Transcript Basic law of heat conduction --Fourier’s Law T Φ  A L T q  k x W  q Φ T k A L Degree Celsius W  m2    Physical mechanism of heat-conduction Conduction may be.

Basic law of heat conduction
--Fourier’s Law
T
Φ  A
L
T
q  k
x
W 
q
Φ
T
k
A
L
Degree Celsius
W
 m2 
 
Physical mechanism of heat-conduction
Conduction may be viewed as the transfer of energy
from the more energetic to the less energetic particles
of a substance due to interactions between the particles
The mechanism in gases: The temperature at any
point can be associated with the energy of gas
molecules in proximity of the point. The energy is
related to the random transitional motion, as well as the
rotational and vibrational motions, of the molecules.
The dependence of
thermal conductivities of
gases on temperature
The faster the molecules move,
the faster they will transport
energy. Therefore the thermal
conductivity of a gas should
be dependent on temperature.
The thermal conductivity of a
gas varies with the square root
of the absolute temperature.
For most gases at moderate pressures the thermal
conductivity is a function of temperature alone.
The mechanism of heat conduction in liquids is
similar to that in gases, but the molecules are more
closely spaced and the interactions between
molecules are more stronger and more frequently.
Two modes of heat conduction in solids:
Lattice vibration: energy transfer may be
attributed to atomic activities in the form of
lattice(晶格，格子) vibrations.
Free electrons transport: large number of free
electrons are moving about in the lattice structure,
and carry thermal energy from higher temperature
region to lower temperature region.
Governing equations
One-dimensional heat conduction equation:
Energy conducted in left face+heat generated
within element=change in internal energy+energy
conducted out right face
T
x
Energy generated within element  qAdx
Energy in left face  q x  kA
T
dx
Change in internal energy  cA
t
T
T  T

q


kA
]


A
[
k
 (k )dx]
Energy conducted out right face
x  dx
x  dx
x
x x x
Combining the above relations gives:
 kA
or
T
T
T  T
 qAdx  cA dx  A[k
 (k )dx]
x
t
x x x
 T
T
(k )  q  c
x x

 T
T
(k )  q  c
x x

For constant thermal conductivity, the above
equation is written
 2T q 1 T
 
2
x
k  
k

c
 is called thermal diffusivity. The larger the value of  ,
the faster heat will diffuse through the material.
3-dimensional heat conduction equations
Cartesian coordinates:
 2T  2T  2T q 1 T
 2  2  
2
k  t
x
y
z
Cylindrical coordinates:
 2T 1 T 1  2T q 1 T

 2 2 
2
r r r 
k  t
r
Spherical coordinates:
1 2
1

T
1
 2T q 1 T
(rT )  2
(sin  )  2 2
 
2
2
r r

k  t
r sin  
r sin  
Newton’s law
Φ  hA(Tw  T ) W
q Φ A

 h(Tw  T f ) W m 2

Emissive power of black body,
Stefan-Boltzmann law
Eb   bT 4
W m 
2
Grey body emissive power
E  bT 4
W m 
2
RADIATION IN AN ENCLOSURE
q   (T14  T24 )
W m2


2-2 THE PLANE WALL
For single layer
kA
q  (T1  T2 )
x
Rth 
x
kA
For multiple layers
q
T1  T4
x A xB xC


k A A k B A kC A
x A xB xC
Rth 


k A A k B A kC A
2-4 RADIAL SYSTEMS
Cylinders
2kL(Ti  To )
q
ln( ro / ri )
ln( ro / ri )
Rth 
2kL
q
2L(T1  T4 )
ln(r2 / r1 ) / k A  ln(r3 / r2 ) / k B  ln(r4 / r3 ) / kC
Rth  ln( r2 / r1 ) / k A  ln( r3 / r2 ) / k B  ln( r4 / r3 ) / kC
Spheres
Consider a shell, the temperatures at the inside and outside walls
are maintained at Ti and To, respectively, the heat flow through
the shell is
A  4r 2
To
2 dT
q   k  4r 
dr
ro
To
dr
q  2  4k  dT
ri r
Ti
Ti
1 1
q (  )  4k (Ti  To )
ri ro
4k (Ti  To )
q
1 / ri  1 / ro
Rth 
1 1 1
(  )
4k ri ro
Convection boundary condition
Convection heat transfer rate is
qconv  hA(Tw  T )
Rearranging gives: qconv
Tw  T

