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Heat andMass
Transfer(ME 209)
"Solved Problems"
Part 1
References
Heat Transfer "A Practical Approach" by Yunus Cengel
Fundamentals of Heat and Mass Transfer by Incropera
Heat Transfer by J.P. Holman
1(Consider a 1.2-m-high and 2-m-wide double-pane window consisting of two 3-mm-thick layers of glass (k =0.78 W/mC) separated by a 12-mm-wide stagnant air
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space (k = 0.026 W/m C). Determine the steady rate of heat transfer through this
double-pane window and the temperature of its inner surface for a day during
which the room is maintained at 24C while the temperature of the outdoors is-5C. Take the convection heat transfer coefficients on the inner and outer surfaces
of the window to be 10 W/m2 C and 25 W/m2
C respectively.2(A 2-m1.5-m section of wall of an industrial furnace burning natural gas is not
insulated, and the temperature at the outer surface of this section is measured to be
80C. The temperature of the furnace room is 30C, and the combined convection
and radiation heat transfer coefficient at the surface of the outer furnace is 10
W/m2C. It is proposed to insulate this section of the furnace wall with glass wool
insulation (k = 0.038 W/mC) in order to reduce the heat loss by 90 percent.
Assuming the outer surface temperature of the metal section still remains at about
80C; determine the thickness of the insulation that needs to be used.
3(Water is boiling in a 25-cm-diameter aluminum pan (k = 237 W/m C) at 95C.Heat is transferred steadily to the boiling water in the pan through its 0.5-cm-thick
flat bottom at a rate of 800 W. If the inner surface temperature of the bottom of the
pan is 108C, determine (a) the boiling heat transfer coefficient on the inner surface
of the pan, and (b) the outer surface temperature of the bottom of the pan.
4(Two 5-cm-diameter, 15cm-long aluminum bars (k = 176 W/mC) with groundsurfaces are pressed against each other with a pressure of 20 atm (h = 11,400
W/m2C). The bars are enclosed in an insulation sleeve and, thus, heat transfer
from the lateral surfaces is negligible. If the top
and bottom surfaces of the two-bar system aremaintained at temperatures of 150C and 20C,
respectively, determine (a) the rate of heat transfer
along the cylinders under steady conditions and
(b) the temperature drop at the interface.
5(An electric current is passed through a wire 1 mm in diameter and 10 cm long. The
wire is submerged in liquid water at atmospheric pressure, and the current is
increased until the water boils. For this situation h = 5000 W/m2C, and the water
temperature will be 100 C. How much electric power must be supplied to the wire
to maintain the wire surface at 114 C?
6(Steam at 320C flows in a cast iron pipe (k = 80 W/m C)
whose inner and outer diameters are 5 cm and 5.5 cm,
respectively. The pipe is covered with 3-cm-thick glass
wool insulation with k = 0.05 W/m C. Heat is lost to the
surroundings at 5C by natural convection and radiation,
with a combined heat transfer coefficient of h2= 18 W/m
C. Taking the heat transfer coefficient inside the pipe to
be h1= 60 W/m2 C, determine the rate of heat loss from
the steam per unit length of the pipe. Also determine the temperature drops across
the pipe shell and the insulation.7(
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8(Consider a large 3-cm-thick stainless steel plate (k = 15.1 W/m C) in which heat is
generated uniformly at a rate of 5 105 W/m3. Both sides of the plate are exposed
to an environment at 30C with a heat transfer coefficient of 60 W/m C. Explain
where in the plate the highest and the lowest temperatures will occur, and
determine their values.9(In a nuclear reactor, 1-cm-diameter cylindrical uranium rods cooled by water from
outside serve as the fuel. Heat is generated uniformly in the rods (k= 29.5 W/m
C) at a rate of 7107 W/m3. If the outer surface temperature of rods is 175C,
determine the temperature at their center.
10) Consider a long resistance wire of radius r1 = 0.2 cm and thermal conductivity
kwire = 15 W/m C in which heat is generated uniformly as a result of resistance
heating at a constant rate of qv = 50 W/m3. The wire is embedded in a 0.5-cm-thick
layer of ceramic whose thermal conductivity is kceramic = 1.2 W/m C. The outer
surface temperature of the ceramic layer is measured to be45C, .and is surrounded by air at 30 C with heat transfer
coefficient is of 10 W/m2C. Determine the temperatures at
the center of the resistance wire and the interface of the wire
and the ceramic layer under steady conditions.
