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7/27/2019 (5) Transient Conduction - Lumped Cap Mtd - S1 2013-2014 (1)
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Dr. K.C. Leong, 2006
Lecture 2:Radiation & Conservation of Energy Requirement
(5) Trans ient Conduc t ion : Lumped
Capacitance Methodby
Assoc Prof Leong Kai ChoongSchool of Mechanical and Aerospace Engineering
MA3003/MP3003/AE3003 Heat TransferSemester 1, AY 2013-2014
Read Chapter 4: Section 4-1 of the
textbook before these lecture
slides
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2
At the end of these lectures, you should be able to
assess when the spatial variation of temperature of a
solid is negligible and temperature varies nearly
uniformly with time, making the lumped capacitance
analysis applicable,
calculate the time taken by a solid to reach some
temperature Tor the temperature reached by a solid
at some time t,
calculate the amount of heat transferred (in J) during
the transient conduction process, and perform general lumped capacitance calculations.
Learning Objectives
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Transient Conduction
Solid body suddenly subjected to change in thermal
conditions at the boundary temperature varies with
space and time until steady-state.
Unsteady, time-dependent ortransient heat
conduction occurs in the interim period.
We can solve general heat conduction equation for
T(x,y,z,t) in solid subject to initial and boundary
conditions.
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t
Tce
z
Tk
zy
Tk
yx
Tk
xpgen
For example, solve
or solve
with 6 boundary conditions and 1 initial condition
T(x,y,z,t)
1
2
2
t
T
x
T
with two boundary conditions and one initialcondition T(x,t)
semi-infinite solid
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5
Analytical solutions may be obtained for simple
geometries and conditions.
Such complicated solutions may be obtained
from advanced heat transfer texts such as
Carslaw, H.S. and J.C. Jaeger, Conduction ofHeat in Solids, 2nd ed., Oxford UniversityPress, London, 1959.
Ozisik, M.N., Heat Conduction, 2nd ed., Wiley,New York, 1993.
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Lumped Capacitance Method
Sudden change in thermal environment e.g.quenching of hot metal forging in cooler liquid
(Figure 1)
Temperature of solid spatially uniform during
transient process i.e. no temperature gradients in
solid
Approximated if resistance to solid conduction is
small compared to resistance to convectionbetween solid and fluid.
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7
Fig. 1 Cooling of a hot metal forging
Source: Incropera et al. (2007)
convQ
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8
Temperature of forging cannot be obtained by solving heat
conduction equation. Take overall energy balance on solid
(3)exp
(2)ln
Write
0
tVc
hA
TT
TT
hA
Vct
dtd
hA
Vc
TT
s
ii
i
s
t
si
convQ
dt
dTVcTThA
EE
s
stout
(1)
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9
Fig. 2 Transient temperature response of lumped
capacitance solids for different time constants
Source: Incropera et al. (2007)
1/e
t
Vc
hAs
i
exp
s
i
hA
Vc
texp
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10
Electrical analogy can be used if we define a
thermal time constant
)4(1
tt
st
CRVchA
Analogous to voltage decay,Ewhen charged
capacitor,Cdischarges through a resistor,R in a
RCcircuit (See figure on next slide)
(3)Recall
tVc
hA
TT
TT s
ii
exp
%8.361
At
e
t
i
t
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11
Equivalent thermal circuit for a lumpedcapacitance solid
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0
RC
E
dt
dE
ForRC electric circuit,
Solution with initial conditionE(0) =E0isRCteEE /
0
For transient conduction (lumped capacitancemodel), total energy transferQin J occurring up
to some time tcan be calculated by
ti
tt
sconv
tVc
dthAdtQQ
/exp1
00
ttCRt
ie/
analogous to
s
t hA
Vc
N.B.
f f
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13
QE
EQ
EEEE
st
st
stoutgin
or00
Alternatively, use conservation of energy over finite
time interval, t= t- 0 = t
tistti
iist
tVcEQ
tVc
VcTTVcE
/exp1Hence,
/exp1
For quenching,Q
is positive (Eout
)
For heating, Q is negative. (Ein
)
tt
ie /
Whi h h l b l b
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Which thermal system below can bemodelled as a lumped capacitance system?