1 / hA
Convection resistance  1/hA
2-6 CRITICAL THICKNESS OF INSULATION
Consider the right tube. The inner
wall is maintained at Ti; the outer
surface is exposed to a convection
environment. From the thermal
network the heat transfer is
q
2L(Ti  T )
ln( ro / ri ) 1

k
ro h
The maximum condition is
dq
0
dro
1
1
 2)
kro hro
ln( r / r ) 1
[ o i  ]2
k
ro h
 2L(Ti  T )(
k
Which gives ro 
h
2-7 HEAT SOURCE SYSTEMS
Plane wall with heat source
The heat source is uniformly distributed in the plane wall,
calculate the temperature distribution in the plane wall.
Governing equation is
d 2T q
 0
2
k
dx
With boundary condition:
T  Tw
at
x  L
General solution is
q 2
T   x  C1 x  C2
2k
The temperature on each side is the same
C1=0
C2 is the temperature at the midplane
To  C2
Therefore
q 2
T  To 
x
2k
or
T  To
x 2
( )
Tw  To
L
How to get To?
Total heat generated must equal to the heat lost at the faces.
dT 
2( kA  )  qA2 L
dx  x  L
By differentiating equation
T  To
x 2
( )
Tw  To
L
gives
dT 
2x 
2

(
T

T
)(
)

(
T

T
)
w
o
w
o
2 
dx  x  L
L
L  x L
then
 k (Tw  To )
and
qL
To 
 Tw
2k
2
2
 qL
L
Alternative form of temperature distribution
T  Tw
x 2
 1 ( )
T0  Tw
L
T  Tw
x 2
 1 ( )
T0  Tw
L
dT
qx  k
dx
L
Tw
x
T
q  xdx  k  dT
q 2
( L  x 2 )  k (T  Tw )
2
q 2
L  k (To  Tw )
2
T  Tw
x 2
 1 ( )
T0  Tw
L
2-8 CYLINDER WITH HEAT SOURCES
d dT
 qr
(r ) 
dr dr
k
Governing equation
d 2T 1 dT q

 0
2
r dr k
dr
(A)
Boundary condition
T  Tw
at
rR
Heat generated equals heat
lost at surface:
qR 2 L   k 2RL
dT 
dr  r  R
At the center of the cylinder
dT
0
dr
at
r 0
d 2T dT
qr
r



Rewrite (A)
k
dr 2 dr
2
Note that r
d T dT d dT

 (r )
2
dr dr dr
dr
Integration yields
dT  qr 2
r

 C1
dr
2k
 qr 2
and T 
 C1 ln r  C2
4k
 qR  qR C1
From red dT 




dr  r  R
2k
2k
R
thus C1  0
 qR 2
T  Tw 
 C2
4k 2
qR
C2  Tw 
4k
Final solution
q 2 2
T  Tw  ( R  r )
4k
at r  R
T  Tw
r 2
Dimensionless form
 1 ( )
T0  Tw
R
To is the temperature of center
qR 2
To 
 Tw
4k
For a hollow cylinder
T  Ti
at r  ri (inside _ surface)
T  To
at r  ro (outside _ surface )
The general solution is
qr 2
T 
 C1 ln r  C2
4k
Using boundary conditions yields
q
r
2
2
T  To 
(ro  ri )  C1 ln
4
ro
where
Ti  To  q(ri 2  ro 2 ) / 4k
C1 
ln( ri / ro )
2-11 THERMAL CONTACT RESISTANCE
T2 B  T3
T1  T2 A T2 A  T2 B
q  kA A

 kB A
x A
1 / hc A
xB
T1  T3
q
x A / k A A  1 / hc A  xB / k B A
1 / hc A —thermal contact resistance
hc —contact coefficient
T2 A  T2 B
T2 A  T2 B T2 A  T2 B
q
 k f Av