11) Steam in a heating system flows through tubes whose outer
diameter is 3 cm and whose walls are maintained at a temperature of 120C. Circular
aluminum fins (k=180 W/m C) of outer diameter 6 cm and constant thickness of 2
mm are attached to the tube. Thespace between the fins is 3 mm, and thus there are
200 fins per meter length of the tube. Heat is transferred to the surrounding air at25C, with a combinedheat transfer coefficient of h = 60 W/m2 C. Determine the
increase in heat transfer from the tube per meter of its length as a result of adding fins.
(Fin efficiency = 95%).
12)A hot surface at 100C is to be cooled by attaching 3-cm-long, 0.25-cm-diameter
aluminum pin fins (k =237 W/m C) to it, with a center-to-center distance of 0.6 cm.
The temperature of the surrounding medium is 30C, and the heat transfer coefficient
on the surfaces is 35 W/m2C. Determine the rate of heat transfer from the surface for
a 1-m 1-m section of the plate. Also determine the overall effectiveness of the fins. .
(Fin efficiency = 95.9%).13) Consider steady two-dimensional heat transfer in a
long solid body whose cross section is given in the
figure. The temperatures at the selected nodes and the
thermal conditions at the boundaries are as shown.
The thermal conductivity of the body is k= 45 W/m
C, and heat is generated in the body uniformly at a
rate ofqv= 6 106 W/m3. Using the finite difference
method with a mesh size of x = y= 5.0 cm,
determine the temperatures at nodes:
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14) Consider steady two-dimensional heat transfer in a long solid body whose cross
section is given in the figure. The measured temperatures at
selected points of the outer surfaces are as shown. The
thermal conductivity of the body is k = 45 W/m C, and
there is no heat generation. Using the finite differencemethod with a mesh size of x = y= 2.0 cm, determine
the temperatures at the indicated points in the medium.
15) Consider steady two-dimensional heat transfer in a
long solid bar whose cross section is given in the Figure 1
(a) and (b). The measured temperatures at selected points of the outer surfaces are as
shown. The thermal conductivity of the body is k = 20 W/m C, and there is no heat
generation. Using the finite difference method with a mesh size of x = y= 1.0 cm.
Determine the temperatures at the indicated points in the medium
FIGURE (1)
16) Consider steady two-dimensional heat transfer in a
long solid body whose cross section is given in the
figure. The temperatures at the selected nodes and the
thermal conditions on the boundaries are as shown.
The thermal conductivity of the body is k = 180
W/mC, and heat is generated in the body uniformly
at a rate ofqv= 107 W/m3. Using the finite difference
method with a mesh size of x = y= 10 cm,
determine the temperatures at nodes 1, 2, 3, and 4.
17) Consider steady two-dimensional heat transfer in an L-shaped solid body whose
cross section is given in the figure. The thermal conductivity of the body is k = 45
W/m C, and heat is generated in the body at a rate of qv= 5 106 W/m3. The right
surface of the body is insulated, and the bottom surface is maintained at a uniform
temperature of 120C. The entire top surface is
subjected to convection with ambient air at T = 30C
with a heat transfer coefficient of h = 55 W/m2 C,
and the left surface is subjected to heat flux at a
uniform rate of 8000 W/m2. The nodal network of the
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problem consists of 13 equally spaced nodes with x = y= 1.5 cm. Five of the nodes
are at the bottom surface and thus their temperatures are known. Obtain the finite
difference equations at the remaining eight nodes.
Solutions
1) A double-pane window consists of two 3-mm thick layers of glass separated by a 12-mm wide
stagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss
through the window and the inner surface temperature of the window are to be determined.
Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor
temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since
any significant temperature gradients will exist in the direction from the indoors to the outdoors.
3Thermal conductivities of the glass and air are constant. 4 Heat transfer by radiation is
negligible.
Properties The thermal conductivity of the glass and air are given to be kglass = 0.78 W/mC
and kair = 0.026 W/mC.
