Heat and

Mass
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-mmthick layers of glass (k =0.78 W/mC) separated by a 12-mm-wide stagnant air
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 ground
surfaces 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 are
maintained 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) Consider a 2-m-high electric hot water heater that has a diameter of 40 cm and
maintains the hot water at 55C. The tank is located in a small room whose average
temperature is 27C, and the heat transfer coefficients on the inner and outer
surfaces of the heater are 50 and 12 W/m2C, respectively. The tank is placed in
another 46-cm-diameter sheet metal tank of negligible thickness, and the space
between the two tanks is filled with foam insulation (k = 0.03 W/m C). The
thermal resistances of the water tank and the outer thin sheet metal shell are very
small and can be neglected. Determine the heat loss from the tank. If 3 cm thick
fiber glass insulation is used to wrap the entire tank with K = 0.035C what will be
the heat loss
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 be
45C, .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. The space 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 at
25C, with a combined heat 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 of qv = 6 106 W/m3. Using the finite difference
method with a mesh size of x = y= 5.0 cm,
determine the temperatures at nodes:
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 difference method 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 of qv = 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
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. 3 Thermal 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

A (12
. m) (2 m) 2.4 m2
1
1

0.0417 C/W
2
h1 A (10 W/m .C)(2.4 m 2 )
L
0.003 m
R1 R3 Rglass 1
0.0016 C/W
k1 A (0.78 W/m.C)(2.4 m 2 )
L
0.012 m
R2 Rair 2
0.1923 C/W
k2 A (0.026 W/m.C)(2.4 m 2 )
1
1
Ro Rconv,2

0.0167 o C/W
h2 A (25 W/m2 .o C)(2.4 m 2 )
Rtotal Rconv,1 2 R1 R2 Rconv,2 0.0417 2(0.0016) 0.1923 0.0167
Ri Rconv,1

Air

0.2539 C/W
The steady rate of heat transfer through window glass then becomes
T T
[24 (5)]C
Q 1 2
114 W
Rtotal
0.2539C/W

Ri
T1
The inner surface temperature of the window glass can be determined from

R1

R2

R3

Ro
T2

T T
Q 1 1
T1 T1 Q Rconv,1 24 o C (114 W)(0.0417C/W) = 19.2C
Rconv,1

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. 2 Thermal conductivities are constant. 3
The furnace operates continuously. 4 The given heat transfer coefficient accounts for the radiation effects.
Properties The thermal conductivity of the glass wool insulation is given to be k = 0.038 W/mC.
Analysis The rate of heat transfer without insulation is

A (2 m)(1.5 m) 3 m2

Insulation

Q hA(Ts T ) (10 W / m2 . C)(3 m2 )(80 30) C 1500 W

In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be

Rinsulation

Q 010
. 1500 W 150 W
T
T (80 30) C
Q

Rtotal

0.333 C / W
Rtotal
150 W
Q

Ts

Ro

and in order to have this thermal resistance, the thickness of insulation must be

Rtotal Rconv Rinsulation

1
L

hA kA

(10 W/m .C)(3 m )

L 0.034 m 3.4 cm
2

L
(0.038 W/m.C)(3 m 2 )

0.333 C/W

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

As

D 2
4

(0.25 m) 2
4

0.0491 m 2
95C

Q hAs (Ts T )
Q
800 W
h

1254 W/m2 .C
As (Ts T ) (0.0491 m 2 )(108 95)C

108C
600 W

0.5 cm

(b) The outer surface temperature of the bottom of the pan is

Q kA
Ts ,outer

Ts ,outer Ts ,inner

L
Q L
(800 W)(0.005 m)
Ts ,inner1
108C +
108.3C
kA
(237 W/m.C)(0.0491 m 2 )

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 direction since the lateral surfaces of
both cylinders are well-insulated. 3 Thermal 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/m2C (Table 3-2).
Ri
T
1
Analysis (a) The thermal resistance network in this case consists of
two conduction resistance and the contact resistance, and are
determined to be
Rcontact
Rplate

1
1

0.0447 C/W
2
hc Ac (11,400 W/m .C)[ (0.05 m) 2 /4]

L
015
. m

0.4341 C / W
kA (176 W / m. C)[ (0.05 m) 2 / 4]

Then the rate of heat transfer is determined to be

Interface

Bar

Bar

Rglass

Ro
T2

(150 20)C
T
T
Q

142.4 W
Rtotal Rcontact 2 Rbar (0.0447 2 0.4341) C/W

Therefore, the rate of heat transfer through the bars is 142.4 W.

