ENERGY, ENVIRONMENT AND SUSTAINABILITY 2ND EDITION BY SAEED MOAVENI SOLUTIONS MANUAL by QuentinAxel

SOLUTIONS MANUAL

ENERGY, ENVIRONMENT AND SUSTAINABILITY 2ND EDITION BY SAEED MOAVENI SOLUTIONS MANUAL Contents Chapter 1 .......................................................................................................................................1-1 Chapter 2 .......................................................................................................................................2-1 Chapter 3 .......................................................................................................................................3-1 Chapter 4 .......................................................................................................................................4-1 Chapter 5 .......................................................................................................................................5-1 Chapter 6 .......................................................................................................................................6-1 Chapter 7 .......................................................................................................................................7-1 Chapter 8 .......................................................................................................................................8-1 Chapter 9 .......................................................................................................................................9-1 Chapter 10 ................................................................................................................................... 10-1 Chapter 11 ...................................................................................................................................11-1 Chapter 12 ...................................................................................................................................12-1 Chapter 13 ...................................................................................................................................13-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 1: Introduction to Energy, Environment, and Sustainability

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

1.18 A gallon (3.8 liters) of gasoline that weighs 6.3 pounds (2.85 kg) can produce 20 pounds (9.1 kg) of carbon dioxide. Yes, 20 pounds (9.1 kg) of carbon dioxide! Assume 100 million people with cars (with 20 miles/gallon (10.6 km/liters) gasoline consumption rates) decide to walk 3 miles (4.8 km) a day (approximately an hour) instead of driving their cars. What would be the reduction in pounds of carbon dioxide released into the atmosphere on a yearly basis?

Solution

3 miles

365 days 1 gallon 20 pounds of CO2 × × day · person year 20 miles gallon pounds of CO 2 = 109.5 × 109 year

100,000,000 people ×

1-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 2: Fundamental Dimensions and Systems of Units

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 2: Fundamental Dimensions and Systems of Units

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

2.1

Convert the information given in the accompanying table from SI units to U.S. Customary units. Show all steps of your solutions. See Example 2.3.

Solution

 km  1000 m   3.28 ft   1 mile   74.5 miles/h 120   1 km 1m 5280 ft  h     



  km  1000 m   3.28 ft   1 h   109.3 ft/s 120   1 km 1m 3600 s  h      3

 3.28 ft    100 m    3529 ft 3

 1 m 

2-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 2: Fundamental Dimensions and Systems of Units

 2.2046 lbm    176.4 lbm 80 kg  1 kg 



 224.809 10-3 lbf    202.3 lbf 900 N   1N  

 

ft  m  3.28 ft  9.81  2    32.2 2 s 1m s   

   2.2

Convert the information given in the accompanying table from U.S. Customary to SI units. Show all steps of your solutions. See Example 2.3.

Solution

  5280 ft   miles   1 km   1 m   104.6 km/h  65 h 1 mile 1 000 m 3.28 ft      



  km  1000 m   1 h   29 m/s 104.6   1 km 3600 s  h      lbm  120  

  3.28 ft 3     1,920 kg/m3 3  ft  2.2046 lbm   1 m  



1 kg





 1 kg   90.7 kg 200 lbm    2.2046 lbm 

2-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 2: Fundamental Dimensions and Systems of Units

1N    890 N 200 lbf    -3  224.809 10 lbf 

   

2.4

A house has a given floor space of 2,000 ft2. Convert this area to m2.

Solution

 1 m 2 2 A  (2000 ft )    185.9 m  3.28 ft  2

2.5

Calculate the volume of water in a large swimming pool with dimensions of 50 m × 25 m × 2 m. Express your answer in liters, m3, gallons, and ft3.

Solution

V  (50 m)(25 m)(2 m)  2,500 m3

 100 cm 3  1 liter  V  (2,500 m3 )      2,500,000 liters 3  1 m   1000 cm   3.28 ft 3 3 V  (2,500 m3 )    88,219 ft 1 m gallons  7.48   V  (88,219 ft3 )    660,000 gallons 1 ft3      

2.6

A 500-sheet ream of copy paper has thickness of 2.25 in. What is the average thickness of each sheet in mm?

Solution

 Thickness  2.25 in.  25.4 mm   0.1143 mm /sheet 500 sheets 1 in.   

3-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 2: Fundamental Dimensions and Systems of Units

2.7

A barrel can hold 42 gallons of oil. How many liters of oil are in the barrel?

Solution

42 gallons ×

2.8

3.78 liters gallon

= 158.76 liters

Express the kinetic energy [½ (mass)(speed)2] of a car with a mass of 1,200 kg moving at a speed of 100 km/h. First, you need to convert the speed from km/h to the fundamental units of m/s. Show the conversion steps. (Note: We explain the concept of kinetic energy in Chapter 5.)

Solution

 km   1 h  1 000 m  speed  (100)      27.7 m/s h 3600 s 1 km      2 1 m m 5 5 5 K. E. = ( ) (1200 kg) (27.7 ) = 4.6 × 10 kg⋅ 2 ⋅m = 4.6 × 10 N⋅m = 4.6 × 10 J 2 s s

1 ft⋅lbf ) = 3.4 × 105ft⋅lbf K.E. = (4.6 × 105 J) ( 1.3558 J

2.9

A machine shop has a rectangular floor shape with dimensions of 30 ft by 50 ft. Express the area of the floor in ft2, m2, in2, and cm2. Show the conversion steps.

Solution

A  (30 ft)(50 ft)  1500 ft 2 A  (1500 ft 2 )

 1m 2  139.4 m2    3.28 ft  2

 12 in  2 A  (1500 ft )   216,000 in2  1 ft 

4-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 2: Fundamental Dimensions and Systems of Units

 1 cm 2 4 2 A  (1500 ft )    139.4 10 cm  0.0328 ft  2

2.10

A trunk of a car has a listed luggage capacity of 18 ft3. Express the capacity in in3, m3, and cm3. Show the conversion steps.

Solution

 12 in 3 3 V  (18 ft )    31,104 in  1 ft  3

 1 m 3 3 V  (18 ft )    0.51 m  3.28 ft  3

 1 cm 3 4 3 V  (18 ft )    5110 cm  0.0328 ft  3

2.11

An automobile has a 3.5-liter engine. Express the engine size in in3. Show the conversion steps. Note that 1 liter is equal to 1,000 cm 3.

Solution

  1000 cm3  1 in 3 3 V  (3.5 liters)     214 in  1 liter   2.54 cm 

 

2.12

The density of air that we breathe at standard room conditions is 1.2 kg/m3. Express the density in U.S. Customary units. Show the conversion steps.

Solution

Density = 1.2

lbm kg 1m 3 1 lbm = 0.075 ( 3) ( ) ( ) 3.28 ft 0.4536 kg m ft3

5-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 2: Fundamental Dimensions and Systems of Units

2.13

On a summer day in Phoenix, Arizona, the inside room temperature is maintained at 68F while the outdoor air temperature is a sizzling 110F. What is the outdoor– indoor temperature difference in degrees (a) Fahrenheit or (b) Celsius?

Solution

(a) Toutdoor  Tindoor  110∘ F  68∘ F  42∘ F

(∘ C) 

(b) T

(∘ F)  32) 

(110  32)  43.3∘ C

outdoor

9 9 5 5 T (∘ C)  (T (∘ F)  32)  (68  32)  20∘ C indoor 9 indoor 9 Toutdoor  Tindoor  43.3∘ C  20∘ C  23.3∘ C

2.14

A person who is 180 cm tall and weighs 750 newtons is driving a car at a speed of 90 kilometers per hour over a distance of 80 kilometers. The outside air temperature is 30°C and has a density of 1.2 kg/m3. Convert all of the values given from SI to U.S. Customary units.

Solution

Person's height, H

H  (180 cm)(

1 ft )  5.9 ft 30.48 cm

Person's weight, W

W  (750 N)(

1 lbf )  168.6 lbf 4.448N

Speed of the car, S = 90 km/h = 90,000 m/h

S  (90,000

m h

)(

1 mile )  55.9 (miles/h) 1 ft )( 0.3048 m 5280 ft

6-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 2: Fundamental Dimensions and Systems of Units

Distance traveled, D

D  (80 km)(

1000 m

)(

1 km

1 mile )  49.7 miles 1 ft )( 0.3048 m 5280 ft

Temperature of air, T

9 T (C)  (30)  32  86F 5

Density of air, 

 0.3048 m 3 3   (1.2 )( )   0.075(lbm/ft ) m3 0.453 kg  1 ft  kg

1 lbm



   

2.15



Convert the given values: (a) area A = 16 in2 to ft2 and (b) volume V = 64 in3 to ft3.

Solution

(a)

1 ft 2

(16 in )(

(b) (64 in3 )(

2.16



144 in2

)  0.1111 ft

1 ft 3 )  0.037 ft3 12 in

The acceleration due to gravity g is 9.81 m/s2. Express the value of g in U.S. Customary units. Show all conversion steps.

Solution

(9.81 m/s2 )(

1 ft )  32.18 ft/s2 0.3048 m

7-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 2: Fundamental Dimensions and Systems of Units

2.17

Atmospheric pressure is the weight of the column of air over an area. For example, under standard conditions, the atmospheric pressure is 14.7 lbf/in2. This value means that the column of air in the atmosphere above a surface with an area of 1 in2 will exert a force of 14.7 lbf. Convert the atmospheric pressure in the given units to the requested units: (a) 14.7 lbf/in2 to lbf/ft2, (b) 14.7 lbf/in2 to Pa, (c) 14.7 lbf/in2 to kPa, and (d) 14.7 lbf/in2 to bars. Show all of the conversion steps. [Note: One Pascal (1 Pa) is equal to one newton per meter squared (1 Pa = 1 N/m2) and 1 bar = 100 kPa.]

Solution lbf (a)

14.7(

144 in 2

lbf )( )  2,117( ) in2 ft2 1 ft2

(b) 14.7 lbf

( 2) ( in

6,895 Pa 1 lbf

) = 101,357 Pa

2 in

2.18

The density of water is 1,000 kg/m3. Express the density of water in lbm/ft3 and lbm/gallon. (Note: 7.48 gallons = 1 ft3)

Solution

1000 (

)( 3

1m 3.28 ft

) (

2.20 lbm 1 kg

) = 62.34 lbm/ft

lbm ft3 62.34 ( 3 ) ( ) = 8.33 lbm/gallon 7.48 gallons ft

8-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: Evidence-Based Data Analysis

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

3.1

A driver wants to know how much CO2 their car might emit over its lifespan. The relationship between miles driven and pounds of CO2 emitted is shown in the accompanying table. What is the equation for the given data? How many pounds of CO2 will be emitted if the owner of this car drives it for 64,000 miles during a period of 4 years?

Solution

By plotting the data, you will see the linear relationship between the miles driven by this hybrid car and the pounds of CO2 emitted. Also, notice when the miles driven is zero (0), the CO2 emission is also zero (0).

3-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

𝑎 = 𝑠𝑙𝑜𝑝𝑒 =

∆𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑜𝑢𝑛𝑑𝑠 𝑜𝑓 𝐶𝑂2 = ∆𝑥 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑚𝑖𝑙𝑒𝑠 𝑑𝑟𝑖𝑣𝑒𝑛

20 500

= 0.04

b = y-intercept = 0 Equation is 𝑦 = 0.04 𝑥 Using this equation, when the car drives 64,000 miles, 𝑝ounds of CO2 emitted = 0.04 × 64,000 = 2,560 pounds

3.2

The relationship between kilowatt-hours of energy consumed at home and kilograms of CO2 emitted in a power plant is shown in the accompanying table. What is the equation for the given data? How many kilograms of CO2 will be emitted if the homeowner consumes 41,000 kilowatthours of energy during a period of 5 years?

