fundamentals of heat and mass transfer solutions manual phần 3 pdf

PROBLEM 4.33
KNOWN:
Dimensions and thermal conductivities of a heater and a finned sleeve. Convection
conditions on the sleeve surface.
FIND:
(a) Heat rate per unit length, (b) Generation rate and centerline temperature of heater, (c)
Effect of fin parameters on heat rate.
SCHEMATIC:
ASSUMPTIONS:
(1) Steady state, (2) Constant properties, (3) Negligible contact resistance
between heater and sleeve, (4) Uniform convection coefficient at outer surfaces of sleeve, (5) Uniform
heat generation, (6) Negligible radiation.
ANALYSIS:
(a) From the thermal circuit, the desired heat rate is
()
ss
t,o tot
cond 2D
TT TT
q
RRR
∞∞
−−
′
==
′′′
+
The two-dimensional conduction resistance, may be estimated from Eq. (4.27) and Case 6 of Table
4.2
()

()()
()
4
cond 2D
ss
ln 1.08w / D ln 2.16
1
R 5.11 10 m K/ W
S k 2 k 2 240W/m K
ππ
−
′
== = =× ⋅
′
⋅
The thermal resistance of the fin array is given by Eq. (3.103), where
η
o
and A
t
are given by Eqs.
(3.102) and (3.99) and
η
f
is given by Eq. (3.89). With L
c
= L + t/2 = 0.022 m, m = (2h/k
s
t)
1/2

= 32.3
m
-1
and mL
c
= 0.710,
c
f
c
tanh mL
0.61
0.86
mL 0.71
η
===
()( )
tfb
A NA A N 2L t 4w Nt 0.704m 0.096m 0.800m
′′′
=+= ++−= + =
() ()
f
of
t
NA 0.704m
1 1 1 0.14 0.88
A 0.800m
η
η
′

=− − =− =
′
()
(
)
1
1
23
t,o o t
R h A 0.88 500W / m K 0.80m 2.84 10 m K / W
η
−
−
−
′′
==×⋅×=×⋅
()
(
)
43
300 50 C
q 74,600W /m
5.11 10 2.84 10 m K / W
−−
−°
′
==
×+× ⋅
<
(b) Eq. (3.55) may be used to determine

q,

if h is replaced by an overall coefficient based on the
surface area of the heater. From Eq. (3.32),
PROBLEM 4.33 (Cont.)
()( )
11
ss s tot
U A U D R 3.35m K/ W 298W/m K
π
−−
′′
== = ⋅ = ⋅
()
2
s
U 298W /m K / 0.02m 4750W /m K
π
=⋅×= ⋅
()
(
)
()
283
ss
q 4U T T / D 4 4750W / m K 250 C /0.02m 2.38 10 W/ m
∞
=−= ⋅° =×

<

From Eq. (3.53) the centerline temperature is
()
()
()
()
22
83
s
h
q D/ 2 2.38 10 W/m 0.01m
T 0 T 300 C 315 C
4k 4 400W / m K
×
=+= +°=°
⋅

<
(c) Subject to the prescribed constraints, the following results have been obtained for parameter
variations corresponding to 16
≤
N
≤
40, 2
≤
t
≤
8 mm and 20
≤
L
≤

40 mm.
N t(mm)
L(mm)
f
η
()
qW/m
′
16 4 20 0.86 74,400
16 8 20 0.91 77,000
28 4 20 0.86 107,900
32 3 20 0.83 115,200
40 2 20 0.78 127,800
40 2 40 0.51 151,300
Clearly there is little benefit to simply increasing t, since there is no change in
t
A
′
and only a
marginal increase in
f
.
η
However, due to an attendant increase in
t
A,
′
there is significant benefit to
increasing N for fixed t (no change in
f

η
) and additional benefit in concurrently increasing N while
decreasing t. In this case the effect of increasing
t
A
′
exceeds that of decreasing
f
.
η
The same is
true for increasing L, although there is an upper limit at which diminishing returns would be reached.
The upper limit to L could also be influenced by manufacturing constraints.
COMMENTS:
Without the sleeve, the heat rate would be
()
s
q Dh T T 7850W /m,
π
∞
′
=−=
which is well below that achieved by using the increased surface area afforded by the sleeve.
PROBLEM 4.34
KNOWN:
Dimensions of chip array. Conductivity of substrate. Convection conditions. Contact
resistance. Expression for resistance of spreader plate. Maximum chip temperature.
FIND:
Maximum chip heat rate.
SCHEMATIC:

ASSUMPTIONS:
(1) Steady-state, (2) Constant thermal conductivity, (3) Negligible radiation, (4)
All heat transfer is by convection from the chip and the substrate surface (negligible heat transfer
from bottom or sides of substrate).
ANALYSIS:
From the thermal circuit,
()
hh
hsp
h,cnv t,c sp,cnv
tsp
TT TT
qq q
RRRR
∞∞
−−
=+ = +
++
()
(
)
()
1
1
1
2
22
h,cnv s,n
h
R h A hL 100W / m K 0.005m 400K / W

−
−
−

=== ⋅ =


()
()()
357
rrrr
tsp
sub h
1 1.410A 0.344A 0.043A 0.034A 1 0.353 0.005 0 0
R 0.408K / W
4k L 4 80W / m K 0.005m
−+ + + −+++
===
⋅
()
42
t,c
t,c
22
h
R
0.5 10 m K / W
R 2.000K / W
L
0.005m

−
′′
×⋅
== =
()
()
1
1
222
sp,cnv sub s,h
R h A A 100W / m K 0.010m 0.005m 133.3K / W
−
−
=− = ⋅ − =





()
70 C 70 C
q 0.18W 0.52W 0.70W
400K / W 0.408 2 133.3 K / W
°°
=+ =+=
++
<
COMMENTS:
(1) The thermal resistances of the substrate and the chip/substrate interface are much
less than the substrate convection resistance. Hence, the heat rate is increased almost in proportion to

the additional surface area afforded by the substrate. An increase in the spacing between chips (S
h
)
would increase q correspondingly.
(2) In the limit
()
r
tsp
A0,R
→
reduces to
1/2
sub h
2kD
π
for a circular heat source and
sub h
4k L
for a square source.
PROBLEM 4.35
KNOWN: Internal corner of a two-dimensional system with prescribed convection boundary
conditions.
FIND: Finite-difference equations for these situations: (a) Horizontal boundary is perfectly insulated
and vertical boundary is subjected to a convection process (T
∞
,h), (b) Both boundaries are perfectly
insulated; compare result with Eq. 4.45.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant
properties, (4) No internal generation.

