, FD = x - u2 x area of plate (for one side) f 2 P ...[ Eqn. (7.31)]

6th = !?.975 * 0.01234 = 0.01355 m or 13.55 mm (Ans.)

(0.699)"~

1 or FD = 0.01596 x y x 956.8 x 0.12 x (4.5 x 1 ) = 0.3436 N per meter width

(iii) The local heat transfer coefficient at the trailing edge, h, (at x = L): h x

Nu, = = 0.332 (~e,)'" (pr)'I3 k \-- I

The Nusselt number at x = x, is given by ln pr)113 NIL = 0.332 (Re,) ( ...[ Eqn. i'1.65)I

= 0.332 x (6923)In (902.77)'" = 266.98 . . . [Eqn. (7.65)]

or h, = 266.98 - - 266'98 0a213 = 12.64 wlm2 OC (Ans,) X 4.5 h X x , Nu, X k

Nu, = or h, = - Put k XC (iv) The heat transfer rate, Q:

Q = z A, (t, - t,) - where h = 2h, = 2 x 12.64 = 2 5 . 2 8 ~ l m ~ 0 ~ Eqn. (7.6711 ,. Q = 25.28 X (4.5 x 1) (95 - 25) = 7963.2 W (Am.) Example 7.23. Air at 20°C and at atmospheric pressure jlows at a velociry of 4.5 m/s past

a jlat plate with a sharp leading edge. The entire plate sur$ace is maintained at a temperature of 60°C. Assuming that the transition occurs at a critical Reynolds number of 5 x 16, find the distance from the leading edge at which the flow in the boundaly layer changes from laminar to turbulent, At the location, calculate the following :

(i) Thickness of hydrodynumic layer; (ii) Thickness of therml boundary layer;

Average heat transfer coefficient,

x=Lf hxak xc - - -

I h = 2hc = 2 x 3.05 = 6.1 w/m2'C (Am.)

(iv) H a t transfer rate from both sides for unit width of the plaki Q : Q = z (us) At =. 6.1 (2 x 1.88 x 1) (60 - 20) = 917.U W (Am)

Heat and MassqTransfer

(v) Mass entrainment in the boundary layer, in:

Here, 6, = 0 at x = 0 and 6, = 0.01234 m at x = x, = 1.88 m

. . 5 m = - x 1.128 X 4.5 (0.01234 - 0 ) = 0.039 kgls or 140.4 kglh (Ans.) 8

(vi) The skin friction coefficient, C':

...[ Eqn. (7.24)1

0.646 CB = d- = 9.136 X lo4 (Ans.)

5 x 10' Example 7.24. A stream of water at 20°C ( p = 1.205 kg/m3, p = 0.06533 kg/hm)Jows at

a velocity of 1.8 4 s over a plate 0.6 m long and placed at zero angle of incidence. Using exact solution, calculate:

(i) The stream wise velocity component at the midpoint of the boundary layer, (ii) The maximum boundary layer thickness, and

(iii) The maximum value of the normal component of velocity at the trailing edge of the plare.

Solution. Given t, = 20°C, p = 1.205 kg/m3, p = 0.06533 kghm, L = 0.6 m, U = 1.8 m/s (i) The stream wise velocity component at the midpoint of the boundary layer, u: The boundary layer thickness by exact solution is given by

Obviously, the midpoint of the boundary layer y = - occurs at i 3 The stream wise velocity component is obtained form the Blasius solution in tabular form

(Refer table 7.1).

At U 11 = y $ = 2.5, we get - = 0.736 U

or u = 0.736 U = 0.736 x 1.8 = 1.325 m/s (Am.) (ii) The maximum boundary layer thickness, 6,: The maximum boundary layer thickness occurs at x = 0.6 m. Thus.

Re, = puL 1.205 x 1.8 x 0.6 - - p (0.06533/3600)

= 71713, hence flow in laminar.

The boundary layer thickness at the trailing edge, '

5L 5 x 0.6 a,=---- 6-m- 0.01 12m or 11.2 mrn (Am.) -

(iii) The inaxilduni dalue of the normal component of velocity at the trailing edge, v : !% < $ $ t The maximum value of the normal cornfionent of velocity occurs at the outer edge of the

Forced Convection

u boundary layer where u = U. Hence for - = 1, we have U

" = 0.86 (Refer Table 7.1) U

7.2 LAMINAR TUBE FLOW

7.2.1. Development of Boundary Layer In case of a pipe flow, the development of boundary layer proceeds in a fashion similar to

that for flow along a flat plate. A fluid of uniform velocity entering a tube is retmded near the walls and a boundary layer begins to develop as shown in Fig. 7.12 by dotted lines. n e thickness of the boundary layer is limited to the pipe radius because of the flow being within a confined passage. Boundary layers from the pipe walls meet at the centre of the pipe and the entire flow acquires the characteristics of a boundary layer. Once the boundary layer thickness becomes equal to the radius of the tube there will not be any further change in the velocity distribution, this invariant velocity distribution is called fully developed v e l c p r o ie. , poiseulle flow (parabolic distribution).

Boundary Layer -1 ~ d l v developed

c-- k- Entrance length (L,) -I Fig. 7.12. The development of a laminar velarity profile in the intake region Le of a tube.

According to Langhar (1942), the entrance length (L,) is expwscd : 5 = 0.0575 Re where

D represents the inside diameter of the pipe.

7.2.2 Velocity Distribution Fig. 7.13 shows a horizontal circular pipe of radius R, having laminar flow of fluid through

it. Consider a small concentric cylinder (fluid element) of radius r and length dx as a free body. If z is the shear stress, the shear force F is given by

F = z x 2 n r ~ d u

Let p be the intensity of pressure at left end and the intensity of pmsum at the dght end

' Thus the forces acting on the fluid element are:

1. The shear force, T x 2zr x dn on the surface of fluid element.

-

4m

2. The pressure force, p x n? on the left-end.

T


( 3. The pressure force, p + - . dr n? on the right end. ( E ] For steady flow, the net force on the cylinder must be zero.

Integrating the above equation w.r.t. 'r', we get --- -0-- @ 1 Pipe 4 p dX

Where C is the constant of integration and its value is obtained from the boundary condition:

Substituting this. value of C in Eqn. (7.97), we get

it--.------ x2 -------3:

Fig. 7.13. Laminar flow through a circular pipe.

Equation 7.98 shows that the velocity distribution curye is a parabola (see Fig. 7.13). The maximum velocity occurs at the centre and is given by

From equation 7.98 and 7.99, we have a

II ' - Equation (7.95) shows that flow will occur only if pressure gradient exists in the direction - \ , -

Equation 7.100 is the most commonly used equation for the velocity distribution for laminar flow through pipes. This equation can be used to calculate the discharge as follows:

The discharge through an elementary ring of thickness dr at radial distances r is given by

of flow. *.

The negative sign shows that pressure decreases in the direction of flow. - Equation (7.95) indicates that the shear stress varies linearly across the section (see Fig.

7.14). Its value is zero at the eentre of pipe ( r = 0 ) and maximum at the pipe wall given by

I

I ... [7.95 (a)]

Ffom Newton's law of viscosity, 1

I du 7 = p . - dv ...f i)

Total discharge Q = ! ~ Q

--, In this equation, the distance

y is measured fmm the boundary. The radial distance r is related to distance y by the relation

Shear stress Velocity distribution distribution curve

y = R - r or d y = - d r The eqn. (i) becomes

" R2 - Q u,, Average velocity of flow, u = - A = =-

nR2 2

Comparing two values of T frm Eqns. 7.95 and 7.96, we have

Fig. 7.14. Shear stress and velocity distribution across a section.

..-.

402 Heat and Mass Transfer Forced Convection

Equation (7.101) shows that the average velocity is one-halfthe maximum velocity. Substituting the value of u,, from Eqn. (7.99), we have

ap The pressure gradient - is usually expressed in terms of a friction factor f, defined as ax

pU where - is dynamic pressure of the mean flow and D is the tube diameter.

2 From Eqn. (7.102) and (7.103), we get the friction factor as a simple function of Reynolds

number,

which is valid for laminar tube flow, Re < 2300 Further, Eqn. (7.102) can be written as,

8p U -ap = -.ax R2

The pressure difference between two sections 1 and 2 at distances, x, and x2 (see Fig. 7.12), is given by

Obviously the head loss hL over a length of pipe varies directly as the first power of the rate of discharge Q and inversely as the fourth power of the pipe diameter.

7.2.3. Temperature Distribution In order to estimate the distribution of temperature let us consider the flow of heat through

an elementary ring of thickness dr and length dx as shown in Fig. 7.15. Considering the radial conduction (neglecting axial conduction) and axial enthalpy transport in the annular element, we have:

Heat conducted into the annular element,

dQr = - at k (271r.d~) -

a r Heat conducted out of the annular element,

, - Net heat convected out of the annular element,

Elementary rine

~dnular Qr + dr

element

Fig. 7.15. Analysis of energy in the tube flow.

Considering energy balance on the annular element, we obtain (Heat conducted in),,, = (Heat convected out),,,

dQr - dQr+dr = (dQmn,)ne, dt

= p (2nrdr)u c - dx ax at at

- k (2nr.h) - + k (271 (r + dr)} dx [$ + $ dr] = p (2nr.dr)u c - dx

a r ax at

+ k ( 2 n d r d x ) - a r

at

Neglecting second order terms, we get at

&.dr = p r u c --dx.dr ax 1 a

or ...( 7.106)

Inserting the value of u from eqn. (7.100), we get,

...( 7.107)

at or ...( 7.108)

at Let us consider the case of uniform heat flux along the wall, where we can take - as a ax

constant. Integrating Eqn. (7.108) we have

404 Heat and Mass TrAnsfer Forced. Convection 405 .

Integrating again, we have

(where C; and C2 a& the constants of integration). The boundary conditions are:

At r = 0, at - = 0 ar

At r = R, t = ts Applying the above boundary conditons, we get

Substituting the values of C, and C2 in Eqn. (7.109), we have

For determining the heat transfer coeficient for fully developed pipe flow, it is imperative to define a characteristic temperature of the fluid. It is the bulk temperature (tb) or the mixing up temperature of the fluid which is an average taken so as to yield the total energy carried by the fluid and is defined as the ratio of flux of enthalpy at a cross-section to the product of the mass flow rate and the specific heat of the fluid. Thus,

l R p ( 2 n r . d r ) u c p t

tb = O

...( 7.111)

lR p (2 nr . dr) u cp 0

For an incompressible fluid having constant density and specific heat

l R u r r d r

tb = O

...( 7.112) I R u r d r o

The averagetmean velocity (i) also known as the bulk mean velocity is calculated from the following definition:

- 2 R u = 3 In u r d r

"

Substituting this value of u in eqn. (7.1 12), we get

Substituting the value of u from eqns. 7.1 10 and 7.101 and that of t from eqn. (7.1 lo), we get

2 2 t b = s f o 2; [I - s ] b s - k a ' a x k { g - < + L } ] r d r 16 4 16R2

4

at 11 U * m R 2 - or tb = t s - - - 96 a ax

The heat transfer coeflcient is calculated from the relation

From eqn. (7.1 lo), we have

where D is the diameter of the tube. The Nusselt number is given by

This shows that the Nusselt number for the fully developed laminar tube flow is constant and is independent of the Reynolds number and Prandtl number.

The first analytical solution for laminar flow for constant wall temperature was formulated at

by Graetz in 1885. Since - ax is not constant, therefore, the analysis of constant wall temperature

is quite cumbersome. The final result comes out to be

Example 7.25. For laminar flow in a circular tube of 120 mm radius, the velocity and temperature distribution are given by the relations:


u = (2.7r - 3.2 ?) ; t = 85 ( I - 2.2r)OC where the distance r is measured from the tube surfnce. Calculate the following:

( i ) The average velocity and the mean bulk temperature of the fluid; (ii) The heat transfer coeficient based on the bulk mean temperature if the tube su$ace is

maintained at a constant miform temperature of 90°C and there occurs a heat loss of 1000 kJ/h per metre length of the tube.

Solution. Given : u = (2.7 r - 3.2 3) ... Velocity distribution t = 85 (1 - 2.2 r)OC ... Temperature distribution.

( i ) Average velocity ( Z ) and mean bulk temperature (t,) : The average velocity is obtained by equating the volumetric flow to the integrated flow

through an elementary ring of radius r and thickness dr.

= f R (2.7 r - 3.2 ?) rdr R2 0

Substituting R = 0.12 m, we have

. . . ( i )

Now,

u = 1.8 x 0.12 - 1.6 x 0.12' = 0.193 m/s (Am.) The mean bulk temperature is given by

lR u t r d r tb = 0

lR u r d r 0

j R u t r d r = J~ (2.7r - 3 . 2 q x 85(1-2 .2r)rdr 0 n .

... From eqn. (i)]

Forced Convection

Substituting, R = 0.12 m and ii = 0.193 mls, we get

170 (0.9 x 0.12 - 2.285 X 0.12' + 1.408 x 0.12)) = 68.290C (Ans.) tb = 0.193

(ii) Heat transfer coefficient, h: Q = hA (t, - tb) -

looo looo = 277.77 Jls ; t, = 90°C ... Given where Q = 1000 Mh per metre = 3600

277.77 h = = 16.97 W / m 2 OC (Am.)

(271 x 0.12 x 1 ) (90 - 68.29)

Example 7.26. Lubricating oil at a temperature of 60°C enters I cm diameter tube with a velocity of 3 d s . The tube su$ace is maintained at 40°C. Assuming that the oil has the following average properties calculate the tube length required to cool the oil to 45OC.

p = 865 kg/&; k = 0.140 W/m K: cp = 1.78 kJ/kg°C.

Assume flow to be laminar (and fully developed) - Nu = 3.657 (AMIE Summer, 1997)

Solution. Given : ti = 60°C, to = 45'C; D = 1 cm = 0.01 m; U = 3 m/s, t, = 40°C;

p = 865 kg/m3; k = 0.14 W/m K; cp = 1.78 kJ/kg°C.

40°C

Fig. 7.16

Length required, L : Q = m cp (ti - to)

= (PA# cp 0, - to)

(where U = average velocity, Af= flow area)

I Also, ~ = h A 0 , j where A = heat transfer area = nDL, and


... ( Given)

Now, Q=5441.7=51.2XmLx 10.82

. . L = 5441.7 = 312.7 m (Ans.) 51.2 x n x 0.01 x 10.82

Example 7.27. When 0.5 kg of water per minute is passed through a tube of 20 mm diameter, it is found to be heated from 20°C to 50°C. m e heating is accomplisted by condensing steam on the sulface of the tube and subsequently the surface temperature of the tube is maintained at 85OC.Detennine the length of the tube required for fully developed flow.

Take the thermo-physical properties of water at 60°C as :

p = 983.2 kg/m3, cp = 4.178 kJ/kgK, k = 0.659 W/m°C, v = 0.478 x 1@ m2/s

Solution. Given : m = 0.5 kglmin, D = 20 mm = 0.02 m, ti = 20°C, to = 50°C

Length of the tube required for fully developed flow, L:

'The mean film temperature, rj = f /8fi +

2o + 50) = 6OOC 2 ,

Let us first determine the type of the flow. /

Reynolds number, D . U 0.02 x 0.0269 R e = - - - = 1125.5 V 0.478 x lov6

Since Re < 2000, hence the flow is laminar. With constant wall temperature having fully developed flow,

Nu = - hD = 3.65 ...[ Eqn. (7.11 8)] k

The rate of heat transfer,

Q = A, h (t, - t,) = m cp (to - ti) Here, t, = 20 + 50

2 = 35°C = t,

or L = -- - 2.76 rn (Ans.) 377.8

B. TURBULENT FLOW

7.3. INTRODUCTION The flow in the boundary layer, in majority of practical applications in the convective heat

transfer, is turbulent rather than laminar. In a turbulent flow the irregular velocity fluctuations are mainly responsible for heat as well as momentum transfer. As the mixing in the turbulent flow is on a macroscopic scale with groups of particles transported in a zig-zig path through the fluid, the exchange mechanism is many times more effective than in laminar flow. Consequently. in turbulent flow, the rates of heat and momentum transfer and the associated friction and heat transfer coefficients are several times larger than that in laminar. Since the nature of turbulent flow is complex, therefore, it is difficult to solve the problems relating turbulent flow analytically. The heat transfer data can best be calculated by laboratory experiments; the other method of study is the analogy between heat and momentum transfer.

A. Forced Convection-Flow over a Flat Plate

7.3.1. Turbulent Boundary Layer Refer Fig. 7.17. As compared to laminar boundary layers, the turbulent boundary layers are

thicker. Further in a turbulent boundary layer the velocity distribution is much more uniform, than in a laminar boundary layer, due to intermingling of fluid particles between different layers of the fluid. The velocity distribution in a turbulent boundary layer follows a logarithmic law i.e. u - log y, which can also be represented by a power

u - log y

0.99 U law of the type

19)

1 o7 Laminar sublayer

$ = ~ 1 " ...( 7.120) . . Smooth flat plate

1 where n = - (approx) for Re < 7 but > 5 x 105

This is known as one-seventh Fig. 7.17. Turbulent boundary layer. power law.

The Eqn. (7.120). however, cannot be applied at the boundary itself because at y = 0,

- = - ~ 6 - l ' ~ y4'7 = OJ. This difficulty is circumvented by considering the velocity in the viscous [El ; \-' /

laminar sublayer to be linear and tangential to the seventh-root profile at the point, where the laminar sublayer merges with the turbulent part of the boundary layer.

Blasius suggested the following relation for viscous shear stress:

(for Re ranging from 5 x 10' to lo7)

Let us now find the values of 6, z,, Cfi, for the velocity distribution given by Eqn. (7.120) in ,

i . , ) 1

41 0 Heat and Mass Transfer Forced Convection

(i) Boundary layer thickness, 6: U Substituting the value of in Von Karman integral eqn. [7.28 (a)], we have

[In the expression, above, the limits have been taken from 0 to 6 instead of 6' to 6 since the laminar sublayer (6') is very thin]

Now equating the eqns. (7.122) and (7.121), we have

1/4 1 or -- d6 - - 0.0225 [A] x -

72 dx P U @)'I4 (cancelling puZ on both sides).

or d6 = 0.0225 x

Integrating both sides, we have

4 - 6'14 = 0.23 14 - 5 (;J + c (where C = constant of integration)

Let boundary layer be assumed to be turbulent over the entire length of plate. \Hence, at x = 0, 6 = 0 :. C = 0

1/5 0.37 1~ = 0.0371 [&) x x = -

(Re,)'"

6 0.371 or - - --

X (Re,)'"

(ii) Shear stress, 70 :

Substituting the value of 6 from eqn. (7.123), we get P

= 0.0225 pu2 [ 371i p ux-

(Re,) 'I5

b

...[ Eqn. (7.12111

(iii)'~ocal skin friction (drag) coefficient, CB: . c

We know

Also

[Eqn. (7.124)l

1 r0 = cfX x 1. p d [Eqn. (7.30)l

Now equating the eqns. (7.124) and (7.30), we have

(iv) Average value of skin friction (drag) coefficient, Cf :


This is valid for 5 x lo5 < Re, < lo7. - For Reynolds number between lo7 and lo9 the following relationship suggested by Prandtl

and Schlichting holds good,

7.3.2. Total Drag Due to Laminar and Turbulent Boundary Layers When the leading edge is not very rough, the turbulent boundary layer does not begin at

the leading edge, it is usually preceded by the laminar boundary layer. The point of transition from laminar to turbulent layer depends upon the intensity of turbulence. The distance xc (Fig. 7.18) of the transition from the leading edge can be obtained from critical Reynolds number which normally ranges from 3 x lo5 to 3 x lo6. - I

___+

u, t, + I

pi- ~ a m i y layer I

Leading edge Turbulent layer b

Fig. 7.18. Drage due to laminar and turbulent boundary layers.

Drag force (FD = F) for the turbulent boundary layer can be estimated from the following relation:

Fturb. = (Fturb.)total - (Fturb.) x,

where (Fturb,)total = the drag which would occur if a turbulent boundary extends along the entire length of the plate, and

(Ftu,.)xc = the drag due to fictitious turbulent.bundary layer from the leading

edge to a distance x,. Let us assume that the pikeis long enough so that Reynolds number is greater than lo7,

then the turbulent drag is given by

0.455 pu2 0072 pu2 x - x ( L x B ) - -

Fturb. = Ooglo ~ e ~ ) ~ ~ ~ ~ 2 115 x 2 x (xc x B) . . . ( i ) (Re,)

where L = length of the plate, B = width of the plate, and U = free stream velocity.

The laminar boundary layer prevails within the length xc and its contribution to drag force is given by

Forced Cohvection 41 3

...( ii)

1.328 x, 0.455 L +

( loglo ~"4 . '~ - (Re,)"'

Assuming that transition occurs at Re, = 5 x 10';

1 - LBpu2 Also, Ftotal = c x - ~ A U ~ = C f x -

f 2 2

[where cf = average value of skin friction (drag) coefficient]

Equating the above two equations, we have 1670 0.455 ...( 7.129)

cf = (loglo ~ e , ) ' . ~ ~ Re,

The above relation in general may be rewritten as

Fig. 7.19. Skin friction (drag) coefficient for smooth flat plates.


where the value of A depends upon the value of critical Reynolds number Re, (at which the laminar boundary layer transforms to turbulent boundary layer).

The values of A for Re,, lo5, 5 x lo5 and 10, are 360, 1670, 3300 respectively.

Fig. 7.19 shows a log-log plot of cf versus Reh

Example 7.28. A flat plate 5 m long and 0.75 m wide is kept parallel to the flow of water which is flowing at a velocity of 5 d s . If the average drdg coeficient for turbulent flow past

this flat plate is expressed as: cf = 0'455 2,58 find the drag force on both sides of the plate. (log,, Re,)

Take v = 0.011 x I@ m2/s

Solution: Given : L = 5 m, B = 0.75 m, U = 5 d s , v = 0.01 1 x 10'' m2/s Drag force on both sides of the plate: The Reynolds number at the end of the plate is given by

This confirms that the nature of flow is turbulent. The average drag coefficient is expressed as

c - 0.455 - [log,, ~ e d ~ . ~ ~

:. The drag force on both sides of the plate,

FD = 2 x c f x [ ;~AU' ) [where A = area of one side]

1 = 2 x 2.642 x x IT x 1000 x (5 x 0.75) x 5'1 = 247.68 N (Ans.) L J

Example 7.29. A submarine can be assumed to have cylindrical shape with rounded nose. Assuming its length to be 50 m and diameter 5.0 m, determine the total Dower reauired to overcome boundary friction if it cruises at 8 rn/s velocity in sei water at 20°C (p = 1030 kg/m3), v = I x lo4 m2/s.

Solution. Length of submarine, L = 5 0 m Diameter of submarine, D = 50 m Velocity of submarine, U = 8 d s Density of sea water, p = 1030 kg/m3 Kinematic viscosity of sea water, v = 1 x lo6 m2/s

Total power required to overcome boundary friction, P :

Reynolds number.

The length over which boundary layer wilf be laminar is given by

Ux - = 5 x 1 0 5 0 r x = 5 x 1 6 ~ ~ v ' U

Forced Convection 41 5

This being very small, contribution to total drag from laminar boundary layer is negligible;

hence cf is given by

Area, A = ?rDL = n x 5 x k = 785.4 m2

:. Drag force, 1 I

FD = cf x 5 p~d = 0.001765 x x 1030 x 785.4 x 8'

Hence total power required to overcome boundary friction,

Example 7-30. Find the ratio of friction drag on the front half and rear half of the f2ot plate kept at zero incidence in a stream of uniform velocity, i f the boundary layer is turbulent over the whole plate.

Solution. The average coefficient of drag (cf) for turbulent boundary layer is given by

UL For the entire plate, ReL = - v

Ux UL For the first half of the plate, Re, = - = - v 2v

Drag force per unit width for the entire plate is, - pu2

FD = Cf x - x area per unit width 2

\ /

Similarly the drag force per unit width for the front halfportion of the plate is

FD, -

:. Drag force for the rear half portion of the plate is

Hence,

1 FD,

?x(2)l" - - - - - 0S74 = 1.347 (Ans.)

1 FD2 1 - - 1 5 - 0.574 2

Example 7.31. A stream-lined train is ZOO m long with a typical cross-section having a perimeter of 9 m above the wheels. If the kinematic viscosity of air at the prevailing tempemture is 1.5 x m2/s and density 1.24 kg/m3, determine the su$ace drag (friction drag) of the train when running at 90 km/h.

41 6 Heat and Mass Transfer

Make allowance for the fact that boundary layer changes from laminar to turbulent on the train surface.

Solution. Length of the train, L = 200 m

Perimeter of cross-section of the train above wheels, P = 9 m

:. Surface area, A = L x P = 2 0 0 x 9 = 1800m2 Kinematic viscosity of air, v = 1.5 x lo-' m2/s Density of air, p = 1.24 kg/m3

Free stream velocity, U = 90 kmh = 9ox 1000 = 25 ds 3600

Friction drag, FD : The Reynolds number with length of the train as the characteristic length,

uL. 25 200 = 3.333 x 108 - Re, = - -

v 1.5x10-~

Obviously the boundary layer is turbulent.

Assuming that the abrupt transition from laminar to turbulent flow occurs at a Reynolds number of 5 x lo5, the average coefficient of drag,

- - 0.455 1670 [log,, (3.333 x 108)12." - 3.333 x lo8

= 0.001807 - 5.01 x = 0.0018 The approximate friction drag over the train surface,

- 1 1 FD = Cf x 2 ~ A V ~ = 0.0018 x - X 1.24 x 1800 x 25'= 1255.5 N (Ans.) 2

Example 7.32. A barge with a rectangular surfne 30 m long x 10 m wide is travelling down a river with a velocity of 0.6 m/s. A laminar bouna'ary layer exists upto a Reynolds number equivalent to 5 x IO' and subsequently abrupt transition occurs to turbulent boundary layer. Calculate:

(i) The maximum distance from the leading edge upto which laminar boundary layer persists and the maximum boundary layer thickness at that point.

lii) The total drag force on the flat bottom s u ~ a c e of the barge, and (iii) R e power required to push the bottom surfoce through water at the given velocity.

For water p = 998 kg/m3 and v = I x I@ m2/s. Solution. Length of the bottom surface, L = 30 m Width of the bottom surface, B = 10 m .'. Area, A = L x B = 3 0 x 10=300m2 Velocity, U = 0.6 d s

Density of water, p = 998 k g h 3 Kinematic viscosity, v = 1 x 10" m2/s ( i ) The maximum distance up to which laminar boundary layer persists, x,:

( ~ t ? ) ~ ~ = - "C - - 5 x lo5 V

Forced Convection 41 7

Maximum boundary layer (laminar) thickness, 8:

(ii) The total drag force FD :

The average coefficient of drag, ...[ Eqn. (7.1291

:. Drag force on the bottom 1 surface of the barge, 1

FD = Cf x 5 ~ A U ~ = 0.002M6 x 5 x 998 x 300 x (0.6)~

(iii) The power required, P: The power required to push the bottom surface through water at the given velocity,

P = FD x U 142.6 x 0.6 = 85.56 W (Am.)

Example 7.33. Ambient air at 2oDC flows at a velocity of 10 m h parallel to a wall 5 m wide and 3 m high. Calculate the heat transfer rate if the wall is maintained at 40°C. The critical Reynolds number is equal to 5 x 1@. f ie properties of air at the mean film temperature may be taken as : k = 0.0263 W/m K, v = 15.89 x 1 @ m2/s and Pr = 0.707.

B the entire boundary layer is assumed turbulent, what will be the percentage error in the computation of heat transfer rate ? Comment on the comparable values of the two results.

Appropriate correlation from the following may be used :

- N = 0.0375 R e f 8 pr'13 -

NU = 0.0375 [~ep~ - 232001 pr113 (AMIE, Summer, 1999)

Solution. Given : t- = 20°C; U = 10 m/s; L = 5 m; B = 3 m; t, = 40°C; R,, = 5 X lo5

k = 0.0263 W/m K; v = 15.89 x lom6 m2/s; Pr = 0.707.

Wind Wall (t, = 40°C)

I t, = 2O0C, B=3m

U = 10 mfs I (+- L = 5 m ------*I

Fig. 7;2G

41 8 Heat and Mass Transfer Forced Convection

Percentage error in computation of heat transfer rate :

i. e., > R,,

Using the following equation, we get - hL k = - = 0.0375 - 23200 PI.'" k [ I

...( Combination of laminar and turbulent flow)

. . k - 23200 pr'l3 I

= 0.00019725 (157860.4 - 23200) x 0.8908 = 23.66 w/m2 OC Heat transfer rate,

ha (ts-t,) = 2 3 . 6 6 ~ ( 5 ~ 3 ) x (40-20) =To98 w If entire boundary is assumed turbulent,

- - hL Nu = - = 0.0375 Rep8 ~ r ' "

k

. Percentage error = 8322 - 7098 x 100 = 17.24% (Am.) 7098 Comments. The difference between the two values is nearly 17%. Further the correlation

for turbulent heat transfer are empirical, which may have a variation of + 25%. Hence, the rate of heat transfer in such cases may be calculated for all purposes, by assuming the boundary layer to be entirely turbulent.

7.3.3. Reynolds Analogy Reynolds analogy is the inter-relationsh@ between f l u i d w o n and Newton's law of viscosity.

du We know, zo = p - dy . . . ( i )

[Heat flow along Y-direction ... Fourier equation]

. . . (iii) when Pr is unity temperature and velocity

[profiles are identical (For most of the gases, 0.6 < Pr < 1.0)] By combining eqns. (i) , (ii) and (iii), we get

dt Q = - c p A 2 O - du

Separating the variables and integrating within the limits: At the plate surface: u = 0 and t = t, At the outer edge of boundary layer; u = U and t = t,

But

1 and zo = C, x 5 p d ..[Eqn. (7.23)l

Making these substitutions in eqn. (7.132), we 1 get e c

h,= C i , x ~ p d x $ = $ ( P c p U )

...( 7.133)

... dimensionless fom.

- hx is called the Stanton number SG. It represents the Nusselt number divided by the P c p u

~roduct o f the Reynolds and Prandlt numbers, i.e., Nu, -- - st, = 5

Re,. Pr 2

Equation (7.134) is called the Reynolds analogy. By using this interrelationship we can infer heat transfer data from shear stress measurement.

h x b - Note: The physical significance of Stanton number is : St = - - p cpU At

actual heat flux to the fluid heat flux capacity of the fluid flow

Further in case of laminar boundary layer on a flat plate, we have hxx Nu, = - = 0.332 ( ~ e , ) " ~ (pr)'" k

...[ Eqn. (7.94)]

Dividing both sides of the above equation by Re, (pr)lt3, we get

Nux - 0.332 - 5 --- Re, (pr)'13 ( ~ e , ) " ~

The L.H.S. of the equality can be rewritten as

- Nu, - - . ( P ~ J ~ ~ ~ = q ( ~ r ) ~ ~ ~ Re, (pr)'13 Re Pr

420 Heat and Mas6 Transfer

~ ~ r o e d Convection

Assuming that the transition occurs at critical Reynolds number, Re, = 5 X lo5, we get k h = : ( ~ r ) ' " 10.664 (5 x lo5)'" + 0.036 (R~J"' - 0.036 (5 X 10')~.'1

... interrelationship bemeen heat and momentum transfer. Equation (7.137) has been designated as Colburn analogy. For Pr = 1 the Reynolds and

Colburn analogies are the same. Heat transfer parameters for turbulmt flow: The heat transfer parameters for turbulent flow may be derived, by using Colbum analopv.

as follows: w.,

From Colburn analogy

and

Example 7.34. Airflows over a heated plate at a

IL

velocity of 50 d s . The local skin friction

co-egident at a point on a plate is 0.004. Estimate the local heat transfer coefficient at this point. The following property data for air are given :

Nu, - ( ~ r ) ~ = 0.0288 (Re,)-'" Pr Re, L

Nu, = 0.0288 (ReJ4" (pr)ln

or

The average value of heat transfer coefficient is given by, - 1 .L

C , s t prl" = $ (U.P.S.C., 1993)

Solution. Given : U = 50 m/s; cp = 0.004; P = 0.88 kg/m3; P = 2.286 x lo-' kg m/s;

cP = 1.001 W/kg K: k = 0.035 W/m K.

Local heat transfer coefficient, h, :

p. C 2.286 x X (1.001 X 1000) = o,654 ~randtl number, Pr = - = k 0.035

...( Given) or = 0.036 (i) (Re3U5 ( ~ r ) ~

- - and NU = hL. = 0.036 (Red" (Prim

k Heat transfer parameters for combination of laminar and turbulent flow: An expression for the average heat transfer coefficient over a plate of length L when bnh

laminar and turbulent boundary layers are present can be derived by evaluating the following integral:

Example 7.35. The crankcase df an LC. engine measuring 80 cm x 20 crn may be idealised as a flat plate. The engine runs at 90 kmh and the crankcase is cooled by the air flowing past it at the same speed. Calculate the heat loss from the crank surfoce maintained at 8S°C, to the ambient air at 1S0C. Due to road induced vibrafion, the boundary layer becomes turbulent from the leading edge itselJ:

90 x 1000 Solution. Given : = 90 k f i = L- = 25 mls, t, = 85OC, t.. = 15OC, L = 80 cm = 3600

85 + 15 The properties of air at tf = - = 50°C are: 2

k = 0.02824 Wlm°C, v = 17.95 x lo4 m2is, Pr = 0.698 ... (From tables)


Heat loss from the crankcase, Q :

uL 25x0.8 =1.114x1o6 The Reynolds number, Re, = - =

v 17.95 x lo4

Since Re, > 5 x lo5, the nature of flow is turbulent.

For turbulent boundary layer,

- hZ Nu = = 0.036 ( ~ e , ) ' . ~ (~r) ' . "~ = 0.036 (1.1 14 x lo6)'.' (0 .698) ' .~~~ = 2196.92

k or h = - x 2196.92 = - L O 02824 x 2196.92 = 77.55 w/m2'c

0.8

:. Q = h~ (t, - $3 = 77.55 x (0.8 x 0.2) (85 - 15) = 868.56 W (Am.) Example 7.36. Air at 20°C and 1.013 bar flows over a flat plate at 40 m/s. The plate is

I m long and is maintained at 60°C. Assuming unit depth, calculate the heat transfer from the plate. Use the following correlation: (M.U.)

NuL = (~r)' .~' [0.037 ( ~ e , ) ' , ~ - 8501

Solution. Given : t , = 20°C, U = 40 qds, L = Im, B = I m, t, = 60°C,

Properties of air at (60 + 20)/2 = 40°C, from the tables:

p = 1.128 kg/m3, c, = 1.005 kJkg°C, k = 0.0275 W/m°C, v = 16.96 x 10d m2/s, Pr = 0.699 Heat transfer from the plate, Q:

Reynolds number,

. . Q = i; A, (5 - I.) = 92.55 x ( 1 x 1 ) (60 - 20) = 3702 W or 3.702 kW (Ans.)

Example 7.37. In a gas turbine system hot gases at 95PCflow at 70 id s over the sugace of a combustion chamber which is at a uniform temperature of 280°C. Determine the heat loss from the gases to the combustion chamber which can be idealised as a flat plate measuring I20 cm x 80 cm. The flow is parallel to the 120 cm side and transition Reynolds number is equal to 5 x 1 6 .

Take the properties of gas as: p = 0.494 kg/m3, k = 0.075 W/m°C, v = 95 X 10d m2/s, Pr = 0.625. Solution. Given: L = 120 cm = 1.20 m, B = 80 cm = 0.8 m, Re, = 5 x lo5, U = 70 m/s. Heat loss from the gases to the combustion chamber: The average Nusselt number for a flow over ?,flat plate when both laminar and turbulent

boundary layers are present is given by

I f transition occurs at Re, = 5 x lo5, then the above expression reduces to

Eu = [0.036 (wO,' - 8361 (~r)'. '~'

Forced Convection

The Reynolds number at the end of the plate is UL 70x1.2 = 8 . 8 4 ~ 1 0 5 - Re, = - - v 9 5 x 10-6

This shows that both laminar and turbulent boundary layers are present and,

Hence, i?u = L0.036 (8.84 x 10')'.~ - 8361 (0.625) ' .~~~ = 1045

Heat loss, Q = ;A At

= 65.31 x (1.2 x 0.8) x (950 - 280) = 42007.4 W = 42 kW (Ans.)

Example 7-38, Air at 20°C and 1.013 bar jlows over a rectangular container, with top suIface 750 mm long in direction of flow and 1 m wide, at 35 nu's. Determine the heat transfer from the top suface maintained at 60°C Use the following properties of air at average temperature 9f 40%

p = 1.906 x I@ kg/ms, c,, = 1.007 kJAg0C and k = 0.0272 W/m°C, and the following co-relations for finding average heat transfer coeficient:

i u = 0.664 ( R ~ J ~ * ~ ( ~ r ) ~ ~ ~ if ReL 5 5 x I @ . . . (i)

i u = I0.037 ( ~ e f l ' - 8501 ( P ~ ) O * ~ ~ ff ReL > 5 X I @ . . .(ii) (P*U

Solution. Given : p = 1.013 bar, L = 750 mm = 0.75 m, B = 1 m U = 35 d s , Mean film -

60 + 20 temp. t, = 60°C, t. = 20°C, t( = = 4O0C

Heat transfer from the to6 surface, Q:

Using the gas equation for air, we have __+

Air

p = p R T o r p = $ flow - Since Re > 5 x lo5, we shall use eqn. (ii) for finding average heat transfer coefficient.


= 0.03627 (3489.96 - 850) x 0.8915 = 85.36 w / m 2 ~ c . . Q = h, 0, - t,)

= 85.36 x (0.75 x 1) (60 - 20) = 2560.8 W (Ans.) Example 7.39. A flat plate 1 m wide and 1.5 m long is to be maintained at 90°C in air

when free stream temperature is ](PC. Determine the velocity at which air must flow over the plate so that the rate of energy dissipation from the plate is 3.75 kW.

1 - hL Use : fiu = - = 0.664 ( ~ e d ' ' ~ ( ~ r ) " ~ k ... for laminar flow

- N U = - - - hL - [0.036(ReL)Od - 8361 (pr)'I3

k ... for turbulent flow

Take the following air properties at 50°C:

p = 1.0877 kg/m3, k = 0.02813 W/m°C, c, = 1007.3 J7kg°C, p = 2.029 x I@' kg/ms and Pr = 0.703. (pou)

Solution. Given: L = 1.5 m. B = 1 m, !, = W0C, t, = 10°C, Q = 3.75 kW, Mean film temperature, t, = (90 + 10)/2 = 50°C Velocity o f flow, U:

Q = h, (5 - t,) = (L x B) (t, - t,) - 3.75 x lo3 = h x (1.5 x 1) x (90 - 10)

The value indicates that the flow must be turbulent. Considering the flow to be parallel to the length of the plate, we have

Re, = ( ~ ~ ~ ) 1 ' 0 ' 8 - = 1246790

Re, = * = 1246790 C1

. , Example 7-40. An aeroplaneflies with a speed of 450 k M at a height where the surrounding

air has a temperature of 1°C and pressure of 65 cm- of Hg. The aeroplane Wing idealised as a jlat plate 6m long, 1.2 m wide is maintained at IQC. If thexow is made parallel to the 1.2 m width calculate:

(i) Hear loss from the wing;

(ii) Drag force on the wing.

m e properties of air at 10°C

(i) Heat loss from the wing: 65

The pressure of the air at flight altitude = p = 5 x 1.013 = 0.866 bar

From the characteristic gas equation, we have

The Reynolds number for the entire wing is

Assuming that the critical Reynolds number is 5 x lo5, we have Uxc 125xxc

Re, = 5 x lo5 = -;- = 14.16 x lo4

5 x 10' x 1 4 . 1 6 ~ l a 6 = 0.0566 or xc = 125

Thus we assume the flow is either a combination of laminar and turbulent or only turbulent. Considering the former, we have

- - ;. Heat lost, Q = h A At = 263.89 x (6 x 1.2) x(19 - 1) = 34200 W = 343 k W (Ans.) (ii) Drag force on the wing, FD: The average friction coefficient is given by

- 0.072 167 0.072 - ] = 2.676 x ' f = (=-~e,"] ' [(lO.S x l ~ ~ ) ~ . ~ (10.59 x lo6)

Drag force on one side of the wing, 1

FD = cf x ;i PA@ (where A = one side area of the wing)

Example 7.41. A square plate maintained at95OC experiences a force of 10.5 N when forced &- \a t 2S°C flows over it at a velocity of 30 d s . Assuming the flow to be turbulent and using Colburn analogy calculate: -

( i ) The heat transfer coeficient; (ii) The heat loss from the plate surface.

42B Heat and Mass Transfer Forced Convection 427

Properties of air are: .

p = 1.06 kg/m3, cp = 1.005 W/kg K, v = 18.97 x 1@ m2/s, PP, = 0.696.

Solution. Given : FD = 10.5 N , t, = 95OC, t, = 25OC, U = 30 m/s (i) The heat transfer coefficient, % : For turbulent flow, the drag force is given by

or L = (g)1'1'8 = 2.478 m

The Reynolds number at the end of the plate,

- 0072 Average skin friction coefficient; Cf = - (Red0.'

From Colburn analogy, we have

...[ Eqn. (7.137)]

- p c u C h = - x - j f

(pr12"

= 70.32 w/rnZOc (Ans.) (ii) Heat loss from the plate surface, Q :

Q = KA At = 70.32 x (2.478 x 2.478) x (9T- 25) = 30226 W = 30,226 kW (Ans.)

Example 7.42. Air at 20°CJ20ws past a 800 mm long plate at a velocity of 45 nds. If the surjace of the plate is maintained at 300°C, determine:

( i ) The heat rransferred from the entire plate length to &ir taking into consideration both laminar and turbulent portions of the boundqry layer.

(ii) The percentage error if the boundary layer is assumed to be of turbulent nature from the very leading edge of the plate.

Assume unit width of the plate and critical Reynolds number to be 5 x ld.