Source: engel, Y.A. and Ghajar, A.J., Heat and Mass Transfer:
Fundamentals and Ap pl ications, 4th Ed. (SI Units), McGraw-Hill, 2011.
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Validity of Lumped Capacitance Method
Assume steady-state conditions of Fig. 3 to developthe conditions under which the lumped capacitance
method may be used.
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Fig. 3 Effect of Biot number on steady-state temperature
distribution in a plane wall with surface convection
Source: Incropera et al. (2007)
convQ
condQ TThATTL
kAsss 2,2,1,
Bi
/1
/
2,
2,1,
khL
R
R
hA
kAL
TT
TT
conv
cond
s
ss
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Dimensionless numberBiot(pronounced as Bee-oh)
numbercan be interpreted as
conv
cond
R
R
hA
kAL
k
hL
/1/
Bi
Note that kis the thermal conductivity of the solid, not thefluid as in Nusselt number,
Nu hL/kto be introduced in convection.Bi
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Fig. 4 The Biot number can be viewed as the ratio of
convection (not convection resistance) at the
surface to conduction (not conduction resistance)
within the body
T
L
Ak
ThA
k
hL
or
R
R
hA
kAL
k
hL
conv
cond
Bi
/1
/Bi
Illustration from engel and Ghajar (2011)
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Born 21 April 1774, Paris
Died 3 February 1862, Paris
Nationality
French
physicist, astronomer, and
mathematician.
Fields
Professor of Mathematics at
Beauvais (1797) and
Professor of Physics at
Collge de France, Paris
(1800)
Doctoral student William Ritchie
Known for Biot-Savart law
Influences Louis Pasteur
Jean-Baptiste Biot
Fig 5 Transient temperature distributions for different Biot
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Fig. 5 Transient temperature distributions for different Biot
numbers in a plane wall cooled by convection
Source: Incropera et al. (2007)
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For arbitrarily-shaped bodies, characteristic
length is defined as
s
cA
VL
?lengthaxialof
cylindersizefiniteandsideofcubeaforisWhat
spherefor3/
cylinderlongfor2/
wallplanefor2/
L
LL
rL
rL
LL
c
c
c
c
Show that
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)5(FoBiexp
TTTT
ii
c
kydiffusivitthermal
LtmberFourier nu
L
t
k
hL
Vc
thA
c
c
cs
,and
timeessdimensionlisFo,where
FoBi
2
2
Dimensionless temperature solution of Equation
(3) can be re-written as
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Example 1: Transient Response of a Thermocouple
Thermocouple used to measure temperature of a gas
stream flowing in a duct
Approximate junction as a sphere (D= 1 mm).
Actual thermocouples come in standard wire sizes denoted
as American Wire Gage (a.w.g.), e.g.
a.w.g. 18 denotes 1.024 mm diameter wire
a.w.g. 24 denotes 0.5106 mm diameter wire
a.w.g. 30 denotes 0.1270 mm diameter wire
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Thermocouple in this example is Type Kmade of
chromel and alumel wires
c= 0.4 kJ/kgK, k= 30 W/ mK, = 8600 kg/m3
Convection coefficient, h = 280 W/m2K
chromelalumel
Gas
stream
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Find:
(a) time required for thermocouple to reach 98% of
the applied temperature difference, i.e.
TTTtTi
100
2)(
(b) time constant or response time of thermocouple
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Solution:
Must check Bito see whether we can use lumped
capacitance approach.
m1067.1
3100.5
34
3
4
4-
3-
2
3
r
r
r
A
VL
s
c
1.01056.130 1067.1280Bi3-
-4
k
hL c
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s03.8
2
100ln
280
104.01067.18600
lnln
34
ici
sh
cLhAVct
(a) Time required to reach 98% of applied
temperature difference is given by
(b) Time constant is
s05.2
280
104.01067.18600 34
h
cL
hA
Vcc
s
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Magnitude of time constant indicates how sensitive
the thermocouple is to a change in temperature.
A smaller bead diameter implies a more sensitive
thermocouple.
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Example 2: Annealing of Steel Balls
Carbon steel balls ( = 7833 kg/m3, k= 54 W/mK,c =
465 J/kgK, and = 1.474 x 10-6 m2/s) 8 mm indiameter are annealed by heating them first to 900C in
a furnace and then allowing them to cool slowly to
100C in ambient air at 35C.
a) If the average heat transfer coefficient is 75 W/m2K,determine how long the annealing process will take.
b) If 2500 balls are to be annealed per hour, determine
the total rate of heat transfer from the balls to the
ambient air.
