157
Chapter 13: Energy and Power
13.1 How much natural gas do you need to burn to heat twenty gallons of water?
SOLUTION
13.3 An elevator has a rated capacity of 2200 lb. It can transport people at the rate
capacity between the first and the fifth floors, with a vertical distance of 15 ft
between each floor, in 7 s. Estimate the power requirement for such an elevator.
SOLUTION
hp 34.3)
sec
.ftlb
550
)(
sec
(18857power
f
13.4 Determine the gross force needed to bring a car which is traveling at 120 km/hr to
a full stop in a distance of 100 m. The mass of the car is 2000 kg. What happens
to the initial kinetic energy, where does it go or to what form of energy does the
kinetic energy convert to?
158
© 2020 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website,
in whole or in part.
SOLUTION
We begin the analysis by first changing the units of speed to m/sec and then use
Equation (13.1) to analyze the problem.
sec
m
33.3)
km
1
m 1000
)(
sec
3600
hr 1
)(
hr
km
( 120
1V
V2 = 0
2
1
2
221
2
1
2
1
stance)(Force)(di mVmVWork
2
)
sec
3.33)( 2000(
2
1
0m) 0(Force)(10 m
kg
Force = 11089 N
When the brakes are applied, the kinetic energy changes into heat.
13.5 A centrifugal pump is driven by a motor. The performance of the pump reveals
the following information:
Power input to the pump by the motor (kW):
0.5, 0.7, 0.9, 1.0, 1.2
Power input to the fluid by the pump (kW):
0.3, 0.55, 0.7, 0.9, 1.0
Plot the efficiency curve. The efficiency of a pump is a function of the flow rate.
Assume that the flow rate readings corresponding to power data points are equally
spaced.
SOLUTION
6.0
5
.
0
7
.
0
159
© 2020 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website,
in whole or in part.
77.0
9
.
0
7.0 corresponding to Q3 9.0
1
9.0 corresponding to Q4
1.0
0.83
1.2
corresponding to Q5
13.6 A power plant has an overall efficiency of 30%. The plant generates 30 MW of
electricity and uses coal from Montana (See Table 11.10) as fuel. Determine how
much coal must be burned to sustain the generation of 30 MW of electricity.
SOLUTION
kg
kW.h 7.8
0.4
0.6
0.8
1
160
13.12 An air conditioning unit has a cooling capacity of 24,000 Btu/hr. If the unit has a
rated energy efficiency ratio (EER) of 11, how much electrical energy is
consumed by the unit in 1 h? If a power company charges 14 cents per kW-hr
usage, how much would it cost to run the air conditioning unit for a month (31
days), assuming the unit runs 8 hrs a day? What is the coefficient of performance
(COP) for the given air conditioning unit?
SOLUTION
161
13.14 Calculate and plot the percentage of each fuel used in generating electricity for
each year shown in the table.
SOLUTION
0%
10%
20%
30%
40%
50%
60%
1970 1980 1990 2000 2010 2020 2030 2040
The percentage of each fuel used in generating electricity
Coal Petroleum Natural Gas Nuclear Renewable/Other
13.15 Calculate and plot the percentage of increase in coal consumption for the data
shown in the table.
SOLUTION
% Increase
Coal Year
37.23 1990
75.70 2005
0
50
100
150
200
250
(%)
Increase in Consumption of Coal
163
13.16 Convert the data given in the table from kilowatt-hours to Btu.
SOLUTION
Results in Billions of Btu
Coal Petroleum Natural Gas Nuclear Renewable/Other Year
5440360 432157.814 1272247.63 1968829 1219253.635 1990
6965636 393950.303 2565957.91 2641910 1282831.682 2005
8548835 364098.499 3763726.71 2971692 1758214.8 2020
13.17 Assuming an average 35% efficiency for power plants and a heating value of
approximately 7.5 MJ/kg, calculate the amount of coal (in kg) required for
generating electricity for each year shown in the table.
164
© 2020 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website,
in whole or in part.
SOLUTION
Energy
Produced 109
kWh
Energy
(considering
efficiency)
109 kWh
Energy
(considering
efficiency)
109 MJ Amount of Coal Needed kg
1161.562 3318.748571 11947.49486 1.593E+12
1594.011 4554.317143 16395.54171 2.18607E+12
1966.265 5617.9 20224.44 2.69659E+12
2040.913 5831.18 20992.248 2.79897E+12
2217.555 6335.871429 22809.13714 3.04122E+12
2504.786 7156.531429 25763.51314 3.43514E+12
3380.674 9659.068571 34772.64686 4.63635E+12
13.18 How many kilograms of coal could be saved if we were to increase the average
efficiency of power plants by 1% to 36%?
165
13.19 Assuming an average 35% efficiency for power plants and a heating value of
approximately 1000 Btu/ft3 (22,000 Btu/lbm), calculate the amount of natural gas
required in ft3 and lbm for generating electricity for each year shown in the table.
13.20 How many ft3 and pounds of natural gas could be saved if we were to increase the
average efficiency of power plants by 1% to 36%?
166
© 2020 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website,
in whole or in part.