ANALYSIS: Consider the nodal network shown above and also as Case 2, Table 4.2. Having
defined the control volume – the shaded area of unit thickness normal to the page – next identify the
heat transfer processes. Finally, perform an energy balance wherein the processes are expressed
using appropriate rate equations.
(a) With the horizontal boundary insulated and the vertical boundary subjected to a convection process,
the energy balance results in the following finite-difference equation:
inout123456
EE0 qqqqqq0
−=+++++=
&&
( )
( )
m-1,nm,nm,n-1m,n
m,n
TTTT
xy
ky1k1h1TT
x2y2
∞
−−
∆∆

∆⋅+⋅+⋅−

∆∆

( )
m+1,nm,nm,n+1m,n
TTTT
y

0k1kx10.
2xy
−−
∆

++⋅+∆⋅=

∆∆

Letting ∆x = ∆y, and regrouping, find
( ) ( )
m-1,nm,n+1m+1,nm,n-1m,n
hxhx
2TTTTT6T0.
kk
∞
∆∆

++++−+=


<
(b) With both boundaries insulated, the energy balance would have q
3
= q
4
= 0. The same result would
be obtained by letting h = 0 in the previous result. Hence,
(
)

(
)
m-1,nm,n+1m+1,nm,n-1m,n
2TTTT6 T0.
+++−=
<
Note that this expression compares exactly with Eq. 4.45 when h = 0, which corresponds to insulated
boundaries.
PROBLEM 4.36
KNOWN: Plane surface of two-dimensional system.
FIND: The finite-difference equation for nodal point on this boundary when (a) insulated; compare
result with Eq. 4.46, and when (b) subjected to a constant heat flux.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction with no generation, (2) Constant
properties, (3) Boundary is adiabatic.
ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2)⋅∆y, and using the
conduction rate equation, it follows that
inout123
EE0 qqq0
−=++=
&&
(1,2)
( )
m-1,nm,nm,n-1m,nm,n+1m,n
TTTTTT
xx
ky1k1k10.
x2y2y
−−−
∆∆


∆⋅+⋅+⋅=

∆∆∆

(3)
Note that there is no heat rate across the control volume surface at the insulated boundary.
Recognizing that ∆x =∆y, the above expression reduces to the form
m-1,nm,n-1m,n+1m,n
2TTT4T0.
++−=
(4) <
The Eq. 4.46 of Table 4.3 considers the same configuration but with the boundary subjected to a
convection process. That is,
( )
m-1,nm,n-1m,n+1m,n
2hxhx
2TTTT22T0.
kk
∞
∆∆

+++−+=


(5)
Note that, if the boundary is insulated, h = 0 and Eq. 4.46 reduces to Eq. (4).
(b) If the surface is exposed to a constant heat flux,
o
q,

′′
the energy balance has the form
123o
qqqqy0
′′
+++⋅∆= and the finite difference equation becomes
o
m-1,nm,n-1m,n+1m,n
qx
2TTT4T.
k
′′
∆
++−=− <
COMMENTS: Equation (4) can be obtained by using the “interior node” finite-difference equation,
Eq. 4.33, where the insulated boundary is treated as a symmetry plane as shown below.
PROBLEM 4.37
KNOWN: External corner of a two-dimensional system whose boundaries are subjected to
prescribed conditions.
FIND: Finite-difference equations for these situations: (a) Upper boundary is perfectly insulated and
side boundary is subjected to a convection process, (b) Both boundaries are perfectly insulated;
compare result with Eq. 4.47.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant
properties, (4) No internal generation.
ANALYSIS: Consider the nodal point configuration shown in the schematic and also as Case 4, Table
4.2. The control volume about the node – shaded area above of unit thickness normal to the page –
has dimensions, (∆x/2)(∆y/2)⋅1. The heat transfer processes at the surface of the CV are identified as
q
1

, q
2
⋅⋅⋅. Perform an energy balance wherein the processes are expressed using the appropriate rate
equations.
(a) With the upper boundary insulated and the side boundary subjected to a convection process, the
energy balance has the form
inout1234
EE0 qqqq0
−=+++=
&&
(1,2)
( )
m-1,nm,nm,n-1m,n
m,n
TTTT
yxy
k1k1h1TT00.
2x2y2
∞
−−
∆∆∆

⋅+⋅+⋅−+=

∆∆

Letting ∆x = ∆y, and regrouping, find
m,n-1m-1,nm,n
hx1hx
TTT21T0.

k2k
∞
∆∆

++−+=


(3) <
(b) With both boundaries insulated, the energy balance of Eq. (2) would have q
3
= q
4
= 0. The same
result would be obtained by letting h = 0 in the finite-difference equation, Eq. (3). The result is
m,n-1m-1,nm,n
TT2T0.
+−=
<
Note that this expression is identical to Eq. 4.47 when h = 0, in which case both boundaries are
insulated.
COMMENTS: Note the convenience resulting from formulating the energy balance by assuming
that all the heat flow is into the node.
PROBLEM 4.38
KNOWN: Conduction in a one-dimensional (radial) cylindrical coordinate system with volumetric
generation.
FIND: Finite-difference equation for (a) Interior node, m, and (b) Surface node, n, with convection.
SCHEMATIC:
(a) Interior node, m (b) Surface node with convection, n
ASSUMPTIONS: (1) Steady-state, one-dimensional (radial) conduction in cylindrical coordinates,
(2) Constant properties.