Lg / 2k A Ac  Lg / 2k B Ac
Lg
1 / hc A
Ac—the contact area
Av—the void area
Lg—the thickness of the void space
Kf—thermal conductivity of the fluid which fills the void space
1 Ac 2k A k B Av
hc  (
 kf )
Lg A k A  k B A
3-2 MATHEMATICAL ANALYSIS OF TWODIMENSIONAL HEAT CONDUCTION
Problem: determine temperature
distribution in a rectangular plate
T  T1
at
y0
T  T1
at
x0
at x  W
x
T  Tm sin( )  T 1 at
W
(4)
T  T1
yH
Tm is the amplitude of the sine function
Approach: separation of variables method.（分离变量法）
T  XY
where
X  X ( x)
Y  Y ( y)
(5)
Substituting T=XY into the Laplace equation gives
 2T  2T
d 2T
d 2T

 0Y 2  X 2  0
x 2 y 2
dx
dy
1 d 2 X 1 d 2Y


X dx 2 Y dy 2
(6)
Note that each side of the equation is independent of the other
d2X
2


X 0
2
dx
(7)
d 2Y
2


Y 0
dy 2
(8)
 —Separation constant, determined from the boundary conditions.
3 possible solutions
 0
2
X  C1  C2 x
Y  C3  C4 y
d2X
2


X 0
2
dx
d 2Y
2


Y 0
2
dy
T  (C1  C2 x)(C3  C4 y)
(7)
(8)
it is impossible
2  0
X  C5e x  C6ex
Y  C7 cos y  C8 sin y
T  (C5e x  C6ex )(C7 cos y  C8 sin y)
it is impossible
2  0
X  C9 cos x  C10 sin x
Y  C11e y  C12e y
T  (C9 cos x  C10 sin x)(C11e y  C12ey )
possible solution
T  (C9 cos x  C10 sin x)(C11e y  C12ey )
Substitution
  T  T1
  (C9 cos x  C10 sin x)(C11ey  C12ey )
Boundary conditions
 0
 0
at
at
 2  2
 2 0
2
x
y
y0
x0
 0
at x  W
x
  Tm sin
at y  H
W
Appling these conditions, we have
0  (C9 cos x  C10 sin x)(C11  C12 )
0  C9 (C11e y  C12ey )
0  (C9 cos W  C10 sin W )(C11e y  C12ey )
x
Tm sin  (C9 cos x  C10 sin x)(C11e H  C12e H )
W
Accordingly,
C11  C12
C9  0
(a)
(b)
(c)
(d)
  (C9 cos x  C10 sin x)(C11ey  C12ey )
0  C10C12 sin W (ey  e y )
From ©
This require
n
W
Final solution form
thus
sin W  0

e y  e  y
双曲函数 sinh y 
2
nx
ny
  T  T1   Cn sin
sinh
n 1
W
W

Appling the final solution gives
Tm sin
x
W

  Cn sin
n 1
nx
nH
sinh
W
W
Which requires that Cn=0 for n>1, therefore the final solution is
T  Tm
T  Tm sin(
x
W
)  T1
sinh(y / W )
x
sin( )  T1
sinh(H / W )
W
at
yH
4x
T  Tm sin(
)  T1
W
at
yH
Problem: determine the temperature distribution in a rectangular plate.
T  T1
T  T1
T  T1
T  T2
at y  0
at x  0
at x  W
at y  H
Using the first 3 boundary conditions, obtain the solution

T  T1   Cn sin
n 1
nx
ny
sinh
W
W
Appling the fourth boundary condition gives

nx
nH
T2  T1   Cn sin
sinh
W
W
n 1
Expanding T2-T1 into Fourier series gives
(1) n1  1 nx
T2  T1  (T2  T1 ) 
sin
 n1
n
W
2

