Solution manual heat and mass transfer a practical approach 3rd edition cengel CH03 3

3-23

Thermal Contact Resistance
3-40C The resistance that an interface offers to heat transfer per unit interface area is called thermal contact
resistance, Rc . The inverse of thermal contact resistance is called the thermal contact conductance.
3-41C The thermal contact resistance will be greater for rough surfaces because an interface with rough
surfaces will contain more air gaps whose thermal conductivity is low.
3-42C An interface acts like a very thin layer of insulation, and thus the thermal contact resistance has
significance only for highly conducting materials like metals. Therefore, the thermal contact resistance can
be ignored for two layers of insulation pressed against each other.
3-43C An interface acts like a very thin layer of insulation, and thus the thermal contact resistance is
significant for highly conducting materials like metals. Therefore, the thermal contact resistance must be
considered for two layers of metals pressed against each other.
3-44C Heat transfer through the voids at an interface is by conduction and radiation. Evacuating the
interface eliminates heat transfer by conduction, and thus increases the thermal contact resistance.
3-45C Thermal contact resistance can be minimized by (1) applying a thermally conducting liquid on the
surfaces before they are pressed against each other, (2) by replacing the air at the interface by a better
conducting gas such as helium or hydrogen, (3) by increasing the interface pressure, and (4) by inserting a
soft metallic foil such as tin, silver, copper, nickel, or aluminum between the two surfaces.

3-46 The thickness of copper plate whose thermal resistance is equal to the thermal contact resistance is to
be determined.
Properties The thermal conductivity of copper is k = 386 W/m⋅°C.
Analysis Noting that thermal contact resistance is the inverse of thermal contact conductance, the thermal
contact resistance is determined to be
Rc =

1
1
=
= 5.556 × 10 −5 m 2 .°C/W

hc 18,000 W/m 2 .°C

L
where L is the thickness of
k
the plate and k is the thermal conductivity. Setting R = R c , the equivalent thickness is determined from the
relation above to be

For a unit surface area, the thermal resistance of a flat plate is defined as R =

L = kR = kRc = (386 W/m ⋅ °C)(5.556 ×10 −5 m 2 ⋅ °C/W) = 0.0214 m = 2.14 cm
Therefore, the interface between the two plates offers as much resistance to heat transfer as a 2.14 cm thick
copper. Note that the thermal contact resistance in this case is greater than the sum of the thermal
resistances of both plates.

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3-24

3-47 Six identical power transistors are attached on a copper plate. For a maximum case temperature of
75°C, the maximum power dissipation and the temperature jump at the interface are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer can be approximated as being onedimensional, although it is recognized that heat conduction in some parts of the plate will be twodimensional since the plate area is much larger than the base area of the transistor. But the large thermal
conductivity of copper will minimize this effect. 3 All the heat generated at the junction is dissipated
through the back surface of the plate since the transistors are covered by a thick plexiglass layer. 4 Thermal
conductivities are constant.
Properties The thermal conductivity of copper is given to be k = 386 W/m⋅°C. The contact conductance at
the interface of copper-aluminum plates for the case of 1.17-1.4 μm roughness and 10 MPa pressure is hc =
49,000 W/m2⋅°C (Table 3-2).

Analysis The contact area between the case and the plate is given to be 9 cm2, and the plate area for each
transistor is 100 cm2. The thermal resistance network of this problem consists of three resistances in series
(contact, plate, and convection) which are determined to be
R contact =
R plate =

1
1
=
= 0.0227 °C/W
2
hc Ac (49,000 W/m ⋅ °C)(9 × 10 − 4 m 2 )

L
0.012 m
=
= 0.0031 °C/W
kA (386 W/m ⋅ °C)(0.01 m 2 )

Rconvection =

Plate
L

1
1
=
= 3.333 °C/W
ho A (30 W/m 2 ⋅ °C)(0.01 m 2 )

Q&

The total thermal resistance is then

R total = Rcontact + Rplate + Rconvection
= 0.0227 + 0.0031 + 3.333 = 3.359 °C/W
Note that the thermal resistance of copper plate is
very small and can be ignored all together. Then
the rate of heat transfer is determined to be
(75 − 23)°C
ΔT
=
= 15.5 W
Q& =
R total 3.359 °C/W

Rcontact

Rplate

Rconv

Tcase

T∞

Therefore, the power transistor should not be operated at power levels greater than 15.5 W if the case
temperature is not to exceed 75°C.
The temperature jump at the interface is determined from

ΔTinterface = Q& Rcontact = (15.5 W)(0.0227 °C/W) = 0.35°C
which is not very large. Therefore, even if we eliminate the thermal contact resistance at the interface
completely, we will lower the operating temperature of the transistor in this case by less than 1°C.

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3-25

3-48 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 wellinsulated. 3 Thermal conductivities are constant.

Interface

Bar

Properties The thermal conductivity of aluminum bars
is given to be k = 176 W/m⋅°C. 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 they are determined to be
Rcontact =

R plate =

Ri
T1

Bar

Rglass

Ro
T2

1
1
=
= 0.0447 °C/W
2
hc Ac (11,400 W/m ⋅ °C)[π (0.05 m) 2 /4]

0.15 m
L
=
= 0.4341 °C/W
kA (176 W/m ⋅ °C)[π (0.05 m) 2 /4]

Then the rate of heat transfer is determined to be
(150 − 20)°C
ΔT
ΔT
=

=
= 142.4 W
Q& =
R total Rcontact + 2 R bar (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

