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Chapter 4 第四章
Unsteady-State (Transient) Conduction
非稳态导热
4-1 Introduction
1. Basic conception
Unsteady-state (or transient) conduction
Temperature distribution in a system varies with time
Steady-state conduction
Temperature distribution in a system does not varies with time
2. Example
The shell of steam turbine
Before start tf1=tw1=tw2=tf2
Admission of steam tf1
Inside of the shell
q1=h1(tf1-tw1)
At a particular time
h1A1(tf1-tw1)=h2A2(tw2-tf2)
Steady-state conduction
penetration time 穿透时间:Thermal layer 穿透深度
3. Problems to be solved
1). Temperature distribution at a given time
thermal stress ( 热应力 )
2). The time to reach a given temperature distribution
or steady-state
quenching ( 淬火过程 )
3). The heat transferred
3. Infinite plate subjected to sudden cooling of surfaces
Lxx i 20,0at 0, 0,0at 0 x
t
ax
t 12
2
Set1tti
ax
12
2
0,2at 0 Lx
xXx,
2
02 aCead
d
xCxCXdx
Xd sincos0 212
2
aexCxC2
)sincos( 21
00),0( 11
2
CeC a
02sin02sin),2(2
2 LLeCL a
,2,12
nL
n
1
]2/[
2sin),(
2
n
aLnn L
xneCx
inn
ni nC
L
xnCx
4
2sin)0,(
1
1
]2/[
2sin
14 2
n
aLn
i L
xne
n
Graphical form for calculation
t
ax
t 12
2
itxt 0,
0,00
xx
tx
ftthx
tLx
,
ftt Set
Math model
Third boundary condition no heat generation,
h=const. Tf =const.
2
2
xa
00, x
00
xx
hx x
Or
xCxCeCx sincos, 321
2
xBxAex sincos,2
From the first boundary condition
0,0
x
x 0cossin
2
xBxAe
x
Separating variables gives general solution
0 02
B B e xAex cos),(
2
Lx LAheLAe cossin22
LhL cossin
From the second boundary condition
L
Bi
L
hLhL
L
L
tancos
sin
h
LhLBi
/1
/
There are infinite number solutions to the equationThe final series form of the solution is
xeAx nn
nn cos,2
1
From initial condition
1
0 cos0,n
nn xAx
nnn
ni
nnn
nin LLL
LA
cossin
sin2
cossin
sin2
Lnn
nnn
nn
n
L
i
Lxe
x n
cossin
/cossin2
,
1
22
BiBinn tan
x
BiFofx
,,,
0
FoL
a 2
4-2 lumped-heat-capacity system
Generally t=f(x,y,z, )
• If internal conduction resistance 0 , uniform in temperature
• Temperature field t=f(x,y,z, ) reduces to t=f(), 0-D problem
• In fact, impossible for internal conduction resistance to be zero,
but if it is small enough , we believe t=f().
• This method is called lumped method or heat conduction with
negligible internal conduction resistance
( 集总参数法 或 忽略物体内部导热热阻的简化分析方法 )
1. Physical problem
k=const. Uniform T distribution Bi 0 , h=const. Find T=f()
2. Mathematical model
cz
t
y
t
x
ta
t
2
2
2
2
2
2
cd
dt
tthAq
ttV
hA
V
No boundary condition. Convection heat transfer taken as source item
Then
0at 0
tt
tthAd
dtVc
Vc
tthA
d
dt
tt
0at 0
Vc
hA
d
d
3. Solve
0 0d
cV
hAd
00ln
cV
hA
cV
hA
0
ln
cV
hA
e
0
Then
cVChA
RCRcV
hAthth
thth
11
BiFocV
hA
ee
00
BiFoAV
a
A
Vh
cV
A
A
hV
cV
hA
22
2
)/(
Characteristic dimension of solid (V/A=s)
)/( AVh
Bi 2/ AV
aFo
4. Heat transfer rate
fehAehAhAtthA cVhABiFo 00
Heat transferred from 0 to
000 00 cV
hAde
hA
cVhAdehAqd cV
hA
cV
hA
cV
hA
cVhAcV
hA
ecVecVecV
11 00
0
0
5. Biot number
resistance convection
resistance conduction internal
/1
/
h
LhLBi
•Bi L/λ has decisive effect, convection waits for conduction
tw t , become first kind of boundary condition
•Bi 0 L/λ is very little , temperature tends towards uniform