Analysis The area of the window and the individual resistances are
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A = =( . ( .12 2 2 4m) m) m2
C/W2539.0
0167.01923.0)0016.0(20417.02
C/W0167.0)m4.2(C).W/m25(
11
C/W1923.0)m4.2(C)W/m.026.0(
m012.0
C/W0016.0
)m4.2(C)W/m.78.0(
m003.0
C/W0417.0)m4.2(C).W/m10(
11
2,211,
o
2o22
2,o
22
22
21
1glass31
221
1,i
=
+++=+++=
====
=
===
=
====
=
===
convconvtotal
conv
air
conv
RRRRR
AhRR
Ak
LRR
Ak
LRRR
AhRR
The steady rate of heat transfer through window glass then becomes
W114=
=
=
C/W2539.0
C)]5(24[21
totalR
TTQ
The inner surface temperature of the window glass can be determined from
C19.2==
= =C/W)W)(0.0417114(C24o1,11
1,
11conv
conv
RQTTR
TTQ
2) An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the
heat loss through that section of the wall by 90 percent. The thickness of the insulation that
needs to be used is to be determined. Also, the length of time it will take for the insulation to
pay for itself from the energy it saves will be determined.
Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2Thermal
conductivities are constant. 3The furnace operates continuously. 4The given heat transfercoefficient accounts for the radiation effects.
Properties The thermal conductivity of the glass wool insulation is given to be k= 0.038
W/mC.
AnalysisThe rate of heat transfer without insulation is
A = =(2 3m)(1.5 m) m2
( ) ( . ( )(80 )Q hA T T s= = = 10 3 30 1500W / m C) m C W2 2
In order to reduce heat loss by 90%, the new heat transfer rate and thermal
resistance must be
.
( ).
Q
QT
RR
T
Qtotaltotal
= =
= = =
=
010 1500 150
80 30
1500333
W W
C
WC / W
and in order to have this thermal resistance, the thickness of insulation must be
cm3.4==
=
+
=
+=+=
m034.0
C/W333.0)mC)(3W/m.038.0()mC)(3.W/m10(
1
1
222
conv
L
L
kA
L
hARRR insulationtotal
Air
R1
R2
R3
Ro
Ri
T1
T2
Insulation
Ro
T
Rinsulation
Ts L
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3) Heat is transferred steadily to the boiling water in an aluminum pan. The inner surface
temperature of the bottom of the pan is given. The boiling heat transfer coefficient and the
outer surface temperature of the bottom of the pan are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the
thickness of the bottom of the pan is small relative to its diameter. 3 The thermal conductivity
of the pan is constant.Properties The thermal conductivity of the aluminum pan is given to be k= 237 W/mC.
Analysis (a) The boiling heat transfer coefficient is
222
m0491.04
m)25.0(
4===
DAs
C.W/m1254
2 =
=
=
=
C)95108)(m0491.0(
W800
)(
)(
2TTA
Qh
TThAQ
ss
ss
(b) The outer surface temperature of the bottom of the pan is
C108.3=
=+=
=
)mC)(0.0491W/m.237(
m)005.0W)(800(+C108
21,,
,,
kA
LQTT
L
TTkAQ
innersouters
innersouters
4) Two cylindrical aluminum bars with ground surfaces are pressed against each other
in an insulation sleeve. For specified top and bottom surface temperatures, the rate of
heat transfer along the cylinders and the temperature drop at the interface are to be
determined.
Assumptions 1 Steady operating conditions exist. 2
Heat transfer is one-dimensional in the axial directionsince the lateral surfaces of both cylinders are well-
insulated. 3Thermal conductivities are constant.
Properties The thermal conductivity of aluminum bars
is given to be k= 176 W/mC. The contact
conductance at the interface of aluminum-aluminum
plates for the case of ground surfaces and of 20 atm 2
MPa pressure is hc= 11,400 W/m2
C (Table 3-2).
Analysis (a) The thermal resistance network in this
case consists of two conduction resistance and the
contact resistance, and are determined to be
C/W0447.0/4]m)(0.05C)[.W/m400,11(
1122
ccontact =
==
cAhR
RL
kAplate 2
m
(176 W/ m. C)[ (0.05 m) / 4]C / W= =
=
01504341
..
