Conduction 1

7/24/2019 Conduction 1

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Heat Conduction

Areas to be covered:

General heat conduction equation

Boundar conditions

1D heat conduction through flat plate, cylinder and sphere.

Contact thermal resistance.

Heat conduction through compound slabs, cylinders and spheres.

Heat transfer between fluids inside and outside lane walls and i es.Overall heat transfer coefficient.

Heat conduction with dissipation Heat transfer to and from finned surfaces


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Heat Conduction

Learning outcomes:

Able to derive heat conduction equations from the first principle;

Able to solve the e uation

Able to calculate heat transfer through flat plate, cylinder and sphere;

Able to calculate heat transfer throu h com ound slabs c linders and

spheres. Able to calculate heat transfer between fluids inside and outside plane

walls and i es

Able to analyse heat transfer with dissipation;

Able to anal se heat transfer to and from finned surfaces

Able to analyse heat transfer using thermal resistance.


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General 3-D Heat Conduction

Net heat transfer into element by conduction

The net heat transfer into the element

by conduction inx-direction is

= ( )

x x dx

xx x

QQ Q dx

=- x

x

Qdx

(1)

Similarly,

The net heat transfer into element by conduction in direction =- yQ

y dyy

The net heat transfer into element by conduction in direction =- zQz dzz


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From Fouriers law we have

T x x

x

whereAx is the area normal tox-direction,Ax =dydz

Substituting Fouriers equation into Eq.(1), the net heat

transfer into the element inx-direction is

( )xQ T

dx dydz dxx x x

Similarly

Net hear tranfer into element in y direction = Tdxdz dy

y y

Net heat transfer into element in z direction = Tdxdy dz z z


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Adding all these, the total net heat transfer rate into element is

T T Tdxdydz

x x y y z z

where dxdydz=dV= volume of element

Total heat transfer rate into element = rate of increase of internal

energy

The rate of increase of internal energy is

TdV cdV

t t

( , =specific heat capacity, J/(kg K))p vc c c c


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Thus, the general 3-D heat conduction equation is

cx x y y z z t

(2)

If =constant, the general equation becomes

2 2 2T T T c T

2 2 2x y z t

,

2 2 2

0T T T

x y z


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For 2-D heat conduction, the general equation is

cx x y y t


2 2T T c T 2 2

x y t

,

2 2

0T T

x y


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The general 1-D heat conduction equation is

T T


x x t

(4)

2

2

T c T

x t

For steady state case, the general equation is2T

2

xFor 1-D, steady state case, T= function ofx only (Note: no dissipation

Initial condition

, , , , ,

uniform temp.o

x y z x y z

T T


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Boundary Conditions at the Surface

1. Constant surface temperature

, s

2. Constant surface heat flux

(a) Finite heat flux

T

sq

0

s

xx

(b) Adiabatic or insulated surface

0T

x

3. Convection surface condition

0 0,T T tx


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General 3-D Heat Conduction cylindrical coordinates

2

r cr r r r z z t


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General 3-D Heat Conduction spherical coordinates

2 2 2 2

sinsin sin

r cr r r r r t


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1-D Heat Conduction in Flat Plate, Cylinder and Sphere

Three examples below show how to solve problems usingthe general heat conduction equation and three kinds of

boundar conditions


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Example I: Heat conduction in an infinite flat plate (1st BC)

Steady, 1-D heat conduction

If =constant the eneral heat conduction e uation becomes

2

2 0T

2

2 0 0T

(5)

Integrating Eq.(5) w.r.t.x, we have

Isothermal surface1 0dT C

dx

Integrating again gives

1 2 0T C x C The boundary conditions are

at

o o

i iT T x x

(Isothermal boundary condition)


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en, we ave

( )( )i oo oT T

T T x x

(6)i o

i.e. Tvaries linearly withx

rate of HT

Area

dTq

dx

Differentiating Eq.(6) gives

i o

i odx x x

i oT Tq

i ox x

We can see qis constant.