ANALYSIS: (a) The network has nodes spaced at equal ∆r increments with m = 0 at the center;
hence, r = m∆r (or n∆r). The control volume is
(
)
V2 rr2mrr.
ππ=⋅∆⋅=∆∆⋅
ll
The energy
balance is
ingab
EEqqqV0
+=++=
&&
&
( )
m-1mm+1m
rTTrTT
k2rk2r+q2mrr0.
2r2r
πππ
∆−∆−


−++∆∆=



∆∆



&
lll
Recognizing that r = m∆r, canceling like terms, and regrouping find
2
m-1m+1m
11qmr
mTm+T2mT0.
22k
∆

−+−+=


&
<
(b) The control volume for the surface node is
(
)
V2 rr/2.
π
=⋅∆⋅
l The energy balance is
ingdconv
EEqqqV=0.+=++
&&
&
Use Fourier’s law to express q
d
and Newton’s law of cooling for
q

conv
to obtain
[ ]
( ) ( )
n-1n
n
rTTr
k2rh2 rTTq2nr0.
2r2
πππ
∞

∆−∆

−+−+∆=


∆


&
lll
Let r = n∆r, cancel like terms and regroup to find
2
n-1n
11hnrqnrhnr
nTnTT0.
22k2kk
∞
∆∆∆


−−−+++=




&
<
COMMENTS: (1) Note that when m or n becomes very large compared to ½, the finite-difference
equation becomes independent of m or n. Then the cylindrical system approximates a rectangular one.
(2) The finite-difference equation for the center node (m = 0) needs to be treated as a special case.
The control volume is
( )
2
Vr/2π=∆ l and the energy balance is
2
10
inga
TT
rr
EEqqVk2q0.
2r2
ππ
−
∆∆
+=+=+=
∆










&&
&&
ll
Regrouping, the finite-difference equation is
2
o1
qr
TT0.
4k
∆
−++=
&
PROBLEM 4.39
KNOWN: Two-dimensional cylindrical configuration with prescribed radial (∆r) and angular (∆φ)
spacings of nodes.
FIND: Finite-difference equations for nodes 2, 3 and 1.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in cylindrical
coordinates (r,φ), (3) Constant properties.
ANALYSIS: The method of solution is to define the appropriate control volume for each node, to
identify relevant processes and then to perform an energy balance.
(a) Node 2. This is an interior node with control volume as shown above. The energy balance is
inabcd
Eqqqq0.

′′′′
=+++=
&
Using Fourier’s law for each process, find
(
)
( )
(
)
( )
( )
( )
( )
( )
5232
i
i
i212
i
i
TTTT
3
krrkr
2rrr
TTTT
1
krrkr0.
2rrr
φ
φ

φ
φ
−−


+∆∆+∆+


∆+∆∆


−−


++∆∆+∆=


∆+∆∆


Canceling terms and regrouping yields,
( )
( )
( )
( )
( )
( )( )
( )
22
i2i531ii

22
i
i
rr
131
2rrTrrTTTrrT0.
rr22
rr
φφ
∆∆
−+∆+++∆++++∆=
+∆
∆+∆∆







(b) Node 3. The adiabatic surface behaves as a symmetry surface. We can utilize the result of Part
(a) to write the finite-difference equation by inspection as
( )
( )
( )
( )
( )
( )( )
22
i3i62ii

22
i
i
r2r
131
2rrTrrTTrrT0.
rr22
rr
φφ
∆∆
−+∆+++∆+⋅++∆=
+∆
∆+∆∆







(c) Node 1. The energy balance is
abcd
qqqq0.
′′′′
+++=
Substituting,
(
)
( )
(

)
( )
4121
i
i
TTTT
3
krrkr
22rrr

φ
φ
−−
∆

+∆+∆+


∆+∆∆


(
)
( )( )
i1
i1
TT
1
krrhrTT0
22r

φ
∞
−
∆

++∆+∆−=


∆


<
This expression could now be rearranged.
PROBLEM 4.40
KNOWN: Heat generation and thermal boundary conditions of bus bar. Finite-difference grid.
FIND: Finite-difference equations for selected nodes.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant
properties.
ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2)(∆y/2)⋅1, find the FDE
for node 1,
( )
( )
(
)
( )
( )
( ) ( )( )
( )
( )

o1
u121
t,c
61
t,cou26
ky/21
TT x
h1TTTT
R/y/212x
kx/21
TTqx/2y/210
y
x/kRThx/kTTT
∞
∞
∆⋅
− ∆

+⋅−+−

′′
∆∆

∆⋅

+−+∆∆=

∆
′′
∆+∆++

&
( )
(
)
( )
2
t,cu1
qx/2kx/kRhx/k2T0.