Then the rate of heat transfer is determined to be
W142.4=+
=
+
=
=
C/W)4341.020447.0(
C)20150(
2 barcontacttotal RR
T
R
TQ
Therefore, the rate of heat transfer through the bars is 142.4 W.
95C
108C600
W0.5
cm
Ri
Rglass
Ro
T1
T2
Bar Bar
Interface
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(b) The temperature drop at the interface is determined to be
C6.4=== C/W)W)(0.04474.142(contactinterface RQT
6)
7(An electric hot water tank is made of two concentric cylindrical metal sheets with
foam insulation in between. The fraction of the hot water cost that is due to the heat
loss from the tank and the payback period of the do-it-yourself insulation kit are to be
determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with
time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the
center line and no variation in the axial direction. 3Thermal conductivities are
constant. 4The thermal resistances of the water tank and the outer thin sheet metal
shell are negligible. 5 Heat loss from the top and bottom surfaces is negligible.
Properties The thermal conductivities are given to be k= 0.03 W/mC for foam
insulation and k= 0.035 W/mC for fiber glass insulation
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Analysis We consider only the side surfaces of the water heater for simplicity, and
disregard the top and bottom surfaces (it will make difference of about 10 percent).
The individual thermal resistances are
2m89.2m)2(m)46.0( === LDA oo
C/W029.0)m89.2(C).W/m12(
1122
=
==oo
oAh
R
C/W40.037.0029.0
C/W37.0)m2(C).W/m03.0(2
)20/23ln(
2
)/ln(2
12
=+=+=
=
==
foamototal
foam
RRR
kL
rrR
The rate of heat loss from the hot water tank is
(55 )
Q
T T
R
w
total=
=
=
2 27
70
C
0.40 C / W W
The amount and cost of heat loss per year are
Q Q t= = = ( . . 0 07 6132kW)(365 24 h / yr) kWh / yr
17.5%===
==
1752.0280$
056.49$
056.49$kWh)/08.0($kWh)2.613(=cost)itenergy)(UnofAmount(EnergyofCost
f
If 3 cm thick fiber glass insulation is used to wrap the entire tank, the individual
resistances becomes
2m267.3m)2(m)52.0( === LDA oo
C/W026.0)m267.3(C).W/m12(
11 o2o2===
oo
oAh
R
C/W676.0279.0371.0026.0
C/W279.0)m2(C).W/m035.0(2
)23/26ln(
2
)/ln(
C/W371.0)m2(C).W/m03.0(2
)20/23ln(
2
)/ln(
22
23
21
12
=++=++=
===
=
=
=
fiberglassfoamototal
fiberglass
foam
RRRR
Lk
rr
R
Lk
rrR
The rate of heat loss from the hot water heater in this case is
W42.41C/W0.676
C)2755(2=
=
=
total
w
R
TTQ
Tw
Ro
T2
Rfoam
Tw
Rfibergla
ss
Ro
T2
Rfoam
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8) Both sides of a large stainless steel plate in which heat is generated uniformly are exposed to
convection with the environment. The location and values of the highest and the lowest
temperatures in the plate are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2
Heat transfer is one-dimensional since the plate is large relative to its thickness, and there is
thermal symmetry about the center plane 3Thermal conductivity is constant. 4 Heat generation
is uniform.
PropertiesThe thermal conductivity is given to be k=15.1 W/mC.
Analysis The lowest
temperature will occur at
surfaces of plate while the
highest temperature will occur
at the midplane. Their values
are determined directly from
C155=
+=+=
C.W/m60
m)015.0)(W/m105(C30
2
35
h
LgTTs
C158.7=
+=+=
C)W/m.1.15(2
m)015.0)(W/m105(C155
2
2352
k
LgTT so
9) A nuclear fuel rod with a specified surface temperature is used as the fuel in a nuclear
reactor. The center temperature of the rod is to be determined.
Assumptions 1 Heat transfer is steady since there is no
indication of any change with time. 2 Heat transfer is one-
dimensional since there is thermal symmetry about the center
line and no change in the axial direction. 3Thermal
conductivity is constant. 4 Heat generation in the rod is
uniform.
PropertiesThe thermal conductivity is given to be k= 29.5
W/mC.