ΔTinterface = Q& Rcontact = (142.4 W)(0.0447 °C/W) = 6.4°C

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3-26

3-49 A thin copper plate is sandwiched between two epoxy boards. The error involved in the total thermal
resistance of the plate if the thermal contact conductances are ignored is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the plate is
large. 3 Thermal conductivities are constant.
Properties The thermal conductivities are given to be k = 386 W/m⋅°C for copper plates and k = 0.26
W/m⋅°C for epoxy boards. The contact conductance at the interface of copper-epoxy layers is given to be
hc = 6000 W/m2⋅°C.
Analysis The thermal resistances of different
layers for unit surface area of 1 m2 are
Rcontact =
R plate =
Repoxy

Copper

plate
Epoxy
Epoxy

1
1
=
= 0.00017 °C/W
hc Ac (6000 W/m 2 ⋅ °C)(1 m 2 )

0.001 m
L
=
= 2.6 × 10 −6 °C/W
kA (386 W/m ⋅ °C)(1 m 2 )

Q&

0.005 m
L
=
=
= 0.01923 °C/W
kA (0.26 W/m ⋅ °C)(1 m 2 )

5 mm

5 mm

The total thermal resistance is

R total = 2 Rcontact + R plate + 2 Repoxy
= 2 × 0.00017 + 2.6 × 10 − 6 + 2 × 0.01923 = 0.03880 °C/W

Then the percent error involved in the total thermal
resistance of the plate if the thermal contact
resistances are ignored is determined to be
2 Rcontact
2 × 0.00017
%Error =
× 100 =
× 100 = 0.88%
0.03880
R total

Rplate

Repoxy

Repoxy

T1
Rcontact

T2

Rcontact

which is negligible.

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3-27

Generalized Thermal Resistance Networks
3-50C Parallel resistances indicate simultaneous heat transfer (such as convection and radiation on a
surface). Series resistances indicate sequential heat transfer (such as two homogeneous layers of a wall).
3-51C The thermal resistance network approach will give adequate results for multi-dimensional heat
transfer problems if heat transfer occurs predominantly in one direction.
3-52C Two approaches used in development of the thermal resistance network in the x-direction for multidimensional problems are (1) to assume any plane wall normal to the x-axis to be isothermal and (2) to
assume any plane parallel to the x-axis to be adiabatic.

3-53 A typical section of a building wall is considered. The average heat flux through the wall is to be
determined.
Assumptions 1 Steady operating conditions exist.
Properties The thermal conductivities are given to be k23b = 50
W/m⋅K, k23a = 0.03 W/m⋅K, k12 = 0.5 W/m⋅K, k34 = 1.0 W/m⋅K.
Analysis We consider 1 m2 of wall area. The thermal resistances are
R12 =

t12
0.01 m
=
= 0.02 m 2 ⋅ °C/W
k12 (0.5 W/m ⋅ °C)

R 23a = t 23

La

k 23a ( La + Lb )
0.6 m
= 2.645 m 2 ⋅ °C/W
(0.03 W/m ⋅ °C)(0.6 + 0.005)
Lb
k 23b ( La + Lb )

= (0.08 m)
R 23b = t 23

= (0.08 m)
R34 =

0.005 m
= 1.32 × 10 −5 m 2 ⋅ °C/W
(50 W/m ⋅ °C)(0.6 + 0.005)

t 34
0.1 m
=
= 0.1 m 2 ⋅ °C/W
k 34 (1.0 W/m ⋅ °C)

The total thermal resistance and the rate of heat transfer are
⎛ R R
⎞
R total = R12 + ⎜⎜ 23a 23b ⎟⎟ + R34
⎝ R 23a + R 23b ⎠
⎛
1.32 × 10 −5

= 0.02 + 2.645⎜⎜
−5
⎝ 2.645 + 1.32 × 10
q& =

⎞
⎟ + 0.1 = 0.120 m 2 ⋅ °C/W
⎟
⎠

T4 − T1
(35 − 20)°C
=
= 125 W/m 2
2
R total
0.120 m ⋅ C/W

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3-28

3-54 A wall consists of horizontal bricks separated by plaster layers. There are also plaster layers on each
side of the wall, and a rigid foam on the inner side of the wall. The rate of heat transfer through the wall is
to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer
through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is
disregarded.

Properties The thermal conductivities are given to be k = 0.72 W/m⋅°C for bricks, k = 0.22 W/m⋅°C for
plaster layers, and k = 0.026 W/m⋅°C for the rigid foam.
Analysis We consider 1 m deep and 0.33 m high portion of wall which is representative of the entire wall.
The thermal resistance network and individual resistances are
R3
Ri

R1

R2

T∞1

R4

R6

R7

R5

T∞2

1
1
=
= 0.303 °C/W
h1 A (10 W/m 2 ⋅ °C)(0.33 × 1 m 2 )
L
0.02 m

=
=
= 2.33 °C/W
kA (0.026 W/m ⋅ °C)(0.33 × 1 m 2 )

Ri = Rconv,1 =
R1 = R foam

L
0.02 m
=
= 0.275 °C/W
kA (0.22 W/m ⋅ °C)(0.33 × 1 m 2 )
L
0.18 m
=
=
= 54.55 °C/W
ho A (0.22 W/m ⋅ °C)(0.015 × 1 m 2 )

R 2 = R6 = R plaster =
side

R3 = R5 = R plaster
center

L
0.18 m
=
= 0.833 °C/W

kA (0.72 W/m ⋅ °C)(0.30 × 1 m 2 )
1
1
=
=
= 0.152 °C/W
h2 A (20 W/m ⋅ °C)(0.33 × 1 m 2 )

R 4 = Rbrick =
Ro = Rconv, 2
1
R mid

=

1
1
1
1
1
1
+
+
=
+
+
⎯
⎯→ R mid = 0.81 °C/W
R3 R 4 R5 54.55 0.833 54.55

Rtotal = Ri + R1 + 2 R 2 + R mid + Ro = 0.303 + 2.33 + 2(0.275) + 0.81 + 0.152 = 4.145 °C/W