Assumptions:
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min2.7s163ln
0013.0
validisanalysisecapacitanclumped
1.00018.0Bi
TT
TT
hA
Vct
A
VL
k
hL
i
s
s
c
c
kW543
J/h1,952,500
J/ball781balls/h2500
ballperJ781
QnQ
TTmcEQ
ball
ifst
Assumptions:
(1) Constant properties (2) Constant and uniform ha)
b)
Source: engel and Ghajar (2011)
Example 3: Freezing of Biological Tissue
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Example 3: Freezing of Biological Tissue
A 1-cm cube of biological tissue is to be frozen in a
freezer at 80C. The properties of the tissue has thesame properties as liquid water at 25C. The convective
heat transfer coefficient is 10 W/m2K, and the tissue is
initially at 25C.
Determine the time it takes to freeze the tissuecompletely.
The freezing temperature is taken as 0C. The properties
of liquid water at 25C may be taken as: Thermal
conductivity of 0.6 W/mK, density of 997 kg/m3
, andspecific heat of 4.18 kJ/kgK.
Schematic:
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validisModeleCapacitancLumped
1.0028.06
Bi
k
hak
hLc
ANALYSIS:
min3.15s189ln
i
shA
Vct
66 2
3 a
a
a
A
VL
s
c
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General Lumped Capacitance Analysis
Transient conduction commonly initiated by
convection heat transfer - simple lumped
capacitance analysis.
In general, thermal conditions in solid may besimultaneously influenced by convection, radiation,
surface heat fluxand internal heat generation as
shown in Fig. 6.
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Fig. 6 Control surface for general lumped
capacitance analysis
sq
radq
convq
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Applying conservation of energy at any instant t:
)7(),(44, dtdTVcATTTThEAq rcssurghss
Eq. (7) - non- linear, first-order, non-
homogeneous ODE with no exact solution.
)6(),(,dt
dTVcAqqEAq rcsradconvghss
Simplified cases:
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Simplified cases:
Case 1: Radiation exchange only
dtdT
VcTTA surrs
44
,
T
T
t
rs
i surTT
dTdt
Vc
A44
0
,
(8)
Time required to reach temperature T
isur
isur
sur
sur
rs TT
TT
TT
TT
TA
Vct
sur
lnln4 3,
sur
i
sur T
T
T
T 11
tantan2(9)
Cannot obtain Tin terms oftexplicitly.
For T = 0 i e radiation to deep space
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ForTsur
0 i.e. radiation to deep space,
(10)11
3 33,
iTTA
Vct
rs
Case 2: Convection, surface heat flux and heat
generation without radiation where h is assumedto be independent of time.
Exact solution to the linear, first-order non-
homogeneous differential equation of the form
TT
VcEAqbVchAawhere
badt
d
ghsscs
and
,/
0
,,
tiontransformagintroducinbyityinhomogeneEliminate
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)exp()/(
)/(
,forngSubstituti
).exp(
),(to0fromgintegratinandvariablesSeparating
0
ngSubstituti
tiontransformagintroducinbyityinhomogeneEliminate
'
'
'
''
''
'
atabTT
abTT
at
tot
adt
d
a
b
i
i
i
Hence
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Hence,
)11(exp1/
exp at
TT
abat
TT
TT
ii
When b = 0, Eq. (11) reduces to Eq. (3) introduced
previously and yields T = Tiat t= 0.
)3(exp
tVc
hA
TT
TT s
i
./becomesEq.(11),As abTTt See Example 5.2 of Incropera et al. (2007)
Some Review Questions
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1. For which solid is the lumped capacitance analysis more
likely to be applicable: an actual apple or a golden apple of
the same size? Why?
2. For what kind of bodies made of the same material is the
lumped capacitance analysis more likely to be applicable:
slender ones or well-rounded ones of the same volume?
3. What is the significance of the thermal time constant?
4. To what electrical circuit is the lumped capacitance systemanalogous?
5. Why is it necessary to employ the concept ofcharacteristic
length in lumped capacitance analysis?
6. What are the physical significances of the Biot and Fouriernumbers?
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