One Dimensional SteadySteady--State State Heat …mazlan/?download=Heat Transfer Chp 3 - One...

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Dr Mazlan - SME 4463

HEAT TRANSFERHEAT TRANSFERSME 4463SME 4463

LECTURER: PM DR MAZLAN ABDUL WAHIDhttp://www.fkm.utm.my/~mazlan

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CChapter hapter 33

One Dimensional One Dimensional SteadySteady--State State

Heat ConductionHeat ConductionPM Dr Mazlan Abdul Wahid

Faculty of Mechanical EngineeringUniversiti Teknologi Malaysiawww.fkm.utm.my/~mazlan

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One-Dimensional Steady-State Conduction

• Conduction problems may involve multiple directions and time-

dependent conditions

• Inherently complex – Difficult to determine temperature distributions

• One-dimensional steady-state models can represent accurately

numerous engineering systems

• In this chapter we will:

� Learn how to obtain temperature profiles for common geometries

with and without heat generation.

� Introduce the concept of thermal resistance and thermal circuits

� Introduce to the analysis of one dimensional conduction analysis

on extended surfaces

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Assumptions

- 1 dimensional heat transfer

- Isothermal surfaces

- Steady state

The Plane Wall

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The Plane Wall

Consider a simple case of one-dimensional conduction in a plane wall, separating two fluids of different temperature, without energy generation

• Temperature is a function of x

• Heat is transferred in the x-direction

Must consider– Convection from hot fluid to wall

– Conduction through wall

– Convection from wall to cold fluid

� Begin by determining temperature distribution within the wall

qx

1,∞T

1,sT

2,sT

2,∞T

x

x=0 x=L

11, ,hT∞

22, ,hT∞

Hot fluid

Cold fluid

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Temperature Distribution – Plane Wall

• Heat diffusion equation in the x-direction for steady-state conditions, with no energy generation:

0=

dx

dTk

dx

d

• Boundary Conditions:2,1, )(,)0( ss TLTTT ==

• Temperature profile, assuming constant k:

1,1,2, )()( sss TL

xTTxT +−=

� Temperature varies linearly with x

� qx is constant

(3.1)

t

Tcq

z

Tk

yy

Tk

yx

Tk

x p ∂∂=+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂ •

ρ

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Thermal Resistance

Based on the previous solution, the conduction heat transfer rate can be calculated:

( ) ( )kAL

TTTT

L

kA

dx

dTkAq ss

ssx /2,1,

2,1,

−=−=−=

� Recall electric circuit theory - Ohm’s law for elect rical resistance:

Similarly for heat convection, Newton’s law of cooling applies:

Resistance

e DifferencPotentialcurrent Electric =

hA

TTTThAq S

Sx /1

)()( ∞

∞−=−=

And for radiation heat transfer:

Ah

TTTTAhq

r

surssursrrad /1

)()(

−=−=

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Thermal Resistance

� The temperature difference is the “potential” or driving force for the heat flow and the combinations of thermal conductivity, convection coefficient, thickness and area of material act as a resistance to this flow:

• We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit).

AhR

hAR

kA

LR

rradtconvtcondt

1,

1, ,,, ===

∑∆==

R

Tq overall

Resistance

Force Driving Overall

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Thermal Resistance for Plane Wall

In terms of overall temperature difference:qx

1,∞T

1,sT

2,sT

2,∞T

xx=0 x=L

11, ,hT∞

22, ,hT∞

Hot fluid

Cold fluid

AhkA

L

AhR

R

TTq

tot

totx

21

2,1,

11 ++=

−= ∞∞

Ah

TT

kAL

TT

Ah

TTq ssss

x2

2,2,2,1,

1

1,1,

/1//1∞∞ −

=−

=−

=

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Composite Walls

Express the following geometry in terms of a an equivalent thermal circuit.

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Composite Walls

What is the heat transfer rate for this system?