The steady rate of heat transfer through the wall per 0.33 m2 is
T −T
[(22 − (−4)]°C
Q& = ∞1 ∞ 2 =
= 6.27 W
4.145°C/W
Rtotal

Then steady rate of heat transfer through the entire wall becomes
( 4 × 6) m
Q& total = (6.27 W)
= 456 W
0.33 m 2
2

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3-29

3-55 EES Prob. 3-54 is reconsidered. The rate of heat transfer through the wall as a function of the
thickness of the rigid foam is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
A=4*6 [m^2]
L_brick=0.18 [m]
L_plaster_center=0.18 [m]

L_plaster_side=0.02 [m]
L_foam=2 [cm]
k_brick=0.72 [W/m-C]
k_plaster=0.22 [W/m-C]
k_foam=0.026 [W/m-C]
T_infinity_1=22 [C]
T_infinity_2=-4 [C]
h_1=10 [W/m^2-C]
h_2=20 [W/m^2-C]
"ANALYSIS"
R_conv_1=1/(h_1*A_1)
A_1=0.33*1 "[m^2]"
R_foam=(L_foam*Convert(cm, m))/(k_foam*A_1) "L_foam is in cm"
R_plaster_side=L_plaster_side/(k_plaster*A_1)
A_2=0.30*1 "[m^2]"
R_plaster_center=L_plaster_center/(k_plaster*A_3)
A_3=0.015*1 "[m^2]"
R_brick=L_brick/(k_brick*A_2)
R_conv_2=1/(h_2*A_1)
1/R_mid=2*1/R_plaster_center+1/R_brick
R_total=R_conv_1+R_foam+2*R_plaster_side+R_mid+R_conv_2
Q_dot=(T_infinity_1-T_infinity_2)/R_total
Q_dot_total=Q_dot*A/A_1

Qtotal [W]
634.6
456.2
356.1
292
247.4

214.7
189.6
169.8
153.7
140.4

700

600

Qtotal [W]

Lfoam [cm]
1
2
3
4
5
6
7
8
9
10

500

400

300

200

100
1

2

3

4

5

6

7

8

9

Lfoam [cm]

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10


3-30

3-56 A wall is to be constructed of 10-cm thick wood studs or with pairs of 5-cm thick wood studs nailed
to each other. The rate of heat transfer through the solid stud and through a stud pair nailed to each other,
as well as the effective conductivity of the nailed stud pair are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer can
be approximated as being one-dimensional since it is predominantly in the x direction. 3 Thermal
conductivities are constant. 4 The thermal contact resistance between the two layers is negligible. 4 Heat
transfer by radiation is disregarded.
Properties The thermal conductivities are given to be k = 0.11 W/m⋅°C for wood studs and k = 50 W/m⋅°C
for manganese steel nails.
Analysis (a) The heat transfer area of the stud is A = (0.1 m)(2.5 m) = 0.25 m2. The thermal resistance and
heat transfer rate through the solid stud are

L
0.1 m
=
= 3.636 °C/W
kA (0.11 W/m ⋅ °C)(0.25 m 2 )
8°C
ΔT
=
= 2.2 W
Q& =
R stud 3.636 °C/W

R stud =

Stud
L

Q&

(b) The thermal resistances of stud pair and nails are in parallel
⎡ π (0.004 m) ⎤
2
= 50⎢
⎥ = 0.000628 m
4
4
⎥⎦
⎣⎢
0.1 m
L
=
=
= 3.18 °C/W
kA (50 W/m ⋅ °C)(0.000628 m 2 )
0.1 m
L
=
=
= 3.65 °C/W
kA (0.11 W/m ⋅ °C)(0.25 − 0.000628 m 2 )
1
1
1
1
=
+
=

+
⎯
⎯→ Rtotal = 1.70 °C/W
R stud R nails 3.65 3.18

Anails = 50
R nails
R stud
1
Rtotal

πD 2

2

T1

T2

Rstud
T1

T2

8°C
ΔT
= 4.7 W
Q& =
=
R stud 1.70 °C/W

(c) The effective conductivity of the nailed stud pair can be determined from
(4.7 W)(0.1 m)
Q& L
ΔT
⎯
⎯→ k eff =
=
= 0.235 W/m.°C
Q& = k eff A
ΔTA (8°C)(0.25 m 2 )
L

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3-31

3-57 A wall is constructed of two layers of sheetrock spaced by 5 cm × 12 cm wood studs. The space
between the studs is filled with fiberglass insulation. The thermal resistance of the wall and the rate of heat
transfer through the wall are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer
through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients
account for the radiation heat transfer.
Properties The thermal conductivities are given to be k = 0.17 W/m⋅°C for sheetrock, k = 0.11 W/m⋅°C for
wood studs, and k = 0.034 W/m⋅°C for fiberglass insulation.
Analysis (a) The representative surface area is A = 1× 0.65 = 0.65 m 2 . The thermal resistance network and
the individual thermal resistances are
R2

Ri
R1
R4
R5
T∞1
T∞2

R3
1
1
=
= 0.185 °C/W
hi A (8.3 W/m 2 ⋅ °C)(0.65 m 2 )
0.01 m
L
R1 = R 4 = R sheetrock =
=
= 0.090 °C/W
kA (0.17 W/m ⋅ °C)(0.65 m 2 )
Ri =

0.16 m
L
=
= 29.091 °C/W
kA (0.11 W/m ⋅ °C)(0.05 m 2 )
0.16 m
L
R3 = R fiberglass =
=

= 7.843 °C/W
kA (0.034 W/m ⋅ °C)(0.60 m 2 )