Take the properties of air (at 3Wl 20 = 1 6 0 ~ ~ ) as :

\ /

k = 0.03638 W/m°C, v = 30.08 x lo4 m2/s, Pr = 0.682. Solution. Given: L = 800 mm = 0.8 m, U = 45 m/s, t, = 300°C, t , = 20°C. ( i ) The heat transferred from the entire plate, considering both laminar and turbulent

boundary layers: The Reynolds number for the entire plate,

The critical distance xc from the leading edge at which transition occurs is Re, v

x, = - U

Laminar boundary layer region: Average heat transfer coefficient,

The heat transfer from the laminar portion, - Q,,. = hA (2, - t,)

= 44.99 x (0.3342 x 1 ) (300 - 20) = 4209.98 W (per metre width) Turbulent boundary layer region: Average heat transfer coefficient,

= 0.002812 x 36618.7 x 0.88 = 90.6 w/m2"c The heat transfer from the turbulent portion, -

Qtuh. = hA (ts - t-1 = 90.6 x [(0.8 - 0.3342) x 11 (300 - 20) = 11816.41 W

Hence, total heat transferred from the plate, Qtotal = Qlam. + Qturb. =4209.98 + 11816.41 ~r 16026 W

Alternatively : Overall average heat transfer coefficient,

k h = l0.036 (Re,)" - 8361 ( ~ r ) ' "

- . . Q,,, = hA(t, - t,) = 7 1 . 9 x (0.8 x 1) X (300 - 20) = 16025 W

428 ' Heat and Mass Transfer

(ii) The percentage error: If the boundary layer is turbulent from the very beginning, then average heat transfer

coefficient,

- (Q,td)tuh, = hA (t, - t,) = 105 x (0.8 x 1) (300 - 20) = 23520 W

:. Percentage error = 23520- lm6 x 100 = 46.76 % increase (Ano.) 16026 B. Forced Convection - Internal Flow

7.4. TURBULENT TUBE FLOW Some of the important relations for fully developed turbulent flow (Re > 2300) ~hrough

pipes and conduits are:

( i ) The velocity distribution : - =

where u = local average velocity, and

u,, = velocity at centre line, R = radius of the pipe, and y = distance from the wall = (R - r).

dJ f L i i 2 (ii) The head loss : hL = - P 2g

(where f = friction factor, u = average flow velocity] ... equally valid for turbulent flow.

In case of turbulent flow through tube, it is difficult to derive simple analytical expressions for heat transfer coefficient and Nusselt number. But such expressions can be easily found out by using empirical relations.

The fricrion factor, for turbulent flow, is well represented by the following empirical relations:

f = 0.316 for 2 x lo4 < Re< 8 x lo4 ...( 7.146) f = 0.184 for lo4 < Re < lo5 ...( 7.147) f = 0.005 + 0.396 (Re)'.' for 2 x lo4 < Re < 2 x lo6 ..(7.148)

(iii) The wall shear stress, 2,:

f 2 2, = 8 PUmm ...( 7.149)

(iv) From Colburn analogy (0.5 < P, < 100)

St (mY3 = f 8 ...( 7.150)

Substituting the value off from Eqn. (7.147) in Eqn. (7.150), we get following equations for heat transfer coefficients:

Forced Convectibn 429

- or NUU = 0.023 (pr)'" ...I 7.151)

The above expressions are valid for I

The properties of fluid are evaluated at film temperature. Example 7.43. A tube 5 m long is maintained at 100°C by steam jacketing. A fluid flows

through the tube at the rate of 175 kg/h at 30°C. The diameter ofthe tube is em. Find Out

average heat transfer coeflcient. Take the following properties of the jZuid:

p = 850 kg/m3, c, = 2000 J/kg°C, v = 5.1 x I @ mws and k = 0.12 w'mOC. (MJJ)

Solution: Given: L = 5 m, D = 2 cm = 0.02 m Average heat transfer - coefficient h:

. The mass flow per second is given by

Since Re > 2300, the expression ( i ) holds good. Substituting the values, we get


'l'he thermo-physical properties of water at 30°C are : p = 995.7 kg/m3, cp = 4.174 kJ/kg°C, k = 61.718 x W/m°C, v = 0.805 x lo4 m2/s,

Pr = 5.42. (i) The heat transfer coefficient from the tube surface to the water:

VD 12 ~ 0 . 0 6 Reynolds number, Re = - = - 0.894 x lo6

v 0.805 x lo4 - Since Re > 2500, hence nature of flow is turbulent. For the turbulent flow -

hD Nu = - = 0.023 ( ~ r ) ' . ~ ~ ~ k

...[ Eqn. (7.151)]

= 23959.6 w/rn2'c (Ans.) (ii) The heat trawferr9, Q:

he heat transferred,

= 4230345 W (Am.) (iii) The length of the tube, L:

Q = h A (t, - t,) or 4230345 = 23959.6 x (71 x 0.06 x L) (70 - 30)

L = 4230345 = 23.4 m (Ans.)

23959.6 x 71 x 0.06 (70-30)

Example 7.45. Water IS jlowing at the rate of 50 kg/min through a tube of inner diameter 2.5 cm. The inner surj4ace of the tube is maintained at 100°C. I f the temperature of water increases from 25°C and 55°C find the length of the tube required.

50 Solution. m = 50 kg/min = - = 0.8333 k d s ; Di = 2.5 cm = 0.025 m; 60

The following relation may be used :

Nu = 0.023 ( ~ r ) ' . ~ . The properties of water can be taken from the following table :

Length of the tube required, L :

Q = hAS e m

where A, = nD,L, = n x 0.025 x L = 0.0785 L m2

Also, Q = m cp (to - ti)

The properties of water should be taken at 1

p x l d (kg/m-s)

652

550

470

405

335

For find h we have to use the given empirical relation.

(B.U. Winter, 1998)

G X

(W/m-K)

63.35

64.74

65.90

66.72

67.41

25OC

Fig. 7.22

I

CP

(JAg-K)

4174

4178

4182

4187

4195

t ("c)

40

50

60

70

80

The mass flow rate is given by,

71 71 h = 0.8333 = - D~~ x V X p = - x 0.025~ x V x 977.8 4 4

P @dm3)

992.2,

988.1

983.2

977.8

971.8

. . . (i)

p 405 x x 4187 = 2.54 p r = 9 = k 66.72 x 10-

Now substituting the values in the given empiiical formula, we get

Now substituting the values in (i), we get

104671 = 9268.2 x 0.0785 L x 58.7

L = 104671

= 2.45 m (Ans.) 9268.2 x 0.0785 x 58.7

Example 7.46. Water at 25°C jlows across a horizontal copper tube 1.5 cm OD with a velocity of 2 d s . Calculate the heat transfer rate per unit length if the wall temperature is maintained at 75°C. Given properties of water :

p = 988 kg/m3

k = 0.648 W/m K


(AMIE Summer, 1997)

Solution. Given : Al = 75 - 25 = 50°C; D = 1.5 cm = 0.015 m, U = 2 m/s;

p = 988 kg/m3; k = 0.648 W/m K; p = 549.2 x N d m 2 ;

cp = 4.174 M/kg K.

Heat transfer rate per unit length, &. L'

pUD 988 x 2 x 0.015 = 53969.4 Reynolds number, Re = - =

p 549.2 x

pc 549.2 x x 4 . 1 7 4 ~ lo3 = 3.5376 Prandtl number, Pr = =

k 0.648

NOW Q = 7 i ; x m X ~ t L

= 12938.4 x n x 0.0 15 x 50 = 30485.4 W/m or 30.485 kW/m (Ans.)

Example 7.47. Air entering at 2 bar pressure and bulk temperature of 200°C is heated as it flows through a tube with a diameter of 25.4 mm at a velocity of 10 m/s. Calculate the heat transfer per unit length of the rube if constant hea t f l u condition is maintained at the wall and wall temperature is 20°C above the air temperature all along the length of the tube. How much would the bulk temperature increase over a 3 metre length of the tube?

, Take the properties of air as: p = 1.493 kg&, p = 2.57 x lr5 ~s/m' , k = 0.0386 W/m°C, c,, = 1025 J/kg°C Use the relation : - Nu = 0.023 ( ~ e ) " ~ ( ~ r ) " ~ Solution. Given: D = 25.4 mm = 0.0254 m; U= 10 mls

Heat transfer per unit length of the tube:

Heat transfer per unit length of the tube,

Q = Z A , A ~ = Z X ( n ~ x 1 ) ~t = 62.8 x (n x 0.0254 x 1) x 20 = 100.22 W (Ans.)

Bulk temperature increase (At)b, over a 3 m length of the tube:

Q = mcp (At), = (PAW cp (At),

I (3 x 100.22) = (1.493 x x 0.02, x 10) x 1025 x (At),

3 x 100.22 x 4 or (At)b =

1.493 x n x (0.0254)' x 10 x 1025 = 38.77"C (Ans.)

Example 7.48. A fluid is flowing in a pipe which is 300 mm in diameter, 3.5 m long and whose sulface in maintained at a constant temperature. The temperature of the wall surface at the inlet section of the pipe exceeds the fluid temperature by 40°C. What is the rise in the fluid temperature at the end section of the pipe?

The Reynolds analogy holds good which is given by St = 8 where friction factor f = 0.022.

Solution. Given: D = 300 rnm = 0.3 m, L = 3.5 m, f = 0.022, (t, - ti) = 40°C Rise in fluid temperature at the end section of the pipe (to - ti) : The energy balance yields -

Q = h A (t, - t,) = m cp (to - ti)

or h ( n ~ ~ ) [ t , - y ] =

Dividing throughout by npVc, D, we have

... substituting, St = f 8

Inserting the values, we get


or 0.0048 [(t, - to) + 401 = 0.075 [40 - (t, - to)] or 0.0048 (t, - to) + 0.192 = 3 - 0.075 (t, - to) or (t, - to) (0.0048 + 0.075) = 2.808

:. The rise in the temperature of the fluid at the end, (to - t,) = 0, - ti) - 0, - to)

= 40 - 35.18 = 4.82OC (Ans.) Example 7.49. Air at 2 bar and 40°C is heated as it jlows through tube of diameter 30 mm

at a velocity of 10 rnk Calculate the heat transfzr per unit length of the tube when wall temperature is maintained at 100°C all along length of the tube. How much would be the bulk temperature increase over one metre length of the tube ? Use the following relation :

Nu = 0.023 ~ 8 . ~ P P . ~

The properties air at loo + = 70°C are : 2

~ = 2 0 . 6 x 1 0 - ~ N- dm2; Pr =0.694; cp= 1.009kJ/kg°C; k=0.0297kg/m°C; (M.U., 1992)

Solution. Given : p = 2 bar = 2 x 16 ~/rn ' ; ti = 40°C, D = 30 mm = 0.03 m;

v= 1 0 d s ; t,= 1oo0c

We know that p = pRT or 2- P ' ~ ~

Pr = 0.696 ...( Given) t, (= 40°C)

Nu = 0.023 ( ~ e ) ' , ~ ( ~ r ) ' . ~ Fig. 7.23

= 0.023 (32417)O.~ (0.696)O.~ = 80.79

Q = hA (AMTD) = m x cp (to - ti)

71 71 where m = - D ~ V ~ = - x 0.03~ x 10 x 2.226= 0.0157 kg/s

4 4

or to = 63.06OC

Rise in bulk temperature of air = 63.06 - 40 = 23.06OC (Ans.)

Q = 0.0157 x (1.009 x lo3) x (63.06 - 40) = 365.3 W/m (Ans.)

7.5. EMPRICAL CORRELATIONS FOR FORCED CONVECTION The following dimensionless numbers are used for the usual forced convection problems:

hL (i) Nusselt number, Nu = -; P LV

k (ii) Reynolds number, Re = -

P

PC h (iii) Prandtl number, Pr = (iv) Stanton number, St = -

k ' P cnv r

In order to determine the value of convection coefficient h, the following conventional generalised basic equations are used:

Nu = f, (Re, Pr) = C, (Re)m (Pr)" ; St = f2 (Re, Pr) = C2 (Re)a ( ~ r ) ~ The values (numerical) of the constants and exponents are determined through experiments.

The properties of thefluid are evaluated on the basis of bulk temperature (unless stated otherwise).

A. Laminar Flow

7.5.1. Laminar Flow over Flat Plates and Walls (a) The local value of heat transfer coeficient is given by

h .x Nu, = = 0.332 (~e , ) ' . ~ ( ~ r ) ' . ~ ~ ~

k ... Blasius equation ...( 7.153)

The average value of heat transfer coeficient is given by

where, Ux PC Re, = - , ReL = UL and Pr = v v k

The above equations are valid for the following:

( i ) All fluids (Pr 2 0.6) except liquid metals;

(ii) Reynolds number Re 2 40000;

t, + L (iii) The fluid properties are evaluated at the mean film temperature, tf = -

2 . (b) For liquid metals, the following correlation has been proposed:

Nu, = 0.565 (Pe,) ...( 7.155)

where Re, = Re,. Pr

The above equation is valid for the following :

( i ) Pr I 0.05 (ii) The fluid properties are evaluated at the film temperature.

7.5.2. Laminar Flow Inside Tubes 64

(a) For fully developed flow : f = - Re ..(7.156)

436 Heat and Mass Transfer Forced Convection 437

, (b) For uniform heat flux : Nu = 4.36 ...( 7.157) (c) For constant wall temperature :

Slug flow : Nu = 5.78 ...( 7.158)

Fully developed flow : Nu = 3.66 ...( 7.159)

(4 For average Nusselt number for flow inside tubes, the following correlations, have been developed:

(iii) 0.7 I Pr 5 160 (iv) Fluid properties are evaluated at the mean bulk temperature. 2. Colburn suggested the following correlation :

Nu = 0.023 (pr)" ...( 7.165) St = 0.023 ( ~ e ) - ' . ~ ( ~ r ) ~ ~ ...( 7.166)

... in terms of staton number It is valid for the following :

( i ) St is evaluated at the mean bulk temperature.

(ii) Re and Pr are evaluated at the mean film temperature.

This equation is valid ( i ) for constant wall temperature and fully developed flow,

(ii) when properties are evaluated "at the bulk temperature, (iii) tube length is much greater than- pameter, (iv) when Re and Nu are calculated on the basis of pipe diameter as the length parameter. (v) when 0.5 < Pr < 100.

This equation is valid for

(i) short tubes [$ > 2) (ii) Re < 2100

ii) 0.48 < Pr < 16700 (iv) 0.0044 < (~'CL,) < 9.75 (v) In this equation all properties except p, are evaluated at bulk temperature; pJ is evaluated

at surface temperature.

B. Turbulent Flow

7.5.3. Turbulent Flow Over Flat Plate The local and average Nusselt numbers, from Colburn analogy, are given by:

Nu, = 0.029 (pr)lt3 ...( 7.162)

= 0.036 (pr)ll3

where the properties are evaluated at the mean film temperature. When the flow lies in the transition range,

Nu = 0.036 [ (~e , ) ' . ~ - AVP~)"~ ...( 7.163)

where A = 18700L when Re, 4 x lo5 A = 23100 when Re, = 5 x lo5

7.5.4. Turbulent Flow in Tubes The following correlations for fully developed turbulent flow in circular tubes have been

proposed : 1. Mc Adams has proposed the following general correlation on the basis pf Nu = @ (Re, Pr)

for heating and cooling of all - fluids with some limitations as given bdow : Nu = 0.023 (Pr)" ...( 7.164)

where n = 0.4 ... for heating n = 0.3 ... for cooling

L (i) 1 60 (ii) 1 x lo4 < Re < 12 X 104

L (iii) 5 2 60, Re 2 lo4 , 0.7 < Pr < 160.

, . 3. The Mc Adams and Colburn correlationsl are fairly accurate for small to moderate tem-

perature difference (lO°C in case of liquids and 50°C for gases). For larger temperature differences, Sieder and Tate proposed the ioll~wing correlafion:

= 0.023 (pr)" ...( 7.167)

This equation is valid for the following: L

(i) - > 60, Re 2 lo4, 0.7 I Pr I 16700 D

(ii) All fluid properties except ps are evaluated at the mean bulk temperature, psis evaluated at the surface temperature.

4. Desman and Sams suggested the following equation for very large temperature difference (t, -"tb) with air:

Nu = 0.026 ( ~ r ) ' . ~ ....( 7.168) The equation is valid for the following:

ts ( i ) - upto 3.55 (ii) Re 2 lo4 tb

(iii) Re is evaluated at mean film temperature (iv) Nu and Pr are evaluated at mean bulk temperature.

7.5.5. Turbulent Flow Over Cylinders 1. The following empirical correlation is widely used for turbulent flow over cylinders.

- hD = c ( ~ e ) " (pr)'" Nu = - k

...( 7.169)

where C and n are constants and have the values as given in the table 7.2 below:

Table 7.2. Constants for Eqn. (7.169) for Flow across Cylinder (Hilpert, 1933; Knudsen, 1958)

c

S. No. Re C n

1. 0.4 to 4 0.989 0.330

2. 4 to 40 0.911 0.385

3. 40 to 4 x lo3 0.683 0.466

4. 4 x lo3 to 4 x lo4 0.193 0.618

5. 4 x lo4 to 4 lo5 0.027 0.805

All properties are evaluated at the film temperature.


2. Churchill and Bernstein have suggested the following empirical correlation which covers the entire range of Re and wide range of Pr :

This equation is valid for the following:

( i ) Re . Pr > 0.2 (ii) All properties are to be evaluated at the film temperature.

7.5.6. Turbulent Flow over Spheres 1. For flow of gases over spheres, Mc Adams suggested the following correlation:

- Nu = 0.37(~e)O.~ ... for 25 < Re < 1 x lo5 ...( 7.171)

Fluid properties are to be evaluated at the film temperature. 2. Kramers proposed the following correlation for flow of liquids past spheres: -

Nu = [0.97 + 0.68 ( ~ e ) ~ . ~ ] ( ~ r ) ~ . ~ ... for I < Re < 2000 ...( 7.172)

The fluid properties are to be evaluated at the film temperature. 3. Whitaker has proposed a single equation for flow of gases and liquids past sphere; his

correlation is

for 0.71 < Pr < 300; 3.5 < Re < 7.6 x lo4; 1.0 < (&I&) < 3.2

All properties except c(, are to be evaluated at t,.

Example 7.50. Air at a temperature of 15OC flows at a velocity of 6.5 m/s across a flQt plate maintained at a temperature of 605OC. Calculate the amount of heat transferred per metre width from both sides of the plate over a distance of 350 mm from the leading edge. The following relation holds good in the case of large temperature difference between the plate and the fluid:

Nu, = 0.332 (pr)'I3 (Re) u2 ($"."' where TTF and T, are the absolute temperatures of the plate sugace and free stream respectively and all fluid properties are evaluated at the mean $film temperature.

Solution. Given: T, = 15 + 273 = 288 K, T, = 605 + 273 = 878 K, x = 350 mrn = 0.35 m. Amount of heat transferred per unit width :

The mean film temperature, tf = 605 + 15 = 310°C

2

The thermo-physical properties, at 3 1 O°C are:

p = 0.614 kg/m3; c,, = 1.046 kJ/kg°C

k = 0.04593 W/m°C; p = 29.7 x lo4 kglms; Pr = 0.675

Reynolds number, pux 0.614 x 6.5 x 0.35 Re, = - - = 47032

CL 29.7 x lo4 The Nusselt number is given by

Nu, = 0.332 (pr)'I3 (Re)lJ2


Heat transfer from both sides of the plate, per metre width,

Q = 2 [ZA , (t,-t,)] = 2l18.88 x (0.35 x 1 ) (605 - 15)] = 7797.44 W (Ans.)

Example 7.51. A surface condenser consists of two hundred thin walled circular tubes (each tube is 22.5 mm in diameter and 5 m long) arranged in parallel, through which water flows. I f the mass flow rate of water through the tube bank is 160 kg/s and its inlet and outlet temperatures are known to be 21°C and 29OC respectively, calculate the average heat transfer coeficient associated with flow of water.

Solution. Given: D = 22.5 mm = 0.0225 m, L = 5 m, t, = 21°C, to = 2g°C. Average heat transfer coefficient, h:

ti + to 21 + 29 - 250C The mean bulk temperature, t,, = - - - - 2 - 2

The thermo-physical properties of water at 25°C are:

p = 996.65 kg/m3; p = 0.862 x kglms;

k = 0.6079 W/m°C; c, = 4.178 kJkg°C

Prandtl number,

The mass flow rate of water through each tube,

As Re >> 2500, therefore, the nature of flow is turbulent. Since the water is being heated in condenser, hence we may use McAdams correlation -

Nu = 0.023 ( ~ e ) ' . ~ ( ~ r ) ' . ~ ... [Eqn. (7.164)] = 0.023 (52518.26)O.~ (5.9241O.~ = 279.9 -

or Nu = - hD = 279.9 k

Example 2.52. Water at 20°C with a flow rate of 0.015 kg/s enters a 2.5 cm ID tube which is maintained at a uniform temperature of 90°C. Assuming hydrodynamically and thermally fully developed flow determine the heat transfer coeficient and the tube length required to heat the water to 70°C. [Water properties at 20°C : p = 1000.5 kg /d , cp = 4181.8 J/kg K,

i


The thermo-physical properties o f air at 90°C are :

p = 0.972 kg/m3; cp = 1.009 kT/kg°C

k = 0.03127 W/m°C; v = 22.1 x m2/s

p = 22.14 x lo4 kgtms; Pr = 0.69

Reynolds number,

As Re >> 2500 the flow is turbulent. ( i ) The heat loss from the duct over its 4.5 m length, Q:

Q = m cp At = 0.06 x (1.009 x lo3) (110 - 70) = 2421.6 W (Ans.)

(ii) The heat flux and duct surface temperature at a length of 4.5 m: The heat flux at x = 4.5 m is calculated from the relation:

n ~t ~t

where, h, = 6.5 W / ~ ~ " C hi (inside convection coefficient) is calculated from the correlation

Nu = 0.023 ( ~ r ) ' . ~ ~ ~ = 0.023 (19169.5)0.8 ( 0 . 6 9 ) " ~ ~ ~ = 54.22

...( Given)

1 1 1 1 = -+- = - + - = 0.26 :. Thermal resistance, Z Rth hi ho 9.42 6.5

Hence, heat flux

From the resistance network,

or 250 t = 70 - - = 43.46OC (Ans.) 9.42

Example 7.55. A horizontal tubular 1-1 condenser is used to condense saturated steam at 4

80°C. The condenser is a shell and tube one with brass tubes (k = 110 W/m°C) of 1.59 cm OD and 1.34 cm ID. Steam is outside tubes and cooling water enters the tubes at 20°C with a velocity of 1.4 d s and leaves at 40°C. I f the rate of cooling water supply is 55000 kg/h and the latent heat of condensation of steam at 80°C is 2304 kJ/kg, calculate :

( i ) The number of tubes; (ii) The length of each tube.


For calculating the tube side heat transfer coeficient use the Dittus-Boelter equation and for the shell side heat transfer coeficient, the average value may be taken as 10760 w/m2 K.

Data : Properties of water at 30°C.

k = 0.659 W / m K; p = 979.8 kg/m3

cp = 4.180 kJ/kg K; p=0.4044 XI&' pa s

(AMIE Summer, 1998) Solution. Given : thl = th2 = t,,, = 8 0 V , t,, = 20°C; tc2 = 40°C;

V, = 1.4 m/s; h, = 10760 w / m 2 K

kw = 0.659 W/m°C; p, = 979.8 kg/m3; cp, = 4.18 kJ/kg K ;

pw = 0.4044 x Pas ..... at 30°C

(i) The number of tubes, N : I

Q = mw Cpw (4.2 - tcl)

= 15.277 x 4.18 x lo3 (40 - 20) = 1277157 W

Also, mw = pw x flow area x velocity

. . x (0.0134)' x 1.4

(where N = no. o f tubes) I

N = 15.277 x 4

= 78.97, i.e., 79 (Ans.) 979.8 x n x (0.0134)~ x 1.4

(ii) The length of each tube, L : For two through tubes,

Using Dittus-Boelter equation, we have

-- - 0'659 x 0.023 (45453)O.' (2.565)'" = 8774 w/m2 K (Ans.) 0.0134


NOW, Q = uaoO,

where 0, = ' 1 - '2 - (80 - 20) - (80 - 40) 20 - 49.320C, and

In ( @ , / € I 2 ) - - 0.4055 in[: I Z]

Substituting the values in the above eqn,, we get

1277157 = 4158.7 x (79 x n x,0.0159 x L) x 49.32

L = 1277157

= 1.578 m (Ans.) 4158.7 x 79 x n x 0~0139 X 49.32

Example 7.56. Liquid mercury flows at a rate of 1.6 kg/s through a copper tube of 20 mm diameter. The mercury enters the tube at 15°C and after getting heated it leaves the tube at 35°C. Calculate the tube length for constant heat flux at the wall which is maintained at an average temperature of 50°C.

For liquid metal flowing through a tube, the following empirical correlation is presumed to agree well with experimental results : -

Nu = 7 + 0.025 ( ~ e ) ' . ~ where Pe is the Peclet number; Pe = Pr.Re

Solution. Given : D = 20 mm = 0.02m, m = 1.6 kgls, ti = 15"C, to = 35°C. Tube length, L: -

15 + 35 The bulk temperature, t, = - = 25°C

2

The thermo-physical properties of mercury at 25OC are: p = 13582 kg/m3; k = 8.69 W/m°C

c, = 140 Jlkg°C; v = 1.5 x m2/s; Pr = 0.0248

pVD pAVD m D 4 m - - - - 4 m The Reynolds number, Re=---------

CL "' $ D 2 r n : D ~ E ~ ( P V )

. . Nu = 7 + 0.025 (Pr . ~ e ) ' . ~

= 7 + 0.025 (0.0248 x 49997)0.8 = 14.46

The heat gained by mercury due to convection,

Q = X A , At = 6282.87 x (n x 0.02 x L) (50 - 25) = 9869.11 L . . . ( i )

From ( i ) and (ii), we get 9869.1 1 L = 4480

or L=-- - 0.454 m (Ans.) 9869.11

Example 7.57. A square channel of side 20 mm and length 2.5 m, in a heat transfer problem, carries water at a velocity of 45 d s . The mean temperature of the water along the length of the channel is found to be 30°C while the inner channel sur$ace temperature is 70°C. Calculate the heat transfer coeflcient from channel wall to the water. Use the correlation:

\ " / , i'le thermo-physical properties are evaluated at the mean bulk temperature except Pr, . Pr, is

evaluated at the channel sur$ace temperature. Take 'epui9alent diameter as the characteristic length of the channel. - j J

The thermo-physical properties of water at 309iZ"i&e: *

p = 995.7 kg/m3; k = 0.6175 W/m°C; v = 0,805 x I@ rn2/s; Pr = 5.42;PrS at 700C = 2.55.

Solution. Given: V = 4.5 rnls, tb = 30°C, r, = 70°C Heat transfer coefficient, E: The equivalent diameter of the channel,

(where A = cross-sectional area of channel, P = perimeter of the channel).

Reynolds number

Nusselt number is given by

...( Given)

k or h = - x 573.4 = - x 573.4 = 17703.7 w / I ~ ~ " c (Ans.) D, 0.02

Example 7 3 . Water is heated while flowing through a 1.5 cm x 3.5 cm rectangular tube at a velocity of 1.2 m/s. The entering water temperature is 40°C and tube wall is maintained at 85°C. Determine the length of the tube required to raise the temperature of water by 35°C.

Use the following properties of water : p = 985.5kg/m3; k = 0.653 W/m K; v = 0.517 x m2/s; c, = 4.19 kJ/kg K.

(N.M.U., 1998) I Solution. Given : x = 3.5 cm = 0.035 m; y = 1.5 cm = 0.015 m; V = 1.2 m / s ;

t, = 40°C; to = 75OC. Length of tube required, L :

Q = hA, 8, = mc, (to - ti) . ..(i) where A, is the surface area of the tube.


(where A, = area of cross-section).

l = 3 . 5 cm

Fig. 7.26

= 0.035 x 0.015 x 1.2 x 985.5 = 0.621 kg/s

The heat transfer co-efficient h for the given configuration is given by :

Nu = 5 = 0.023 (I?e)O.' (~r)"" k

[Eqn. (7.151)]

4 4 4 ( 1 x b ) 21b Characteristic length, LC = - = --

P 2 (1 + b) - ( 1 + b)

P C PC 985.5x0.517 x x (4.19 x lo3) p r = - e = A = k k 0.653

= 3.27

Substituting the values in (ii), we get

0'021 = 0.023 (0.487 x 105)0.8 (3.271°." = 191 -23

0.653

Now substituting the values in (i) , we get

Forced Convection

0.621 ~ ( 4 . 1 9 ~ 103)x35 L= = 6.57 m (Ans.)

5946.3 x 0.1 x 23.3

Example 7.59. 3.8 kg of oil per second is heated from 20°C to 40°C by passing through a circular annulus with a velocity of 0.3 m/s. The hot gases at 400°Care passed through the inside tube of 100 mm diameter and are cooled to 100°C. Find the length of the pipe required for the above heat transfer process assuming the gas is flowing in opposite direction to the oil ?

Take the following properties of oil and gases at mean temperature : Oil Gases

P 800 kg/m2 0.8 kg/m3 v 8 x ~ O - ~ m2/s 32.8 x 1 r 6 m2/s

c~ 3350 J/kg K 1050 W/K

k 0.2 W/m°C 0.035 W/m°C

Solution. Given : m, = 3.8 kg/s; t,, = 20°C; tC2 = 40°C; V, = 0.3 m/s;

t,,, = 400°C; th2 = 100°C; d = 100 mm = 0.1 m

[The suffix c stands for cold (i.e., oil) and suffix h stands for hot (i.e., gases)]. Length of the pipe required, L : The diameter of the annulus is calculated as follows :

(P.U.)

L A For oil : For annulus, D,, = (D - d) ...[ Eqn. [6.28)1

Nu = 0.023 (~e)'.' (pr)'l3 ...[ Eqn. [7.151)]

= 0.023 (2764)O.' (107.2)~'~ = 61.9

For gases : Heat lost by gases = heat gained by oil

mh x I050 x (400 - 100) = 3.8 x 3350 x (40 - 20)

. . mh = 0.808 kg/s

The velocity of hot gases (Vh) can be calculated as follows :

448 Heat and Mass Transfer Forced Convection 449

Overall heat transfer coefficient,

Log-mean temperature difference,

Example 7.60. Gasoline at the mean bulk temperature of 27Ocjows inside a circular tube of 19 mm inside diameter. The average bulk velocity is 0.061 ids and the tube is 1.5 m long. The $ow starts at the heated tube inlet (no upstream developing section) and the tube sur$ace temperature is constant at 38OC.

Determine the average heat transfer co-eficient over 1.5 m length of the tube. Are the temperature and velocity profies developing or developed in the 1.5 m length of the tube ?

Take the following fluid properties :

CL, = 5,223 x 1 F 4 Pas; pb= 5.892 x 1 F 4 PU S; kb= 0.1591 w/mOc;

pb = 874.6 kg/m3; cpb = 1757 J/kg K; Ph = 6.5

Use the following formula :

x, (entry length for fully developed velocity profile) = 0.05 Re.D, and f x, (entry length for fully developed temperature profile) = x,, . Ph i

(M.U., 1997)

Solution. Given : D = 19 mm = 0.019 m; V = 0.061 m/s; L = 1.5 m

( 1 Pas = 1 N-s/m2)

x, (velocity depth) = 0.05 ReD= 0.05 x 1720.4 x 0.019 = 1.634 m

xt (temperature depth) = x, x P, = 1.634 x 6.5 = 10.62 m

The tube length L is less than x,, as well as x,, so both depths are developing. (Ans.)

Example 7.61. Liquid metal flows at a rate of 270 kg/min through a 5 cm diameter stainless steel tube. It enters at 415OC and is heated to 440°C as it passes through the tube. The tube wall tenzperature is 20°C higher than liquid bulk temperature and a constant heat flow is maintained along the tube. Calculate the length of the tube required to effect the transfer.

For constant wall temperature, Nu = 5 + 0.025 (pe)08 . . . ( i )

For constant heat f l u ,

Nu = 4.82 + 0.0185 (pe)Og2' . . .(ii) Use the following fluid properties :

p= 1.34 x I O - ~ kg/m-s, cp= 149J/kgK, and

Pr = 0.013, k = 15.6 W/m K. (P.U. Winter, 1997)

Solution. Given : h = u0 60 = 4.5 kg/s; D = 5 cm = 0.05 rn; At = 440 - 415 = 25OC 460°C

Length of the tube required, L : The temperatures of the wall and flowing

440°C metal along the tube are shown in Fig. 7.27.

We have to use the eqn. (ii) as it is a constant heat flux condition to the tube. 435°C

Q=hcpAt=4 .5x 149x25 415°C

= 16762.5 W Fig. 7.27


Now, Nu = 4.82 + 0.0185 (Pe)0.827

= 4.82 + 0.0185 (1094.6)O.~~~ = 10.85

i. e., -- hD - 10.85 k

Further, Q = hA (At) = 16762.5 (calculated above)

z.e., 16762.5 = 3385.2 x (n x 0.05 x L) x 25

Example 7.62. Assuming that a man can be represented by a cylinder 350 mm in diameter and 1.65 m high with a sulface temperature of 28°C. Calculate the heat he would lose while standing in a 30 km/h wind at 12°C.

Solution. Given : D = 350 mm = 0.35 m, L = 1.65 m, t, = 28"C, t, = 12°C.

Heat lost by the man, Q :

28 + 12 - 2ooc The film temperature, t - ----- - f - 2

The properties of air at 20°C are:

k = 2.59 x W/m°C, v = 15.0 x lo4 m2/s, Pr = 0.707.

UD 8.33 0.35 = 1.94 105 Reynolds number, Re = - = v 1 5 . 0 ~ lo4

Using Eqn. (7.169), we have

k h = - . C (Re)" (pr)'I3 D

where C = 0.027 and n = 0.805 (From table 7.2) Substituting the value in the above equation, we get

:. Heat lost by the man, Q = h A, (t, - t,) = 32.15 x (n x 0.35 x 1.65) (28 - 12)

or Q = 933.26 W (Ans.) Example 7.63. A copper bus bar 25 mm diameter is cooled by air (in cross-flow) at 30°C

and flowing past the bus bar with a velocity of 2.5 d s . the sugace temperature of the bar is not to exceed 85OC and resistivity of copper is 0.0175 x 1Qd ohm-m3/m, calculate the following:

(i) The heat transfer coeflcient from the sutjGace to the air; (ii) The permissible current intensity for the bus bar. The following empirical correlations may be applicable for a single cylinder placed in

cross-flow :

For 10 <Re < I d Nu = 0.44 (~e)"' For I d < Re < 2 x ld Nu = 0.22 ( ~ e ) ' . ~ The thermo-physical properties are evaluated at t, (30°C) and are given as: k = 0.02673 W/m°C; v = 16 x I@ m2/s Solution. Given : D = 25 mm = 0.025 m, t, = 30°C, U = 2.5 d s , t, = S ° C , p,,,,, = 0.0175 x lo4 ohm-m31m. ( i ) The heat transfer coefficient from the surface to the air:

Reynolds number

Since Re = 3906, so the second relation will be applicable. Thus,

(ii) The permissible current intensity for the bus bar, I: Heat dissipation to air = h A, (t, - t,)

= 33.6 x (n x 0.025 x 1 ) (85 - 30) = 145.14 Wlm The heat generated in the bus bar

From ( 1 ) and (2), we have 35.65 x lo4 P = 145.14

Example 7.64. Air stream at 24OC is flowing at 0.4 d s across a 100 W bulb at 130°C. & the bulb is approximated by a 65 mm diameter sphere, calculate:

( i ) The heat transfer rate, and (ii) The percentage of power lost due to convection.

Solution. Given: t, = 24°C t, = 130°C, D = 65 min = 0.065 m, U = 0.4 mls (i) The heat transfer rate, Q:

The film temperature,

The properties of air at 77°C are: k = 0.03 W/m°C; v = 2.08 x lo-' m21s, Pr = 0.697

UD 0.4 x 0.065 = 1250 Reynolds number, R e = - = 2.08 x l o 5

Equation (7.171) gives the Nusselt number as - hD % = 0.37 ( ~ e ) ' . ~ = T


-- - 0'03 x 0.37 (1250)'.~ = 12.32 w / ~ " c , 0.065 :. The heat transfer rate,

Q = % A , (t,-t,) = L x 7 1 ~ ~ x (tS-t,)

(ii) The percentage of power lost due to convection:

%age of power lost = - 17'33 x 100 = 17.33% (Am) 100 Example 7.65. Compare the heat transfer coeflcients under the following conditions,

assuming that there is no change in the temperatures of the liquid and the tube wall and the flow through the tube is turbulent in character.

(i) Twofold increase in the diameter of the tube; the flow velocity is mintairied constant by a change in the rate of liquidflow;

(ii) Two-fold increase in the flow velocity, by varying mass flow rate. Solution. In case of turbulent tube flow, the average heat transfer coefficient is given by

- hD

x u = - = 0.023 (Re)08 ( ~ r ) ' . ~ ~ ~ k ...[ Eqn. (7.151)]

- 0.33 ( W O ' ~ h = 0.023 (Pr)

v (D) 0.2 ( i ) When the flow velocity and the fluid properties remain unchanged:

This shows that the heat transfer coeflcient decreases to 0.87 when there is a twofold increase in the diameter of the tube.

(ii) When the tube diameter and fluid properties remain the same: - h .. (uO'

This shows that the heat rransfer is increased to 1.74 times when there is a hvo-fold increase in flow velocity.

Example 7.66. A refrigerated truck is moving on a high way at 90 km/h in a desert area where the ambient air temperature is 50°C. The body of the truck may be considered as a rectangular box measuring 10 m (lengrh) x 4 m (width) x 3 m (height). Assume that the boundary


layer on the four walls is turbulent, the heat transfer takes place only from the four suq4aces and the wall surjGace of the truck is maintained at 10°C. Neglecting heat transfer from the front and back and assuming the flow to be parallel to I0 m longside, calculate the following:

( i ) The heat loss from the four surj5aces; (ii) The tonnage of refrigeration; (iii) The power required to overcome the resistance acting on the four sudaces.

50 + 10 The properties of air (at i) = - 2 = 30°C) are:

p = 1.165 kg/m3, cc = 1.005 kUkg°C, k = 0.02673 w/m2"c, v = 16 x lod m2/s, Pr = 0.701.

Solution. Given : U = 90 kmlh = 90 3600 loo0 = 25; d s , tr., = 50°C, ts = 10°C, L = 10 m ,

B = 4m, H = 3m. ( i ) The heat loss from the four surfaces:

The Reynolds number at the end of the side,

Evidently the boundary layer is turbulent for which - hL

Nu = - = 0.036 (Re)'.' ( ~ r ) ~ . ~ ~ ~ ...[ Eqn. (7.141)] k

:. Heat loss from the four surfaces,

Q = 5 A At = 48.64 [2(4 + 3) x 101 x (50 - 10) = 272384 W = 272.384 kW (Ans.)

(ii) The tonnage of refrigeration (TR) : 272.384 x 3600 = 70 TR

The cooling capacity required = (-: 1 TR = 14000 Wlh) 14000

(iii) The power required to overcome the resistance, P: Average skin friction coefficient

1 Drag force, F, = cf x pAu2

1 = 2.62 x x ~ x 1.165 x [2 (4 + 3) x lo] ~ 2 5 ~

= 133.54 N

. . P = FD x U = 133.54 x 25 = 3338.5 W = 3.3385 kW (Ans.)

Example 7.67. 800 kg/h of cream cheese at 15°C is pumped through a tube 100 mm in diameter, 1.75 m long and maintained at 95°C. Calculate :

( i ) The temperature of cheese leaving the heated section; (ii) The rate of heat transfer from the tube to the cheese. Use the following correlation for larninarflow inside a tube :


L\ A

The thermo-physical properties of cheese are:

p = 1150 kg/m3; p = 22.5 kg/ms

k = 0.42 W/m°C; cp = 2.75 kJ/kg°C Solution. Given: m = 800 kg&, D = 100 rnm = 0.1 m, L = 1.75 m, t, = 95"C, t , = 15°C.

( i ) The temperature of cheese leaving the heated section, tz : Velocity of cheese flowing through the tube,

Reynolds number

Obviously the nature of the flow is laminar and the given correlation is valid for the given flow situation.

Prandtl number,

- t l +tz 15 + t2 Mean bulk temperature of cheese, tb = - - -

2 - 2 Now, heat gained by cheese = convective heat flow from the tube to cheese,

or t2 = 29.82"C (ii) The rate of heat transfer from the tube to the cheese, Q :

Q = 71 A, (t, - tb) = m cp (t2 - t l )

= 9056.67 W (Ans.) Example 7.68. The following data relate to a metallic cylinder of 20 mm diameter and

120 mm in length heated internally by an electric heater and subjected to cross flow of air in a low speed wind tunnel:

Temperature of free stream = 25°C; Velocity of free stream air = 16.5 m/s; Average temperature of cylinder su$ace = 130°C; Power dissipation by heater = 100 W.

I f 12 percent of the power dissipation is lost through the insulated end pieces of the cylinder, calculate the experimental value of the convective heat transfer coeflcient. Compare this value with that obtained by using the correlation :

\ /

where all thermo-physical properties except Pr, are evaluated at the mean bulk temperature lfree stream) of air. Pr, is evaluated at the average temperature of cylinder.

The thermo-physicAl properties of air at 25°C are: k = 0.0263 W/m°C; v = 15.53 x I@ m2/s; Pr = 0.702

Pr, at 130°C = 0.685.

Solution. Given: t, = 25"C, U = 16.5 m/s, Power dissipation by heater = 100 W, D = 20mm = 0.02 m. (a) Heat flow from the heater to the air flowing past it is given by

Q = h ~ , At -

or 100 [1-g) = h x (71 x 0.02 x 0.12) x (130 - 25)

= 111.16 W / ~ ~ " C (Ans.)