AlternativelyUAq

TRR

TUAq

ttot

x

1=∆==

∆=

∑

where U is the overall heat transfer coefficient and ∆T the overall temperature difference.

)]/1()/()/()/()/1[(

11

41 hkLkLkLhARU

CCBBAAtot ++++==

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Example – single layer wall

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Example – multi layer wall

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Example – single layer window

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Example – two layer window

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Thermal Resistance concept - Convection & Radiation

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(a) Surfaces normal to the x-direction are isothermal

(b) Surfaces parallel to x-direction are adiabatic

� For resistances in series: Rtot=R1+R2+…+Rn

� For resistances in parallel:

1/Rtot=1/R1+1/R2+…+1/Rn

Composite Walls – with parallel resistances

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� For resistances in series: Rtot=R1+R2+…+Rn

� For resistances in parallel:1/Rtot=1/R1+1/R2+…+1/Rn


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Parallel heat conduction

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AttentionThe result obtained will be somewhat approximate, since the surfaces of thethird layer will probably not be isothermal, and heat transfer between the firsttwo layers is likely to occur.

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Heat loss through a composite wall

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Contact Resistance

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Contact Resistance

The temperature drop across the interface between materials may be appreciable, due to surface roughness effects, leading to air pockets. We can define thermal contact resistance:

cct hR /1"

, =

",intint

" / ctcx RTThq ∆=∆=

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Contact resistance of transistors

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Composite Walls – with contact resistances

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Problem: Thermal Barrier Coating

Problem 3.30: Assessment of thermal barrier coating (TBC) for protectionof turbine blades. Determine maximum blade temperaturewith and without TBC.

Schematic:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constant properties, (3) Negligible radiation.

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Problem: Thermal Barrier Coating (cont..)

( )-3 -4 -4 -4 -3 2 -3 2tot 10 3.85 10 10 2 10 2 10 m K W 3.69×10 m K W ,wR = =′′ + × + + × + × ⋅ ⋅

( ) ( )3 4 2 5 2400 K 2 10 2 10 m K W 3.52 10 W m 1174 K− −= + × + × ⋅ × =( ) ( )(w) Ins,o ,i i wT = T + 1 h + L k q∞ ′′

5 23 2

tot

1300K3.52 10 W m

3.69 10 m K W

,o ,iw

,w

T - Tq = =

R

∞ ∞−

′′ = ×′′ × ⋅

With a heat flux of

the inner and outer surface temperatures of the Inconel are

( )(w)s,i ,i w iT = T + q h∞ ′′5 2

400 K 1104 K2

3.52 10 W m

500W m K / W= + =

× ⋅

ANALYSIS: For a unit area, the total thermal resistance with the TBC is

( ) ( )1 1tot Zr In

- -,w o t,c iR = h + L k + R + L k + h′′ ′′

<

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Problem: Thermal Barrier Coating (cont..)

( )1 1 3 2tot wo In 3 20 10 m K W

- -, o iR = h + L k + h .

−′′ = × ⋅

The inner and outer surface temperatures of the Inconel are then

( )(wo) 1212Ks,i ,i wo iT T q h∞ ′′= + =

( ) ( )[ ](wo) In1 1293 Ks ,o ,i i woT T h L k q∞ ′′= + + =

Use of the TBC facilitates operation of the Inconel below Tmax = 1250 K.

COMMENTS: Since the durability of the TBC decreases with increasing temperature, which increases with increasing thickness, limits to its thickness are associated with reliability considerations.