R 2 = R stud =

1
1
=
= 0.045 °C/W
2
o
ho A (34 W/m ⋅ C)(0.65 m 2 )
1
1
1
1
=
+
=
+
⎯
⎯→ R mid = 6.178 °C/W
R 2 R3 29.091 7.843

Ro =
1
R mid

Rtotal = Ri + R1 + R mid + R 4 + Ro
= 0.185 + 0.090 + 6.178 + 0.090 + 0.045 = 6.588 °C/W (for a 1 m × 0.65 m section)

T −T
[20 − (−9)]°C
Q& = ∞1 ∞ 2 =
= 4.40 W
6.588 °C/W
Rtotal

(b) Then steady rate of heat transfer through entire wall becomes

(12 m)(5 m)
Q& total = (4.40 W)
= 406 W
0.65 m 2

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3-32

3-58E A wall is to be constructed using solid bricks or identical size bricks with 9 square air holes. There
is a 0.5 in thick sheetrock layer between two adjacent bricks on all four sides, and on both sides of the wall.
The rates of heat transfer through the wall constructed of solid bricks and of bricks with air holes are to be
determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer
through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients
account for the radiation heat transfer.
Properties The thermal conductivities are given to be k = 0.40 Btu/h⋅ft⋅°F for bricks, k = 0.015 Btu/h⋅ft⋅°F
for air, and k = 0.10 Btu/h⋅ft⋅°F for sheetrock.
Analysis (a) The representative surface area is A = (7.5 / 12)(7.5 / 12) = 0.3906 ft 2 . The thermal resistance

network and the individual thermal resistances if the wall is constructed of solid bricks are

R2
Ri

R1

T∞1

R3

R5

Ro

R4

T∞2

1
1
=
= 1.7068 h °F/Btu
2
hi A (1.5 Btu/h ⋅ ft ⋅ °F)(0.3906 ft 2 )
L
0.5 / 12 ft
R1 = R5 = R plaster =
=
= 1.0667 h ⋅ °F/Btu

kA (0.10 Btu/h ⋅ ft ⋅ °F)(0.3906 ft 2 )
Ri =

L
9 / 12 ft
=
= 288 h ⋅ °F/Btu
kA (0.10 Btu/h ⋅ ft ⋅ °F)[(7.5 / 12) × (0.5 / 12)]ft 2
L
9 / 12 ft
=
=
= 308.57 h ⋅ °F/Btu
kA (0.10 Btu/h ⋅ ft⋅ o F)[(7 / 12) × (0.5 / 12)]ft 2

R 2 = R plaster =
R3 = R plaster

L
9 / 12 ft
=
= 5.51 h ⋅ °F/Btu
kA (0.40 Btu/h ⋅ ft ⋅ °F)[(7 / 12) × (7 / 12)]ft 2
1
1
=
= 0.64 h ⋅ °F/Btu
Ro =
ho A (4 Btu/h ⋅ ft 2 ⋅ °F)(0.3906 ft 2 )
1

1
1
1
1
1
1
=
+
+
=
+
+
⎯
⎯→ R mid = 5.3135 h ⋅ °F/Btu
R mid
R 2 R3 R 4 288 308.57 5.51
R 4 = Rbrick =

Rtotal = Ri + R1 + R mid + R5 + Ro = 1.7068 + 1.0667 + 5.3135 + 1.0667 + 0.64 = 9.7937 h ⋅ °F/Btu
T −T
(80 − 30)°F
Q& = ∞1 ∞ 2 =
= 5.1053 Btu/h
9.7937 h ⋅ °F/Btu
Rtotal

Then steady rate of heat transfer through entire wall becomes

(30 ft)(10 ft)
Q& total = (5.1053 Btu/h)

= 3921 Btu/h
0.3906 m 2

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3-33

(b) The thermal resistance network and the individual thermal resistances if the wall is constructed of
bricks with air holes are
R2
Ri

R1

T∞1

R3

R6

Ro

R4

T∞2

R5
Aairholes = 9(1.5 / 12) × (1.55 / 12) = 0.1406 ft 2

Abricks = (7 / 12 ft) 2 − 0.1406 = 0.1997 ft 2
L
9 / 12 ft
=
= 355.62 h ⋅ °F/Btu
kA (0.015 Btu/h ⋅ ft ⋅ °F)(0.1406 ft 2 )
L
9 / 12 ft
=
=
= 9.389 h ⋅ °F/Btu
kA (0.40 Btu/h ⋅ ft ⋅ °F)(0.1997 ft 2 )

R 4 = R airholes =
R5 = Rbrick

1
1
1
1
1
1
1
1
1
=
+
+
+
=

+
+
+
⎯
⎯→ R mid = 8.618 h ⋅ °F/Btu
R mid
R 2 R3 R 4 R5 288 308.57 355.62 9.389
Rtotal = Ri + R1 + R mid + R6 + Ro = 1.7068 + 1.0667 + 8.618 + 1.0677 + 0.64 = 13.0992 h ⋅ °F/Btu
T −T
(80 − 30)°F
Q& = ∞1 ∞ 2 =
= 3.817 Btu/h
13.0992 h ⋅ °F/Btu
Rtotal

Then steady rate of heat transfer through entire wall becomes
(30 ft)(10 ft)
Q& total = (3.817 Btu/h)
= 2932 Btu/h
0.3906 ft 2

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3-34

3-59 A composite wall consists of several horizontal and vertical layers. The left and right surfaces of the
wall are maintained at uniform temperatures. The rate of heat transfer through the wall, the interface
temperatures, and the temperature drop across the section F are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer
through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Thermal contact resistances
at the interfaces are disregarded.
Properties The thermal conductivities are given to be kA = kF = 2, kB = 8, kC = 20, kD = 15, kE = 35 W/m⋅°C.
B

