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TimeTime--Dependent Conduction Dependent Conduction ::The Lumped Capacitance MethodThe Lumped Capacitance Method
Chapter FiveChapter FiveSections 5.1 thru 5.3Sections 5.1 thru 5.3
Transient Conduction• A heat transfer process for which the temperature varies with time,
as well as location within a solid.
• It is initiated whenever a system experiences a change in operating conditions and proceeds until a new steady state (thermal equilibrium) is achieved.
• It can be induced by changes in:– surface convection conditions ( ), ,h T∞
• Solution Techniques– The Lumped Capacitance Method– Exact Solutions– The Finite-Difference Method
– surface radiation conditions ( ),,r surh T– a surface temperature or heat flux, and/or
– internal energy generation.
5.1 The Lumped Capacitance Method• Based on the assumption of a spatially uniform temperature distribution
throughout the transient process.
• Why is the assumption never fully realized in practice?
• General Lumped Capacitance Analysis:
Consider a general case, which includes convection, radiation and/or an applied heat flux at specified surfaces as well as internal energy generation.
( ), , ,, , ,s c s r s hA A A
Hence, .( )( , ) T r t T t→
≈
First Law: stin out g
dE dTc E E Edt dt
ρ= ∀ = − +i i i
• Assuming energy outflow due to convection and radiation and withinflow due to an applied heat flux ,sq′′
( ) ( ), , , gs s h s c r s r surdTc q A hA T T h A T T Edt
ρ ∞′′∀ = − − − − +i
• Is this expression applicable in situations for which convectionand/or radiation provide for energy inflow?
• May h and hr be assumed to be constant throughout the transient process?
• Special Cases (Exact Solutions, ) ( )0 iT T≡
Negligible Radiation ( ), / :T T b aθ θ θ∞ ′≡ − ≡ −
, ,/ /gs c s s ha hA c b q A E cρ ρ⎛ ⎞′′≡ ∀ ≡ + ∀⎜ ⎟⎝ ⎠
i
The non-homogeneous differential equation is transformed into a homogeneous equation of the form:
d adtθ θ′
′= −
d a dtθθ′= −
′
Integrating from t=0 to any t and rearranging,
( ) ( )/, exp 1 expi i
T T b aor at atT T T T
∞
∞ ∞
−⎡ ⎤= − + − −⎣ ⎦− − (5.25)
To what does the foregoing equation reduce as steady state is approached?
How else may the steady-state solution be obtained?
1 1
2 2
2
ln( ) exp( ) exp( ) or / exp( )
at 0, ( ) /Hence, ( ) / [( ) / ] ex
i i
i
at C at CC at b a C at
t T T C T T b aT T b a T T b a
θ θθ θ
∞
∞ ∞
′ ′= − + ⇒ = − +′ = − − = −
= = ⇒ = − −− − = − −
初始條件:
p( )/ / = [1 ] exp( )
i i i
atT T b a b a atT T T T T T
∞
∞ ∞ ∞
−
−+ − −
− − −
Negligible Radiation and Source Terms , 0, 0) :gr s(h h E q ′′>> = =i
( ),s cdTc hA T Tdt
ρ ∞∀ = − − (5.2)
, 0is c
tc d hA
dtθ
θ
ρ θθ
∀= − ∫∫
,s c
i i t
hAT T texp t expT T c
θθ ρ τ
∞
∞
⎡ ⎤ ⎡ ⎤⎛ ⎞−= = − = −⎢ ⎥ ⎢ ⎥⎜ ⎟− ∀⎝ ⎠ ⎣ ⎦⎣ ⎦
The thermal time constant is defined as
( ),
1t
s c
chA
τ ρ⎛ ⎞
≡ ∀⎜ ⎟⎜ ⎟⎝ ⎠
(5.7)
ThermalResistance, Rt
Lumped ThermalCapacitance, Ct
The change in thermal energy storage due to the transient process is
0
toutstE Q E dt∆ ≡ − = −∫i
,0
t
s chA dtθ= − ∫ ( ) 1 expit
tcρ θτ
⎡ ⎤⎛ ⎞= − ∀ − −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦(5.8)
Negligible Convection and Source Terms , 0, 0 :gr sh h E q⎛ ⎞′′>> = =⎜ ⎟⎝ ⎠
i
Assuming radiation exchange with large surroundings,
( )4 4,s r sur
dTc A T Tdt
ρ ε σ∀ = − −
,
4 40 i
s r T
surT
tAc
dTT T
dtε σρ
=∀ −∫∫
3,
1n 1n4
sur sur i
s r sur sur sur i
T T T TctA T T T T Tρ
ε σ⎧ + +∀ ⎪= −⎨
− −⎪⎩ (5.18)
Result necessitates implicit evaluation of T(t).