( )wo tot,wo,o ,iq T T R∞ ∞′′ ′′= − = 4.06×105 W/m2

Without the TBC,

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Radial Systems-Cylindrical Coordinates

Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures

Temperature distribution

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• Heat diffusion equation (eq. 2.5) in the r-direction for steady-state conditions, with no energy generation:

t

Tcq

z

Tk

z

Tk

rr

Tkr

rr p ∂∂=+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂ •

ρφφ2

11

01 =

dr

dTkr

dr

d

r

• Boundary Conditions: 2,21,1 )(,)( ss TrTTrT ==

• Temperature profile, assuming constant k:

2,221

2,1, ln)/ln(

)()( s

ss Tr

r

rr

TTrT +

−= � Logarithmic temperature

distribution

• Fourier’s law: constdr

dTrLk

dr

dTkAqr =−=−= )2( π

01 =

dr

dTr

dr

d

r

for constant k

Temperature Distribution - Thermal resistance for cylinder

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dr

dTrLk

dr

dTkAqr −=−= )2( π

2,221

2,1, ln)/ln(

)()( s

ss Tr

r

rr

TTrT +

−=

)/ln(

)(2

12

21

rr

TTkLq r

−= π

kLrr

TTq r

π2/)/ln( 12

21 −=

kLrrR cyl π2/)/ln( 12=therefore

Temperature Distribution - thermal resistance for cylinder

3.33

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Temperature Distribution - thermal resistance for cylinder

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Thermal resistance for cylinder

Based on the previous solution, the conduction hear transfer rate can be calculated:

( ) ( ) ( )condt

ssssssx R

TT

Lkrr

TT

rr

TTLkq

,

2,1,

12

2,1,

12

2,1,

)2/()/ln()/ln(

2 −=

−=

−=

ππ

� In terms of equivalent thermal circuit:

)2(

1

2

)/ln(

)2(

1

22

12

11

2,1,

LrhkL

rr

LrhR

R

TTq

tot

totx

πππ++=

−= ∞∞

• Fourier’s law: constdr

dTrLk

dr

dTkAqr =π−=−= )2(

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Composite Walls

Express the following geometry in terms of a an equivalent thermal circuit.

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Composite Walls

What is the heat transfer rate?

where U is the overall heat transfer coefficient. If A=A1=2πr1L:

44

1

3

41

2

31

1

21

1

1lnlnln

11

hr

r

r

r

k

r

r

r

k

r

r

r

k

r

h

U

CBA

++++=

alternatively we can use A2=2πr2L, A3=2πr3L etc. In all cases:

∑====

tRAUAUAUAU

144332211

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Spherical Coordinates

Fourier’s law:

dr

dTrk

dr

dTkAqsph

)4( 2π−=

−=

• Starting from Fourier’s law, acknowledging that q r is constant, independent of r, and assuming that k is constant, derive the equation describing the conduction heat transfer ra te. What is the thermal resistance?

Need temperature profile, then differentiate

t

Tcq

Tk

r

Tk

rr

Tkr

rr p ∂∂=+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂ •

ρθ

θθθφφθ

sinsin

1

sin

11222

22

Assume:1) Steady state2) Constant properties3) 1 dimensional coordinate

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Spherical Coordinates

01 22

=

∂∂

∂∂

r

Tkr

rr

2,21,1 )(,)( ss TrTTrT ==

2,221

2,1, ln)/ln(

)()( s

ss Tr

r

rr

TTrT +

−=

)11

(4

1

21 rrkR sph −=

π

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dTkrdrqsph ∫∫ −=2/4/ π

)/1()/1(

)(4

21

21

rr

TsTskq sph −

−= π

)11

(4

1

21 rrkR sph −=

π

Temperature Distribution - thermal resistance for sphere

3.41

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Conduction with Thermal Energy Generation

Thermal energy may be generated or consumed due to conversion from some other energy form.

� If thermal energy is generated in the material at the expense of some other energy form, we have a source: is +ve

– Deceleration and absorption of neutrons in a nuclear reactor

– Exothermic reactions

– Conversion of electrical to thermal energy:

V

RI

V

Eq eg

2

==&

where I is the current, Re the electrical resistance, V the volume of the medium

� If thermal energy is consumed we have a sink: is -ve– Endothermic reactions

q

q

24

.