(b) Rcynolds number

Using the correlation given, we have - Nu = 0.26 ( ~ e ) ' . ~

k or 0'0263 x 90.93 = 119.57 w / ~ ~ ~ c (Ans.) h = - x 90.93 = -

D 0.02

HIGHLIGHTS

Important Formulae A. Laminar Flow

I . Flow over flat plate:

If v < 5 x los ... boundary layer is laminar (velocity distribution is parabolic)

Ux If - > 5 x lo5 ... boundary layer is turbulent on that portion (velocity distribution

v follows log law or a power law)

( i ) Displacement thickness, 6* =

(ii) Momentum thickness, 0 = 0


\ /

The thermo-physical properties are evaluated at the mean bulk temperature except Pr,; Pr, is evaluated at the channel surface temperature. Take equivalent diameter as characteristic length of the channel. The thermo-physical properties of water at 30°C are :

p = 995.7 kg/m3, k = 0.6175 W/m°C; v = 0.805 x lo4 m2/s; Pr = 5.42; Pr, at 70°C = 2.55 [Ans. 25456.6 w/m2 OC]

31. A copper bus bar 20 mm diameter is cooled by air (in cross-flow) at 30°C and flowing past the bus bar with a velocity of 2 m/s. If the surface temperature of the bar is not to exceed 80°C and resistivity of copper is 0.0175 x lo4 ohm-m3/m, calculate the following:

( i ) The heat transfer coefficient from the surface to the air; (ii) The permissible current density for the bus bar.

The following empirical correlations may be applicable for a single cylinder placed in cross-flow : For 10 < Re < lo3 Nu = 0.44 ( ~ e ) ' ~ For lo3 < R e < 2 x lo5 Nu = 0.22 ( ~ e ) ' . ~

The thermo-physical properties are evaluated at t, (30°C) and are given as : k = 0.02673 W/m°C v = 16 x lo4 m2/s [Am. ( i ) 32.14 w/m2 OC (ii) 1346 amp.]

Free Convection 8.1. Introduction. 8.2. Characteristic parameters in free convection. 8.3. Momentum and energy

equations for laminar free convection heat transfer on a vertical flat plate. 8.4. Integral equations for momentum and energy on a flat plate - velocity and temperature profiles on a vertical flat plate - solution of integral equations for vertical flat plate - free convection heat transfer coefficient for a vertical wall. 8.5. Transition and turbulence in free convection. 8.6. Empirical correlations for free convection. 8.7. Simplified free convection relations for air. 8.8. Combined free and forced convection. Typical Examples - Highlights - Theoretical Questions - Unsolved Examples. .

8.1. INTRODUCTION When a surface is maintained in still fluid at a temperature higher or lower than that of the

fluid, a layer of fluid adjacent to the surface gets heated or cooled. A density difference is created between this layer and the still fluid surrounding it. The density difference introduces a buoyant force causing flow of fluid near the surface. Heat transfer under such conditions is known as free or natural convection. Thus "Free or natural convection is the process of heat transfer which occurs due to movement of the jluid particles by density changes,associated with temperature differential in a jluid." This mode of heat transfer occurs very commonly, some examples are given below:

(9 The cooling of transmission lines, electric transformers and rectifiers. (ii) The heating of rooms by use of radiators.

(iii) The heat transfer from hot pipes and ovens surrounded by cooler air. (iv) Cooling the reactor core (in nuclear power plants) and carry out the heat generated by

nuclear fission etc. - In free convection, the flow velocities encountered are lower compared to flow velocities

in forced convection, consequently the value of convection coefficient is lower, generally by one order of magnitude. Hence, for a given rate of heat transfer larger area could be required. As there is no need for additional devices to force the liquid, this mode is used for heat transfer in simple devices which have to be left unattended for long periods.

- The rate of heat transfer is calculated using the general convection equation given below:

Q = h A (t, - t,) ...( 8. 1) where Q = heat transfer, W

h = convection coefficient, w/m2"c A = area, m2, and t, = temperature of fluid at distances well removed from the surface (here

the stagnant fluid temperature).

8.2. CHARACTERISTIC PARAMETERS IN FREE CONVECTION It has been observed that during heat transfer from a heated surface to the surrounding fluid,

the fluid adjacent to the surface gets heated; this results in thermal expansion of the fluid and

464 Heat and Mass Transfer Free Convection

reduction in its density (compared to the fluid away from the surface). Subsequently a buoyancy force acts on the fluid causing it to flow up the surface, and in the neighbourhood of the plate, hydrodynamic and thermal boundary layers are set up. Since here the flow velocity is developed due to difference of temperatures, the two boundary layers are of the same order irrespective of Prandtl number.

- A property that comes into play in free or natural convection is the coeflcient of thermal expansion of the fluid defined by

For an ideal gas

and hence

- Since in free convection heat transfer coefficients are low and Reynolds number is not an independent parameter, a new dimensionless grouping plays the major role (in free convection) which incorporates the coefficient of thermal expansion P in the expression. This dimensionless grouping is called the Grashof number, expressed as

where L = characteristic length A t = ( t , - t,), where t, and t , are the surface temperature and temperature of

the surrounding fluid respectively. The role of Grashof number is the same in free convection as that of Reynolds number in

forced convection. The critical Grashof number for the flow of air over a flat plate has been observed to be

4 x lo8 (approximately). In general, Nu = f (Gr, Pr) = C(Gr)a (prlb ...( 8.4) In several cases, the above relation simplifies to the form

Nu = C (Gr . Pr)m ...( 8.5) Hence, a new dimensionless group is often used called Rayleigh number viz.,

Ra = Gr . Pr The product is also a criterion of laminar or turbulent character of the flow as determined

by its values. Thus lo4 < Gr Pr < lo9 ... for laminar flow G r Pr > lo9 ... for turbulent flow

[f - lw ; l - ) The values of Gr and Pr are evaluated at the mean film temperature t --

8.3. MOMENTUM AND ENERGY EQUATIONS FOR LAMINAR FREE CONVECTION HEAT TRANSFER ON A VERTICAL FLAT PLATE Figure 8.1. shows a free convection boundary layer formed on a flat vertical plate when it

is heated. Here, the velocity profile is different from that of forced convection boundary layer. At the surface 1 wall the velocity in zero, it increases to some maximum value and then decreases to zero again at the edge of the boundary since the free stream is at rest. Initially the boundary layer is laminar in character and at some distance from the bottom edge, it becomes turbulent. This onset of turbulence depends on the fluid properties and the temperature difference (t, - t,).

In order to analyse the problem, let us consider the momentum equation for forced convection.

In free convection, the additional force is the body force pg and hence the momentum equation gets modified to

...( 8.6)

In free convection, outside the boundary layer, (y -, 00) as p tends to p,, whereas u and v approach zero. So Eqn (8.6) becomes

(i.e. the change in pressure on a height ah is equal to the weight per unit area of the fluid element).

Subtituting Eqn. (8.7) in Eqn. (8.6), we get

Fig. 8.1 : Free-convection boundary layer over heated vertical plate.

...( 8.8)

Further the density difference (p, - p) may be expressed in terms of the volume coefficient of expansion p

Thus

so that Eqn. (8.8) beocmes

The energy equation for free convection boundary layer remains the same as forced convection at low velocities, viz.,

8*4* INTEGRAL EQUATIONS FOR MOMENTUM AND ENERGY ON A FLAT PLAm For free convection on a flat plate, the following equations can be derived. Integral momentum equation:

Integral energy equation:

8-41. Velocity and Temperature Profiles on a Vertical Fht Plate Assuming the velocity and temperature profiles to be similar at any x, the temperature profile

466 Heat and Mass Transfer Free Convection 467

may be taken as t - t , -- t, - t- - c1 + ~2 @ ) + ~ 3 @ J f

2m + n C1 = - gp (ts - t,) C# - v -(x)~-" [T) c12 C2 (x) 2m+n-1 3 ...( 8.19) c2

The following boundary conditions apply: (i) At y = 0 t = t, (ii) At y = 6 t = t,

The temperature distribution is, therefore, obtained as t - t , ...( 8.14) ts - tw

The velocity profile may be similarly assumed.

where ux is any arbitrary function with the dimension of velocity. The boundary condition are:

( i ) At y = 0, u = 0 (ii) At y = 6, u = 0

au - 0 (iii) At y = 6, - - ay

The velocity distribution is, therefore, found to be

where

8.4.2. Solution of the Integral Equations for Vertical Flat Plate Substitution of the velocity and temperature distributions from Eqns. (8.14) and (8.16) into

Eqn. (8.12) for momentum integral gives

and substitution into Eqn. (8.13) for energy integral gives

For a similar solution to exist, both sides of these equations must be independent of x. Equating the exponents, we get

2 m + n - 1 = m - n m + n - 1 = - n

1 1 This gives m = - and n = - (Thus u1 = clxin, 6 = c2x1I4) 2 4

Inserting these values into the above equations, we get

Cl2c2 -- c2 Cl - gp (ts - t,) T - v - ...( 8.21)

84 c2

c1C2 2a -- - - ...( 8.22) 40 C2

Solving these two equations for C, and C2, we get

...( 8.23)

20 v + ' / 4 gp (t - t ) -'I4

~ ~ = 3 . 9 3 [ ~ + - - ) [ > [:)-In ...( 8.24)

From Eqn. (8.16), the maximum velocity within the boundary layer is found to be 4 4

Urn, = - U , = - c ~ x ~ ~ ~ 27 27

which on substitution for C1 from Eqn. (8.23) gives

( . ;JIA[gB(r;- tw)]l"(x)l . u,, = 0.766 v 0 952 + -

The resultant expression for boundary layer thickness is 6 - = c2 2-1 = c2x - 314

X

which on substitution for C2 from Eqn. (8.24) yields

... is dimensionless numbers

gp (4 - tw) 2 v (where Gr, = and Pr = - )

v2 a

1 d 0, - t J - (t, - t-) (u, 6 ) = 2a - 30 6

To solve these equations, let us assume u, and 6 as exponential functions of x 6 = C #

8.4.3. Free Convection Heat Transfer Coefficient for a Vertical Wall u, = C I P ;

(where C,, C2, m and n are constants). The heat transfer coefficient may be evaluated from

Substituting these values in the above equation, we get = - ka ($)y;O = hp 0, - t,) ...( 8.28)

468 Heat and Mass Tjansfer

Using the temperature distribution of Eqn. (8.14), we obtain

Substituting this value in Eqn. (8.28), we get

From Eqns. (8.27) and (8.29) we obtain the heat transfer correlation for nz for a vertical flat plate as

Nu, = 0.508 (pr)ln (0.952 + pr)-'I4 ( ~ r , ) " ~

...( 8.30)

ma1 convection

For a given Prandtl number, Nu, varies as (Gr,)'" or h, a x-"~

The local film heat transfer coefficient decreases with x. It is inversely proportional to the fourth root of x. By integration over the distance, the average heat transfer coefficient is found to be

For air with Pr = 0.7, Eqns. (8.31) and (8.33) simplify to

Nu, = 0.378 ( G ~ J " ~ ...( 8.34) - NuL = 0.504 (~t-3"~ ...( 8.35)

These result are only 5 per cent higher than the exact solution of the free convection problem obtained numerically by Schmidt and Beckrnan.

Here all properties are evaluated at 'film temperature'.

8.5. TRANSITION AND TURBULENCE IN FREE CONVECTION A transition from laminar to turbulent condition in both forced and free convection is

caused by thermal and hydrodynamic instabilities. It has been observed that the transition in free convection boundary layer occurs when the value of Rayleigh number approaches 10'. Thus

The analysis of turbulent free convection is very complex. However, the results obtained from various experiments have been used to obtain the pertinent empirical relations. .

8.6. EMPIRICAL CORRELATIONS FOR FREE CONVECTION The following dimensionless numbers apply for the usual fiee convection problems:

hL (i) Nusselt number Nu = --a (ii) Grashof number Gr = L3P@

k ' v2 P C

(iii) Prandtl number Pr = -f

8.6.1. Vertical Plates and Cylinders The commonly used correlations are:

Laminar pow: NUL = 0.59 (Gr . Pr)'" for (lo4 < Gr . Pr < lo9) ...( 8.37) Turbulent flow: = 0.10 (Gr . Pr)'" for (lo9 < Gr . Pr < 1012) ...( 8.38)

All the fluid properties are evaluated at the mean film temperature ff=- I ":"I Churchill and Chu have recommended the following correlations:

- NuL = 0.68 + 0.67 (Gr - ~ r ) " ~

for (Gr . Pr c lo9) ...( 8.39)

2

for (Gr. Pr > lo9) ...( 8.40)

8.6.2. Horizontal Plates 1

In case of an irregular plate, the characteristic length is defined as the surface area divided '& the perimeter of the plate.

(i) The upper surface heated or the lower surface cooled: Laminar flow: NUL = 0.54 (Gr . Pr)'" for (lo5 c Gr. Pr S 2 x lo7) ...( 8.4 1 ) Turbulent flow: NUL = 0.14 (Gr . pr)In for (2 x lo7 c Gr. Pr I 3 x 10") ...( 8.42)

, (ii) The lower sulface heated or upper surface cooled: - Laminar flow: NuL = 0.27 (Gr . Pr)'" for (3 x 16 c Gr. Pr 5 3 x 101°) ...( 8.43) Turbulent flow: NUL = 0.107 (Gr . ~ r ) ' " for (7 x lo6 c Gr. Pr $ 11 x 10l0) ...( 8.44)

, - I 8.6.3. Horizontal Cylinders

' For such a case, the outside diameter is used as the characteristic dimension. Mc Adam has recommended the following correlations: iarninar flow: Nu = 0.53 (Gr . Pr)'I4 for (lo4 Gr . Pr c 109) ...( 8.45) Turbulent flow: Nu = 0.13 (Gr . pr)ln for (lo9 < Gr. Pr c 1012) ...( 8.46) The following general correlation has been suggested by Churchill and Chu for use over a

wide range of Gr.Pr.

0.387 (Gr . pr)'l6 ,, ] for (lo-' c Gr.Prc 1012) (8.47) { I + ( O S S ~ / P ~ ) ~ / ' ~ }

The fluid properties, in all preceding equations, are determined at the mean film temperature

8.6.4. Inclined Plates For this case multiply Grashof number by cos 8, where 8 is the angle of inclination from

the vertical and use vertical plate constants.

8.6.5. Spheres Yuge (1959) has recommended the following correlation for free convection from a sphere of

diameter D. - Nu = 2 + 0.43 (Gr . ~ r ) ' "

for (1 c Gr. Pr c 16) and Pr = 1 ) ...( 8.48)

8.6.6. Enclosed Spaces In the literature, several correlations for heat transfer between surfaces at different temperatures

Heat and Mass Transfer - - -

Free Convection

separated by enclosed fluids are available. In this section, some correlations are presented that are per- Boundin tinent to the most common geometries. wall 9,

(a) Vertical spaces Jacob has suggested the following correlations

for vertical enclosed air spaces shown in Fig. 8.2. Q -

8.6.8. Concentric Sphere Spaces Refer Fig. 8.3. Raithby and Hollands have recommended the following correlations:

Q = ke 7~ (Di Do IL) (ti - to) ...( 8.55) and ke is expressed as

...[ 8.55 (a)]

L RaL where Ra, = (Gr . Pr), = 7/15 5

[(DoDir(Di7/15+DL ) 1'" for (2000 GrL c 2 x 104)' . ...( 8.49)

The above equations are valid ior ( lo2 I Ra, S lo4) a

- ke - (ii) - = Nu = = 0.064 (&dl" [fr* Fig. 8.2 Fne conveaion in vertical enclosure. k k 8.7. SIMPLIFIED FREE CONVECTION RELATIONS FOR AIR

The simplified relation for the heat transfer coefficient from different surfaces to air at atmopheric pressure, in general, is given by for ( 2 x 104 c GrL c ' l i x lo6) ...( 8.50)

where ke = effective thermal conductivity, h = c[$J ...( 8.57)

where C and m are constants depending on geometry and flow conditions; the significant length L is also a function of geometry and flow. Table 8.1 lists the values suggested by Mc Adarns for various geometries, orientations and flow conditions indicated by the magnitude of product (Gr . Pr).

H = height of the air space, and L = thickness of the air space.

Gr is based on thickness of air space, L. (6) Horizontal spaces The following correlation has been suggested by Jacob for hotizontal enclosed air spaces:

-

S. No.

Table 8.1: Simplified Free Convection Relations for Air

Surface and its orientation Laminar Turbulent ke - h L

(i) - = Nu = - = 0.195 (Grd1I4 for ( lo4 < GrL < 4 x l d ) k k ...( 8.31) Vertical plate or cylinder

- h = l A 2 T

(lo4 c Gr. r < lo9)

- h = 1.32 (~ t ) ' "

(lo9 < Gr. Pr < 1 0 ' ~ ) ke - ' L (ii) - = Nu = - = 0.068 (Grdl" for (4 x l d < G p k k ...( 8.52)

...[ 8.52 (a)]

- h = 1.25 (At)'"

(lo9 < Gr. Pr < 1012)

Horizontal cylinder

In the case of liquids contained in horizontal spaces, G1dbe;apd Dropkin have proposed the following correlation:

- NU = - 'L - - 0.069 (Grdl" (Pr)00"07

k for (3 x lo5 c Gr . P r c 7 x lo9) ...( 8.53)

Horizontal plate: Heated surface facing up

- h = 1.32 ($r

(lo5 < Gr. Pr < 2 x lo7)

- h = 1.67 (~t) '"

2 x lo7 < Gr. Pr < 3 x 10")

8.6.7. Concentric Cylinders Spaces Heated surface facing down

Spheres

Raithby and Hollands have recommended the following correlations for long horizontal concentric cylinders:

ke Pr I / I I

- k = 0.386 [0.861 +pry (RaC)lA

for lo2 I Ra, I lo7 ...( 8.54)

[In ( ~ d ~ i ) l ~ where Rac = (Gr. Pr)c = p Dr3/5 + R ~ L T-7 and ke is related as

- 114 k h = [2 + 0.392 (Gr) ]D ( 1 c Gr < 10')

8.8. COMBINED FREE AND FORCED CONVECTION Free convection effects are negligible when Gr cc Re2, at the other extreme, free convection

dominates when Gr >> Re2. There are certain situations where the free as well as the forced convection are of comparable magnitude. One such case is when air is flowing over a heated su#ace at a low velocity. A dimensionless group (GrlRe2) is used to delineate the convection regimes. The various flow regimes are characterised as follows:

(GrlRe2) 1 1 ... Pure free convection

2n ke = ln ( D m (ti - to)

Fig. 83. Long concentric cylinders (or spheres)


(crlRe2) z 1 ... Mixed (free and forced) convection

(GrlRe2) I 1 ... Pure forced convection.

8.8.1. External Flows The local Nusselt number Nu, for mixed convection on vertical plates is given by

Nu, = 0.332 (Re,)'/* ( ~ r ) ' " if (Gr,lRe~ ) I A ...( 8.58)

and Nu, = 0.508 ( ~ r ) ' " (0.952 + ~r)-l" ( ~ r , ) " ~ if (GrxlRex2) > A ...( 8.59)

where A r 0.6 for Pr I 10

and A = 1.0 for Pr = 100

For horizontal plates when (Gr, l~e?') I 0.083 the following equation for forced convection may be used

Nu, = 0.332 ~ e , ' ~ prln ...( 8.60)

8.8.2. Internal Flows (i) For mixed convection in laminar flow, Brown and Gauvin recommended a correlation of -

the form as

(r-3)

where Gz = Graetz number = Re Pr (D/L)

and &, CL, = Viscosities of the fluid at the bulk mean temperature and surface temperature respectively.

(ii) For mixed convection with turbulent flow in horizontal tubes, Metais and Eckert (1966) suggest

= 4.69 ( ~ e ) ' . ~ ~ (pr)O.'l (Gr)'.07 (D/L)P.~~ ... (8.62)

Example 8.1. A vertical cylinder 1.5 m high and 180 mm in diameter is maintained at 100°C in an atmosphere environment of 20°C. Calculate heat loss by free convection from the surface of the cylinder. Assume properties of air at mean temperature as, p= 1.06 kg/m3, V = 18.97 x 10- ' m2/s, C, = 1.004 k.//kg°C and k = 0.1042 kJ/mh°C. (AMIE Summer, 2000)

Solution. Given : L = 1.5 m; D = 180 mm = 0.18 m, ts = 100°C;

Heat loss by free convection, Q : p = pv = 1.06 x (18.97 x x 3600) = 0.07239 kg/rnh

Free Convection

Gr Pr = 2.208 x 101° x 0.6975 = 1.54 x 101° For this value of Gr Pr (tubulent range) - -

Nu - hL = 0.10 (Gr.pr)lB (for lo9 < Gr.Pr < 1012) L- k ...[ Eqn. (8.38)]

:. Range of heat loss, Q = hA (t, - t,) = 17.283 x (IT X 0.18 x 1'S) ~ ( 1 0 0 - 20)

= 1172.8 kYh (Ans.) Example 8.2. A cylindrical body of 300 mm diameter and 1.6 m height is maintained at a

constant temperature of 36.PC. The surrounding temperature is 13S°C. Find out the amount of heat to be generated by the body per hour if p= 1.025 kg/m3; cD = 0.96kJAg°C;

I v = 15.06 x 1 r 6 m2/s; k = 0.0892 W/m-h-'C and P = - K I . Assume Nu = 0.12 (Gr.pr)lA (the 298

symbols have their usual meanings). (AMIE Winter, 1997) Solution. Given : D = 300 mm = 0.03 m; L = 1.6 m; t, = 36S°C; t, = 13S°C;

1 k = 0.0892 Wm-h-"C; = - K- I ; Nu = 0.12 (Gr.pr)113 298

The amount of heat to be generated :

Grashof number, Gr = L3gP 0 s - t,)

,J

CLc PVC 1.025 x (15.06 x 10- x 3600) x 0.96 Prandtl number, Pr = --L = 2 = k k 0.0892

\

hL /

Nusselt number, Nu = - = 0.12 (Gr.pr)'I3 k

Heat lost from the surface by natural convection,

Q = hA (t, - t,) = 13.478 x (n x 0.3 x 1.6) (36.5 - 13.5) = *467.5 W h (Ans.)

* This is the amount of heat to be generated.

Example 8.3. A hot plate 1.2 m wide, 0.35 m high and at 115°C is exposed to the ambient still air at 2S°C. Calculate the following:

A' (i) Maximum velocity at 180 mm from the leading edge of the plate; (ii) The boundary layer thickness at 180 mm from the leading edge of the plate;


(iii) Local heat transfer coefficient at 180 mm from the leading edge of the plate; (iv) Average heat transfer coefficient over the suface of the plate; (v) Total mass flow through the boundary;

(vi) Heat loss from the plate; (vii) Rise in temperature of the air passing through the boundary.

Use the approximate solution.

Solution. Given : t, = 115OC, t, = 25OC, x = 180 mm = 0.18 m 4 + t 115 + 25 = 700C The mean film temperature, tf = - - 2 - * 2

The thermo-physical properties of air at 70°C are : p = 1.029 kg/m3; k = 0.0293 w/m°C; v = 20.02 x lo4 m2/s

Gr, = 2 gP 0 s - tw)

Grashof number v2

and GrL = 3.745 x lo7 x

(i) Maximum velocity at 180 mm from the leading edge of the plate, u-:

( , [""'a- "1'" urn = 0.766 v 0 952 + - (4'" ... [Eqn. (8.25)]

9.81 ~0.002915 (115-25) = 0.766 x 20.02 x lo4 (0.952 + 0.694)-In x (20.02 x 10-6)~

] (0.18)'~

V = 0.406 m/s (Ans.) (.; Pr = - a = 0.694 ... as given above)

(ii) The boundary layer thickness at 180 mm from the leading edge of the plate, 6: 6 - = 3.93 (0.952 + pr)'I4 (~r , ) - ' '~ ( ~ r ) - l n . . . [Eqn. (8.27)] X

= 3.93 (0.952 + 0.694)"~ (3.745 x lo7)-'" (0.694)-ln = 0.0683

or 6 = 0.0683 x 0.18 = 0.01229 m = 12.29 mm (Ans.)

(iii) Local heat transfer coefficient at 180 mm from the leading edge of the plate, h,: 2x

hxx - 0.508 (pr)ln (0.952 + pr)-lI4 ( ~ r ~ ) ' ~ = - Nu, =: - - k 6 ...[ Eqn. (8.31)]

" 2k 0'02964 = 4.823 wlmtoc (Ans.) h x = - x - - - - - x 6 - ij - 0.01229

(iv) Average heat transfer coefficient over the surface of the plate, % : k h = ; 10.677 (pr)ln (0.952 + ~r)- '" ( G I J ' ~ ] ...[ Eqn. (8.33)]

-- 0.02964 10.677 (0.694)ln (0.952 + 0.694)-I" (27.532 x 107)114] i - 0.35

Free ConvecX~on 475

(v) 1dtal mass flow through the boundary, mf : Thtal mass flow through the boundary can be calculated from the following formula :

m, = 1.7 p.v [(P:) (Pr + 0.952)

or m, = 0.00478 kg/s (Ans.) (vi) Heat loss from the plate, & :

Heat lost from the plate will be from both the sides, hence

Q = 2 Z A , (t,-t,) .

= 2 x 5.43 x (0.35 x 1.2) (115 - 25 ) = 410.5 W (Ans.) (vii) Rise in temperature of the air passing through the boundary, At:

Heat lost, Q = m, c, At or 410.5 = 0.00478 x (1.005 x lo3) x At

or At = 410.5 0.00478 x (1.005 x lo3)

= 85.4S°C (Ans.)

Example 8.4. A 350 mm long glass plate is hung vertically in the air at 24OC while its temperature is maintained at 80°C. Calculate the boundry layer thickness at the trailing edge of the plate. I f a similar plate is placed in a wind tunnel and air is blown over it at a velocity of 5 m/s, find the boundary layer thickness at its trailing edge.

Also determine the average heat transfer coeflcient, for natural and forced convection for the above mentioned data.

Solution. Given: L = 350 mm = 0.35 m, t, = 24OC, ts = 80°C, U = 5 rn/s

ts + t, Film temperature, !f=---- 80 + 24 - s2oC 2 - 2

The properties of air at 52OC are:

Boundary layer thickness, 6: Free convection:

Grashof number, Gr, = L3 gfl ( 4 - 1-1 (0.35)3 x 9.81 x 3.07 x 1W3 (80 - 24) - - v2 (18.41 x 1 0 ~ ) ~ = 2.133 x lo8

. Rayleigh number, RaL = Gr,. Pr = 2.133 x lo8 x 0.7 = 1.493 x lo8 This value of the Rayleigh number, according to Eqn. (8.36), indicates a laminar boundary layer. Using the Eqn. (8.27), we get

or 6 = 0.35 [ 3.93 (0.952 + 0.7 )'I4 (2.133 x 10')-'I4 (0.7)-'~*] = 0.0154 m = 15.4 mm (Ans.)

Forced convection:

Reynolds number,

476 Heat and Mass Ttansfer '

So the boundary layer is laminar. 5L a=----- x 0'35 = 0.00567 m = 5.67 mm (Ans.)

.JRe - d9.505 x 10

Thus the boundary layer thickness - in forced convection is less than that in free convection. Heat transfer coefficient h : Free convection : - - hL

Nu - - = 0.677 ( ~ r ) ' " (0.952 + pr)-'I4 ( ~ 1 ) ' ' ~ L - k

...[ Eqn (8.33)]

k or h = - x 60.378 =

L 28'15 lo-' x 60.378 = 4.856 W / ~ ' O C (Ans.)

0.35

Forced convection : -

k or h = - x 181.78 =

L 28'15 x 18133 = 14.62 w l m 2 ' ~ : ( A m ) ,

0.35

Thus it is seen that heat transfer coefficient in forced convect?m is muc in free convection.

Example 8.5. A sheet metal air duct carries air-conditioned air at an average tempe+arure of 10°C. The duct size is 320 mm x 200 mm and length of the duct exposed to the surrounding air at 30°C is 15 m long. Find the heat gain by the air in the duct. Assume 200 mm side is vertical and top su$ace of the duct is insulated. Use the following properties :

Nu = 0.6 ( G r . ~ r ) " ~ j for vertical sugace.

Nu = 0 . 2 7 ( ~ r . ~ r ) " ~ ~ f o r horizontal su$ace.

Take the properties of the air at mean temperature of (30 + 10)/2 = 20°C as given below.

c,,=lOO J /kgK; p=1.204kg/m3; p= 18.2 x 1 r 6 ~ - s / m ~ ;

v = 15.1 x 10- m2/s; k = 0.256 W/m K and Pr = 0.71. (M.U. Winter, 1998)

Solution. Refer Fig. 8.4.

There will be two vertical surfaces and one horizontal surface from which the heat Air conditioned

will be gained by natural convection. There F / air is no heat transfer from the top surface as it is insulated.

Now, (Gr), = g w 3 (At) v2

9.81x( 273 + 20 ) x 0 . 2 ~ x ( 3 0 - 1 0 )

Free Conve~tion 477

h, H (Nu), = - -

k - 0.6 (16.68 x 1 0 ~ ) " ~ ~ = 38.34

Qv = hv x A, x (At) = 49.07 x (2 x 15 x 0.2) x (30 - 10) = 5888.4 W 1

h v x B (Nu), = --

k - 0.27 (6.83 x 107)0.25 = 24.54

Qh = hh xAh x (At) = 19.63 x (15 x 0.32) x (30 - 10) = 1884.5 W

:. Total heat gain, Q = Qv + Qh = 5888.4 + 1884.5 = 7772.9 W (Am.)

Example 8.6. Airflows through long rectangular heating duct of width and height of 0.75 m and 0.3 m respectively. The outer surjoce temperature of the duct is maintained at 45OC. If the duct is exposed to air at 15OC in a cramp-space beneath a home, what is the heat loss from the duct per metre length ?

use' the following correlations : (i) Top sugace :

Take B LC= -= 0.375 m 2 -

NuL = 0.54 (Ra$25 if 104 5 Rat 5 107

= 0.15 ( ~ a 3 " ~ ~ if lo7 5 Ra, 5 10" (ii) Bottom surface :

Take B LC -- = 0.375 nz - - 2 N U ~ = O . ~ ~ ( R ~ J ~ . ~ ~ , 105 I R q S 101°

(iii) Sides of the duct :

Use the following properties of air :

v = 1 6 . 2 ~ 1 0 - ~ r n ~ / s , a = 2 2 , 9 ~ 1 0 - ~ m ~ / s

k=0.0265 W/mK, p=0.0033 K-' and Pr=0.71. b

(N.M.U. Winter, 1995) Solution. The configuration of the duct is shown in Fig. 8.5

...[ Eqn. (8.36)] I1 I

I Fig. 8.4


a - -

inside the L

Fig. 8.5

- - 9.81 x 0.0033 X (45 - 15) L: = 2.62 x lo9 (L:)

(22.9 x x (16.2 x For two sides :

Lc=H=0.3m

. . Ra, = 2.62 x lo9 (0.31~ = 7.07 x lo7

For top surface :

For bottom surface :

The rate of heat loss per unit length of the duct is given by,

Q=Qs+Qt+Qb = [2 x h, x (0.3 x 1) + h, x (0.75 x 1) + h,, x (0.75 x 1) (ts - fa)

- I = [2 x 4.23 x 0.3 + 4.127 x 0.75 x 1 + 1.75 x 0.75 x 11 (45 - 15)

= 208.4 W/m (Ans.)

Free Convection 479

Example 8.7. A vertical plate measuring 180 mm x 180 mm and at 50°C is exposed to atmosphere at IO°C. Compare the free convection heat transfer from this plate with that which would result due to forced convection over the plate at a velocity equal to twice the maximum velocity which would occur in free convection boundary layer.

Solution. Free convection : Velocity 50 + 10 profile -

The film temperature, tf = - = 30°C. 2

The thermo-physical properties at 30°C are :

1 I 0.0033 K-I

= (++ 273) - 303 -

- - (0.1 8)3 x 9.81 x 0.0033 x (50 - 10) (16 x 1 0 ~ ) ~

= 2.95 x lo7

Since Gr < lo9, hence the flow in laminar.

Boundary layer

For laminar flow, using the following relation, we get, Fig. 8.6 - - hL

NU, = = 0.677 (~r)'' (0.952 + ~r)- '" (~r,)"'

The heat lost from one side of the plate, - QM = h A, At = 5.47 x (0.18 X 0.18) x (50 - 10) = 7.089 W

Forced convection :

\ /

The average heat transfer coefficient with forced convection if velocity is assumed equal to 2u,, :

EL = 0.664 (Re)'' (pr)ln

480 Heat and Mass frmsf&r

Heat lost due to forced convection, - Qfirced = h A, At = 7.039 x (0.18 x 0.18) (50 - 10) = 9.12 W

$Free Convection 481

Solution. Given : A = 1 x 0.5 = 0.5 m2; tp (= t,) = 130°C; t, = 20°C;

m = 20 kg; cp = 400 J/kg K - Plate

- 7.089 = 0.777 (Ans.) Hence - - Qforced 9-l2

Vertical Example 8.8. Two vertical plates, each ,,late

120 mm high and at 85OC are placed in a tank \ ofwuter at 15°C. Calculate the minimum spacing which will prevent interference of the free convection boundary layers.

Solution: Let 6 = boundary layer thickness at the trailing edge of the plate.

L = 26 = the minimum spacing required.

85 + 15 The film temperature, tf = ----- = 50°C

2 The thermo-physical properties of water at

50°C are: k = 0.674 Wm°C; v = 0.556 x m2/s

Fig. 8.7

G r . Pr = 11.88 x lo9 x 3.54 = 42.055 x lo9

Since Gr. Pr > lo9, hence the character of the flow is turbylent. For turbulent flow, using the following relation, we get

6 = 0.0306 x 0.12 = 0.003672 m = 3.672 mm L = 26 = 2 x 3.672 = 7.344 mrn ( ~ n s . j

Example 8.9. A hot plate 1 m x 0.5 m at 130°C is kept vertically in still air at 20°C. Find :

( i ) Heat transfer coeflcient,

(ii) Initial rate of cooling the plate in OC/min.

(iii) Time required for cooling plate from 180°C if the heat transfer is due to convection d y . Mass of the plate is 20 kg and cp = 409 J/kg K.

Assume 0.5 m side is vertical and that the heat transfer coeflcient calculated in ( i ) above remains constant and convection takes place from both sides of the plate.

130 + 20 Take properties of air at - = 75OC as :

2

c, = 1007J/kgOC, p = 1.07 m2/s; k = 0.029 J /kg K; V = 19.1 x m2/s (N.M.U., 5998) /

= 1.06 x lo9

:. Gr.Pr = 1.06 x 109 x 0.709 = 0.75 x lo9

The heat transfer coefficient for the vertical plate is given by

EL = 0.59 ( G r . ~ r ) ' / ~ for ( lo4 < Gr.Pr < lo9) ...[ Eqn. (8.37)] Convection

Substituting the - values in the above eqn., we get cuttents - hL Nu - - = 0.59 (0.75 x 1 0 ~ ) " ~

L - k Fig. 8.8

(ii) Initial rate of cooling the plate in OCImin. :

Heat lost from both sides of the plate is given by

Q = 2 h x A (ts - t,) = 2 [5.66 x ( 1 x 0.5) x (130 - 20)] = 622.6 W [ I Also, Q = mc,(At); where At is the rate of cooling.

. . At= 622'6 = O.O778OC/s = 4.668OUmin (Am.) 20 x 400

(iii) Time required for cooling plate from 180°C to 80°C, z : The instantaneous heat lost by the plate is given by

- 20 x 400 80 - 20 . . Z= 2 x 5 . 6 6 ~ (1 ~ 0 . 5 ) 1n(180-20)

= - 1413.43 x (- 0.9808) = 1386 s or 0.385 h (Am.) Example 8.10. A square plate 40 cm x 40 cm maintained at 400 K is suspended vertically

in atmospheric air at 300 K. ( i ) Determine the boundary layer thickness at trailing edge of the plate. (ii) Calculate the average heat transfer coeflcient using a relation


Nu = 0.516 (Gr, . ~ r ) " ' ~ Take the following properties of air :

v = 20.75 x lo4 m2/s; k = 0.03 W/m°C; P = 2.86 x 1 r 3 K-*; Pr = 0.7 (P.U> Solution. Given : t, = 400 K, t, = 300 K ( i ) Boundary layer thickness, 6: The boundary layer thickness is given by

6 = x 13.93 (0.952 + pr)'I4 (Gr)-'I4 (pr)-lJ2 ] ...[ Eqn. (8.27)]

Gr, = 2 go (At) - (0.4)'~ 9.81 x (2.86 x x (400 - 300) - = 4.17 x lo8

v2 (20.75 x 10")~ Substituting the values in the above equation, we get

6 = 0.4 [ 3.93 (0.952 + 0.7)'" (4.17 x 10' )-'I4 (0.7)-'"1 = 0.0149 m = 14.9 mm (Ans.)

(ii) Average heat transfer coefficient, h : Gr. Pr = (4.17 x 10') (0.7) = 2.919 x lo8

- h L Nu = - = 0.5 16 (Gr . ~r) ' .~ '

k

or . - k h = - x 0.516 ( Gr. ~ r ) ' . ~ ~ L

= $ x 0.5 16 (2.919 x 10')'.~' = 5.058 WII~"C (Ans.)

Example 8.11. A nuclear reactor with its core constructed of parallel vertical plates 2.2 m high and 1.4 m wide has been designed on free convection heating of liquid bismith. The maximum temperature of the plate su$aces is limited to 960°C while the lowest allowable temperature of bismith is 340°C. Calculate the maximum possible heat dissipation from both sides of each plate.

For the convection coeflcient, the appropriate correlation is Nu = 0.13 ( G r . ~ r ) ' . ~ ~ ~

where diferent parameters are evaluated at the mean film temperature.

Solution. The mean film temperature, tf = 960 + 340 = 650°C 2

The thermo-physical properties of bismith are:

p = lo4 kg/m3 ; p = 3.12 kglm-h; c, = 150.7 Jkg°C; k = 13.02 W/m°C

Gr = L ~ ~ ~ ~ P A ~ (2.21~ x ( 1 0 ~ ) ~ 9.81 x 1.08 x x (960 - 340) - -

cl' (3.1 2/3600)2 = 9.312 x 10''

Gr.Pr = 9.312 x 1015 x 0.01 = 93.12 x lo1* Using the given correlation, we get

N U = - - hL - 0.13 ( ~ r . ~ r ) ' ~ ~ ~ ' k

Free Convection

-- - 13'02 x 0.13 ( 93.12 x 10 '~ )"~ = 34500 W I ~ ~ " C 2.2

:. Heat dissipation from both sides of each plate, Q = 2 h A , At

= 2 x 34500 x (2.2 x 1.45) x (960 - 340)

= 136.47 x lo6 W = 136.47 MW (Ans.)

Example 8.12. Find the convective heat loss from a radiator 0.6 m wide and 1.2 m high maintained at a temperature of 90°C in a room at 14°C. Consider the radiator as a vertical plate.

Solution. Given : L = 1.2 m, B = 0.6 m, t, = 90°C. t, = 14°C.

Convective heat loss, Q :

Film temperature

The thermo-physical properties of air at 52°C are:

k = 28.15 x W/m°C; v = 18.41 x 10" m2/s, Pr = 0.7;

L3g (3 (4 - t J Rayleigh number = Ra, = GrL.Pr = . Pr

v2

= 11.69 x lo9

From Eqn. (8.36) it follows that the flow becomes turbulent on the radiator plate. Using Eqn. (8.40), we have

The convective heat Kss, Q = h A, At = 6.187 x (1.2 x 0.6) (90 - 14) = 338.55 W (Ans.)

Example 8.13. A gas at 195°C is flowing through a thin walled vertical duct which is in the form of circular cross-section having diameter of 450 mm. The ambient air at 15"C, which may be considered still, surrounds the duct. Find the rate of heat transfer using the following simplified relation for air for laminar flow.

where, L is the length of the duct in metres.

484 Heat and Mass Transfw

Solution. Given : At = 195 - 15 = 180°C, D = 450 mm = 0.45 m

\ 1 :. The rate o f heat transfer,

Q = A, At = 5.055 x (n x 0.45 x 1 ) x 180 = 1286.34 Whn length. (Ans.) Example 8.14. A horizontal heated plate measuring -1.5 m x 1.1 m and at 215"C, facing

upwards, is placed in still air at 25°C. Calculated the heat loss by natural convection. The convective film coeficient for free convection is given by the following empirical relation:

h = 3.05 (T,)'" w/m2"c where, T f is the mean jilm temperature in degrees kelvin.

Solution. Given: A, = 1.5 x 1.1 = 1.65 m2, t, = 215"C, t, = 25°C. Mean film temperature,

. . h = 3.05 (393)'" = 13.58 w/rnZOc Rate of heat loss, by natural convection,

Q = h A, (t, - t,) = 13.58 x 1.65 x (215 - 25) = 4257.33 W (Ans.) Example 8.15. A transfonner is cooled by immersing in an oil bath which is housed in a

cylindrical tank which is 0.8 m in diameter and 1.3 m long. If the electrical loss is 1.3 kW, calculate the surface temperature of the tank; the entire loss of electrical energy may be assumed to be due to free convection from the bottom of the tank. For this case the following simplified relations for the boundary layer are applicable :

... For a cylindrical plane;

A 0.25 h = 1.45 ... For n vertical plane.

