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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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Conduction with Thermal
Energy Generation
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Conduction With Thermal
Energy Generation (contd)
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Conduction with Thermal
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.htm7/28/2019 Hmt Kkk Unit 1, 2
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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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Extended Surface Analysis
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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Extended Surface Analysis
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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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Fin Effectiveness (contd...)
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 Effectiveness (contd...)
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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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Lumped Parameter Analysis
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
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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