(b) The temperature drop at the interface is determined to be
Tinterface Q Rcontact (142.4 W)(0.0447 C/W) 6.4C

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. 3 Thermal conductivities are constant. 4 The 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

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
Ao Do L (0.46 m)(2 m) 2.89 m2

Rfoam

1
1
Ro

0.029 C/W
ho Ao (12 W/m2 .C)(2.89 m 2 )

Ro
T2

Tw

ln(r2 / r1 )
ln(23 / 20)

0.37 C/W
2kL
2 (0.03 W/m2 .C)(2 m)
Ro R foam 0.029 0.37 0.40 C/W

R foam
Rtotal

The rate of heat loss from the hot water tank is

T T
(55 27) C
Q w 2
70 W
Rtotal
0.40 C / W

The amount and cost of heat loss per year are

Q Q t (0.07 kW)(365 24 h / yr) 6132
. kWh / yr
Cost of Energy ( Amount of energy)(Un it cost) = (613.2 kWh)($0.08 / kWh) $49.056
f

$49.056
0.1752 17.5%
$280

If 3 cm thick fiber glass insulation is used to wrap the entire tank, the individual resistances becomes
Ao Do L (0.52 m)(2 m) 3.267 m 2
1
1
Ro

0.026 o C/W
2
o
ho Ao (12 W/m . C)(3.267 m 2 )

Rfoam
Tw

ln(r2 / r1 )
ln(23 / 20)

0.371 C/W
2k1 L
2(0.03 W/m2 .C)(2 m)
ln(r3 / r2 )
ln(26 / 23)
R fiberglass

0.279 C/W
2k 2 L
2(0.035 W/m2 .C)(2 m)
R foam

Rtotal Ro R foam R fiberglass 0.026 0.371 0.279 0.676 C/W

The rate of heat loss from the hot water heater in this case is
T T 2 (55 27)C
Q w

41.42 W
Rtotal
0.676 C/W

Rfiberglass

Ro
T2

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 onedimensional since the plate is large relative to its thickness, and there is thermal symmetry about the center plane 3
Thermal conductivity is constant. 4 Heat generation is uniform.
Properties The thermal conductivity is given to be k =15.1 W/mC.
Analysis The lowest temperature will
k
occur at surfaces of plate while the
highest temperature will occur at the
T =30C
g
midplane. Their values are
determined directly fromh=60 W/m2.C 2L=3 cm
gL
(5 10 5 W/m3 )(0.015 m)
Ts T
30C
155C
h
60 W/m2 .C
gL2
(5 10 5 W/m3 )(0.015 m) 2
To Ts
155C
158.7C
2k
2(15.1 W/m.C)

T =30C
2

h=60 W/m .C

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. 3 Thermal
conductivity is constant. 4 Heat generation in the rod is uniform.
Properties The thermal conductivity is given to be k = 29.5 W/mC.
Analysis The center temperature of the rod is determined from
To Ts

10)

gro 2
(7 10 7 W/m3 )( 0.025 m) 2
175C
545.8C
4k
4(29.5 W/m.C)

175C
g

Uranium rod

11)

12) The number of fins, finned and unfinned surface areas, and heat transfer rates from those areas are
n

1 m2
27777
(0.006 m)(0.006 m)

(0.0025) 2
D 2
2
Afin 27777 DL
6.68 m
27777 (0.0025)( 0.03)
4
4

D 2
1 27777 (0.0025) 0.86 m 2
Aunfinned 1 27777
4
4

Q
Q
hA (T T )
finned

fin

fin,max

fin

fin

0.959(35 W/m .C)( 6.68 m )(100 30)C

15,700 W
2

Q unfinned hAunfinned(Tb T ) (35 W/m2 o C)( 0.86 m 2 )(100 30)C 2107 W

Then the total heat transfer from the finned plate becomes

Q total,fin Q finned Q unfinned 15,700 2107 1.78 10 4 W 17.8 W

The rate of heat transfer if there were no fin attached to the plate would be

Ano fin (1 m)(1 m) 1 m2

Q no fin hAno fin (Tb T ) (35 W / m2 . C)(1 m2 )(100 30) C 2450 W
Then the fin effectiveness becomes

fin

Q fin
17800

7.27

2450
Qno fin

13) Tleft Ttop Tright Tbottom 4Tnode

g node l 2
0
k

where

g nodel 2 g 0 l 2 (8 10 6 W/m3 )(0.05 m) 2

93.5C
k
k
214 W/m C

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:
g l 2
290 T1
l 240 T1
l 325 T1
kl
k
hl (T T1 ) 0 0
2
l
l
2
l
2k
2
g l
350 290 325 290 - 4 T2 0 0
k
g 0 l 2
260 290 240 200 - 4T3
0
k

Node 1 ( convection ) : k
Node 2 (interior) :
Node 3 (interior) :
where

k 45 W/m.C, h 50 W/m2 .C, g 8 10 6 W/m3 , T 20C

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

element, m

hA

surface, m (Tm

T )

h(l / 2)( 200 T ) hl (240 T ) hl (T1 T ) h(l / 2)(325 T )

