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HEAT TRANSFER
Many heat transfer problems require the understanding of
the complete time history of the temperature variation. For
example, in metallurgy, the heat treating process can be
controlled to directly affect the characteristics of the
processed materials. Annealing (slow cool) can soften
metals and improve ductility. On the other hand,
quenching (rapid cool) can harden the strain boundary and
increase strength. In order to characterize this transient
behavior, the full unsteady equation is needed:
2 21, or
kwhere = is the thermal diffusivity
c
T Tc k T T
t t
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“A heated/cooled body at Ti is suddenly exposed to fluid at T with a
known heat transfer coefficient . Either evaluate the temperature at a
given time, or find time for a given temperature.”
Q: “How good an approximation would it be to say the annular cylinder is
more or less isothermal?”
A: “Depends on the relative importance of the thermal conductivity in the
thermal circuit compared to the convective heat transfer coefficient”.
Fig. 5.1
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Biot No. Bi •Defined to describe the relative resistance in a thermal circuit of
the convection compared
Lc is a characteristic length of the body
Bi→0: No conduction resistance at all. The body is isothermal.
Small Bi: Conduction resistance is less important. The body may still
be approximated as isothermal
Lumped capacitance analysis can be performed. Large Bi: Conduction resistance is significant. The body cannot be treated as
isothermal.
surfacebody at resistance convection External
solid withinresistance conduction Internal
/1
/
hA
kAL
k
hLBi cc
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Transient heat transfer with no internal resistance: Lumped Parameter Analysis
Solid
Valid for Bi<0.1
Total Resistance= Rexternal + Rinternal
GE: dT
dt
hA
mcpT T BC: T t 0 Ti
Solution: let T T , therefore
d
dt
hA
mcp
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Lumped Parameter Analysis
ln
hA
mct
i
tmc
hA
i
pi
ii
p
p
eTT
TT
e
tmc
hA
TT
Note: Temperature function only of time and not of space!
- To determine the temperature at a given time, or
- To determine the time required for the
temperature to reach a specified value.
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)exp(T0
tcV
hA
TT
TT
tL
BitLLc
k
k
hLt
cV
hA
ccc
c
2
11
c
k
Thermal diffusivity: (m² s-1)
Lumped Parameter Analysis
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Lumped Parameter Analysis
tL
Foc
2
k
hLBi C
T = exp(-Bi*Fo)
Define Fo as the Fourier number (dimensionless time)
and Biot number
The temperature variation can be expressed as
thickness2La with wallaplane is solid the when) thickness(half cL
sphere is solid the whenradius) third-one(3cL
cylinder.a is solid the whenradius)- (half2or
cL, examplefor
problem thein invloved solid theof size the torealte:scale length sticcharacteria is c Lwhere
L
or
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Spatial Effects and the Role of Analytical Solutions
The Plane Wall: Solution to the Heat Equation for a Plane Wall with
Symmetrical Convection Conditions
iTxT )0,(
2
21
x
TT
a
00
xx
T
TtLTh
x
Tk
lx
),(
k
x*=x/L x=+L
T∞, h
x= -L
T∞, h
T(x, 0) = Ti
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The Plane Wall:
Note: Once spatial variability of temperature is included, there is existence of seven different independent variables.
How may the functional dependence be simplified?
•The answer is Non-dimensionalisation. We first need to understand the physics behind the phenomenon, identify parameters governing the process, and group them into meaningful non-dimensional numbers.
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Dimensionless temperature difference:
TT
TT
ii
*
Dimensionless coordinate: L
xx *
Dimensionless time: FoL
tt
2
*
The Biot Number: solidk
hLBi
The solution for temperature will now be a function of the other non-dimensional quantities
),,( ** BiFoxf
Exact Solution:
*
1
2* cosexp xFoC nn
nn
BiC nn
nn
nn
tan
2sin2
sin4
The roots (eigenvalues) of the equation can be obtained from tables given in standard textbooks.
