Steady State Conduction
=0
Heat Diffusion Equation
dE dE g
dW
Energy balance equation:

 qin  qout  E1  E 2 
dt
dt
dt
The above equation states:
Rate of change of total energy (dE/dt) = Rate of energy generated (dEg/dt) +
Rate of Heat Transfer in (qin) - Rate of Heat Transfer out (qout)
Apply this equation to a solid undergoing conduction heat
transfer as shown where:
E = mcpT= (rV)cpT = r(dxdydz)cpT
dx
dy
T
qin = q x   KA
x
y
x
qx
Heat diffusion equation – mod 11/27/01
qx+dx
qout = q x  dx
x
T
  K ( dydz )
x
q x
 q x  dq x  q x 
dx
x
x
Heat Diffusion Equation (cont’d, page 2)
Substituting in the Energy Balance equation, we obtain:
dEg/dt
EnergydE/dt
Storage = Energy
Generation
+
qin -qout
Net Heat Transfer

r c pT  dxdydz  qdxdydz  qx  qx  dx

t
qx
T
rcp
dxdydz  qdxdydz  qx  (qx 
dx)
t
x

T
 qdxdydz  (k
)dxdydz
x
x
T

T

T
 T
 q  (k
)  (k
)  (k
)
=> r c p
t
x x
y y
z z
Note: partial differential operator
is used since T = T(x,y,z,t)
Generalized to three-dimensions
Heat Diffusion Equation (cont’d, page 3)
T

T
 T
 T

rc p
 q  (k )  (k )  (k )
t
x x
y y
z z
Special case1: no generation q = 0
Special case 2: constant thermal conductivity k = constant
T
2T 2T 2T
rc p
 k ( 2  2  2 )  q  k 2 T  q,
t
x
y
z
2
2
2



where  2  2  2  2 is the Laplacian operator
x
y
z

Special case 3:
 0 and q = 0
t
2
2
2

T

T

T
2
 T  2  2  2  0, The famous Laplace' s equation
x
y
z
1-D, Steady State Conduction
Assume steady and no generation, 1 - D Laplace' s equation
d 2T
 0, T ( x, y, x, t )  T ( x ), function of x coordinate alone
2
dx
Note: ordinary differential operator is used since T = T(x) only
The general solution of this equation can be determined by integting twice:
dT
 cons
tan t  C1 .
Constant
dx
Integrate again T(x) = C 1 x  C2
First integration leads to
Second order differential equation: need two boundary conditions to
determine the two constants C 1 and C 2 .
Example:
T(x=0)=100°C=C2
T(x=1 m)=20°C=C1+C2, C1=-80°C
T(x)=100-80x (°C)
T
100
20
x
1-D Heat, Steady State Heat Transfer
Thermal Resistance (Electrical Circuit Analogy)
Recall Fourier' s Law:
dT
q = -kA
, If the temperature gradient is a constant
dx
Constant (heat transfer rate is a constant)
q = constent
T1
dT T2  T1

, where T ( x  0)  T1 , T ( x  L)  T2
dx
L
T1  T2
T1  T2
dT
T2
q   kA
 kA

dx
L
( L / kA)
x
L
T1  T2
L
q
, where R 
:thermal resistance
R
kA
q (I)
T1 (V1)
This is analogous to an electrical circuit
T2 (V2)
R (R)
I = (V1-V2)/R
Thermal Resistance - Composite Wall Heat Transfer
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 for
steady state heat transfer with no heat generation.
Example: Consider a composite wall made of two different materials
R1=L1/(k1A)
T
T1 k1
k2
T1
T2
R2=L2/(k2A)
T2
T1  T2 T1  T2T
T1  T2
q


R
R1  R2  L1   L2 



k
A
k
A
 1   2 
L1
L2
 L1 
T1  T
Also, q=
, T  T1  qR1  T1  q 

R1
k
A
 1 
Composite Wall Heat Transfer (cont’d)
•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 1
and from material 2 to the second fluid (on right).
• Similar to conduction resistance, we can determine the convection resistance,
where Rconv = 1/hA
R1=L1/(k1A)
Rconv,1= 1/(h1A)
T
T1 k1
k2
T2
T,1
where
T,1
T1
Q 
T,2

L1
R2=L2/(k2A)
L2
T
T , 2  T, 2
Rtot
T  T2
R2

Rconv,2= 1/(h2A)
T,2
T2

T , 2  T1
Rconv ,1

T1  T
R1
T 2  T , 2
Rconv , 2
This approach can be extended to much more complex geometries
(see YAC, 8-4 and 8-5)
“R” value of insulations
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 ft2 °F)/Btu (Note: 1 Btu=1055 J)
R-11 for wall, R-19 to R-31 for ceiling.
Q: Why is the Rvalue given 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:
T
T
x
R



T
q" k
k
x
Example:
The typical space inside the residential frame wall is 3.5 in. Find the R-value if
the wall cavity is filled with fiberglass batt. (k=0.046 W/m.K=0.027 Btu/h.ft.R)
x
0.292 ft
R

 10.8( R. ft 2 . h / Btu)  R  11
k
0.027 Btu / h. ft. R