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SOLUTIONS MANUAL
PHOTOVOLTAIC SYSTEMS ENGINEERING
5TH EDITION
CHAPTER 1: BACKGROUND
PROBLEM SOLUTIONS
1.1 A common rating of a typical size PV panel (also called module) is 400 Watts. An array or a
cluster of 10 solar panels therefore generates 4000 Watts of power during peak solar conditions. If
peak solar conditions are 4.2 hours per day for a specific water pumping application in northern
Oklahoma, how much energy can be expected to be collected in one day in kWh units? Convert
this answer to kJ.
The average daily energy collected will be 4 kW × 4.2 hr = 16.8 kWh.
Using 1 W = 1 J/sec, 1 kW = 1 kJ/sec. Thus 1 kWh = 1 kJh/sec × 3600 sec/hr = 3600 kJ
Thus, 16.8 kWh = 16.8 kWh × 3600 kJ/kWh = 60,480 kJ
1.2 Solar PV panels are labeled at the manufacturing facility with a sticker that indicates the Power
Rating of each module. These ratings, however are based on a test of the module in a “solar test
chamber” where the module is subjected to an industry standard 1000 Watts/m2 simulated solar
light intensity. As will be discussed in later chapters, a more realistic solar energy intensity on the
surface of the earth is closer to 800 Watts/m2 maximum except in higher elevations with low
humidity conditions. Based on this information, recalculate the energy collected in Problem 1.1.
, Page |2
The only thing that changes will be the sunlight intensity. Thus, the result of Problem 1 needs to be
multiplied by 800/1000. Thus, the new daily collection becomes 0.8 × 60,480 = 48,384 kJ.
1.3 If 20 kWh of energy is stored in batteries from solar PV generation, and only 80% of this stored
energy can be used to run the electric loads of a remote microwave repeater station, what is the
maximum power that can be used for a 24-hour constant load?
Available energy is 80% of 20 kWh = 16 kWh. Discharged uniformly over a 24 hr period will produce
16 kWh ÷ 24 hr = 0.667 kW, or 667 W.
1.4 The CO2 emissions plot (Figure 1.2) shows an increasing slope from 1960 to 2020. Extrapolating
the graph, what should the slope be after 2020 to keep atmospheric CO2 below 430 ppm by 2050?
The answer depends upon the method. If only one slope is used, it will be lower overall. If it is recognized
that there will probably not be an instantaneous change in slope, but, at least, hopefully, a gradual decrease,
this would produce a different result. It is interesting to look at official graphs and projections. One good
source is metoffice.gov.uk/research.
Extrapolating the data to 2050 should result in an answer of 1 ppm to 1.2 ppm per year increase as a limit.
1.5 The Paris Climate accord of 2015 among 196 countries, sets a limit of 1.5°C on the average
global temperature rise. Figure 1.3 shows the relationship between the global temperature rise and
the global carbon dioxide levels measured in ppm. How close is the global temperature rise to this
1.5°C limit? Is there still time to avoid reaching this level?
The 1.5°C temperature has essentially been reached based upon the latest data on Figure 1.3. This means
that CO2 levels must be kept at current levels and gradually reduced.
, Page |3
1.6 Based on the brief discussion of the second law of thermodynamics, calculate the heat loss of
an ideal engine to the environment for a 50 MW (Net output) coal plant with a 50% ideal efficiency.
If an actual coal plant runs at 30% efficiency, how much more heat will be rejected to the
environment?
Based upon the Carnot ideal engine, η = W/QH. Thus, for an efficiency of 50% and 50% efficiency, the
heat input, QH = 100 MW and thus, the difference between heat in and work out is the heat loss, which is
50 MW.
Thus, at 30% efficiency, maintaining constant work out, the input heat now becomes 50 MW/0.3 = 166.7
MW heat in. Thus, the heat loss will be 166.7 − 50 = 116.7 MW instead of 50 MW. Note that this is a 133%
increase in heat loss.
1.7 An engineer is suggesting the increase in the temperature of the boiler in a power plant from
500°C to 600°C, and claims that this change could result in reducing the heating of the local
atmosphere by at least 20%. Assume the ambient temperature is 300 K. Is she correct?