Assume the ambient air temperature as 20°C. Solution. Given : D = 0.8 m, L = 1.3 m. The total rate o f heat transfer from the tank,

Qtoml = Qsge + Qtop ...( i )

Qside = hside X Aside X At

Substituting the values in eqn. (i), weTget

1.3 x lo3 = 4.13 ( t - 20)'." + 0.77 (t - 2 0 ) ' ~ ~ ~ = 4.9 (t - 2 0 ) l . ~

1.3 x 10 3 1A.25

or (t - 20) = [ 4.9 ] = 86.89

or t = 86.89 + 20 = 106.89°C (Am.) Example 8.16. A steam pipe 7.5 em in diameter is covered with 2.5 cm thick layer of

insulation which has a suface emissivity of 0.9. The surface temperature of the insulation is

R&e Convection 485

80°C and {he pipe is placed in atmospheric air at 20°C. Considering heat loss both by radiation and natural convection, calculate:

(i) The heat loss from 6m length of the pipe; (ii) The overall heat transfer coeficient and the heat transfer coeficient due to radiation

.5 + 2 x2.5 = 12.5 cm=0.125 m, E =0.9, t,= 80°C, t,= 20°C

The mean film temperature, ' f= 80 + 20 = 50°C. 2

The thermo-physical properties of air at 50°C are: p = 1.092 kg/m3; c, = 1007 J/kg°C; p = 19.57 x 10" kglms; k = 27.81 x WIm°C

Using

(i) Heat loss from 6m length of pipe: Heat lost by convection, -

Q,, = h A, At = 6.24 x (n x 0.125 x 6) x (80 - 20) = 882.16 W Heat lost by radiation,

Q , . ~ , = E OA ( T ~ ~ - T ~ ~ ) = 0.9 x (5.67 x lo-') x (n x 0.125 x 6 ) [(80 + 27314 - (20 + 273)4] = 980.81 W

Total heat loss, Qt = Q,, + Qrd, = 882.16 + 980.81 = 1862.97 W (Ans.)

(ii) Heat transfer coefficients, overall (hJ and due to radiation alone (h,,) : Q, = h,. A At l

I . . h, = ht - h,, = 13.17 - 6.24 = 6.93 W/mZOc (Am)

Example 8.17. Calculate the heat transfer from a 60W incandescent bulb at 115°C to 1 ! ambient air at 25°C. Assume the bulb as a sphere of 50 mm diameter. Also, find the percentage of power lost by free convection.

The correlation is given by: Nu = 0.60 ( ~ r . ~ r ) " I '

Solution. Given : t, = 1 10°C, t.=., = 25"C, D = 50 mm = 0.05 m.

486 Heat and Mass Transfer Free Convection 487

The film temperature, +=-= < + 1 1 5 + 2 5 = 7 0 0 C 2 2

The thermo-physical properties of air at 70°C are:

k = 2.964 x W/m°C, v = 20.02 x lo4, Pr = 0.694,

('.' The characteristic length L = D in this case) Using given correlation, we get

- Heat transfer = h A , At = 9.72 x (x x 0.05~) x (1 15 - 25) = 6.87 W (Ans.)

%age of power lost by free convection = 6.87 x LOO = 11.45 Z (Ana) 60 Example 8.18. A horizontal high pressure steam pipe of 10 cm outside diameter passes

through a large room whose walls and air are at 23°C. The pipe outside suface temperature is 465°C and it emissivity is 0.85. Estimate the heat loss from the pipe per unit length. Use the following correlation for the calculation of film coeflcient

where Ra is known as Rayleigh Number and is given by

v a Take the following properties of air :

a = 3 2 . 8 ~ 1 0 - ~ r n ~ / s , Pr=0.697, P = 2 . 7 2 5 x l ~ ' ~ ~ - ' . (N.M.U. Wincdr, 1995)

Solution. Given : D = 10 cm = 0.1 m; t, = 23OC; t, = 165°C; E = 0.85

Properties of air : k = 0.0313 W/m K; v = 22.8 x m2/s;

a = 32.8 x m2/s, Pr = 0.697, P = 2.725 x K- ' Heat loss from the pipe per unit length, Q : Heat lost by convection and radiation from the sphere is given by :

...( where L = 1 m) = 1.514 [368.04 + 76.761 = 441 W/m

QCon,. = h x (t, - t,) where h is calculated by using the given relation.

Ra = Gr.Pr = Pg (At) (L,)' 2.725 x 10-~,x 9.81 X (165 - 23) X (0.1)' = - - va 22.8 x x 32.8 x

.. Q,,,,, = 7.2 x (x x 0.1 x 1) (165 - 23) = 321.2 W, when L = 1 m

Hence, Q = 441 + 321.2

= 762.2 W/m (Ans.) Example 8.19. A vertical slot of 20 mm thickness is formed by two 2 nr x 2 m square plates.

I f the temperatures of the plates are 115OC and 25°C respectively, calculate the following: ( i ) The effective thermal conductivity;

(ii) The rate of heatflow through the slot.

Solution. Given: L = 20 mm = 0.02 m; t1 = 115°C; t2 = 25°C (i) The effective thermal conductivity, k, :

The mean temperature between the plates = 115 + 25 = 700C 2

The properties of air at 70°C are:

k = 0.0295 W/m°C; v = 2.0 x lo-' m2/s

The Grashof number based on the gap L, is

= 5.147 x lo4 I

The effective thermal conductivity is given by

...[ Eqn. (8.50)]

Q = Qconv. + Qrad.

Heat and ~ a & ~iarisfer Frpe Co 488 489

(ii) The rate of heat flow through the slot Q :

Q = ke . A . (F)

Example 8.20. Two horizontal panels separated by a distance of 30 mm contain air at atmaspheric pressure. The temperatures of lower and upper panels are 55OC and 20.6OC respectively. Calculate the free convection heat transfer per m2 of the panel surface.

Solution. Given. L = 30 mm = 0.03 m, tl F S ° C , t2 = 20.6"C

The mean temperature of two panels = 55 + 20.6 = 37.80C 2

The properties of air at 373°C are: p = 1.121 kglm3 ; v = 0.171 x 104m2/s

t e

k = 29.2 x low3 W/m°C; Pr = 0.70

= 3.22 x K-' ' = 37.8 + 273

Free convection heat transfer per m2 of the panel surface, q:

- a.

Nu = 0.195 (Gr$" ...[ Eqn. (8.51)1

or NU = 0.195 (10.03 x lo4)lI4 = 3.47

Also, L Z L - Nu - k 0 1 - t2)

(where q = QlA) ...[ Eqn. 8.52 (a)]

Example 8.21. (Mixed convection) Air at 200C and 1 atmosphere is forced throvgh a 25 mrn diameter tube 400 mm long, at an average velocity of 0.33 d s . Calculate the rate of heat transfer if the tube wall is maintained at 1800C.

Solution. Given: tb = 20°C, t, = 180°C, V = 0.33 m/s, L = 400 mm = 0.4 m, D = 25 mm = 0.025 m.

Rate of heat transfer :

180 + 20 = ]OoOC t *

The mean film temperature, tf = 2

The properties of air at 100°C are: k = 3.208 x 1w2 W/mOG, v..= 23.13 x 1@ ~ s / m ~ , Pr = 0.688

" *

= 2.68 x 1p3 ' = 100 + 273

For the bulk temperature of 20°C, we have lb = 18.15 x ~ s / m ~

Again, for the surface temperature of tube wall (i.e., 180°C) .

& (viscosity at the wall) = 25.19 x lo4 ~ s / m ~

Reynolds number

Thuy'the nature of flow is laminar.

drashof number, - 038 P 0 s - t,) I

Gr = 3

Since [ G r / ( ~ e ) ~ ] t: 1, hence it is a case of mired (free and forced) convection and the following equation may be used for calculation of Nu.

...[ Eqn. (8.61)]

where, Gz = Graetz number = Re.Pr = 356.68 x 0.688 I"::) - = 15.34

- k- h = - N u = D

3'208 x 7.7 = 9.88 w/m2"c 0.025

Q = h A, At = 9.88 x (n X 0.025 x 0.4) x (180 - 20) = 49.66W (Ana.)

TYPICAL EXAMPLES Example 8.22. Derive a relation between Grashof and Reynolds numbers assuming the heat

transfer coeficients over vertical plates for pure forced and free convection are equal in laminar flow.

Solution. For pure forced convection: Zu = 0.664 (Re)" (pr)ln - ...( i ) ' For pure natural convection: Nu = 0.677 (~r)l) l" (0.952 + (Gr)'" ...( ii)

Equating equations (i) and (ii), we get 0.664 (Re)ln ( ~ r ) ' " = 0.677 (pr)In (0.952 + pr)-'I4 (Gr)'I4

or ( ~ r ) ~ = (pr12 (0.952 + ~r)-l (Gr)

Example 8.23. Two horizontal steam pipes having diameters I00 mm and 300 mm are so laid in a boiler house that the mutual heat transfr m a y be neglected. The surface temperature of each of the steam pipes is 475OC. If the temperature of the ambient air is 35OC, calculate the mtio of heat transfer coeficients and heat losses per metre length of the pipes.

Solution. Given : Dl = 100 mm = 0.1 m D2 = 300 mm = 0.3 m, t, = 475OC, t- = 35°C The steam mains are located in the boiler house where the ambient air is stationary, thus

his is a case of free convection for which the following relation applies :


Since, the surface temperature of both pipes is the same and both are exposed to the same ambient conditions, therefore, thermo-physical properties are the same.

. . Nu = (D3)lI4 = ( D ) ~ ' ~

:. The ratio of heat transfer coefficients,

h 1 1/4 1 / 4

% = (2) = (z) = 1.316 (Ans.)

Also. Q, - h, Dl

:. The ratio of heat losses, - - - = 1.316 x = 0.438 (Ans.) Q2 h2 0 2

Example 8.24. A rqom is heated by initiating fire at the fireplace. The exfitration of room air through a chimney ih reduced by the use of a glass-door fire screen of height 0.8 m and width of 1 .I m. If the ambient air temperature is 2J°C and the sulface temperature attained by the glass is 175'C, calculate the free convectin heat transfer rate from the fire place to the room. For this case the followibg correlation given by Churchill and Chu is valid:

Solution. Given: t, = 17S°C, t, = 2S°C, L = 0.8m, B = 1.1 m

The film temperature,

The thermo-physical properties of air at 100°C are: k = 3.208 x WIIII~~C, v = 23.13 x lo4 m2/s, Pr = 0.688

1 p = - = 1 = 2.68 x Tf (100 + 273)

Grashof number,

Free Convection

Free convection heat transfer rate, -

Q = h A, (t,, - t_) = 6. = 871.2W (Ans.)

Example 8.25. A 2-stroke motor cycle petrol engine cylinder consists of I5 fins. I f the outside and inside diameters of each fin are 200 mm a d 100 mm respectively, the overage fin su$ace temperature is 475OC and atmosphe;ic air temperature is 2S°C, calculate the heat transfer rate from the fins for the following cases:

(i) When the motor cycle is stationary; (ii) When the motor cycle is running at a speed of 60 kmh. The fin may be idealised as single horizontal flat plate of the same area. Solution. Case (i) When the motor cycle is stationary the heat transfer will be due to free

convection.

The mean film temperature, tf =

475 + 25 2 = 250°C.

The thermo-physical properties of air at 250°C from the tables are:

k = 4.266 x 1W2 W/m°C; v = 40.61 x 106 m2/s; Pr = 0.677

where significant length L = 0.9 D

Gr.Pr = 2.985 x lo7 x 0.677 - 2 x lo7

Since Gr.Pr c lo9, hence the nature of flow is laminar. -

Since the heat transfer is from both sides of the fins, therefore,

[ n

Q = A, At = 8.56 2 x 15 x - x (0.2~ - 0.13 [475 - 251 4

= 2722.82 W (Ans.) 1 Case (ii) : When the motor cycle is running the heat transfer will be due to forced convection. The velocity of air on the fin = speed of the motor cycle

Reynolds number

Hence the flow is turbulent. For turbulent flow, for cooling, -

Heat and W S s transfer Free Convection 493

:. The heat transfer from the cylinder,

= 18713 W (Ans.)

Example 8.26. Gas at 350°C is conveyed through 250 mm diameter pipe laid in an atmosphere of quiescent air at 20°C.The convection heat transfer coeficient from a hot cylindrical su$ace ht 0.25

freely exposed to still air is prescribed by the relation: h = 1.55 w/mZOc where At is the ., - -

temperature difference and D is the diameter of the cylinder is metres. ( i ) Calculate the heat loss per metre length of the bare pipe. (ii) Estimate the percentage reduction in heat loss $the pipe is covered with 75 mm thick

laver o f material whose thermal conductivi~ is 0.072 W/maC. Any temperature drop in thk meial may be neglected.

Solution. Convective heat transfer coefficient, 350 - 0.25

h = 1.55 [ 0.25 20] = 9.34 w l m 2 ~ c

(i) Heat loss from the bare pipe: Convective heat loss per metre length of the bare pipe.

0 = h A. At = 9.34 x (n x 0.25 x 1 ) x (350 - 20) = 2420.75 W (Ans.) - (ii) Percentage reduction in heat loss when pipe is lagged :

Dl = 250 mm = 0.25 m, 4 = 250 + (2 x 75) = 400 mm = 0.4 m

On attainment of steady state, the heat conducted through the lagging equals the convective heat loss from the outer su$ace of the lagging. Thus

271 k L (tl - t ) = h x ( X D2 L) (t - tm)

In (r2/r1>

where t is the temperature at the outer surface of the lagging. Substituting the appropriate values, we get

2n 0.072 (350 - ') ; 1.55 (27 x (n x 0.4 x 1) (t - 20) In (0.2/0.125)

Or 0.962 (350 - = 2.449 (t - or 350 - t = 2.545 (t - 2 0 ) ' . ~ ~

By trial and error, we get, t z 64°C

:. Heat loss from the lagged pipe, 2 n k L ( t l - 1) - - --275.28w

= ln (r2/r1) In (0.2/0.125)

2420.75 - 275.28 . Percentage reduction in loss = 2420.75

HIGHLIGHTS 1. Free or natural convection is the process of heat transfer which occurs due to movement

of the fluid particles by density changes associated with temperature differential in a fluid. 2. List of formulae :

I . Flow over vertical plates (Laminar $ow)

(a) Exact solution:

- (i) Nu, = 0.676

- (ii) Nu = 0.902

(b) Approximate solution:

(where Gr, = gP 0 s - L ) X 3 v 3 and Pr = - )

a P (vi) m = - 6 . u, 12 .

(viii) Total mass, m = 1.7 pv [ ( ~ r - ) ' (Pr + 0.952) h x 2x

(ix) Nu, = A = 0.508 ( ~ r ) ' " (0.952 + Pr)-'I4 (Grx)'I4 = - k 6

- - hkL - Nu, = 0.677 (Pr)" (0.952 + Pr)-'" (GrL)lM (x) NuL=- - -

11. Flow over vertical plates (Turbulent jlow):

6 (xi) - = 0.565 ( ~ r ) - ~ ' ~ [

1 + 0.494 ( ~ r - ) ~ ~ ~ ] O" X G rx

(xii) Nu, = 0.0295 ( ~ r ) ~ / ' ~ [ I + 0.494 (Pr)"

UNSOLVED EXAMPLES


1. A hot plate 1.1 m wide, 300 mm high and at 120°C is exposed to the ambient still air at 20°C. Calculate : ( r ) Maximum velocity boundary layer thickness and local heat transfer coefficient at 150 mm from the leading edge of the plate, (ii) Total mass flow through the boundary, (iii) Heat loss from the plate and (iv) Rise in temperature of the air passing through the boundary.

[Ans. (i) 0.39 mls, 11.4 mm, 5.199 W/m2"C, (ii) 0.004372 kgls,]

2. A 300 rnm long glass plate is hung vertically in the air at 27OC while its temperature is maintained at 77°C. Calculate the boundary layer thickness at the trailing edge of the plate. If a similar plate is placed in a wind tunnel and air is blown over it at a velocity of 4 mls, find the boundary layer thickness at its trailing edge. Also determine the heat transfer coefficient, for natural and forced convection for the above mentioned data. [Ans. 15.2 mm, 5.67 mm; 4.9 W/m2"C, 14.11 WlmZ0CI

3. A nuclear reactor with its core constructed of parallel vertical plates 2.25 m high and 1.5 m wide has been designed on free convection heating of liquid bismuth. The maximum temperature of the plate surfaces is limited to 975°C while the lowest allowable temperature of bismuth is 325°C. Calculate the maximum possible heat dissipation from both sides of each plate. For the convection coefficient, the appropriate correlation is

Nu = 0.13 (Gr . pr)' 333

where different parameters are evaluated at the mean film temperature. (Ans. 153 MW)

4. Find the convective heat loss from a radiator 0.5 m wide and 1 m high maintained at a temperature of 84°C in a room at 20°C. Consider the radiator as a vertical plate. (Ans. 110.08 W)

5. A steam pipe 60 mm in diameter is covered with 20 mm thick layer of insulation which has a surface emissivity of 0.92. The surface temperature of the insulation is 75OC and the pipe is placed in atomspheric air at 25OC. Considering heat loss both by radiation and natural convection, calculate : (i) The heat loss from 5 m length of the pipe; (ii) The overall heat transfer coefficient and the heat transfer coefficient due to radiation alone.

(Ans. (i) 1049.83 W, (ii) 7.07 W / ~ ~ " C )

6. Calculate the heat transfer from a 60 W incandescent bulb at 125°C to ambient air at 25°C. Assume the bulb as a sphere of 50 mm diameter. Also, find the percentage of power lost by free convection. The correlation is given by : Nu = 0.60 (Gr. pr)'I4 (Ans. 7.71 W; 19.278%)

7. A vertical slot of 15 mm thickness is formed by two 2m x 2m square plates. If the temperature of the plates are 120°C and 20°C respectively, Calculate the following: (i) The effective thermal conductivity; (ii) The rate of heat flow through the slot. [Ans. (i) 0.0317 W/m°C, ( i i ) 845.3 W]

8. Two horizontal plates separated by a distance of 25.4 mm contain air at atmospheric pressure. The temperatures of lower and upper panels are 60°C and 15.6"C respectively. CaIculate the free convection heat transfer per mZ of the panel surface.

(Ans. 167.2 w/m2)

Boiling and Condensation (Heat Transfer with Phase Change)

I 9.1. Introduction, 9.2. Boiling heat transfer : General aspects - Boiling regimes - Bubble Laps

and size consideration - Bubble growth and collapse - Critical diameter of bubble - factors affecting nucleate boiling - Boiling correlations. 9.3. Condensation heat transfer: General aspects - Laminar flow condensation on a vertical plate - Turbulent film condensation - Film condensation on horizontal tubes - Film condensation inside horizontal tubes - Influence of the presence of noncondensable gases - Highlights - Theoretical Questions - Unsolved Exam~les.

I I 9.1. INTRODUCTION

We have considered in the preceding2hapter b convective heat transfer, homogeneous single phase systems only. However. there are specific convection processes as boiling a n d condensation which a re associated with change of phase. Whereas boiling involves change from liquid to vapour phase of a fluid substance, condensation refers to a change from the vapour to a liquid phase,

The mode of heat transfer with change of phase (i.e. boiling and condensation processes) finds wide applications as mentioned below:

(i) Cooling of nuclear reactors and rocket motors; (ii) Steam power plants (Boilers and condensers); (iii) Refrigerating and airconditioning systems (Evaporators and condensers); (iv) Melting of metal in furnaces; (v) Refineries and sugar mills (Heat exchangers); (vi) Process heating and cooling etc.

Boiling and condensation processes entail the following unique features:

(i) As a consequence of phase change in these processes, the heat transfer to or from the fluid can occure without influencing the fluid temperature.

(ii) The heat transfer coeficient and rates, due to latent heat associated with phase change, are generally much higher compared with the normal convection process (i.e., without phase change)

(iii) High rate of heat transfer is achieved with small temperature direrenee.

The phenomena associated with boiling and condensation are much more complex (than the normal convection process) due to the following being very significant:

(i) Latent heat effects; (ii) Surface tension; (iii) Surface characteristics and other properties of two phase systems.

9.2. BOILING HEAT TRANSFER

9.2.1. General Aspects

Boiling is the convective heat transfer process that involves a phase change from liquid to

Boiling and Condensation Heat and Mass Transfer 497

vapour state. Boiling is also defined as evaporation at a solid-liquid surface. This is possible only when the temperature of the surface (t,) exceeds the saturation temperature corresponding to the liquid pressure (tsa,). Heat is transferred from the solid surface to the liquid according to the law

Q = h A, (t, - t,,,) = h A, A t, ...( 9.1)

where At, = (t, - tsa$ is known as excess temperature. The boiling process finds applications in the following cash:

(i) Steam production (for generation of power and for indps&al processes and space heating) in steam and nuclear power plants;

(ii) Heat absorption in refrigeration and air conditioning systems; (iii) Distillation and refining of liquids; (iv) Concentration, dehydration and drying of foods and materials, (v) Cooling the machines like nuclew reactors and rocket motors where the large quantities

of heat are released in relatively small volume (dissipation rates are as high as lo8 w/m2; the maximum heat transfer rate in modem boiler is about 2 x lo5 w/m2).

The boiling heat transfer phenomenon may occurs in the following forms: 1. Pool boiling In this case the liquid above the hot surface is essentially stagnant and its motion near the

surface is due to free convection and mixing induced by bubble growth and detachment. The pool boiling occurs in steam boilers involving natural convection. 2: Forced convection boiling This refers to a situation where the fluid motion is induced by external means (and also by

free convection and bubble induced mixing). The liquid is pumped and forced to flow. This type of boiling occurs in water tube boilers involving forced convection. 3. Sub-cooled or local boiling In this case the liquid temperature is below the saturation temperature and bubbles are formed

in the vicinity of heat surface. These bubbles after travelling a short path get condensed in the liquid which has a temperature less than the boiling point.

4 . Saturated boiling Here, the liquid temperature exceeds the saturation temperature. The vapour bubbles formed

at the solid surface (liquid-solid interface) are then propelled through the liquid by buoyancy effects and eventually escape from a free surface (liquid-vapour interface).

9.2.2. Boiling Regimes The process of boiling depends upon the nature of the surface, thermo-physical properties

of thefluid and vapour bubble dynamics. Due to involvement V of large number of variables, general

Vapour bubbles

Fig. 9.1. Pool boiling with liquid-vapour interface.

equations describing the boiling process are not available. Nonetheless, considerable progress has been made in arriving at a physical understanding of the boiling mechanism.

Figure 9.1 shows the temperature distribution in saturated pool boiling with a liquid-vapour interface. It is evident from the figure that although there is a sharp decline in the liquid temperature close to the solid surface, the temperature.through most of the liquid remains slightly above saturation. Consequently bubbles generated at liquid-solid interface rise to and are transported acrlpss the liquid-vapour interface. Whether the boiling phenomenon corresponds to pool boiling or forced circulation boiling, there are three definite regimes of boiling (Interface evaporation, nucleate boiling and film boiling ) associated with progressively increasing heat flux, as shown in Fig. 9.2. This specific curve has been obtained from an electrically heated platinum wire submerged in a pool of water (at saturation temperature) by varying its surface temperature and measuring the surface heat flux q,.

1. Interface evaporation Interface evapo&n (evaporation process with no bubble formation) exists in region I,

called the free convection zone. Here the excess temperature, A t,, is very small and 5 5°C. In this region the liquid near the surface is superheated slightly, the convection currents circulate the liquid I

t h

"E s p*

rd evaporation takes place at the liquid surface.

Interface +evaporation or-+

Free convection

Heat transferred by superheated liquid rising to the liquid-vapour interface where evaporation takes place

- Nucleate - Film boiling boiling -1

Excess temperature At, = t, - t,, - Fig. 9.2. The boiling curve for water.

2. Nucleate boiling This type of boiling exists in regions 11 and III. With the increase in At, (excess temperature)

the formation of bubbles on the surface of the wire at certain localised spots commences. The bubbles condense in the liquid without reaching the liquid surface. In fact it is the region 11 where nucleate boiling starts. With further increase in At, the bubbles are formed more rapidly and rise to the surface of the liquid resulting in rapid evaporation, as indicated in the region III. The nucleate b i I k g - i s thus characterised by formation of bubbles at the nucleation sites and the resulting liquid agitation. The bubble agitation induces considerablefluid mixing and that promotes substantial increase in the heatflux and the boiling heat transfer coefficient (The equipment used tor boiling should be designed to operate in this region only).

Nucleate-boiling exists upto At, = 50°C. The maximum heat flux, known as the critical hear $m, occurs at point A (see Fig. 9.2) and is of v & d e r of l h 4 ~ 1 m ~ .

498 Boiling and Condensation

3. Film boiling Film boiling comprises of regions N, V and VZ. The trend of increase of heat flux with

increase in excess temperature observed upto region ZZZ is reversed in region N (called film boiling region). This is due to the fact that the bubble formation is very rapid and the bubbles blanket the heating surface and prevent the incomingfresh liquid from taking their place. Eventually the bubbles coalesce and form a vapour film which covers the surface completely. Since the thermal conductivity of vapour film is much less than that of the liquid the heat flux drops with growth in At,. Within the temperature range 50°C < At, < 15O0C, conditions oscillate between nucleate and film boiling and the phase is referred to as transition boiling, unstable jilm boiling or partial film boiling (region IV). With further increase in At, #he vapour film is stablised and the heating surface is completely covered by a vapour blanket and the heat flux is the lowest as

.shown in region V. The surface temperatures required to maintain a stable film are high and under these conditions a sizeable amount of heat is lost by the surface due to radiation, as indicated in the region VZ. The phenomenon of stable film boiling can be observed when a drop of water falls on a red hot stove. The drop does not evaporate immediately but dances a few times on the stove; this is due to the formation of a stable steam film at the interface between the hot surface and the liquid droplet.

Critical heat flux or burnout point: The critical heat flux or burnout point (Point A in Fig. 9.2) is the point of w i m u m heat flux on the boiling curve at which transition from nucleate to film boiling initiates. This point is also called the boiling crisis because the boiling process beyond that .point is unstable unless of course, point B is reached. The temperature at point B is extremely high and normally above the melting of the solid. So if the heating of the metallic surface is not limited to point A, it is possible that the metal may get damaged or it may even melt (For this reason, point A is often termed as boiling crisis or burnout point). Thus we may be interested to operate the equipment close to this value and not beyond it.

9.2.3. Bubble Shape and Size Consideration The heat transfer rate in nucleate boiling is greatly influenced by the nature and condition

of the heating suvace and su@ace tension at the solid-liquid interjke (shape, size or inclination of bubbles, however, do not have much effect on the heat transfer rate). The surface tension signifies wetting capability of the surface with the liquid (i.e., low surface tension, highly wetted surface) and that influences the angle of contact between the bubble and the solid surface. If the surface is contaminated, its wetting characteristics are affected which eventually influence the size and shape of the vapour bubbles.

If the surface tension of the liquid is low, it tends to wet the surface, so that the bubble is readily pushed by the liquid and rises. The liquid shear-off the bubbles causing them to become globular or oval as shown in Fig. 9.3 ( i ) (for totally wetted surface). In case of liquids having intermediate surface tension (partially wetted surface) a momentary balance may exist between the bubbles and solid surface so that it is necessary to form larger bubbles before the buoyant force can free them from the surface; the shape of the bubble is shown in Fig. 9.3 (ii).

, Q /\c Solid

Totally wetted Partially wetted Unwetted ( i ) ( i i ) ( i i i )

Fig. 9.3. Typical shapes of steam bubbles.

On the unwetted. surface [Fig. 9.3 ( i i i ) ] , the bubbles spread out; forming a wedge between the water and heating surface, thereby allowing hydrostatic forces to resist the action of buoyancy.

Heat and Mass Transfer 499

The formation of bubble as sgaun in Fig. 9.3 ( i ) gives high heat transfer rate compared. with the bubble shapes shown in Fig. 9.3 (ii) and (iii).

Addition of agents for reducing the sut$ace tension was found to have the same effect as providing of wettable sudace and to give increased rates of heat transfer.

9.2.4. Bubble Growth and Collapse - - spherical

From Experiments it has been observed that the bubbles are bubble not always in thermodynamic equilibrium with surrotmding liquid. Liquid The vapour inside the bubble is not necessarily at the same temperature as the liquid. Consider the forces acting on a sP&cal vapour bubble as shown in Fig. 9.4; the pressure forces on-the bubble must be Vapour balanced by the surface tension a t vapour-liquid interface. Thus

d (pV - p,) = 2m.o

or 2 0 PV - PI = 7 ...( 9.2)

where pv = vapour pressure inside the bubble - u pl = liquid pressure over the surface of bubble

Fig. 9.4. Force balance on a o = surface tension of vapour-liquid interface. spherical vapou bubble. The vapour may be considered as a perfect gas for which the

Clayperon equation may be used, which is given below:

where hfg = latent heat of vapourisation.

From perfect gas law: e- RT - Pv

[whers R = gas or vapour constant; pv = density of vapour formed] Substituting the above equation in Eqn. (9.3) and rearranging, we get

Pv - Pl hfg Pv P . hfg -=-- -- Tv - Tsar Tsar R Tm2

where Tv = vapour temperature inside the bubble T, = saturation temperature of vapour inside the bubble at p,.

From Eqn. (9.2) and (9.4), we get

.- - The above equation suggests that if (TI - Tsar) > (Tv - T,), the bubble of radius r will grow

otherwise it will collapse. Here T, is the temperature surrounding the bubble.

9.2.5. Critical Diameter of Bubble Refer figure 9.5. The maximum diameter of the bubble formed on the heating surface depends

on the following parameters:

olv = tension between liquid and vapour

ols = tension between liquid and solid surface

ovs = tension between vapour and solid surface


P = angle formed by the bubble as shown in Fig. 9.5

dc = maximum or critical diameter of bubble.

g(pI - pv)= buoyancy force.

By the use of the dimensional analysis

Bubble

technique, we get I

- . ) (9 .6) Fig. 9.5. Critical diameter of bubble. dc -

i? (PI - 4)

where C is constant_which is generally calculated by experimental results. The value of C = 0.0148 for water bubbles ... suggested by Fritz.

9.2.6. Factors Affecting Nucleate Boiling The nucleate boiling is affected by the following factors: 1. Material, shape and condition of the heating surface The boiling heat transfer coefficient depends greatly on the material of the heating surface;

under identical conditions of pressure and temperature difference, it is different for different metals (viz oopper has high value than steel, zinc and chromium).

The heat transfer rates are also influenced by the condi<ion of heating sur$ace. A rough surface gives 9 better heat transmission than when the surface is either smooth or has been coated (smoothness weakens the metal tendency to get wetted).

The shape of the heating surface also affects transmission of heat. 2. Liquid properties Through experiments it has been observed that the size of the bubble increases with the

dynamic viscosity of the liquid. With increase in bubble size, frequency of bubble formation decreases which results in reduced heat transfer.

Further, high thermal conductivity of the liquid improves the rate of heat transfer. 3. Pressure The pressure influences the rate of bubble growth and in turn also affects the temperature

difference (t, - t,) causing heat flow. For a boiling liquid, the maximum allowable heat flux first increases with pressure until critical pressure is reached and thereafter it declines.

4. Mechanical agitation Experiments have shown that the heat transfer rate increases with the increase in degree of

agitation. . 9.2.7. Boiling Correlations

In boiling heat transfer, a driving force is the excess temperature, which is given by:

At, = ts - tsar ...( 9.7)

For the boiling process the governing equation is

where h is the boiling jilm coeficient. Since no analytical solution is av for boiling heat transfer due to difficult fluid

behaviour, empirical relations are used for engineering calculations. some of them are given in folldwing subsections.

\


9.2.7.1. Nucleate pool boiling (i) For nucleate pool boiling, Rosenhow has recommended the following correlation:

wnere q, = surface heat flux, w/m2; -----

liquid viscosity, kg/ms; ,* -.#- ; 4 sc*,<!,\sl > d L L E : enthalpy of vaporisation, Jkg; t';yy& , ': ': t l j$ . "i.'

density of saturated liquid, kg/m3; D[:'2T. C? !$%:c;( *,:.fi" .&.l- r:ts-;l;SG. density of the saturated vapour, kg/m3;

surface tension of the liquid-vapour inte 1.q.. .. . ...

S. No.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

cpl = specific heat of saturated liquid, Jkg K; P F T - L K E ~ ~ C E ONLY -_____I-=

At, = (t, - t,,) = excess temperature; Csl = surface fluid constant (determined from experimental data);

n = another constant which depends upon the liquid and the surface; c, for water n = 1, while for other liquids n F 1.7.

The value of C,, are given Table 9.1.

Table 9.1. Values of C, for Pool Boiling

1 I Liquid-su~ace

Water - copper Water - brass Water - platinum Water - ground and polished stainless steel Water - mechanically polished stainless steel Benzene - chromium Ethanol - chromium n-pentane - chromium n-butanol - copper Isopropyl alcohol - copper

(ii) Jacob has proposed the following correlation for nucleate boiling at atmospheric pressure on a flat plate and with low heat fluxes

Nu = 0.16 (Gr . ...( 9.9) (iii) For the nucleate boiling on a vertical flat plate, Jacob correlation is of the form:

I Nu = 0.6 1 (Gr . ~ r ) ~ . ~ ~ ...( 9.10)

9.2.7.2. Critical heat flux foq nucleate pool boiling On the boiling curve the critical heat flux is an important point. It is always desirable to

operate a boiling process close to this point. Zuber (1958) predicted the following expression, for such a case:

qsc = 0.18 (pV)ln hfg [go (PI -PV )ill4 ...( 9.11) I The expression given above is independent of fluid viscosity, conductivity and specific heat.

9.2.7.3. Film pool boiling In stable film boiling, the heat transfer is due to both convection and radiation. Bromley

(1950) has suggested the following correlation for film boiling from the outer surface of horizontal tubes:

(h)413 = (hconv.)4n + hd . (h)'" ...( 9.12)

Boiling and Condensation

The equation (9.12) being tedious to solve, could be written within zk 5% of error as 3

h = h,". + q hrd ...( 9.13)

The convective coefficient, h,,, (in the absence of radiation), is given by

where D is the outer diameter of the tube. The vapouk properties in the above equation are evaluated at the arithmetic mean of the surface and saturation temperatures.

Radiative heat transfer coefficient

5.67 x lo4 & (T: - hrud =

('8 - Tsar,)

J where E is the emissivity of solid. Example 9.1. A wire of 1.2 mm diameter and 200 mm length, is submerged horizontally in

water at 7 bar. The wire carries a current of I35 A with an applied voltage of 2.18 V. If the su@ace of the wire is maintained at 200°C, calculate :

(i) The heat jlux, and (ii) The boiling heat transfer coeficient. Solution. Given : d = 1.2 mm = 0.0012 m, 1 = 200 mm = 0.2 m, I = 135 A, V = 2.18 V, t, = 200°C. ( i) The heat flux, q : The electrical energy input to the wire is given by

Q = VI = 2.18 x 135 = 294.3 W

Surface area of the wire, A = d l = IG x 0.0012 x 0.2 = 7.54 x lo4 m2

(ii) The boiling heat transfer coefficient, h: Corresponding to 7 bar, t,,, = 164.97"C, and

9 = h (1s - tsu,)

,/* Example 9.2. An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at atmbspheric pressure. The wire has an applied voltage of 18 V and carries a current of 45 amperes. Calculate:

( i) he heat jlux, and (ii) The excess temperature. The following correlation for water boiling on horizontal submerged su@ace holds good:

h = 1.58 (fr5 = 5.62 ( ~ t , ) ~ , w/m2"c

Solution. Given : d = 1.25 mm = 0.00125 m, 1 = 250 mm = 0.25 m, V = 18 V , I = 45 A. (i) The heat flux, q : Electrical energy input to the wire, Q = VI = 18 x 45 = 810 W

Surface area of the wire, A, = d l = IG x 0.00125 x 0.25 = 9.817 x lo4 m2

Haat and Mass Transfer 503

. . L 810 = 0.825 x lo6 w/m2 = 0.825 M W I ~ ~ = A 9.817 x lo4

(ii) The excess temperature, At, : Using the correlation,

1.58 [fr = 5.62 (At,)' ...g iven

At, = 1.58 (0.825 x 106)0.75 0'333 [ 5.62

] = 19.68OC

ExampIe 9.3. A nickel wire of I mm diameter and 400 mm long, carrying current, is submerged in a water bath which is open to atmospheric pressure. Calculate the voltage at the burnout point if at this point the wire carries a current of 190 A.

Solution. Given : d = 1 rnm = 0.001 m; 1 = 400 mm = 0.4 m, I = 190 A

The thermo-physical properties of water and vapour at 100°C are: pI = (pf) = 958.4 kg/m3, p, = 0.5955 kg/m3, hfg = 2257 Mlkg, a = 58.9 x ~ j m . Voltage at the burnout point, Vb: At burnout i.e., the points of critical heat flux, the correlation is

qsc = 0.18 (pV)ln hB [ go (PI -pV)l1" ...[ Eqn. (9.1 I)]

= 0.18 (0.5955)'~ x 2257 x lo3 [9.81 x 58.9 x (958.4 - 0.5955)11"

= 1.52 x lo6 w/m2 = 1.52 M W / ~ ~ Electric energy input to the wire,

Q = V b x I

or Vb = 10.05 V xample 9.4. Water is boiled at the rate of 25 kg/h in a polished copper pan, 280 mm in r, at atmospheric pressure. Assuming nucleate boiling conditions, calculate the temperature

of the bottom surjace of the pan. Solution. Given: m = 25 kgh; D = 280 mm = 0.28 m The properties of water at atmospheric pressure are:

tsar = 100°C; p, = 958.4 kg/m3; pv = 0.5955 kg/m3; cpl = 4220 Jkg K; p, = 279 x lo4; Pr, = 1.75; hfg = 2257 kJkg; o = 58.9 x lo-' N/m; n = 1 (for water) The temperature of the bottom surface, t,: Excess temperature At, = t, - tsa For nucleate boiling (assumed), the following correlation holds good:

For polished copper pan, Csl = 0.013

...[ Eqn. (9.811

50a Boiling and Condensation

Here mh

q, = surface heat flux = 2 = --f (where m = rate of water evaporation)

\ . J

254544 } 0.5]0.333 r 1 3 x 22;i2; 10 . At, = [

279 x 1@ x 2257 x 1 d

i. e. At, = t, - tsa = 12.2

or t, = 12.2 + tsat = 12.2 + 100 = 112.2OC

?/" Example 9.5. Water at atmospheric pressure is to be boiled in polished copper pan. The diameter of the pan is 350 mm and is kept at llS°C. Calculate the following:

(i) Power of the burner; (ii) Rate of evaporation in kgh; (iii) Critical heat flux for these conditions. Solution. Given : D = 350 mm = 0.35 m, ts = 115OC. tsa, = 100°C .The thermo-physical properties of water (from table) at 100°C are: p, (= pf) = 958.4 kg/m3; pv = 0.5955 hglm3; cpl (= cpf) = 4220 Jkg K; & = (14) = 279 x 106 Ns/m2 Pr, (= Prf) = 1.75; hfg = 2257 Wkg; n = 1; o = 58.9 x Nlm

The excess temperature, At, = t, - tsa = 115 - 100 = 15OC (i) Power of the burner to maintain boiling: As per boiling curve, for At, = 15°C nucleate pml boiling will occur and for this the

following correlation holds good: - ...[ Eqn (9.8)l

For polished copper pan, CsI = 0.013 ... Refer table 9.1

Substituting the values in the above Eqn. we get 9.8 1 (958.4 - 0.5955) 4220 x 15

9s = 279 x 106 x (2257 x 1 ~ ~ ) [ 58.9 ]O1 [0.013 x 2257 x lo3 x 1.75

= 629.7 x 399.4 x 1.873 = 471.06 x ld w1m2 = 47 1.06 kw1m2

The boiling heat transfer rate (power of the burner) is given by

(ii) Rate of evaporation, m,: Under steady state conditions, all the heat added to the pan will result in evaporation of

water. Thus


(iii) Critical heat flux, q,,:

qSc = 0.18 (P,)'" hfg [ go ( Pl- Pv )I'" ...[ Eqn. 9.1 1)3 = 0.1 8 (0.5955)'" x 2257 x lo3 [9.81 x 58.9 x low3 (958.4 - 0.5955)]"~ = 1.52 x lo6 w1m2 = 1.52 M W / ~ ~

'-yA~xample 9.6 A metal-clad heating element of 10 mm diameter and of emissivity 0.92 is submerged in a water bath horizontally. Zf the surface temperature of the metal is 260°C under steady boiling conditions, calculate the power dissipation per unit length for the heater. Assume that water is exposed to atmospheric pressure and is at a uniform temperature.

Solution. Given = D = 10 mm = 0.01 m, E = 0.92, tS = 260°C The thermo-physical properties of water at 100°C from table art? : pl = pf = 958.4 kg/m3; h,, = 2257 kJkg

- - The thermo-physical properties of vapour at 260°C from table are: pv = 4.807 kglm3, Cpv = 2.56 Ulkg K, K = 0.0331 WImK; pv = pg = 14.85 x lo4 Ns/m2 ' a , -

Power dissipation per unit length for the heater: The excess temperature At, = t, - tsat = 260 - 100 = 160°C As per boiling curve, at At, = 160°C. there exists a film pool boiling conditions. In this

case, the heat transfer is due to both convection and radiation. The heat transfer coefficient, h (approximate) is calculated from the equation:

3 h = ~ C O ~ V + 7 hrd ...[ Eqn. (9.13)]

The convective heat transfer coefficient,

...[ Eqn. (9.14)]

or h,, = 395.84 w / m 2 ~ c The radiation heat transfer coefficient,

5.67 x lo4 E (T: - T,:) brad = ...I Eqn.(9. l5)]

(Ts - TS'J

- - 5.67 x lo-* x 0.92 [(260 + 273)4 - (100 + 273)4] [(260 + 273) - (lo0 + 273)]

or h, = 20 W I ~ ~ O C . . h = 395.84 + 20 = 415.84 w/m2~c Hence the power dissipation per unit length for the heater

= h x (xD x 1) x (260 - 100) = 415.84 x A x 0.01 x 160 = 2090 Wlm = 2.09 kW/m

9.3. CONDENSATION HEAT TRANSFER

93.1. General Aspects The condensation process is the reverse of boiling process.Thecondensation sets iq, whenever

r saturation vapour comes in contact with a surface whose temperature is lower than the saturation

Boiling and Condensation Heat and Mass Transfer 507

temperature comsponding to the vapour pressure. As the vapour condenses, latent heat is liberated and there is flow of heat to the surface. The liquid condensate may get somewhat sub-cooled by contact with the cooled surface and that may eventually cause more vapour to condense on the exposed surface or upon the previously formed condensate.