.

.

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The Plane Wall

Consider one-dimensional, steady-state conduction in a plane wall ofconstant k, with uniform generation, and asymmetric surface conditions:

• Heat diffusion equation (eq. 2.21) :

02

2

=+k

q

dx

Td

• Boundary Conditions:

2,1, )(,)( ss TLTTLT ==−

• General Solution:

212

2CxCx

k

qT ++−=

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Temperature Profile

22

)(1

2)( 2,1,1,2,

2

22ssss TT

L

xTT

L

x

k

qLxT

++

−+

−=

� Profile is parabolic.

� Heat flux not independent of x

What happens when:

(3.46)

?0 increases, ,0 <= qqq

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.

. . .

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Symmetrical Distribution

• When both surfaces are maintained at a common temperature, Ts,1= Ts,2 = Ts

sTL

x

k

qLxT +

−=

2

22

12

)(

What is the location of the maximum temperature?

2

max

max)(

=−

−∴L

x

TT

TxT

s

27

.(3.48)

(3.49)

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Symmetrical Distribution

� Note that at the plane of symmetry:

0q" 00

0

=⇒=

=

=x

xdx

dT

� Equivalent to adiabatic surface

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Calculation of surface temperature T s

In equations (3.48) and (3.49) the surface temperature, Ts is needed.

� Boundary condition at the wall:

)( ∞=

−=− TThdx

dTk s

Lx

Substituting (dT/dx)x=L from equation(3.47):

h

qLTTs += ∞

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Example

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m.K having uniform volumetric heat generation of 0.3 MW/m3 is insulated on one side, while the other side is exposed to a fluid at 92°C. The convection heat transfer coefficient between the wall and the fluid is 500 W/m2.K. Determine the maximum temperature in the wall.

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Radial Systems

Cylindrical (Tube) Wall Spherical Wall (Shell)

Solid Cylinder (Circular Rod) Solid Sphere

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Radial Systems• Heat diffusion equation in

the r-direction for steady-state conditions:

01 =+

k

q

dr

dTr

dr

d

r

• Boundary Conditions: sor

TrTdr

dT ===

)( ,00

• Temperature profile:

so

o Tr

r

k

qrrT +

−=

2

22

14

)(

L

hT ,∞

))(2()( 2

∞−= TTLrhLrq soo ππ h

qrTT o

s 2+= ∞

• Calculation of surface temperature:

and

• General Solution:

33

212 ln

4CrCr

k

qT ++−=

.

.

.

Cylinder

..

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Example

A cylindrical shell of inner and outer radii ri and ro, respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate. The inner surface is insulated, while the outer surface of the shell is exposed to a fluid with a convection coefficient h.

a) Obtain an expression for the steady-state temperature distribution T(r) in the shell.

b) Determine an expression for the heat rate q’ (ro) at the outer radius of the shell in terms of the heat generation rate and the shell dimensions

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CRITICAL THICKNESS OF INSULATION

For smaller diameter tubes it is sometimes possible to increase heat loss by adding insulation on the outer surface

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Critical radius of insulation

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Optimum thickness of insulation

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Heat Transfer from Extended SurfacesHeat Transfer from Extended Surfaces-- FinsFins

There are two ways to increase the rate of heat transfer: to increase the convection heat transfer coefficient h or to increase the surface area A .

Fins enhance heat transfer from a surface by exposing a larger surface area toconvection and radiation.

qconv =

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Extended surfacesExtended surfaces-- FinsFins

An extended surface (also know as a combined conduction-convection system or a fin) is a solid within which heat transfer by conduction is assumed to be one dimensional, while heat is also transferred by convection (and/or radiation) from the surface in a direction transverse to that of conduction

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Extended surfacesExtended surfaces-- FinsFins

Extended surfaces may exist in many situations but are commonly used as fins to enhance heat transfer by increasing the surface area available for convection (and/or radiation). They are particularly beneficial when h is small, as for a gas and natural convection.

