978-0073398198 Chapter 8 Part 8

subject Type Homework Help
subject Pages 9
subject Words 3387
subject Authors Afshin Ghajar, Yunus Cengel

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8-134
8-144E Water flows through a concentric annulus tube with constant inner surface temperature and insulated outer surface,
the length of the annulus tube is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Constant inner tube surface temperature. 4
Insulated outer tube surface. 5 Fully developed flow.
Properties The properties of water at Tb = (Ti + Te)/2 = 120°F: cp = 0.999 Btu/lbm∙R, k = 0.371 Btu/h∙ft∙R,
= 3.744 10−4
lbm/ft∙s,
= 61.71 lbm/ft3, and Pr = 3.63 (Table A-9E).
Analysis The Reynolds number is
)lbm/s 3600/396(4
)(
4
))(4/(
)(
)(
Re
22
avg
+
=
=
=
io
io
io
io
DD
m
DD
DDm
DDV
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8-135
Fundamentals of Engineering (FE) Exam Problems
8-145 Internal force flows are said to be fully-developed once the ____ at a cross-section no longer changes in the direction
of flow.
(a) temperature distribution (b) entropy distribution (c) velocity distribution
(d) pressure distribution (e) none of the above
8-146 The bulk or mixed temperature of a fluid flowing through a pipe or duct is defined as
(a)
=
c
Ac
c
bTdA
A
T1
(b)
=
c
Acb VdAT
m
T
1
(c)
=
c
Acb VdAh
m
T
1
(d)
=
c
Ac
c
bhdA
A
T1
(e)
=
c
Acb VdATT
V
1
c
Acb VdAT
m
1
8-147 Water (
= 9.010-4 kg/ms,
= 1000 kg/m3) enters a 4-cm-diameter, 3-m-long tube whose walls are maintained at
100oC. The water enters this tube with a bulk temperature of 25oC and a volume flow rate of 3 m3/h. The Reynolds number
for this internal flow is
(a) 29,500 (b) 38,200 (c) 72,500 (d) 118,100 (e) 122,900
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8-136
8-148 Water enters a circular tube whose walls are maintained at constant temperature at a specified flow rate and
temperature. For fully developed turbulent flow, the Nusselt number can be determined from Nu = 0.023 Re0.8 Pr0.4. The
correct temperature difference to use in Newton’s law of cooling in this case is
(a) the difference between the inlet and outlet water bulk temperature
(b) the difference between the inlet water bulk temperature and the tube wall temperature
(c) the log mean temperature difference
(d) the difference between the average water bulk temperature and the tube temperature
(e) None of the above.
8-149 Air (cp = 1007 J/kg°C) enters a 17-cm-diameter and 4-m-long tube at 65ºC at a rate of 0.08 kg/s and leaves at 15ºC.
The tube is observed to be nearly isothermal at 5ºC. The average convection heat transfer coefficient is
(a) 24.5 W/m2ºC (b) 46.2 W/m2ºC (c) 53.9 W/m2ºC (d) 67.6 W/m2ºC (e) 90.7 W/m2ºC
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8-139
8-154 Water enters a 5-mm-diameter and 13-m-long tube at 15ºC with a velocity of 0.3 m/s, and leaves at 45ºC. The tube is
subjected to a uniform heat flux of 2000 W/m2 on its surface. The temperature of the tube surface at the exit is
(a) 48.7ºC (b) 49.4ºC (c) 51.1ºC (d) 53.7ºC (e) 55.2ºC
(For water, use k = 0.615 W/m°C, Pr = 5.42, =0.80110-6 m2/s)
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8-140
8-155 Water enters a 5-mm-diameter and 13-m-long tube at 45ºC with a velocity of 0.3 m/s. The tube is maintained at a
constant temperature of 8ºC. The exit temperature of water is
(a) 4.4ºC (b) 8.9ºC (c) 10.6ºC (d) 12.0ºC (e) 14.1ºC
(For water, use k = 0.607 W/m°C, Pr = 6.14, =0.89410-6 m2/s, cp = 4180 J/kg°C,