3-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

Solution

By plotting the data, you will see the linear relationship between the energy consumed and the kilograms of CO2 emitted. Also, notice when the energy consumption is zero (0), the CO2 emission is also zero (0).

𝑎 = 𝑠𝑙𝑜𝑝𝑒 =

(6,400 − 5,600) ∆𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝐶𝑂2 = = 0.8 = ∆𝑥 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 (8,000 − 7,000)

b = y-intercept = 0 Equation is 𝑦 = 0.8 𝑥 Using this equation, when the homeowner consumes 41,000 kWh of energy, 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚𝑠 of CO2 emitted = 0.8 × 41,000 = 32,800 kg

3-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

3.3

The relationship between the usage of a hot water heater in months and kilograms of CO2 emitted by burning natural gas is shown in the accompanying table. What is the equation for the given data? How many kilograms of CO2 will be emitted by the hot water heater during a period of 60 months?

Solution

By plotting the data, you will see the linear relationship between the use of a hot water heater in months and the kilograms of CO2 emitted. Also, notice when the usage of the hot water heater is zero (0), the CO2 emission is also zero (0).

𝑎 = 𝑠𝑙𝑜𝑝𝑒 =

∆𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝐶𝑂2 = ∆𝑥 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑢𝑠𝑎𝑔𝑒 𝑜𝑓 𝑎 ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟 ℎ𝑒𝑎𝑡𝑒𝑟 (𝑚𝑜𝑛𝑡ℎ𝑠)

3-4

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

(481.25 − 412.50) (35 − 30)

= 13.75

b = y-intercept = 0 Equation is 𝑦 = 13.75 𝑥 Using this equation, when the hot water heater is used for 60 months, 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚𝑠 of CO2 emitted = 13.75 × 60 = 825 kg

3.4

In order to predict the life expectancy of light bulbs, a manufacturer conducted a series of experiments on 135 light bulbs and gathered the data shown in the table. Plot the data and calculate the mean and standard deviation.

Solution

3-5

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

To calculate the mean and standard deviation, refer to the following table.

Mean(𝑥) =

Sum of hours of all bulbs Total number of light bulbs tested

128,500 = 952 ℎ/bulb 135

∑𝑛 (𝑥 − 𝑥)2 3,037,040 𝑠 = � 𝑖=1 =� = �22,664.48 = 150.55 ℎ/bulb 𝑛−1 (135 − 1)

3.5

For the data given in Problem 3.4, calculate the probability distribution and plot the probability distribution curve. Solution

3-6

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

3.6

A drinking water-fountain manufacturer collected the following data in order to test the performance of their system. Plot the data and calculate the mean and standard deviation.

3-7

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

Solution

To calculate the mean and standard deviation, refer to the following table.

Milliliters of water in the bottle 485 490 495 500 505 510 515

Frequency

Sum of milliliters

13 6,305 17 8,330 25 12,375 40 20,000 23 11,615 18 9,180 15 7,725 Total number of Sum of milliliters water bottles of all bottles tested 151

Mean(𝑥) =

Difference between each milliliter value and the mean value x-15 -10 -5 0 5 10 15

(x -

)2 225 100 25 0 25 100 225

(x-

)2∙frequency 2,925 1,700 625 0 575 1,800 3,375

Sum of the square of the difference between each value and the mean value

75,530

Sum of milliliters Total number of water bottles tested

11,000

75,530 = 500 milliliters/bottle 151

3-8

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

∑𝑛 (𝑥 − 𝑥)2 11,000 = √73.33 = 8.56 milliliters/bottle 𝑠 = � 𝑖=1 =� (151 − 1) 𝑛−1

3.7

For the data given in Problem 3.6, calculate the probability distribution and plot the probability distribution curve.

Solution

Milliliters of water in the bottle 485 490 495 500 505 510 515

Frequency

Probability (p ) 13 17 25 40 23 18 15

13/151 0.086 17/151 0.113 25/151 0.166 40/151 0.265 23/151 0.152 18/151 0.119 15/151 0.099 Sum of probabilities = 1

3-9

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

3.8

The scores of a test for a class of 30 students are shown here. Organize the data in a manner similar to Table 3.5 and create a histogram. Scores: 57, 94, 81, 77, 66, 97, 62, 86, 75, 87, 91, 78, 61, 82, 74, 72, 70, 88, 66, 75, 55, 66, 58, 73, 79, 51, 63, 77, 52, 84

Solution

Scores

Range

Frequency

57, 55, 58, 51, 52

50-59

66, 62, 61, 66, 66, 63

60-69

77, 75, 78, 74, 72, 70, 75, 73, 79, 77

70-79

81, 86, 87, 82, 88, 84

80-89

94, 97, 91

90-100

3-10

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

3.9

For the data given in Problem 3.8, calculate the cumulative frequency and plot a cumulativefrequency polygon.

Solution Range

Frequency

Cumulative Frequency

50-59

60-69

5 + 6 = 11

70-79

5 + 6 + 10 = 21

80-89

5 + 6 + 10 + 6 = 27

90-100

5 + 6 + 10 + 6 + 3 = 30

3-11

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

3.10

For the data given in Problem 3.8, calculate the mean and standard deviation of the class scores.

Solution

To calculate the mean and standard deviation, refer to the following table.

Scores (x)

Difference between each score and the mean score

57 94 81 77 66 97 62 86 75 87 91 78 61 82 74 72 70 88 66 75 55 66 58 73 79 51 63 77 52 84

x-16.23 20.77 7.77 3.77 -7.23 23.77 -11.23 12.77 1.77 13.77 17.77 4.77 -12.23 8.77 0.77 -1.23 -3.23 14.77 -7.23 1.77 -18.23 -7.23 -15.23 -0.23 5.77 -22.23 -10.23 3.77 -21.23 10.77

Sum of scores 2,197

Mean(𝑥) =

(x -

)2 263.413 431.393 60.373 14.213 52.273 565.013 126.113 163.073 3.133 189.613 315.773 22.753 149.573 76.913 0.593 1.513 10.433 218.153 52.273 3.133 332.333 52.273 231.953 0.053 33.293 494.173 104.653 14.213 450.713 115.993

Sum of the square of the difference between each score and the mean score 4,549.37

Sum of scores Total number of scores

2,197 = 73.23 30

∑𝑛 (𝑥 − 𝑥)2 4,549.37 � 𝑠 = 𝑖=1 =� = √156.87 = 12.52 𝑛−1 (30 − 1)

3-12

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 3: EvidenceBased Data Analysis

3.11

For the data given in Problem 3.8, calculate the probability distribution and plot the probability distribution curve.

Solution

Range of scores

Frequency

Probability (p )

50-59

5/30

0.167

60-69

6/30

0.200

70-79

10/30

0.333

80-89

6/30

0.200

90-100

3/30

0.100

Sum of probabilities = 1

3-13

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

End of Section Problem Solutions 4.1

Countries in Asia use the Celsius temperature scale. The annual average temperature for the city of Tokyo, Japan, during the period of 1970 through 2020 is shown in the accompanying table. Use Excel to convert the given temperatures from degrees Celsius to degrees Fahrenheit.

Solution

As seen in the figure below, input the formula = (B2*9/5)+32 in the cell C2, and the result would be 59.4. Then apply Fill command to cells C3 through C12.

4-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

4.2

In Europe, common garbage bag sizes are 4, 8, 13, 25, 33, 42, 45, 55, and 60 liters. Use Excel to convert these sizes from liters to gallons.

Solution

As seen in the figure below, input the formula = A2*0.264 in the cell B2, and the result would be 1.1. Then apply Fill command to cells B3 through B10.

4-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

4.3

In the U.S., an apartment size is expressed in square feet, whereas in Europe and Asia, the size is given in square meters. Use Excel to convert the apartment sizes shown in the accompanying table from square feet to square meters.

Solution

As seen in the figure below, input the formula = A2*0.3048^2 in the cell B2, and the result would be 46.5. Then apply Fill command to cells B3 through B9.

4-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

4.4

Typical water bottle sizes in Asia are shown in the accompanying table. Use Excel to convert these sizes from milliliters and liters to fluid ounces and gallons.

Solution

As seen the figure below, to convert milliliters to fluid ounces, input the formula = A2*0.033814 in the cell C2, and the result would be 9.5. Then apply Fill command to cells C3 and C4. To convert litters to fluid ounces, input the formula = A5*1000*0.033814 in the cell C5, then apply Fill command to cells C6 through C8. To convert milliliters to gallons, input the formula = A2*0.000264 in the cell D2, and the result would be 0.074. Then apply Fill command to cells D3 and D4. To convert liters to gallons, input the formula = A5*1000*0.000264 in the cell D5, then apply Fill command to cells D6 through D8.

4-4

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

4.5

Plot the data shown in Problem 4.1 and show the trendline with an equation that best fits the given set of data.

Solution

The Year zero represents Year 1970, and Year 10 represents Year 2020. Also, note that the temperature trendline suggests an increase of 0.1373 °C each year.

4-5

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

4.6

The shipment of iPhone units worldwide during 2012 through 2019 is shown in the accompanying table. Plot the data in the given table.

Solution

4-6

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

4.7

The trend in adults being overweight in the U.S. during the period of 1999 through 2018 is shown in the accompanying table. Plot this data and show the trendline with an equation that best fits the given set of data.

Solution

Year zero represents (1999-2000) and year 9 represents (2017-2018). Also, note that the trendline suggests an increase of 1.22% per year.

4-7

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

4.8

The monthly total rainfall in 2020 for Singapore is shown in the accompanying table. Use Excel to convert the data from millimeters to inches.

Solution

As seen in the figure below, input the formula = B2*0.03937 in the cell C2, and the result would be 3.5. Then apply Fill command to cells C3 through C13.

4.9

Plot the rainfall data given in Problem 4.8 and show the trendline with an equation that best fits the given set of data.

4-8

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

Solution

Month zero represents January and Month 11 represents December. The linear trendline suggests an increase of 7.91 mm per month. Also, note the linear model is not the best fit for this set of data.

4.10

The total and per capita of U.S. Municipal Solid Waste (MSW) generation from 1960 to 2012 are shown in the accompanying table. Use the steps discussed in Example 4.7 to plot the total MSW and the per capita generation rates on the same chart.

4-9

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

Solution

First, plot the Total MSW Generation versus year as shown below.

With the mouse pointer in the chart area, click the right mouse button and choose Select Data . . . , click on the Add button, then type in the series name and choose the Series values, as shown below.

4-10

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

With the mouse pointer over the Per Capita Generation curve, double-click the left mouse button. Choose Format Data Series…, then under Series Options, turn on the Secondary Axis. Add the secondary axis label.

4.11

The total and per capita of U.S. Municipal Solid Waste (MSW) recycling from 1960 to 2012 are shown in the accompanying table. Use the steps discussed in Example 4.7 to plot the total MSW and the per capita recycling rates on the same chart.

4-11

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

Solution

First, plot the Total MSW Recycling versus year as shown below.

4-12

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

With the mouse pointer over the Percent Recycling curve, double-click the left mouse button. Choose Format Data Series…, then under Series Options, turn on the Secondary Axis. Add the secondary axis label.

4.12

As shown in the accompanying table, coal has been China’s primary source of energy production. Use the steps discussed in Example 4.7 to plot China’s total energy production and the percentages of production by coal on the same chart.