As per the 2nd Law of Thermodynamics, the temperatures first need to be converted to K. Thus, 500°C =
500 + 273.15 = 773.15 K, and 600°C = 873.15 K.
The ideal efficiency at 500°C is 1 − 300/773 = 61.2%. Thus, QH = 1/0.612 = 1.634
The ideal efficiency at 600°C is 1 − 300/873 = 65.64%. Thus, QH = 1/0.6564 = 1.523
Thus, at 600°C, QL = 0.52 and at 500°C, QL = 0.63, and 0.63/0.52 = 1.21, or 21% more, provided that the
useful work out is the same for both systems.
One can also note that this heat does not include any greenhouse gases that may have been emitted by
the source of QH.
, Page |4
1.8 An engineering intern is asked to compare the theoretical thermal environmental impact of a
PWR (Pressurized Water Reactor) nuclear power plant with a combined cycle natural gas-fired
power plant. Both plants are to have a net electrical power output of 1 GW and operate for an
average of 80% of the time on an annual basis. PWR plants are designed to run at a maximum 647
K for safety reasons, and the combined cycle gas turbine runs at 1500 °F at its highest temperature
point in the gas turbine section. Using these temperatures as TH for each cycle and assuming 300
K as TL for both cycles, determine:
a. The Carnot efficiency of each plant.
b. The total energy input for each plant on an annual basis, assuming a year of power
generation to be 80% of 8,760 hours.
c. The thermal impact of each plant on the environment for a year, and for the expected life
cycle of 30 years.
a. The Carnot efficiency of the PWR is 1 − 300/647 = 53.6%. For the efficiency of the combined cycle unit,
need to first convert TH in °F to K. The result is (1500 − 32)(5/9) = 815.55 °C, which becomes 815.55 +
273.15 = 1088.7 K. Thus, the Carnot efficiency is 1 − 300/1088.7 = 72.4%.
b. 80% of 8,760 = 7008 hr of annual operation at 1 GW. Thus, the total annual energy output is 7,008 × 1
GW = 7,008 GW for each plant. Thus, the total input energy for the PWR plant is 7008/0.536 = 13,065
GWh and the total energy input for the combined cycle plant is 7008/0.724 = 9680 GWh.
c. The direct impact on the environment for each plant is the “waste heat” from each. For the PWR plant,
this is 13,065 − 7,008 = 6057 GWh/yr, or 30 × 6057 = 181,710 GWh over the 30-year life of the plant. For
the combined cycle plant, this is 9,680 − 7,008 = 2672 GWh/yr, or 80,160 over the 30-yr lifetime. Since 1
kWh = 3600 kJ, it is left as an exercise for the reader to convert these GWh values into kJ or any other
more familiar units of energy.
SOLUTIONS MANUAL
PHOTOVOLTAIC SYSTEMS ENGINEERING
5TH EDITION
CHAPTER 1: BACKGROUND
PROBLEM SOLUTIONS
1.1 A common rating of a typical size PV panel (also called module) is 400 Watts. An array or a
cluster of 10 solar panels therefore generates 4000 Watts of power during peak solar conditions. If
peak solar conditions are 4.2 hours per day for a specific water pumping application in northern
Oklahoma, how much energy can be expected to be collected in one day in kWh units? Convert
this answer to kJ.
The average daily energy collected will be 4 kW × 4.2 hr = 16.8 kWh.
Using 1 W = 1 J/sec, 1 kW = 1 kJ/sec. Thus 1 kWh = 1 kJh/sec × 3600 sec/hr = 3600 kJ
Thus, 16.8 kWh = 16.8 kWh × 3600 kJ/kWh = 60,480 kJ
1.2 Solar PV panels are labeled at the manufacturing facility with a sticker that indicates the Power
Rating of each module. These ratings, however are based on a test of the module in a “solar test
chamber” where the module is subjected to an industry standard 1000 Watts/m2 simulated solar
light intensity. As will be discussed in later chapters, a more realistic solar energy intensity on the
surface of the earth is closer to 800 Watts/m2 maximum except in higher elevations with low
humidity conditions. Based on this information, recalculate the energy collected in Problem 1.1.
, Page |2
The only thing that changes will be the sunlight intensity. Thus, the result of Problem 1 needs to be
multiplied by 800/1000. Thus, the new daily collection becomes 0.8 × 60,480 = 48,384 kJ.
