ch05
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ch05


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= fd_1d_int(T02,T03,T01,km,qdot,deltax)
/* Node 19: surface node (e-orientation); transient conditions; w labeled 18. */
rhob * cb * der(T19,t) = fd_1d_sur_e(T19,T18,kb,qdot,deltax,Tinf19,h19,q''a19)
q''a19 = 0 // Applied heat flux, W/m^2; zero flux shown
Tinf19 = 30
h19 = 3000
// Finite-difference equation from energy balance on CV, Node 13
km*(T12 - T13)/deltax + kb*(T14 - T13)/deltax = (rhom*cm + rhob*cb) *(deltax/2)*der(T13,t)
(5) The screen from the FEHT analysis is shown below. It is important to use small time steps in the
integration at early times. Use the View|Temperatures command to find the temperature of the
mixture surface at tres = 50 s.
PROBLEM 5.121
KNOWN: Thin, circular-disc subjected to induction heating causing a uniform heat
generation in a prescribed region; upper surface exposed to convection process.
FIND: (a) Transient finite-difference equation for a node in the region subjected to induction
heating, (b) Sketch the steady-state temperature distribution on T-r coordinates; identify
important features.
SCHEMATIC:
ASSUMPTIONS: (1) Thickness w << ro, such that conduction is one-dimensional in r-
direction, (2) In prescribed region, q\ufffd is uniform, (3) Bottom surface of disc is insulated, (4)
Constant properties.
ANALYSIS: (a) Consider the nodal point arrangement
for the region subjected to induction heating. The size of
the control volume is mV 2 r r w.\u3c0= \u22c5\u2206 \u22c5 The energy
conservation requirement for the node m has the form
in out g stE E E E\u2212 + =\ufffd \ufffd \ufffd \ufffd
with a b conv stq q q qV E .+ + + = \ufffd\ufffd
Recognizing that qa and qb are conduction terms and qconv is the convection process,
p p p p
m mm-1 m+1
m m
T T T Tr rk 2 r w k 2 r w
2 r 2 r
\u3c0 \u3c0
\u2212 \u2212\uf8ee \uf8f9 \uf8ee \uf8f9\u2206 \u2206\uf8ee \uf8f9 \uf8ee \uf8f9
\u2212 + +\uf8ef \uf8fa \uf8ef \uf8fa\uf8ef \uf8fa \uf8ef \uf8fa\u2206 \u2206\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f0 \uf8fb \uf8f0 \uf8fb
[ ]( ) [ ] [ ] p+1 pp m mm m m p m T Th 2 r r T T q 2 r r w c 2 r r w .t\u3c0 \u3c0 \u3c1 \u3c0\u221e \u2212+ \u22c5\u2206 \u2212 + \u22c5\u2206 \u22c5 = \u22c5\u2206 \u22c5 \u2206\ufffd
Upon regrouping, the finite-difference equation has the form,
2
p pp+1 p
m mm-1 m+1
m m
r r r q r r
T Fo 1 T 1 T Bi T 1 2Fo Bi Fo T
2r 2r w k w\u221e
\u2206 \u2206 \u2206 \u2206 \u2206
= \u2212 + + + + + \u2212 \u2212 \u22c5
\uf8ee \uf8f9\uf8ee \uf8f9 \uf8ee \uf8f9 \uf8ee \uf8f9 \uf8ee \uf8ee \uf8f9\uf8f9\uf8ef \uf8fa\uf8ef \uf8fa \uf8ef \uf8fa \uf8ef \uf8fa \uf8ef \uf8ef \uf8fa\uf8fa\uf8f0 \uf8fb \uf8f0 \uf8f0 \uf8fb\uf8fb\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8ef \uf8fa\uf8f0 \uf8fb
\ufffd <
where 2Fo t/ r Bi h r/k.\u3b1= \u2206 \u2206 = \u2206
(b) The steady-state temperature distribution has
these features:
1. Zero gradient at r = 0, r0
2. No discontinuity at r1, r2
3. Tmax occurs in region r1 < r < r2
Note also, distribution will not be linear anywhere;
distribution is not parabolic in r1 < r < r2 region.
PROBLEM 5.122
KNOWN: An electrical cable experiencing uniform volumetric generation; the lower half is well
insulated while the upper half experiences convection.
FIND: (a) Explicit, finite-difference equations for an interior node (m,n), the center node (0,0), and
an outer surface node (M,n) for the convective and insulated boundaries, and (b) Stability criterion for
each FDE; identify the most restrictive criterion.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional (r,\u3c6), transient conduction, (2) Constant properties, (3)
Uniform q.\ufffd
ANALYSIS: The explicit, finite-difference equations may be obtained by applying energy balances
to appropriate control volumes about the node of interest. Note the coordinate system defined above
where (r,\u3c6) \u2192 (m\u2206r, n\u2206\u3c6). The stability criterion is determined from the coefficient associated with
the node of interest.