� Solutions for various fin geometries can be found in the literature (see for example in textbook).

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A general conduction analysis for an extended surfaces

Applying the conservation of energy

Using,

Then, the heat equation becomes:

General form of the energy equation for an extended surface

Eq. (3.66)

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Heat transfer from extended surface

The Fin Equation� Assuming 1-D case, steady state conduction

in an extended surface, constant k, uniform cross sectional area, negligible generation and radiation.

� Cross section area, Ac is constant and fin surface area, As = Px, this mean dAc/dx = 0 and dAs/dx = P

� General equation becomes:

Eq. (3.67)

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� To simplify the equation, we define an excess temperature ( the reduced temperature) as:

� The previous equation becomes:

where,

* m also known as fin parameter

Eq. (3.68)

Eq. (3.70)

P is the fin perimeter

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Extended Surface: Fin equation

0)(2

2

=−− ∞TTkA

hp

dx

Td

c

022

2

=− θθm

dx

d

2

ckA

hpm =

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)( 21mxmx eCeCx −+=θ

Two unknowns, C1 and C2, therefore

require 2 BC to solve

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By referring to Table 3.4 : at different case of heat transfer analysis

• Temperature distribution, θ/θb

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D

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B

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A

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C

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Proper length of a fin

41

Problem: Turbine Blade CoolingProblem: Turbine Blade Cooling

Problem 3.126: Assessment of cooling scheme for gas turbine blade.Determination of whether blade temperatures are lessthan the maximum allowable value (1050°C) for prescribed operating conditions and evaluation of bladecooling rate.

Assumptions: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3)Adiabatic blade tip, (4) Negligible radiation.

Analysis: Conditions in the blade are determined by Case B of Table 3.4.

(a) With the maximum temperature existing at x = L, Eq. 3.80 yields

Schematic:

( ) 1

coshb

T L - T

T - T mL∞

∞=

( ) ( )1/22 4 2250 W/m K 0.11m/20W/m K 6 10 m1/ 2cm hP/kA −= = ⋅ × ⋅ × × = 47.87 m-1

mL = 47.87 m-1 × 0.05 m = 2.39

From Table B.1 (or by calculation), Hence, cosh 5.51.mL =

and, subject to the assumption of an adiabatic tip, the operating conditions are acceptable.

Eq. 3.81 and Table B.1 yield

Hence,

Comments: Radiation losses from the blade surface contribute to reducing the blade temperatures, but what is the effect of assuming an adiabatic tip condition? Calculatethe tip temperature allowing for convection from the gas.

( ) o o o1200 C (300 1200) C/5.51 1037 CT L = + − =

(b) With ( ) ( ) ( )1/22 4 2 o1/2 250W/m K 0.11m 20W/m K 6 10 m 900 C 517Wc bM hPkA θ −= = × × ⋅ × × − = −⋅ ,

( )tanh 517W 0.983 508Wfq M mL= = − = −

508Wb fq q= − =

Problem: Turbine Blade Cooling (cont.)Problem: Turbine Blade Cooling (cont.)

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Fin effectiveness

Fin Performance

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Fin efficiency

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Performance ParametersPerformance Parameters Fin Performance• Fin Efficiency:

, max

f ff

f f b

q q

q hAη

θ≡ = (3.91)

How is the efficiency affected by the thermal conductivity of the fin?Expressions for are provided in Table 3.5 for common geometries.fη

( )1/ 2222 / 2fA w L t = +

( )/ 2pA t L=( )( )

1

0

21

2f

I mL

mL I mLη =

• Fin Effectiveness:

Consider a triangular fin:

,

ff

c b b

q

hAε

θ≡

• Fin Resistance:

with , and /f ch k A Pε ↑ ↓ ↑ ↓(3.86)

,

1bt f

f f f

Rq hA

θη

≡ = (3.97)

where 0 1fη≤ ≤

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Efficiency of fins (I)

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Efficiency of fins (II)

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Dr Mazlan - SME 4463 93

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(cont.)