SOLUTION
Energy
Produced
109 Btu
Energy
(considering
efficiency)
109 Btu
Amount of
Natural Gas
Needed lbm
Amount of
Natural Gas
Needed ft3
Natural Gas
Saved (lbm)
Natrual Gas
Saved (ft3)
1181716.779 3282546.608 1.49207E+11 3.28255E+12 4263047542 93787045929
1272247.628 3534021.188 1.60637E+11 3.53402E+12 4589637906 1.00972E+11
2051343.377 5698176.046 2.59008E+11 5.69818E+12 7400228631 1.62805E+11
2565957.906 7127660.849 3.23985E+11 7.12766E+12 9256702402 2.03647E+11
2641059.264 7336275.734 3.33467E+11 7.33628E+12 9527630823 2.09608E+11
3763726.706 10454796.41 4.75218E+11 1.04548E+13 13577657670 2.98708E+11
3388326.058 9412016.827 4.27819E+11 9.41202E+12 12223398477 2.68915E+11
13.26 Calculate the amount of natural gas that you need to burn to heat twenty gallons
of water from room temperature at 70°F to 120°F to take shower if the water
heater has an efficiency of: (a) 78 percent, (b) 85 percent, (c) 90 percent.
SOLUTION
13.27 In January and June, how much solar radiation (in kWh/m2/day) on average is
intercepted by a surface (with an effective area of 2 m2) that is tilted at an angle
equal to the latitude of the location for the following states: Georgia, Michigan,
and New Mexico? State your assumptions.
167
© 2020 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website,
in whole or in part.
SOLUTION
For Georgia in January, from 3 to 5 kWh/m2/day of solar radiation could strike
the surface with an effective area of 2 m2 that is tilted at an angle equal to the
latitude of the location. Assuming 4 kWh/m2/day of solar radiation, we get
total solar energy intercepted by the surface
= (2 m2)4 kWh
m2∙day(31 days) = 248 kWh
= (248 kWh)3,412 Btu
1kWh = 846,176 Btu
And for Georgia in June, assuming 5 kWh/m2/day,
total solar energy intercepted by the surface
= (2 m2)5 kWh
m2∙day(30 days) = 300 kWh
= (300 kWh)3,412 Btu
1kWh = 1,023,600 Btu
For Michigan in January, assuming 3 kWh/m2/day,
total solar energy intercepted by the surface
= (2 m2)3 kWh
m2∙day(31 days) = 186 kW
= (186 kWh)3,412 Btu
1kWh = 634,632 Btu
For Michigan in June, assuming 6 kWh/m2/day,
total solar energy intercepted by the surface
= (2m2)6 kWh
m2∙day(30 days) = 360 kWh
= (360 kWh)3,412 Btu
1kWh = 1,228,320 Btu
For New Mexico in January, assuming 6 kWh/m2/day,
total solar energy intercepted by the surface
= (2 m2)6 kWh
m2∙day(31 days) = 372 kW
= (372 kWh)3,412 Btu
1kWh =1,269,264 Btu
For New Mexico in June, assuming 7 kWh/m2/day,
total solar energy intercepted by the surface
168
© 2020 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website,
in whole or in part.
= (2 m2)7 kWh
m2∙day(30 days) = 420 kWh
= (420 kWh)3,412 Btu
1kWh = 1,433,040 Btu
13.28 In the southern region of Arizona, how much solar radiation is intercepted on
average by two flat plate collectors (with dimensions 1 m × 1.5 m) that are tilted
at an angle equal to the latitude of its location in the month of January as
compared to June?
13.29 For a solar system located in Colorado, how many flat-panel solar collectors with
dimensions of 1 m × 1.5 m and an efficiency of 58% would be required to heat up
80 gallons of water from 65°F to 120°F during the month of January? State your
assumptions.
169
SOLUTION
13.30 What is the efficiency of a photovoltaic module with the following specifications?
maximum power output
= 250 W (at illumination of 1 kW/m2)
A = 900 mm
B = 1400 mm
SOLUTION
13.31 How much electricity is generated by a photovoltaic system consisting of 14
modules? The system has an efficiency of 14%, and each module has an effective
area of 1.4 m2. The photovoltaic system is located in a region with an average
solar radiation of 4.5 kWh/m2/day.
SOLUTION
13.32 Assume a photovoltaic system is located in a region with an average solar
radiation of 6.0 kWh/m2/day. How much electricity is generated by a photovoltaic
system consisting of 12 modules? The system has an efficiency of 13%, and each
module has an effective area of 1.2 m2.
13.33 How much electricity is generated at wind speeds of 8 m/s, 10 m/s, 12 m/s, and 14
m/s by a wind turbine that has a blade length of 20 m? Assume an efficiency of
35% for the system and an air density of 1.2 kg/m3.
13.34 A wind turbine manufacturer states that one of its systems with a blade length of
31 m can generate 1.3 MW of electricity when the wind speed is 14 m/s. What is
the efficiency of this system? Note: The density of air is 1.2 kg/m3.
172
13.35 How much electricity is generated at wind speeds of 8 m/s, 10 m/s, 12 m/s, and 14
m/s by a wind turbine that has a blade length of 50 m? Assume an efficiency of
37% for the system and an air density of 1.2 kg/m3.
13.37 The Hoover Dam generates more than 4 billion kWh a year. How many 18.5 W
LED light bulbs could be powered in a year by the Hoover Dam’s power plant?
SOLUTION
13.38 How much coal must be burned in a steam power plant with a thermal efficiency
of 34% to generate enough power to equal the 4 billion KWh a year generated by
the Hoover Dam?
SOLUTION
173
© 2020 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website,
in whole or in part.
amount of coal needs to be burned = 4×109×3,412 Btu
0.34 1 pound
104 Btu
= 4,014 million pounds
174