3-61
𝑅3=1
ℎ𝐴or 𝐴𝑅3=1
ℎ=1
5 W/m2
K= 0.2 m2∙K/W
To determine the temperatures that the copper-silicon plate experiences, we need to find T1 and Ti1, which can be calculated
using the heat flux and the individual thermal resistance:
𝑞̇ = 𝑇𝑖2 −𝑇2
𝐴𝑅2 → 𝑇𝑖2 = 𝑇2+𝑞̇𝐴𝑅2=116.7℃+(750 W/m2)(0.0007692 m2∙K/W)=117.3℃
3-62
Heat Conduction in Cylinders and Spheres
3-78C When the diameter of cylinder is very small compared to its length, it can be treated as an infinitely long cylinder.
3-63
3-81 A steam pipe covered with 3-cm thick glass wool insulation is subjected to convection on its surfaces. The rate of heat
transfer per unit length and the temperature drops across the pipe and the insulation 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 contact resistance at the interface is negligible.
Properties The thermal conductivities are given to be k = 15 W/m°C for steel and k = 0.038 W/m°C for glass wool
insulation
Analysis The inner and the outer surface areas of the insulated pipe per unit length are
2
m 157.0m) 1(m) 05.0(
===
LDA
ii
3-64
3-65
3-66
3-67
3-85 Hot liquid flows in a pipe with FEP lining on the inner surface. The pipe outer surface is subjected to uniform heat
flux. The liquid mean temperature and convection heat transfer coefficient are given. Thermal contact resistance exists in the
interface between the FEP lining and the steel surface. Determine the temperatures at the FEP lining T1, and at the pipe outer
surface T2. Does the FEP lining comply with the recommendation of the ASME code?
Assumptions1 Heat transfer is steady. 2 One dimensional heat conduction. 3 Uniform surface temperatures. 4 Thermal
properties are constant.
Properties The thermal conductivity for the pipe wall is k = 15 W/m·K.
Analysis The heat transfer rate through the pipe is
𝑄̇= 𝑞̇𝐴2= 𝑞̇𝜋𝐷2𝐿 = (1200 W/m2)𝜋(0.027 m)(1 m)=101.79 W
3-68
3-86 Hot water flows in an ABS pipe. The pipe outer surface is subjected convection with air. Determine the maximum
water temperature, such that the ABS pipe is operating at the recommended temperature or lower by the ASME Code for
Process Piping.
Assumptions1 Heat transfer is steady. 2 One dimensional heat conduction. 3 Uniform surface temperatures. 4 Thermal
properties are constant.
Properties The thermal conductivity for the pipe wall is k = 0.1 W/m·K.
Analysis The inner and outer surface areas of the pipe are
𝐴1= 𝜋𝐷1𝐿 = 𝜋(0.022 m)(1 m)= 0.06912 m2
The total thermal resistance between T∞1 and T∞2 is
𝑅tot = 𝑅1+𝑅2+𝑅3= 0.2894 K/W + 0.3259 K/W + 1.1790 K/W = 1.7943 K/W
The maximum water temperature is limited by the temperature at the inner pipe surface at T1 = 80°C. Thus, the heat transfer
rate through the thermal circuit is
3-69
3-70
3-71
3-72
Do
[in]
Ltube
[ft]
0.5
0.525
0.55
0.575
0.6
0.625
0.65
0.675
0.7
0.725
0.75
0.775
0.8
0.825
0.85
0.875
0.9
0.925
0.95
0.975
1
2387
2386
2385
2384
2383
2383
2382
2381
2381
2380
2380
2380
2379
2379
2379
2378
2378
2378
2378
2377
2377
0.5 0.6 0.7 0.8 0.9 1
2376
2378
2380
2382
2384
2386
2388
2390
Do [in]
Ltube [ft]
3-73
3-90 An electric wire is tightly wrapped with a 1-mm thick plastic cover. The interface temperature and the effect of doubling
the thickness of the plastic cover on the interface temperature 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 centerline and no variation in the axial direction. 3 Thermal properties
are constant. 4 The thermal contact resistance at the interface is negligible. 5 Heat transfer coefficient accounts for the
radiation effects, if any.
Properties The thermal conductivity of plastic cover is given to be k = 0.15 W/m°C.
Analysis In steady operation, the rate of heat transfer from the wire is equal to the heat generated within the wire,
W104)A 13)(V 8( ==== IWQ eV
The total thermal resistance is
)1.1/1.2ln(
)/ln(
C/W 2256.0
m)] m)(14 (0.0042C)[. W/m24(
11
12
2
conv
=
==
rr
Ah
R
oo
Rconv
T2
Rplastic
T1
3-74
3-75
3-92 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.
Analysis The rate of heat transfer without the insulation is
W4096C7)–C)(8J/kg kg/s)(4180 (0.98
old === TcmQ p
The total resistance in this case is
C/W0.005493
C)5.730(
W4096 total
total
total
old
=⎯→⎯
−
=
−
=
R
R
R
TT
Qw
The convection resistance on the outer surface is
C/W 004421.0
m) m)(200 04.0(C) W/m9(
11
2=
==
oo
oAh
R
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
C/W 001072.0004421.0005493.0
ototal1 =−=−= RRR
The rate of heat transfer with the insulation is
W1024C)C)(0.25J/kg kg/s)(4180 (0.98
new === TcmQ p
The total thermal resistance with the insulation is
C/W0.02234
C)]2/)25.77(30[
W1024 newtotal,
newtotal,newtotal,
new =⎯→⎯
+−
=⎯→⎯
−
=R
RR
TT
Qw
It is expressed by
)m 200(C) W/m05.0(2
)04.0/ln(
m) 200(C) W/m9(
1
001072.0C/W02234.0
2
)/ln(
1
2
2
2
ins
12
1insnewo,1newtotal,
+
+=
++=++=
D
D
Lk
DD
Ah
RRRRR
oo
Solving this equation by trial-error or by using an equation solver such as EES, we obtain
R1
T1
Rins
T2
Ro
Water
L
Insulation
r1
r2
3-76
3-93E A steam pipe covered with 2-in thick fiberglass insulation is subjected to convection on its surfaces. The rate of heat
loss from the steam per unit length and the error involved in neglecting the thermal resistance of the steel pipe in calculations
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 contact resistance at the interface is negligible.