Interior Node (m,n). The control volume
for an interior node is
mV r r\u3c6= \u2206 \u22c5\u2206 \u22c5\ufffd
(with mr m r, 1= \u2206 =\ufffd ) where \ufffd is the
length normal to the page. The
conservation of energy requirement is
in out g stE E E E\u2212 + =\ufffd \ufffd \ufffd \ufffd
( ) ( )
p+1 p
m,n m,n
1 2 3 4r
T T
q q q q qV cV
t\u3b8
\u3c1
\u2212
+ + + + =
\u2206
\ufffd
( )
p p pp p p
m,n m,n m,nm-1,n m+1,n m,n+1T T T T T T1 1k m r k m r k r
2 r 2 r m r
\u3c6 \u3c6 \u3c6
\u2212 \u2212 \u2212
\u22c5 \u2212 \u2206 \u22c5\u2206 \u22c5 + \u22c5 + \u2206 \u22c5\u2206 \u22c5 + \u22c5 \u2206 \u22c5
\u2206 \u2206 \u2206 \u2206
\uf8ee \uf8f9 \uf8ee \uf8f9\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb
( ) ( ) ( )
p p p+1 p
m,nm,n-1 m,n m,nT T T Tk r q m r r c m r r
m r t
\u3c6 \u3c1 \u3c6\u3c6
\u2212
\u2212
+ \u22c5\u2206 \u22c5 + \u2206 \u22c5\u2206 \u22c5\u2206 = \u2206 \u22c5\u2206 \u22c5\u2206 \u22c5
\u2206 \u2206 \u2206
\ufffd (1)
Define the Fourier number as
2 2
k t tFo
 c r r
\u3b1
\u3c1
\u2206 \u2206
= \u22c5 =
\u2206 \u2206
(2)
and then regroup the terms of Eq. (1) to obtain the FDE,
( ) ( )p+1 p p p p 2m,n m-1,n m+1,n m,n+1 m,n-12m 1/2 m 1/2 1 qT Fo T T T T rm m km \u3c6
\uf8f1 \uf8fc
\u2212 +\uf8f4 \uf8f4
= + + + + \u2206\uf8f2 \uf8fd
\u2206\uf8f4 \uf8f4\uf8f3 \uf8fe
\ufffd
( )
p
m,n2
2Fo 2 1 T .
m \u3c6
\uf8f1 \uf8fc\uf8ee \uf8f9\uf8f4 \uf8f4\uf8ef \uf8fa+ \u2212 + +\uf8f2 \uf8fd\uf8ef \uf8fa\u2206\uf8f4 \uf8f4\uf8f0 \uf8fb\uf8f3 \uf8fe
 (3) <
Continued \u2026..
PROBLEM 5.122 (Cont.)
The stability criterion requires that the last term on the right-hand side in braces be positive. That is,
the coefficient of pm,nT must be positive and the stability criterion is
( )2Fo 1/2 1 1/ m \u3c6\u2264 + \u2206\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb (4)
Note that, for m >> 1/2 and (m\u2206\u3c6)2 >>1, the FDE takes the form of a 1-D cartesian system.
Center Node (0,0). For the control volume,
( )2V r/2 1.\u3c0= \u2206 \u22c5 The energy balance is
in out g st in nE E E E where E q .\u2032 \u2032 \u2032 \u2032 \u2032 \u2032\u2212 + = = \u3a3\ufffd \ufffd \ufffd \ufffd \ufffd
p p 2N
o1,n
n 0
T Tr r
k q
2 r 2
\u3c6 \u3c0
=
\u2212\u2206 \u2206
\u22c5 \u2206 \u22c5 +
\u2206
\uf8ee \uf8f9 \uf8ee \uf8f9\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb\u2211 \ufffd
p+1 p2
o oT Tr
 c
2 t
\u3c1 \u3c0
\u2212\u2206
= \u22c5
\u2206
\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb (5)
where N = (2\u3c0/\u2206\u3c6) - 1, the total number of qn. Using the definition of Fo, find
( )
N
pp+1 2 p
o o1,n
n 0
1 q
T 4Fo T r 1 4Fo T .
N 1 4k
=
= + \u2206 + \u2212
+
\uf8f1 \uf8fc\uf8f4 \uf8f4\uf8f2 \uf8fd\uf8f4 \uf8f4\uf8f3 \uf8fe\u2211
\ufffd
 <
By inspection, the stability criterion is Fo \u2264 1/4. (7)
Surface Nodes (M,n). The control volume
for the surface node is V = (M - ¼)\u2206r\u2206\u3c6\u22c5\u2206r/2.1.
