Heat Diffusion Equation
All go to zero
dE dE g
dW


Energy balance equation:

 qin  qout  E1  E2 
dt
dt
dt
Apply this equation to a solid undergoing conduction heat transfer:
E=mcpT=(rV)cpT=r(dxdydz)cpT
dx
dy
qx+dx
qx
T
q x   KA
x
y
x
q x  dx
x
T
  K ( dydz )
x
q x
 q x  dq x  q x 
dx
x
x
Heat Diffusion Equation (2)
Energy Storage = Energy Generation + 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
rcp
 q  (k
)  (k
)  (k
)
t
x x
y y
z z
Note: partial differential operator
is used since T=T(x,y,z,t)
Generalized to three-dimensional
Heat Diffusion Equation (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 Heat Transfer
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
First integration leads to
 cons tan t  C1 .
dx
Integrate again T(x) = C 1 x  C2
Second order differential equation: need two boundary conditions to
determine the two constants C 1 and C 2 .
T
100
20
T(x=0)=100°C=C2
T(x=1 m)=20°C=C1+C2, C1=-80°C
T(x)=100-80x (°C)
x
1-D Heat Transfer (cont.)
Recall Fourier' s Law:
dT
q = -kA
, If the temperature gradient is a constant
dx
q = constent (heat transfer rate is a constant)
dT T2  T1

, where T ( x  0)  T1 , T ( x  L)  T2
dx
L
T1
T2
x
L
T1  T2
T1  T2
dT
q   kA
 kA

dx
L
( L / kA)
T1  T2
L
q
, where R 
:thermal resistance
R
kA
q (I)
T1 (V1)
Electric circuit analogy
T2 (V2)
R (R)
I = (V1-V2)/R
T
T1 k1
Composite Wall Heat Transfer
k2
T2
R1=L1/(k1A)
T1
L1
L2
R2=L2/(k2A)
T2
T
T1  T2 T1  T2
T1  T2
q


R
R1  R2  L1   L2 



k
A
k
A
 1   2 
 L1 
T1  T
Also, q=
, T  T1  qR1  T1  q 

R1
k
A
 1 
Note: In the US, insulation materials are often specified in terms of
their thermal resistance in (hr ft2 °F)/Btu ----> 1 Btu=1055 J.
R-value = L/k, R-11 for wall, R-19 to R-31 for ceiling.
R value
The thermal resistance of insulation material can be characterized
by its R-value. R is defined as the temperature difference across the
insulation by the heat flux going through it:
T
T
x
R


q" k T
k
x
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)
R
x
0.292 ft

 10.8( R. ft 2 . h / Btu)  R  11
k
0.027 Btu / h. ft. R