Depending upon the condition of cool surface, condensation may occur in two possible ways: Film condensation and dropwise condensation. 1. Film Condensation /

If the condensate tends to wet the surface and thereby forms a liquidfilm, then the condensation process is known as 'jilm condensation'. Here, the heat from the vapour to the cooling medium is transferred through the film of the condensate formed on the surface. The liquid flows down the cooling surface under the action of gravity and the layer continuously grows in thickness because of newly condensing vapours. The continuous film offers thermal resistance and checks further transfer of heat between the vapour and the surface.

Further, the heat transfer from the vapour to the cooling surface takes place through the film formed on the surface. The heat is transferred from the vapour to the condensate formed on the sur$ace by 'convection' and it isfurther transferred from the condensate film to the cooling surfoee by the 'conduction'. This combined mode of heat transfer by conduction and convection reduces the rates of heat transfer considerably (compared with dropwise condensation). That is the reason that heat transfer rates of filmwise condensation are lower than dropwise condensation. Figure 9.6 ( i ) shows the film condensation on a vertical plate.

drops

0 0 0

tv tsar

f- Condensate film

( i ) Film pondensation (ii) Dropwise condensation

Fig. 9.6. Film and dropwise condensations on a vertical surface

2. Dropwise condensation In 'dropwise condensation' the vapour condenses into small liquid droplets of various sizes

which fall down the surface in random fashion. The drops form in cracks and pits on the surface. grow in size, break away from the surface, knock off other droplets and eventually run off the surface, without forming a film under the influence of gravity. Fig. 9.6 (ii) shows the dropwise condensation on a vertical plate.

In this type of condensation, a large portion of the area of sdid surface is directly exposed to vapour without an insulating film of condensate liquid, consequently higher heat transfer rate (to the order of 750 kwlm2) are achieved. Dropwise condensation has been observed to occur either on highly polished surfaces, or on surfaces contaminated with impurities like fatty acids and organic compounds. This type of condensation gives coefficient of heat transfer generally 5 to 10 times larger than with film condensation. Although dropwise condensation would be prefened to filmwise condensation yet it is extremely difficult to achieve or maintain. This is because most

surfaces become 'wetted' after being exposed to condensing vapours ovkq a period of time. Dropwise condensation can be obtained under controlled conditions with the help of certain additives to the condensate and various surface coatings but its commercial viability has not yet been approved. For this reason the condensing equipment in use is designed on the basis of filmwise condensation.

(.

9.3.2. Laminar Film Condensation on a Vertical Plate I

An analysis for filmwise condensation on a vertical plate can be made on lines prepared by Nusselt (1916). Unless the velocity of the vapour is very high or the liquid film very thick, the motion of the condensate would be laminar. The thickness of the condensate film will be a function of the rate of condensation of vapour and the rate at which the condensate is removed from the surface. The film thickness on a vertical surface will increase gradually from top to bottom as shown in Fig. 9.7.

Condensate film

Element

L P$ f b h dy) pig f b h dy)

(b) Force balance m I

(a) Film growth, velocity (c) Mass and heat balance and temper ture profiles a

Fig. 9.7. Film condensation on a flat vertical plate. Nusselt's analysis of film condensation makes the following simplifying assumptions.

1. The film of the liquid formed flows under the action of gravity. 2. The condensate flow is laminar and the fluid properties are constant. 3. The liquid film is in od thermal contact with the cooling surface and, therefore,

temperature at the inside o ,the film is taken equal to the surface temperature t,. Further,

the prevailing pressure.

\ the temperature at the liquid-ttapour interface is equal to the saturation temperature t,,, at

4. Viscous shear and gravitational rorces are assumed to act on the fluid; thus normal viscous force and inertia forces are neglected.

5. The shear stress at the liquid-vapour interface is negligible. This means there is no velocity

gradient at the liquid-vapour interface [ i.e.,

6. The heat ansfer across the condensate layer is by pure conduction and temperature dis tr ibut6 is linear.


7. The condensing vapour is entirely clean and free from gases, air and non-condensing impurities.

8. Radiation between vapour and liquid film; horizontal component of velocity at any point in the liquid film; and curvature of the film are considered negligibly small.

Consider the process of film condensation occuring on the surface of a flat vertical plate as shown in Fig. 9.7. The coordinate system has also been depicted on the figure. The origin '0' is at the upper end of the plate, the X-axis lies along the vertical surface with the positive direction of X measured downward and Y-axis is perpendicular to it. The vertical plate height I, width b, and 6 denotes the thickness of the film at a distance x from the origin. The liquid film thickness which is zero at the upper end of plate gradually increases as further condensation occurs at the liquid-vapour interface and attains its'maximum value at the lower end of the plate.

Let, p, = density of liquid film, .

pv = density of vapour, hfg = latent heat of condensation,

k = conductivity of liquid film,

p = absolute viscosity of liquid film, ts = surface temperature, and

tSa = saturation temperature of vapour at prevailing pressure. (a) Velocity distribution

o find an expression for velocity distribution u as a function of distance y from the wall , let us consider the equilibrium between gravity and viscous forces on an elementary

volume (b dx dy) of the liquid film. Gravitational force on the element = plg (b dx dy) - pv g (b dx dy) . . . ( i )

Viscous shear force on the element du . . . (ii)

Equating ( i ) and (ii), we get

1

d2u ( P ~ - P V ) ~ - = - ...( 9.16) d>" c1

Upon integration, we have

Integrating again, we get

CL The relevant boundary conditions are:

Using these boundary conditions, we get the following values of C, and C2.


Substituting the values of C, and C2 we get the velocity profile

Equation (9.18) is the required velocity profile. The mean flow velocity urn, of the liquid film at a distance y is given by

(b) Mass flow rate The mass flow rate of condensate through apy x position of the film is given by : Mass flow rate (m) = mean flow velocity (urn) x flow area x density

or m = (PI - PV) 8 . a2 x b . 6 x p l = ~ l ( ~ l - ~ v ) g . b .

3 P 3 Y 0 )

The mass flow is thus a function of x; this is so because the film thickness 6 is essentialfy dependent upon x.

As the flow proceeds from x to (x + dx) the film grows from 6 to (6 + d6) because of additional condensate. The mass of condensate added between x and (x + dx) can be worked out by differentiating Eqn. (9,.20) with respect to x (or 6).

(c) Heat flux The heat flow rate into the film (dQ) equals the rate of energy release due to condensation

at the surface. Thus

According to our assumption the heat transfer across the condensate layer is by pure conduction, hence

Combining Eqns. (9.22) and (9.23), We have

51 0 Boiling and Condensation

Integrating the above equation, we get

Substitution of the boundary condition : 6 = 0 at x = 0 yields C, = 0. Hence

4k p (tsa - is) x 1'" ...( 9.24)

P I ( P I - Pv) g hfg The equation (9.24) depicts that the heat film thickness increases as the fourth root of the

distance down the surface; the increase is rather rapid at the upper end of the vertical surface and slows thereafter.

(d) Film heat transfer coefficient According to Nusselt assumption the heat flow from the vapour to the surface is by conduction

through the liquid film. Thus

The heat flow can also be expressed as

dQ = hx (b &) (tsat - ts) . . .(ii) where hx is the local heat transfer coefficient From ( i ) and (ii), we get

Equation (9.25) depicts that at a definite point on the heat transfer surface, the film coefficient hx is directly proportional to thermal conductivity k and inversely proportional to thickness offilm 6 at that point.

Substituting the value of 6 from equation (9.24), we get

Local heat transfer coefficient at the lower end of the plate, i.e., x = 1

Evidently the rate of condensation heat transfer is higher at the upper end of the plate than that at the lower end.

The average value of heat transfer can be obtained by integrating the local value of coefficient [Eqn (9.26)] as follows :


where hL is the local heat transfer coefficient at the lower edge of the plate. , - 4 . *

This shows that the average heat transfer coefficient is - times the local heat transfer 3 coefficient at the trailing edge of plate.

Equation 9.28 is usually written in the form

The Nusselt solution derived above is an approximate one because experimental results have shown that it yields results which are approximately 20 percent lower than the measured values. McAdams proposed to use a value of 1.13 in place of coefficient 0.943. Hence

h = 1.13 L P L (tsar - ts) J

While using above equation, it may be noted that, all liquid properties are to be evaluated

at the temperature ("'i - ls) and hfg should be evaluated at tSap

The total heat transfer to the surface, Q = h As (tsat - ts) ...( 9.31)

The total condensation rate,

Figure 9.8. shows the variation of film thickness and film coefficient with plate height Y

t Film u 5 Y e 0 .g zi k 3 8 .Y s 3 3 C4 iZ

Plate height -+ Fig. 9.8. Film thickness and film coefficient Fig. 9.9. Condensation on an inclined surface.

vs. plate height.

Heat and Mass Transfer 51 3 and ( Boiling

[graphical representation of Eqns. (9.24), (9.26) and (9.28)]. The film thickness increases with the increase of plate height. Heat transfer rate decreases with the increase of plate height since thermal resistance increases with the film thickness.

e. Inclined flat plate surface For inclined flat surfaces, the gravitational acceleration g in equation (9.30) is replaced

g sine where 0 is the angle between the surface and horizontal (Refer Fig. 9.9). The Eqn. (9.3

For the plate, A = L x B and P = B, where L and B are height and width of plate, respectively. Thus,

When the value of Re exckds 1800 (approximately), the turbulence will appear in the liquid film. For Re > 1800, the following correlation is used: is modified as :

, 1 or hinclined = hwnical X ...( 9.34) I /~ ' I . Equation (9.34) is applicable only for cases where 0 is small; is not at all applicable for

9.3.4. Film Condensation on Horizontal Tubes NusseIt's analysis for laminar filmwise condensation on horizontal tubes leads to the following

relations: ; (1 horizontal plate. 1 Condensation

* 1 9.3.3. Turbulent Film Condensation surface ', When the plate on which condensation occurs

is quite long or when the liquid film is vigorous &ugh, the condensate flow may become turbulent.

... For single horizonal tube. I

The turbulent results in higher-heat transfer rates because heat is now transferred not only by con- densdtion but also by eddydifision. The-transition criterion may be expressed in terms of Reynolds number defined as

PI Urn Dh Re = - Fig. 9.10

C.. 1s of film

4 cndens

A v. "W.W...

film

I ;ation on a

... For horizontal tube bank with N tubes placed directly over one another in the vertical direction,

where D = outer diameter of the tube.

9.3.5. Film Condensation Inside Horizontal Tubes Condensation of vapour inside the tubes finds several engineering applications such as

condensers used in refrigeration and airconditioning systems and several chemical and petrochemical industries. The phenomena insides tubes are very complicated because the overall flow rate of vapour strongly affects the heat transfer rate and also the rate of condensation on the walls.

Chato (1962) has recommended the following correlation for low velocities inside horizontal tubes (condensation of refrigerants):

- - -

PI vekcal surface.

where D,, = hydraulic diameter cross-sectionalareaof fluid flow

= 4 x wetted perimeter

1 urn = mean or average velocity of flow

I

I where m = pA urn For a vertical plate of unit depth, P = 1, the Reynolds number is sometimes expressed in where

terms of the mass flow rate per unit depth of plate r, so that Equation (9.43) is restricted to low vapour Reynolds number such that

R~~ = (" ") < 3500

where Rev is evaluated at inlet conditions to the tubes. with r = 0, at the top of the plate and r increasing with x. The Reynolds number may also be related to heat transfer coefficient as follows: -

Q = h As (tsar - ts) = m hjR 9.3.6. Influence of the Presence of Noncondensable Gases

The presence of noncondensable gas such as air in a condensing vapour produces a detrimental effect on the heat transfer coefficient. It has been observed that even with a few percent by volume of air in steam the condensation heat transfer coefficient is reduced by more than fifty percent. This is owing to the fact that when a vapour containing noncondensable gas condenses, the noncondensable gas is left at the surface. Any further condensation at the surface will occur only after incoming vapour has diffused through this noncondensable gas


collected in the vicinity of the surface. The noncondensable gas adjacent to the sui$ace acts as a thermal resistance to the condensation process. The rate of condensation decreases greatly when the condensable vapour is contaminated with even very small amounts of noncondensable gases.

As the presence of noncondensable gas in a condensing vapour is undesirable, the general practice in the design of a condenser should be to vent the noncondensable gas to the maximum extent possible.

Example 9.7. Discuss the diflerent types of processes for condensation of vapours on a solid su$ace. (AMIE Summer, 1998)

Solution. Whenever a saturated vapour comes in contact with a surface at a lower temperature condensation occurs.

There are two modes of condensation.

Filmwise - in which the condensation wets the surface forming a continuous film which covers the entire surface;

Dropwise - in which the vapour condenses into small droplets of various sizes which fall down the surface in a random fashion.

Filmwise condensation generally occurs on clean uncontanimated sugaces. In this type of condensation the film covering the entire surface grows in thickness as it moves down the surface by gravity. There exists a thermal gradient in the film and so it acts as a resistance to heat transfer.

In dropwise condensation a large portion of the area of the plate is directly exposed to the vapour, making heat transfer rates much higher (5 to 10 times) than those in filmwise condensation.

Although dropwise condensation would be preferred to filmwise condensation yet it is extremely d@cult to achieve or maintain. This is because most surfaces become "wetted" after being exposed to condensing vapours over a period of time. Dropwise condensation can be obtained under controlled conditions with the help of certain additives to the condensate and various surface coatings, but its commercial viability has not yet been proved. For this reason the condensing equipments in use are designed on the basis of filmwise condemation.

Example 9.8. Saturated steam at tsa, = 90°C (p = 70.14~kPa) condenses on the outer surface of a 1.5 m long 2.5 m OD vertical tube maintained at a uniform temperature T, = 70°C. Assuming film condensation, calculate :

( i ) The local transfer coeficient at the bottom of the tube, and (ii) The average heat transfer coeficient over the entire length of the tube.

Properties of water at 80' C are : p, = 974 kg/m3, kt = 0.668 W/m K, p, = 0.335 xld kg/ms,

hfg = 2309 kJ/kg, p, << p,. (AMIE Winter, 1998) Solution. Given : t,,, = 90°C (3 = 70.14 kPa); L = 1.5 rn;

D = 2.5 cm= 0.025 m; t, = 70°C.

Properties of water at 80°C ; pl = 974 kg/m3;

k = 0.668 W/m K; y = 0.335 x lo3 kg/ms; hfg = 2309 H/kg {pv << pl) ( i ) The local heat transfer coefficient, h, : With usual notations, the local heat transfer coefficient for film condensation is given as

...[ Eqn. (9.26)]


:. Local heat transfer coefficient at the bottom of the tube, x= 1.5 m, is

(ii) Average heat transfer coefficient, h : - 4 4 h = h, =, x 3552.9 = 4737.2 w/& O C (Ans.)

J J

Example 9.9. Saturated steam at 120°C condenses on a 2 cm OD vertical tube which is 20 cm long. The tube wall is maintained at a temperature of 119°C. Calculate the average heat transfer coeflcient and the thickness of the condensate film at the base of the tube. Assume ,, Nusselt's solution is valid. Given :

I

p, = 1.985 bar; pw = 943 kg/m3; hfg = 2202.2 kJ/kg;

kw=0.686 W/m K; p=237 .3x lu6 Ns/m2, (AhfIE Summer, 1997)

Solution. From Nusselt's solution, we have

neglecting p,, in comparison to pl

= 1.92 x 1013 'I4 = 5. I x 10-5 m or 0.051 nun (An.)

Now, k h --=

L- o'686 = 13451

6, 0.051 x

:. Average heat transfer coefficient,

Example 9.10. A vertical cooling jin approximating a flat plate 40 cm in height is exposed to saturated steam at atmospheric pressure (t,,, = IOOOC, A,- = 2257 kJ/kg). The fin is maintained at a temperature of 90°C. Estimate the following :

(i) Thickness of the film at the bottom of the fin; (ii) Overall heat transfer coeficient; and (iii) Heat transfer rate after incorporating McAdam's correction. The relevant fluid properties are :

p, = 965.3 kg/m3

k, = 0.68 W/m°C

p, = 3.153 x N s/m2

The following relations hay be used :


1/4 4kA (tsar. - t x 4=[ ghfe Pl (PI - PV) ]

(AMIE, summer, 1999)

Solution. Given : L = 60 cm = 0.6 m; tSa, = 400°C; hfg = 2257 kJ/kg;

t, = 90°C; p, = 965.3 kg/m3; kl = 0.68 W/m°C;

pl = 3.153 x N s/m2

( i ) Thickness of film at the bottom edge of the fin, 6L:

= 0.0001 136 m = 0.1 136 mm (Ans.) (ii) Overall heat transfer coefficient, I; :

(iii) Heat transfer rate with McAdam's correction : With McAdam's correction, the value of 7; is 20 percent higher. Hence heat transfer rate

after incorporating McAdam's correction for unit width, is : Q = 1 . 2 ~ 7 9 8 1 . 2 2 ~ ( 0 . 4 ~ l)x(l00-90)

= 38309.8 W/m or 38.3098 kW per m width (Ans.)

Example 9.11. A vertical plate 500 mm high and maintained at 30°C is exposed to saturated steam at atmospheric pressure. Calculate the following:

( i ) The rate of heat transfer, and (ii) The condensate rate per hour per metre of the plate width for film condensation. The properties of water film at the mean temperature are:

p = 980.3 kg/m3; k = 66.4 x W/nt°C; p = 434 x 1Od kg/ms and hfg = 2257 kmg. Assume vapour density is small compared to that of the condensate.

Solution. Given: L = 500 mm = 0.5 m; B = lm; t, = 30°C; 2 ( i ) The rate of heat transfer per metre width, Q:

. . [Eqn. (9.29)]

r p,'Pg% l lA = 0.943 1 I ... neglecting p,[(p, << pi ...g iven)]

* L P L (tsar - J


- (980.3)~ x (66.4 x I O - ~ ) ~ x 9.81 x (2257 x l d ) or h = 0.943 [ 434 x lo4 x 0.5 (loo - 30)

(ii) The condensate rate per meter width, m:

w = - = 534S9 lo3 = 236.86 kg& (Ans.) hfg 2257

Example 9.12. A vertical plate 350 mm high and 420 mm wide, at 40°C is exposed to saturated steam at 1 atm. Calculate the following:

( i ) The Jilm thickness at the bottom of the plate; (ii) The maximum velocity at the bottom of the plate; (iii) The total heat flux to the plate. Assume vapour density is small compared to that of the condensate.

Solution. Given: t, = 40°C; tsar = 100°C, L = 350 mm = 0.35 m, B = 420 mm = 0.42 m. The properties will be evaluated at the film temperature, i.e., the average of tsar and t,;

t f= loo +40 = 70°C; further hfL is evaluated at 100°C. 2

The properties at 70°C are: p, = 977.8 kg/m3; P = 0.4 x kgtms; k = 0.667 W/m°C and hfg = 2257 kl/kg.

( i ) The film thickness at the bottom of the plate, 6:

6 = g P, (PI - P,) hfg

...[ Eqn. (9.24)l

Neglecting p,, pv << PI ...(g hen)

4 x 0.667 x 0.4 x lo-) (100 - 40) x 0.35 = 104 = 0.18 mm or 9.81 x (977.81~ x 2257 x lo3

I 1/4

(': x = 1 = 0.35 m in this case ) (ii) The maximum velocity at the bottom of the plate, u,, :

u = (p1-pv)g (Sy-$] ...[ Eqn. (9.17)] P

- - " (.-$] ..neglecting p, P

At y = 6, u = urnax., therefore,

p1gS2 977.8~9.81x(1.8Xl@)~ =0,38slnls - urn=-- 2P 2 ~ 0 . 4 ~


(iii) The total heat flux to the plate, Q:

Boiling and Condensation 51 9

... [Eqn. (9.29)]

The total heat flux is given by Q = % A (tsar - t,) = L x (L x B) (t,,, - ts)

= 4931.35 x 0.35 X 0.42 X (100 - 40) = 43494 W or 43.494 k W (Ans.)

Example 9.13. Vertical f i t plate in the form of fin is 600 m in height and is exposed to steam at atmospheric pressure. if surface of the plate is maintained at 60°C, calculate the following:

(i) The film thickness at the trailing edge of the film, (ii) The overall heat transfer coeficient, (iii) The heat transfer rate, and (iv) The condensate mass flow rate. Arsume laminarflow conditions and unit width of the plate.

Solution. Given : L = 600 mm = 0.6 m; ts = 100°C; The properties of vapour at atmospheric pressure are: tsar = l0O0C, hfg = 2257 Hkg ; pv = 0.596 kglm3.

The properties of saturated vapour at the mean film temperature tf = loo + 6o

= 80°C are: 3 rr

p, = 971.8 kg/m3, k = 67.413 x lo-' W/m°C, p = 355.3 x lod Ns/m2 or kglms

( i ) The film thickness at the trailing edge of the plate, 6 (at x = L = 0.6 m):

(ii) The overall heat transfer coefficient, %:

...[ Eqn. (9.24)l

Using McAdam's correction which is 20% higher than Nusselt's result. - h = 4938.68 x 1.2

= 5926.4 w/rn2"c (Ans.)

(iii) The heat transfer rate, Q : - Q = h A, (I,,, - t,) = h x ( L x - ty)

= 5926.4 x (0.6 x 1) (100 - 60) = 142233.6 W (iv) The condensate mass flow rate, m :

- - 142233'6 = 0.063 kgls or 226.8 k g h (Am.) 2257 x lo3

Let us check whether the flow is laminar or not

. . [Eqn. (9.35)]

This shows that the assumption of laminar flow is correct. Example 9.14. A vertical tube of 60 mm outside diameter and 1.2 m long is exposed to

steam at atmospheric pressure. The outer surface of the tube 50°C by circulating cold water through the tube. Calculate th

( i ) The rate of heat transfer to the coolant, and (ii) The rate of condensation of steam.

Solution. Given : D = 60 mm = 0.06 m, L = 1.2 m, t, = 50°C Assuming the condensation film is laminar and noncondensable gases in steam are absent.

The mean film temperature tf = loo + 50 = 75°C

2

The thermo-physical properties of water at 75°C are: p, = 975 kg/m3, pl = 375 x lo4 Ns/m2, k = 0.67 W/m°C.

The properties of saturated vapour at tsa = 100"C\ are :

p, = 0.596 kg/m3, hfg = 2257 Hkg. ( i ) The rate of heat transfer, Q: For laminar condensation on a vertical surface

...[ Eqn. (9.30)]

= 4627.3 w/m2"c - Q = h A, (tsar - t,) = h (n DL) (t,,, - ts)

= 4627.3 x (71 x 0.06 X 1.2) (100 - 50) = 52333.5 = 52.333 kW (Ans.)

(ii) The rate of condensation of steam, m : The condensation rate is given by


Let us check the assumption of laminar film condensation by calculating, Re.

...[ Eqn. (9.3511

' Since Re (= 13 12.85) < 1800, hence the flow is laminar. Example 9.15. A horizontal tube of outer diameter 20 mm is exposed to dry steam at 100°C.

The tube surfQce temperature is maintained at 84°C by circulating water through it. Calculate the rate of formation of condensate per metre length of the tube.

Solution. Given: D = 20 mm = 0.02 m, ts = 84°C; t,, = 100°C

The mean film temperature t, = loo + 84 = 9 2 0 ~ 2

The properties of saturated liquid at 92°C are: p, = 963.4 kg/m3, CL, = 306 x lo4 ~ s l m ~ ; k = 0.677 W/m°C The properties of saturated vapour at tsar = 100°C are:

I p, = 0.596 kg/m3, hfg = 2257 Wkg Rate of formation of condensate per metre length of the tube, m: The average heat transfer coefficient is given by

...[ Eqn. (9.40)]

= 11579.7 w / m 2 ~ c The heat transfer per unit length is

= 11579.7 X 71 x 0.02 x (100 - 84) = 11641.2 W Rate of formation of condensate per metre length of the tube,

- =-- "' 11641'2 = 5.157 x kgls = 18.56 kgh (Ans.) L hfg - EKiiY

Example 9.16. A steam condenser consisting of a square array of 625 horizontal tubes, each 6mm in diameter, is installed at the exhaust hood of a steam turbine. The tubes are exposed to saturated steam at a pressure of 15 kPa. If the tube sudace temperature is maintained at 25°C calculate:

( i) The heat transfer coeficient, and (ii) The rate at which steam is condensed per unit length of the tubes. Assume film condensation on the tubes and absence of noncondensable gases.

Solution. Given: D = 6 mm = 0.006 m, ts = 25°C. Corresponding to 15 kPa pressure, the properties of vapour (from the table) are:

tsar = 54OC, p, = 0.098 kg/m3, hfg = 2373 kJkg.

54 + 25 The properties of saturated water at film temperature t f = - = 39.5"C are : 2

Boiling and Condensation 521

p, = 992 kg/m3; j~ = 663 x lo4 ~ s / m ~ ; k = 0.631 W/m°C Since the tubes are arranged in square array, therefore, the number of horizontal tubes in

vertical column is : N = a = 25 ( i ) The heat transfer coefficient, x: The average heat transfer coefficient for steam condensing on bank of horizontal tubes is

given by

...[ Eqn. (9.41)]

(ii) The rate at which steam is condensed per unit length, m The rate of condensation for the single tube of the array per metre length is

The rate of condensation for the complete array is m = 625 x m, = 625 x 1.1 16 x lom3 = 0.6975 kg1s.m (Ans.)

TYPICAL EXAMPLES Example 9.17. A 750 mm square plate, maintained at 28OC is exposed to steam at 8.132

kPa. Calculate the following: ( i ) The film thickness, local heat transfer coeficient and mean jlow velocity of condensate

at 400 mm from the top of the plate, (ii) The average heat transfer coeficient and total heat transfer from the entire plate, (iii) Total steam condensation rate, and

(iv) The heat transfer coeficient if the plate is inclined at 25" with the horizontal plane.

Solution. Given: L = B = 750 mm = 0.75 m, t, = 28OC, x = 400 mm = 0.4 m Assme laminar flow film condensation. Properties of saturated vapour at 8.132 kPa (or 0.08132 bar) are: t,,, = 42°C; p, = 0.0561 kg/m3; hfg = 2402 kJkg

42 + 28 The mean film temperature t, = - = 35°C

2

The properties of saturated water at 35°C are:

p1 = 993.95 kg/m3; k = 62.53 x W/m°C, p = 728.15 x lo6 kg/ms.

( i) 6,, h,, u, at 400 mm from the top of the plate: The film thickness at a distance x from the top edge of the plate is given by:

...[ Eqn. (9.24)]

HBat and Mass Transfer

- At x = 0.4 m,

6, = 1.819 x 104x (0.4)'"- 1.45 x 1 0 4 m = 0.145 mm (Ans.)

SL = 1,819 x (0.75)"~ = 1.69 x lo4 m = 0.169 mm The local heat transfer coeficient,

The mean flow velocity of condensate,

(ii) Average heat transfer coefficient (g):

(where = the film thickness at the bottom of the plate). Using McAdarns correction, - -

h = 1.2 x 4933.33 = 5920 W I ~ ~ O C The total heat transfer from the entire plate, Q:

Q = i; A, (tsar - t,) = ( L x B) (tsar - 2,)

PY = 5920 x (0.75 x 0.75) (42 - 28 ) = 46620 W (Ans.)

(iii) Total steam condensation rate, m:

m = 46620 = 0.0194 kgls or 69.87 kglh (Ans.) 2402 x lo3

(iv) f i e heat transfer coefficient when the plate is inclined 25" with the horizontal, hinc~in,:

hind,,, = hvertica, x (sin 0)'" ... [Eqn. (9.34)] = 5920 x (sin 250)"~ = 4773.2 w / ~ ~ " c (Ans.)

Let us check the type of flow [using Eqn. (9.35)],

Hence assumption is correct.

Example 9.18. A vertical plate 3.2 m high maintained at 54"C, is exposed to saturated steam at atmospheric pressure. Calculate the heat transfer rate per unit width.

Solution. Given: L = 3.2 m; B = 1 m, t, = 54°C; tsa, = 100°C. Heat transfer rate per unit width : In order to determine whether the condensate film is laminar or turbulent; the Reynolds

number must be checked.

Boiling and Gondensation

The mean film temperature tf = 100 + 54 = 7 7 0 ~ 2

The properties of the condensate at 77°C are :

& = 365 x 10" ~ s / m ~ ; k = 668 x W/m°C;

The properties of saturated vapour at tSat = 100°C

pv = 0.596 kg/m3; hfg = 2257 kJkg. Assuming the flow to be turbulent the relevant equations are :

and

4 h L (t,, - t,) Re =

hfg PI ...[ Eqn. (9.38)]

- 1/3 '' ( " - lZg] (~e)'.' ...[ Eqn. (9.39)] h = 0.0077 1 L P? J

Eliminating from these equations, we get the condition that the flow will be turbulent if

Substit-lting the values, we get

Thus the film is turbulent as assumed and Re = 4144.8

= 0.0077 x (2.0797 x 10'~)'" x 27.99 = 5866.62 w/m2 "C Heat transfer rate per unit width,

Q = h As (t,,, - 4 ) = 5866.62 x (3.2 x 1 ) (100 - 54) = 863566 W/m = 863.566 kW/m (Am)

Example 9.19. A condenser is to be designed to condense 1800 kg/h of dry and saturated steam at a pressure of 10 kPa. A square array of 400 tubes, each of 8 mm in diameter, is to be used. If the tube su$ace temperature is to be maintained at 24"C, calculate the following:

( i ) The heat transfer coeficient, and (ii) The length of each tube assuming single pass.

Solution. Given: m = 1800 kgh; D = 8 mm = 0.008 m, t, = 24°C. ( i ) The heat transfer coefficent, h: Corresponding to 10 kPa (0.1 bar), from table, the properties of dry and saturated vapour

are :

tsar = 458°C; pv = - = 0.0676 kg/m3; hf, = 2393 Wlkg (J The properties of saturated vapour at the mean film temperature,

Heat and Mass Transfer Boiling and Condensation 525

pl = 993.95 kg/m3; k = 62.53 x W/m°C; J.L = 728.15 x lo4 kglms. ,

As the tubes are to be arranged in an array, therefore, the number of horizontal tubes in the vertical column is, N = = 20.

The average heat transfer coefficient for steam condensing on bank of horizontal tubes is given by :

...[ Eqn. (9.41)]

(ii) The length of each tube, assuming single pass, L : The heat transfer rate,

Q = 4 (tsa, - ts) - or mhfg = h x (400 x nDL) (tsa, - t,)

L = 196500 = 1.09 m (Ans.) 1092147.3

Example 9.20. The outer sutface of a cylindrical drum 350 mm diameter is exposed to ' ~ tu ra t ed steam at 2.0 bar for condensation. If the sutface temperature of the drum is maintained at 80°C calculate the following:

( i ) The length of the drum; (ii) The thickness of the condensate layer to condense 70 kg/h of steam.

Solution. Given : D = 35 mm = 0.35 m; t, = 80°C; m = 70 kg / h Assuming film condensation and laminar flow : Corresponding to 2.0 bar, from the table, the properties of the saturated vapour are:

The properties of saturated water at the mean film temperature 9 = 1202 + 80 = loooC 2

p, = 956.4 kg/m3; k = 68.23 x W/m°C; p = 283 x lo4 kgtms ( i ) The length of the drum, L: The film at the bottom edge of the drum is given by

...[ Eqn. (9.24)]


Using McAdam's correlation, we have - h = 1.2 x 3432.09 x (L)-'I4 = 4118.5 x (L)- ' /~

The heat transfer rate is given by Q = i; A, (tsa, - t,) = m hfg

70 x (2201.6 x lo3) or 41 18.5 x (L)-'" (n x 0.35 x L) (120.2 - 80) =

or 182046.8 (Ll3I4 = 42808.88

(ii) The thickness of condensate layer, 6: 6 = 1.988 x 10-4 x (L)'I4

= 1.988 x lo4 x (0.1452)'" = 1.227 x lo4 m = 0.1227 mm (Am.)

Let us check whether flow is lamniar or not.

As Re (= 249.9) < 1800, hence assumuption is correct.

HIGHLIGHTS 1. Boiling is the convective heat transfer process that involves a phase change from liquid

to vapour state. 2. The boiling heat transfer phenomenon may occur in the follows forms:

( i ) Pool boiling (ii) Forced convection boiling (iii) Sub-cooled or local boiling (iv) Saturated boiling.

3. The three regimes of boiling are: ( i ) Interface evaporation (ii) Nucleate boiling

(iii) Film boiling. 4. The condensation process is reverse of boiling process. Condensation may occur in two

possible ways: ( i ) Film condensation (ii) Dropwise condensation.

If the condensate tends to wet the surface and thereby forms a liquid film, the condensation process is known as 'film condensation'. In 'dropwise condensation' the vapour condenses into small liquid droplets of various sizes which fall down the surface in random fashion. Summary of formulae : A. Boiling


Nu = 0.16 ( ~ r . P r ) q . ~ ~ ... for nucleate boiling at atmospheric pressure on a flat plate with low heat fluxes.

Nu = 0.61 ( ~ r . P r ) ~ . ~ ' ... For the nucleate boilng on a vertical flat plate.

qsc = 0.18 (pv)l0 hfg rg 0 ( P I - m 1 / 4

... critical heat flbx for nucleate pool hoiling (h)"I3= (hconv.)413 + brad . (h)lJ3

3 h ' hem. + 2 hn2d ... within f 5% ertor

B. Condensation :

10. hinclinad = (h)venical x (sin 8)lJ4

... for Re > 1800


... for single horizontal tube

- r P, ( P, - pV)pg hfg]'I4 ' 13. h = 0.725 / ... for horizontal tube bank with N

L N Pr (tsat - ts)D J tubes placed directly over one another in the vertical direction.

where D = outer diameter of the tube.

3 where hig = hfg + g c,, ( t , - t,)

THEORETICAL QUESTIONS 1. Define the term 'boiling'. 2. Enumerate the applications of boiling heat transfer. 3. Explain briefly the physical mechanism of boiling. 4. Differentiate between pool boiling and forced convection boiling. 5. Explain briefly the various regimes of saturated pool boiling. 6. What is burnout point? 7. Explain briefly the condensation mechanism. 8. Differentiate between the mechanism of filmwise and dropwise condensation. 9. Derive the Nusselt theory of laminar flow film condensation on a vertical palte.

10. Drive the following relation for laminar film condensation on vertical plate.

UNSOLVED EXAMPLES Boiling Heat Transfer

1. Water at atmospheric pressure is to be boiled in polished copper pan. The diameter of the pan is 300 mm and is kept at 11 1 "C. Calculate the following : ( i) Power of the burner to maintain boiling;

(i i) Rate of evaporation in kglh. Take the properties of water at 100°C as follows: pi = 958 kg/m3; pv = 0.597 kg/m3; ~ l f = 278 x lod kg/ms; cfl= 4216 JkgK; hfg = 2257 kJ&g ; Pr = 1.723, o = 58.9 x Nlm [Ans. ( i) 13.664 kW, (i i) 21.8 kgh]

2. A wire of 1 mm diameter gnd 150 mm length is submerged horizontally in water at 7 bar. The wire carries a current of 131.5 A with an applied voltage of 2.15 V. If the surface of the wire is maintained at 180°C, calculate: ( i ) The heat flux, and (i i) The boiling heat transfer coefficient.

[Ans. ( i ) 0.6 MW/m2, (ii) 39920 ~ l r n * O C ] 3. An electric wire of 1.5 mm diameter and 200 mm long is'laid horizontally and submerged in water at

atmospheric pressure. The wire has an applied voltage of 16 V and carries a current of 40 amperes. Calculate: ( i ) The heat flux, and (i i) The excess temperature. [Ans. ( i ) 0.679 MWI~' , ( i i ) 18.52OC)

4. A nickel wire of 1.5 mm diameter and 500 mm long, carrynig current, is submerged in a water bath which is open to atmospheric pressure. Calculate the voltage at the burnout point if at this point the wire carries a current of 200 A. [Ans. 17.9 V (uppx.))

5. A metal-clad heating element is of 8 mm diameter and of emissivity 0.95. The element is horizontally immersed in a water bath. The surface temperature of the metal is 260°C under steady state bolllng


conditions. Calculate the power dissipation per unit length for the heater if water is exposed to atmospheric pressure and is at uniform temperature. [Am. 1.75 kW/m]

Condensation Heat Transfer 6. A vertical plate 450 mm high and maintained at 30°C is exposed to staturated steam at atmospheric

pressure. Calculate: (i) The rate of heat transfer, and (ii) The condensate rate per hour per metre for plate width film condensation. The properates of water film at the mean temperature are: p = 980.3 kg/m3 ; k = 66.4 x W/m°C; p = 434 x lo4 kglms; and hf, = 2256.9 kJkg

[Am. 439.9 x lo3 kJ/h, 218.8 kg/hl 7. A vertical plate in the form of fin is 500 mm in height and is exposed to steam at atmospheric pressure.

If the surface of the plate is maintained at 60°C, calculate: (i) The film thickness at the trailing edge of the film, (ii) The overall heat transfer coefficient, (iii) The heat transfer rate, and (iv) The condensate mass flow rate. Assume laminar flow conditions and unit width of the plate.

[Ans. (i) 0.1732 mm, (ii) 6227.52 W/m°C (McAdam's), (iii) 124550 W, (iv) 0.055 kg/s] 8. A vertical plate 2.8 m high is maintained at 5 4 T in the presence of saturated steam at atmospheric

pressure. Calculate the heat transfer rate per unit width. [Ans. 700 kW/m) 9. A vertical tube of 50 mm outside diameter and 2 m long is exposed to steam at atmospheric pressure.

The outer surface of the tube is maintained at a temperature of 84°C by circulating cold water through the tubes. Determine: (i) The rate of heat transfer to the coolant, and (ii) the rate of condensation of steam. [Ans. (i) 179 kW, (ii) 28.6 kgh]

10. A horizontal tube of outer diameter 25 mm is exposed to dry steam at 100°C. The tube surface temperature is maintained at 84°C by circulating water through it. Calculate the rate of formation of condensate per metre length of the tube. [Ans. 21.94 kg/h]

11. A condenser is to be deigned to condense 2250 kg/h of dry and saturated steam at a pressure of 15 kPa. A square array of 400 tubes each of 6 mm in diamter, is to be used. If the tube surface temperature is to be maintained at 2G°C, calculate the heat transfer coefficient and the length of each tube, assuming single pass. [Ans. 5205.3 w/m2"C; 1.35 ml

Heat Exchangers 10.1. Introduction. 10.2. 'Qpes of heat exchangers. 10.3. Heat exchanger analysis. 10.4. Logarithmic

mean temperature difference (LMTD). 10.5. Overall heat transfer coefficient. 10.6. Correction factors for multipass arrangements. 10.7. Heat exchanger effectiveness and number of transfer units (NTU). 10.8. Pressure drop and pumping power; rnical Examples-Highhghts-Theoretical Questions-Unsolved Examples.

10.1. INTRODUCTION A 'heat exchanger' may be defined as an equipment which transfers the energy from a hot

fluid to a cold fluid, with maximum rate and minimum investment and running costs. In heat exchangers the temperature of each fluid changes as it passes through the exchangers,

and hence the temperature of the dividing wall between the fluids also changes along the length of the exchanger.

. Examples of heat exchangers: (i) Intercoolers and preheaters; (ii) Condensers and boilers in steam plant; (iii) Condensers and evaporators in (iv) Regenerators;

refrigeration units; (v) Automobile radiators; (vi) Oil coolers of heat engine;

I (vii) Milk chiller of a paste- (viii) Several other industrial processes. urising plant;

10.2. TYPES OF HEAT EXCHANGERS In order to meet the widely varying applications, several types of heat exchangers have been

developed which are classified on the basis of nature of heat exchange process, relative direction of fluid motion, design and constructional features, and physical state of fluids.

1. Nature of heat exchange process Heat exchangers, on the basis of nature of heat exchange process, are classified as

follows: ( i ) Direct contact (or open) heat exchangers. (ii) Indirect contact heat exchangers.

(a) Regenerators. (b) Recuperators. (i) Direct contact heat exchangers In a direct contact or open heat exchanger the exchange of heat takes place by direct mixing

of hot and cold fluids and transfer of heat and mass takes place simultaneously. The use of such units is made under conditions where mixing of two fluids is either harmless or desirable. Examples: ( i ) Cooling towers; (ii) Jet condensers; (iii) Direct contact feed heaters.

Figure 10.1 shows a direct contact heat exchanger in which steam mixes with cold water, gives its latent heat to water and gets condensed. Hot water and non-condensable gases leave the container as shown in the figure.

(ii) Indirect contact heat exchangers

330 Heat and Mass ~ransfer

In this type of heat exchanger, the heat transfer between two fluids could be carried out by transmission Hot , through wall which separates water the two fluids. This type in- Container cludes the following:

(a) Regenerators. (b) Recuperators or sur-

face exchangers. (a) Regenerators: In a

regenerator type of heat exchanger the hot and coMfluids pass alternately through a

t Cold

space containing solid parti- water

cles (matrix), these particles Fig. 10.1. Direct contact or open heat exchanger. providing alternately a sink and a source for heat flow.

Examples : (i) I.C. engines and gas turbines; (ii) Open hearth and glass melting furnaces; (iii) Air heaters of blast furnaces.

A regenerator generally operates periodically (the solid matrix alternately stores heat extracted from the hot fluid and then delivers it to the cold fluid). However, in some regenerators the matrix is made to rotate through the fluid passages arranged side by side which makes the heat exchange process continuous.

The performance of these regenerators is affected by the following parameters: (i) Heat capacity of regenerating material, (ii) The rate of absorption, and (iii) The release of heat. Advantages 1. Higher heat transfer coefficient; 2. Less weight per kW of the plant; 3. Minimum pressure loss; 4. Quick response to load variation; 5. Small bulk weight; 6. Efficiency quite high. Disadvantages 1. Costlier compared to recuperative heat exchangers. 2. Leakage is the main trouble, therefore, perfect sealing is required. (b) Recuperators: 'Recuperator' is the most important type of heat exchanger in which the

flowing fluids exchanging heat are on either side of dividing wall (in the form of pipes or tubes generally). These heat exchangers are used when two fluids cannot be allowed to mix ix., when the mixing is undesirable.