′′
+∆−∆+∆+=

&
<
(b) Performing an energy balance on the control volume, (∆x)(∆y/2)⋅1, find the FDE for node 13,
(
)
(
)
(
)
(
)
(
)
( )( )( ) ( )( )( ) ( )
l131213
8131413
hx1TTk/xy/21TT
k/yx1TTk/xy/21TTqxy/210

∞
∆⋅−+∆∆⋅−
+∆∆⋅−+∆∆⋅−+∆⋅∆⋅=
&
( ) ( ) ( ) ( )
2
l12814l13
hx/kT1/2T2TTqx/2khx/k2T0.
∞
∆++++∆−∆+=
&
<
COMMENTS: For fixed T
o
and T
∞
, the relative amounts of heat transfer to the air and heat sink
are determined by the values of h and
t,c
R.
′′
PROBLEM 4.41
KNOWN: Nodal point configurations corresponding to a diagonal surface boundary subjected to a
convection process and to the tip of a machine tool subjected to constant heat flux and convection
cooling.
FIND: Finite-difference equations for the node m,n in the two situations shown.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties.
ANALYSIS: (a) The control volume about node m,n has triangular shape with sides ∆x and ∆y while
the diagonal (surface) length is 2 ∆x. The heat rates associated with the control volume are due to

conduction, q
1
and q
2
, and to convection, q
c
. Performing an energy balance, find
( ) ( )
( )
( )
inout12c
m,n-1m,nm+1,nm,n
m,n
EE0 qqq0
TTTT
kx1ky1h2 x1TT0.
yx
∞
−=++=
−−
∆⋅+∆⋅+∆⋅−=
∆∆
&&
Note that we have considered the tool to have unit depth normal to the page. Recognizing that ∆x =
∆y, dividing each term by k and regrouping, find
m,n-1m+1,nm,n
hxhx
TT2 T22 T0.
kk
∞

∆∆

++⋅−+⋅=


<
(b) The control volume about node m,n has triangular shape with sides ∆x/2 and ∆y/2 while the lower
diagonal surface length is
( )
2x/2.
∆ The heat rates associated with the control volume are due to
the constant heat flux, q
a
, to conduction, q
b
, and to the convection process, q
c
. Perform an energy
balance,
( )
inoutabc
m+1,nm,n
om,n
EE0 qqq0
TT
xyx
q1k1h2TT0.
22x2
∞
−=++=

−
∆∆∆

′′
⋅⋅+⋅⋅+⋅⋅−=

∆

&&
Recognizing that ∆x = ∆y, dividing each term by k/2 and regrouping, find
m+1,nom,n
hxxhx
T2Tq12T0.
kkk
∞
∆∆∆

′′
+⋅⋅+⋅−+⋅=


<
COMMENTS: Note the appearance of the term h∆x/k in both results, which is a dimensionless
parameter (the Biot number) characterizing the relative effects of convection and conduction.
PROBLEM 4.42
KNOWN: Nodal point on boundary between two materials.
FIND: Finite-difference equation for steady-state conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant
properties, (4) No internal heat generation, (5) Negligible thermal contact resistance at interface.

ANALYSIS: The control volume is defined about nodal point 0 as shown above. The conservation of
energy requirement has the form
6
i123456
i1
qqqqqqq0
=
=+++++=
∑
since all heat rates are shown as into the CV. Each heat rate can be written using Fourier’s law,
102030
AAA
304010
BBB
TTTTTT
yy
kkxk
2xy2x
TTTTTT
yy
kkxk0.
2xy2x
−−−
∆∆
⋅⋅+⋅∆⋅+⋅⋅
∆∆∆
−−−
∆∆
+⋅⋅+⋅∆⋅+⋅⋅=
∆∆∆

Recognizing that ∆x = ∆y and regrouping gives the relation,
( ) ( )
AB
01234
ABAB
1k1k
TTTTT0.
42kk42kk
−++++=
++
<
COMMENTS: Note that when k
A
= k
B
, the result agrees with Eq. 4.33 which is appropriate for an
interior node in a medium of fixed thermal conductivity.
PROBLEM 4.43
KNOWN: Two-dimensional grid for a system with no internal volumetric generation.
FIND: Expression for heat rate per unit length normal to page crossing the isothermal boundary.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, (3) Constant
properties.
ANALYSIS: Identify the surface nodes (T
s
) and draw control volumes about these nodes. Since
there is no heat transfer in the direction parallel to the isothermal surfaces, the heat rate out of the
constant temperature surface boundary is
abcdef
qqqqqqq

′′′′′′′
=+++++
For each
i
q,
′
use Fourier’s law and pay particular attention to the manner in which the cross-
sectional area and gradients are specified.
( ) ( ) ( )
( ) ( ) ( )
1s2s3s
5s6s7s
TTTTTT
qky/2kyky
xxx
TTTTTT
kxkxkx/2
yyy
−−−
′
=∆+∆+∆
∆∆∆
−−−
+∆+∆+∆
∆∆∆
Regrouping with ∆x = ∆y, find
[
]
123567s
qk0.5TTTTT0.5T5T.

′
=+++++−
<
COMMENTS: Looking at the corner node, it is important to recognize the areas associated with
cd
q and q
′′
(∆y and ∆x, respectively).
PROBLEM 4.44
KNOWN: One-dimensional fin of uniform cross section insulated at one end with prescribed base
temperature, convection process on surface, and thermal conductivity.
FIND: Finite-difference equation for these nodes: (a) Interior node, m and (b) Node at end of fin, n,
where x = L.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction.
ANALYSIS: (a) The control volume about node m is shown in the schematic; the node spacing and
control volume length in the x direction are both ∆x. The uniform cross-sectional area and fin
perimeter are A
c
and P, respectively. The heat transfer process on the control surfaces, q
1
and q
2
,
represent conduction while q
c
is the convection heat transfer rate between the fin and ambient fluid.
Performing an energy balance, find
( )
inout12c

m-1mm+1m
ccm
EE0 qqq0
TTTT
kAkAhPxTT0.
xx
∞
−=++=
−−
++∆−=
∆∆
&&
Multiply the expression by ∆x/kA
c
and regroup to obtain
22
m-1m+1m
cc
hPhP
TTxT2xT0 1<m<n
kAkA
∞

++⋅∆−+∆=


<
Considering now the special node m = 1, then the m-1 node is T
b
, the base temperature. The finite-

difference equation would be
22
b21
cc
hPhP
TTxT2xT0 m=1
kAkA
∞

++∆−+∆=


<
(b) The control volume of length ∆x/2 about node n is shown in the schematic. Performing an energy
balance,
( )
inout34c
n-1n
cn
EE0 qqq0
TTx
kA0hPTT0.
x2
∞
−=++=
−∆
++−=
∆
&&
Note that q