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The One-Term Approximation 2.0Fo
Variation of mid-plane temperature with time
)0( * xFo
FoCTT
TT
i
2
11
*
0 exp
From tables given in standard textbooks, one can obtain 1C and 1
as a function of Bi. Variation of temperature with location )( *x and time ( Fo ):
*
1
*
0
* cos x
Change in thermal energy storage with time:
TTcVQ
QE
i
st
0
*
0
1
10
sin1
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Numerical Methods for Unsteady Heat Transfer
Unsteady heat transfer equation, no generation, constant k, one-
dimensional in Cartesian coordinate:
The term on the left hand side of above eq. is the storage term,
arising out of accumulation/depletion of heat in the domain under
consideration. Note that the eq. is a partial differential equation as a
result of an extra independent variable, time (t). The corresponding
grid system is shown in fig. on next slide.
Sx
Tk
xt
Tc
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PP WW EE
xx ww ee
(d(dx)x)ww (d(dx)x)ee
tt
∆∆xx
Integration over the control volume and over a time interval
gives
tt
t CV
tt
t cv
tt
t CV
dtSdVdtdVx
Tk
xdtdV
t
Tc
tt
t
tt
t we
e
w
tt
t
dtVSdtx
TkA
x
TkAdVdt
t
Tc
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If the temperature at a node is assumed to prevail over the whole
control volume, applying the central differencing scheme, one obtains:
tt
t
tt
t w
WPw
e
PEe
old
PPnew dtVSdt
x
TTAk
x
TTAkVTTc
Now, an assumption is made about the variation of TP, TE and Tw with
time. By generalizing the approach by means of a weighting
parameter f between 0 and 1:
tfftdt
tt
t
old
P
new
PPP
1
xSx
TTk
x
TTkf
x
TTk
x
TTkfx
t
TTc
w
old
W
old
Pw
e
old
P
old
Ee
w
new
W
new
Pw
e
new
P
new
Ee
old
P
new
P
)1(
Repeating the same operation for points E and W,
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Upon re-arranging, dropping the superscript “new”, and casting the
equation into the standard form
bTafafa
TffTaTffTaTa
old
PEWP
old
EEE
old
WWWPP
)1()1(
)1()1(
0
0
PEWP aaaa t
xcaP
0
w
wW
x
ka
e
eE
x
ka
xSb
;
;
;
;
The time integration scheme would depend on the choice of the
parameter f. When f = 0, the resulting scheme is “explicit”; when
0 < f ≤ 1, the resulting scheme is “implicit”; when f = 1, the
resulting scheme is “fully implicit”, when f = 1/2, the resulting
scheme is “Crank-Nicolson”.
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t
T
TP old
t TP
new t+Dt
f=0
f=1
f=0.5
Variation of T within the time interval ∆t for different schemes
Explicit scheme
Linearizing the source term as and setting f = 0
u
old
PEWP
old
EE
old
WWPP STaaaTaTaTa )(0
0
PP aa t
xcaP
0
w
wW
x
ka
e
eE
x
ka
For stability, all coefficients must be positive in the discretized
equation. Hence, 0)(0 PEWP Saaa
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0)(
e
e
w
w
x
k
x
k
t
xc
x
k
t
xc
2
k
xct
2
)( 2
The above limitation on time step suggests that the explicit
scheme becomes very expensive to improve spatial accuracy.
Hence, this method is generally not recommended for general
transient problems.
Crank-Nicolson scheme
Setting f = 0.5, the Crank-Nicolson discretisation becomes:
bTaa
aTT
aTT
aTa PWE
P
old
WWW
old
EEEPP
00
2222
PPWEP Saaaa2
1)(
2
1 0 t
xcaP
0
w
wW
x
ka
e
eE
x
ka
old
ppu TSSb2
1
;
;
;
;
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For stability, all coefficient must be positive in the discretized
equation, requiring
2
0 WEP
aaa
k
xct
2)(
The Crank-Nicolson scheme only slightly less restrictive than the
explicit method. It is based on central differencing and hence it is
second-order accurate in time.
The fully implicit scheme
Setting f = 1, the fully implicit discretisation becomes:
old
PPWWEEPP TaTaTaTa 0
PWEPP Saaaa 0
t
xcaP
0
w
wW
x
ka
e
eE
x
ka
;
;
;
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General remarks:
A system of algebraic equations must be solved at each time
level. The accuracy of the scheme is first-order in time. The time
marching procedure starts with a given initial field of the scalar
0. The system is solved after selecting time step Δt. For the
implicit scheme, all coefficients are positive, which makes it
unconditionally stable for any size of time step. Hence, the
implicit method is recommended for general purpose transient
calculations because of its robustness and unconditional stability.