(50 W/m2 C)(0.05 m 1 m)[(200 - 20)/2 (240 - 20) (280.9 - 20) (325 - 20)/2]C 1808 W

14) Tleft Ttop Tright Tbottom 4Tnode

g nodel 2
0 Tnode (Tleft Ttop Tright Tbottom) / 4
k

There is symmetry about the horizontal, vertical, and diagonal lines

passing through the midpoint, and thus we need to consider only
150
th

1/8 of the region. Then,

180

200

180

150

T1 T3 T7 T9
T2 T4 T6 T8

180

Therefore, there are there are only 3 unknown nodal temperatures,

T1 , T3 , and T5 , 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

200
difference equations for the interior nodes.

150
180

180

200

180

180

200

150
180

Node 1 (interior) :

T1 (180 180 2T2 ) / 4

Node 2 (interior) :

T2 (200 T5 2T1 ) / 4

Node 3 (interior) :

T5 4T2 / 4 T2

Solving the equations above simultaneously gives

T1 T3 T7 T9 185C
T2 T4 T5 T6 T8 190C

15) Tleft Ttop Tright Tbottom 4Tnode

g nodel 2
0 Tnode (Tleft Ttop Tright Tbottom) / 4
k

(a) There is symmetry about the insulated surfaces as well as about the diagonal line. Therefore T3 T2 , and

T1 , T2 , and T4 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.

Node 1 (interior) :

T1 (180 180 T2 T3 ) / 4

Node 2 (interior) :

T2 (200 T4 2T1 ) / 4

Node 4 (interior) :

T4 (2T2 2T3 ) / 4

T3 T2

Also,

150

180

Insulated

180

200

Solving the equations above simultaneously gives

T2 T3 T4 190C
T1 185C

Insulated

(b) There is symmetry about the insulated surface as well as the diagonal line. Replacing the symmetry lines by
200
insulation, and utilizing the mirror-image concept, the finite difference equations for the interior nodes can be written
as

Node 1 (interior) :

T1 (120 120 T2 T3 ) / 4

Node 2 (interior) :

T2 (120 120 T4 T1 ) / 4

Node 3 (interior) :

T3 (140 2T 1 T4 ) / 4 T2

Node 4 (interior) :

T4 (2T2 140 2T3 ) / 4

100

120

120

Solving the equations above simultaneously gives

T1 T2 122.9C

120

T3 T4 128.6C

120

100

Insulated

140

140

16) Tleft Ttop Tright Tbottom 4Tnode

g nodel 2
0
k

There is symmetry about a vertical line passing through the middle of the region, and thus we need to consider only
half of the region. Then,

T1 T2 and T3 T4
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.

gl 2
0
k
gl 2
150 200 T1 T4 4T3
0
k

100 120 T2 T3 4T1

Node 1 (interior) :
Node 3 (interior) :

100

100

100

120

0.1 m

150
200
200

200

Noting that T1 T2 and T3 T4 and substituting,

(10 7 W/m3 )(0.1 m) 2

0
180 W/m C
(10 7 W/m3 )(0.1 m) 2
350 T1 3T3
0
180 W/m C
220 T3 3T1

The solution of the above system is

T1 T2 411.5C
T3 T4 439.0C

17) Tleft Ttop Tright Tbottom 4Tnode

120

100

150
200

g 0 l 2
0
k

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: q L

l
l
l T2 T1
l T4 T1
l2
h (T T1 ) k
k
g 0
0
2
2
2
l
2
l
4

Node 2: hl (T T2 ) k

T T2
l T1 T2
l T3 T2
l2
k
kl 5
g 0
0
2
l
2
l
l
2

Node 3: hl (T T3 ) k

l T2 T3
l T6 T3
l2
k
g 0
0
2
l
2
l
4

Node 4: q L l k

T T4
l T1 T4
l 120 T4
l2
k
kl 5
g 0
0
2
l
2
l
l
2

Node 5: T4 T2 T6 120 4T5

g 0 l 2
0
k

Node 6: hl (T T6 ) k

T T6
120 T6
l T3 T6
l T7 T6
3l 2
kl 5
kl
k
g 0
0
2
l
l
l
2
l
4

Node 7: hl (T T7 ) k

120 T7
l T6 T7
l T8 T7
l2
k
kl
g 0
0
2
l
2
l
l
2

Node 8: h

l
l T7 T8
l 120 T8
l2
(T T8 ) k
k
g 0
0
2
2
l
2
l
4

2
where g 0 5 10 6 W/m3 , q L 8000 W/m2 , l = 0.015 m, k = 45 W/mC, h = 55 W/m C, and T =30C. This system

of 8 equations with 8 unknowns is the finite difference formulation of the 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