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Combining with Eq.(6) gives

( )o oq

T T x x

o i

i o

T Tq

x x

The thermal resistance is

i ox

x xR


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Example II: Heat conduction in 1-D flat plate (2nd BC)

2

2 0

d

d

T

x

ato oT T x x

ati iq q xx


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2d T

2

ato oT T x x

dx

ati iq q xx

Integrating the general equation twice gives

dT1

1 2

0

dx

T C x C

0dT

C

Applying one boundary condition gives

1 0i

dx

q C 1

iC


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2d T

2

ato oT T x x

dx

ati iq q xx

2 0iq

T x C Applying another boundary condition gives

2o o

i

x

qC T x

0i io oq q

T x T x

( )io oqT T x x


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Example III: Heat conduction in 1-D flat plate (3rd BC)

2

20

d T

atT T x x

( ) ati ix

TT T x

xx


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2

20

d T

xd

ato oT T x x

T

i

i i

x xx

Integrating the general equation twice gives

1 0

0

Cdx

T C x C

( )ix x

TT T

x

Applying a boundary condition gives

1 ( )iC T T


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0T T T x C

Applying another boundary condition gives

2( ) 0o i oT T T x C

2( )

o i oC T T T x

1 2 0T C x C

Substituting C1 and C2 into

gives

( ) ( ) 0

( )( )

i o i oT T T x T T T x

T T T T x x


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Heat flux is

( )i o ii o

q T Tx x

( )1

oi

T Tq T T

x x


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Heat conduction in a cylinder (1)

Uniform, steady, radial (1-D) heat transfer No longitudinal heat transfer

Concentric c lindrical isothermal surface exist

i.e. T = function of ronly

The net heat transfer into the element in radial direction is

r

r r

dQQ Q dr

dr

rdQ dQdr dr dr dr


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From Fouriers equation

dQ d dT r

dr dr dr

r

2dQ d dT

dr rL dr

2 d dT

L r dr

For steady state, Heat transfer into element = Heat transfer out of

element, Q=constant

0 (but 0 because 0)rdQ dq dA

dr dr dr

0

d dT

r drdr dr (7)


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For =constant:

0r drdr dr

0r drdr dr

. . .

1 0dT

r C (8)

1 0

r

dT C

Integrate again

1 2ln( ) 0T C r C (9)


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From Fouriers equation we have

dT dT

rdr dr

dT Qr r

Substituting into Eq.(8) gives

C Q

1 .

ln 0Q

T r C 10


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Put , then Eq.(10) becomesat ro oT T r

21ln /

o o

o

T T T T Q L

r r

2

o

o

L

T T

ln /2 od dL

Thermal resistance is

1ln /r r

2 L


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Conduction in hollow cylinder (tube)

Uniform , steady, radial (1-D) heat transfer

0d dT

rdr dr

at

at

o o

i i

T T r

T r r

r

T

The solution is

2

ln /

i o

o i

T TQ L

r r

Thermal resistance is

ln /

2r o ir r

L


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Heat conduction in a sphere (1)

Steady, radial (1-D) conduction

Net heat transfer rate into element

r r dr Q Q

rr rQ Q dr

dr

r dr dr dr dr

From Fouriers e uation

For steady state, Q = const., 0dr

0rdQ d dT

Adr dr dr

2d dT

2where 4r r dr dr


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For =constant the eneral e uation is

2

0

d dT

rdr dr

Integrating twice w.r.t. r we have

2

1 0dTr Cdr

12 0

CT C

r


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24 dT

Q rdr

2 4

dT Qr

dr

14

C

21 0

4QT C

r

2

10

4o

o

QT C

r

2 4

o

o

C Tr


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1 1Q Q

4 4

1 1

o

or r

4

o

or r

T T

4 oo

o

rr

r r

T T

4

o

o

r r

rr

4

or

o

r rR

rr


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Heat conduction in hollow sphere

For a hollow sphere with constant, uniform , steady, radial(1-D) heat transfer,

2 0d dT

rdr dr

at

at

o o

i i

T T r

r

r

T T r

The solution is

4

o iT T

Q

Heat transfer rate

o ir r

Thermal resistance

1 1 1

4r o iR r r

Conduction 1

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