4
= 0 since the end (x = L) is insulated. Multiplying by ∆x/kA
c
and regrouping,
22
n-1n
cc
hPxhPx
TT1T0.
kA2kA2
∞

∆∆
+⋅−⋅+=



<
COMMENTS: The value of ∆x will be determined by the selection of n; that is, ∆x = L/n. Note that
the grouping, hP/kA
c
, appears in the finite-difference and differential forms of the energy balance.
PROBLEM 4.45
KNOWN: Two-dimensional network with prescribed nodal temperatures and thermal conductivity of
the material.
FIND: Heat rate per unit length normal to page,
q.
′
SCHEMATIC:
Node T

i
(°C)
1 120.55
2 120.64
3 121.29
4 123.89
5 134.57
6 150.49
7 147.14
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, (3) No internal
volumetric generation, (4) Constant properties.
ANALYSIS: Construct control volumes around the nodes on the surface maintained at the uniform
temperature T
s
and indicate the heat rates. The heat rate per unit length is
abcde
qqqqqq
′′′′′′
=++++
or in terms of conduction terms between nodes,
123457
qqqqqqq.
′′′′′′′
=+++++
Each of these rates can be written in terms of nodal temperatures and control volume dimensions using
Fourier’s law,
1s2s3s4s
5s7s
TTTTTTTT
x

qkkxkxkx
2yyyy
TTTTy
kxk.
y2x
−−−−
∆
′ =⋅⋅+⋅∆⋅+⋅∆+⋅∆
∆∆∆∆
−−∆
+⋅∆+⋅⋅
∆∆
and since ∆x =∆y,
(
)
(
)
(
)
(
)
( ) ( ) ( )( )
1s2s3s
4s5s7s
qk[1/2TTTTTT
TTTT1/2TT].
′
=−+−+−
+−+−+−
Substituting numerical values, find

(
)
(
)
(
)
(
)
( ) ( ) ( )( )
q50 W/mK[1/2120.55100120.64100121.29100
123.89100134.571001/2147.14100]
′
=⋅−+−+−
+−+−+−
q6711 W/m.
′
=
<
COMMENTS: For nodes a through d, there is no heat transfer into the control volumes in the x-
direction. Look carefully at the energy balance for node e,
e57
qqq,
′′′
=+ and how
57
q and q
′′
are
evaluated.
PROBLEM 4.46

KNOWN:
Nodal temperatures from a steady-state, finite-difference analysis for a one-eighth
symmetrical section of a square channel.
FIND:
(a) Beginning with properly defined control volumes, derive the finite-difference equations for
nodes 2, 4 and 7, and determine T
2
, T
4
and T
7
, and (b) Heat transfer loss per unit length from the channel,
q
′
.
SCHEMATIC:
Node
T(
°
C)
1 430
3 394
6 492
8,9 600
ASSUMPTIONS:
(1) Steady-state conditions, (2) Two-dimensional conduction, (3) No internal
volumetric generation, (4) Constant properties.
ANALYSIS:
(a) Define control volumes about the nodes 2, 4, and 7, taking advantage of symmetry
where appropriate and performing energy balances,

in out
EE 0
−=

, with
∆
x =
∆
y,
Node 2:
abcd
qqqq0
′′′′
+++=
()() ()
32 62
12
2
TT TT TT
hxT T k y2 kx k y2 0
xy x
∞
−− −
∆−+∆ +∆ +∆ =
∆∆ ∆
() ()
2136
T 0.5T 0.5T T h x k T 2 h x k
∞


=+++∆ +∆

(
)
[]
2
2
T 0.5 430 0.5 394 492 50W m K 0.01m 1W m K 300 K 2 0.50

=×+×++ ⋅× ⋅ +


2
T 422K
=
<
Node 4:
abc
qqq0
′′′
++=
()( ) ()
34
4
TT
hx2T T 0ky2 0
x
∞
−
∆−++∆ =

∆
() ()
43
TThxkT1hxk
∞

=+∆ +∆

[][]
4
T 394 0.5 300 K 1 0.5 363K
=+× +=
<
Continued
PROBLEM 4.46 (Cont.)

Node 7: From the first schematic, recognizing that the diagonal is a symmetry adiabat, we can treat node
7 as an interior node, hence
()( )
73366
T 0.25 T T T T 0.25 394 394 492 492 K 443K
=+++= +++=
<
(b) The heat transfer loss from the upper surface can be expressed as the sum of the convection rates
from each node as illustrated in the first schematic,
cv1234
qqqqq
′′′′′
=+++
()( ) ( ) ( )()( )

cv 1 2 3 4
q h x2 T T hxT T hxT T h x2 T T
∞∞∞ ∞
′
=∆ −+∆ −+∆ −+∆ −
()()()()
2
cv
q 50W m K 0.1m 430 300 2 422 300 394 300 363 300 2 K
′

=⋅× −+−+−+−

cv
q 156W m
′
=
<
COMMENTS:
(1) Always look for symmetry conditions which can greatly simplify the writing of the
nodal equation as was the case for Node 7.
(2) Consider using the IHT Tool, Finite-Difference Equations, for Steady-State, Two-Dimensional heat
transfer to determine the nodal temperatures T
1
- T
7
when only the boundary conditions T
8
, T
9

and (
T
∞
,h)
are specified.
PROBLEM 4.47
KNOWN: Steady-state temperatures (K) at three nodes of a long rectangular bar.
FIND: (a) Temperatures at remaining nodes and (b) heat transfer per unit length from the bar using
nodal temperatures; compare with result calculated using knowledge of
q.
&
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties.
ANALYSIS: (a) The finite-difference equations for the nodes (1,2,3,A,B,C) can be written by
inspection using Eq. 4.39 and recognizing that the adiabatic boundary can be represented by a
symmetry plane.
( )
2
73
2
2
neighborsi
510 W/m0.005m
qx
T4Tqx/k0 and 62.5K.
k20 W/mK
×
∆
−+∆===
⋅