4-13

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

Solution

First, plot the Total Energy Production versus year as shown below.

4-14

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

With the mouse pointer over the Percentage of Production by Coal line, double-click the left mouse button. Choose Format Data Series…, then under Series Options, turn on the Secondary Axis. Add the secondary axis label.

4.13

The annual average price of regular-grade gasoline in the U.S. during the period of 2010 through 2020 is shown in the accompanying table. Use Excel’s logic function to create a table that shows if the price of gas has gone up or down, when compared to the previous year.

4-15

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

Solution

In cell C3, type the formula =IF(B3=B2, “Same”, IF(B3>B2, “Up”, “Down”)), as shown below and use the Fill command to copy the formula in cells C4 through C12.

4.14

The Body Mass Index (BMI) is calculated based on the height and weight of a person to show if an individual is considered underweight (BMI values less than 18.5), normal weight (BMI values between 18.5 and 24.9), or overweight (BMI values greater than 25). The BMI values for 10 people are shown in the accompanying table. Use Excel’s logic functions to create a table that shows, if the person is underweight, normal weight, or overweight.

4-16

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

Solution

In cell C2, type the formula =IF(B2<18.5,"underweight",IF(B2<=24.9,"normal weight", "overweight")), as shown below and use the Fill command to copy the formula in cells C4 through C12.

4.15

The Wet Bulb Globe Temperature (WBGT) is an index that shows the level of the stress on the body from heat. To prevent heat-related illnesses, the Japanese government provides the guidelines for exercise shown in the accompanying table.

The data in the following table shows the WBGT at 2 p.m. in Tokyo from June 1 to October 15, 2019. Use Excel’s logic functions to create a table that shows the level of warning for each day.

4-17

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 4: Electronic Spreadsheets

Solution

In cell C3, type the formula =IF(B3<21,"Almost safe", IF(B3<25,"Caution",IF(B3<28,"Warning",IF(B3<31,"Severe warning", "Danger")))), as shown below and use the Fill command to copy the formula in cells C4 through C12.

4-18

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 5: Energy and Power

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 5: Energy and Power

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

5.1

If you were to push a lawn mower with a constant horizontal force of 10 N, how much work would you do while pushing the lawn mower a total distance of 5 m on level ground?

Solution work = 10 N × 5 m = 50 N∙m = 50 N∙m (

5.2

0.225 lbf

3.28 ft )(

) =36.9 ft⋅lbf

Which of the following requires more work: to change the speed of a car from 0 to 40 mph (0 to 64 km/h) or from 40 to 65 mph (64 to 105 km/h)?

Solution Work (which is equal to the change in kinetic energy) required when the speed of a car goes from 0 to 40 mph (0 to 64 km/h) is calculated as follows:

1 km 1000 m h )) � �( ) (masscar) ((64 ) ( 1 km )( 3600 s 2 h final position 1

–� ( ) (masscar)(0)2� 2

= (158.02) (masscar) initial position

Similarly, work required when the speed of a car goes from 40 to 65 mph (64 to 105 km/h) is calculated as follows:

5-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 5: Energy and Power 2

1 km 1000 m h �( )(masscar) ((105 ) ( 1 km )(3600 s)) � 2 h final position 2

1 km 1000 m h – �( )(mass car ) ((64 )) � ) ( 1 km ) ( 2 3600 s h initial position

=(425.35)(masscar) ⎼ (158.02)(masscar) = (267.32)(masscar)

Therefore, changing the speed of a car from 40 to 65 mph (64 to 105 km/h) requires more work.

5.4

An elevator has a rated capacity of 2,000 pound mass (907 kg). It can transport people at the rated capacity between the first and the fifth floors in 7 seconds with a vertical distance of 15 ft (4.6 m) between each floor. Estimate the power requirement for such an elevator.

Solution

m work = change in potential energy = (907 kg) (9.8 2) (4.6 m × 4) = 163,550 J s work 163,550 J = = 23,364 W power = time 7 seconds

or in the U.S. Customary units, 2,000 lbm

work = change in potential energy= (

power =

ft ) (32.2 2) (15 ft × 4) = 120,000 lbf∙ft s 32.2 lbm⋅ft lbf⋅s2

work 120,000 lbf∙ft lbf∙ft lbf∙ft = = 17,143 )( = (17,143 time 7 seconds s s

1 hp lbf∙ft 550 s

) =31.1 hp

5-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 5: Energy and Power

5.5

Determine the gross force needed to bring a car that is traveling at 110 km/h to a full stop in a distance of 100 m. The mass of the car is 2,100 kg. What happens to the initial kinetic energy? Where does it go or to what form of energy does the kinetic energy convert?

Solution 1

(force)(distance) = [0]𝑓𝑖𝑛𝑎𝑙⎼ �( ) (2,100 kg) ((110

)(

1,000 m km

)(

1h 3600 s

)) � 𝑖𝑛𝑖𝑡𝑖𝑎𝑙

(force)(100 m) = 0 ⎼ 980,324 N∙m Force = -9,803 Newtons Therefore, the force needed to stop the car is -9,803 N. The negative sign indicates the force must be applied in the opposite direction to motion. The kinetic energy is converted into heat.

5.6

A power plant has an overall efficiency of 30%. The plant generates 20 MW of electricity and uses coal as fuel. Determine how much coal must be burned to sustain the generation of 20 MW of electricity.

Solution net energy generated by the power plant 30% = energy input from burning coal in the plant

0.30 =

20 MW energy input from burning coal in the plant

energy input from coal =

20 MW

= 66.67 MW = 66.67 × 106 J/s

0.3

amount of coal required=

66.67×106 J/s 10.5×106J 1 lbm

= 6.35 lbm/s

5-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 5: Energy and Power

5.12 An air-conditioning unit has a cooling capacity of 18,000 Btu/h. If the unit has a rated energy efficiency ratio (EER) of 11, how much electrical energy is consumed by the unit in one hour? If a power company charges 12 cents per kWh for usage, how much would it cost to run the air-conditioning unit for a month (31 days), assuming the unit runs eight hours a day?

Solution heat removal from the air-conditioning unit (Btu) 11 = energy input to the air-conditioning unit (Watt-hour) 18,000 Btu 11 = energy input to the unit (Wh)

energy input to the unit = 1,636 Wh = 1.636 kWh

The cost to run the air-conditioning unit for a month for a period of 8 hours a day is calculated in the following matter. cost to operate the unit = (

5.14

days 1.636 kWh 8 hours $0.12 )( )( )(31 ) = $48.70/month 1 hour day kWh month

In Example 5.10, we calculated the amount of natural gas that you would burn to heat 20 gallons of water from room temperature at 70°F to 120°F to take a shower. Determine the additional amount of gas that must be burned if the water heater has an efficiency of (a) 80 percent, (b) 85 percent, and (c) 90 percent.

Solution

efficiency =

(a)

desired output required input

0.80 =

8,340 Btu

required input =

8,340 Btu

required input

0.8

= 10,425 Btu

10,425 Btu = (amount of natural gas in cubic feet) (1,000

Btu cubic feet

)

amount of natural gas in cubic feet = 10.43 ft3

5-4

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 5: Energy and Power

additional amount of natural gas in cubic feet = 10.43 ft3 − 8.34 ft3 = 2.09 ft3 (b)

0.85 =

required input =

8,340 Btu

8,340 Btu 0.85

required input

= 9,812 Btu Btu

9,812 Btu = (amount of natural gas in cubic feet) (1,000

) cubic feet

amount of natural gas in cubic feet = 9.81 ft3 additional amount of natural gas in cubic feet = 9.81 ft3 − 8.34 ft3 = 1.47 ft3 (c)

0.90 =

required input =

8,340 Btu

required input

0.9

= 9,267 Btu Btu

9,267 Btu = (amount of natural gas in cubic feet) (1,000

) cubic feet

amount of natural gas in cubic feet = 9.27 ft3 additional amount of natural gas in cubic feet = 9.27 ft3 − 8.34 ft3 = 0.93 ft3

5.15

For Example 5.12, how much fuel would be saved if the efficiency of the power plant is increased from 36 to 40 percent?

Solution For Example 5.12, we did not state what losses were assumed for the transmission lines; for this problem, let’s assume a 6% loss in the transmission lines, then the overall efficiency would be ((1-0.06) x 40%) = 37.6% net energy generated by the power plant efficiency = 0.376 =

energy input from burning the fuel in the boiler

660 GJ energy input from burning the fuel in the boiler

660 GJ energy input from fuel =

0.376

= 1,755 GJ = 1.755 TJ

5-5

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 5: Energy and Power

1.755×1012 ≈ 234,000 kg amount of coal required = 6 J 7.5×10 kg

amount of coal that could be saved is, therefore, 293,000 kg - 234,000 kg = 59,000 kg

5.16

Estimate the amount of fuel that could be saved if the efficiency of 10 million cars is increased by five percent. Assume an annual driving distance of 12,000 miles (19,300 km) and an average gasoline consumption of 25 miles per gallon (10.6 km/liter).

Solution Difference in amount of gasoline required for 10 million cars to drive 12,000 miles each = ((

12,000 miles 25 miles 1 gallon

12,000 miles

) - ( 25 miles×(1 + 0.05) ))(10 ×106 cars) 1 gallon

= (480 gallons - 457.14 gallons)(10 × 106 cars) = 228.57×106 ≈ 229 million gallons In SI units, Difference in amount of gasoline required for 10 million cars to drive 12,000 miles each = ((

19,300 km 10.6 km 1 liter

19,300 km

) - ( 10.6 km×(1 + 0.05))) (10 ×106 cars) 1 liter

=(1,820.75 liters - 1,734.05 liters )(10 × 106 cars) = 867.03×106 ≈ 867 million liters

5.19

For Example 5.11, estimate the amount of natural gas that would be consumed by a furnace with an efficiency of 90% during a cold spell lasting one week.

Solution

efficiency =

desired output required input

5-6

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 5: Energy and Power

Btu day 0.90 = required input 1,200,000

1,200,000 required input =

0.90

Btu day

= 1,333,333

Btu 1,333,333

day

Btu day

cubic feet = (amount of natural gas in

day

) (1,000

Btu

) cubic feet

amount of natural gas/day = 1,333.33 cubic feet/day amount of natural gas/week = (1,333.33 cubic feet/day)(7 days/week) =9,333.31 cubic feet/week

5.20

Calculate the power requirement in horsepower for an escalator in a mall that transports 50 people with an average of weight of 170 lbf (756 N) a vertical distance of 30 feet (9.1 m) in 25 seconds.

Solution (50 people)(170 lbf)(30 ft) change in potential energy lbf∙ft =( ) = 10,200 power = time 25 seconds s = 10,200 (

lbf∙ft s

)(

1 hp ) = 18.5 hp lbf∙ft 550 s

5-7

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 5: Energy and Power

5-8

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 6: Electricity

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 6: Electricity

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

6.1

If a 1500-W hair dryer is connected to a 120-V line, what is the maximum current drawn?

Solution P = VI

1500 W = (120 V)(I)

current I = 12.5 amperes (A)

6.2

A toaster connected to a 120-V line draws 7 amps. What is the power consumption of the toaster?

Solution P = VI = (120 V)(7 A) = 840 W

6.3

A 20-MW coal-fired power plant has an efficiency of 40%. Estimate the annual consumption of coal burned in this plant. Assume a heating value of approximately 7.5 MJ/kg (megajoules per kilogram) for coal.