= 997 kg/m3.)
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8-141
8-156 Water enters a 5-mm-diameter and 13-m-long tube at 45ºC with a velocity of 0.3 m/s. The tube is maintained at a
constant temperature of 5ºC. The required length of the tube in order for the water to exit the tube at 25ºC is
(a) 1.55 m (b) 1.72 m (c) 1.99 m (d) 2.37 m (e) 2.96 m
(For water, use k = 0.623 W/m°C, Pr = 4.83, =0.72410-6 m2/s, cp = 4178 J/kg°C,
= 994 kg/m3.)
"The properties of water at (45+25)/2 = 35 C are (Table A-9)"
rho=994 [kg/m^3]
c_p=4178 [J/kg-C]
k=0.623 [W/m-C]
mu=0.720E-3 [kg/m-s]
Pr=4.83
Re=(rho*V*D)/mu "The calculated Re value is smaller than 2300. Therefore the flow is laminar."
L_t=0.05*Re*Pr*D "We assume that the entire flow remains in the entry region. We will check this after calculating
total length of the tube"
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8-142
8-157 Air enters a 7-cm-diameter, 4-m-long tube at 65ºC and leaves at 15ºC. The duct is observed to be nearly isothermal at
5ºC. If the average convection heat transfer coefficient is 20 W/m2ºC, the rate of heat transfer from the air is
(a) 491 W (b) 616 W (c) 810 W (d) 907 W (e) 975 W
8-158 Air (cp = 1000 J/kgK) enters a 16-cm-diameter and 19-m-long underwater duct at 50°C and 1 atm at an average
velocity of 7 m/s, and is cooled by the water outside. If the average heat transfer coefficient is 35 W/m2°C and the tube
temperature is nearly equal to the water temperature of 5°C, the exit temperature of air is
(a) 6°C (b) 10°C (c) 18°C (d) 25°C (e) 36°C
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8-143
8-159 Water enters a 2-cm-diameter, 3-m-long tube whose walls are maintained at 100oC with a bulk temperature of 25oC
and volume flow rate of 3 m3/h. Neglecting the entrance effects and assuming turbulent flow, the Nusselt number can be
determined from Nu = 0.023 Re0.8 Pr0.4. The convection heat transfer coefficient in this case is
(a) 4140 W/m2K (b) 6160 W/m2K (c) 8180 W/m2K (d) 9410 W/m2K (e) 2870 W/m2K
(For water, use k = 0.610 W/m°C, Pr = 6.0,
= 9.010-4 kg/ms,
= 1000 kg/m3)
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8-145
8-161 Air at 10ºC enters an 18-m-long rectangular duct of cross section 0.15 m 0.20 m at a velocity of 4.5 m/s. The duct is
subjected to uniform radiation heating throughout the surface at a rate of 400 W/m2. The wall temperature at the exit of the
duct is
(a) 58.8ºC (b) 61.9ºC (c) 64.6ºC (d) 69.1ºC (e) 75.5ºC
(For air, use k = 0.02551 W/m°C, Pr = 0.7296, = 1.56210-5 m2/s, cp = 1007 J/kg°C,
= 1.184 kg/m3.)
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8-146
8-164 A computer is cooled by a fan blowing air through the case of the computer. The flow rate of the fan and the diameter
of the casing of the fan are to be specified.
Assumptions 1 Steady flow conditions exist. 2 Heat flux is uniformly distributed. 3 Air is an ideal gas with constant
properties.
Properties The relevant properties of air are (Tables A-1 and A-15)
/kg.KkPa.m 287.0
CJ/kg. 1007
3
=
=
R
cp
Analysis We need to determine the flow rate of air for the worst case
scenario. Therefore, we assume the inlet temperature of air to be 50C, the
atmospheric pressure to be 70.12 kPa, and disregard any heat transfer from
the outer surfaces of the computer case. The mass flow rate of air required
to absorb heat at a rate of 80 W can be determined from
J/s 80
Q
Cooling
air

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