8-122
8-132 Hot water enters a cast iron pipe whose outer surface is exposed to cold air with a specified heat transfer coefficient.
The rate of heat loss from the water and the exit temperature of the water are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the pipe are smooth.
Properties We assume the water temperature not to drop significantly since the pipe is not very long. We will check this
assumption later. The properties of water at 90C are (Table A-9)
96.1Pr
CJ/kg. 4206 /s;m 10326.0/
C W/m.675.0 ;kg/m 3.965
26–
3
=
===
==
p
c
k
Analysis (a) The mass flow rate of water is
m) (0.04
2
3
Water
90C
1.2 m/s
10C
Di = 4 cm
Do = 4.6 cm
8-124
8-134 Hot exhaust gases flow through a pipe. For a specified exit temperature, the pipe length is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The inner surface of the pipe is smooth. 3 For hot gases, air properties are
used. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm.
Properties The properties of air at 1 atm and the bulk mean temperature of (450+250)/2 = 350C are (Table A–15)
6937.0Pr
CJ/kg. 1056
/sm 10475.5
C W/m.04721.0
kg/m 5664.0
25–
3
=
=
=
=
=
p
c
k
Analysis The Reynolds number is
m) m/s)(0.15 (7.2
avg =
DV
Exhaust
gases
450C
7.2 m/s
250C
Ts = 180C
D = 15 cm
8-125
8-135 Water is heated in a heat exchanger by the condensing geothermal steam. The exit temperature of water and the rate of
condensation of geothermal steam are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the tube are smooth. 3 The surface temperature of
the pipe is 165C, which is the temperature at which the geothermal steam is condensing.
Properties The properties of water at the anticipated mean temperature of 85C are (Table A-9)
kJ/kg 5.2066
/sm 1044.3
kg/m 968.1
kg/m.s 10333.0
08.2Pr
CJ/kg. 4201
C W/m.673.0
kg/m 1.968
C165@
27–
3
3
3
=
=
==
=
=
=
=
−
fg
p
h
c
k
Analysis The velocity of water and the Reynolds number are
m/s 6576.0
4
m) 04.0(
)kg/m 1.968(kg/s 8.0 avgavg
2
3
avg =⎯→⎯=⎯→⎯= VVAVm
465,76
/sm 1044.3
m) m/s)(0.04 (0.6576
Re 27
avg =
== −
DV
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly
m 0.4=m) 04.0(1010 =DLL th
which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow in the entire
duct, and determine the Nusselt number from
7.248)08.2()465,76(023.0PrRe023.0 4.08.04.08.0 ==== k
hD
Nu
Heat transfer coefficient is
C. W/m4184)7.248(
m 04.0
C W/m.673.0 2=
== Nu
D
k
h
Next we determine the exit temperature of air,
2
m 1.759=m) m)(14 04.0(
== DLAs
C148.8=−−=−−= −
−)4201)(8.0(
)759.1)(4184(
)/( )20165(165)( eeTTTT ps cmhA
isse
The logarithmic mean temperature difference is
C77.58
20165
8.148165
ln
208.148
ln
lm =
−
−
−
=
−
−
−
=
is
es
ie
TT
TT
TT
T
The rate of heat transfer can be expressed as
W500,432)C77.58()m 759.1)(C. W/m4185( 22
lm === ThAQs
The rate of condensation of steam is determined from
kg/s 0.209=⎯→⎯=⎯→⎯= mmhmQ fg
kJ/kg) 5.2066(kW 5.432
14 m
Water
20C
0.8 kg/s
4 cm
Ts = 165C
8-126
8-136 Cold-air flows through an isothermal pipe. The pipe temperature is to be estimated.
Assumptions 1 Steady operating conditions exist. 2 The inner surface of the duct is smooth. 3 Air is an ideal gas with
constant properties. 4 The pressure of air is 1 atm.
Properties The properties of air at 1 atm and the bulk mean temperature of (5+19)/2=12C are (Table A-15)
7331.0Pr
CJ/kg. 1007
/sm 10444.1
C W/m.02454.0
kg/m 238.1
25–
3
=
=
=
=
=
p
c
k
Analysis The rate of heat transfer to the air is
m) 12.0(
2
3
20 m
Air
5C
2.5 m/s
12 cm
Ts
19C
8-127
8-137 Crude oil is heated as it flows in the tube-side of a multi-tube heat exchanger. The rate of heat transfer and the tube
length are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The surface temperature is constant and uniform. 3 The inner surfaces of the
tubes are smooth. 4 Heat transfer to the surroundings is negligible.
Properties The properties of crude oil are given to be = 950 kg/m3, cp = 1.9 kJ/kgK, k = 0.25 W/mK, µ = 12 mPas.