∑
&
&
Node A (to find T
2
):
2
2BA
2T2T4Tqx/k0
+−+∆=
&
( )
2
1
T2374.64398.062.5K390.2K
2
=−×+×−= <
Node 3 (to find T
3
):
2
c2B3
TTT300K4Tqx/k0
+++−+∆=
&
( )
3
1
T348.5390.2374.630062.5K369.0K
4

=++++=
<
Node 1 (to find T
1
):
2
C21
3002TT4Tqx/k0
++−+∆=
&
( )
1
1
T3002348.5390.262.5362.4K
4
=+×++= <
(b) The heat rate out of the bar is determined by calculating the heat rate out of each control volume
around the 300K nodes. Consider the node in the upper left-hand corner; from an energy balance
inoutgaa,ingg
EEE0 or qqE where EqV.
′′
−+==+=
&&&&&
&
Hence, for the entire bar
barabcdef
qqqqqqq,
′′′′′′′
=+++++ or
bar

C1
abc
C3 B
de
T300yT300xyxxy
qkqkyqyq
2x22x222
T300T300yyxT300xy
kxqxkxqxkq
y2y22y22
−∆−∆∆∆∆∆
′
=+⋅+∆+⋅∆+⋅+
∆∆
−−∆∆∆−∆∆
∆+∆⋅+∆+∆⋅++⋅
∆∆∆












&&&

&&&
f
.

Substituting numerical values, find
bar
q7,502.5 W/m.
′
= From an overall energy balance on the bar,
( ) ( )
bar
2
73
g
qEqV/q3x2y510W/m60.005m7,500 W/m.′′
===∆⋅∆=××=
&
&&
l <
As expected, the results of the two methods agree. Why must that be?
PROBLEM 4.48
KNOWN:
Steady-state temperatures at selected nodal points of the symmetrical section of a flow
channel with uniform internal volumetric generation of heat. Inner and outer surfaces of channel
experience convection.
FIND:
(a) Temperatures at nodes 1, 4, 7, and 9, (b) Heat rate per unit length (W/m) from the outer
surface A to the adjacent fluid, (c) Heat rate per unit length (W/m) from the inner fluid to surface B,
and (d) Verify that results are consistent with an overall energy balance.
SCHEMATIC:

ASSUMPTIONS:
(1) Steady-state, two-dimensional conduction, (2) Constant properties.
ANALYSIS:
(a) The nodal finite-difference equations are obtained from energy balances on control
volumes about the nodes shown in the schematics below.
Node 1
abcdg
qqqqE0
′′′′ ′
++++ =

() () ()
3121
TT
TT
0 k y/2 k x/2 0 q x y/4 0
xy
−−
+∆ +∆ ++∆⋅∆ =
∆∆

()
2
123
TTT/2qx/2k
=+ +∆

() ()( )
63 62
1

T 95.47 117.3 C / 2 10 W / m 25 25 10 m / 2 10 W / m K 122.0 C
−
=+°+ ×× × ⋅=°
Node 4
abcdef g
qqqqqqE 0
′′′′′′ ′
++++++ =

() ()
()
()( )
24
i,i4i 4
TT
k x/2 h y/2 T T h x/2 T T
y
∞∞
−
∆+∆−+∆−+
∆
Continued …
PROBLEM 4.48 (Cont.)
() () ()
54 84 34
TT TT TT
ky/2 kx ky
xyx
−−−
∆+∆+∆

∆∆∆
()
q3x y/4 0
+∆⋅∆ =

()
(
)
2
42358 i ,i
T T 2T T 2T 2 h x/k T 3q x /2k
∞

=++++∆ +∆



()
i
62hx/k

+∆

4
T
=
94.50 C
°
<
Node 7

abcdg
qqqqE0
′′′′ ′
++++ =

() () ()
()
()
37 87
o,o7
TT TT
k x/2 k y/2 h x/2 T T 0 q x y/4 0
yx
∞
−−
∆+∆+∆−++∆⋅∆=
∆∆

() ( )
2
738o ,o o
TTThx/kT qx/2k2hx/k
∞

=++∆ +∆ +∆



7
T 95.80 C

=°
<
Node 9
abcdg
qqqqE0
′′′′ ′
++++ =

() () ()
()
5 9 10 9
o,o9
TT T T
kx ky/2 h xT T
yy
∞
−−
∆+∆ +∆−
∆∆
() ()
89
TT
ky/2 qxy/2 0
x
−
+∆ +∆⋅∆ =
∆

() ( )
2

95 8 10o ,o o
T T 0.5T 0.5T h x / k T q x / 2k / 2 h x / k
∞

=+ + +∆ +∆ +∆



9
T 79.67 C
=°
<
(b) The heat rate per unit length from the outer surface A to the adjacent fluid,
A
q,
′
is the sum of the
convection heat rates from the outer surfaces of nodes 7, 8, 9 and 10.
()
()()()
()
()
Ao 7,o 8,o 9,o 10,o
qh x/2TT xTT xTT x/2TT
∞∞∞ ∞
′
=∆ −+∆−+∆−+∆ −


()( )( )

2
A
q 250 W / m K 25/ 2 95.80 25 25 87.28 25
′

=⋅ −+−

()()()
3
25 79.67 25 25/ 2 77.65 25 10 m K
−

+−+ −×⋅

Continued …
PROBLEM 4.48 (Cont.)
A
q 1117 W / m
′
=
<
(c) The heat rate per unit length from the inner fluid to the surface B,
B
q,
′
is the sum of the
convection heat rates from the inner surfaces of nodes 2, 4, 5 and 6.
()
()
()