6-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 6: Electricity

Solution

net energy generated

power plant efficiency =

20×106 0.40 =

energy input from fuel

J 3600 s 24 h 365 days (s) ( 1 h ) (1 day) ( 1 year ) energy input from coal 20×106

energy input from coal =

3600 s 24 h 365 days (s) ( 1 h ) (1 day) ( 1 year ) 0.4

6.3072×1014(J/year) =

= 1.5768×10

0.4

J ( ) year

J 1.5768×1015 ( ) year amount of coal required = =210,240,000 kg/year J 6 7.5×10 ( ) kg

or in terms of pound mass and tons as

amount of coal required = (210,240,000

kg year

amount of coal required = (462,528,000

6.4

)(

2.2 lbm

) = 462,528,000 lbm/year

1 kg

lbm

) ( 1 ton ) = 231,264 tons year 2,000 lbm year

A 2,000-W dishwasher is run for approximately one hour for 122 days during a year. If the electric power company charges 12 cents per kWh, calculate the annual electric energy consumption of the dishwasher and the associated cost.

6-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 6: Electricity

Solution

1 h 122 days ) annual electric energy consumption = (2,000 W) ( )( day year = 244,000 Wh/year = 244 kWh/year

annual cost = (244 kWh) ( $0.12 ) = $29.28 1 kWh

6.5

A 500-W clothes washing machine is run for approximately two hours a week for 52 weeks during a year. If the electric power company charges 12 cents per kWh, calculate the annual electric energy consumption of the clothes washer and the associated cost.

Solution annual electricity consumption = (500 W) (

2 hours 52 weeks )( year ) week kWh Wh = 52,000 = 52 year year

annual cost = (52 kwh) ( $0.12 ) = $6.24 1 kWh

6.6

A 1,200-W hair dryer is run for approximately 5 minutes every day. If the electric power company charges 12 cents per kWh, calculate the annual electric energy consumption of the hair dryer and the associated cost.

Solution

5 minutes

annual electricity consumption = (1,200 W) (

day

)(

)( 60 minutes

365 days

year

)

kWh

Wh = 36,500

year

= 36.5

year

$0.12 annual cost = (36.5 kwh) (

1 kWh

) = $4.38

6-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 6: Electricity

6.7

A TV set consumes 150 W and is left on for approximately 5 hours every night. If the electric power company charges 12 cents per kWh, calculate the annual electric energy consumption of the TV and the associated cost.

Solution 5 hours 365 days )( ) day year kWh Wh = 273,750 = 273.75 year year

annual electricity consumption = (150 W) (

$0.12 annual cost = (273.75 kWh) ( ) = $32.85 1 kWh

6.8

A relatively small house in the United States has an annual electrical energy consumption of 4,000 kWh. Estimate the annual consumption of coal burned in a power plant to generate the required electrical energy. Assume a heating value of approximately 7.5 MJ/kg (megajoules per kilogram) for coal, 40% efficiency at the power plant, and 6% loss for the power transmission lines.

Solution efficiency =

desired energy output required energy input

kWh year (0.40 × 0.94) = required energy input 4,000

4,000 kWh required energy input =

amount of coal required =

0.376

J 10,638 × 103 ( ) (1 h) (3600 s) s 1h = = 10,638 year year

10,638×36×105J 7.5×106J kg

kWh

= 5,106 kg

6-4

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 6: Electricity

or in terms of pound mass and tons as amount of coal required = 5,106 (

2.2 lbm )( ) = 11,233 lbm 1 kg year year

amount of coal required = 11,233 (

6.9

lbm year

)(

1 ton tons ) = 5.61 year 2,000 lbm

Rework Problem 6.8 for a larger house with an annual electrical energy consumption of 10,000 kWh.

Solution kWh year (0.40 × 0.94) = required energy input 10,000

10,000 kWh kWh required energy input = = 26,595 = year 0.376

amount of coal required =

26,595×36×105J 7.5×106J kg

J 3600 s 26,595×103 ( ) (1 h) ( ) s 1h year

= 12,765 kg

or in terms of pound mass and tons as lbm kg 2.2 lbm amount of coal required = 12,765 ( )( ) = 28,083 year 1 kg year

amount of coal required = 28,083 (

lbm year

)(

1 ton tons ) = 14.04 year 2,000 lbm

6-5

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 6: Electricity

6.12

According to Sylvania (a light bulb manufacturer), its 40-W CFL light consumes 9 W and produces 495 lumens. What is the efficacy for this light?

Solution efficacy =

6.13

light produced (lumens) energy consumed by the lamp (watts)

495 =

= 55

The Sylvania Super Saver 75-W light uses 20 W, produces 1,280 lumens, and costs $4.49. As an alternative, a generic 75-W incandescent light bulb costs $0.40 and produces 1,200 lumens. a. Compare the performance of each light by calculating the efficacy for each light. b. Estimate how much it would cost to run each light for 4 hours a day for 300 days in a year. Assume electricity costs 9 cents per kWh.

Solution (a) For the Sylvania Super Saver 75-W light, Efficacy =

1,280 lumens 20 W

= 64

For the generic 75-W incandescent light bulb, 1,200 lumens Efficacy =

75 W

= 16

(b) For the Sylvania Super Saver 75-W light, 4 h 300 days $0.09 $0.09 cost = (20 W) ( ) ( )( ) =(24,000 Wh) ( ) = $2.16 day year kWh 1,000 Wh For the generic 75-W incandescent light bulb, 4 h 300 days $0.09 $0.09 cost = (75 W) ( ) ( )( ) =(90,000 Wh) ( ) = $8.10 day year kWh 1,000 Wh

6-6

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 6: Electricity

6.19

Estimate the amount of electrical energy consumed during a year for a case when a garage door is lifted four times each day. Assume the garage door opener has a ½hp-rated motor and it takes 14 s to lift the garage door.

Solution work or energy power = time

Energy required to lift the garage door = (power)(time)

1 0.746 kW 1h 4 times 365 days ) =2.1 kWh = (hp) ( ) (14 s) ( )( )( 2 1 hp 3,600 s day year year

6.20

Estimate the annual electrical energy consumption of a laptop computer that consumes approximately 65 W. Assume the laptop is used on average for 8 hours per day and for 220 days during a year.

Solution energy consumed = (65 W) (

1 kW kWh 8 hours 220 days )( )( ) = 114.4 year year 1000 W day

6-7

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

7.1

Calculate the R-value for the following materials: (a) 4-in.-thick brick, and (b) 12-in.thick concrete slab.

Solution (a)

Referring to Table 7.1, you will find the thermal conductivity of brick: k = 1.0 (

)

m∙º C

R-factor or R-value for the 4-in.-thick brick is 4 in. 0.0254 m 2 ( 1 in. ) 𝐿 R= = = 0.1016 m ∙ºC W 𝑘 W 1.0 (m∙ºC)

(b)

Similarly, you get the thermal conductivity of concrete: k = 1.4 (

)

m∙C

R-factor or R-value for the 12-in.-thick concrete is 12 in. 0.0254 m 2 ( 1 in. ) L R= = = 0.2177 m ∙ºC w k W 1.4 ( ) m∙ºC

7.2

Calculate the R-value (film resistance) for a situation when the wind is blowing over a wall, with h = 4.0 Btu/h·ft2·°F.

7-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

Solution film resistance R =

1 h

1 Btu

4.0 (

)

h ∙ ft2·°F = 0.25 Btu

h·ft ·°F

7.3

Calculate the overall U-value for a 12-in.-thick concrete wall with the wind blowing over its outside surface (h = 5.88 Btu/h·ft2·°F), and still air over its inside surface (h = 1.47 Btu/h·ft2·°F).

Solution For the situation where wind is blowing over the outside surface, 1 h∙ft2∙ºF = 0.17 R= = Btu h 5.88 Btu 2 h∙ft ∙ºF 1

For still air over its inside surface, 1

1 R=

1.47 Btu h∙ft2∙ºF

= 0.68

h∙ft2∙ºF Btu

2∙ºC

For 12-in.-concrete, R = 0.2177 m

(see Problem 7.1)

W h∙ft2∙ºF m2∙C 1.0 (m2∙ºC ) (0.2177 )( Btu ) = 1.23 Btu W 5.88 2 h∙ft ∙ºF

U-factor for the wall is equal to U=

1 = sum of resistances

1 h∙ft2∙ºF (0.68 + 1.23 + 0.17)

= 0.48 Btu h∙ft2∙ºF

Btu

7-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

7.4

If a window has a total R-value of 1.7, what is its U-value?

Solution U=

7.5

1 =

sum of resistances

1.7

Btu = 0.588

h∙ft2∙ºF

Calculate the heat transfer rate from a 1,000 ft2, 6-in.-thick concrete wall with inside and outside surface temperatures of 70ºF and 40ºF, respectively.

Solution

T1 − T2

heat transfer rate q = kA

A = 1,000 ft = 1,000 ft (

L 0.3048 m 2 ) = 92.90 m2 1 ft

5 T1 = 70ºF =

(70 - 32) = 21.1ºC

5 T2 = 40ºF =

(40 - 32) = 4.4ºC

L = 6 in. = 6 in. (

0.0254 m ) = 0.1524 m 1 in.

(21.1 − 4.4)ºC = 14,252 W q = 1.4 ( W ) ) (92.90 m2) ( 0.1524 m m∙ºC

7.6

Calculate the heat transfer rate from a 1,000 ft2, 9-in.-thick concrete wall with inside and outside surface temperatures of 70ºF and 40ºF, respectively.

7-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

Solution heat transfer rate q = kA

T1 − T2 L

0.3048 m 2 ) = 92.90 m2 A = 1,000 ft = 1,000 ft ( 1 ft 5 T1 = 70ºF = (70 - 32) = 21.1ºC 9 5 T2 = 40ºF = (40 - 32) = 4.4ºC 9 0.0254 m L = 9 in. = 9 in. ( ) = 0.2286 m 1 in. (21.1 − 4.4)ºC = 9,501 W q = 1.4 ( W ) ) (92.90 m2) ( 0.2286 m m∙ºC 2

7.7

Calculate the heat transfer rate from a 2,000-ft2 ceiling of a house with a total Rvalue of 41.0 h·ft2·°F/Btu. Assume an inside room temperature of 68ºF and an attic air temperature of 10ºF.

Solution T − Toutside q = �inside �(area) = total resistance

7.8

⎡

⎤ Btu (68 − 10)ºF 2 h∙ft2∙ºF (2,000 ft ) = 2,829 h ⎢41.0 ( )⎥ Btu ⎦ ⎣

Calculate the heat loss through the walls of a building with a net surface area of 1,400 ft2 and a total R-value of 25 h·ft2·°F/Btu. Assume an inside room temperature of 68ºF and an outside air temperature of 5ºF.

Solution T − Toutside q = �inside �(area) = total resistance

⎡

⎤ Btu (68 − 5)ºF 2 2 (1,400 ft ) = 3,528 h∙ft ∙ºF h ⎢25 ( ⎥ ) Btu ⎦ ⎣

7-4

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

7.9

Calculate the heat transfer rate through a 24-ft2 door with U = 0.73

Btu

h·ft 2 ·℉

. The

indoor and outdoor temperatures are 68ºF and 10ºF, respectively.