Analysis The rate of heat transfer is
Heat transfer coefficient is
C W/m.25.0 2=
k
m 2.78=⎯→⎯=⎯→⎯=
LLDLA
s
m) 01.0(100m 8.719
2
8-129
8-139 Geothermal water is supplied to a city through stainless steel pipes at a specified rate. The electric power consumption
and its daily cost are to be determined, and it is to be assessed if the frictional heating during flow can make up for the
temperature drop caused by heat loss.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully
developed. 3 The minor losses are negligible because of the large length-to-diameter ratio and the relatively small number of
components that cause minor losses. 4 The geothermal well and the city are at about the same elevation. 5 The properties of
geothermal water are the same as fresh water. 6 The fluid pressures at the wellhead and the arrival point in the city are the
same.
Properties The properties of water at 110C are = 950.6 kg/m3, = 0.25510-3 kg/ms, and cp = 4.229 kJ/kgC (Table A-
9). The roughness of stainless steel pipes is 210-6 m (Table 8-3).
Analysis (a) We take point 1 at the well-head of geothermal resource and point 2 at the final point of delivery at the city, and
the entire piping system as the control volume. Both points are at the same elevation (z2 = z2) and the same velocity (V1 = V2)
since the pipe diameter is constant, and the same pressure (P1 = P2). Then the energy equation for this control volume
simplifies to
22 upump,turbine2
2
22
upump,1
2
11
LL hhhhz
g
V
g
P
hz
g
V
g
P=→++++=+++
That is, the pumping power is to be used to overcome the head losses due
to friction in flow. The mean velocity and the Reynolds number are
7
3
3
avg
2
3
2
avg
10186.1
skg/m 10255.0
m) m/s)(0.60 305.5)(kg/m 6.950(
Re
m/s 305.5
4/m) (0.60
/sm 1.5
4/
=
==
====
−
DV
D
A
V
c
VV
which is greater than 10,000. Therefore, the flow is turbulent. The relative roughness of the pipe is
1033.3
m 60.0
m 102
/6
6−
−
=
=D
The friction factor can be determined from the Moody chart, but to avoid the reading error, we determine it from the
Colebrook equation using an equation solver (or an iterative scheme),
+
−=→
+−=
−
fff
D
f7
6
10187.1
51.2
7.3
1033.3
log0.2
1
Re
51.2
7.3
/
log0.2
1
It gives f = 0.00829. Then the
pressure drop, the head loss, and the required power input become
kPa2218
kN/m 1
kPa 1
m/skg 1000
kN 1
2
m/s) 305.5)(kg/m 6.950(
m 0.60
m 000,12
00829.0
222
23
2
avg =
==V
D
L
fP
kW 5118=
=
== /smkPa 1
kW 1
0.65
)kPa 2218)(/sm (1.5
3
3
motor–pumpmotor–pump
upump,
elect
PV
W
W
Therefore, the pumps will consume 5118 kW of electric power to overcome friction and maintain flow.
(b) The daily cost of electric power consumption is determined by multiplying the amount of power used per day by the unit
cost of electricity,
kWh/day 832,122h/day) kW)(24 5118(Amount inelect, === tW
$7370/day=== 0.06/kWh)kWh/day)($ 832,122(costUnit AmountCost
(c) The energy consumed by the pump (except the heat dissipated by the motor to the air) is eventually dissipated as heat due
to the frictional effects. Therefore, this problem is equivalent to heating the water by a 5118 kW of resistance heater (again
except the heat dissipated by the motor). To be conservative, we consider only the useful mechanical energy supplied to the
water by the pump. The temperature rise of water due to this addition of energy is
kJ/s) (51180.65
inelect,motor–pump
W
L = 12 km
D = 60 cm
Water
1.5 m3/s
2
1
8-130
8-140 Geothermal water is supplied to a city through cast iron pipes at a specified rate. The electric power consumption and
its daily cost are to be determined, and it is to be assessed if the frictional heating during flow can make up for the
temperature drop caused by heat loss.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully
developed. 3 The minor losses are negligible because of the large length-to-diameter ratio and the relatively small number of
components that cause minor losses. 4 The geothermal well and the city are at about the same elevation. 5 The properties of
geothermal water are the same as fresh water. 6 The fluid pressures at the wellhead and the arrival point in the city are the
same.
Properties The properties of water at 110C are = 950.6 kg/m3, = 0.25510-3 kg/ms, and cp = 4.229 kJ/kgC (Table A-
9). The roughness of cast iron pipes is 0.00026 m (Table 8-3).