•Bi = some certain value , L/λ and 1/h play an important role
6. Applicability of lumped-capacity Analysis
1.0)/(
AVh
Bi
Characteristic dimension of solid (V/A=s)
) thicknesshalf ( Plate LLA
AL
A
Vs
(radius) Cylinder rs
(radius) Sphere rs
•Plane wall BiV= Bi Cylinder BiV= Bi/2 Sphere BiV= Bi/3
•M= 1 (for plane wall), ½ (for cylinder) and 1/3 (for sohere)•Reason M= BiV /Bi
MAVh
BiV 1.0)/(
) thicknesshalf ( wallPlane LLA
AL
A
V
22Cylinder
2 R
Rl
lR
A
V
34
3/4Sphere
2
3 R
R
R
A
V
Characteristic dimension of solid (V/A=s)
In our country
7. Time constantceeΘ cV
hA
0
constant timeis hA
cVc
36.8%
0
5%
cc3
The rate of temperature change
tan1
0
ced
d
c
2/,tan,0 c
1,0,0tan, Θc 02/,0 c c rate
4-3 Transient heat flow in a semi-infinite solid
1. Physical problem
i
i
ttx
ttx
tt
t
ax
t
,
,0
,0
1
0
2
2
i
i
x
x
ax
,
0,0
,0
12
2
Introduction of a new variable
mCx
2. Mathematical model
0tt
ditermined be to & constants where mC
d
dm
d
dCxm m
1
d
dC
xxm
2
222
2
2
d
dCd
dCd
d
xxxxmm
Substitute into the eq.
2
2122
d
dac
d
dm m
To eliminate , set 2m +1=0, that is m = -1/2
Then 02
122
2
d
d
aCd
d
a
xCx
aC m
4 have we
41set
022
2
d
d
d
d
12ln2 C
d
dd
dddd
d
mCx
20 11
22
, CdeCeCd
d
0,0,0,0 2 Cx From initial condition
0,,0
0
11010
2
2
2
CCdeC
Then
a
xerfde
ax
4
2 4/
00
2
functionerror is 4 a
xerf
2 known when is it fig.2 From
%5error ,9953.0 ,00
TT
3. Two important parameters:
0,4 ax
xa
xo at ,,
16
2
4. Heat flow
erfx
ttAx
tA ix
0
aettA a
x
i2
12 20
a
ttA i )( 00
da
ttAd i
0
0
0 0
)( ittcA 02
qckq ,
5. Constant heat flux on semi-infinite solid
Governing eq. and initial condition are the sameBoundary condition
0for 0
0
xx
t
A
a
xerf
A
x
a
x
A
att i
21
4exp
/2 02
0
6. Energy pulse at surface
Instantaneous pulse of energy Q0/A
a
x
acAtt i 4
exp2
0
as allfor 0 xtt i
4-4 Convection boundary conditions
1.Convection boundary For the semi-infinite-solid problem, the boundary condition
0for )(0
0
x
x x
ttthA
The solution
ah
Xerfahhx
erfXtt
tt
i
i 1exp12
2
uret temperatenvironmen
solid of re temperatuinitial
2/
t
t
axX
i
2. Important case
1-D solids suddenly subjected to convection environment at T∞
Infinite plate
nnn
nn
n
L
i
Lxe
x n
cossin
/cossin2
,
1
22
x
BiFof ,,
0
,,r
rBiFof
i
0hr
Bi 20r
aFo
Infinite cylinder and sphere
• the nature of series: the first term is the largest, then the 2nd, 3rd …
when Fo is large enough the decrease is rapid.
• when Fo>0.2, the the error of the single-term approximation of the
series is less than <0.1% 。
111
11
cossin
cossin2
, 221
L
x
ex L
i
The ratio of (x,) to 0 is
L
xxx1
0
cos),0(
,,
It depends on the position and boundary condition, is independent of .
Initial condition has no effect on temperature distribution
Take infinite plate as an example
This region is called regular regime ( 正规热状况 )
or fully developed regime ( 充分发展阶段 )
0
0,
ii
x Fig. 4-13
ii cVTTcVQ 0
Fig. 4-7
Chart solution
3. Boit number and Fourier number
/hsBi
22 css
aFo
4. Applicability of the Heisler Charts
2.0Fo
4-5 Multidimensional systems
1. Physical problem: Infinite rectangular bar
2. Mathematical model
cz
t
y
t
x
ta
t
2
2
2
2
2
2
TThy
tLy
y
ty
tthx
tLx
x
tx
tt
y
T
x
Ta
T
i
,
0,0
,
0,0
,0
2
1
2
2
2
2
Set dimensionless temperature difference
tt
ttΘ
ii
Substitute into eq.