Analysis The center temperature of the rod is determined
from
C545.8=
+=+=
C)W/m.5.29(4
m)025.0)(W/m107(C175
4
2372
k
rgTT oso
10)
T
=30C
h=60
W/m2.C
2L=3 cm
k
gT
=30C
h=60
W/m2.C
g175
C
Uranium
rod
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11)
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12) The number of fins, finned and unfinned surface areas, and heat transfer rates from those
areas are
/n = =1
0 006 0 00627777
m
m) m)
2
( . ( .
W2107C)30100)(m86.0)(CW/m35()(
W700,15
C)30100)(m68.6)(C.W/m35(959.0
)(
m86.04
)0025.0(277771
4277771
m68.64
)0025.0()03.0)(0025.0(277774
27777
2o2unfinnedunfinned
22
finfinmaxfin,finfinned
222
unfinned
222fin
===
==
==
=
=
=
=
+=
+=
TThAQ
TThAQQ
DA
DDLA
b
b
Then the total heat transfer from the finned plate becomes
W17.8==+=+= W1078.12107700,15 4unfinnedfinnedfintotal, QQQ
The rate of heat transfer if there were no fin attached to the plate would be
A
Q hA T T b
no fin2
no fin no fin2 2
m m m
W / m C m C W
= =
= = =
( )( )
( ) ( . )( )( )
1 1 1
35 1 100 30 2450
Then the fin effectiveness becomes
7.27===2450
17800
finno
finfin
Q
Q
13) T T T T T g l
kleft top right bottom nodenode
+ + + + =4 0
2
where
C5.93CW/m214
)m05.0)(W/m108( 2362
02
node=
==
k
lg
k
lg
The finite difference equations for boundary nodes are obtained by applying an energy
balance on the volume elements and taking the direction of all heat transfers to be
towards the node under consideration:
04-200240290260:(interior)3Node
04-290325290350:(interior)2Node
02
)(325
2
290240
2:)convection(1Node
20
3
20
2
20
1111
=++++
=++++
=++
+
+
k
lgT
k
lgT
k
lgTThl
l
Tlk
l
Tkl
l
Tlk
where C20,W/m108C,.W/m50C,W/m.45 362 ==== Tghk
Substituting, T1 = 280.9C, T2 = 397.1C, T3 = 330.8C,
(b) The rate of heat loss from the bottom surface through a 1-m long section is
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W1808=+++=
+++=
==
C20)/2]-(32520)-(280.920)-(24020)/2-m)[(2001mC)(0.05W/m50(
)325)(2/()()240()200)(2/(
)(
2
1
surface,element,
TlhTThlThlTlh
TThAQQ
m
mm
m
m
14) 4/)(04 bottomrighttopleftnode
2node
nodebottomrighttopleft TTTTTk
lgTTTTT +++==++++
There is symmetry about the horizontal, vertical, and
diagonal lines passing through the midpoint, and thus
we need to consider only 1/8th of the region. Then,
8642
9731
TTTT
TTTT
===
===
Therefore, there are there are only 3 unknown nodal
temperatures, 531 and,, TTT , and thus we need only
3 equations to determine them uniquely. Also, we can
replace the symmetry lines by insulation and utilize the
mirror-image concept when writing the finite difference
equations for the interior nodes.
225
152
21
4/4:(interior)3Node
4/)2200(:(interior)2Node
4/)2180180(:(interior)1Node
TTT
TTT
TT
==
++=
++=
Solving the equations above simultaneously gives
C190
C185
=====
====
86542
9731
TTTTTTTTT
15) 4/)(04 bottomrighttopleftnode
2node
nodebottomrighttopleft TTTTTk
lgTTTTT +++==++++
(a) There is symmetry about the insulated surfaces as well as about the diagonal line. Therefore
23 TT = , and 421 and,, TTT are the only 3 unknown nodal temperatures. Thus we need only 3
equations to determine them uniquely. Also, we can replace the symmetry lines by insulation
and utilize the mirror-image concept when writing the finite difference equations for the interior
nodes.