Properties The thermal conductivities are given to be k = 8.7 Btu/hft°F for steel and k = 0.020 Btu/hft°F for fiberglass
insulation.
Analysis The inner and outer surface areas of the insulated pipe are
2
2
ft 094.2ft) 1(ft) 12/8(
ft 916.0ft) 1(ft) 12/5.3(
===
===
LDA
LDA
oo
ii
F/Btuh 65.5
which is insignificant.
Ri
T1
Rinsulation
Ro
T2
Rpipe
3-77
3-94 Hot water is flowing through a 15-m section of a cast iron pipe. The pipe is exposed to cold air and surfaces in the
basement. The rate of heat loss from the hot water and the average velocity of the water in the pipe as it passes through the
basement 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
properties are constant.
Properties The thermal conductivity and emissivity of cast iron are given to be k = 52 W/m°C and = 0.7.
Analysis The individual resistances are
2
2
m 168.2m) 15(m) 046.0(
m 885.1m) 15(m) 04.0(
===
===
LDA
LDA
oo
ii
C/W 00003.0
)m 15(C) W/m.52(2
)2/3.2ln(
2
)/ln(
C/W 00442.0
)m 885.1(C). W/m120(
11
1
12
22
=
==
=
==
Lk
rr
R
Ah
R
pipe
ii
i
The outer surface temperature of the pipe will be somewhat below the water temperature. Assuming the outer surface
temperature of the pipe to be 80°C (we will check this assumption later), the radiation heat transfer coefficient is determined
to be
.K W/m167.5283)+353]()K 283()K 353)[(.K W/m1067.5)(7.0(
))((
222428
2
22
2
=+=
++=
−
surrsurrrad TTTTh
Since the surrounding medium and surfaces are at the same temperature, the radiation and convection heat transfer
coefficients can be added and the result can be taken as the combined heat transfer coefficient. Then,
C/W 02732.002287.000003.000442.0
C/W 02287.0
)m 168.2(C). W/m17.20(
11
C. W/m17.2015167.5
22
2
2,
=++=++=
=
==
=+=+=
opipeitotal
ocombined
o
convradcombined
RRRR
Ah
R
hhh
The rate of heat loss from the hot water pipe then becomes
W2928=
−
=
−
=
C/W 0.02732
C)1090(
21
total
R
TT
Q
For a temperature drop of 3C, the mass flow rate of water and the average velocity of water must be
m/s 0.186===⎯→⎯=
=
=
=⎯→⎯=
4
m) 04.0(
)kg/m 1000(
kg/s 2335.0
kg/s 2335.0
C) C)(3J/kg. (4180
J/s 2928
2
3
c
c
p
p
A
m
VVAm
Tc
Q
mTcmQ
Discussion The outer surface temperature of the pipe is
C77.0=
C/W0.00003)+(0.00442
C)90(
W928 2
1→
−
=→
+
−
=
s
s
pipei
sT
T
RR
TT
Q
which is close to the value assumed for the surface temperature in the evaluation of the radiation resistance. Therefore, there
is no need to repeat the calculations.
Ri
Rpipe
Ro
T1
T2
3-78
3-95 Liquid flows in a pipe with PVDC lining on the inner surface. The pipe outer surface is subjected to convection and
radiation heat transfer. The pipe outer surface temperature is known. Determine the temperature at the PVDC lining T1, and
the temperature of the liquid T∞1. Does the PVDC lining comply with the recommendation of the ASME code?
Assumptions1 Heat transfer is steady. 2 One dimensional heat conduction. 3 Uniform surface temperatures. 4 Thermal
properties are constant. 5 Negligible contact resistance.
Properties The thermal conductivity for the pipe wall is k = 12 W/m·K. The emissivity of the outer pipe surface is 0.3.
Analysis The inner and outer surface areas of the pipe are
𝐴1= 𝜋𝐷1𝐿 = 𝜋(0.022 m)(1 m)= 0.06912 m2
3-79
The temperature of the liquid T∞1 in the pipe is
3-80
3-96 To avoid condensation on the outer surface, the necessary thickness of the insulation around a copper pipe that carries
liquid oxygen is to be determined.
Assumptions1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Thermal contact
resistance is negligible.
Properties The thermal conductivities of the copper pipe and the insulation are given to be 400 W/m ∙ °C and 0.05 W/m ∙ °C,
respectively.
Analysis From energy balance and using the thermal resistance concept, the following equation is expressed:
combined
,
convccond,icond,combined
,,
R
TT
RRRR
TT soio −
=
+++
−
Ah
TT
hALk
DD
Lk
DD
Ah
TT so
ci
io
combined
,
12
23
combined
,,
1
1
2
)/ln(
2
)/ln(
1
−
=
+++
−
TT
TT so
io
,
,,
−
−