Examples: (i) Automobile radiators, (ii) Oil coolers, intercoolers, air preheaters, economisers, superheaters, condensers and surface feed heaters of a steam power plant, (iii) Milk chiller of pasteurising plant, (iv) Evaporator of an ice plant:

Advantages 1. Easy construction; 2. More economical; 3. ore surface area for heat transfer; 4. Much suitable for stationary plants. Disadvantages 1. Less heat transfer coefficient; 2. Less geneuting capacity; 3. Heavy and sooting problems.

Free Convection 531

The flow through direct heat exchangers and recuperators may be treated as steady state while through regenerators the flow is essentially transient.

2. Relative direction of fluid motion According to the relative directions of two fluid streams the heat exchangers are classified

into the following three categories: (i) Parallel flow or unidirection flow (ii) Counte-flow (iii) Cross-flow. (i) Parallel flow heat exchangers In a parallel flow heat exchanger, as the name suggests, the two fluid streams (hot and

cold) travel in the same direction. The two streams enter at one end and leave at the other end. The flow arrangement and variation of temperatures of the fluid streams in case of parallel flow heat exchangers, are shown in Fig. 10.2. It is evident from the Fig.l0.2(b) that the temperature diference between the hot and cold fluids goes on decreasing .from inlet to outlet. Since this type of heat exchanger needs a large area of heat transfer, therefore, it is rarely used in practice.

r (temperature)

- L (length)

Fig. 10.2. Parallel flow heat exchanger.

Examples: Oil coolers, oil heaters, water heaters etc. As the two fluids are separated by a wall, this type of heat exchanger may be called parallel

flow recuperator or surface heat exchanger. (ii) Counter-flow heat exchangers In a counter-flow heat exchanger, the two fluids flow in opposite directions. The hot and

cold fluids enter at the opposite ends. The flow arrangement and temperature distribution for such a heat exchanger are shown schematically in Fig. 10.3. The temperature difference between the

I I

rc 1

(4 (b )

Fig. 10.3. Counter-flow heat exchanger


two fluids remains more or less nearly constant. This type of heat exchanger, due to counter flow, gives maximum rate of heat transfer for a given surface area. Hence such heat exchangers are most favoured for heating and cooling of fluids.

(iii) Cross-flow heat exchanger In cross-flow heat exchangers, the two fluids (hot and cold) cross one another in space,

usually at right angles. Fig. 10.4 shows a schematic diagram of common arrangements of cross-flow heat exchangers.

Cold fluid (in) (unmixed stream)

Cold fluid (in) Tube- (Mixed hot stream)

E3

e Hot fluid Hot (in) e + fluid (Unmixed (out) stream) c3

Cold fluid (out) (a)

Free Convection

Cold fluid (out) (6)

Fig. 10.4. Cross-flow heat exchangers

Refer Fig. 10.4 (a): Hot fluid flows in the the separate tubes and there is no mixing of the fluid streams. The cold fluid is perfectly mixed as it flows through the exchanger. The temperature of this mixed fluid will be uniform across any section and will vary only in the direction of flow.

Examples: The cooling unit of refrigeration system etc. Refer Fig. 10.4(b) : In this case each of the fluids follows a prescribed path and is unmixed as it flows through heat exchanger. Hence the temperature of the fluid leaving the heater section is not uniform.

Examples: Automobile radiator etc. In yet another arrangement, both the fluids are mixed while they travel through the exchanger; consequently the temperature of both the fluids is uniform across the section and varies only in the direction in which flow takes place. I

3. Design and constructional features

Hot fluid (out)

Cold fluid (out)

I Cold fluid

(in) (a) One-shell pass and two-tube pass heat exchanger

Shell fluid +

A I Baffles I

I I I .- I

- \ I I \ .I , / I I Tube fluid -b I I \A \A/ I I

I * (6) Two-shell pass and four-tube pass heat exchanger

On the basis of design and constructional features, the heat exchangers are classified as under: Fig. 10.5. Shell and tube heat exchangers. ( i ) Concentric tubes In this type, two concentric tubes are used, each carrying one of the fluids. The direction

of flow may be parallel or counter as depicted in Fig. 10.2 (a) and Fig. 10.3 (a). The effectiveness of the heat exchanger is increased by using swirling flow.

(ii) Shell and tube In this type of heat exchanger one of the fluids flows through a bundle of tubes enclosed

by a shell. The other fluid is forced through the shell and it flows over the outside surface of the tubes. Such an arrangement is employed where reliability and heat transfer effectiveness are important. With the use of multiple tubes heat transfer rate is amply improved due to increased surface area.

(iii) Multiple shell and tube passes Multiple shell and tube passes are used for enhancing the overall heat transfer. Multiple

shell pass is possible where the fluid flowing through the shell is re-routed. The shell side fluid is forced to flow back and forth across the tubes by baffles. Multiple tube pass exchangers are those which re-route the fluid through tubes in the opposite direction.

(iv) Compact heat exchangers These are special purpose heat exchangers and have a very large transfer surface area per

unit volume of the exchanger. They are generally employed when convective heat transfer coefficient associated with one of the fluids is much smaller than that associated with the other fluid.


Example : Plate-fin, flattened fin tube exchangers etc. (4) Physical state of fluids Depending upon the physical state of fluids the heat exchangers are classified~as follows: (i) Condensers (ii) Evaporators (i) Condensers: In a condenser, the condensing fluid remains at constant temperature throughout

the exchanger while the temperature of the colder fluid gradually increases from inlet to outlet. The hot fluid loses latent part of heat which is accepted by the cold fluid (Refer Fig. 10.6).

Fig. 10.6. Temperature distribution in a condenser. Fig. 10.7. Temperature distribution in an evaporator.

(ii) Evporators: In this case, the boiling fluid (cold fluid) remains at constant temperature while the temperature of hot fluid gradually decreases from inlet to outlet. (Refer Fig. 10.7).

10.3. HEAT EXCHANGER ANALYSIS For designing or predicting the performance of a heat exchanger it is necessary that the total

heat transfer may be related with its governing parameters : (i) U (overall heat transfer coefficient due to various modes of heat transfer, (ii) A total surface area of the heat transfer, and (iii) tl, t2 (the inlet and outlet fluid temperatures). Fig. 10.8 shows the overall energy balance in a heat exchanger.

Let m mass flow rate, kgts.

cp = specific heat of fluid at constant pressure J/kg°C

t = temperature of fluid, "C

At = temperature drop or rise of a fluid across the heat exchanger. Heat exchanger

Hot fluid - Q,

- Heat transfer area

Fig. 10.8. Overall energy balance in a heat exchanger.

Subscripts h and c refer to the hot and cold fluids respectively; subscripts band 2 correspond to the inlet and outlet conditions respectively.

Free Convection

Assuming that there is no heat loss to the surroundings and potential and kinetic energy changes are negligible, from the energy balance in a heat exchanger, we have:

Heat -given up by the hot fluid, Q = mh Cph (th~ - th2) ...( 10. 1)

Heat picked up by the cold fluid, Q = mc cpc <tc2 -tCl) ...( 10.2)

Total heat transfer rate in the heat exchanger, Q = U A 8, ...( 10.3)

where U = overall heat transfer coefficient between the two fluids, A = effective heat transfer area, and

0, = appropriate mean value of temperature difference or logrithmic mean temperature difference (LMTD).

10.4. LOGARITHMIC MEAN TEMPERATURE DIFFERENCE (LMTD) Logarithmic mean temperature difference (LMTD) is defined as that temperature difference

which, if constant, would give the same rate of heat transfer as actually occurs under variable conditions of temperature difference.

In order to derive expression for LMTD for various types of heat exchangers, the following assumptions are made:

1. The overall heat transfer coefficient U is constant. 2. The flow conditions are steady. 3. The specific heats and mass flow rates of both fluids are constant. 4. There is no loss of heat to the surroundings, due to the heat exchanger being perfectly

insulated. 5. There is no change of phase either of the fluid during the heat transfer. 6. The changes in potential and kinetic energies are negligible. 7. Axial conduction along the tubes of the heat exchanger is negligible.

10.4.l.Logarithmic Mean Temperature Difference for "Parallel Flow" Refer Fig. 10.9, which shows the flow arrangement and distribution of temperature in a

single-pass parallel flow heat exchanger. Let us consider an elementary area dA of the heat exchanger. The rate of flow of heat

through this elementary area is given by dQ = UdA (th - tc) = u.dA.At

As a result of heat transfer dQ through the area dA, the hot fluid is cooled by dh whereas the cold fluid is heated up by dtc. The energy balance over a differential area dA may be written as

dQ = - mh Cph . dth = hc Cpc . dtc = U. dA . (th - tc) ...( 10.4)

(Here dth is -ve and dtc is +ve)

and dQ dQ dtc = - = - ;., cpc c c

where Ch L i th cph = heat capacity or water equivalent of hot fluid, and

Cc = riz, cpC = heat capacity or water equivalent of cold fluid.

?ith and in, are the mass flow rates of fluids and cph and cp, are the respective specific horlr.

Annulus surrounding /- the pipe

Heat and Mass Transfer Free Convection

II Pipe ---+ ----+ Cold ---+ ---+ /

K

4 Hot 4 4 - - Cold - - (a) Flow arrangement

2 Area -+

(b) Temperature distribution

Subscri~ts h. c refer to : hot and cold fluids ~ubscriDt 1. .2 refer to : inlet and outlet conditions.

Fig. 10.9. ~alculaGon bf LMTD for a parallel flow heat exchanger. Substituting the value of dQ from Eqn. (10.4) the above equation becomes

Integrating between inlet and outlet conditions (i.e. from A = 0 to A = A), we get

Now, the total heat transfer rate between the two fluids is given by

Q = ch (thl - th2) = Cc (tc2 - tcl)

1 1 Substituting the values of - and - into Eqn. (10.6), we get

Ch cc

The above equation may be written as

Q = U A 8 ,

...[ 10.7 (a)]

...[ 10.7 (b)]

8, is called the logarithmic mean temperature differencr (LMTD).

10.4.2. Logarithmic Mean Temperature Difference for "Counter-flow" Refer Fig. 10.10, which shows the flow arrangement and temperature distribution in a

single-pass counter-flow heat exchanger. Annulus surrounding

/- the pipe II Pipe

t t Cold t - + Hot - - Cold - -

(a) Flow arrangement

I I

Area -+ (b) Temperature distribution

Free Convection 538 Heat and Mass Transfer

Let us consider an elementary area dA of the heat exchanger. The rate of flow of heat through this elementary area is given by

dQ = U.dA (th - tc) = U.dA.At ...( 10.11)

In this case also, due to heat transfer dQ through the area dA, the hot fluid is cooled down by dth whereas the cold fluid is heated by dt,. The energy balance over a differential area dA may be written as

dQ = - mh . Cph . dlh = - mc . cpc . dtc ...( 10.12)

In a counter-flow system, the temperatures of both the fluids decrease in the direction of heat exchanger length, hence the -ve signs.

A special case arises when 0 , = e2 = 0 in case of a counter-flow heat exchanger. In such , a case, we have

This value is indeterminate. The value of 8, for such a case can be found by applying L' Hospital's rule:

and d tc=- -=- - dQ dQ mc Cpc cc Let (8J0,) = R. Therefore, the above expresion can be written as

lim 0 (R- 1)

~ - 1 1 ln(R) Differentiating the numerator and denomenator with respect to R and taking limits, we get

0 lim - - - 8

(R* 1) (1/R) Inserting the value of dQ from Eqn. (10.11), we get

de = - U dA (th - t,) - - - [ i h id Hence when 0 , = 0, Eqn. (10.3) becomes

Q = U A 8 0, (LMTD) for a counter-flow unit is always greater than that for a parallel flow unit;

hence counter-flow heat exchanger can transfer more heat than parallel-flow one; in other words a counter-flow heat exchanger needs a smaller heating surface for the same rate of heat transfer. For this reason, the counter-flow arrangement is usually used.

Integrating the above equation from A = 0 to A = A, we get When the temperature variations of the fluids are relatively small, then temperature variation curves are approximately straight lines and adequately accurate results are obtained by taking the arithmatic mean temperature dijference (AMTD).

Now, the total heat transfer rate between the-two fluids is given by Q = ch (thl - t h ~ ) = Cc (tc2 ' tcl) ...( 10.15)

However, practical considerations suggest that the logarithmic mean temperature ...[ 10.15 (a)] 01 difference (8,) should be invariably used when - > 1.7. 82

10.5. OVERALL HEAT TRANSFER COEFFICIENT In a heat exchanger in which two fluids are separated by a plane wall as shown in the Fig.

10.1 1 , the overall heat transfer coefficient is given by 1 1

substituting the values of - and - into Eqn. (10.14), we get Ch cc

U = 1 ...( 10.18)

1 L 1- -+-+- hi k ho

If the fluids are separated by a tubewall as shown in Fig. 10.12 the overall heat transfer coefficient is given by:

Inner surface

Since Q = U A 8 ,

Heat and Mass Transfer Heat ~xchangers* 543

S. No. Fluid combination

Water to water

Water to oil

Steam condensers(water in tubes)

Alcohol condensers (water in tubes)

Feed water heaters

Air-condensers

Air to various gases

Air to heavy tars and liquids

Air to low viscosity liquids

Finned-tube heat exchanger (water in tubes, air in cross-flow)

Distilled water

Treated boiler feed water

Worst water used in heat exchangers

Fuel oil and crude oil

Industrial liquids

Transformer or lubricating oil

Engine exhaust and fuel gases

Steam (non-oil bearing)

Refrigerant liquids, brine or oil-bearing

Fouling processes : 1. Precipitation or crystallization fouling. 2. Sedimentation or particulate fouling. 3. Chemical reaction fouling or polymerisation. 4. Corrosion fouling. 5. Biological fouling. 6. Freeze fouling. Parameters affecting fouling :

Velocity

--

0.0001

0.0001 - 0.0002

< 0.0002

0.0009

0.0002

0.0002

0.002

0.0001

0.0002

Temperature

Table 10.2. Representative values of overall heat transfer coefficient (U)

u ( w/m2~c) 850 - 1170

110 - 350

1000 - 6000

250 - 700

110 - 8500

350 - 780

60 - 550

As low as 45

As high as 600

25 - 50

Water chemistry

Tube material. Prevention of fouling : The following methods may be used to keep fouling minimum : 1. Design of heat exchanger. 2. Treatment of process system. 3. By using cleaning system.

Properties to be censidered for&lectiod:of materials for heat exchaneers : Physical properties

~echanical ~opert ies

Climatic properties

Chemical environment

Quality of surface finish

Service life

Freedom from noise

Reliability.

Common failures In heat exchangers :

Chocking of tubes either expected or extraordinary.

Excessive transfer rates in heat exchanger.

Increasing the pump pressure to maintain throughout.

Failure to clean tubes at regularly scheduled intervals.

Excessive temperatures in heat exchangers.

Lack of control of heat exchangers atmosphere to retard scaling.

Increased product temperature over a safe design limit.

Unexpected radiation from refractory surfaces.

Unequal heating around the circumference or along the length of tubes.

01 Example 10.1. For what value of end temperature dflerencis ratio -, is the arithmatic 92

mean temperature difference 5 per cent higher than the lbg-mean temperature diference? Solution: The arithmetic mean temperature difference (8) and log-mean temperature difference

(0,) ratio may be written as

f0, + 0,) I

It is given' that 0 is to be 5 percent higher than 0,

By hit and trial method, we get

Thus the simple arithmetic mean temperature difference gives results to within 5 percent when end temperature differences vary by no more than a factor of 2.2.

Example 10.2. (a) Derive an expression for the effectiveness of a parallelflow heat exchanger in terms of the number of transfer units, NTU, and the capacity ratio Cmi,,/Cm,


' (b) h a parallel flow double-pipe heat exchanger water flows through the inner pipe and is heated from 20°C to 70°C. Oil flowing through the annulus is cooled from 200°C to 100°C. It is desired to cool the oil to a lower exit temperature by increasing the length of the heat exchanger. Determine the minimum temperature to which the oil may be cooled.

(U.P.S.C.. 1995) Solution. (a) Refer Article 10.7.

(b) Using subscripts.h and c for oil and water respectively, we have

Let 't' be the lowest temperature to which oil may be cooled and this will be the highest temperature of water too (Refer Fig. 10.13).

Hence, mhcph (200 - t ) = m,cPc (t - 20)

Areallength - Fig. 10.13

=2( t -20 )

200-t=2t-40

t = 80°C (Ans.) 4' Eample 10.3. The flow rates of hot and cold water streams running through a parallel flow

heat exchanger are 0.2 kg/. and 0.5 kg/s respectively. The inlet temperatures on the hot and cold sides are 75°C and 20°C respectively. The exit temperature of hot water is 45°C. If the individual heat transfer coeflcients on both sides are 650 w/m2"c, calculate the area of the heat exchanger.

Solution. Given: mh = 0 2 kgls; mc = 0.5 kgls; thl = 750 C; th2 = 4 5 0 ~ ; tc, = 2 0 0 ~ ; hi = h, = 650 w/m2'c.

The area of heat exchanger, A: The heat exchanger is shown diagrammatically in Fig. 10.14.

The heat transfer rate, Q = 4 cph ( - th2)

= 0.2 x 4.187 x (75 - 45) = 25.122 WS Heat lost by hot water = heat gained by cold water

mh cph (thl - th2) = me cpc (re- - tcl)

0.2 x 4.187 X (75 - 45) = 0.5 X 4.187 x (tc2 - 20)

. . tC2 = 32°C

Logarithmic mean temperature difference (LMTD) is given by

...[ Eqn. (10.9)]

Heat Exchangers 545

tcl = 20°C 4 Cold water - A tc2

thl = 75°C 4 ~ o t water 4 4 4 5 ' ~ (th2)

tcl = 20°C 4 Cold water 4 4 tc2



Fig. 10.14. Parallel flow heat exchanger.

Overall heat transfer coefficient U is calculated from the relation 1 1 1 * - U =-+- hi

h, ,

Also, Q = U A 9 ,

Q - 25'122 lo* = 2.65 ( A = - - ' Ans.) U 8, 325 x 29.12

n-

&xample 10.4. The following data relate to a parallel flow heat exchanger in which air is heated by hot exhaust gases.

Heat transferred per hour ... 155450 kJ

Inside heat transfer coeflcient ... 120 w/rn2"c

Outside heat transfer coeflcient ... 195 w/rn20c

Inlet and ouilet temperatures of the hot fluid ... 450°C and 2.50% respectively

546 Heat and Mass Transfer Heat Exchangers

Inlet and outlet temperatures of the cold fluid ... 60°C and 120°C, respectively Inside and outside diameters of the tube . .SO mm and 60 mm, respectively. Calculate the length of the tube required for the necessary heat transfer to occur. Neglect

the tube resistance. Solution. Given : Q = 155450 kJ/h; h, = 120 w/m2'c; ho = 195 W I ~ ~ ~ C ; th, = 450°C;

th2 = 250°C; tcl = 60°C; tc2 = 120°C; di = 50 mm = 0.05m; do = 60 mm = 0.06 m.

Length of each tube, L: Logarithmic mean temperature differenqe (LMTD) is given by

The overall heat transfer coefficient, U is given by

. . U = 66.09 w / m 2 ~ c Total heat transfer rate is given by

Q = U A 8,= U x ( n d 0 L ) x 8 ,

155450 x (1000/3600) L = Q - - = 14.65 m (Ans.) U x 7t do x 8, 66.09 x 7t x 0.06 x 236.66

Example 10.5. A hot fluid at 2OO0C enters a heat exchanger at a mass flow rate of 10* kg/h. Its specijc heat is 2000 Jkg K. It is to be cooled by another fluid entering at 25OC with a mass flow rate 2500 kg/h and specijc heat 400 J/kg K. The overall heat transfer coejficient based on outside urea of 20 m2 is 250 w/m2 K. Find the exit temperature of the hot fluid when the fluids are in parallel flow. (GATE, 1998)

Solution. Given : thl = 200°C; mh = - lWW - - 2.78 kg/s; c,,,, = 2000 J/kg K; t,, = 25OC; 3600

Exit temperature of the hot fluid, tm :

Heat lost by the hot fluid, Q = mhcph (thl - th2)

t h ~ = 2ooocc Hot fluid

th2

Cold fluid t,, = 25OC

I I Fig. 10.15.

Heat lost by the cold fluid, Q = 0.694 x 400 (tc2 - 25) = 277.6 (tc2 - 25)

Equating (i) and (ii), we have 5560 (200 - th2) = 277.6 (tc2 - 25)

Also, heat transferred is given by, Q = UA8,

where,

Here 8, = thl - tc, = 200 - 25 = 175OC; and e2 = tm - tc2

175 - ( h 2 - tc2) . . th2 - 4 2 "=m

Substituting the values in the above equati*, we get

. . . (ii)

. . .(iii)

...( iv)

L J

Substituting the values of tc2 from (iii) in (iv), we get 175 - Itu - (4025 - 20 th2)l

Q = 2 5 0 ~ 2 0 [ 1

Equating (i) and (v), we get 175 - (2 1 th, - 4025)

5560 (2W - lh2) = 5000[ { 175 1 ] 21 th2-425

Using hit and trial method, the value of th2 may be found out. -

J Example 10.6. In a certain double pipe heat exchanger hot water flows at a rate of 5VW

-

kgih and gets cooled from 95'C to 65OC. At the,same time 50000 kg/h of cooling water at 30°C enters the heat exchanger. The flow conditions are such that overall heat transfer coejficient remains constant at 2270 w/m2 K. Determine the heat tmsfer area required and the effectiveness, assuming two streams an in pamllel flow. Assume for the both the s t m s cp= 4.2 kJAg K.

(GATE, 1997)

50000 - 13.89 k g / ~ ; fhl = 95OC; th2 = 6S°C; Solution. Given : mh = - - 3600

Heat and Mass Transfer Heat Exchangers

Solution: Refer Fig. 10.9, for parallel flow heat exchanger. The heat flow rate through an elementary strip of area dA is

dQ = - mh cph dt,, = mc cpc dlc = -Ch dth = Cc dtc . . . (i) cph = cpc = 4.2 kJ/kg or 4200 J/kg K.

Q = heat lost by hot water = heat gained by cold water. dQ dQ and dr, = - dth = - - Ch cc

...( ii)

From eqns. (i) and (ii), we get

d (th - tc) = d (0) = - dQ [& + -&] ...( iii)

cold fluid

(water) Also d~ = u.dA.0 = (a + be) dA.0

Integrating eqn. (iv) within limits 1 and 2, we get

91 - 92 Q = 1

[G + $1 1 1 (31 - 92 or -+- = -

Ch cc Q From eqns. (iv) and (v) and using eqn. (vi), we have

- d0 - - -de Qde (a + b e ) dA.0 = =-

8 , -8, 02-01 [i + k] Integrating eqn. (viii) between limits 1 and 2, we get

Length ----+ Fig. 10.16. Parallel-flow heat-exchanger.

...( vi)

...( vii)

...( viii)

mh Cph (thl - thl ) = lj2c Cpc (tc2 - t ~ l )

or 13.89 x 4200 x (95 - 65) = 13.89 x 4200 x (tc2 - 30) . . tc2 = 60°C

Log mean temperature difference,

LMTD, 0, = (91 - 92)

In ('31/92)

Also, Q = UA9,

or 13.89 x 4200 x (95 - 65) = 2270 x A x 23.4 :. Heat transfer area, A = 32.95 m2 Also, Qacrual = mh~ph (thl - th2) and Qm. = mh cph (thl - tcl) :. Effectiveness of the heat exchqger, The constant a may be expressed by solving UI = a + beI and

U2 = a + beZ which results in

a = u192 - u2 01 0, - 0 , Example 10.7. Prove that the rate of heat t rangr Q for a double parallel flow heat

exchanger in which the overall heat trans$er coeficient varies linearly with temperature direrence i.e., U = a + be, where a and b are constants, is given by

Q = u1e2 - u2 01 A

ln Wle2/U2 011 Here su& I and suffix 2 represent inlet and outlet of the heat exchanger respectively.

ibstituting this value of a in eqn. (ix), we get

4 0 2 - u201 * Q = ... Proved.

In t(Ule2)/(U2 % ) I


Example 10.8. In a counter-jlow double pipe heat exchunger, water is heated fmm 25°C to 65OC by an oil with a specijic heat of 1.45 W/kg K and mars flow rate of 0.9 kgh. 73e oil is cooled from 230°C to 160DC. If the overall heat tran.$er coescient is 420 W@C, calculate the following:

(i) The rate of heat transfer, (ii) The mass flow rate of water, and (iii) The surjGace area of the hat exchanger.

Solution. Given: tcl = 25°C; tc2 = 65°C. cph = 1.45 Hlkg K; mh = 0.9 kgls;

thl = 230°C; th2 = 160°C, LI = 420 w/m2"c.

( i ) The rate of heat transfer, Q:

or Q = 0.9 x (1.45) x (230 - 160) = 91.35 kJ1s (Ans.) (ii) The mass flow rate of water, m,':

Heat lost by oil (hot fluid) = heat gained by water (cold fluid)

mh cph (thl - th2) = cpc x (tc2 - tcl)

91.35 = ri?, x 4.187 (65 - 25)

m, = 91'35 = 0.545 kgls (Am.) 4 . 1 8 7 ~ (65-25)



Fig. 10.17. Counter-flow heat exchanger.

(iii) The surface area of heat exchanger, A: Logerithmic mean temperature difference (LMTD) is given by

Heat Exchangers

Also, Q = U A O ,

Q - 91'35 lo3 = 1-45 m2 (Am.) A = - - U 0, 420 x 149.5 ,..

~J'Example 10.9. An oil cooler for a lubrication system has to cool 1000 kg/h of oil (cp = 2.09 kJ/7cg°C) from 8PC to 4O"C by using a cooling wafer flow of 1000 kg/h at 30°C. Give your choice for a parallel flow or counter-jlow heat exchanger, with reasons. Calculate the surface area of the heat exchanger, if the overall heat transfer coeflcient is 24 w/m2"c. r

Take cp of water = 4.18 kJ/kg°C. 1000

Solution. Given: Ah = 1000 kgls; cph = 2.09 W k g 0 ~ ; ' c p , = 4.18 Wlkg°C; kc = - kgh; 3600 3600

Surface area of heat exchanger, A: I,@ subscripts h and c stand for hot and cold fluids respectively.

Since tC2 > th2, counter-flow arrangement must be used.

Again, 8, = ( 3 1 - 82

In (81/(32)

Fig. 10.18.


Also, Q = U A 0 ,

Example 10.10. Show that in a double-pipe counter flow heat exchanger if &,ch = 4 c , the temperature projiles of the two fluids along its length are parallel straight lines.

(GATE, 1996) Solution. For heat exchanger, dQ = - mh ch dth = mc c, dtc

= - Ch dt, = cc dt,

where ih,, = mass flow rate of hot fluid, and

m, = mass flow rate of cold fluid.

Due to heat exchanger, temperature of hot fluid decreases by dth and that of cold fluid increases by dt,.

d Also, C, = heat capacity of cold fluid, E

and F Ch = heat capacity of hot fluid.

mhch = mc cC ...( Given)

or Ch = cc In counter-jlow system, temperature of

both the fluids decrease in the direction of heat exchanger length, therefore

Length

dQ dt,, = - - and dtc= -- dQ Ch cc

dth-dtc=dO=-dQ --- Id,' d,l

Fig. 10.19.

L -1

Since Ch = Cc, :. d0 = 0 or 0 = constant. Thus, both the straight lines showing the variation of temperatures along the length are

parallel lines. ... Proved J' Example 10.11. A counter-jlow double [email protected] exchanger using superheated steam is used to heat water at the rate of 10500 kgh. T k steam enters the hear exchanger at 180°C and lehves at 13O0C The inlet and exit temperatures of water are 30°C and 80°C respectively. I f overall heat transfer coeficient from steam to water is 814 w/mZ0c, calculate the heat transfer area. What would be the increase in area if the fluid flows were parallel?

Heat Exchangers 553

( i ) When the flow is counter:

Om = 01 - 02 In W 0 2 )

In this 0, = = O2 = 100°C

I I

nr

th, = 130°C t , = 80°C tc2 = 80°C

t,, = 30°C + = 30°C I I 'c l I

(a) Counter-flow (b) Parallel flow

Fig. 10.20.

The heat transfer rate is given by Q = U A 0 ,

or mC x cpC x (tc2- tcl) = U A Om

or 2.917 x 4.187 x (80 - 30) = 814 x A x 100 2.917 x 4.187 x lo3 (80 - 30) = m2

or A = 814 x 100'

Again, Q = U A 0 ,

or 2.917 x (4.187 x lo3) x (80 - 30) = 814 x A x 91 2.917 x (4.187 x lo3) x (80 - 30) = 8.24 m2

A = 814 x 91

8.24 - 7 S = 0.0987 or 9.87% (Ans.) . Increase in area = , , 1 ..I

Example 10.12. A counter-flow heat exchanger, through which passes 12.5 kg/. of air to be cooled from 540°C to 146OC, contains 4200 tube, %dch having a diameter of 30 mm. The inlet and outlet temperatures of cooling water are 25OC and 75OC respectively. I f the water side resistance to flow is negligible, calculate the tube lehgNt required for this duty.

0.8 p4 For turbulent flow inside tubes : Nu = 0.023 Re P

Properties of the air at the average temperature are as follows:

p = 1.009 kg/m3; cp = 1.0082 kJ/kg°C; p = 2.075 x I @ kg/ms (lVs/m2) and k = 3.003 x 1 r 2 W/m°C.

Solution. Given : A, = 12.5 kgls, thl = 540°C; th2 = 146OC; tC1 = 25OC; t, = 7S°C; N = 4200.

554

Tube length, L:


Reynolds number, Re = f?L!.! P

Mass flow m = NAVp, therefore,

P c 2.075 x 10" x 1.0082 x lo3 = 0.6966 Prandtl number, Pr = = k 3.003 x lo-2

h d Nusselt number, Nu = - = 0.023 k

Since the water side resistance to flow is negligible

. . - - - - - I I I or u = 21.22 w / ~ ~ ~ c U - h - 21.22


em = 01 -82 In (el/@,)

Further, the rate of heat transfer,

Q = mh x cph x (th2-thl) = U A em = U x(Nn d L ) X%

L = Ah X Cph ( 4 2 - th,) - 12.5 x (1.0082 x lo3) x (540 - 146) - U x N n d x 8 , 21.22 x 4200 x n x 0.03 x 255.5 = 2.31 m (Ans.)

J Example 10.13. Steam enters a cowter-flow heat exchanger, dry saturated at 10 bar and leaves at 350°C. The man flow of steam is 800 kg/min. The gas enters Ske heat exchanger at

" 650°C and mass flow rate is 1350 kgimin. If the tubes are 30 mm diameter and 3 m long, determine the number of tubes required. Neglect the resistance offered by metallic tubes. Use the following data :

For steam : t,,, = 18O0C (at 10 bar); cps = 2.71 W/kg°C; hs = 600 w/m2O€

For gas : cpg = 1 kJ/kg°C; h, = 250 w/m2Oc. (P.U) . 800 1350 Solution. Given: 4 = mc = - = 13.33 kg/s; ritp t &,, = - - 60 60 - 22.5 kg/~; thl = 650°C;

tcl (= tsar) = 180°C; tc2 = 350°C; d = 30 mm = 0.03 m; L = 3 m.

Number of tubes required, N: Heat lost by gases = heat gained by steam

mh x cph X (thl - th2) = mc Cpc (tc2 - t c l )

22.5 x 1 x (650 -th2) = 13.33 x 2.71 x (350 - 180)

. . th2 = 377°C

fhl =650T 6 4 Gas 4 4 th2 (= 377OC)

tc2 = 3S0°C t + Steam t + tcl (= 180°C)

th, = 6s0°C 4 4 Gas C3 4 thl (= 377Q

I (a) Flow arrangement

I Area ---b


Fig. 10.21. I Counter-flow heat exchanger.

Overall heat transfer doeffici{nt is given by

Total heat transfer rate is given by Q = U A 8 ,

where A = N x ( ~ d L ) = N x n x 0 . 0 3 x 3 = 0 . 2 8 2 7 ~ m ~

Substituting the values in e4n. ( i ) , we get 6142.5 x lo3 = 176.5 x 0.2827 N x 244.9

N = 6142'5 = 503 tubes (Am.) 176.5 x 0.2827 x 244.9

...(g iven)


Example 10.14. In a shell and tube counter-jlow heat exchanger waterflows through a copper tube 20 mm I.D. (internal diamet8i-) and 23 mm O.D. (outer diameter), while oil flows through the shell. Water enters at 20°C and comes out at 30"C, while oil enters at 75°C and comes out at 60°C. The water and oil side jilm coeflcients are 4500 and I250 w/m2"c respectively. 'The thermal conductivity of the tube wall is 355 W/m°C. 7Xe fouling factors on the water and oil sides may be taken to be 0.0004 and 0.001 respectively. If the length of the tube is 2.4 m, calculate the following:

( i ) The overall heat transfer coeflcient; (ii) The heat transfer rate.

Solution. Given : di = 20 mm = 0.02 m; do = 23 mm = 0.023 m; tcl = 20°C; tc2 = 30°C;

thl = 75°C; th2 = 60°C, hi = 4500 W/m20C; ho= 1250 W/m2"C;

k = 355 W/m°C; Rfi = 0.0004; Rfo = 0.001; L = 2.4 m ( i ) The overall heat transfer coefficient, U,: The overall heat transfer coefficient based on outer surface of inner pipe is given by,

1 ro 1 ro - '0 1 = - X - + - R , +-In (rdri) + Rfo i - - Uo ri hi ri f k h~ ...[ Eqn. (10.28)]

(Rf stands for fouling factor)

. . 1 u, = - 0.00252

= 396.8 w / m 2 ~ c thl = 7 5 ' ~

(ii) The heat transfer rate, Q: Area, A, = n d o L

= n x 0.023 x 2.4 tc2 = 30°C

= 0.1734 m2 Logarithmic mean temperature dif-

ference (LMTD) is given by,

Area ---b

Fig. 10.22.

The heat transfer rate is given by

Q = u04e, or = 396.8' x 0.1734 x 42.15 = 2900 W (Ans.) Example 10.15. In a counter-flow double pipe heat exchanger waterflows through a copper

tube (19 mm O.D. and I6 mm I.D.), at a flow rate of 1.48 m/s. The oil flows through the annulus formed by inner copper tube and outer steel tube (30 mm O.D. and 26 mm I.D.). The steel ttrbe is insulated from outside. The oil enters at 0.4 kg/s and is cooled from 65°C to 500C whereas water enters at 32OC. Neglecting the copper tube wall thermal resistance, calculate the length of the tube required.

eat Exchangers

Date given :

Nu = 0.023 ( ~ e ) ' . ~ ( ~ r ) ' . ~

Fouling factor, water side = 0.0005 m2 K / W Fouling factor, oil side = 0.0008 m2 K / W

(U.P.S.C., 1998)

Water and oil properties :

Solution. Given : Inner dia. of the copper tube (di) = 16 mm = 0.016 m;

Outer dia. of the copper tube (do), = 19 mm = 0.019 m; Inner dia. of the steel tube, (di)$ = 26 mm = 0.026 m;

Outer dia. of the steel tube, (do), = 30 mm = 0.03 m;

tcl = 30°C; fhl = 65°C; th2 = 50°C

n ic = - 4 ~ ( 0 . 0 1 6 ) ~ x 1.48 x 995 = 0 796 k g s ; mh = 0.4 kg/s

t,z (32

fluid (waer)

(4 (b)

Fig. 10.23. Counter-flow double pipe heat exchanger.

Length of the tube, L : The rate of heat transfer is given by,

Q = mh Cph (thl - fh2) = k cpc (tc2 - tcl)

0.4 x 1.89 (65 - 50) = 0.296 X 4.187 (tc2 - 30)

Water

995

4.187

0.615

4.18xlo-'

Property

P (kg/m3)

~p ( k J h g K)

k (W/m K)

v (m2/s)

Also, Q = 0 . 4 ~ 1 . 8 9 ~ ( 6 5 - 5 0 ) = 11.34kW Reynolds number for the flow of water through copper tube,

4 m Re=-= 4 x 0.296 = 56826 (*: P=Q.V) n(di),p nx0 .016x (995x418x 10-3

'i

Now, Nu = 0.023 (~e)'.' ( ~ r ) ' . ~ ...(G ha)

Oil

850

1.89

0.138

7.44 x 1 0 - ~


w , (Since, Pr = and water is heated)

k

Now oil flows through annulus diameter, therefore hydraulic diameter,

Dh=(di), -(do),=0.026-0.019=0.007 m Reynolds number through the annulus,

- = 1790

Since Re c 2500, hence annulus flow is laminar. Heat transfer coefficient at the inner surface of the annulus,

NU--- h" Dh - 0.023 (~e)'.' ( ~ r ) ' , ~ k

Overall heat transfer coefficient referred to outer diameter of inner tube is given by :

(Rf stands for fouling factor)

- 1 - 0.002203 + 0.000594 + 0.005309 + 0.0008 + 0.01466

= 42.43 w/m2 K Logarithmic mean temperature difference is given by,

The heat transfer rate is given by,

Q = Ua,em = 42.43 x (n x 0.019 x L) = 20.86

*/ Example 10.16. A counter-flow, concentric tube heat exchanger is used to cool the lubricating oil for a large industrial gas turbine engine. The flow rate of cooling water through the inner tube (d, = 20 mm) is 0.18 kg/s, while theflow rate of oil through the outer annulus ld, = 40 mm) is 0.12 kg/s. The inlet and outlet temperatures of oil are 95OC and 65°C respectively. The water enters at 30°C to the exchanger. Neglecting tube wall thermal resistance, fouling factors and heat loss to the surroundings, calculate the length of the tube.

Take the following properties at the bulk mean tempeature:

Engine oil at 80°C : c, = 2131 J/kg°C; )I = 0.0325 ~s /m ' ; k = 0.138 W/m°C:

Water at 35OC : c, = 41 74 J/kg°C, p = 725 x I@ ~s /m ' ; k = 0.625 W/m°C, Pr = 4.85 Solution: Given: di = 20 mm = 0.02 m; do = 40 mm = 0.04 m, m, = m, = 0.18 kg/s;

m,, = mh = 0.12 kg/s; tll, = 95OC, th2 = 65OC; t,, = 30°C. Annulus

Fig. 10.24. The counter-flow, concentric tube heal exchanger.

Length of the tube, L: The rate of heat transfer is given by,

t,,, = 65°C

r,, = 30°C

Heat Exchangers 560 Heat and Mass Transfer

,/ Example 10.17. The following data pertain to an oil cooler of the form of thl = 803C tubular heat exchanger, where oil is cooled by a large pool of stagnant water: -.-- * --

Temperature of stagnant water = 20°C (assumed constant),

Inlet and outlet temperatures of oil = 80°C and 30°C respectively,

Inside diameter and length of the tube carrying oil = 20 mm and 30 m, respectively,

th2 = 3O0c

Cold fluid (water)

LMTD is given by,

The overall heat transfer coefficient U is given by the relation

Fig. 10.25 . . ( i )

Reynolds number for flow of water through the tube is Specific heat and specific gravity of oil = 2.5 W/kg°C and 0.85 respectively, Average velocity of oil = 0.55 ids. Calculate the overall heat transfer coeflcient obtainable from thk system. Solution. Given : tcl = tC2 = 20°C; thl = 80°C; th2 = 30°C; d = 20 mm = 0.02 m; L = 30 m; Since the flow is turbulent, hence

cph = 2.5 kJ/kg°C; sp.gr = 0.85; V = 0.55 d s . Overall heat transfer coefficient, Ui : The mass flow rate of hot fluid (oil),

Heat lost by the hot fluid,

Q = hh Cph (thl - th2) = 0.1468 x (2.5 x LO') x (80 - 30) = 18350 w = 3088.7 w / m 2 OC

The oil flows through the annulus, hence the hydraulic diameter Dh = do - di = 0.02 m

Reynolds number through the annulus is

Re = p.umDe - P(do-di) - X mh

P CL X - (d,2 - dl2) P 4

LMTD is given by,

4 h, Re = - - 4 x0.12

= 78.35 7[: (d, + d,) y x (0.04 + 0.02) x 0.0325

The heat transfer rate is given by

Q = Ui Ai 0, = Ui ( X di L) 0, 18350 = Ui x x x 0.02 x 30 x 27.9

Since Re c 2300, hence the annular flow is laminar. Assuming uniform temperatun along the inner surface of the annulus the heat transfer coefficient at the inner surface of the annulus, is

. . . [Eqn. 7.1 O8)]

/&ample 10.18. R e velocity of water flowing through a tube of 22 mm diameter is 2 m/sS Steam condensing at 150°C on the outside su$ace of the tube heats the warerfrom 15°C to 60°C over the length of the tube. Neglecting the tube and steam sidefilm resistance, calculate the following:

( i ) The heat transfer coeficient, and (ii) The length of the tube.

Take the following properties of water at mean temperature :

p = 990 kg/m3; c,, = 4.2 kJ/kg°C, k = 0.5418 W/m°C; p = 700 x 106 kg/ms. (Pm Solution. Given: d = 22 mm = 0.022 m; V = 2 d s , thl = th2 = 150°C; tcl = 15OC; tC2 = 60.C.

