Hmt Kkk Unit 1, 2

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HEAT AND MASS TRANSFER (EME 504)

VTH SEM MECHANICAL

Prepared By:

Krishna Kumar Karothiya (Asstt. Prof.)

Mechanical Engineering DepartmentABES , ENGG. COLLEGE GHAZIABAD

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UNIT-1Introduction to Heat Transfer, Conduction,

Steady State one-dimensional Heat conduction

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What is Heat Transfer

Energy in transit due to temperature difference.

Thermodynamics tells us: How much heat is transferred (dQ)

How much work is done (dW)

Final state of the system

Heat transfer tells us: How (with what modes) dQ is transferred

At what rate dQ is transferred

Temperature distribution inside the body

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MODES

Conduction needs matter

Molecular phenomenon (diffusion process)

Without bulk motion of matter

Convection Heat carried away by bulk motion of fluid

Needs fluid matter

Radiation

Does not needs matter Transmission of energy by electromagnetic waves

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APPLICATIONS OF HEAT

TRANSFER

Energy production and conversion

Steam power plant, solar energy conversion etc

Refrigeration and air conditioning

Domestic applications

ovens, stoves, toaster

Cooling of electronic equipment

Manufacturing / materials processing

welding, casting, soldering, laser machining

Automobiles / aircraft design

Nature (weather, climate etc)

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Conduction

(Needs medium, Temperature gradient)

RATE:

q(W) or (J/s) (heat flow per unit time)

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Conduction

Rate equations (1D conduction):

Differential Form

Difference Form

(negative sign denotes heat transfer in the direction of decreasing temperature)

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CONVECTION

MOVING FLUID

Energy transferred by diffusion + bulk motion of fluid

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Rate equation

(convection)

h=heat transfer coefficient (W /m 2 K)(not a property) depends on geometry ,nature of flow,

thermodynamics properties etc

Ts= surface temp.

T = Ambient temperature

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Convection

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Convection (contd)

Typical values of h (W/m 2 K)

Free convection gases: 2 -25

liquid: 50 -100 Forced convection gases: 25- 250

liquid: 50 -20,000

Boiling/Condensation 2500 -100,000

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RADIATION

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Rate equations

(Radiation)

RADIATION:

Heat Transfer by electromagnetic waves or

photons( no medium required. )

Emissive power of a surface (energy released

per unit area)

e= emissivity

s=Stefan Boltzmann constant

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Rate equations

(Contd.)

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ONE DIMENSIONAL STEADY STATEHEAT CONDUCTION

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Objectives of conduction

analysis To determine the temperature field, T(x,y,z,t), in a body(i.e.

how temperature varies with position within the body)

T(x,y,z,t) depends on: Boundary conditions

initial condition Material properties (k, c p , r )

Geometry of the body (shape, size)

why we need T(x,y,z,t) ? To compute heat flux at any location (using Fouriers eqn.)

Compute thermal stresses, expansion, deflection due to temp. etc.

Design insulation thickness Chip temperature calculation

Heat treatment of metals

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Unidirectional heat

conduction (1D)

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Unidirectional heat

conduction (1D)

First Law (energy balance)

(Ein- Eout)+Egen. = Est

-++ A q=

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Unidirectional heat conduction

(1D)(contd

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Unidirectional heat conduction

(1D)(contd

For T to rise, LHS must be positive (heat input is

positive)

For a fixed heat input, T rises faster for higher

In this special case, heat flow is 1D. If sides were not

insulated, heat flow could be 2D, 3D.

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Boundary and Initial conditions

The objective of deriving the heat diffusion

equation is to determine the temperature

distribution within the conducting body.

We have set up a differential equation, with T

as the dependent variable. The solution will

give us T(x,y,z). Solution depends on boundary

conditions (BC) and initial conditions (IC).

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Boundary and Initial

conditions (contd)

How many BCs and ICs ? Heat equation is second order in spatial coordinate. Hence, 2

BCs needed for each coordinate. 1D problem: 2 BC in xdirection

2D problem: 2 BC in xdirection, 2 in ydirection

3D problem: 2 in xdir., 2 in ydir., and 2 in zdir.

Heat equation is first order in time. Hence one

IC needed

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1Dimensional

Heat Conduction

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Thermal resistance

(electrical analogy)

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Thermal Analogy to Ohms

Law

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1 D Heat Conduction through a

Plane Wall

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Resistance expressions

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Composite Walls

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Overall Heat transfer Coefficient

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Series Parallel

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1 D Conduction(Radial

conduction in a composite

cylinder)

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Critical Insulation

Thickness

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Critical Insulation

Thickness (contd)

MaximumMinimum problem

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Critical Insulation

Thickness (contd)

Minimum q at r 0 =(k/h)=r c r (critical radius)

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I D CONDUCTION IN SPHERE

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Conduction with Thermal

Energy Generation

=

= Energy generation per unit volume

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Energy Generation

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Conduction With Thermal

Energy Generation (contd)

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Energy Generation (cont..)