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49

ArraysArrays Fin Arrays

• Representative arrays of(a) rectangular and(b) annular fins.

– Total surface area:

t f bA NA A= + (3.104)

Number of fins Area of exposed base (prime surface)– Total heat rate:

,

bt f f b b b o t b

t o

q N hA hA hAR

θη θ θ η θ= + ≡ =(3.105)

– Overall surface efficiency and resistance:

( )1 1fo f

t

NA

Aη η= − −

(3.107)

Arrays (Cont.)Arrays (Cont.)

• Equivalent Thermal Circuit:

• Effect of Surface Contact Resistance:

( )( ),

bt t bo c

t o c

q hAR

θη θ= =

( )1

1 1f fo c

t

NA

A C

ηη

= − −

(3.110a)

( )1 , ,1 /f f t c c bC hA R Aη ′′= + (3.110b)

( )( )

,

1t o c

to c

RhAη

= (3.109)

,1b

t ot o t

Rq hA

θη

= = (3.108)

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Problem 3.144: Determination of maximum allowable power for a 20 mm 20 mm electronic chip whose temperature is not to exceedwhen the chip is attached to an air-cooled heat sink with N = 11 fins of prescribed dimensions.

cq85 C,cT = o

Schematic:

Assumptions: (1) Steady-state, (2) One-dimensional heat transfer, (3) Isothermal chip, (4)Negligible heat transfer from top surface of chip, (5) Negligible temperature rise for air flow,(6) Uniform convection coefficient associated with air flow through channels and over outersurface of heat sink, (7) Negligible radiation, (8) Adiabatic fin tips.

Problem: Chip Heat SinkProblem: Chip Heat Sink

×

Analysis: (a) From the thermal circuit,

tot

c cc

t,c t,b t,o

T -T T -Tq = =

R R + R + R∞ ∞

( )22 6 2m K/W, / W 2 10 / 0.02m 0.005 K/Wt,c t cR R −′′= = × =⋅

( )2/ Wt,b bR L k= ( )2W/m K0.003m / 180 0.02m 0.042 K/W⋅= =

From Eqs. (3.108), (3.107), and (3.104)

( )1, 1 1 ,

ft,o o f t f b

o t t

NAR A NA A

h A Aη η

η= = − − = +

Af = 2WLf = 2 × 0.02m × 0.015m = 6 × 10-4

m2

Ab = W2 – N(tW) = (0.02m)

2 – 11(0.182 × 10

-3 m × 0.02m) = 3.6 × 10

-4 m

2

With mLf = (2h/kt)1/2

Lf = (200 W/m2⋅K/180 W/m⋅K × 0.182 × 10

-3m)

1/2 (0.015m) =

1.17, tanh mLf = 0.824 and Eq. (3.94) yields tanh 0.824

0.7041.17

ff

f

mL

mLη = = =

At = 6.96 × 10-3

m2

ηo = 0.719,

Rt,o = 2.00 K/W, and

( )( )

85 20 °C31.8 W

0.005 0.042 2.00 K/Wcq−

= =+ +

Problem: Chip Heat Sink (cont.)Problem: Chip Heat Sink (cont.)

<

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Comments: The heat sink significantly increases the allowable heat dissipation. If it were not used and heat was simply transferred by convection from the surface of the chip with

from Part (a) would be replaced by 2tot100 W/m K, 2.05 K/W= ⋅ =h R

2conv 1 / 25 K/W, yielding 2.60 W.cR hW q= = =

Problem: Chip Heat Sink (cont.)Problem: Chip Heat Sink (cont.)

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One Dimensional SteadySteady--State State Heat …mazlan/?download=Heat Transfer Chp 3 - One...

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