Solution q = U(Tinside − Toutside)(area)

= (0.73

Btu

)(68 - 10)ºF (24 ft2) = 1,016

Btu

h∙ft2∙ºF

7.10

Calculate the heat transfer rate through a 12-ft2 window with U = 0.8

Btu

h·ft 2 ·℉

. The

indoor and outdoor temperatures are 68ºF and 10ºF, respectively. Solution q = U(Tinside − Toutside)(area)

= (0.8

Btu

)(68 - 10)ºF (12 ft2) = 557

h∙ft2∙ºF

7.11

Btu h

Calculate the heat transfer rate through a 12-ft2 window with U = 0.2

Btu

h·ft 2 ·℉

. The

indoor and outdoor temperatures are 68ºF and 10ºF, respectively. Compare the results of this exercise with results of Problem 7.10. How much energy will be saved if three windows with U = 0.8 Btu2 are replaced with three windows with U = Btu

h·ft ·℉

0.2 h·ft 2·℉ for a home in a location with annual degree-days of 6,200?

Solution Heat transfer rate through a 12-ft2 window with U = 0.2 Btu

is calculated as follows. Btu 2 q = U(T −T )(area) = (0.2 Btu ) (68 - 10)ºF (12 ft ) = 139 inside outside h h∙ft2∙ºF h·ft2·℉

As calculated in Problem 7.10, with U = 0.8, q = 557 Btu. h

7-5

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

When replacing three windows with U = 0.8 with the windows with U = 0.2, Btu

Btu h h Btu ) × 24h energy savings ( Btu) × 24h 1,254 h ( h Q = = 519 Btu = DD (68-10)ºF design temperature difference(ºF) DD

energy savings = 3×(557 - 139)

= 1,254

Btu Btu Qyearly =519 ( ) (6,200 DD) = 3,217,800 DD year

7.12

A typical exterior masonry wall of a house consists of the items in the table shown in Figure P7.12. Assume an inside room temperature of 68ºF, an outside air temperature of 10ºF, and an exposed area of 150 ft2. Calculate the heat loss through the wall.

7-6

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

Solution The total resistance to heat flow is given by sum of resistances = 0.17 + 0.44 + 0.1 + 1.72 + 1.28 + 0.45 + 0.68 = 4.84

h∙ft2∙ºF Btu

Therefore, the heat loss is (68 - 10)°F Tinside − Toutside Btu q= � �(area) = ( )(150 ft2) = 1,798 h sum of resistances h∙ft2∙ºF 4.84 Btu

7.13

In order to increase the thermal resistance of a typical exterior frame wall, it is customary to use 2  6 studs instead of 2  4 studs to allow for the placement of more insulation within the wall cavity. A typical exterior (2  6) frame wall of a house consists of the materials shown in Figure P7.13. Assume an inside room temperature of 68ºF, an outside air temperature of 20ºF, and an exposed area of 150 ft2. Determine the heat loss through this wall.

7-7

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

Solution h∙ft2∙ºF sum of resistances = 0.17 + 0.81 + 1.32 + 19.0 + 0.45 + 0.68 = 22.43 Btu Therefore, the heat loss through the wall is (68 - 20)°F Tinside − Toutside Btu q= � �(area) = ( ) (150 ft2) = 321 h sum of resistances h∙ft2∙ºF 22.43 Btu

7.14

A typical ceiling of a house consists of items shown in Figure P7.14. Assume an inside room temperature of 70ºF, an attic air temperature of 15ºF, and an exposed area of 1,000 ft2. Calculate the heat loss through the ceiling.

Solution sum of resistances = 0.68 + 19.0 + 0.45 + 0.68 = 20.81

h∙ft2∙ºF Btu

7-8

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

Therefore, the heat loss though the ceiling is (70 − 15)°F Tinside − Toutside Btu q= � �(area) = ( )(1,000 ft2) = 2,643 h sum of resistances h∙ft2∙ºF 20.81 Btu

7.15

What would be the reduction in heat loss through the ceiling of a house if the insulation R-value of 19 is increased to 40 by adding more insulation? The ceiling area is 2,000 ft2.

Solution Tinside − Toutside q = � �(area) sum of resistances

we compare T

q = �inside

7.17

−Toutside T −T T −T �(2,000) − �inside outside�(2,000) = 55.26 �inside outside� 19 40 40

Calculate the heat loss from a double-pane glass window consisting of two pieces of glass, each having a thickness of 10 mm with a thermal conductivity of k = 1.3 W . m·K

The two glass panes are separated by an air gap of 7 mm. Assume the thermal conductivity of air to be k = 0.022 W . Also, express the total R- and U-values. m·K

Solution L 0.01 m m2∙K = 0.0077 Rglass = = W k 1.3 ( W ) m∙K 0.007 m m2∙K = 0.31818 Rair = = W k 0.022 ( W ) m∙K L

7-9

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

m2∙K R total= Rglass+Rair+Rglass = 0.0077 + 0.31818 + 0.0077 = 0.33358 W

7.18

1 R total

m2∙K 0.33358 ( W )

W m2∙K

A building is located in Baltimore, Maryland, where the annual heating degree days are 4,654. The building has a heating load (heat loss) of 30,000 Btu/h, and a design temperature difference of 52°F (68°F indoor and 16°F outdoor). Estimate the building’s annual energy consumption. If the building is heated with a furnace with an efficiency of 92%, how much gas is burned to keep the home at 68°F? State your assumptions.

Solution Btu 30,000 ( ) ×24 h h Q = = 13,846 Btu DD 52°F DD Btu Qyearly = 13,846 ( ) (4,654 DD) = 64.4×106 Btu/year DD Assuming natural gas with a heating value of 1,000 Btu per cubic foot and the furnace has an efficiency of 92%, the amount of natural gas required to keep the home at 68°F is given by 64.4×106Btu 1 �� = 70,000 ft3/year volume of natural gas burned = � Btu 0.92 1,000 gallon

7-10

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

7.19

Nine old, 12-ft2 windows with a U-value of U = 1.2 windows having U = 0.3

Btu

were replaced with new Btu . Calculate the energy savings on a day during a fiveh·ft 2 ·℉

h·ft2·℉

hour period when Tin = 68ºF and Toutside = 10ºF.

Solution

Btu Btu ) (9×12 ft2)(68 - 10°F) = 5,638 𝑞old−new = (1.2 - 0.3) ( h h·ft2·°F

energy savings for a period of 5 hours = 5,638 Btu/h × 5 h = 28,190 Btu

7.20

For Problem 7.18, calculate the savings in cubic feet of natural gas. Assume the furnace has an efficiency of 98%.

Solution Assuming natural gas has a heating value of 1,000 Btu per cubic foot and the furnace has an efficiency of 98%, the amount of fuel oil required to keep the home at 68°F is given by 64.4×106 Btu 1 3 �� volume of natural gas burned = � �= 65,714 ft /year Btu 0.98 1,000 3 ft

7-11

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

7.21

Calculate the annual degree days for Boston, Massachusetts, using the following given monthly values.

Solution Annual degree days = ∑ monthly values = 5,634.

7.23

Calculate the R-value for the following materials: (a) 10-cm-thick brick, and (b) 30cm-thick concrete slab.

Solution W (a) Referring to Table 7.1, you will find the thermal conductivity of brick: k = 1.0 ( ) m∙º C

R-factor or R-value for the 10-cm-thick brick is 1m 2 10 cm ( ) 100 cm = 0.1 m ∙ºC R= = W W 𝑘 1.0 (m∙ºC) 𝐿

7-12

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 7: Thermal Energy: Heat Loss and Gain in Buildings

(b)

Similarly, you get the thermal conductivity of concrete: k = 1.4 (

)

m∙C

R-factor or R-value for the 30-cm-thick concrete is m) 30 cm (1001 cm 2 L= = 0.2143 m ∙ºC R= w W k 1.4 ( ) m∙ºC

7.24

Calculate the R-value (film resistance) for a situation where the wind is blowing over a wall with h = 25 W/m2∙℃.

Solution

1 m2·℃ = 0.04 film resistance R = = W h W 25 ( 2 ) m ·℃ 1

7.25

Calculate the heat transfer rate through a 2-m2 window with U = 1.8 W/m2∙K. During a 24-hour period, the indoor and outdoor temperatures remain at 20℃ and –5℃, respectively. How much energy will be saved if three of these windows have U-values equal to 1.2 W/m2∙K?

Solution Heat transfer rate through a 2-m2 window with U = 1.8 W

m2·℃

−T

q = U(T inside

outside

)(area) = (1.8

is calculated as follows.

W ) (20 ⎼ (−5))℃ (2 m2) = 90 W 𝑚2∙℃

Similarly, the heat transfer rate through a 2-m2 window with U = 1.2 W2

m ·℃

−T

q = U(T inside

outside

)(area) = (1.2

W ) (20 ⎼ (−5))℃ (2 m2) = 60 W m2∙℃

energy savings for a period of 24 hours = (90-60) W × 24 h = 720 Wh

7-13

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 8: Energy Consumption Rates and Non-Renewable Energy Sources

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 8: Energy Consumption Rates and Non-Renewable Energy Sources

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

8.4

How many gallons of gasoline would be saved during the next 10 years if a driver were to change her existing car with a 25 mpg (10.6 km/liter) fuel efficiency to a more efficient car with an average fuel economy of 40 mpg (17 km/liter)? Assume she drives her car about 12,000 miles (19,300 kilometers) per year. Solution

(

8.5

12,000 miles 12,000 miles (10 years) = 1,800 gallons 25 miles 40 miles ) 1 gallon 1 gallon

Perform a sensitivity cost saving analysis for Problem 8.4. See Example 8.10. Solution

Gasoline cost (in dollars) 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00

Savings (in dollars) 3,600.00 4,050.00 4,500.00 4,950.00 5,400.00 5,850.00 6,300.00 6,750.00 7,200.00

8-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 8: Energy Consumption Rates and Non-Renewable Energy Sources

8.6

How much energy in Btu would be saved during the next five years if a household were to reduce its annual electricity consumption rates from 10,000 kWh to 7,000 kWh? Perform a sensitivity cost saving analysis, assuming the electric utility company could charge between 10 to 20 cents per kWh. Use increments of 2-cent change for your analysis. Solution

kWh

Energy saving = (5 years)(10,000 ⎼ 7,000) (

) = 15,000 kWh year Btu 3.4123 h ) = 51.2×106 Btu = (15,000×103) (Wh) ( W

8.7

Electricity cost

Savings

(in dollars/kWh)

(in dollars)

0.10

1,500.00

0.12

1,800.00

0.14

2,100.00

0.16

2,400.00

0.18

2,700.00

0.20

3,000.00

In 2011, petroleum provided 34% of the world's energy consumption of 519 quadrillion Btu. Assuming petroleum has an average energy content of 130,000 Btu/gallon, how many barrels of petroleum were consumed? Also express your answer in gallons. Solution

amount of petroleum consumed =

519×1015 Btu 3

130×10 Btu gallon

= 3.99×1012 gallons

4×1012gallons = 9.50×1010 barrels 42 gallons 1 barrel

8-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 8: Energy Consumption Rates and Non-Renewable Energy Sources

8.8

How many gallons of gasoline are consumed annually on average by a 25-year-old male driver (see Example 8.8) if he drives a car with a fuel economy of 20 mpg? Solution

From Example 8.8, a 25-year old male would drive 17,976 miles on average. 17,976 miles 20 miles gallon

8.9

= 898.8 gallons

Assume the annual heating energy consumption of a house is 122 million Btu. How many cubic feet of natural gas does it take to keep the house warm during the cold months? Solution

Assuming 100% efficiency, Btu year = 122,000 ft3/year 1,000 Btu 1 ft3

122 ×106 Amount of natural gas required =

8.10

The annual heating energy consumption of a house located in the northeastern part of the United States is 140 million Btu. How many gallons of fuel oil does it take to keep the house warm during the cold months? Solution