0
,,,,
0,,
0,,
0,,
,,
10,,
2
1
2
0
0
1
2
2
2
2
Ly
y
x
Lx
y
yx
hLx
y
yx
x
yx
x
yx
hyL
yx
yxa
t
If equation
0
,,
0),(
10,
1
111
0
1
1
12
1
Lx
x
x
xt
hLt
x
xt
xtx
ta
t
Has a solution T1(x,), and
0
,,
0,
10,
,,
2
222
0
2
2
22
22
Ly
y
y
xt
hLt
y
yt
yt
y
yta
yt
has a solution of T2(y,)
,,then 21 ytxt
is the solution to the equation
Prove:
performing partial
differentiation of
1
22
1
, tt
tt
yx
x
tt
x
1
2x
tt
x
12
22
2
In a similar way
22
2
12
2
y
tt
y
3. Solution
000 2121
21
222
22
122
2
121
2
21
22
1
ttx
ta
tt
y
ta
tt
y
tt
x
tta
tt
tt
(x,y,)= t 1(x, ) t2 (y, ) satisfies the governing equation, and
1110,0,,, 21 ytxtyx also satisfies the initial condition.
00,,
,,,
20
12
0
yt
x
xtyt
x
yx
xx
11
,,,,
,,,, 1
22111LxLx x
xtyt
hytLt
x
yx
hyL
00,,
,, 21
112
1
ytx
xt
hLtyt
Lx
By the same way, it can be proved that (x,y,)= t 1(x, ) t2 (y, ) satisfy other two boundary conditions.
Substituting these relations in the governing equation
By the same way
* Constant 1st and 3rd kinds of boundary conditions
3 solidonintersecti
2 solidonintersecti
1 solidonintersecti
solidcombined
iiii
4. Heat transfer in multidimensional systems
201030102010total0
111Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
4-6 Transient numerical method
•
Space coordinate x 1N Space increment x
1. Mathematical models
2. Discretization of domain (node, grid)
3. Algebraic equations for all the nodes
4. Initial variable fields
5. Solving the algebraic equation
6. Analysis of the solutionTime coordinate 1I Time increment( 时间步长 )
T(p)n is the temperature of node (n,p)
0t
• Procedure is the same as steady-state
1-D, λ=const. problem
2
2
x
ta
t
2
)(1
)()(1
)()1( 2
x
ttta
tt pn
pn
pn
pn
pn
2
)1(1
)1()1(1
)()1( 2
x
ttta
tt pn
pn
pn
pn
pn
Implicit finite difference scheme ( 隐示格式 )
)(2
)(1
)(12
)1( 21 p
np
np
np
n tx
att
x
at
Governing equation
Forward difference
backward difference
Explicit finite difference scheme ( 显示格式 )
)1(2
)1(1
)1(12
)( 21
pn
pn
pn
pn t
x
att
x
at
)(2
)(1
)(12
)1( 21 p
np
np
np
n tx
att
x
at
Explicit formulation
If the time and space increment are chosen so that
22
a
xM
)(1
)(1
)1(
2
1 pn
pn
pn TTT
• If x△ and △ the rate that the solution proceeds the accuracy
• △ depends x△ , If M>2 the coefficient becomes negative,
the condition violates the 2nd law of thermodynamics.
• Stability problem, M<2 stable, M>2 instable
Boundary
)(,
)1(,)(
,
)(,
)(1,
)(,
)(1,
)(,
)(,1
2)(
22pnm
pnmp
nm
pnm
pnm
pnm
pnm
pnm
pnm
tty
xcttyh
y
ttx
y
ttx
x
tty
)(
,
2)(
1,)(
1,)(
,12)1(
, 4222 pnm
pnm
pnm
pnm
pnm t
xh
a
xtttt
xh
x
at
For uniform grid x= y
)(2
)(12
)1( 2222 pm
pm
pm T
k
xh
a
xTT
k
xh
x
aT
For 1-D problem
case D-2 for the )2/(2
case D-1 for the )1/(22
kxh
kxh
a
x
case D-2 for the )2/(2
case D-1 for the )1/(22
kxh
kxh
a
x
Zero coefficient
To insure convergence
Set 2x
aFo
Fourier number in the numerical format
TBiFoTFoFoBiFoTT p
Mp
Mp
M 22221 )(1
)()1(
)()(1
)(1
)1( 21 pm
pm
pm
pm TFotTFoT
Infinite plate
Convection boundary node
Interior node
k
xhBi
Biot number in the numerical format
4-6 Thermal resistance and capacity formulation
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