Chapter 4 Unsteady state conduction.pdf
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Transcript of Chapter 4 Unsteady state conduction.pdf
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Sheu Long Jye Dep. of Mechanical Engineering
Chung Hua University [email protected]
Chapter 4 Unsteady state conduction (Text: J. P. Holman, Heat Transfer, 8th ed., McGraw Hill, NY)
Consider 1-D unsteady state conduction problem in a slab (Fig. 4-1)
2
2
0
1 , 0 , 0
0 , 0
0 0, 0
( ) 0 , 0
T q T x L tx k t
T T in x L tT at x txTk h T T at x L tx
α
∞
∂ ∂+ = ≤ ≤ >
∂ ∂= ≤ ≤ =
∂= = >
∂∂
+ − = = >∂
There are totally 7 parameters: 0, , , , , ,k q L T h and Tα ∞ Nondimensinoalization,
02
02 2
2 2 2
20 0
2
2
2
20
,
1
( ) ( )
.( )
T TxletL T T
T T q TL k t
qL L Tk T T T T t
G
t qLwhere Fourier No G dimensionaless heat sourceL k T T
ξ θ
θξ α
θξ α
θ θξ τ
ατ
∞
∞
∞
∞ ∞
∞
−= =
−
− ∂ ∂+ =
∂ ∂
∂ ∂+ =
∂ − − ∂
∂ ∂+ =
∂ ∂
= = = =−
With initial and boundary conditions
1 0 1, 0
0 0, 0
0 0, 0.
.
in
at
Bi at
hLwhere Bi Biot Nok
θ ξ τθ ξ τξθ θ ξ τξ
= ≤ ≤ =∂
= = >∂∂
+ = = >∂
= =
There are only 2 paramters: G and Bi. The physical meanings of Fo and Bi:
2 3
2 3 3
( / ) ./
t k L L rate of heat conduction across L in L FoL cL t rate of heat storage in Lα
ρ= = =
The larger the Fourier No. is, the deeper is the penetration of heat inot a solid over a given period of time.
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Sheu Long Jye Dep. of Mechanical Engineering
Chung Hua University [email protected]
/1/
/
hL L kA conduction resistanceBik hA convection resistanceh heat transfer coefficient at surface of a solid
k L internal conductance of solid across L
= = =
= =
Biot number is the ratio of the heat transfer coefficient to unit conductance of a solid over the characteristic length. For solids in the shape of a slab, long cylinder, or a sphere with no internal heat generation, 1-D, transient temperature distribution within solid may be considered uniform if
0.1 5%,ss
hL VBi with error Lk A
= < < =
If we analyze systems which are considered uniform in temperature at any instant during the transient conduction, this type of analysis is called lumped heat capacity system. Consider a hot ball immersed in a cool run of water and assume that lumped heat capacity method might be used. The convection heat loss form the body is evidenced as a decrease in internal energy of the body (Fig. 3-2)
0
0
( ) , (0)
, .hA ttcV
dThA T T cV T Tdt
T T cVe e time constantT T hA
ρ τ
ρ
ρτ
∞
− −∞
∞
− = − =
−= = = =
−
When the internal resistance of the body is significient, the temperature may not be assumed uniform and heat equation for 1-D slab problem is
2
2
1T Tx tα
∂ ∂=
∂ ∂
Consider the heat equation
2
2
0
1 , 0 , 0
( ,0) , 0 , 0(0, ) , 0
i
T T x tx t
subject to IC and BCsT x T x tT t T t
α∂ ∂
= < < ∞ >∂ ∂
= < < ∞ =
= >
Solving by Laplace transform defined as
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Sheu Long Jye Dep. of Mechanical Engineering
Chung Hua University [email protected]
0
[ ] ( ) , [ '] (0)stL f f t e dt L f sf f∞ −= = −∫ .
