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Heat Diffusion Equation

Apply this equation to a solidundergoing conduction heattransfer as shown where:

E = mcpT= (rV)cpT = r(dxdydz)cpT

Energy balance equation:

dE

dt

dE

dt

g

q q E E

dW

dtin out

1 2

x

qx qx+dx

q KA T

xK dydz

T

x

q q dq q q

xdx

xx x

x dx x x x

x

( )

= 0

Heat diffusion equation mod 11/27/01

dy

dx

y

Steady State Conduction

The above equation states:

Rate of change of total energy (dE/dt) = Rate of energy generated (dEg/dt) +

Rate of H eat Transfer in (qin) - Rate of Heat Transfer out (qout)

qin =

qout =

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r

r

c T

tq

xk

T

x yk

T

y zk

T

z

c

T

t k

T

x

T

y

T

z q k T q

wherex y z

T T

x

T

y

T

z

p

p

( ) ( ) ( )

( )

,

,

Special case1: no generation q = 0

Special case 2: constant thermal conductivity k = constant

is the Laplacian operator

Special case 3:t

and q = 0

The famous Laplace'2

2

2

2

2

2

2

2

22

2

2

2

2

2

2

2

2

2

2

2

0

0 s equation

Heat Diffusion Equation (contd, page 3)

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1-D, Steady State Conduction

Assume steady and no generation, 1- D Laplace' s equation

function of x coordinate alone

Note: ordinary differential operator is used since T = T(x) only

The general solution of this equation can be determined by integting twice:

First integration leads to

dT

dx

Integrate again T(x) = C

Second order differential equation: need two boundary conditions to

determine the two constants C and C

1

1 2

d T

dx T x y x t T x

cons t C

x C

2

2

1

2

0

, ( , , , ) ( ),

tan .

.

Example:

T(x=0)=100C=C2

T(x=1 m)=20C=C1+C2, C1=-80C

T(x)=100-80x(C)

100

20

T

x

Constant

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Thermal Resistance - Composite Wall Heat Transfer

T2T1

R1=L1/(k1A) R2=L2/(k2A)

T1 2 1 2 1 2

1 2 1 2

1 2

1 11 1 1

1 1

TAlso, q= ,

T T T T T T q

R R R L L

k A k A

T LT T qR T q

R k A

T1 T2

L1 L2

k1 k2T

Use of the thermal resistance concept make the analysis of complex geometries

relatively easy, as discussed in the example below.

However, note that the thermal heat resistance concept can only be applied forsteady state heat transfer with no heat generation.

Example: Consider a composite wall made of two different materials

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Composite Wall Heat Transfer (contd)

R1=L1/(k1A)R2=L2/(k2A)

T2T1 TT1 T2

L1 L2

k1 k2T

Now consider the case where we have 2 different fluids on either sides of the wall at

temperatures, T,1 and T,2 , respectively.

There is heat transfer by convection from the first fluid (on left) to material 1and from material 2 to the second fluid (on right).

Similar to conduction resistance, we can determine the convection resistance,

where Rconv= 1/hA

T,1 T

,2

T,1

T,2

Rconv,1= 1/(h1A)Rconv,2= 1/(h2A)

1

1

1,

12, 2,2,

R

TT

R

TT

R

TTQ

convtot

where

2,

2,

2

2 2

convR

TT

R

TT

This approach can be extended to much more complex geometries

(see YAC, 8-4 and 8-5)

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R value of insulations

Q:Why is the Rvaluegiven by L/k ?

A: From previous slide, we know that the thermal resistance to conduction is

defined as the temperature difference across the insulation by the heat flux going

through it, hence:

R T

q

T

k T x

x

k

"

Example:

The typical space inside the residential frame wall is 3.5 in. Find the R-value ifthe wall cavity is filled with fiberglass batt. (k=0.046 W/m.K=0.027 Btu/h.ft.R)

R x

k

ft

Btu h ft RR ft h Btu R

0 292

0 02710 8 11

2.

. / . .. ( . . / )

In the US, insulation materials are often specified in terms of their thermal

resistance, denoted as R values, where Rvalue= L/k(thickness/thermal conductivity)

In the US, it has units of (hr ft2F)/Btu (Note: 1 Btu=1055 J)R-11 for wall, R-19 to R-31 for ceiling.