From the energy balance,
( ) ( )in out g 1 2 3 4 strE E E q q q q qV E\u3c6\u2032 \u2032 \u2032 \u2032 \u2032 \u2032 \u2032 \u2032\u2212 + = + + + + =\ufffd \ufffd \ufffd \ufffd\ufffd
( ) ( )( ) ( )
p p p p
M,n M,nM-1,n M,n+1p
M,n
T T T Tr
k M 1/2 r h M r T T +k
r 2 M r
\u3c6 \u3c6
\u3c6\u221e
\u2212 \u2212\u2206
\u22c5 \u2212 \u2206 \u22c5 \u2206 + \u2206 \u22c5 \u2206 \u2212 \u22c5 \u22c5
\u2206 \u2206 \u2206
( ) ( ) ( )
p p p+1 p
M,n-1 M,n M,n M,nT T T Tr r rk q M 1/4 r c M 1/4 r .
2 M r 2 2 t
\u3c6 \u3c1 \u3c6
\u3c6
\u2212 \u2212\u2206 \u2206 \u2206
+ \u22c5 \u22c5 + \u2212 \u2206 \u22c5 \u2206 \u22c5 = \u2212 \u2206 \u22c5 \u2206 \u22c5
\u2206 \u2206 \u2206
\uf8ee \uf8f9 \uf8ee \uf8f9\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb\ufffd
Regrouping and using the definitions for Fo = \u3b1\u2206t/\u2206r2 and Bi = h\u2206r/k,
( ) ( ) ( )p p pp+1 2m,n M-1,n M,n+1 M,n-12M 1/2 1 qT Fo 2 T T T 2Bi T rM 1/4 kM-1/4 M \u3c6 \u221e\u2212= + \u2212 + \u22c5 + \u2206\u2212 \u2206
\uf8f1 \uf8fc\uf8f4 \uf8f4\uf8f2 \uf8fd\uf8f4 \uf8f4\uf8f3 \uf8fe
\ufffd
( ) ( )
p
M,n2
M-1/2 M 1
1 2Fo Bi T .
M-1/4 M 1/4 M 1/4 M \u3c6
+ \u2212 + \u22c5 +
\u2212
\u2212 \u2206
\uf8f1 \uf8fc\uf8ee \uf8f9\uf8f4 \uf8f4\uf8ef \uf8fa\uf8f2 \uf8fd\uf8ef \uf8fa\uf8f4 \uf8f4\uf8f0 \uf8fb\uf8f3 \uf8fe
 (8) <
The stability criterion is
( ) ( )2
1 M 1/2 M 1
Fo Bi .
2 M 1/4 M 1/4 M 1/4 M \u3c6
\u2212
\u2264 + +
\u2212 \u2212
\u2212 \u2206
\uf8ee \uf8f9\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb
(9)
To determine which stability criterion is most restrictive, compare Eqs. (4), (7) and (9). The most
restrictive (lowest Fo) has the largest denominator. For small values of m, it is not evident whether
Eq. (7) is more restrictive than Eq. (4); Eq. (4) depends upon magnitude of \u2206\u3c6. Likewise, it is not
clear whether Eq. (9) will be more or less restrictive than Eq. (7). Numerical values must be
substituted.
PROBLEM 5.123
KNOWN: Initial temperature distribution in two bars that are to be soldered together; interface
contact resistance.
FIND: (a) Explicit FDE for T4,2 in terms of Fo and t,cBi x/k R ;\u2032\u2032= \u2206 stability criterion, (b) T4,2
one time step after contact is made if Fo = 0.01 and value of \u2206t; whether the stability criterion is
satisfied.
SCHEMATIC:
PROPERTIES: Table A-1, Steel, AISI 1010 (1000K): k = 31.3 W/m\u22c5K, c = 1168 J/kg\u22c5K, \u3c1 = 7832
kh/m3.
ASSUMPTIONS: (1) Two-dimensional transient conduction, (2) Constant properties, (3) Interfacial
solder layer has negligible thickness.
ANALYSIS: (a) From an energy balance on
the control volume V = (\u2206x/2)\u22c5\u2206y\u22c51.
in out g stE E E E\u2212 + =\ufffd \ufffd \ufffd \ufffd
p+1 p
4,2 4,2
a b c d
T T
q q q q cV .
t
\u3c1
\u2212
+ + + =
\u2206
Note that ( )a t,c cq T/R A\u2032\u2032= \u2206 while the remaining qi are conduction terms,
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )[ ]
p p p p
4,3 4,2 5,2 4,2p p
3,2 4,2
t,c
T T T T1
T T y k x/2 k y
R y x
p p p+1 p
4,1 4,2 4,2 4,2
T T T T
 k x/2 c x / 2 y .
y t
\u3c1
\u2212 \u2212
\u2212 \u2206 + \u2206 + \u2206
\u2032\u2032 \u2206 \u2206
\u2212
\u2212
+ \u2206 = \u2206 \u22c5 \u2206
\u2206 \u2206
Defining ( ) 2 c t,cFo k/ c t/