Analysis (a) We take point 1 at the well-head of geothermal resource and point 2 at the final point of delivery at the city, and
the entire piping system as the control volume. Both points are at the same elevation (z2 = z2) and the same velocity (V1 = V2)
since the pipe diameter is constant, and the same pressure (P1 = P2). Then the energy equation for this control volume
simplifies to
22 upump,turbine2
2
22
upump,1
2
11
LL hhhhz
g
V
g
P
hz
g
V
g
P=→++++=+++
That is, the pumping power is to be used to overcome the head losses due
to friction in flow. The mean velocity and the Reynolds number are
7
3
3
avg
2
3
2
avg
10187.1
skg/m 10255.0
m) m/s)(0.60 305.5)(kg/m 6.950(
Re
m/s 305.5
4/m) (0.60
/sm 1.5
4/
=
==
====
−
DV
D
A
V
c
VV
which is greater than 10,000. Therefore, the flow is turbulent. The relative roughness of the pipe is
1033.4
m 60.0
m 00026.0
/4−
==D
The friction factor can be determined from the Moody chart, but to avoid the reading error, we determine it from the
Colebrook equation using an equation solver (or an iterative scheme),
+
−=→
+−=
−
fff
D
f7
4
10187.1
51.2
7.3
1033.4
log0.2
1
Re
51.2
7.3
/
log0.2
1
It gives f = 0.01623. Then the pressure drop, the head loss, and the required power input become
kPa 4342
kN/m 1
kPa 1
m/skg 1000
kN 1
2
m/s) 305.5)(kg/m 6.950(
m 0.60
m 000,12
01623.0
22
23
2
avg =
==V
D
L
fP
kW 10,020=
=
== /smkPa 1
kW 1
0.65
)kPa 4342)(/sm (1.5
3
3
motor–pumpmotor–pump
upump,
elect
P
W
W
V
Therefore, the pumps will consume 10,017 W of electric power to overcome friction and maintain flow.
(b) The daily cost of electric power consumption is determined by multiplying the amount of power used per day by the unit
cost of electricity,
kWh/day 480,240h/day) kW)(24 020,10(Amount inelect, === tW
y$14,430/da=== 0.06/kWh)kWh/day)($ 480,240(costUnit AmountCost
(c) The energy consumed by the pump (except the heat dissipated by the motor to the air) is eventually dissipated as heat due
to the frictional effects. Therefore, this problem is equivalent to heating the water by a 10,020 kW of resistance heater (again
except the heat dissipated by the motor). To be conservative, we consider only the useful mechanical energy supplied to the
water by the pump. The temperature rise of water due to this addition of energy is
kJ/s) (10,0200.65
inelect,motor–pump
W
cost savings should be compared to the increased cost of larger diameter pipe.
L = 12 km
D = 60 cm
Water
1.5 m3/s
2
1
8-131
8-141 Air enters the underwater section of a duct. The outlet temperature of the air and the fan power needed to overcome the
flow resistance are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermal resistance of the duct
is negligible. 4 The surface of the duct is at the temperature of the water. 5 Air is an ideal gas with constant properties. 6 The
pressure of air is 1 atm.
Properties We assume the bulk mean temperature for air to be 20C since the mean temperature of air at the inlet will drop
somewhat as a result of heat loss through the duct whose surface is at a lower temperature. The properties of air at 1 atm and
this temperature are (Table A-15)
7309.0Pr
CJ/kg. 1007
/sm 10516.1
C W/m.02514.0
kg/m 204.1
25–
3
=
=
=
=
=
p
c
k
Analysis The Reynolds number is
4
avg 10958.3
m) m/s)(0.2 (3
h
DV
Air
25C
3 m/s
River
water
L = 15 m
D = 20 cm
8-132
8-142 Air enters the underwater section of a duct. The outlet temperature of the air and the fan power needed to overcome the
flow resistance are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the duct are smooth. 3 The thermal resistance of the duct
is negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm.
Properties We assume the bulk mean temperature for air to be 20C since the mean temperature of air at the inlet will drop
somewhat as a result of heat loss through the duct whose surface is at a lower temperature. The properties of air at 1 atm and
this temperature are (Table A-15)
7309.0Pr
CJ/kg. 1007
/sm 10516.1
C W/m.02514.0
kg/m 204.1
25–
3
=
=
=
=
=
p
c
k
Analysis The Reynolds number is
4
avg 10958.3
m) m/s)(0.2 (3
h
DV
55.0
3
motor–pumpmotor–pump
fan
Water
25C
3 m/s
River water
15C
L = 15 m
D = 20 cm
Mineral deposit
0.25 mm
8-133
8-143 Oil is heated by saturated steam in a double-pipe heat exchanger. The tube length is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces of the tube are smooth.
Properties The properties of oil at the average temperature of
(15+25)/2=20C are (Table A–13)
skg/m 8374.0
CJ/kg 1881
C W/m145.0
kg/m 1.888 3
=
=
=
=
p
c
k
Analysis The cross-sectional area of the annulus, the mass flow
rate, the rate of heat transfer, and Reynolds number are
2
22
22
m) 03.0(m) 05.0(
−
−
io
DD
Oil
15C
0.8 m/s
5 cm
Ts = 100C
25C
3 cm