()()
()
()
B i ,i 2 ,i 4 ,i 5 ,i 6
q h y/2T T y/2 x/2T T xT T x/2T T
∞∞∞∞
′
=∆ −+∆+∆ −+∆ −+∆ −


()( )( )( )
2
B
q 500 W / m K 25/ 2 50 95.47 25/ 2 25/ 2 50 94.50
′

=⋅ −++−

()()()
3
25 50 79.79 25/ 2 50 77.29 10 m K
−

+− + − × ⋅

B
q 1383 W / m
′
=−
<

(d) From an overall energy balance on the section, we see that our results are consistent since the
conservation of energy requirement is satisfied.
in out gen A B gen
E E E q q E ( 1117 1383 2500)W / m 0
′′ ′ ′′′
−+ =−++ =−−+ =
  
where
[]
63 62
gen
E q 10 W /m 25 50 25 50 10 m 2500 W / m
−
′′
=∀= × + × × =


COMMENTS:
The nodal finite-difference equations for the four nodes can be obtained by using
IHT Tool Finite-Difference Equations | Two-Dimensional | Steady-state. Options are provided to
build the FDEs for interior, corner and surface nodal arrangements including convection and internal
generation. The IHT code lines for the FDEs are shown below.
/* Node 1: interior node; e, w, n, s labeled 2, 2, 3, 3. */
0.0 = fd_2d_int(T1,T2,T2,T3,T3,k,qdot,deltax,deltay)
/* Node 4: internal corner node, e-n orientation; e, w, n, s labeled 5, 3, 2, 8. */
0.0 = fd_2d_ic_en(T4,T5,T3,T2,T8,k,qdot,deltax,deltay,Tinfi,hi,q• a4
q• a4 = 0 // Applied heat flux, W/m^2; zero flux shown
/* Node 7: plane surface node, s-orientation; e, w, n labeled 8, 8, 3. */
0.0 = fd_2d_psur_s(T7,T8,T8,T3,k,qdot,deltax,deltay,Tinfo,ho,q• a7
q• a7=0 // Applied heat flux, W/m^2; zero flux shown

/* Node 9: plane surface node, s-orientation; e, w, n labeled 10, 8, 5. */
0.0 = fd_2d_psur_s(T9, T10, T8, T5,k,qdot,deltax,deltay,Tinfo,ho,q• a9
q• a9 = 0 // Applied heat flux, W/m^2; zero flux shown
PROBLEM 4.49
KNOWN:
Outer surface temperature, inner convection conditions, dimensions and thermal
conductivity of a heat sink.
FIND:
Nodal temperatures and heat rate per unit length.
SCHEMATIC:
ASSUMPTIONS:
(1) Steady-state, (2) Two-dimensional conduction, (3) Uniform outer surface
temperature, (4) Constant thermal conductivity.
ANALYSIS:
(a) To determine the heat rate, the nodal temperatures must first be computed from the
corresponding finite-difference equations. From an energy balance for node 1,
()()() ()
5121
1
TT
TT
hx/21T T ky/21 kx1 0
xy
∞
−−
∆⋅ −+∆⋅ +∆⋅ =
∆∆
12 5
hx hx
3TT2TT0

kk
∞
∆∆

−+ + + + =


(1)
With nodes 2 and 3 corresponding to Case 3 of Table 4.2,
1236
hx 2hx
T2 2T T2T T 0
kk
∞
∆∆

−++++ =


(2)
237
hx hx
T2TTT0
kk
∞
∆∆

−+++ =



(3)
where the symmetry condition is invoked for node 3. Applying an energy balance to node 4, we
obtain
45s
2T T T 0
−++=
(4)
The interior nodes 5, 6 and 7 correspond to Case 1 of Table 4.2. Hence,
14 56s
T T 4T T T 0
+− ++=
(5)
25 67s
TT4TTT0
+− ++=
(6)
367s
T2T4TT0
+−+=
(7)
where the symmetry condition is invoked for node 7. With
s
T50C,T 20C,
∞
=° =°
and
()
2
h x / k 5000W / m K 0.005m / 240 W / m K 0.1042,
∆= ⋅ ⋅=

the solution to Eqs. (1) – (7) yields
1234
T 46.61 C, T 45.67 C, T 45.44 C, T 49.23 C
=°=°=°=°
567
T 48.46 C, T 48.00 C, T 47.86 C
=°=°=°
<
Continued …
PROBLEM 4.49 (Cont.)
The heat rate per unit length of channel may be evaluated by computing convection heat transfer from
the inner surface. That is,
()()()
12 3
q8hx/2TT xTT x/2TT
∞∞ ∞
′

=∆ − +∆ − +∆ −

() ()
2
q 8 5000W / m K 0.0025m 46.61 20 C 0.005m 45.67 20 C
′

=× ⋅ − °+ − °

()
]
0.0025m 45.44 20 C 10,340 W / m

+−°=
<
(b) Since
2
h 5000 W / m K
=⋅
is at the high end of what can be achieved through forced convection,
we consider the effect of reducing h. Representative results are as follows
(
)
2
hW/m K
⋅
()
1
TC
°