Assuming 100% efficiency, Btu year = 1,007 gallons/year 139,000 Btu 1 gallon

140 ×106 Amount of fuel oil required =

8-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 8: Energy Consumption Rates and Non-Renewable Energy Sources

8.11

The annual electricity consumption of a household is 9,800 kWh. How many pounds of coal must be burned in a power plant to address this need? Assume a combined overall efficiency of 30% for the power plant and the transmission lines. Solution

First, we convert kWh to Btu/h. 3

9,800 kWh/year = 9,800×10 (

3.4123

Wh )( year

Btu h ) = 3.344×107 Btu/year

3.344×107 Btu/year 10,000 Btu 1 pound Amount of coal required = 0.3 3,344 pounds/year =

8.12

0.3

= 11,147 pounds/year

In a recent year, California consumed 41 million gallons/day of gasoline. How many Btu of energy were consumed each day? What is California’s gasoline consumption per capita? Assume an approximate population of 37 million. Solution

Assuming the average energy content of gasoline = 124,000 Btu/gallon, The energy consumed by 41 million gallons/day 124,000 Btu = (41 ×106)(gallon/day) ( ) = 5.08×1012 Btu/day gallon Gasoline consumption per capita per day in California (41 ×106)(gallon/day) = = 1.11 gallons/person∙day (37 ×106) people

8-4

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 8: Energy Consumption Rates and Non-Renewable Energy Sources

8.13

In a recent year, 23 trillion ft3 of natural gas was delivered to 70 million customers in the United States. How many rooms with dimensions of 15 ft x 15 ft x 10 ft could be filled with 23 trillion ft3? This problem is intended to give you a visual image of how much natural gas we consume. Solution

number of rooms =

8.14

23×1012 ft3 (15×15×10) ft

= 10.2 × 10 9 = 10.2 billion rooms

In 2010, United States coal mines produced 1,805.3 million tons of coal. What was the coal consumption per capita for the United States? Assume an approximate population of 308 million for that year. Solution

coal consumption per capita =

8.15

1,805.3×106 tons 308×106 people

= 5.86 tons per capita

In the United States, wood and wood waste could account for 2% of energy use. How many Btus of energy are generated from wood and wood waste? State your assumptions. Solution

According to Figure 8.8, 100.2 quadrillion Btu was consumed by major sectors in the U.S. in 2019. Therefore, about 2 quadrillion Btu (2% of 100.2 quadrillion Btu) was generated from wood and wood waste.

8.20 In a recent year, the United States consumed 29,000,000,000,000 kWh of energy. Express this value in MWh, GWh, TWh, and PWh. See Table 2.2 for the list of decimal multiples and prefixes used with SI base units. Solution

29,000,000,000,000 kWh = 29,000,000,000 MWh = 29,000,000 GWh = 29,000 TWh = 29 PWh

8-5

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

9.1

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. Solution

Use Figures 9.10 and 9.11 to find the average daily solar radiation in Georgia, Michigan, and New Mexico for the months of January and June. 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 4 kWh = (2 m2) ( 2 )(31 days) = 248 kWh m ∙day 3,412 Btu = (248 kWh) ( ) = 846,176 Btu 1 kWh And for Georgia in June, assuming 5 kWh/m2/day, total solar energy intercepted by the surface 5 kWh = (2 m2) ( 2 )(30 days) = 300 kWh m ∙day 3,412 Btu = (300 kWh) ( ) = 1,023,600 Btu 1 kWh For Michigan in January, assuming 3 kWh/m2/day, total solar energy intercepted by the surface 3 kWh ) (31 days) = 186 kW m2∙day 3,412 Btu = (186 kWh) ( ) = 63,463 Btu 1 kWh = (2 m2) (

9-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

For Michigan in June, assuming 6 kWh/m2/day, total solar energy intercepted by the surface 6 kWh )(30 days) = 360 kWh m2∙day 3,412 Btu = (360 kWh) ( ) = 1,228,320 Btu 1 kWh = (2 m2) (

For New Mexico in January, assuming 6 kWh/m2/day, total solar energy intercepted by the surface 6 kWh ) (31 days) = 372 kW m2∙day 3,412 Btu = (372 kWh) ( ) =1,269,264 Btu 1 kWh = (2 m2) (

For New Mexico in June, assuming 7 kWh/m2/day, total solar energy intercepted by the surface 7 kWh )(30 days) = 420 kWh m2∙day 3,412 Btu = (420 kWh) ( ) = 1,433,040 Btu 1 kWh = (2 m2) (

9.2

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? Solution

For January, assuming 5 kWh/m2/day of solar radiation could strike the collectors that are tilted at an angle equal to the latitude of the location, total solar energy intercepted by the two collectors is 5 kWh = (2×1.5 m2) ( )(31 days) = 465 kWh m2∙day 3,412 Btu = (465 kWh) ( ) = 1,586,580 Btu 1 kWh

9-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

For June, assuming 8 kWh/m2/day, total solar energy intercepted by the two collectors is 8 kWh = (2×1.5 m2) ( )(30 days) = 720 kWh m2∙day 3,412 Btu = (720 kWh) ( ) = 2,456,640 Btu 1 kWh The amount of solar radiation intercepted in January would be 255 kWh (or 870,060 Btu) less than that in June.

9.3

Assume that the solar collector system of Problem 9.2 has an average efficiency of 60% during the month of January and 68% during the month of June. On average, how many gallons of supply water at 70°F could be heated to 120°F by the system each day during each month? Solution

For January, total solar energy intercepted by the collectors each day 5 kWh kWh kWh 3,412 Btu Btu = (2×1.5 m2) ( ) = 15 = (15 )( ) = 51,180 m2∙day day day 1 kWh day Considering the efficiency of the system is 60% in January, Btu Btu total available energy = (0.6) (51,180 ) = 30,708 day day amount of water required per day Btu (1 lbm)∙(1°F) lbm (30,708 )( ) day 1 Btu = 614.16 = (120 - 70)°F day lbm 1 gallon = 614.16 ( )( ) = 73.6 gallons of water per day day 8.34 lbm For June, total solar energy intercepted by the collectors each day 8 kWh kWh kWh 3,412 Btu Btu = (2×1.5 m2) ( ) = 24 = (24 )( ) = 81,888 day day 1 kWh day m2∙day Considering the efficiency of the system is 68% in June, Btu Btu total available energy = (0.68) (81,888 ) = 55,684 day day amount of water required per day Btu (1 lbm)∙(1°F) lbm (55,684 )( ) day 1 Btu = 1,113.68 = (120 - 70)°F day lbm 1 gallon ) = 133.5 gallons of water per day = 1,113.68 ( )( day 8.34 lbm

9-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

9.4

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. Solution

Energy required to heat up 80 gallons of water from 65°F to 120°F is given by 8.34 lbm 1 Btu (80 gallons) ( ) (120 - 65)°F( ) = 36,696 Btu 1 gallon 1 lbm · 1°F The efficiency of the system is 58%, so the amount of energy required to intercept 36,696 Btu/day =

0.58

= 63,269 Btu/day

Assuming 4 kWh/m2/day of solar radiation could strike the collectors that are tilted at an angle equal to the latitude of the location, the total required size of the flat solar panels is given as follows. Btu 63,269 day = 63,269 m2 = 4.636 m2 13,648 4 kWh 3,412 Btu ( 1 kWh ) 2 m ∙ day Considering the size of each panel is 1.5 m2, the number of the flat solar panels required 4.636 m2 =

1.5 m2

= 3.09

So, four (4) panels are required to heat up 80 gallons of water from 65°F to 120°F every day in January.

9.7

What is the efficiency of a photovoltaic module (Figure P9.7) with the following specifications? maximum power output = 250 W (at illumination of 1 kW/m2) A = 900 mm B = 1,400 mm

9-4

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

Solution

efficiency =

9.8

250 W 1000 W = 0.198 = 19.8% 2 (0.9×1.4 m ) ( ) m2

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 New York with an average solar radiation of 4.5 kWh/m2/day. Solution

amount of electricity generated = (0.14)(14×1.4 m2) (4.5

kWh m2∙ day

) = 12.3

kWh day

= 4,490 kWh/year

9.9

Assume a photovoltaic system is located in Colorado 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. Solution

amount of electricity generated = (0.13)(12×1.2 m2) (6.0

kWh m2∙ day

) = 11.2

kWh day

= 4,088 kWh/year

9-5

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

9.12

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. Solution

sweep area = π(blade length)2 = π(20 m)2 = 1,257 m2 For wind speed = 8 m/s, 1 wind power = (efficiency) ( ) (density)(sweep area)(speed)3 2 1 kg kg∙m2 m3 2 = (0.35) ( ) (1.2 3)(1,257 m ) (8 ) = 135,153 3 = 135,153 W ≈ 135 kW s 2 m s For wind speed = 10 m/s, 1 3 wind power = (efficiency) ( ) (density)(sweep area)(speed) 2 1 kg kg∙m2 m3 2 = (0.35) ( ) (1.2 3)(1,257 m ) (10 ) = 263,970 3 = 263,970 W ≈ 264 kW s 2 m s For wind speed = 12 m/s, 1 wind power = (efficiency) ( ) (density)(sweep area)(speed)3 2 1 kg kg∙m2 m3 2 = (0.35) ( ) (1.2 3)(1,257 m ) (12 ) = 456,140 3 = 456,140 W ≈ 456 kW s 2 m s For wind speed = 14 m/s, 1 3 wind power = (efficiency) ( ) (density)(sweep area)(speed) 2 3 1 kg kg∙m2 m 2 = (0.35) ( ) (1.2 3)(1,257 m ) (14 ) = 724,334 3 = 724,334W ≈ 724 kW s 2 m s

9.13

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. Solution

efficiency =

1.3× 106 W kg 1 m 3 (2) (1.2 3)((π)(31 m)2) (14 s ) m

1.3× 106 W = 4,970,598

kg∙m2 s3

= 0.26 = 26%

9-6

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

9.14

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. Solution

sweep area = π(blade length)2 = π(50 m)2 = 7,854 m2 For wind speed = 8 m/s, 1 wind power = (efficiency) ( ) (density)(sweep area)(speed)3 2 1 kg kg∙m2 m3 2 = (0.37) ( ) (1.2 3)(7,854 m ) (8 ) = 892,717 3 = 892,717 W ≈ 893 kW s 2 m s Similarly, For wind speed = 10 m/s, wind power = 1,743,588 W ≈ 1,744 kW For wind speed = 12 m/s, wind power = 3,012,920 W ≈ 3,013 kW For wind speed = 14 m/s, wind power = 4,784,405 W ≈ 4,784 kW

9.16

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

4×109×103 Wh/year ≈ 24.6 million light bulbs 18.5 W×(24 h/day × 365 days/year) 1 light bulb

9.17

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 9

energy generated by burning coal = = 4 × 1013 Btu

4 × 10 kWh 0.34

3,412 Btu 9 ( 4 × 10 kWh) ( kWh ) 0.34

Assuming coal has an average energy content of 10,000 Btu per pound, 4×109×3,412 Btu amount of coal needed to be burned = (

0.34

) (

1 pound

) 104 Btu

= 4,014 million pounds

9-7

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

9.18

In 2014, in the United States, 23% of the 9.6 quadrillion Btu of renewable energy came from wood and wood byproducts, and waste (e.g., sawdust and scraps). Estimate the number of cords of wood that would have the equivalent energy content. State your assumptions. Solution

energy from wood and wood by-products = 9.6×1015×0.23 = 2.21×1015 Btu Assuming one cord of wood has an average energy content of 20,000,000 Btu, 2.21×1015 Btu 8 amount of wood = = 1.11×10 cords = 111 million cords Btu 2×107 (cord)

