Heat Transfer Lecture 03

8/6/2019 Heat Transfer Lecture 03

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ACRi Infotech India Pvt. Ltd, 2011

Lecture Notes forHeat Transfer - 2

ACRi PMDCertified Course

inComputational Fluid Dynamics


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Convection Boundary Condition

The heat transfer from the surface of a body due to convection is q = hA(T - T ) A is the area of the surface from which convectiontakes place


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h is the heat transfer coefficientDepends on the fluid propertiesDepends on the nature of the flow Reynoldsnumber etc


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One side with convective heat transfer


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Convective heat transfer from both sides


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Homework problem: Given a specified heatflux on one side and convective heat transfercoefficient on the other side, what is thetemperature distribution in the body?Given: L, q, h, T , k; What is T(x)?


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Radiation Boundary Condition

The heat transfer from the surface of a body due to radiation isEmax = Ts4, =5.67*10-8 w/m 2k 4


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Types of Boundary Conditions - Summary

Dirichlet specified temperatureNeumann specified heat fluxIncludes zero flux case perfectly insulated

Convection Surface specified heat transfercoefficient and medium temperatureRadiation bundary condition

Interface boundary conditionContinuity of temperatureContinuity of heat fux


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Black and Grey body

A black body is an idealised surface that emitsradiation at all frequencies. The Stefan-Boltzmann law is for the perfect blackbody In reality, most bodies are gray they only emitradiation at some frequencies


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Grey body

In reality, most bodies are gray they only emitradiation at some frequenciesSo the amount of radiation emitted is less than thatof a blackbody at the same temperatureEgrey = I Ts4, =5.67*10-8 w/m 2k 4

The fraction I is called emissivity of the surfaceIt is a measure of how efficiently a body emitsradiation compared to a perfect blackbody


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Absorbtion, transmittance

Surfaces generally absorb some incidentradiation, reflect back some radiation and allow some fraction of the radiation to pass throughthem completely

Absorbtion characterised by the absorbtivity EReflectance characterised by the reflectivity

Transmittance characterised by the

transmittivity E + + = 1


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Radiation inside a room

Consider a grey body at temperature T andemissivity I in a room with wall temperature T wall


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Radiation inside a room

Radiation heat energy emitted by body isEbody = A I T4Radiation heat incident on body from wall is

G = A I T w 4

The energy balance equation for the body canbe written as

Qnet = -A I T4 +

AI T w 4

= hA(T - T )

What are the assumptions made in this case?


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Heat Diffusion Cylindrical Coordinates

Consider the conduction of heat through a long cylinderChange in the axial direction is considered to benegligible

Symmetry in the direction


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Consider a section along the axisQ(r) Q(r + ( r) + Qg = ( E/ ( t


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So the 1st

law for the annular element can ve written asQ(r) Q(r + ( r) + q q ( V = ( ( V)C ( T/ ( t

( V = A* ( r = 2 rL* ( r After simplification and taking limit as ( r 0,

We get

t T

C qr T

kAr A g x

x!

xx

xx V

1


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So for a material with variable conductivity

For constant conductivity

t q

r kr

r r g xx

xx

xx1

t T

k

q

r T

r r r g

xx

!

xx

xx

E11


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Steady State without heat generationSteady State with heat generation Transient conduction

Similarly for a sphere, we have

t T

C qr T

kr r r g x

x!

xx

xx V22

1


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3D Heat Diffusion Cylindrical Coordinates

In 3D, we have full heat diffusion equation incylindrical coordinates as

t

T C q

z

T k

z

T k

r r

T kr

r r g

x

x!

x

x

x

x

x

x

x

x

x

x

x

x VJ M

2

11


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Ex. 1 - Heat Diffusion in AnnularCylinder Dirichlet BC

Consider an annular cylinder - thermalconductivity k; What is T(r)?


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Ex. 1 - Heat Diffusion in AnnularCylinder Dirichlet BC

For planar wall, area of c/s was constantSo heat flux through each section constantSo temperature gradient between sections was same

For cylinder area of heat transfer increases with increasing radius

Hence heat flux through section of greater radiusless than section of smaller radius

Hence temperature gradient is not a constant


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Ex. 2 - Heat Diffusion in Solid Cylinder Dirichlet BC

Consider a solid cylinder of radius r1 - thermalconductivity k;

Heat generation / unit volume of q q Temperature at outer surface = T1 What is T(r)


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Ex. 2 - Heat Diffusion in Solid Cylinder Dirichlet BC

What are the boundary conditions? At r = r1, T = T1Need one more B.C

Temperature has to be maximum at center of cylinder? Why?

So at r = 0, dT/dr = 0


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Ex. 3 - Wire + insulationConsider a wire of radius r1 and an insulationmaterial with outer radius r2; conductivity k1,k2;

Heat generation / unit volume of q q - electrical Temperature at outer surface = T2 What is T(r)


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Ex. 3 - Wire + insulationSolve in two parts

For the solid wireFor the annular insulation material

Use interface boundary condition


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Summary What is heat transfer coefficient? What are itsunits?

What does convective heat transfer coefficientdepend on?

Heat Transfer Lecture 03

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