Analysis (a) The representative surface area is A = 0.12 × 1 = 0.12 m . The thermal resistance network and
the individual thermal resistances are
R2
R5
R1
R7
R3
T1
T2
2

R4
0.01 m

R6

⎛ L ⎞
R1 = R A = ⎜ ⎟ =
= 0.04 °C/W
⎝ kA ⎠ A (2 W/m ⋅ °C)(0.12 m 2 )
0.05 m
⎛ L ⎞
R 2 = R 4 = RC = ⎜ ⎟ =
= 0.06 °C/W

⎝ kA ⎠ C (20 W/m ⋅ °C)(0.04 m 2 )
0.05 m
⎛ L ⎞
= 0.16 °C/W
R3 = R B = ⎜ ⎟ =
⎝ kA ⎠ B (8 W/m ⋅ °C)(0.04 m 2 )
0.1 m
⎛ L ⎞
R5 = R D = ⎜ ⎟ =
= 0.11 °C/W
⎝ kA ⎠ D (15 W/m⋅ o C)(0.06 m 2 )
0. 1 m
⎛ L ⎞
R6 = R E = ⎜ ⎟ =
= 0.05 o C/W
⎝ kA ⎠ E (35 W/m ⋅ °C)(0.06 m 2 )
0.06 m
⎛ L ⎞
= 0.25 °C/W
R7 = R F = ⎜ ⎟ =
⎝ kA ⎠ F (2 W/m ⋅ °C)(0.12 m 2 )

1
Rmid ,1

=

1
1
1

1
1
1
+
+
=
+
+
⎯
⎯→ Rmid ,1 = 0.025 °C/W
R2 R3 R4 0.06 0.16 0.06

1
1
1
1
1
=
+
=
+
⎯
⎯→ Rmid , 2 = 0.034 °C/W
Rmid , 2 R5 R6 0.11 0.05
Rtotal = R1 + Rmid ,1 + Rmid , 2 + R7 = 0.04 + 0.025 + 0.034 + 0.25 = 0.349 °C/W
T − T∞ 2 (300 − 100)°C
=
= 572 W (for a 0.12 m × 1 m section)
Q& = ∞1
0.349 °C/W

Rtotal
Then steady rate of heat transfer through entire wall becomes
(5 m)(8 m)
Q& total = (572 W)
= 1.91 × 10 5 W
2
0.12 m
(b) The total thermal resistance between left surface and the point where the sections B, D, and E meet is
Rtotal = R1 + R mid ,1 = 0.04 + 0.025 = 0.065 °C/W
Then the temperature at the point where the sections B, D, and E meet becomes
T −T
Q& = 1
⎯
⎯→ T = T1 − Q& Rtotal = 300°C − (572 W)(0.065 °C/W) = 263°C
Rtotal
(c) The temperature drop across the section F can be determined from
ΔT
Q& =
→ ΔT = Q& R F = (572 W)(0.25 °C/W) = 143°C
RF

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3-35

3-60 A composite wall consists of several horizontal and vertical layers. The left and right surfaces of the
wall are maintained at uniform temperatures. The rate of heat transfer through the wall, the interface
temperatures, and the temperature drop across the section F are to be determined.

Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer
through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Thermal contact resistances
at the interfaces are to be considered.
Properties The thermal conductivities of various materials used are given to be kA = kF = 2, kB = 8, kC = 20,
kD = 15, and kE = 35 W/m⋅°C.
B

Analysis The representative surface area is A = 0.12 × 1 = 0.12 m 2

R2
R5

R1

R3

R4

R7

R8

R6

(a) The thermal resistance network and the individual thermal resistances are
0.01 m
⎛ L ⎞
R1 = R A = ⎜ ⎟ =
= 0.04 °C/W
kA

⎝ ⎠ A (2 W/m ⋅ °C)(0.12 m 2 )
0.05 m
⎛ L ⎞
R 2 = R 4 = RC = ⎜ ⎟ =
= 0.06 °C/W
⎝ kA ⎠ C (20 W/m ⋅ °C)(0.04 m 2 )
0.05 m
⎛ L ⎞
= 0.16 °C/W
R3 = R B = ⎜ ⎟ =
⎝ kA ⎠ B (8 W/m ⋅ °C)(0.04 m 2 )

0.1 m
⎛ L ⎞
R5 = R D = ⎜ ⎟ =
= 0.11 °C/W
⎝ kA ⎠ D (15 W/m⋅ o C)(0.06 m 2 )
0.1 m
⎛ L ⎞
R6 = R E = ⎜ ⎟ =
= 0.05 o C/W
2
⎝ kA ⎠ E (35 W/m ⋅ °C)(0.06 m )
0.06 m
⎛ L ⎞
= 0.25 °C/W
R7 = R F = ⎜ ⎟ =
kA
⎝ ⎠ F (2 W/m ⋅ °C)(0.12 m 2 )
R8 =

0.00012 m 2 ⋅ °C/W
0.12 m 2

1
R mid ,1

=

= 0.001 °C/W

1
1
1
1
1
1
+
+
=
+
+
⎯
⎯→ R mid ,1 = 0.025 °C/W
R 2 R3 R 4 0.06 0.16 0.06

1
1
1
1

1
=
+
=
+
⎯
⎯→ R mid , 2 = 0.034 °C/W
R mid , 2 R5 R6 0.11 0.05
Rtotal = R1 + R mid ,1 + R mid , 2 + R7 + R8 = 0.04 + 0.025 + 0.034 + 0.25 + 0.001
= 0.350 °C/W
T −T
(300 − 100)°C
= 571 W (for a 0.12 m × 1 m section)
Q& = ∞1 ∞ 2 =
0.350 °C/W
Rtotal