9.19

In 2014, in the United States, 4% of the 9.6 quadrillion Btu of renewable energy came from solar systems. How much coal is saved (not consumed) because of the solar energy segment? Assume coal has an average energy content of 10,000 Btu per pound and a coal-fired power plant has an efficiency of 36% and a 6% loss in the transmission lines. Solution

energy from solar systems = 9.6×1015×0.04 = 3.84×1014 Btu pound (3.84×1014 Btu) ( ) 10,000 Btu amount of coal saved = (0.36 - 0.06) = 1.28×1011 pounds = 128 billion pounds

9.20

In 2014, in the United States, 26% of the 9.6 quadrillion Btu of renewable energy came from hydroelectric plants. How much coal is saved (not consumed) because of the hydroelectric plants? Assume coal has an average energy content of 10,000 Btu per pound and a coal-fired power plant has an efficiency of 36% and a 6% loss in the transmission lines. Solution

energy from hydroelectric plants = 9.6×1015×0.26 = 2.5×1015 Btu

pound (2.5×1015 Btu) ( ) 10,000 Btu amount of coal saved = (0.36 - 0.06) = 8.32×1011 pounds = 832 billion pounds

9-8

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

9.21

Estimate how much electricity is generated by a south-oriented photovoltaic system consisting of 16 modules located in the southern part of Spain. The system has an efficiency of 13%, and each module has an effective area of 1.4 m2. Solution

Assuming the solar radiation available for the photovoltaic system in the southern part of Spain is about 2,100 kWh/m2/year, amount of electricity generated = (0.13)(16×1.4 m2) (2,100

kWh

kWh ) = 6,115 m2∙ year year

kWh year kWh )( = 6,115.2 ( ) = 16.8 day year 365 days

9.22

For the system given in Problem 9.21, how much electricity will be generated if the photovoltaic system is located in the northern part of Sweden? Solution

Assuming the solar radiation available for the photovoltaic system in the northern part of Sweden is about 1,000 kWh/m2/year, amount of electricity generated = (0.13)(16×1.4 m2) (1,000

kWh

kWh ) = 2,912 m2∙ year year

kWh year kWh )( = 2,912 ( ) = 8.0 day year 365 days

9.23

For the system given in Problem 9.21, how much electricity will be generated if the photovoltaic system is located in the northern part of France? Solution

Assuming the solar radiation available for the photovoltaic system in the northern part of France is about 1,400 kWh/m2/year, amount of electricity generated = (0.13)(16×1.4 m2) (1,400

kWh

) = 4,077 m2∙ year

year

kWh year kWh )( = 4,077 ( ) = 11.2 day year 365 days

9-9

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

9.24

Estimate how much electricity is generated by a south-oriented photovoltaic system consisting of 20 modules located in the northern part of Poland. The system has an efficiency of 14%, and each module has an effective area of 1.4 m2. Solution

Assuming the solar radiation available for the photovoltaic system in the northern part of Poland is about 1,300 kWh/m2/year, amount of electricity generated = (0.14)(20×1.4 m2) (1,300

kWh

kWh ) = 5,096 m2∙ year year

kWh year kWh )( = 5,096 ( ) = 14.0 day year 365 days

9.25

For the system given in Problem 9.24, how much electricity will be generated if the photovoltaic system is located in the northern part of Italy? Solution

Assuming the solar radiation available for the photovoltaic system in the northern part of Italy is about 1,700 kWh/m2/year, amount of electricity generated = (0.14)(20×1.4 m2) (1,700

kWh

) = 6,664 m2∙ year

year

kWh year kWh )( = 6,664 ( ) = 18.3 day year 365 days

9.26

According to the United Nations, China had a population of 1,439,323,776 in 2020 (mid-year). What fraction of electricity demand from China’s population is met by solar energy, wind energy, hydropower, and bioenergy? Solution

According to Table 6.2, China’s electricity consumption is 4.91 MWh/capita, therefore electricity demand in China = 1,439,323,776 (𝑝𝑒𝑜𝑝𝑙𝑒) 4.91 (

MWh person ∙ year

)

= 7,067,079,740 MWh/year

9-10

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

According to Figure 9.34, China’s installed capacity of solar energy is 254,355 MW; therefore, electricity that could be generated by solar energy = 254,355 (MW) (

365 days 24 h )( ) year day

= 2,228,149,800 MWh/year electricity demand can be fulfilled by solar energy =

2,228,149,800 MWh/year (%) 7,067,079,740 MWh/year

= 31.5% According to Figure 9.35, China’s installed capacity of wind energy is 281,993 MW; therefore, electricity that could be generated by wind energy = 281,993 (MW) (

365 days 24 h )( ) year day

= 2,470,258,680 MWh electricity demand can be fulfilled by wind energy =

2,470,258,680 MWh/year (%) 7,067,079,740 MWh/year

= 35.0% According to Figure 9.36, China’s installed capacity of hydropower is 339,840 MW; therefore, electricity that could be generated by hydropower = 339,840 (MW) (

365 days 24 h )( ) year day

= 2,976,998,400 MWh/year electricity demand can be fulfilled by hydropower =

2,976,998,400 MWh/year (%) 7,067,079,740 MWh/year

= 42.1% According to Figure 9.37, China’s installed capacity of bioenergy is 18,687 MW; therefore, electricity that could be generated by bioenergy = 18,687 (MW) (

365 days 24 h )( ) year day

= 163,698,120 MWh/year electricity demand can be fulfilled by hydropower =

163,698,120 MWh/year (%) 7,067,079,740 MWh/year

= 2.3%

9-11

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

9.27

According to the United Nations, Germany had a population of 83,783,942 people in 2020 (midyear). What fraction of electricity demand from Germany’s population is met by solar energy, wind energy, hydropower, and bioenergy? Solution

According to Table 6.2, Germany’s electricity consumption is 6.85 MWh/capita, therefore, electricity demand in Germany = 83,783,942 (𝑝𝑒𝑜𝑝𝑙𝑒) 6.85 (

MWh person ∙ year

)

= 573,920,003 MWh/year According to Figure 9.34, Germany’s installed capacity of solar energy is 53,783 MW; therefore, electricity that could be generated by solar energy = 53,783 (MW) (

365 days year

)(

24 h ) day

= 471,139,080 MWh/year electricity demand can be fulfilled by solar energy =

471,139,080 MWh/year (%) 573,920,003 MWh/year

= 82.1% According to Figure 9.35, Germany’s installed capacity of wind energy is 62,184 MW; therefore, electricity that could be generated by wind energy = 62,184 (MW) (

365 days 24 h )( ) year day

= 544,731,840 MWh/year electricity demand can be fulfilled by wind energy =

544,731,840 MWh/year (%) 573,920,003 MWh/year

= 94.9% Figure 9.36 does not show Germany’s installed capacity of hydropower, assuming it is around 11,000 MW, electricity that could be generated by hydropower = 11,000 (MW) (

365 days 24 h )( ) year day

= 93,360,000 MWh electricity demand fulfilled by hydropower =

96,360,000 MW (%) = 16.8% 573,920,003 MW

According to Figure 9.37, Germany’s installed capacity of bioenergy is 10,367 MW; therefore, electricity that could be generated by bioenergy = 10,367 (MW) (

365 days 24 h )( ) year day

9-12

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

electricity demand can be fulfilled by hydropower =

9.28

90,814,920 MW (%) = 15.8% 573,920,003 MW

The United States population is estimated at 331,000,000 people. What fraction of electricity demand from the U.S. population is met by solar energy, wind energy, hydropower, and bioenergy? Solution

According to Table 6.2, the U.S.’s electricity consumption is 13.10 MWh/capita, therefore electricity demand in the U. S. = 331,000,000 (𝑝𝑒𝑜𝑝𝑙𝑒) 13.10 (

MWh person ∙ year

)

= 4,336,100,000 MWh/year According to Figure 9.34, the U.S.’s installed capacity of solar energy is 75,572 MW; therefore, electricity that could be generated by solar energy = 75,572 (MW) (

365 days year

)(

24 h ) day

= 662,010,720 MWh/year electricity demand can be fulfilled by solar energy =

662,010,720 MWh/year (%) 4,336,100,000 MWh/year

= 15.3% According to Figure 9.35, the U.S.’s installed capacity of wind energy is 117,744 MW; therefore, electricity that could be generated by wind energy = 117,744 (MW) (

365 days 24 h )( ) year day

= 1,031,437,440 MWh/year electricity demand can be fulfilled by wind energy =

1,031,437,440 MWh/year (%) 4,336,100,000 MWh/year

= 23.8% According to Figure 9.36, the U.S.’s installed capacity of hydropower is 83,790 MW; therefore, electricity that could be generated by hydropower = 83,790 (MW) (

365 days 24 h )( ) year day

= 734,000,400 MWh/year electricity demand can be fulfilled by hydropower =

734,000,400 MWh/year (%) 4,336,100,000 MWh/year

= 16.9%

9-13

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 9: Renewable Energy

According to Figure 9.37, the U.S.’s installed capacity of bioenergy is 12,372 MW; therefore, electricity that could be generated by bioenergy = 12,372 (MW) (

365 days 24 h )( ) year day

= 108,378,720 MWh/year electricity demand can be fulfilled by hydropower =

108,378,720 MWh/year (%) 4,336,100,000 MWh/year

= 2.5%

9-14

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 10: Air and Air Quality Standards

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 10: Air and Air Quality Standards

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems. 10.1

We can reduce the amount of CO2 released into the atmosphere by designing cars with improved fuel economy ratings. What would be the reduction in pounds of CO2 released into the atmosphere by a car with an improved fuel economy rating of 10 mpg (~4 km/L)? Assume the car is driven 12,000 miles (~19,000 km) annually. Solution Let x represent the fuel economy of a car and x + 10 the improved fuel economy: 12,000 miles reduction in the amount of CO2 = (

1 gallon )(

year x miles 20 pounds of CO2 − )( ) 1 gallon (x + 10) miles 1 gallon

1 1 = 240000 ( − ) pounds of CO 2 per year x miles (x+10) miles For example, comparing two cars with 25 mpg and 35 mpg, we get: 1 1 240,000 ( − ) = 2,743 reduction in pounds of CO2 per year 25 35 Or, comparing two cars with 20 mpg and 30 mpg, we get: 1 1 240,000 ( − ) = 4,000 reduction in pounds of CO2 per year 20 30

10.2

We can reduce the amount of CO2 released into the atmosphere by driving our cars less. What would be the reduction in pounds of CO2 released into the atmosphere by a car that is driven 10,000 miles (16,000 km) per year instead of 12,000 miles (19,000 km) per year? State your assumption. Solution Assuming an average fuel economy of 25 mpg,

10-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 10: Air and Air Quality Standards

reduction in the amount of CO2 = (12,000 - 10,000) (

miles year

)(

gallon 20 pounds of CO2 )( ) 25 miles gallon

= 1,600 pounds of CO2 per year

10.3

For Example 10.3, what would be the addition in pounds of CO2 released into the atmosphere if the hot water heater had an efficiency of 85%? Solution additional amount of CO2 released into the atmosphere 365 = ( ) − 365 = 429 − 365 = 64 pounds of CO2 0.85

10.4

For Example 10.4, what would be the reduction in pounds of CO2 released into the atmosphere if the TV consumed 150 W? Solution reduction in the amount of CO2 = 340 − (