1 12 tan tan i
sur sur
TTT T
− −⎫⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎪+ − ⎬⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎪⎣ ⎦⎭
5.2 Validity of The Lumped Capacitance Method
• The Biot Number: The first of many dimensionless parameters to be considered.
Definition:
/cBi Biot number = hL k≡
convection or radiation coefficienth →
thermal conductivity of t soe dh lik →
of the solid ( / or coordinate associated with maximum spa
characteristtial temperature difference)
ic leng th c sL A→ ∀
Physical Interpretation:/
/1/
c s cond solid
s conv solid fluid
L kA R TBihA R T
∆=
∆∼ ∼
Criterion for Applicability of Lumped Capacitance Method: 1Bi <<
Physical Interpretation: , ,2
,2
/ = 1/
s i s s cond
s s conv
T T L kA R hL BiT T hA R k∞
−= = =
−
,1 ,2 ,2kA: ( ) ( )L s s sT T hA T T∞− = −穩態下
0.1 ( Lumped Heat Capacitance )
/: ;
: / 2 ; : / 3 ( )
c
c s
c
c o c o c o
hLBik
L V APlate L LCylinder L r Sphere L r L r
≡ <
=
=
= = =
使用 方法之條件
保守考慮下:
[ ]
, ,
2
2
2
( )( )
/( )( ) ( )( ) ( )( )
s c s cc
i c
c c
c
c
c
hA A khLT T exp t = exp tT T c k cL
hL hLk c t exp t
tFo = Fourier numberL
exp exp Bi Fo k L k L
α
ρ ρ
ρ α
∞
∞
⎡ ⎤ ⎡ ⎤⎛ ⎞−= − −⎢ ⎥ ⎢ ⎥⎜ ⎟− ∀ ∀⎝ ⎠ ⎣ ⎦⎣ ⎦
⎡ ⎤ ⎡ ⎤= − = − ≡ −⎢ ⎥ ⎢
≡
⎥⎣ ⎦ ⎣ ⎦
(5.6)
暫態下:
Problem 5.12: Charging a thermal energy storage system consistingof a packed bed of aluminum spheres.
KNOWN: Diameter, density, specific heat and thermal conductivity of aluminum spheres used in packed bed thermal energy storage system. Convection coefficient and inlet gas temperature.
FIND: Time required for sphere at inlet to acquire 90% of maximum possible thermal energy and the corresponding center temperature.
Schematic:
ASSUMPTIONS: (1) Negligible heat transfer to or from a sphere by radiation or conduction due to contact with other spheres, (2) Constant properties.
ANALYSIS: To determine whether a lumped capacitance analysis can be used, first compute Bi = h(ro/3)/k = 75 W/m2⋅K (0.025m)/150 W/m⋅K = 0.013 <<1.
Hence, the lumped capacitance approximation may be made, and a uniform temperature may be assumed to exist in the sphere at any time.
From Eq. 5.8a, achievement of 90% of the maximum possible thermal energy storage corresponds to
( )stt
i
E0.90 1 exp t /
cVτ
ρ θ∆
− = = − −
( )tt ln 0.1 427s 2.30 984sτ= − = × =
3
t s 2
2700 kg / m 0.075m 950 J / kg KVc / hA Dc / 6h 427s.
6 75 W / m Kτ ρ ρ
× × ⋅= = = =
× ⋅
From Eq. (5.6), the corresponding temperature at any location in the sphere is
( ) ( ) ( )g,i i g,iT 984s T T T exp 6ht / Dcρ= + − −
( )( )2 3
T 984s
300 C 275 C exp 6 75 W / m K 984s / 2700 kg / m 0.075m 950 J / kg K= ° − ° − × ⋅ × × × ⋅
If the product of the density and specific heat of copper is (ρc)Cu ≈ 8900 kg/m3 × 400 J/kg⋅K = 3.56 × 106 J/m3⋅K, is there any advantage to using copper spheres of equivalent diameter in lieu of aluminum spheres?
Does the time required for a sphere to reach a prescribed state of thermal energy storage change with increasing distance from the bed inlet? If so, how and why?
( )T 984s 272.5 C= °