The Laplace transform of the governing equations gives
2
2
0
/0
1 10
0
(0, )
, /
[ ] ( ) , [ ]2 2
2( ) 1 ( ) 1 , ( 1)
qx qxi
s xi i
s x
i i
x
dT s TTdx
TT ss
TT Ae Be q ss
The BCs givesT T TT es s
x e xT L T T T T erfc L erfcst t
where erfc x erf x e d Table A
α
α
η
α α
α
α α
ηπ
−
−
−
− −
−
− = −
=
= + + =
−= +
⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟= = + − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟
⎝ ⎠
= − = − −∫
∵
2
0
0
0
0 4
( ) 1 ( )2 2
( )2
"
i
i
i
xi t
T T x xerfc erfT T t tT T xerfT T t
T qq kx A
T Tk et
α
α α
α
πα
−
−= = −
−−
=−
∂= − =
∂
−=
If heat flux boundary condition is applied,
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Sheu Long Jye Dep. of Mechanical Engineering
Chung Hua University [email protected]
2
2
"0
0
/
"0
0
1 1/ 2 4
" "1 1/ 20 04
[ ] 2( )2
2[ ] ( )2
x
s xi
x
s x xt
xt
i
Tk qxTT Aes
qTkx s
e t xL e xerfcs ts
q q xt xT L T T e erfck k t
α
αα
α
απ α
α
απ α
=
−
=
−−−
−−
∂− =
∂
= +
∂− =
∂
⎛ ⎞⎜ ⎟⎛ ⎞⎜ ⎟= − ⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎝ ⎠
⎛ ⎞= = + − ⎜ ⎟⎝ ⎠
∵
If convective boundary condition is applied, the solution is
2
00
00
1
( )
( )( ) ,( ) ( / )
1 1[ ]( ) 2 2
2
xx
qxi
xx
iqxi i
qxx t
h xi k
i
Tk h T TxTT Aes
hTTk hTx s
h T Th T T T kA T es kq h s s q h k
e x xL erfc e erfc ts q t t
T T xerfc eT T t
β αβ β αβ β βα α
α
∞ ==
−
∞
==
∞−∞
−− − +
− +
∞
∂− = −
∂
= +
∂− = −
∂
−−= = +
+ +
⎛ ⎞⎛ ⎞ ⎛ ⎞= − +⎜ ⎟⎜ ⎟ ⎜ ⎟+ ⎝ ⎠ ⎝ ⎠⎝ ⎠
− ⎛ ⎞= −⎜ ⎟− ⎝ ⎠
∵
2
2
2
,2
h tk
i
i
x h terfckt
T T x h tfT T kt
α αα
αα∞
⎛ ⎞+⎜ ⎟⎜ ⎟
⎝ ⎠⎛ ⎞−
= ⎜ ⎟⎜ ⎟− ⎝ ⎠
Laplace transform on infinite slab.
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Sheu Long Jye Dep. of Mechanical Engineering
Chung Hua University [email protected]
2
2
0
1 , , 0
( ,0) ,
0
( )
i
x
T T L x L tx t
subject to IC and BCsT x T L x L
Tx
Tk h T T at x Lx
α
=
∞
∂ ∂= − < < >
∂ ∂
= − < <
∂=
∂∂
− = − =∂
2
2
0
1 , , 0
( ,0) ,
0
i
x
let T T
L x L tx t
subject to IC and BCsx L x L
x
k h at x Lx
θ
θ θα
θ θθ
θ θ
∞
=
= −
∂ ∂= − < < >
∂ ∂
= − < <
∂=
∂∂
− = =∂
By Laplace transform
2
2
22
2
( )
cosh sinh ,
cosh( sinh cosh )
1 .....2 2 2
i
i
i i
h h tL xk k
i
qx s
sA qx B qx qs
h qxs s kq qL h qL
L x L x h L xerfc erfc e erfc tkt t t
α
θθ θ
θθα
θ θθ
θ θ αα α α
− +
∂− = −
∂
= + + =
= −+
⎧ ⎫⎡ ⎤− + −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪ ⎪= − + + + +⎨ ⎬⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎪ ⎪⎩ ⎭
The solutions of 1-D transient heat conduction in infinite slab, cylinder, and sphere are summarized and figured by Heilser chart. (Fig. 4-5 to 4-16) Multidimensional system Consider an infinite rectangular bar (Fig.)