Substituting the values in eqn. (i) , we get 1

. . I u=-- 0.0399

- 25.06 w/m2'c

Also, Q = hh Cph ( f h l - fh2) = UA 8, = u x (7[: di L) x 8,

0.12 x 2131 x (95-65) = 25.06 x 7[: x 0.02 x L x 44.2 ( i ) The heat transfer coefficient, h : L =

0.12 X 2131 (95 -65) = 110.23 m (Ans.) 25.06 x ~r: x 0.02 x 44.2


steam r d = 2 2 m m

= 62228 1 -+ Water -+ -+ 1

Hot fluid (steam)

(ii) The length o f the tube, L: I I

The mass flow rate (water), Fig. 10.26

Heat gained by water passing through the tube,

Q = mc X Cpc X (tcz-tc1)

= 0.753 x (4.2 x lo3) x (60 - 15) = 142317 W Q is also given by,

Q = U A en,

Substituting the values in eqn. (i) , we get

142317 = 6771.7 x ( X X 0.022 x L) x 1 1 1 or L = 2.74 m (Ans.)

.... ( i )

(Here, U = h)

Example 10.19. Saturated steam at 100°C is condensing on the shell side of a shell-and-tube heat exchanger. The cooling water enters the tube at 30°C and leaves at 70°C. Calculate the mean temperature d~fference if arrangement is ( i ) parallel flow, (ii) counter flow.

(AMIE, Winter, 1997) Solution. The mean temperature difference will be the same for parallel flow or counter

flow arrangements as is evident from the diagrams (Fig. 10.27) of temperature variations in both the arrangements. The mean temperature difference may be logarithmic mean (LMTD), or arithmetic mean temperature difference (AMTD).

( i ) LMTD = (100 - 30) - (100 - 70) =-- 40 - 4 7 f l o c (Ans.) (100 - 30)

$00-7011 ln($)

(100 - 30) + (100 - 70) 70 + 30 - 500C (ii) AMTD = ---

2 - 2

(i) Parallel-flow (ii) Counter-flow

Fig. 10.27.

Example 10.20. The amount of F12 used in compression refrigeration system is 4 tonnesntour.

The brine, flowing at 850 kg/min. with inlet temperature of 12°C is cooled in the evaporator. Assuming F12 entering and leaving the evaporator as saturated liquid and saturated vapour respectively, determine the area of evaporator required. Take the following properties :

For FI2 : Saturation temperature : - 23°C; cp = 1.1 7 kJ/kdC; hfg = 167.4 kJ/kg

c, (brine) = 6.3 kJ/kg°C; U = 8368 kJ/m2hoc. (AMIE Summer, 2000)

Solution. Making energy balance, we have

:. Exit temperature of brine, th2 = 12 - 11160 ,9.90c 850 x 6.3

- 23°C

01 - 92 - (thl - tcl) - (th2 - tcd LO^ mean temperature, 9, =

In (e1 /e2) - I th l - In -

I 41 = 4 2 I .

w 1

Cold fluid (F,,) I I I

Fig. 10.28.


Now, Q = UAem (where A = area o f evaporator)

A = 60 = 2.357 I$ (Ans.) 8368 x 33.94

500 hfg(organic vapour) = 600 kl/kg, m, = mh = - 60 - - 8.33 kg/s; m, = rit, = 60 kg/s.

-:' Example 10.21. A heat exchanger is to be designed to condense an organic vapour at a rate of 500 kg/min which is available at its saturation temperature 355 K. Cooling water at 286 K is available at a flow rate of 60 kg/s.

(i) The number of tubes required, N :

The overall heat transfer coeficient is 475 w/m2'c. Latent heat of condensation of the organic vapour is 600 W . g . Cal-

Heat lost by vapour = heat gained by water

fhl = th2 = tmt. = 82°C

- Hot fluid (vapour)


culate: tc2 (= ?I (i) The number of tubes required,

if25 mm outer diameter, 2mm thick and 4.87 m long tubes are available, and

(ii) The number of tube passes, if 'c'

the cooling water velocity (tube (= 130C) side) should not exceed 2 m.4.

(M.U.) Fig. 10.29

Solution. Given : do = 25 mm = 0.025 m; di = 25 - 2 x 2 = 21 mm = 0.021 m; L = 4.87 m; v = 2 tcl = 286 - 273 = 13OC, thl = th2 = to , = 355 - 273 = 82OC; v = 475 w/mOc,

Heat transfer rate is given by,

Q = mhxhfg = v A e m = v(7 tdoLN) x 8,

8.33 x 600 x lo3 = 475 x ( 7t x 0.025 x 4.87 x N) x 58.5 . . N = 470 tubes (Ans.) (ii) The number of tube passes, v : . - The cold water flow mass passing through each pass (assume p are number of passes) is

given by,

eat Exchangers

where

n kc = ( - d ? x V X p) x Np 4

Np = number o f tubes in each pass (N = p x Np)

n 60 = 7 x (0.021)~ x 2 x 1000 x Np

N 470 Number of passes, p = - = - = 4.91 = 5 (Ans.)

N, 95.5

Example 10.22. (a) A steam condenser consists of 3000 brass tubes of 20 mm diameter. Cooling water enters the tubes at 20°C with a mean pow rate of 3000 kgh. The heat transfer coescient on the inner sugace is 11270 w/m2'c, and that for condensation on the outer sutjize is 15500 w/m2 OC. The steam condenses at 50°C, and the condenser load is 230 MW. The latent heat of steam is 2380 kJ/kg. Assuming counter flow arrangement, calculate the tube length per pass if two tube passes are used.

(b) Explain why in steam condensers the LMTD is independent of flow arrangement ? (AMIE Summer, 1999)

Solution. (a) Given : Np = 3000 per pass; d = 20 rnm = 0.02 m; tcl = 20°C; mc = 3000 kgls;

h, = 11270 w / m 2 OC; ho = 15500 w / m 2 "C, thl = th2 = 50°C; condenser load = 230 MW (or

230 x lo3 kW); hfg = 2380 kJ/kg; number o f passes = 2.

Tube length per pass, L : h F t,,, = th2 = 50°C . Assuming the tubes to be thin, the Steam

overall heat transfer coefficient : 1 ,yo=-- 1

1 1 - 1 -+- - 1 +- hi h, 11270 15500

= 6525.4 w/m2'c

Heat exchanger load t,, (= 20°C)

= mc cpc (tc2 - tcl) Fig. .10.30

i.e., 230 x lo3 = 3000 x 4.187 (tc2 - 20)

:. Water outlet temperature, tc2 = 38.31°C

Log-mean temperature difference,

In 150 - 38.311

Now, Q = VJ9m

230 x lo6 = 6525.4 x (n dL) x (2N,,) x 19.43


L= 230x lo6 6525.4 x n x 0.02 x (2 x 3000) x 19.43 = 4.812 m (Ans.)

(b) In steam condensers, the temperature o f hot fluid (steam) is the same at inlet and exit. Hence the terminal temperature difference would not depend upon the arrangement. The effectiveness of heat exchanger for all the arrangements in this case (the heat capacity ratio for condensation being zero) would be

E = 1 - e-NTU

The temperature variations for parallel-flow and counter-flow are as shown in Fig. 10.31 and Fig. 10.32 respectively.

Steam . r

I I

Fig. 10.31. Parallel-flow. Fig. 10.32. Counter-flow.

Example 10.23. A two-pass sudace condenser is required to handle the exhaust from a turbine developing 15 MW with specijic steam consumption of 5 kg/kWh. The condenser vacuum is 660 mm of Hg when the barometer reads 760 mm of Hg. The mean velocity of water is 3 d s , water inlet temperature is 24°C. The condensate is saturated water and outlet temperature of cooling water is 4OC less than the condensate temperature. The quality of exhaust steam is 0.9 dry. The overall heat transfer coeflcient based on outer area of tubes is 4000 w/m2"c. The water tubes are 38.4 mm in outer diameter and 29.6 mm in inner diameter. Calculate the following:

( i ) Mass of cooling water circulated in kg/min,

(ii) Condenser sur$ace area,

Steam in (x = 0.9)

Tho-pass surface

Water \

1 Condensate out

760-660 x 1.0133 = 0.133 bar Ps = 760

The properties of steam at p, = 0.133 bar, from steam table, are: tsar = 51°C; hh = 2592 kJkg

. . tc2 = 51 - 4 = 47OC

The steam condensed per minute,

( i ) Mass of cooling water circulated per minute, m, (= m,): Heat lost by steam = heat gained by water

mh x ( X . hfg) = mc X cpc X (tc2 - tC,)

1250 x (0.9 x 2592) = m, x 4.187 (47 - 24)

. . mc (= m,) = 30280 kglmin (Ans.)

(ii) Condenser surface area, A : ms x (x . hfR)

'= 60 = UA9, . . .(i)

where

Substituting the values in eqn. (i), we get

- 1250 60 x (0.9 x 2592 x lo3) = 4000 x A x 12.04

or A = 1009.1 m2 (Ans.) (iii) Number of tubes required per pass, N, :

30280 x 4 Np =

= 244.46 say 245 (Ans.) 60 x n x (0.0296)~ x 3 x 1000

( Total number o f tubes required, N = 2Np = 2 x 245 = 490) (iv) Tube length, L:

A = (n do L) x (2Np)

I 1009.1 = n x 0.0384 x L x (2 x 245)

(iii) Number of tubes required per Fig. 10.33. A two-pass surface condenser. pass, and

(iv) Tube length. (P.U Solution. Given : di = 29.6 mm = 0.0296 m; do = 38.4 mm = 0.0384 m; U = 4000 w/m2"c;

V = 3 mls; tc, = 24°C; x (dryness fraction) = 0.9. The pressure o f the steam in the condenser,

L = 1009.1 =. 17.1 m (Ans.)

n x 0.0384 x 2 x 245

Example 10.24. A feed water heater which supplies hot water to a boiler comprisr 8 ahd and tube heat exchanger with one-shell pass and two-tube passes. One hundred t h i n - ~ l l d dCW each of 20 mm diameter and length of 2m per pass are used. Under normal operating water enters the tubes at 10 kg/s and 17°C and is heated by condensing saturated s- 1 l a on the outer sudace of the tubes. The convection coeflcient of the saturated steam is 10 k

%8 Heat and Mass Transfer

Determine the water exit temperature. Use the following properties of water: c,, = 4.18 kJ/kg°C; p = 0.596 x lod3 N S / ~ ~ . k = 0.635 W/m°C and Pr = 3.93. (M.U.) Solution. Given : p (number of Steam in

tube passes) = 2, N (total number of tubes) = 200, d = 20 mm = 0.02 m; L (length per pass) = 2m, m, = riz, = 10 kg/s, tcl = 17OC;

Water Water exit temperature, tc2 :

. n m c = - d 2 x v x p x ~ ,

4 [where V = velocity of water; N, = in

N number of tubes per pass = - = P

n Condensate out or 10 = - ~ 0 . 0 2 ~ x v x 10OOx 100

4 thl = th2 = fsat = loo"c 10x4 2 :. v = Hot fluid (steam)

n x 0.02~ x 1000 x 100 82 T tc2 = ?

= 0.318 m/s Using non-dimensional heat

transfer equation to water side, we get t,, = 17°C

-. hi d

Nlc = - = 0.023 (Re)'.' ( ~ r ) ' , ~ ~ -

k Fig. 10.34. One-shell pass and two-tube passes condenser.

Substituting the values in eqn. (i), we get

h, = - 0'635 x 0.023 (10671)'~~3.93)'~~~ = 1915 w/m2'c 0.02

The overall heat transfer coefficient is given by the relation,

Further, 91=t l , l - t c l=100-17=830C 1 e2 = tIl2-tc2 = 100 - tc2 4

:. Arithmatic mean temperature difference,

Heat Exchangers

9, + 02 - 83 + (100 - tc2) AMTD = - - 2 = 91.5 2

The heat transfer rate is given by Q = m, c, (t, - tcl) = U A, (AMTD) = U x (n d L x N) (AMTD)

(where A, = surface area of all the tubes in both passes)

or 10 x (4.18 x lo3) (t, - 17) = 1607.7 x ( n x 0.02 x 2 x 200) x (91.5 - 0.5 t,)

41800 (tc2 - 17) = 40406 (91.5 - 0.5 t,)

or 40406 (91.5 - 0.5 tc2 ) = 0.966 (91.5 - 0.5 t,) tc2 - 17 = - 41800 = 88.39 - 0.483 tc2

or t,, = 71°C (Ans.)

Example 10.25. A one ton window airconditioner removes 3.5 W/s from a room and in the process rejects 4.2 U/s in the air cooled condenser. The ambient temperature is 30°C whereas condensing temperature of the refrigerant is 4PC. For the condenser the product of overall heat transfer coe8cient and corresponding area is 350 W/K. Calculate the temperature rise of the air as it jlows over the condenser tubes. (M.U., Winter, 1998)

Solution. Given : tcl = 30°C; thl = th2 = 45°C; UA = 350 W/K. Temperature rise of air, tc2 :

The arrangement and temperature distribution are shown in Fig. 10.35.

Condenser refrigerant

4.2 kJ/s in

3.5 kJ/s

(a) Arrangement (b) Temperature distribution

Fig. 10.35.

Heat lost by the refrigerant in condenser = heat trankferred to air

4.2 x lo00 = UA9, = UA 91 - 92

In (01/02)


Substituting the values in eqn. (2), we get

Using trial and error method, we get, tn = 35°C

:. Temperature rise of air = 35 - 30 = S°C (Ans.) Example 10.26. A rectangular tube, 30 mm x 50 mm, carries water at a rate of 2 kg/s.

Determine the length required to heat water from 30°C to 50°C i f the wall temperature is Now, putting the values - in eqn. (I), we have Q = UA8, = hA8, (where A = surface area)

166960 = 5082.4 x [2(0.03 + 0.05) x L] x 49.3 or L = 4.16 m (Ans.) Example 10.27. Carbon dioxide (C02) for a gas cooled reactor is used to generate steam.

A flow rate of 90000 kgh of C02 at 4 bar enters the tubes of a shell and tube type steam generator at 500°C. The C02 leaves the generator at 330°C and steam saturation temperature is 250°C. Assume that the steam formed is dry and satwated. Using 25 mm inner diameter (I.D.) copper tubes, 2 mm wall thickness and designed for C02 mass flow rate of 350000 kg/m2h, calculate the length and number of tubes to be used neglecting steam side thermal resistance.

Take the following properties of C02:

c,, = 1.1 72 kJ/kg°C; p = 0.0000298 Ns/m2: k = 0.043 W/m°C; p = 3.26 kg/m3. Take k (for copper) = 384 W/m°C. (P.U.)

Water in (cold fluid)

maintained at 90°C.

Use the following properties of water at 40°C:

p = 992.2 kg/m3; k = 0.634 W/m°C; C, = 4.174 kJ/kg°C; /.k = 6.531 X Ns/m2.

(P.U.) Solution: Given: h, = rit, = 2 kgts; 61 = 3O0c

tcl = 30°C; tc2 = 50°C; thl = th2 = 90°C. Length required, L: I

Fig. 10.36. Heat gained by cold fluid (water),

Q = m c x C,c X (t,2 - t,l) = 2 x (4.174 x lo3) x (50 - 30) = 166960 W

The heat transfer Q is also given by

Q = UAe,

where 8, = 01 - 92 - (90 - 30) - (90 - 50) In (8,/82) - In [ ( g o - 30)/(90 - 50)]

C02 (hot fluid) C02 out

in (t,, = 500 '~ ) (th2 = 330 '~ ) Steam out (tc2 = 250°C)

The heat transfer coefficient 7; (=U is this case) is given by

4Ac 4 X (0003 X 0.05) - 0.0375 where de (equivalent diameter) = - - p 2 (0.03 + 0.05)

(A, = cross-sectional area, and P = perimeter)

P VLC Reynolds number, Re = - , where V is the velocity of water flowing through the tube. CL

m, (= m,) = A,. V. p

2 = (0.03 x 0.05) x V x 992.2 n

Cold fluid (steam)

Fig. 10.37. Gas cooled reactor.

Solution: Given: mCC2 = 90000 25 kgls; t,, = 500°C; t, = 330°C, t,, = rc2= 250°C m h = 3 6 0 0 =

di = 25 mm = 0.025 m; do = 25 + 2 x 2 = 29 mm = 0.029 m.


Number of tubes (N) and length of each tube (L): Considering the flow of C02 through tubes : Mass flow, m = A.V.p (where V is the velocity of CO, in the tubes).

Again, G = ~ v = - 350000 (given) 3600 350000 1 v=- x - = 29.82 d s (:. p = 3.26 kg/m3 ...g iven) 3600 3.26 m . . G = - (mass flow per unit area per unit time) A

The mass flow rate of CO, through the tubes is given by

P Vd G.d - Reynolds number, Re = - - - P P

(where G is in kg/m2.s and p in ~ s l m ~ )

(where N, number of tubes)

or N = 523.89 say 524 (Am.) Now Ai = (x diL) x N

106.06 = x x 0.025 x L x 524 I Prandtl number P r = A = PC 0.0000298 x (1.172 x 1000) k 0.043 = 0.812

L = 106'06 2.58 m (Am.) x x 0.025 x 524 Nusselt number, N U = - - hi di - 0.023 (Re)'.' (PI)'."

k Example 10.28. The tubes (k = 106 I4'/m0C) of a single pass condenser are of 30 mm

outside diameter and 25 mm inside diameter. The condenser is required to handle 20000 kgh of dry and saturated steam at 50°C. The inlet and outlet temperatures of water are 15°C and 25°C respectively. If the average velocity of water in each tube is 2.5 m/s and steam side plm heat transfer coeficient is 5150 w/m2"c, calculate the outside tube area.

The properties of water at mean temperature are given as under:

p = 998.2 kg/m3, c,, = 4.182 W/kg K: p = 1004.5 x lod ~s /m ' ; v = 1.006 x 1od m2/s; and k = 0.598 W/m°C.

The latent heat (hh) at 50°C = 2374 W/kg. For turbulent flow inside tubes: Nu = 0.023 Rea8 ~ 9 . ~ . Solution. Given : k = 106 W/m°C; do = 30 mm = 0.03 m; di = 25 mm = 0.025 m; m, =

20000 kglh; h, = 5150 w / ~ ~ " c ; V = 2.5 d s . Outside tube area, A:

= 3 13.76 w/m2 "C .

The temperature distribution during the flow is shown in Fig. 10.37. It is assumed that the water enters into the boiler at saturated condition and comes out as saturated steam, so the temperature during the generation of steam remains constant at 250°C.

The logarithmic mean temperature difference (LMTD) is given by

V di 2.5 X 0.025 = 62127 Reynolds number, R e = - - -

4 v 1.006 x lom6

Further, Q = m, x cpl, x - t,,2)

= 25 x (1.172 x lo3) (500 - 330) = 4.981 x lo6 W Negletcting the steam side thermal resistance, we can write Prandlt number,

1 1 - - ' i +- u; - % kc,,,, In (ro 1 r, ) ...[ Eqn. (10.27)] hw di

k - 0.023 Re0.' = 0.023 x (62127)O.~ x (7.02)'.~ = 282 Nusselt number, Nu = - -

or u. = = 314.76 w/m2 "C 0.003 177

Also, Q = U; A, 8, :. 4.981 x lo6 = 314.76 x A; x 149.2

The overall heat transfer coefficient is given by

where h, = heat transfer coefficient on the water side, and

h, = heat transfer coefficient on the steam side. Substituting the values, we get


AMTD (arithmatic mean temperature difference) = (50 - 15) + (50 - 25) = 300c

Heat transfer rate is given by,

Q = UA (AMTD) m, hh = UA (AMTD)

Solution. Given: N = 100; di = 25 mm = 0.025 m: do = 29 rnm = 0.029 m; m, = m, = 500 kglmin; I,., = 30°C; tc2 = 70°C; h, (steam side) = 5000 w /~ 'oc ; thl = th2 = IOOOC. R~ = 0.0002 ~ * " c / w .

Overall heat transfer coefficient (U,); tube length (L): Arrangement o f the heat exchanger is shown in Fig. 10.38 Heat gained by water,

Q = mc X Cpc X (tc2 - tCl)

Example 10.29. A single pass shell and tube heat exchanger, consisting of a bundle of 100 tubes (inner diameter = 25 mm and outer diameter = 29 mm) is used for heating 500 kg/min of water from 30°C to 70°C with the help of steam condensing at atmospheric pressure on the shell rhl = rh2 = 100°C

The Q transferred from the steam to the water is given by

Q = u, A,, 0, ...( 1 ) where U, = overall heat transfer coefficient based on inner surface o f the tubes,

A,, = inner surface area o f the tubes.

0, = logarithmic mean temperature difference (LMTD)

side. Calculate the overall heat transfer coeflcient based on the inner area and length of the tube

Heat Exchangers

Ui is given by

bundle if the condensing side heat transfer coefl- ( tcz)

cient is 5000 w /m2"~ . Take the fouling factor on the water side to be 0.0002 m2"c7W per tube. Neglect the effect of fouling on the shell side and thermal resistance of the tube wall.

30°C Take the following properties of water at the (tcl)

mean temperature of 50°C: Area / length -+ p = 988.1 kg/m3; cp = 4174 J/kg°C; k =

0.6474 W/m°C; p = 550 x I@ kg/ms; v = 0.555 Fig. 10.38. x I@ m2/s, Pr = 3.54.

Hot fluid (steam)

.(2) [Eqn. (10.27)l

70°C

where hi, ho= inside and outside heat transfer coefficients respectively, and Rfi = inside thermal resistance due to fouling.

In order to find out hi we shall use the following relation: h. d.

Nu = k = 0.023 ( ~ r ) ' . ~ ~

v di where Re = -, v

V being the average velocity of the water flow which can be calculated by using the following equation.

Substituting the values in eqn. (3), we get

hi = - x 0.023 (77481O.~ x (3.54)0.33 = 1168 wlm2"C 0.025

Further, inserting the values in eqn. (2), we get

= 0.000856 + 0.0002 + 0.000172 = 0.001228

. . Ui = 814.3 ~ 1 r n ~ " C

Now, substituting all the values in eqn. ( I ) , we get Q = U i x ( N x ~ x d i x L ) x O ,

1.39 x lo6 = 814.3 x (100 x n; x 0.025 x L) x 47.2 1.39 x lo6

L = = 4.6 m (Ans.) 814.3 x 100 x n x 0.025 x 47.2

Example 10.30. A multipass heat exchanger has two passes on shell side and four passes on the tube side. The oil is passed through the tubes and cooled from 135°C to 52°C. The cooling water passing through shells enters at 13°C and leaves at 31°C. Calculate the heat tranSfcr mfe using the following data :


h, (oil) = 270 w/m2 "C, ho (water) = 965 w/& "C

h (scale on water side) = 2840 w/m2 "C Number of tubes per pass = 120 Length and outer diameter of each tube are 2 m and 2.54 cm respectively. Thickness of the tube = 1.65 mm LMTD correction factor = 0.98 Neglect the tube wall resistance. (A.U. Winter, 1998) Solution. Given : No. of passes : Two on shell side and four on tube side.

tcl = 13°C; tc2 = 31°C; thl = 135°C; thz = 52°C; hi (oil) = 270 w/rn2"c;

h, (water) = 965 w/m2"c; h (scale on water side) = 2840 w/m2 "C, Np = 120;

do = 2.54cm = 0.0254 m; L = 2 m; di = 0.0254 - 2 x 0.00165= 0.0221 m;

LMTD correction factor, F = 0.98

Water in

Water out

Oil in :

Oil out 4

(a) Arrangement

h v

1 L T

Rate of heat transfer, Q :


Fig. 10.39.

66.3 x F = 66.3 x 0.98 = 65°C Theoverall heat transfer coefficient,

Substituting the values, we have r,, do) I

:. Rate of heat transfer, Q = U J , (0m)actual = 175.4 x [nD& x (N, X 4)l X 65

= 175.4 x [n x 0.0254 X 2 X (120 X 4)l X 65

= 873369 W or 873.369 kW (Ans.)

10.6. CORRECTION FACTORS FOR MULTI-PASS ARRANGEMENTS -

The expression Om = (01 - 02) for LMTD is essentially valid for single-pass heat exchangers. 13 (Ol/O2)

The analytical treatment of multiple pass shell and tube heat exchangers and cross-flow heat exchangers is much more difficult than single pass cases; such cases may be analysed by using the following equation:

Q = UAF Om ...( 10.31)

where F is the factor; the correction factors have been published in the form of charts

by Bonman, Mueller and Nagle and by TEMA. Correction factors for several common arrangements have been given in Figs. 10.40 to 10.43.

The data is presented as a function of two non-dimensional variables namely the temperature - - ratio P and the capacity ratio R.

Temperature ratio, P: It is defined as the ratio of the rise in temperature of the cold fluid to the difference in the inlet temperatures of the two fluids. Thus:

4 2 - tCl P - - ...( 10.32) * -

thl - fel

where subscripts h and c denote the hot and cold fluids respectively, and the subscripts 1 and 2 refer to the inlet and outlet conditions respectively.

The temperature ratio P indicates cooling or heating effectiveness and it can vary from zero for a constant temperature of one of the fluids to unity for the case when inlet temperatun of the hot fluid equals the outlet temperature of the cold fluid.

Capacity ratio R: The ratio of the products of the mass flow rate times the heat capacity of the fluids is termed as capacity ratio R. Thus:

k . Cnc - ... ( 10.33)


Fig. 10.40. Correction factor plot for heat exchanger with one shell pass and two, four or any multiple of tube passes.

Fig. 10.41. Correction factor plot for heat exchanger with two shell passes and four, eight or any multiple of tube passes.

Fig.

0

tc2 - tc1 p=- 'hl - lcl

10.42. Correction factor plot for single cmss-flow heat exchanger with both fluids

Heat Exchangers

temperature drop of the hot fluid = ltemoeratursise in the cold fluid I L .

For thew correction factor plots it is immaterial whether the hot fluid flows in the shell or the tubes. The value of the correction factor indicates the p e d o m n c e level of a given arrangement for the given terminal fluid temperatures. The correction factor F is always less than univ as no arrangement can be more eflective than the conventional counter flow.

Example 10.31. Calculate for the following cases, the surfoce area required for a heat exchanger which is required to cool 3200 kgh of benzene (c, = 1.74 W/kg°C) from 72°C to 42°C The cooling water (c, = 4.18 kUXg°C) at 15°C has a flow rate of 2200 kgh.

( i ) Single pass counter-flow, (ii) 1-4 exchanger (one-shell pass and four-tube passes), and

(iii) Cross flow single pass with water mixed and benzene unmixed.

For each configuration, the overall heat transfer coeficient may be taken as 0.28 k~/m'"C.

Sohtion Given: hh = 32W = 0.889 kgls; cph = 1.74 Wkg°C; thl = 72"C, th2 = 42°C; 3600

Surface area required, A: Using energy balance on both the fluids, we have


' h 'ph (lhl - th2) = cpc (t& - tcl) 0.889 X 1.74 (72 - 42) = 0.61 1 X 4.18 (tc2 - 15)

. . tc2 = 33.2OC

An energy balance on the hot fluid yields the total heat transfer,

Q = mh cph (thl - th2) = 0.889 X 1.74 (72 - 42) = 46.4 kW (i) Single-pass counter-flow:

Q :. Area of the exchanger, A = - = 46'4 = 5.1 m2 (Ans.) U 8, 0.28 x 32.5 (ii) 1-4 exchanger:

Since the number of passes is more than one hence 8, ( W D ) needs correction factor, F. To know the constion factor we have to know fvsl P (temperature ratio) and (capacity ratio) R.

Fig. 10.43. Comclion factors plot for single-pass cross-flow heat exchanger, one fluid mixed and the other unmixed.

Heat Exchangers 3 .

Using P = 0.32 and R = 1 .65 the correction factor F from Fig. 10.40 is read as

e= 46.4 :. Area of the exchanger, A = = 5.66 m2 F U 8, 0.9 x 0.28 x 32.5

(iii) Cross-flow single-pass with water mixed and benzene unmixed: Using P = 0.32 and R = 1.65 the cometion factor F from Fig. 10.43 is read as

F 1 0.92

e= 46.4 :. Area of the exchanger, A = F U 8, 0.92 x 0.28 x 32.5 = 5.54 m2 (Ans.)

Example 10.32. it is required to design a shell-and-tube heat exchanger for heating 2.4 kg/s of water from 20°C to 9PC by hot engine oil (cp = 2.4 W/'kg°C) flowing through the shell of the heat exchanger. The oil makes a single pass entering at 145'C and leaving 9PC with

an average heat transfer coeflcient of 380 w/m2'c. The waterflows through 12 thin-walled tubes of 25 mm diameter with each-tube making 8-passes through the shell. The heat transfer coefiient on the water side is 2900 w/m2"c. Calculate the length of the tube required for the heat exchanger to accomplish the required water heating.

Solution. Given: m, = mc = 2.4 kgls, tcl = 20°C, tc2 = 90°C; thl = 145OC. th2 = 90°C,

cPh = 2.4 kJ/kg°C, d = 25mm = 0.025 m, N = 12; hi = 2900 Wlm°C, h, = 380 w1m2OC

Length of the tube L: The overall heat transfer coefficient, neglecting thermal resistance of the tube, is given by

The parameters required to get the correction factors are:

Form Fig. 10.40, F 1 0.82

For the conventional counter-flow arrangement; el = thl - tc2 = 145 - 90 = 55OC

8, = th2 - tcl = 90 - 20 = 70°C

The heat transfer rate is given by Q = ' , Cpc (tc2 - tcl)

= 2.4 x 4.18 x lo3 (90 - 20) = 702240 W

Heat and Mass Transfer Heal Exchangers

Also, Q = F U A Om, where A = heating surface as such it is inconvenient to combine them in a graphical or tabular form. However, by compiling a non-dimensional grouping, E can be expressed as a function of three non-dimensional parameters. This method is known as NTU method. This methodlapproach facilitates the comparison between the various types of heat exchangers which may be used for a particular application. The effectiveness expresssions for the parallel flow and counter-flow cases can be derived as follows:

But,

(i) Effectiveness for the Parallel flow heat exchanger : Refer Fig. 10.8. The heat exchange dQ through an area M of the heat exchanger is given by

dQ = U.dA (th - tc) . . . ( i )

= - m.cph. dth = m,.cpc.dtc

= - C,.dth = C,.dtc ...( ii)

The shell length = - 26'1 - - 3.26 m (Ans.) 8

10.7. HEAT EXCHANGER EFFECTIVENESS AND NUMBER OF TRANSFER UNITS (NTU)

From expression (ii), we have A heat exchanger can be designed by the LMTD (logarithmic mean temperature difference) when inlet and outlet conditions are specified. However, when the problem is to determine the inlet or exit temperatures for a particular heat exchanger, the analysis is performed more easily, by using a method based on effectiveness of the heat exchanger (concept first proposed by Nusselt) and number of transfer units (NTU).

-dQ and dth = - dQ dt, = - ch cc

The heat exchanger effectiveness (E) is defined as the ratio of actual heat transfer to the maximum possible heat transfer. Thus Substituting the value of dQ from expression ( i ) and rearranging, we get

( f h - = - u.dA [i + $1 (th - tc)

E = actual heat transfer -2 maximum possible heat transfer - Q,,,

Thei actual heat transfer rate Q can be determined by writing an energy balance over either side of tye heat exchanger.

Upon integration, we get

Q = mh Cp,l (thl - th2) = mc Cpc (tc2 - tcl) ...( 10.36) The product of mass flow rate and the specific heat, as a matter of convenience, is defined

as the fluid capacity rate C: mh cph = Ch = hot fluid capacity rate

m, cpc = C, = cold fluid capcity rate

,Cm, = the minimum fluid capacity rate (Ch or Cc) C,, = the maximum fluid capacity rate (Ch or Cc).

The maximum rate of heat transfer for parallel flow or counter-flow heat exchangers would occur if the outlet temperature of the fluid with smaller value of Ch or Cc Le., Ch were to be equal to the inlet temperature of the otherfluid. The maximum possible temperature change can be achieved by only one ofjuids, depending upon their heat capacity rates. This maximum change cannot be obtained by both the fluids except in the very special case of equal heat capacity rates. Thus :

From eqn. (10.38), we have the expressions for effectiveness

E = Ch (thl - th2) - Cc (tc2 - t c l ) - (thl ' t c l ) emin (thl '

Hence,

Eliminating th2 and tc2 from eqn. (10.40) with the help of eqns. (10.41) and (10.42), we get

.Once the effectiveness is known, the heat transfer rate can be very easily calculated by using th<equation \ /

1 - exp [ - (UA / Ch) {l + (Ch / c~)}] or E = ...( 10.43)

cmin (i >+) If Cc > Ch then Cmin = Ch and C,, = Cc, hence eqn. (10.43) becomes

Q = E Cm,, (t,, 1 - f, 1 ) ...( 10.39) Number of transfer units (NTU) method :

It is obvious from Eqn. (10.38) that effectiveness & is a function of several variables and

584 Heat and Mqss Transfer Heat Exchangers

If Cc < Ch thed Cm, = Cc and CM, = Ch, hence eqn. (10.43) becomes ...( iii)

...( iv)

1 - exp [-(UAIC-)I1 + (Cm,/Cm,)}]' " E = ...( 10.45)

1 + (Cmin 1 Cmm) By rearranging eqns. (10.44) and (10.45), we get a common equation

E = 1 - exP 1- (UA I CmJ (1 + <Cmin/ Cmax)]I

1 + (Cminl Cmm) where Cmin and C,, represent the smaller and larger of the two heat capacities Cc and Ch.

The grouping of the terms (UA)ICmin is a dimensionless expression called the number of transfer units NTU; NTU is a measure of qfectiveness of the heat exchanger. Cmhl C,, isi the second dimensionless parameter and is called the capacity ratio R. The last dimensionless parameter is theflow arrangement, i.e., parallel flow, counter-flow, cross-flow and so on.

Substituting these values in e4n. (10.471, we get,

[lhl - Cmb ( fh l - tc l)]

ch - tcl , = exp [(UAIC,) I 1 - (Cc I Ch) 11 E Cmin (thl - tcl)]

thl - k c l + cc

Thus the effectiveness of a parallel flow heat exchanger is given by

1 - exp [- NTU (1 + (Cm, / cmAt] E = ...( 10.46)

1 + (Cm, CmA E . Cmin 1 -- or

Ch = exp [(UNC,) I 1 - (Cc I Ch) 11 ...( 10.48) E . Cmin 1 - 7

E = 1 - exp [- NTU(1 + R)]

l + R .., l10.46 (a)]

(ii) Counter-flow heat exchanger : Refer Fig. 10.9. The heat exchange dQ through an area dA of the heat exchanger is given by

dQ = U.dA (th - tc) ...( i ) = - m cPh dth = - m cCpc dtc = - Ch dth = - Cc dt, . . . (ii)

From expression (ii), we have

LC

Assume Cc < Ch, Cc = Cmin and C,, = C,,. Substituting these values is eqn. (10.48), we get,

E - Cmin 1 ---;;---

dth = - - dQ and dQ Ch

dtc = - - c c

E . Cmin 1 -, . . d(th - tc) = - dQ [l - l] = dQ [l - -y

ch cc cc ch

Substituting the value of dQ from expression (i), we get,

Upon integration, we get

- - exP [(UA /Cmin) ( 1 - (Cmin /cm,)}I Cmm

- - ~ X P [(UA /Cmin) {l - (Cmin /CrnaJ\I - 1 Cmin

exP [(UA /Cmin) ( 1 - (cmin /cma>}I - - Cm,

1 - exp [(- UA /Cmi,J { I - (Cmin /c,,))] or E =

From Eqn. (10.38), we have the expressions for effectiveness


Since Cmin / C,, = R and UA/C,, = NTU, therefore,

1 - exp [ -NTU(1 - R)] E =

1 - R exp [- NTU(1 - R)]

Heat Exchangers

We find that effectiveness of parallel flow and counter-flow heat exchangers is given by the following expressions:

- 1 - exp [- NTU(1 + R)] (~)pru/lei flow - ...( 1 ) 1 + R

- 1 - exp [- NTU ( 1 - R)] (~)counter~ow - . . . (2)

1 - R exp [ -NTU(1 - R)],

here R = (C ,,,,, 1 C,,) Let us discuss two limiting cases of eqns. ( 1 ) and (2) Case I: When R - 0 ... Condensers and evaporators (boilers)

By using the above case, we arrive at the following common expression for parallel flow as well as counter-flow heat exchangers

E = 1 - exp (- NTU) ...( 10.51) Such cases are found is condensers and evaporators in which onejuid remains at constant

temperature throughout the exchanger. Here C,,,, = and thus R = - 0 . (29 Obviously, no matter how large the exchanger is or how large the overall transfer coefficient

is the maximum effectiveness for parallel flow heat exchanger is 50%. For counter-flow, this limit is 100%. For this reason, a counter flow is usually more advantageous for a gas turbine heat exchangers.

Case I1 : When R = 1 . . . Typical regenerators (i) In case of parallelflow heat exchanger using R = 1 , we get

1 - exp (- 2NTU) E = 2

(ii) In case of counter-jlow heat exchanger using R = 1 we get an expression for effectiveness which is indeterminate. We can find the value of E by applying L, Hospital's rule :

1 - exp [- NTU ( 1 - R)] lim

R . + I 1 - Rexp[ -NTU(1 - R)]

lim exp[NTU( l - R)] - 1

-+ 1 exp [NTU ( 1 - R)] - R Differrentiating the numerator and the denomenator with respect to R and taking the limit,

we get,

lim exp [NTU(l - R) (- NTU)] - - NTW

...( 10.53) R+ 1 exp [NTU(l - R)] (- NTU) - 1 1 + NTU

The NTU is a measure of the heat transfer size of the exchanger; the larger the value of NTU, the closer the heat exchanger approaches its thermodynamic limit.

The effectiveness of various types of heat exchangers in the form of graphs (prepared by

Kays and London) for values of R (=%I and NTU are shown in Fig. 10.44 to 10.49. Cmu

10.8. PRESSURE AND PUMPING POWER An important consideration in heat exchanger design, besides the heat transfer requirements,

-ycr Number of transfer units, NTU,,- Cmin

Fig. 10.45. Effectiveness for counter-flow heat exchanger.

Heat transfer Hot fluid , 3 surface

Y/7////711///7/////1////7/////7n - Cold f lu~d

UA Number of transfer units, NTU,, = -

Cmin

Fig. 10.44. Effectiveness for parallel flow heat exchanger.

Heat transfer Hot fluid ,I> r surface -

Cold fluid


Shell fluid 1 4 1 /-Tube fluid

I

One shell One Shell Pass

UA Number of transfer units, NTU-=- Cmin

Fig. 10.46. Effectiveness for 1-2 parallel counter-flow heat exchanger.

Shell fluid

2 Shells

n o shell passes 4,8, 12 etc, Tube passes

UA Number of transfer units, NTU,, = - Cmin

Fig. 10.47. c

Heat Exchangers

Cold fluid 4

CIA Number of transfer units, mu-= -

Cmin

Fig. 10.48. Effbctiveness for cross-flow heat max exchanger with both fluids unmixed.

1

Mixed fluid

) Unmixed fluid

Number of transfer units, NTU- = - em&#

Fig. 10.49. Effectiveness for cross-flow heat exchanger with one fluid mixed and other &ud,


is the pressure drop pumping cost. The heat exchanger size can be reduced by forcing the fluid through it at higher velocities thereby increasing the overall heat transfer coefficient. But due to higher velocities there will be larger pressure drops resulting in larger pumping costs. The smaller diameter pipe, for a given flow rate, may involve less initial capital cost but definitely higher pumping costs for the life of the exchanger.

We know that, Ap = m2 ...( i)

where Ap = pressure drop of an incompressible fluid flowing through the pipes, and m = mass flow rate.

In order to pump fluid in a steady state, the power requirement is given by m

power = vdp = -dp z h3 . . .(ii) P

This indicates that the power requirement is proportional to the cube of the mass flow rate of the fluid and it may be further increased by dividing it by the pump (fan or compressor) efficiency. Thus, we find that the pumping cost increases greatly with higher velocities, hence, a compromise will have to be made between the larger overall heat transfer coefficient and corresponding velocities.

Example 10.33. Steam condenses at atmospheric pressure on the external sulface of the tubes of a steam condenser. The tubes are 12 in number and each is 30 mm in diameter and 10 m long. The inlet and outlet temperatures of cooling water flowing inside the tubes are 25°C and 60°C respectively. I f the flow rate is 1.1 kgh, calculate the following:

( i ) The rate of condensation of steam,

(ii) The mean overall heat transfer coefficient based on the inner sur- t,l = 2 5 0 ~ face area,

(iii) The number of transfer units, and

(iv) The ejfectiveness of the Fig. 10.50.

condenser.

Solution. Given : N = 12; di = 30 mm = 0.03 m; L = 10 m; tcl = 25°C tc2 = 60°C;

thl = th2 = 100°C; m, = m, = 1.1 kgh

( i ) The rate of condensation of steam, & (= &)

Heat lost by steam = heat gained by water

liz, x hf , = mc x cPc (tc2 - tcl)

where hf, (latent heat of steam) at atmospheric pressure = 2257 kJ/kg. Substituting the values, we get,

riz, x 2257 = 1.1 x 4.187 x (60 - 25)

or iit, = 0.0714 kgls = 257 kglh (Ans.) (ii) The mean overall heat transfer coefficient, U: Total heat transfer rate is given by

Q = m, x cp, x (tc2 - t,l)

, Heat Exchangers 591

where . 'V,

and A = N x ( n d L ) = 12 X 7~ 11'0.03 x 10 = 11.31 m2

Substituting the values in the above equation, we get 161 199.5 = U x 11.31 x 55.68

or U = 255.9 w/m2"c (Ans.)

(iv) The effectiveness of the condenser, E:

E = 1 - exp (- NTU)

or E = 1 - e-0.628 = 0.47 (Ans.) ...[Eqn. (10.51)]

Solution. Given: d = 25 mm = 0.025 m; m, = & = 0.05 Ws, tc , = 1s0c, tc2 = 70°C;

u = 230 w/m2Oc ; thl = 100°C.

( i ) The effectiveness of the heat exchanger, E : Throughout the condenser the hot fluid (i.e., steam), remains at

temperature. Hence cm, is infinity and thus Cmin is obviously for cold fluid (i.e., water) ~h Cmin . U S 7 ; - z O .