Then steady rate of heat transfer through entire wall becomes
(5 m)(8 m)
Q& total = (571 W)
= 1.90 × 10 5 W
2
0.12 m

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3-36

(b) The total thermal resistance between left surface and the point where the sections B, D, and E meet is
Rtotal = R1 + R mid ,1 = 0.04 + 0.025 = 0.065 °C/W

Then the temperature at the point where The sections B, D, and E meet becomes
T −T
Q& = 1
⎯
⎯→ T = T1 − Q& Rtotal = 300°C − (571 W)(0.065 °C/W) = 263°C
Rtotal

(c) The temperature drop across the section F can be determined from
ΔT
Q& =
⎯
⎯→ ΔT = Q& R F = (571 W)(0.25 °C/W) = 143°C
RF

3-61 A coat is made of 5 layers of 0.1 mm thick synthetic fabric separated by 1.5 mm thick air space. The
rate of heat loss through the jacket is to be determined, and the result is to be compared to the heat loss
through a jackets without the air space. Also, the equivalent thickness of a wool coat is to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer
through the jacket is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients
account for the radiation heat transfer.
Properties The thermal conductivities are given to be k = 0.13 W/m⋅°C for synthetic fabric, k = 0.026
W/m⋅°C for air, and k = 0.035 W/m⋅°C for wool fabric.
Analysis The thermal resistance network and the individual thermal resistances are
R1

R2

R3

R4

R5

R6

R7

R8

R9

Ro
T∞2

Ts1
0.0001 m
L
=
= 0.0006 °C/W
kA (0.13 W/m ⋅ °C)(1.25 m 2 )
L
0.0015 m
R air = R 2 = R 4 = R6 = R8 =
=
= 0.0462 °C/W
kA (0.026 W/m ⋅ °C)(1.25 m 2 )
1

1
Ro =
=
= 0.0320 °C/W
2
hA (25 W/m ⋅ °C)(1.25 m 2 )
Rtotal = 5 R fabric + 4 R air + Ro = 5 × 0.0006 + 4 × 0.0462 + 0.0320 = 0.2198 °C/W

R fabric = R1 = R3 = R5 = R7 = R9 =

and
T −T
(28 − 0)°C
= 127 W
Q& = s1 ∞ 2 =
Rtotal
0.2198 °C/W

If the jacket is made of a single layer of 0.5 mm thick synthetic fabric, the rate of heat transfer would be
T − T∞ 2
Ts1 − T∞ 2
(28 − 0)°C
Q& = s1
=
=
= 800 W
Rtotal
5 × R fabric + Ro (5 × 0.0006 + 0.0320) °C/W
The thickness of a wool fabric that has the same thermal resistance is determined from
L

1
R total = R wool + Ro =
+
kA
hA
fabric
0.2198 °C/W =

L
(0.035 W/m ⋅ °C)(1.25 m 2 )

+ 0.0320 ⎯
⎯→ L = 0.0082 m = 8.2 mm

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3-37

3-62 A coat is made of 5 layers of 0.1 mm thick cotton fabric separated by 1.5 mm thick air space. The rate
of heat loss through the jacket is to be determined, and the result is to be compared to the heat loss through
a jackets without the air space. Also, the equivalent thickness of a wool coat is to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer
through the jacket is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients
account for the radiation heat transfer.
Properties The thermal conductivities are given to be k = 0.06 W/m⋅°C for cotton fabric, k = 0.026 W/m⋅°C
for air, and k = 0.035 W/m⋅°C for wool fabric.
Analysis The thermal resistance network and the individual thermal resistances are

R1

R2

R3

R4

R5

R6

R7

R8

R9

Ro
T∞2

T1
L
0.0001 m
=
= 0.00133 °C/W
kA (0.06 W/m ⋅ °C)(1.25 m 2 )
L
0.0015 m
R air = R 2 = R 4 = R6 = R8 =

=
= 0.0462 °C/W
kA (0.026 W/m⋅ o C)(1.25 m 2 )
1
1
Ro =
=
= 0.0320 °C/W
2
hA (25 W/m ⋅ °C)(1.25 m 2 )
Rtotal = 5R fabric + 4 R air + Ro = 5 × 0.00133 + 4 × 0.0462 + 0.0320 = 0.2235 °C/W

Rcot ton = R1 = R3 = R5 = R 7 = R9 =

and
T −T
(28 − 0)°C
= 125 W
Q& = s1 ∞ 2 =
Rtotal
0.2235 °C/W

If the jacket is made of a single layer of 0.5 mm thick cotton fabric, the rate of heat transfer will be
T − T∞ 2
Ts1 − T∞ 2
(28 − 0)°C
Q& = s1
=
=
= 724 W

Rtotal
5 × R fabric + Ro (5 × 0.00133 + 0.0320) °C/W

The thickness of a wool fabric for that case can be determined from
R total = R wool + Ro =
fabric

0.2235 °C/W =

L
1
+
kA hA

L
(0.035 W/m ⋅ °C)(1.25 m 2 )

+ 0.0320 ⎯
⎯→ L = 0.0084 m = 8.4 mm

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3-38

3-63 A kiln is made of 20 cm thick concrete walls and ceiling. The two ends of the kiln are made of thin
sheet metal covered with 2-cm thick styrofoam. For specified indoor and outdoor temperatures, the rate of
heat transfer from the kiln is to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer

through the walls and ceiling is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer
coefficients account for the radiation heat transfer. 5 Heat loss through the floor is negligible. 6 Thermal
resistance of sheet metal is negligible.
Properties The thermal conductivities are given to be k = 0.9 W/m⋅°C for concrete and k = 0.033 W/m⋅°C
for styrofoam insulation.
Analysis In this problem there is a question of which surface area to use. We will use the outer surface area
for outer convection resistance, the inner surface area for inner convection resistance, and the average area
for the conduction resistance. Or we could use the inner or the outer surface areas in the calculation of all
thermal resistances with little loss in accuracy. For top and the two side surfaces:

Ri

Rconcrete

Ro

Tin

Tout

1
1
=
= 0.0071× 10 − 4 °C/W
hi Ai (3000 W/m 2 ⋅ °C)[(40 m)(13 − 1.2) m]
L
0.2 m
=
=
= 4.480 × 10 − 4 °C/W

kAave (0.9 W/m ⋅ °C)[(40 m)(13 − 0.6) m]

Ri =
Rconcrete

Ro =

1
1
=
= 0.769 × 10 − 4 °C/W
ho Ao (25 W/m 2 ⋅ °C)[(40 m)(13 m)]

Rtotal = Ri + Rconcrete + Ro = (0.0071 + 4.480 + 0.769) × 10 − 4 = 5.256 × 10 − 4 °C/W
and

T − Tout
[40 − (−4)]°C
=
= 83,700 W
Q& top + sides = in
Rtotal
5.256 × 10 − 4 °C/W

Heat loss through the end surface of the kiln with styrofoam:
Ri

Rstyrofoam

Ro

Tin

Tout

1
1
=
= 0.201× 10 − 4 °C/W
hi Ai (3000 W/m 2 ⋅ °C)[(4 − 0.4)(5 − 0.4) m 2 ]
0.02 m
L
=
=
= 0.0332 °C/W
kAave (0.033 W/m ⋅ °C)[(4 − 0.2)(5 − 0.2) m 2 ]

Ri =
R styrofoam

Ro =

1
1
=
= 0.0020 °C/W
ho Ao (25 W/m 2 ⋅ °C)[4 × 5 m 2 ]

Rtotal = Ri + R styrpfoam + Ro = 0.201 × 10 − 4 + 0.0332 + 0.0020 = 0.0352 °C/W

and

T − Tout [40 − (−4)]°C
=
= 1250 W
Q& end surface = in
0.0352 °C/W
Rtotal

Then the total rate of heat transfer from the kiln becomes
Q& total = Q& top + sides + 2Q& side = 83,700 + 2 × 1250 = 86,200 W

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3-39

3-64 EES Prob. 3-63 is reconsidered. The effects of the thickness of the wall and the convection heat
transfer coefficient on the outer surface of the rate of heat loss from the kiln are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
width=5 [m]
height=4 [m]
length=40 [m]
L_wall=0.2 [m]
k_concrete=0.9 [W/m-C]
T_in=40 [C]
T_out=-4 [C]
L_sheet=0.003 [m]

L_styrofoam=0.02 [m]
k_styrofoam=0.033 [W/m-C]
h_i=3000 [W/m^2-C]
h_o=25 [W/m^2-C]
"ANALYSIS"
R_conv_i=1/(h_i*A_1)
A_1=(2*height+width-6*L_wall)*length
R_concrete=L_wall/(k_concrete*A_2)
A_2=(2*height+width-3*L_wall)*length
R_conv_o=1/(h_o*A_3)
A_3=(2*height+width)*length
R_total_top_sides=R_conv_i+R_concrete+R_conv_o
Q_dot_top_sides=(T_in-T_out)/R_total_top_sides "Heat loss from top and the two side
surfaces"
R_conv_i_end=1/(h_i*A_4)
A_4=(height-2*L_wall)*(width-2*L_wall)
R_styrofoam=L_styrofoam/(k_styrofoam*A_5)
A_5=(height-L_wall)*(width-L_wall)
R_conv_o_end=1/(h_o*A_6)
A_6=height*width
R_total_end=R_conv_i_end+R_styrofoam+R_conv_o_end
Q_dot_end=(T_in-T_out)/R_total_end "Heat loss from one end surface"
Q_dot_total=Q_dot_top_sides+2*Q_dot_end

Lwall [m]
0.1
0.12
0.14
0.16
0.18

0.2
0.22
0.24
0.26
0.28
0.3

Qtotal [W]
151098
131499
116335
104251
94395
86201
79281
73359
68233
63751
59800

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educators for course preparation. If you are a student using this Manual, you are using it without permission.


3-40

ho [W/m2.C]
5
10
15

20
25
30
35
40
45
50

Qtotal [W]
54834
70939
78670
83212
86201
88318
89895
91116
92089
92882

160000

Qtotal [W]

140000
120000
100000
80000
60000
0.08

0.12

0.16

0.2

0.24

0.28

0.32

Lwall [m]
95000
90000

Qtotal [W]

85000
80000
75000
70000
65000
60000
55000
50000
5

10

15

20

25

30

35

40

45

50

2

ho [W/m -C]

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3-41

3-65E The thermal resistance of an epoxy glass laminate across its thickness is to be reduced by planting
cylindrical copper fillings throughout. The thermal resistance of the epoxy board for heat conduction across
its thickness as a result of this modification is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the plate is one-dimensional. 3
Thermal conductivities are constant.
Properties The thermal conductivities are given to be k = 0.10 Btu/h⋅ft⋅°F for epoxy glass laminate and k =
223 Btu/h⋅ft⋅°F for copper fillings.
Analysis The thermal resistances of copper fillings and the epoxy board are in parallel. The number of
copper fillings in the board and the area they comprise are
Atotal = (6 / 12 ft)(8 / 12 ft) = 0.333 ft 2
n copper =

0.33 ft 2
= 13,333 (number of copper fillings)
(0.06 / 12 ft)(0.06 / 12 ft)