(a)
Comparing (a) and (b) gives
1
(1) n1  1
Cn  (T2  T1 )

sinh(nH / W )
n
2
(b)
Final solution is
T  T1 2  (1) n1  1 nx sinh( ny / W )
 
sin
T2  T1  n1
n
W sinh( nH / W )
3-3 GRAPHICAL ANALYSIS(作图法）
q  kx(1)
T
y
Toverall
N
M
M
q  kToverall  k (T2  T1 )
N
N
T 
S
M
—Conduction shape factor
N
Methods of constructing flux plots（通量图）
1. Trial-and error method(试错法）
2. Experimental measurement
2E 2E
 2 0
2
x
y
3-4 THE CONDUCTION SHAPE FACTOR
Definition of conduction shape factor
q  kSToverall
The calculation of inverse hyperbolic cosine
cosh 1 x  ln(x  x 2  1)
Separate shape factors of 3-dimensional wall
A
L
 0.54 D
S wall 
Sedge
Scorner  0.15L
A = area of wall
L = wall thickness
D = length of edge
3-5 NUMERICAL METHOD OF ANALYSIS
Finite difference form of heat equation
 2T  2T
 2 0
2
x
y
Tm1,n  Tm,n
T 


x  m1/ 2,n
x
Tm,n  Tm1,n
T 


x  m1/ 2,n
x
Tm,n1  Tm,n
T 

y  m,n1/ 2
y
Tm,n  Tm,n1
T 


y  m,n1/ 2
y
 2T 
x 2  m,n
T 
T 

x  m1/ 2,n x  m1/ 2,n Tm1,n  Tm1,n  2Tm,n


x
(x) 2
 2T 

2
y  m ,n
T 
T 

y  m ,n1/ 2 y  m ,n1/ 2
y
Tm,n1  Tm,n1  2Tm ,n

(y ) 2
Tm1,n  Tm1,n  2Tm,n Tm,n1  Tm,n1  2Tm,n

0
2
2
(x)
(y )
If
x  y
Tm1,n  Tm1,n  Tm,n1  Tm,n1  4Tm,n  0
Conclusion: the net heat flow into any node is zero at steady state conditions
The finite difference scheme of governing equation with heat source.
Tm1,n  Tm1,n  2Tm,n Tm,n1  Tm,n1  2Tm,n q

 0
2
2
k
(x)
(y )
If
x  y
q(x) 2
Tm1,n  Tm1,n  Tm,n1  Tm,n1 
 4Tm,n1  0
k
Convection boundary nodal equations
Equations for plane surface nodes
 ky
Tm,n  Tm1,n
x Tm,n  Tm,n1
x Tm,n  Tm,n1
k
k
 hy (Tm,n  T )
x
2
y
2
y
If
x  y
Tm,n (
hx
hx
1
 2) 
T  (2Tm1,n  Tm,n1  Tm,n1 )  0
k
k
2
External corner with convection boundary
k
If
y Tm,n  Tm1,n
x Tm,n  Tm,n1
x
k
 h (Tm,n  T )  h
2
x
2
y
2
x  y
2Tm,n (
hx
hx
 1)  2
T  (Tm1,n  Tm,n1 )  0
k
k
yh
x
(Tm,n  T )
2
(Tm,n  T )
2
3-6 NUMERICAL FORMULATION IN TERMS OF
RESISTANCE ELEMENTS
qi  
j
T j  Ti
Ri j
0
Tm1,n  Tm,n Tm1,n  Tm,n Tm,n1  Tm,n Tm,n1  Tm,n



 x 2 q  0
x / ky
x / ky
y / kx
y / kx
q(x) 2
Tm1,n  Tm1,n  Tm,n1  Tm,n1 
 4Tm,n1  0
k
3-7 GAUSS-SEIDEL ITERATION （高斯－赛德尔迭代）
from
qi  
j
We obtain
T j  Ti
Ri j
 0  qi  
j
Tj
Rij
 Ti 
j
1
0
Rij
qi   (T j / Rij )
Ti 
j
 (1 / Rij )
j
The solution can be obtained by Gauss_Seidel Iteration:
1. An initial set of values for the Ti is assumed.
2. Using the most recent values of Tj, the new values of Ti are
calculated from the above equation
3. The process continues until the following requirements are
satisfied.
Ti
n 1
 Tin   
or

Tin 1  Tin
Tin
Biot number
hx
 Bi
k
By setting Bi=0, the convection boundary
can be converted into insulated boundary.
Heat sources and boundary radiation exchange
qi / k

qi  q V
For radiation exchange at boundary node
''
qi  qrad
,i  A
Net radiation transferred to node i per unit area
''
qrad
,i   (Tr  Ti )
4
4