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Sheu Long Jye Dep. of Mechanical Engineering
Chung Hua University [email protected]
2 2
1 1 2 22 2
12
1 11 12
22
2 22 22
1 2
1
1 , , , 0 (*)
( , )
1
( , )
1
(*)
L x L L z L tx z t
If x t satisfies
in L x Lx t
and x t satisfies
in L z Lz t
Then satisfiesif at the surface the medium subjects to
k h at xx
θ θ θα
θ
θ θα
θ
θ θα
θ θ θ
θ θ
∂ ∂ ∂+ = − < < − < < >
∂ ∂ ∂
∂ ∂= − < <
∂ ∂
∂ ∂= − < <
∂ ∂=
∂− =
∂ 1
2 2
,L and
k h at z Lzθ θ
=
∂− = =
∂
θ can be obtained as
1 22
1 1 11 1 12
22 2 2
2 2 22
,
1 ,
1 ,
where
k h at x Lx t x
k h at z Lz t z
θ θ θ
θ θ θ θα
θ θ θ θα
=
∂ ∂ ∂= − = =
∂ ∂ ∂∂ ∂ ∂
= − = =∂ ∂ ∂
Heat transfer in multidimensional systems for 2-D system:
0 0 0 01 2 1
1total
q q q qq q q q
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
For 3-D
0 0 0 0 0 0 01 2 1 3 1 2
1 1 1total
q q q q q q qq q q q q q q
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Examples: Transient Finite Difference Method Consider a 2-D body within which the heat flow is governed by
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Sheu Long Jye Dep. of Mechanical Engineering
Chung Hua University [email protected]
2 2
2 2
21, 1, ,
2 2
2, 1 , 1 ,
2 2
1, ,
( )
2,
( )2
,( )
m n m n m n
m n m n m n
p pm n m n
T T Tk cx y t
T T TTx x
T T TTy y
T TTt t
ρ
+ −
+ −
+
∂ ∂ ∂+ =
∂ ∂ ∂+ −∂
≈∂ Δ
+ −∂≈
∂ Δ
−∂≈
∂ Δ
If the nodal temperature derivatives are evaluated at time p:
11, 1, , , 1 , 1 , , ,
2 2
1, 1, 1, , 1 , 1 ,
2
2 2 1( ) ( )
( ) (1 4 )
( )
p p p p p p p pm n m n m n m n m n m n m n m n
p p p p p pm n m n m n m n m n m n
T T T T T T T Tx y t
if x yT Fo T T T T Fo T
twhere Fox
α
α
++ − + −
++ − + −
+ − + − −+ =
Δ Δ ΔΔ = Δ
= + + + + −
Δ=
Δ
If the system is 1-D in x (Fig.)
11 1( ) (1 2 )p p p p
m m m mT Fo T T Fo T++ −= + + −
The above expression are called explicit formulations Stability criterion: For 1-D system: 1/ 2Fo ≤ . For 2-D system: 1/ 4Fo ≤ . Example: (Fig.) Δx and Δt are such that
1
1.121.1 1.1(2 2) (1 2 ) 2.1 2.2 0.21 2( 2 )2 2
p ndm
Fo
T break Law+
=
= + + − = − <i i
The boundary nodes: Consider the transient energy balance at node (m,n) on a boundary surface, (Fig)
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Sheu Long Jye Dep. of Mechanical Engineering
Chung Hua University [email protected]
11, , , 1 , , 1 , , ,
,
1, 1, , 1 , 1 ,
2
( )2 2 2
1{2 2 [ 2 4] }
,( )
p p p p p p p pm n m n m n m n m n m n m n m np
m n
p p p p pm n m n m n m n m n
T T T T T T T Tx x xk y k k h y T T c yx y y t
if x y
T Fo BiT T T T Bi TFo
t h xFo Bix k
ρ
α
+− + −
∞
+∞ − − +
− − − −Δ Δ ΔΔ + + + Δ − = Δ
Δ Δ Δ ΔΔ = Δ
= + + + + − −
Δ Δ= =
Δ
Corresponding 1-D relation is
11
1{2 2 [ 2 2] }p p pm m mT Fo BiT T Bi T
Fo+
∞ −= + + − −
Stability criterion:
For 1-D system: 1 ,1 2 2 02
Fo Fo BiFo≤ − − ≥ .
For 2-D system: 1 ,1 4 2 04
Fo Fo BiFo≤ − − ≥ .
Implicit formulations:
1 1 1 1 1 1 11, 1, , , 1 , 1 , , ,
2 2
1 1 1 1 1, 1, 1, , 1 , 1 ,
1
2 2 1( ) ( )
(1 4 ) ( )
p p p p p p p pm n m n m n m n m n m n m n m n
p p p p p pm n m n m n m n m n m n
T T T T T T T Tx y t
if x yFo T Fo T T T T T
linear equations systemAX B X A B
α
+ + + + + + ++ − + −
+ + + + ++ − + −
−
+ − + − −+ =
Δ Δ ΔΔ = Δ
− − + + + =
= ⎯⎯→ =