When Ch > C,, then effectiveness is given by L m ~

For

or

Cmin = mc cW = 0.05 x 4.18 = 0.209 ulK cmtn - (= R) z 0 ctn~

e = 1 - exp (-NTU) 0.647 = 1 - e-NTU ...[ Eqn (10.51)]


or e-NTu = 1 - 0.647 = 0.353 or - NTU = In (0.353) = - 1.04 . . NTU = 1.04

But NTU = - UA - - U X n d L Cmin c m in

L = N T u X Cmin 1.04 X (0.209 x 1000) - -

U n d 230 x n x 0.025 = 12 m (Ans.)

(iii) The rate of steam condensation, m,,:

Using the overall energy balance, we get

or &, = 0.00509 kgls or 18.32 k@ (Ans.) / Example 10.35. A counter-jl~w heat exchanger is employed to cool 0.55 kg/s (c, = 2.45 kJLkg°C) of oil from l lJ°C to 40°C by the use of water. The intel and outlet temperatures of cooling water are I f C and 7j°C, respectively. The oyerag heat transfer co@cient is expected to be 1450 w / ~ ~ ~ c . Using NTU method, calculate the following:

(i) The mass flow rate of water; (ii) The effectiveness of the heat exchanger;

(iii) The sulfQce area required.

5 Fig. 10.51.

Solution. Given: kt,,,, = 4, = 0.55 kgls; cph = 2.45 I L J A C ~ O C ; thI = I lS0C, th2 = 40°C; tcl = 15OC, tc2 = 75OC;

( i ) The mass flow rate of water, mc (= m,):

The mass flow rate of water can be found by using the overall energy balance

mh Cph (thl - th2) = m, Cpc (tc2 - tcl) 0.55 X 2.45 (115 - 40) = mc x 4.18 (75 - 15)

. . i+ = 0.4 kg/s (Ans.)

(ii) The effectiveness of the heat exchanger, E:

The thermal capacity of cold stream (water), C, = in, cpc = 0.4 x 4.18 = 1.672 kW -

Heat Exchangers 5@3

The thermal capacity of hot stream (oil), Ch = mh cph = 0.55 x 2.45 = 1.347 kW

Since Cc > Ch, hence the effectiveness of the heat exchanger is given by

E = actual heat transfer = & = thl - th2

maximum heat transfer Qm, thl - tcl

E = 115 - 40

= 0.75 (Ans.) 115 - 15

(iii) The surface area required, A: Here em,, = Ch = 1.347 kW; Cm, = Cc = 1.672 kW, hence

Cmin - - - R = - = c m , 347 0.806

1.672

For counter-flow heat exchanger,

...[ Eqn. (10.35)]

1 - exp [- NTU(1- R)] E =

1 -Rexp [-NTU(1 - R ) After rearrangement, we get

...[ Eqn. (10.50)]

E - 1 = exp [- NTU (1 - R)]

(a - 1 ) 0.75 - 1

or = exp '[- N W (1 - 0.806)] (0.75 x 0.806 - 1 )

or 0.632 = exp [- NTU x 0.1941 or In 0.632 = - 0.194 NTU or NTU = 2.365

[This value of NTU may also be obtained h m Fig. 10.45 for R = % = 0.806 and E = 0.751 e m , Also

UA NTU = - c m i n

Example 10.36. 16.5 kgh of the product at 650°C (cp = 3.55 kl/kg°C), in a chemical plant, are to be used to heat 20.5 kgh of the incoming fluid from IOO°C (cp = 4.2 kl/kgO~).' If the overall heat transfer coefficient is 0.95 k ~ / m ' " ~ and the installed heat transfer surface is 44 m2, calculate the fluid outlet temperatures for the counter-flow and parallel flow arrangements.

Solution. Given: tih = 16.5 kgls, th, = 650°C; cph = 3.55 kT/kg°C; m, = 20.5 kgfs; t,,

= 100"; c, = 4.2 kJkg°C; U = 1.2 kw/m2Oc; A = 44 m2.

Fluid outlet temperatures: Case I. Counter-flow arrangement: Thermal capacity of hot fluid, Ch = l jZh x cph = 16.5 x 3.55 = 58.6 k W K

Thermal capacity of cold fluid, C, = m, x c, = 20.5 x 4.- 86.1 k W K

The cold fluid is the maximum fluid, whereas the hot fluid is the minimum fluid. Therefore,

594 Heat and Mass Transfer Heat Exchangers ,595

Cmin = R = - - i::: - 0.68 c,,

UA - 0.95 x 44 Number of transfer units, NTU = - - = 0.71

Cmin 58.6

The value of E (effectiveness) for counter-flow arrangement is given by

E = 1 - exp [- NTU(1 - R)] 1 - R ~ X ~ ' [ - N T U ( ~ - R)]

...[ Eqn. (10.50)]

1 - exp [- 0.71 (1 - 0.68)] 1 - e-0.2272 - - - -

1 - 0.68 X exp [- 0.71 (1 - 0.68)] 1 - 0.68 x e-0.2272

Further, E = ch (thl - 'h2) ...[ Eqn. (10.38)] Cmin (fh~ - tcl)

Because, the hot jluid is minimum, we have

or thz = 650 - 0.443(650 - 100) = 406.3S°C (Ans.)

Also, E = c c (tc2 - tcl) ...[ Eqn. (10.38)] Cmin (ih~ - tc~)

. . tc2 = 2658°C (Ans.) Case 11. Parallel jlow arrangement:

The value of & for parallel flow arrangement is given by

1 - exp [-NTU(1 + R)] E = ... [Eqn. 10.46 (a)]

l + R

- - 1 - exp [- 0.71 (1 + 0.68)J - 1 - e-1.1928 - 1.68

= 0.415 (1 + 0.68)

...[ Eqn. (10.28)]

or tc2 = 255.4OC (Ans.)

Example 10.37. Oil (cp = 3.6 kJ/kg°C) at 100°Cjlows at the rate of 30000 kgh and enters into a parallel flow heat exchanger. Cooling water (c,, = 4.2 kJ/kg°C) enters the heat exchanger at 10°C at the rate of 50000 kgh. The heat transfer area is 10 m2 and U = 1000 w/rn2"c. Calculate the following:

(i) The outlet temperatures of oil, and water; (ii) The maximum possible outlet temperature of water. (P.U.)

30000 Solution. Given: moil = mh = - 3600

= 8.333 kgls; cph = 3.6 kJ/kg°C; thl = 100°C; m,,,

. 50000 = m, = - = 13.89 kgls, cp, = 4.2 kJ/kg°C, t,, = 10°C; U = 1000 w/m2~c ; A = 10 m2. 3600

(i) The outlet temperature of oil and water, th2. tc2:

th = IOOOC

t,, = loOc

UA - 1000 X 10 NTU = - - = 0.33

cmin 3 0 x 1 0 ~

Fig. 10.52.

' ( i i ) II

Lmin For the calculated values of - = 0.514 and NTU = 0.33, from the Fig. 10.44, we get c m a x

E = 0.32

. . t,,, = 100 - 0.32 (100 - 10) = 71.2OC (Ans.)

and 4 2 = 0'32 (loo - lo) + 10 = ZQ8OC (Am.) 1.945 (ii) The maximum possible outlet temperature of water, td: When maximum possible outlet temperature of water exists then,

th2 = tC2 and under this case

cph (thl - tc2) = cpc (tc2 - tcl) or 30 x lo3 (100 - tc2) = 58.34 x lo3 (tc2 - 10) or 100 - tc2 = 1.945 (tc2 - 10) = l.ecw'tc2 - 19.45

tcz = 40.S°C (Ans.)


Example 10.38. The following data is given for counter-flow heat exchanger: The total heat transfer rate, Q: mh = 1 kg/s; m, = 0.25 kg/s As outlet temperatures of both fluids are not known, we have to use NTU method for solving cph = 1.045 kJ/kg°C; cp, = 4.18 kJ/kg°C the problem.

th, = 1000°C ; tC2 = 8500C; U = 88.5 w/m2"c; A = 10 in2 (P.U.) Calculate th2 and t,,.

Solution. Ch = mh cph = 1 x (1.045 x lo3) = 1045

Cc = m, c,,, = 0.25 x (4.18 x lo3) = 1045

. . Cmin = C,, = Ch = C, = 1045 The effectiveness E is given by the relation:

t,,, = 10oo0c C, E = c c (thl - th2)

Cmin (thl -

UA - 88.5 x 10 NTU = - - = 0.85

Cmin 1045 L

Fig. 1053.

For the known values of CmiJCm, and NTU for the counter-flow, we get from Fig. 10.45,

E = 0.47

Substituting this value in eqn. (i), we get

0.47(1000 - t,,) = 850 - tCl or 470 - 0.47 tCl = 850 - t,, tC1 - 0.47 tcl = 850 - 470

Example 10.39. Water (cpc = 4200 J/kg°C) enters a counter-flow double pipe heat exhanger at 3PCflowing at 0.076 kg/s. It is heated by oil (c, = 1880 J/kg°C)flowing at the rate of 0.152 kg/s from an inlet temperature of 116°C. For an area of 1m2 and U=340 w/mZOc, determine the total heat transfer rate,

Fig. 10.54. Counter-flow heat exchanger.

The effectiveness E of heat exchanger is given by

E = ch(thl - th2) - - cc 0c2 - tcl) Cmin (thl - tcl) Cmin (thl - tcl)

Ch = mh cph = 0.152 x 1880 = 258.8 = Cmin

C, = m, cp, = 0.076 x 4200 = 319.2 = C,,

. . . (i)

c,n For the calculated values of - = 0.895 and NTU = 1.19, from the Fig. 10.45, we get c m , E rr 0.53

Substituting the values in eqn (i), we get

and

The total rate of heat transfer is given by Q = U A 8 ,

Solution: Given: m, = lit, = 0.076 kgls, cDC = 4200 J/kg°C; t,, = 38°C; lit,,,, = lith = 0.152

kgls, cph = 1880 J/kg°C; thl = 116°C; U = 340 W/m2"c; A = 1 m2. , /hxample 10.40. The overall temperature rise of the coldfludin a cross-jlow heat exchanger is 20°C and overall temperature drop of hot-fluid is 30°C, The effectiveness of heat exchanger


is 0.6. The heat exchanger area is lm2 and overall heat transfer coeflcient is 60 W / ~ ~ O C . Find out the rate of heat transfer. Assume both jluids are unmixed.

Solution. Given: tc2 - tcl = 20°C; thl - th2 = 30°C; E = 0.6; A = lm2; u = 60 W / m 2 ~ ~ ,

Rate of heat transfer, Q: Heat lost by hot fluid = heat gained by water

mh cph (thl - th2) = cp, (tc2 - tcl)

. . mc cpc = C,, and mh cph = Cmin

- - - - 'mi' - I - 0.67 Cm, 1.5

Now from the graph, for the given values of

But

E = 0.6 and Cmin / C,, = 0.67, we get NTU = 1.4 (From Fig. 10.48)

UA m u = - Cmin UA 60 x 1 Cmin = - = - -

NTU 1.4 - 42.86 = Ch;

Cmin 42.86 c,,=-=---- 0.67 0.67 - 63.97 = Cc

. . Q = mh cph (thl - th2) = ch (thl - th2) = 42.86 x 30 = 1285.8 W (Ans.)

Example 10.41. Dejhe the terms NTUand effectiveness. Derive an expression for effectiveness of a counter-flow heat exchanger in terms of NTU and capacity ratio. (U.P.S.C., 1996)

Solution. Refer Article 10.7. J

Example 10.42. Two fluids, A and B exchange heat in a counter-current heat exchanger. Fluid A enters at 4200C and has a mass flow rate of I kg/s. Fluid B enters at 20°C and has a mass flow rate of I kg/.. Effectiveness of heat exchanger is 75%.

Determine : (i) The heat transfer rate;

(ii) The exit temperature of Jluid B. Specz3c heat of Jluid A is I kJ/kg K and that of fluid B is 4 kJkg K. (GATE, 1999)

Heat Exchangers

Now, Q = E Cmin (thl - tcl) ...[ Eqn. (10.39)]

= 0.75 x mh cph (thl - tcl)

= 0.75 x 1 x 1 x (420 - 20) = 3200 kJ (Ans.)

(ii) The exit temperature of fluid B, tc2 :

Q = mc cpc x ( 4 2 - 41) or 300 = 1 x 4 (tc2 - 20)

. . 300 tc2 = - + 20 = 9S°C (Ans.) 4

Example 10.43. Water at the rate of 0.5 kg/s is forced through a smooth 25 mm ID tube of 15 m length. The inlet water temperature is 10°C and the tube wall is at a constant temperature of 40°C. What is the exit water temperature ?

Average properties of water are :

c, = 4180 J/kg°C; p = 0.8 X 10- Pa S; k = 0.57 W/m 'C. (AMIE Summer, 1998)

Solution. Given : m, = 0.5 kg/s; D = 25 mm = 0.025 m; L = 15 m

ti = 10°C; t* = 40°C; cp = 4180 J/kg°C; p = 0.8 x Pas;

k = 0.57 W/m°C

Exit water temperature, to :

We know that, m, = pVA

pVD 1000 X 1.086 X 0.025 = 3,183 104 :. Reynolds number, Re = - = P 0.8 x low3

i.e., flow is turbulent (since Re > 2300) Using the relation,

= 0.023 ( ~ e ) ' . ~ ( ~ r ) " ~ ' . . . [Eqn. (7.1 5) ] (for t, > tf)

Solution. Given : tAl = thl = 420°C; mh = 1 kg/s; = 1 kg/s; Substituting the values in the above eqn., we get -

cpB = cpc = 4 kJ/kg K .

(i) The heat transfer rate, Q :

Heat transfer area, A = nDL = x 0.025 x 15 = 1.1781 m2

Since the surface temperature is constant,

- UA hA 3785 x 1.1781 = 2.133 NTu=-=-- Cmin Cmin - 2090


Effectiveness, E = 1 - e-NTU ...[ Eqn. (10.51)]

= 1 - (e)- 2.133 = 0.8815

Now,

. . to = 0.88 15 (40 - 10) + 10 = 3644°C (Ans.) ,/ Example 10.44. A counterflow heat exchanger is to heat air entering at 400°C with aflow rate of 6 kg/s by Hot fluid (Gas)

the exhaust gas entering at 800°C with a flow rate of 4 kg/s. The overall heat thz

transfer coefficient is 100 w/m2 K and *c2

the outlet temperature of the air is 551S°C. Specific heat at constantpres- t C l

sure for both air and exhaust gas can I (= 400°C)

be taken as 1100 Jkg K. Calculate : Fig. 10.55. (i) The heat transfer area needed;

(ii) The nhmber of transfer units. (GATE, 1995) Solution. Given : m, = 6 kg/s; tcl = 400°C; tc2 = 551 S°C; thl = 800°C;

Now, C, = m, c,, = 6 x 1 100 = 6600, and

C , = m,, c,,,, = 4 x 1 100 = 4400

Hence, Ch < C,

. . Cmln = C,, = 4400 Now, heat transferred to cold air = heat transferred from hot gases

- - ( i ) The heat transfer area needed, A :

Q = UA 8,

where

Ltth2 - ' c l ) J

- - (800 - 551.5) - (572.75 - 400) 75.75 - --- o.3636 - 208.33"C (800 - 55 1.5) (572.75 - 400)

Substituting the various values, we get 999900 = 100 x A x 208.33

. . A 2 48 m2 (Ans.)

eat Exchangers 601

(ii) The number o f transfer units, (NTU) : UA 100 x 48

NTU = - = - = 1.09 (Ans.) Cmin 4400

L Example 10.45. A chemical having specijk heat of 3.3 kJ/kg Kflowing at the rate of 20000 kgh enters a parallel flow heat exchanger at 120°C. The flow rate of cooling water is 50000 kgh with an inlet temperature of 20°C. The heat transfer area is 10 m2 and the overall heat transfer coefficient is 1050 w/m2 K.

Find : (i) The effectiveness of the heat exchanger. (ii) The outlet temperature of water and chemical.

Take for water, specijk heat = 4.186 kJ/kg K. (U.P.S.C., 1992)

Solution. Given : cph = 3.3 kJ/kg K; h --= 5.56 kg/s; - 3600

t,, = 20°C; A = 10 m2; U = 1050 w / m 2 K.

( i ) The effectiveness o f the heat exchanger, E :

~ o t fluid capacity rate, Ch = 6th cph

= 5.56 ~ 3 . 3 = 18.35 i

---* Water +

Fig. 10.56.

Cold fluid capacity rate, Cc = m, c,

= 13.89 x 4.186 = 58.14

We find Ch c Cc

Heat lost by hot fluid = heat gained by cold fluid .'. 5.56 x 3.3 x (120 - th2) = 13.89 x 4.186 x (tc2 - 20)

or ( 1 20 - th2) = 3.17 (tC2 - 20) ...( i)

Now,

Heat and Mass Transfer Heat Exchangers 603

Effectiveness, E = 1 - exp [- NTU(l+ R)]

l + R

Cmin 18.35 - o.316 where R (capacity ratio) = - = - - C,,, 58.14

1 - exp [- 0.572 (1 + 0.316)] 1 - 0.471 E = - -

(1 + 0.316) 1.316 = 0.402 (Ans.)

(ii) The outlet temperatures of water (tc2) and chemical, (th2) :

(ii) The exit temperatures of hot and cold fluids,Lh2, tc2 : When mh is increased from 10 kglmin to 20 kglmin hi will become h,' (considering hot fluid

is inside).

(120 - t,) 0.402 =

(120 - 20) or th2 = 120 - 0.402 (120 - 20) = 79.S°C (Ans.)

Substituting the value of th2 = 79J°C in eqn, (i) , we get

(120-79.8) = 3.17 (tc2- 20) h,' = 60 x 1.74 = 104.4 w/m2 OC

+ 20 = 32.70, (Ans.) Cc2= 3.17

.,#' Example 10.46. A parallel flow heat exchange has hot and cold water streams running through it and has the following data:

mh = I0 kg. / min ; mc = 25 kg / min; I cph = cpc = 4.18 kJ/kg OC; thl = 70°C;

th2 50°c8 fc lX 25OC thl = 7ooc Individual heat transfer coeficent on

both sides = 60 w/m2 OC. Calculate the th2 = 50°c UA 38.1 x 0.0161 - NTU = - -

1.39 = 0.44 cmin

following: ( i ) The area of the heat exchanger; (ii) The exit temperatures of hot and

cold fluids if hot water pow rate is doubled.

(P.U.)

Cmin For the calculated value of - = 0.799 and NTU = 44, from the fig. 10.44, we get cm, Fig. 10.57

Solution: (i) The area of heat exchanger, A: Heat lost by hot fluid = heat gained by cold fluid

. . mit, cph (thl - th2) = mc cpc (tc2 - tc,) 10 x 4.18 (70 - 50) = 25 x 4.18 (tc2 - 25)

Also,

Log-mean temperature difference, t,, = 70 - 0.31 (70 - 25) = 56.0S°C (Ans.) 0.3 1 (70 - 25)

t c 2 = 1.25 + 25 = 36.16OC (Ans.)

Example 10.47. An oil is cooled to 375 K in concurrent heat exchanger by transferring its heat to the cooling water that leaves the cooler at 300 K. However, it is required that oil must be cooled down to 350 K by lengthening the cooler while the oil and waterpow rates, their inlet temperatures and other dimensions of the cooler remaining unchanged. The inlet temperatures of the cooling water and oil being 288 K and 425 K respectively. The overall heat transfer coefficient is given by,

TI- - 604 Heat and Mass Transfer Heat Exchangers 605

If the length of the original cooler was I m, calculate the following: ( i ) The outlet temperature of cooling water of the new cooler, and

(ii) The length of the new cooler. (M.U.) Solution: Given: thl = 425 - 273 = 152°C; th2 = 375 - 2'73 = 102°C; tcl = 288 - 273 = 15°C;

tc2 = 300 - 273 = 27°C; L, = lm ... Case I 4, = 350 - 273 = 77OC; tc2 = ? ... Case I1

Case I1 : (After increasing the length) The effectiveness of the parallel flow for the second case is given by

1 - exp [-(NTU), (1 + R)] E =

l + R Cmin where R = - cm

1 - exp [-0.486 L2 (1 + 0.24)] 1 - (e)-0.6Lz E = - -

1 + 0.24 1.24 .. .(i)

The effectiveness E for the second case is also given by

Lmin ['.' Ch = C,,, Cc = Cm, and R = - = 0.241

Cm,

...( ii)

152 - 77 ''' = (152 - 15) x 0.24 (152 - 15) + 15 = 33OC (Ans.)

Fig. 10.58. (ii) The length of the new cooler, L, : The outlet temperature of the cooling water of the new cooler, 4, :

Case I : (Before increasing the length)

Equating eqns. ( i ) and (ii), we get. 1 - (e)-0.6% 152 - 77 - -

1 .24 152 - 15

. . Cm in C,, = hh cp, and - - - R = 0.24 cm,

Heat transfer rate is given by or L,=-- 134 - 1.89 m (Ans.)

0.6 This indicates 89% increase in length of the cooler. Example 10.48. A simple counter-flow heat exchanger operates under the following conditions:

Fluid-A, inlet and outlet temperatures 80°C and 40°C;

or (152 - 102) = (NTO), = (NTO), x 102.9

- lo2) = 0.486 Or WTUh =

102.9

Fluid-B, inlet and outlet temperatures 20°C and 40°C. The exchanger is cleaned, causing an increase in the overall heat transfer coeflcient by

10% and inlet temperature offluid B is changed to 30°C. What will be new outlet temperatures of fluid A and of fluid B. Assume heat transfer coeflcients and capacity rates are unalered by temperature changes.

Solution. Given: thl = 80°C. th2 = 40°C. tcl = 20°C tc2 = 40°C ... Case I

thl = 80°C, th2 = ?, tcl = 30°C, tc2 = ? ... Case I1 U2 = 1.1 U,

As outlet temperatures of both fluids are to be calculated, so we have to use NTU method to find th2 and tc2 for the new inlet condition of the cold fluid after cleaning the exchanger.

Further the area of heat exchanger and mass flow rates in both cases remain same.

UA, UCL, (NTU), = - = - , where C = n d

Cmin c m i n

. . UC - 0.486 - - - = 0.486 [': L, = lm (given)] Cmin LI


Again E =

where Cmin R = capacity ratio = -7,

It is obvious that m h cph = cmin

= 47.4"C (Ans.)

Example 10.49. In a large steam power plant, a shell and tube lype steam condenser is employed which has the following data:

Heat exchange rate ... 2100 MW

Number of shell passes ... one

Number of tubes (thin walled) ... 31500, each executing two passes

Diameter of each tube ... 25 mm

Mass jlow rate of water through the tubes ... 3.4 x 104 kg/s

The condensation temperature of steam ... 50°C.

(The steam condenses on the outer suface of the tubes) The heat transfer coeflcient on the steam side ... 11400 w/~**c

I I Case I ( U l = U ) Case I ( U2 = 1.1 U 1 )

Fig. 10.59

We can also write The inlet temperature of water Using LMTD correction factor method

and NTU method, calculate: (i) The outlet temperature of cooling

water, and (ii) The length of tube per pass. Take the following properties of water

(at tbulk = 27'C):

cp = 4.18 Wkg°C, p = 855 x 1@ ~ s / m ~ , k = 0.613 W/m°C and Pr = 5.83

The thermal resistance of tube material and fouling effects may be neglected.

Solution. Given : Q = 2300 x lo6 W ; N, = 31500; d = 25 mm = 0.025 m; hw = hc = 3.4 X lo4 thl = th2 = 50°C;

tcl = 20°C; h, = 11400 ~ l m ~ ~ c .

Steam (mh) I I

Case 11:

as U2 = 1.1 U1 and A and Cm, both the variables remain same in both cases. The effectiveness of the counter-flow heat exchanger for the case I1 is given by (m,)

(a) Flow arrangement thl = th2 = 5O0C

tC1 = 20°C


1 - exp [- (NTU), ( 1 - R)] E =

1 - R exp [- (NTU)2 ( 1 - R)] where R = 0.5 (calculated earlier)

- 1 - exp [ ( - I .52) ( 1 - OS)] 1 - e- 0.76 - - - 1 - 0.5 exp [(-I S 2 ) ( 1 - 0.5)] 1 - 0.5 x e-0.76

= 0.695

The effectiveness E is also given by (i) The water outlet temperature, 4 2 :

In order to obtain water outlet temperature, using overall energy balance, we get

Q = mc Cpc ($2 - tCl)

...[ Eqn. (10.381

Fig. 10.60. Shell and tube type condenser. E = ' h ( * h ~ - 'h2) - ',in (thl - th2) fhl - th2 - - --

Cmin ( th~ - t c ~ ) Cmin (thl - tcl) thl - tcl

2100 x lo6 = 3.4 x lo4 x (4.18 X 1000) (tc2 - 20)

. . tc2 = 34.77OC (Ans.)

(ii) The length of the tube per pass, L: (1) LMTD correction factor method:

or = 80 - 0.696 (80 - 30) = 452°C (Ans.)


The total rate of heat transfer is given by,

Q = FUAO, . . . ( i ) where F = correction factor,

U = overall heat transfer coefficient,

A = 2Np n d L, where Np is the number of tubes per pass, and

Om = log-mean temperature difference. Tn find F we have to find P(temperature ratio) and R (capacity ratio).

'hl - th2 50 - 50 R=-- - = 0 tc2 - t,, 37.44 - 20

With the values of P = 0.492 and R = 0, we get F = 1 To find the value of U, hi has to be calculated first.

Here, mass flow rate through each tube = riz = 3.4 x 104

31500 = 1.079 kgls

:. Reynokis number, Re = PDV = * - - 4 X 1.079 p n d p n x 0 . 0 2 5 ~ 8 5 5 ~ lo6 = 6.43 x 104

Since Re > 2300, hence the flow is turbulent. -

Now, 1 1 1 - = -+- U h, hi

... neglecting thermal resistance of tube material and fouling effects.

Further, LMTD is given by,

Substituting the values in eqn. ( i ) , we get

Q = F U ( 2 N P n d L ) 8 , 2100 x lo6 = 1 x 4707.2 [ 2 x 31500 x n x 0.025 x L] x 21.8

. . L = 4.136 m (Ans.) (2) NTU method: Since the heat exchanger is a condenser, hence Ch = Cmm = - and C,,,,, = Cc = m,cpc = 3.4 x lo4 x (4.18 x 1000) = 14212 x lo4

Since Ch > Cc, hence

Heat Exchangers

Cmin Here - = 0 .-.

But NTU = - U A - - U ( 2 N p n d L )

Cmin Cmin

0.677 x 14212 x 104 . . L = = 4.131 m (Ans.)

4707.2 x (2 x 31500 x n x 0.025)

Example 10.50. In a double pipe heat exchanger ih cph = 0.5 inc cpc, The inlet temperatures of hot and cold fluids are thl and tcl. Deduce an expression in t e n s of thl, tcl and th2 for the ratio of the area of the counter-flow heat exchanger to that of parallel flow heat exchanger which will give the same hot fluid outlet temperature th2. Find this ratio if thl = 150°C, tcl = 30°C and th2 = 90°C. I (U.P.S.C., 1994)

(a) Parallel flow arrangement

I (b) Counter-flow arrangement

The effectiveness of the heat exchanger & is given by Ch (thl - th2) - Cc (tc2 - ' c l )

E = - . . i n h - c Cmin (thl - tc l )

In this case Cmi, = Ch and Ch = 0.5 Cc and Cc = C,, = 2Ch

This equation is true for counter flow as well as parallel flow. Let A, = area of parallel flow heat exchanger, and

A, = area of counter-flow heat exchanger. As (th2), = (th2), therefore, the heat lost by hot fluid in both the cases is same.


. . Q = uAp (ern), = u A c (ern),, since U is not dependent on the direction of flow.

. . Ac - = -

From eqn. (I), we get tc2 in terms of thl, th2 and tc1 as

tc2 = tcl + 0 3 (thl - th2) (a) Parallel flow:

1.5 th2 - 0.5 thl - tcl (b) Counter-flow: I

L (th2 - tcl) 1 Substituting the values of ( i ) and (ii) in eqn. (2), we get

The given data is : thl : 150°C, tc, = 30°C and th2 = 90°C

. . . (i)

Heat Exchangers 61 1

A x a m p l e 10.51. 8000 kgh of air at l0O0C is cooled by passing it through a single-pass cross-flow heat exchanger. To what temperature is the air cooled if water entering at 15OCflows through the tubes unmixed at the rate of 7500 kgh?

Take : t.J = 500 k.l/h-mZ0c and A = 20 mZ

c ( a ) = 1 kr/kg°C and cp (water) rcz = ?

= 4.2 kJkg°C Treat both fluids are unmixed.

Solution: Given: rhh = = 3600 = 2.22 kgls; cph = 1

UA - 1389x20 = NTU = - -

Cm in 2220

Cmin For the calculated values of - = 0.254

Cmm

and NTU = 1.25, from the Fig. 10.48, we get E = 0.63

. . th2 = 100 - 0.63 (100 -15) = 46.45"C

0'63 (loo - 15) + 15 = 28.60C and t"2 = 3.935

The air is cooled to a minimum temperature of 46.4S°C. (Ans.)

dExample 10.52. It is required to design a finned tube, cross-flow heat exchanger to heat pressurized water by means of hot exhaust gases, which enter the exchanger at 310°C and leave it at llO°C, respectively. The water flowing at the rate of 1.4 kg/s enters the exchanger at 3oOC and leaves at 130°C. The hot exhaust gases heat tranCr coeficient based on the gas side is 105 w/rn2'c. Using NTU method, calculate the following:


(i) The efectiveness, and (ii) The gas side surface area. Take the following properties:

Exhaust gas: cp = 1 kJ/kg°C: water (at tbUIk = 80°C) : cp = 4.2 U/kg°C

Solution. Given: m, = i tc = 1.4 kgh; tcl = 30°C; m = 130°C; cpc = 4.2 Wlkg°C;

thl = 3 lo°C; th2 = 1 10°C; cPh = 1 kl/kg°C; Uh = 105 w /m2 OC.

' h I

Fig. 10.63. Fixed tube, cross-flow heat exchanger.

( i ) The effectiveness, & : The he'at capacity o f cold fluid,

The value o f Ch (heat capacity o f hot fluid) may be obtained by writing overall energy balance (since mh is not given). Hence

Hence Cm, = Cc = 5.88 kWI0C and Cmin = Ch = 2.94 kWI0C When Cc > Ch, then effectiveness E is given by

lhl - th2 310 - 110 & = - - - = 0.714 (Ans.) thl- tc l 3 1 0 - 3 0

(ii) The gas side surface area, Ah :

Cmin Corresponding to - = 0.5 and E = 0.714 from Fig. 10.48, we get ~ m ,

- But NTU = - 41 Ah ,

Cmin

A, = 1.8 x 2.94 x 1000

105 = 50.4 m2 (Ans.) Example 10.53. In a gas turbine power plant heat is being transferred in a heat exchanger

from the hot gases leaving the turbine at 450°C to the air r,, = 4 5 0 " ~

leavhg the 'c~m~ressor at 1 70°C. The air flow rate is 5000 kgh and fuel-air ratio is 0.015 kg&. The overall heat transj2r coeficient fhr the heat exchanger is 52.33 w/m2 OC. tC2 = '!

The surface area is 50 m2 and arrangement is cross-flow (both fluids unmixed). Calculate the following:

(i) The exit tempeatures on the air and gas sides, and thZ = ?

(ii) The rate of heat transfer in the exchanger.

Take cph = cpc = 1.05 W k g OC Fig. 10.64

Solution: Given : th, = 450°C; tcl = 170°C; A = 50 m2. ( i ) The exit temperatures on the air and gas sides tc2, th2 :

50M) = 1.4 1 kgls as 1.01 5 kg of gases are formed per kg of air mh = 3600

"fl - 50 = 1.795 NTU = - - cm in 1457.4

Cmin For the calculated values o f = 0.984 and NTU = 1.795, From the Fig. 10.48, we get

The effectiveness & is given by

tc2 - tcl - tc2 - 170 and 0.52 = - -

thl - tcl 450 - 170

. . tc2 = 0.52 (450 - 170) + 170 = 315.6OC (Ans.)

(ii) The rate of heat transfer in exchanger, Q: Q = UA(0m)coun,er = FUA (0m)counter

where F = correction factor


0 Basket vertical tube

tc2-tcl 315.6-170 Temperature ratio, P = - - - = 0.52 thl - tCl 450 - 170

41 - th2 450 - 306.6 Capacity ratio, R = - - - = 0.985 4 2 - tcl 315.6 - 170

Using the values of P and R, from Fig. 10.48, we get

F = 0.76 Now substituting the values in eqn. (i), we get

Q = 0.76 x 52.33 x 50 x 135.5 = 269447 W or 269.45 kW (Ans.) 10.9. EVAPORATORS

10.9.1. Introduction Evaporation is a process in which the solvent is removed from a solution of a non-volatile

solute and a volatile solvent.

Examples. Brine when heated with water (solvent) is evaporated and salt (solute) is produced.

Other typical examples of evaporation of water are concentration aqueous solutions of sodium hydroxide: glycerol, sugar, juices and glue.

In certain cases where the primary object is to get crystals by evaporation, this special process is termed crystallisatlon.

On account of changes ili concentration and properties of the solution the following special problems may creep up :

1. Concentration of liquid 2. Solubility and crystal formation 3. Foaming 4. Pressure and temperature 5. Temperature sensitivity of materials 6. Entrainment of drops 7. Scale deposition and materials of construction.

10.9.2. Classification of Evaporators Evaporators may be classified as follows :

A. On the basis of the fluids to be evaporated : 1. Power plant evaporators : ( i ) Make up evaporators for boiler feed

(ii) Process evaporarors (iii) Salt water evaporators (iv) Heat transfer evaporators. In all above evaporators water is the solvent. 2. Chemical evaporators : Chemical evaporators may be natural circulation type or forced circulation type. Natural

circulation may fall into the following four main classes : Horizontal tube Vertical tube

0 Long tube vertical.

B. On the basis of the method of evaporation : 1. Single effect evaporators :

( i ) Batch evaporators. Here filling, evaporating and emptying are consecutive steps. (ii) Semi-batch. Here feed is continuously added to maintain a constant level until entire change

reaches a final density. (iii) Continuous evaporators. Here feed and discharge are continuous, and concentration of

both feed and discharge remain substantially constant. 2. Multiple effect evaporators :

( I ) . Forward feed Here raw feed is introduced in the first effect, and passed from effect to effect in parallel to the steam flow. The product is withdrawn from the last feed.

(ii) Backward feed, Here the raw feed enters the last (coldest) effect, the discharge from this effect becomes the feed to next to the last effect and so on, until the product is discharged from this first effect,

The following characteristics are important in design and operation of evaporators ' 1. The boiling heat transfer coefficient. 2. The boiling point. - The heat transfer coefficient (i.e,, boiling film coefficient) depends upon viscosity, and

surface tension; p (boiling coefficient) increases with increase in viscosity and varies inversely with surface tension.

- h increases with increase in boiling point.

- h .; pa, where n lies between 0.3 to 0.78 depending upon the solution.

Types of Evaporators : The general types of evaporators are : 1. Open kettle or pan. 2. Horizontal tube natural circulation evaporator. 3. Vertical type natural circulation evaporator. 4. Long tube vertical type natural circulation evaporator. 5. Falling film type vertical natural circulation evaporator. 6. Forced circulation type evaporator. 7. Agitated film evaporator. Example 10.54. (a) Between counter-current and co- current heat exchangers which is more

preferable, and why ? (b) A triple effect evaporator is concentrating a liquid that has no appreciable elevation in

boiling point. The temperature of the steam to the jirst effect is 108OC, the boiling point of the solution in the last effect is 52OC The overall heat transfer co-eflcients in w/m2'c 2500 in

the fint effect, 2000 in the second effect and 1000 in the third effect. At whal tempemtun will the l iquidbi l in the first and second effects ?

(AMIE Winter, 1996)

Solution. (a) Counter-current heat exchanger is more preferable to co-current heat exchanger because of the following reasons :

1. The counter-current heat exchanger is thermodynamically the most eficient. 2. For given heat flow rates, and given inlet and outlet fluid temperatures, a counter-current

heat exchanger requires the minimum heat transfer area and a cwcurrent heat exchanger requires the maximum heat transfer area.


3. In co-current flow (i.e.. parallel flow), the lowest temperature theoretically attainable by the hot fluid is that of the outlet temperature of the cold fluid. If this temperature were attained, the WMTD would be zero, requiring infinite heat transfer surface. Further, inability of the hot fluid in a co-current flow to fall below the outlet temperature of the cold fluid has a marked effect upon the ability of the co-current apparatus to recover heat. Hence, co-current heat flow heat exchanger are rarely used.

(b) Assuming that the multiple effect evaporator is forward feed type, the arrangement is drawn below :

I Feed I

I I I - -

- - - - - Steam 1 08°C .

Drip

Steam to ejector --3

Product

Heat Exchangers 61 7

space pressure is 100 mm Hg (absolute). Find out the steam consumption, steam economy and the heat transfer area if the following data is available.

Feed temperature = 37'C; BPE = 52OC;

(cP)feed= 0.92

(cp)product= O' 75

Overall heat transfer co-eflcient, Uo = 1200 w/m20c.

Solution. (a) Refer Article 10.9.2. mv -

(vapour at 100 mrn Hg)

(mf= 15000 kg/h, 20% solid)

Fig. 10.65. Multiple effect evaporator (Forward feed type).

For the above arrangement, if equal surfaces are employed in each effect (as is the usual practice), the difference in pressures between tbeffects, &I, will be nearly equal. From steam . .

tables, Correspondingto 108"C, pressure = 1.3390 bar,

Corresponding to 52"C, pressure = 0.1360 bar;

Average pressure difference = 1.3390 - 0.1360 3 = 0.401 bar

Break-up of the total pressure difference is given in the table below : I I 1

Steam chest, first effect

Steam chest, second effect

Steam chest, third effect

Example 10.55. (a) How are evaporators dasszj?ed ? What are the characteristics of the solution which are important in design and operation of evaporators and how do these characteristics effect the evaporation process ?

(b) A single effect evaporator is used to concentrate 15000 kg/h of a 20% solution of caustic soda to 60% concentration. Heating medium is dry and saturated steam at 125°C. The vapour

Vaccuum to ejector

Heating steam s Pressure

(bar)

1.3390

0.9380

0.5370

(AMIE Winter, 1997)

So, the boiling temperatures of liquid are : First effect = 98°C (Ans.)

Second effect = 83°C (Ans.)

0.1360

4 Product (m, = 60%)

AP (bar)

( ... 0.401

0.401

Fig. 10.66. Single effect evaporator.

(b) Caustic soda balance : 15000 x 0.2 = mp x 0.6

Steam or vapour "C

108

98

83

0.401

0e2 = 5000 l@, :. Mass of product, mp : 0.6

Total mass balance, 1500 = m, + mv = 5080 + m (where mv = mass of vapour)

52

. . m, = 15000 - 5000 = 10000 k g h

hh of heating steam (at 25°C) from steam tables = 2188 kJ/kg

Vapour pressure = 100 mm Hg

= (13.6 x 1000) x 9.81 x E loo0 x bar = 0.13342 bar

h, = enthalpy of vapour = 2595 Wkg (from steam tables)

Assuming that only latent heat of steam is used for heating, an energy balance of the evaporator is made as :

mf x ( c ~ ) ~ ~ ~ ~ x 37 + m, x hl, = mv x hv + mp x ( ~ ~ ) ~ ~ d ~ ~ t x 52 = mass of feed,

m, = mass of steam [ZY mf

15000 x 0.92 x 37 + m, x 2188 = 10000 x 2595 + 5000 x 0.75 x 52

I :. The steam consumption, m, = 11716 kg/h (Ans.)


Steam economy is defined as kg of solvent

I 125°C evaporated per kg of steam used : ...( iii)

Steam economy = - mv = - loooo - - 0.8535 Am. m, 11716

11716 x 2188 = 3600

dB = dQ.Z = U.dA.0.Z ... from (ii) and (i)

Again integrating, we get

in (02/81) = U.A.Z = UA LMTD

I

Fig. 10.67. .,,from (iii)

= U A 8, = U A (LMT.03 (Am.)

(b) Given : d = 25 mm = 0.025 m; L = 35 m; tc = 25OC; thl = 85°C;

th2 = 35°C; V = 0.6 m/s; cph = 2.51 kJ/kg K; specific gravity of oil = 0.8 Q 7121 x 1000 Heat transfer area = - = 73.93 d (Ans.) U8, 1200 x 80.27

Overall heat transfer co-efficient, U : - tc (= 25°C) Example 10.56. (a) Derive an expression for LMTD of an evaporator. (b) The tube of an oil cooler is submerged in a large pool of stagnant water at temperature

I of 25°C. The inside diameter of the tube is 25 mm and its length is 35 m. Estimate the overall

I heat transfer coeficient of this system if the temperature of oil drops from 85°C to 35°C and the average velocity of oil is 0.6 m/s. Assume 0 for oil specific heat = 2.51 HAE K and

th, spec@ gravity = 0.8. (AMIE Summer, 1995)

Solution. (a) Derivation of expression for LMTD of an evaporator :

Let the temperature in the evapo- th2 rator be t,, the hot fluid (heat capacity

Fig. 10.69.

m = pAf V

I

(where A!= area of flow)

25 mm 4

tc 3 I rate mh cph) enters the evaporator at thl 4

and leaves at th2. i I

The temperature distribution along & W-dA I

the heat exchanger area is as shown in Area, A --+ Fig. 10.68.

For an elementary area, dA Fig. 10.68. Temperature distribution of an evaporator.

_ - - - _ - _ - - - - - _ - - - - - - - - - - - - - - - -'- - Water - - _ - - - _ - _ - - - - - _

1 where Z = - - = a constant. mh Cph

8 = t , - t ,

d8 = dt, = dQ.2 On integration, we get

2 2

. . . (ii) 9 (..a Surface area, A, = dL)

= 385.53 w/m2'c (Ans.)

12908714 Heat and Mass Transfer R K Rajput (2)

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Transcript of 12908714 Heat and Mass Transfer R K Rajput (2)