Acopper = n

πD 2
4

= 13,333

Aepoxy = Atotal − Acopper

π (0.02 / 12 ft) 2

= 0.0291 ft 2

4
= 0.3333 − 0.0291 = 0.3042 ft 2

Rcopper

The thermal resistances are evaluated to be
0.05 / 12 ft
L
=
= 0.00064 h ⋅ °F/Btu
kA (223 Btu/h ⋅ ft ⋅ °F)(0.0291 ft 2 )
0.05 / 12 ft
L
=
=
= 0.137 h ⋅ °F/Btu
kA (0.10 Btu/h ⋅ ft ⋅ °F)(0.3042 ft 2 )

Rcopper =
Repoxy

Repoxy

Then the thermal resistance of the entire epoxy board becomes
1
1
1
1
1
=
+
=
+
⎯

⎯→ Rboard = 0.00064 h ⋅ °F/Btu
Rboard
Rcopper Repoxy 0.00064 0.137

Heat Conduction in Cylinders and Spheres
3-66C When the diameter of cylinder is very small compared to its length, it can be treated as an infinitely
long cylinder. Cylindrical rods can also be treated as being infinitely long when dealing with heat transfer
at locations far from the top or bottom surfaces. However, it is not proper to use this model when finding
temperatures near the bottom and the top of the cylinder.
3-67C Heat transfer in this short cylinder is one-dimensional since there will be no heat transfer in the axial
and tangential directions.
3-68C No. In steady-operation the temperature of a solid cylinder or sphere does not change in radial
direction (unless there is heat generation).

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educators for course preparation. If you are a student using this Manual, you are using it without permission.


3-42

3-69 Chilled water is flowing inside a pipe. The thickness of the insulation needed to reduce the
temperature rise of water to one-fourth of the original value 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 centerline and no variation in the axial
direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is
negligible.
Properties The thermal conductivity is given to be k = 0.05 W/m⋅°C for insulation.
Insulation

Analysis The rate of heat transfer without the insulation is

Q& old = m& c p ΔT = (0.98 kg/s)(4180 J/kg ⋅ °C)(8 - 7)°C = 4096 W
r2

The total resistance in this case is
T − Tw
Q& old = ∞
R total
4096 W =

Water

(30 − 7.5)°C
⎯
⎯→ R total = 0.005493°C/W
R total

L

R1

Ro

Rins
T∞2

T∞1

The convection resistance on the outer surface is
Ro =

r1

1
1
=
= 0.004716 °C/W
ho Ao (9 W/m 2 ⋅ °C)π (0.05 m)(150 m)

The rest of thermal resistances are due to convection resistance on the inner surface and the resistance of
the pipe and it is determined from
R1 = R total − Ro = 0.005493 − 0.004716 = 0.0007769 °C/W

The rate of heat transfer with the insulation is
Q&
= m& c ΔT = (0.98 kg/s)(4180 J/kg ⋅ °C)(0.25 °C) = 1024 W
new

p

The total thermal resistance with the insulation is
T − Tw
[30 − (7 + 7.25) / 2)]°C
Q& new = ∞
⎯
⎯→ 1024 W =
⎯
⎯→ R total, new = 0.02234°C/W
R total, new
R total, new

It is expressed by
R total,new = R1 + R o, new + Rins = R1 +
0.02234°C/W = 0.0007769 +

ln( D 2 / D1 )
1
+
ho Ao
2πk ins L
1

(9 W/m ⋅ °C)πD 2 (150 m)
2

+

ln( D 2 / 0.05)
2π (0.05 W/m ⋅ °C)(150 m)

Solving this equation by trial-error or by using an equation solver such as EES, we obtain
D 2 = 0.1265 m

Then the required thickness of the insulation becomes
t ins = ( D 2 − D1 ) / 2 = (0.05 − 0.1265) / 2 = 0.0382 m = 3.8 cm

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educators for course preparation. If you are a student using this Manual, you are using it without permission.


3-43

3-70 Steam flows in a steel pipe, which is insulated by gypsum plaster. The rate of heat transfer from the
steam and the temperature on the outside surface of the insulation are 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 centerline and no variation in the axial
direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is
negligible.
Properties (a) The thermal conductivities of steel and gypsum
plaster are given to be 50 and 0.5 W/m⋅°C, respectively.

Insulation

Analysis The thermal resistances are

Ri
Ti

Rsteel

Rins

Steam

Ro
To

1
1
=
= 0.0003316°C/W

hi Ai (800 W/m 2 ⋅ °C)π (0.06 m)(20 m)
ln( D 2 / D1 )
ln(8 / 6)
=
=
= 0.0000458°C/W
2πk steel L
2π (50 W/m ⋅ °C)(20 m)

L

Ri =
Rsteel

Rins =
Ro =

ln( D3 / D 2 )
ln(16 / 8)
=
= 0.011032°C/W
2πk ins L
2π (0.5 W/m ⋅ °C)(20 m)
1
1
=
= 0.0004974°C/W
2
ho Ao (200 W/m ⋅ °C)π (0.16 m)(20 m)

The total thermal resistance and the rate of heat transfer are
R total = Ri + Rsteel + Rins + Ro = 0.0003316 + 0.0000458 + 0.011032 + 0.0004974 = 0.011907°C/W
T − To
(200 − 10)°C
=
Q& = i
= 15,957 W
R total
0.011907 m 2 ⋅ C/W

(b) The temperature at the outer surface of the insulation is determined from
(Ts − 10)°C
T − To
Q& = s
⎯
⎯→ 15,957 W =
⎯
⎯→ Ts = 17.9°C
Ro
0.0004974 m 2 ⋅ °C/W

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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