150 W × 4 hours night

)(

1 kW

)( 1000 W

200 nights year

)(

1.7 pounds of CO2 ) 1 kWh

= 340 − 204 = 136 pounds of CO2 per year

10.5

Estimate the annual reduction in pounds (or kilograms) of CO2 released into the atmosphere if a home that uses 1,500 kWh for lighting replaces the current system with one that is 20% more efficient. Solution reduction in the amount of CO2 = (0.2) (

1,500 kWh 1.7 pounds of CO2 )( ) 1 year 1 kWh

= 510 pounds of CO2 per year

10-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 10: Air and Air Quality Standards

0.45 kg = (510 pounds) ( ) = 230 kg of CO2 per year 1 pound

10.6

Homes in the United States consume on average anywhere between 4,000 kWh and 10,000 kWh annually. Calculate the amount of CO2 released into the atmosphere for a million homes with an annual consumption of 7,000 kWh. Solution amount of CO2 released into the atmosphere = (1 million homes) (7,000

kWh year

)(

1.7 pounds of CO2 ) 1 kWh

= 11.9×109 pounds of CO2 per year ≈ 12 billion pounds of CO2 per year

10.7

What would be the reduction in pounds of CO2 released into the atmosphere if 100 million people walked 3 miles (~5 km) a day instead of driving their cars? Present your results on daily, weekly, monthly, and yearly basis. Solution Assuming a fuel economy of 25 mpg, the reduction in the amount of CO2 by walking 3 miles instead of driving miles 1 gallon 20 pounds of CO2 = (100×106 people) (3 )( )( ) day 25 miles 1 gallon = 2.4×108 pounds of CO2 per day 2.4×108 (

2.4×108(

pounds day pounds

)(

7 days

) = 16.8×108 pounds of CO per week

week average 30.5 days

) = 7.3×109 pounds of CO per month

2 month 365 days 2.4×108 ( )( ) = 87.6×109 pounds of CO per year 2 day year

day pounds

10-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 10: Air and Air Quality Standards

10.8

The U.S. Federal Highway Administration's recent data regarding the average annual miles driven per driver by age group is shown in the accompanying table.

Assume an average fuel economy of 25 miles per gallon, and calculate the amount of CO2 released into the atmosphere by gender and age group. Present your results in a similar tabular form. Solution The amount of CO2 released into the atmosphere, in pounds, is given by the following formula: 1 gallon 20 pounds of CO2 (average annual miles) ( )( ) 25 miles

Age 16–19 20–34 35–54 55–64 65+ Average

10.9

1 gallon

Male 6,565 14,381 15,086 12,687 8,243 13,240

(unit: pound) Female Average 5,498 6,099 9,603 12,078 9,171 12,233 6,224 9,578 3,828 6,117 8,114 10,781

In the northeastern section of the United States, many homes during the winter months are heated by furnaces that burn fuel oil. Calculate the amount of CO2 released into the atmosphere by 100,000 homes if each house consumes on average 800 gallons of fuel oil during this period. Solution amount of CO2 released = (100,000 homes) (

800 gallons 22.5 pounds of CO2 )( ) 1 home 1 gallon

= 1.8 billion pounds of CO2

10-4

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 10: Air and Air Quality Standards

10.10

In many areas where natural gas lines are not available, homes are heated during the winter months by furnaces that burn propane. Calculate the amount of CO2 released into the atmosphere by 100,000 homes if each house consumes on average 600 gallons of propane during this period. Solution 600 gallons 12.4 pounds of CO2 )( ) 1 home 1 gallon = 744 million pounds of CO2

amount of CO2 released = (100,000 homes) (

10-5

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 11: Water Resources, Consumption Rates, and Quality Standards

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 11: Water Resources, Consumption Rates, and Quality Standards

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

11.1

Assuming that a large family uses a dishwasher that uses 10 gallons of water per wash three times a week, what is the total amount of water consumed by the dishwasher in a year? Solution amount of water consumed per year = (

10 gallons 3 washes 52 weeks )( )( ) 1 wash week year = 1,560 gallons per year

11.2

An old showerhead delivers approximately 3 gallons of water per minute. Assuming that a person showers twice a day for 5 minutes each time, what is the total amount of water that is consumed for this activity annually? Solution Assuming that the person takes a shower twice a day every day, amount of water consumed per year 3 gallons 5 minutes 2 times 365 days = ( )( )( )( ) = 10,950 gallons 1 minute 1 time 1 day 1 year

11.4

A car-wash system consumes 60 gallons of water per wash. Assuming that a person washes her car once every week, what is the total amount of water consumed annually by this activity? Solution amount of water consumed per year 60 gallons 52 weeks =( )( ) = 3,120 gallons per year 1 week 1 year

11-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 11: Water Resources, Consumption Rates, and Quality Standards

11.6

Assuming a household water consumption of 70 gallons per day per capita, what was the total amount of water that was consumed during 2010 by all of the people in the United States? (Note: The population of the United States in 2010 was about 309 million people.) How much water would have been saved if the per capita consumption was reduced to 60 gallons per day? Solution total household water consumption in the U.S. in 2010

70 gallons (1 person)(day)

)(309 ×10 6 people) (

365 days

)

1 year

= 7.9×1012 gallons ≈ 8 trillion gallons

Reduction in the total household water consumption

(70 − 60) gallons 365 days ) (309 ×106 people) ( ) (1 person)(day) 1 year

= 1.1×1012 gallons ≈ 1 trillion gallons

11.18

Imagine 100 million adults who will live another 65 years. Also assume that each person would use 0.27 m3 of water every day on average. How much water would be consumed by this population over their expected remaining lives? Solution amount of water consumed 0.27 m3 365 days 6 )( = (100 ×10 people) ( ) (65 years) day 1 year 1,000 liters = 6.4 ×1011 m3 of water = (6.4 ×1011) (m3) ( ) = 6.4×1014 liters of water 3 m 1 gallon 14 12 (liters) = (6.4 ×10 ) ( ) = 170 ×10 gallons 3.78 liters = 170 trillion gallons of water

11-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 11: Water Resources, Consumption Rates, and Quality Standards

11.19

According to the USGS in 2010, about 161 billion gallons of water per day were consumed in the thermoelectric power category in the United States. How much water does this volume represent in terms of square miles of land covered to a depth of 4 ft each year? Solution amount of water consumed per year 365 days gallons 1 ft3 9 ) ( )( ) = 7.86×10 12 ft3 /year = (161 ×10 day 7.48 gallons 1 year 7.86×1012 ft3 = (x) ft2×(4 ft) Solving for x, we get the area of the land covered by that amount of water = 1.965×1012 ft2 12

= (1.965×10

11.20

1 mile

) = 70,485 square miles )ft ( 5,280 ft

According to the USGS in 2010, about 355 billion gallons of water per day were consumed to address various needs in the United States. Show that this volume is equal to 397,000 thousand acre-feet per year. (Note: An acre-foot is the amount of water required to cover 1 acre to a depth of 1 ft, 1 acre is equal to 43,560 ft2, and 1 ft3 = 7.48 gallons.) Solution amount of water consumed per year = (355 ×109 gallons) ( day

1 ft3 7.48 gallons

1.73 × 1013 ft3 = (x) acre (

365 days

)( 1 year ) = 1.73×1013 ft3/year

43,560 ft2 ) × (1 ft) 1 acre

Solving for x, we get x = 397,153,352 ≈ 397,000 acre-feet per year Therefore, the amount of water consumed per year is 1.73 ×1013 ft3 and this volume is equal to 397,000 thousand acre-feet.

11-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 12: Understanding the Materials We Use in Our Daily Lives

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 12: Understanding the Materials We Use in Our Daily Lives

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems. 12.1

Identify and list at least five different materials that are used in making a car. Solution Examples include:  Steel  Aluminum  Plastic  Glass  Rubber  Fiberglass  Leather  Copper

12.2

Name at least five different materials that are used in making a refrigerator. Solution Examples include:  Aluminum  Stainless steel  Plastic  Glass  Copper

12-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 12: Understanding the Materials We Use in Our Daily Lives

12.3

Identify and list at least five different materials that are used in making your TV set or computer. Solution Examples include:  Plastic  Glass  Copper  Gold  Silver  Aluminum

12.4

List at least five different materials that are used in making a building envelope (walls, floors, roofs, windows, and doors). Solution Examples include:  Concrete  Wood  Plastic  Glass  Steel  Aluminum

12.5

List at least five different materials used to fabricate window and door frames. Solution Examples include:  Aluminum  Steel  Fiberglass  Vinyl  Wood

12-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 12: Understanding the Materials We Use in Our Daily Lives

12.6

List some of the materials used in the fabrication of LED lights. Solution Examples include:  Glass  Nickel  Lead  Copper  Gallium

12.7

Identify at least five products around your home that contain plastics. Solution Examples include:  TV  Laptop computer  Printer  Dish washer  Clothes washing machine  Dryer

12-3

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 13: Municipal and Industrial Waste and Recycling

Solution and Answer Guide Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 13: Municipal and Industrial Waste and Recycling

End of Section Problem Solutions Note: Some solutions are not provided due to the nature of the problems.

13.1

During the holiday seasons, we tend to produce more waste. For example, we discard more wrapping paper, cards, and shopping bags. Moreover, according to the EPA, each year over 30 million live Christmas trees are sold in North America. Suggest at least five ways to reduce waste during the holidays. Solution Examples include: -

13.2

Plant and grow a Christmas tree outside and decorate it during the holiday season Change Christmas lights to LED lights Send e-cards instead of mailing cards Reuse wrapping paper Exchange useful gifts that could be easily recycled

Consumer electronics and their components (e.g., batteries, printer ink cartridges) are increasingly becoming a major portion of our trash. Suggest at least three ways to reduce waste caused by electronics and their components. Solution Examples include: -

Do not print a paper unless absolutely necessary and instead use a digital document form Turn off a computer when it is not used Download music and/or video data instead of buying CDs and DVDs

13-1

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 13: Municipal and Industrial Waste and Recycling

13.4

Use the tables given in this chapter to look up the following information: total plastics, glass, paper, and paperboard generation in nondurable goods. Solution According to Table 13.1, the generated amount of the total plastics in nondurable goods in 2018 is 7,460 thousand tons, and the total plastics in MSW is 35,680 thousand tons. According to Table 13.2, the generated amount of paper and paperboard in nondurable goods in 2018 is 25,490 thousand tons. According to Table 13.4, the total amount of glass generated is 12,250 thousand tons, in which the amount categorized as durable goods is 2,460 thousand tons and the amount of glass containers is 9,790 thousand tons.

13.7

Use Figure 13.5 to determine the amount of municipal solid waste that was recycled in 1960, 1970, 1980, 1990, 2000, and 2010. Solution Figure 13.5 shows those amounts. 1960 1970 1980 1990 2000 2010

13.8

5.6 8.0 14.5 33.2 69.5 85.4

million tons

Estimate the amount of plastic in the water bottles that you throw away each year. State your assumptions. Solution Assuming that you throw away one 20-oz bottle every day and a 20-oz water bottle has a mass of 10 grams, amount of plastics thrown away per year 1 bottle 365 days 10 g )( =( )( ) = 3,650 g = 3.7 kg 1 year day 1 bottle

13-2

Solution and Answer Guide: Moaveni, Energy, Environment, and Sustainability 2e 2023, 9780357676073; Chapter 13: Municipal and Industrial Waste and Recycling

13.9

There are some organic materials that you should not compost. For example, meat and fish bones could attract flies and rodents. Give examples of other organic materials that should not be composted. Solution Examples include diseased plants, cooking oil, rice, and milk products.

13-3