Use boundary conditions to find C1 and C2

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Cylinder with heat source

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Cylinder With Heat Source

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Summary of Electrical Analogy

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SOME ANIMATIONS

Conduction Wall

http://www.rpaulsingh.com/animated%20figures/fig4_16.htm

http://www.rpaulsingh.com/animated%20figures/

fig4_17.htm

http://www.rpaulsingh.com/animated%20figures/

fig4_18.htm
http://www.rpaulsingh.com/animated%20figures/fig4_15.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_16.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_16.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_17.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_17.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_18.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_18.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_18.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_18.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_17.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_17.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_16.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_16.htmhttp://www.rpaulsingh.com/animated%20figures/fig4_15.htm

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UNIT-2

EXTENTED SURFACE (FINS)HEAT

TRANSFER

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EXTENDED SURFACES / FINS

Convection: Heat transfer between a solid surface and a moving fluid is

governed by the Newtons cooling law: q = hA(Ts -T). Therefore, to

increase the convective heat transfer, one can

Increase the temperature difference (Ts -T) between the surface and the

fluid.

Increase the convection coefficient h. This can be accomplished by

increasing the fluid flow over the surface since h is a function of the flow

velocity and the higher the velocity, the higher the h. Example: a coolingfan.

Increase the contact surface area A. Example: a heat sink with fins

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Extended Surface Analysis

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A second-order, ordinary differential equation

Equation Define a new variable so that

Characteristics equation with two real roots: +m & -mThe general solution

is of the form

To evaluate the two constants C1 and C2 we need to specify two boundary

conditions: The first one is obvious: the base temperature is known asT(0)=Tb The second condition will depend on the end condition of the tip

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Temperature distribution for fins of

different configurations

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Example An Aluminium pot is used to boil water as shown below. The handle of the pot is

20-cm long, 3-cm wide, and 0.5-cm thick. The pot is exposed to room air at 250C,

and the convection coefficient is 5 W/m2 0C. Question: can you touch the handle

when the water is boiling? (k for aluminum is 237 W/m 0C).

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Example

We can model the pot handle as an extended surface. Assume that there is

no heat transfer at the free end of the handle. The condition matches that

specified in the fins Table, case B. Use the following data.

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Example

Plot the temperature distribution along the pot handle

As shown in the figure, temperature drops off but not very steeply. This isbecause k of aluminium is fairly high. At the midpoint, T(0.1)=90.40C. At theend T(0.2)=87.30C. Therefore, it should not be safe to touch the end of thehandle.The end condition is insulated, hence the gradient is zero.

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Fin Design

Total heat loss: qf =Mtanh(mL) for an adiabatic

fin, or qf =Mtanh(mLC ) if there is convective

heat transfer at the tip

Use the thermal resistance concept

where Rt .fis the thermal resistance of thefin.For a fin with an adiabatic tip, the fin

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Fin Effectiveness

How effective a fin can enhance heat transfer is characterized by the fin effectiveness (f )Ratio of fin heat transfer and the heat transfer without the fin. For an adiabatic fin:

If the fin is long enough, mL>2, tanh(mL)1, it

can be considered an infinite fin

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Fin Effectiveness (contd...)

To increase f, the fins material should have higher thermalconductivity, k.

It seems to be counterintuitive that the lower convection

coefficient, h, the higherf . But it is not because if h is very

high, it is not necessary to enhance heat transfer by adding heatfins. Therefore, heat fins are more effective if h is low.

Observation: If fins are to be used on surfaces separating gas and

liquid. Fins are usually placed on the gas side. (Why?)

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P/AC should be as high as possible. Use a

square fin with a dimension of W by W as an

example: P=4W, AC=W2, P/AC=(4/W). Thesmaller W, the higher the P/AC, and the higher

f.

Conclusion: It is preferred to use thin andclosely spaced (to increase the total number)

fins.

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Fin Efficiency

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Fin Efficiency (contd)

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Fin Efficiency (cont.)

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Overall Fin Efficiency

Overall fin efficiency for an array of fins:

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Overall Fin Efficiency (contd)

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Heat Transfer from a Fin Array

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Thermal Resistance Concept

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UNSTEADY STATE HEAT TRANSFER

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UNSTEADY HEAT TRANSFER

Many heat transfer problems require the understanding of the complete

time history of the temperature variation. For example, in metallurgy, the

heat treating process can be controlled to directly affect the characteristics

of the processed materials. Annealing (slow cool) can soften metals andimprove ductility. On the other hand, quenching (rapid cool) can harden the

strain boundary and increase strength. In order to characterize this transient

behavior, the full unsteady equation is needed

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UNSTEADY HEAT TRANSFER

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Biot No. BiDefined to describe the relative resistance in a thermal circuit ofthe convection compared.

Lc is a characteristic length of the body

Bi0: No conduction resistance at all. The body is

isothermal.

Small Bi: Conduction resistance is less important. The bodymay still be approximated as isothermal

Lumped capacitance analysis can be performed.

Large Bi: Conduction resistance is significant. The body

cannot be treated as isothermal.

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Lumped Parameter Analysis

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To determine the temperature at a given

time, or

To determine the time required for the

temperature to reach a specified value

Note: Temperature function only of time and not of space!

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Lumped Parameter Analysis Define Fo as the Fourier number (dimensionless time)

The temperature variation can be expressed as

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THANK YOU

Hmt Kkk Unit 1, 2

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