4/)22(:(interior)4Node
4/)2200(:(interior)2Node4/)180180(:(interior)1Node
324
142
321
TTT
TTTTTT
+=
++=
+++=
Also, 23 TT =
Solving the equations above simultaneously gives
C185
C190
=
===
1
432
T
TTT
(b) There is symmetry about the insulated surface as well as the diagonal line. Replacing the
symmetry lines by insulation, and utilizing the mirror-image concept, the finite difference
equations for the interior nodes can be written as
18
0
20
0
180
150180200180
150
150180200180
150
18
0
20
0
18
0
12
3
456
78
9
Insulat
ed
18
0
20
0
150180
200
3
1 2
4
Insulat
ed
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4/)21402(:(interior)4Node
4/)12140(:(interior)3Node
4/)120120(:(interior)2Node
4/)120120(:(interior)1Node
324
243
142
321
TTT
TTTT
TTT
TTT
++=
=++=
+++=
+++=
Solving the equations above simultaneously gives
C128.6
C122.9
====
43
21
TT
TT
16) 042
nodenodebottomrighttopleft =++++
k
lgTTTTT
There is symmetry about a vertical line passing through the middle of the region, and thus weneed to consider only half of the region. Then,
4321 and TTTT ==
Therefore, there are there are only 2 unknown nodal temperatures, T1 and T3, and thus we need
only 2 equations to determine them uniquely. Also, we can replace the symmetry lines by
insulation and utilize the mirror-image concept when writing the finite difference equations for
the interior nodes.
04200150:(interior)3Node
04120100:(interior)1Node
2
341
2
132
=++++
=++++
k
lg
TTT
k
lgTTT
Noting that 4321 and TTTT == and substituting,
0CW/m180
m))(0.1W/m10(3350
0CW/m180
m))(0.1W/m10(3220
237
31
237
13
=
++
=
++
TT
TT
The solution of the above system is
C439.0
C411.5
==
==
43
21
TT
TT
100
120
140
120120
3
1 2
4
Insulat
ed
100
120
140
100100100
100
12
0
15
0
3
1 2
0.1
m
g
200200200
200
12
0
15
0
4
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17) 042
0nodebottomrighttopleft =++++
k
lgTTTTT
We observe that all nodes are boundary nodes except node 5 that is an interior node. Therefore,
we will have to rely on energy balances to obtain the finite difference equations. Using energy
balances, the finite difference equations for each of the 8 nodes are obtained as follows:
Node 1: 0422
)(22
2
01412
1 =+
+
++ l
gl
TTlk
l
TTlkTT
lh
lqL
Node 2: 0222
)(2
0252321
2 =+
+
+
+l
gl
TTkl
l
TTlk
l
TTlkTThl
Node 3: 0422
)(2
03632
3 =+
+
+l
gl
TTlk
l
TTlkTThl
Node 4: 02
120
22
2
045441
=+
+
+
+l
gl
TTkl
l
Tlk
l
TTlklqL
Node 5: 04120
2
05624 =++++k
lgTTTT
Node 6: 04
3
2
120
2)(
2
06766563
6 =+
+
+
+
+l
gl
TTlk
l
Tkl
l
TTkl
l
TTlkTThl
Node 7: 02
120
22)(
2
077876
7 =+
+
+
+l
gl
Tkl
l
TTlk
l
TTlkTThl
Node 8: 04
120
22)(
2
2
0887
8 =+
+
+l
gl
Tlk
l
TTlkTT
lh
where ,W/m8000,W/m105 2360 == Lqg l = 0.015 m, k= 45 W/mC, h = 55 W/m2
C,
and T
=30
C. This system of 8 equations with 8 unknowns is the finite difference formulation ofthe problem.
(b) The 8 nodal temperatures under steady conditions are determined by solving the 8
equations above simultaneously with an equation solver to be
T1 =163.6C, T2 =160.5C, T3 =156.4C, T4 =154.0C, T5 =151.0C, T6
=144.4C,
T7 =134.5C, T8 =132.6C
100100100
100
12
0
15
0
3
1 2
0.1
m
g
200200200
200
12
0
15
0
4
100100100
100
12
0
15
0
3
1 2
0.1
m
g
200200200
200
12
0
15
0
4
100100100
100
12
0
15
0
3
1 2
0.1
m
g
200200200
200
12
0
15
0
4