()
2
TC
°

()
3
TC
°

()
4

TC
°

()
5
TC
°

()
6
TC
°

()
7
TC
°
()
qW/m
′
200 49.84 49.80 49.79 49.96 49.93 49.91 49.90 477
1000 49.24 49.02 48.97 49.83 49.65 49.55 49.52 2325
2000 48.53 48.11 48.00 49.66 49.33 49.13 49.06 4510
5000 46.61 45.67 45.44 49.23 48.46 48.00 47.86 10,340
There are two resistances to heat transfer between the outer surface of the heat sink and the fluid, that
due to conduction in the heat sink,
()
cond 2D ,
R
and that due to convection from its inner surface to the

fluid,
conv
R.
With decreasing h, the corresponding increase in
conv
R
reduces heat flow and
increases the uniformity of the temperature field in the heat sink. The nearly 5-fold reduction in
q
′
corresponding to the 5-fold reduction in h from 1000 to 200
2
W/m K
⋅
indicates that the convection
resistance is dominant
()
()
conv cond 2D
RR .
>>
COMMENTS:
To check our finite-difference solution, we could assess its consistency with
conservation of energy requirements. For example, an energy balance performed at the inner surface
requires a balance between convection from the surface and conduction to the surface, which may be
expressed as
()
()
() ()
51

62 73
TT
TT TT
qkx1 kx1 kx/21
yy y
−
−−
′
=∆⋅ +∆⋅ +∆ ⋅
∆∆ ∆
Substituting the temperatures corresponding to
2
h 5000 W / m K,
=⋅
the expression yields
q10,340W/m,
′
=
and, as it must be, conservation of energy is precisely satisfied. Results of the
analysis may also be checked by using the expression
()
()
()
scond2Dconv
q TT/R R ,
∞
′′′
=− +
where, for
()

2
conv
h 5000 W / m K, R ,
3
1/ 4hw 2.5 10 m K/ W
′
=⋅
−
==×⋅
and from Eq. (4.27) and Case 11 of
Table 4.1,
()
[]
4
cond
R 0.930 ln W / w 0.05 / 2 k 3.94 10 m K / W.
π
−
′
=−=×⋅
Hence,
()
()
34
q 50 20 C / 2.5 10 3.94 10
mK/W
−−
′
=−° × + × ⋅

10,370 W / m,
=
and the agreement with the
finite-difference solution is excellent. Note that, even for
2
conv
h 5000 W / m K, R
′
=⋅>>

()
cond 2D
R.
′
PROBLEM 4.50
KNOWN: Steady-state temperatures (°C) associated with selected nodal points in a two-dimensional
system.
FIND: (a) Temperatures at nodes 1, 2 and 3, (b) Heat transfer rate per unit thickness from the
system surface to the fluid.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.
ANALYSIS: (a) Using the finite-difference equations for Nodes 1, 2 and 3:
Node 1, Interior node, Eq. 4.33:
1neighbors
1
TT
4
=⋅
∑
( )

1
1
T172.9137.0132.8200.0C160.7C
4
=+++=
o
o
<
Node 2, Insulated boundary, Eq. 4.46 with h = 0, T
m,n
= T
2
( )
2m-1,nm+1,nm,n-1
1
TTT2T
4
=++
( )
2
1
T129.445.82103.5C95.6C
4
=++×=
o
o
<
Node 3, Plane surface with convection, Eq. 4.46, T
m,n
= T

3
( )
3m-1,nm,n+1m,n-1
hx2hx
22T2TTTT
kk
∞
∆∆
+=+++



2
hx/k50W/mK0.1m/1.5W/mK3.33∆=⋅×⋅=
( ) ( )
3
23.332T2103.545.867.0C23.3330C
+=×+++××°
o
( )
3
1
T319.80199.80C=48.7C
10.66
=+°°
<
(b) The heat rate per unit thickness from the surface to the fluid is determined from the sum of the
convection rates from each control volume surface.
convabcd
qqqqq

′′′′′
=+++
(
)
iii
qhyTT
∞
=∆−
( )
conv
2
W0.1
q50m 45.830.0C
2
mK
′
=−°+
⋅



(
)
( )
( )
0.1m 48.730.0C
0.1m 67.030.0C
0.1m
200.030.0C
2

−°+
−°+
+−°



(
)
conv
q39.593.5185.0425 W/m743 W/m.
′
=+++=
<
PROBLEM 4.51
KNOWN:
Nodal temperatures from a steady-state finite-difference analysis for a cylindrical fin of
prescribed diameter, thermal conductivity and convection conditions (
T
∞
, h).
FIND:
(a) The fin heat rate, q
f
, and (b) Temperature at node 3, T
3
.
SCHEMATIC:
T
0
= 100.0

°
C
T
1
= 93.4
°
C
T
2
= 89.5
°
C
ASSUMPTIONS:
(a) The fin heat rate, q
f
, is that of conduction at the base plane, x = 0, and can be
found from an energy balance on the control volume about node 0,
in out
EE 0
−=

,
f 1 conv f 1 conv
q q q 0 or q q q
++ = =−−
.
Writing the appropriate rate equation for q
1
and q
conv

, with A
c
=
π
D
2
/4 and P =
π
D,
()( ) ()() ( )
2
10
fc 0 10 0
TT
kD
qkA hPx2TT TT 2DhxTT
x4x
π
π
∞∞
−
=− − ∆ − =− − − ∆ −
∆∆
Substituting numerical values, with
∆
x = 0.010 m, find
()
()
()
2

f
2
15W m K 0.012m
q 93.4 100 C
4 0.010m
0.012m 25W m K 0.010m 25 100 C
2
π
π
×⋅
=− −
×
−× × ⋅× −
()
f
q 1.120 0.353 W 1.473W
=+ =
. <
(b) To determine T
3
, derive the finite-difference equation for node 2, perform an energy balance on the
control volume shown above,
in out
EE 0
−=

,
cv 3 1
q qq0
++=

()
32
12
2c c
TT
TT
hP x T T kA kA 0
xx
∞
−
−
∆−+ + =
∆∆
[]
2
2
312 2
c
hP x
T T 2T x T T
kA
∞
∆
=− + − ∆ −
Substituting numerical values, find
2
T 89.2 C
=
<
COMMENTS:

Note that in part (a), the convection heat rate from the outer surface of the control
volume is significant (25%). It would have been poor approximation to ignore this term.