1.3 If 20 kWh of energy is stored in batteries from solar PV generation, and only 80% of this stored
energy can be used to run the electric loads of a remote microwave repeater station, what is the
maximum power that can be used for a 24-hour constant load?
Available energy is 80% of 20 kWh = 16 kWh. Discharged uniformly over a 24 hr period will produce
16 kWh ÷ 24 hr = 0.667 kW, or 667 W.
1.4 The CO2 emissions plot (Figure 1.2) shows an increasing slope from 1960 to 2020. Extrapolating
the graph, what should the slope be after 2020 to keep atmospheric CO2 below 430 ppm by 2050?
The answer depends upon the method. If only one slope is used, it will be lower overall. If it is recognized
that there will probably not be an instantaneous change in slope, but, at least, hopefully, a gradual decrease,
this would produce a different result. It is interesting to look at official graphs and projections. One good
source is metoffice.gov.uk/research.
Extrapolating the data to 2050 should result in an answer of 1 ppm to 1.2 ppm per year increase as a limit.
1.5 The Paris Climate accord of 2015 among 196 countries, sets a limit of 1.5°C on the average
global temperature rise. Figure 1.3 shows the relationship between the global temperature rise and
the global carbon dioxide levels measured in ppm. How close is the global temperature rise to this
1.5°C limit? Is there still time to avoid reaching this level?
The 1.5°C temperature has essentially been reached based upon the latest data on Figure 1.3. This means
that CO2 levels must be kept at current levels and gradually reduced.
, Page |3
1.6 Based on the brief discussion of the second law of thermodynamics, calculate the heat loss of
an ideal engine to the environment for a 50 MW (Net output) coal plant with a 50% ideal efficiency.
If an actual coal plant runs at 30% efficiency, how much more heat will be rejected to the
environment?
Based upon the Carnot ideal engine, η = W/QH. Thus, for an efficiency of 50% and 50% efficiency, the
heat input, QH = 100 MW and thus, the difference between heat in and work out is the heat loss, which is
50 MW.
Thus, at 30% efficiency, maintaining constant work out, the input heat now becomes 50 MW/0.3 = 166.7
MW heat in. Thus, the heat loss will be 166.7 − 50 = 116.7 MW instead of 50 MW. Note that this is a 133%
increase in heat loss.
1.7 An engineer is suggesting the increase in the temperature of the boiler in a power plant from
500°C to 600°C, and claims that this change could result in reducing the heating of the local
atmosphere by at least 20%. Assume the ambient temperature is 300 K. Is she correct?
As per the 2nd Law of Thermodynamics, the temperatures first need to be converted to K. Thus, 500°C =
500 + 273.15 = 773.15 K, and 600°C = 873.15 K.
The ideal efficiency at 500°C is 1 − 300/773 = 61.2%. Thus, QH = 1/0.612 = 1.634
The ideal efficiency at 600°C is 1 − 300/873 = 65.64%. Thus, QH = 1/0.6564 = 1.523
Thus, at 600°C, QL = 0.52 and at 500°C, QL = 0.63, and 0.63/0.52 = 1.21, or 21% more, provided that the
useful work out is the same for both systems.
One can also note that this heat does not include any greenhouse gases that may have been emitted by
the source of QH.
, Page |4
1.8 An engineering intern is asked to compare the theoretical thermal environmental impact of a
PWR (Pressurized Water Reactor) nuclear power plant with a combined cycle natural gas-fired
power plant. Both plants are to have a net electrical power output of 1 GW and operate for an
average of 80% of the time on an annual basis. PWR plants are designed to run at a maximum 647
K for safety reasons, and the combined cycle gas turbine runs at 1500 °F at its highest temperature
point in the gas turbine section. Using these temperatures as TH for each cycle and assuming 300
K as TL for both cycles, determine:
a. The Carnot efficiency of each plant.
b. The total energy input for each plant on an annual basis, assuming a year of power
generation to be 80% of 8,760 hours.
c. The thermal impact of each plant on the environment for a year, and for the expected life
cycle of 30 years.
a. The Carnot efficiency of the PWR is 1 − 300/647 = 53.6%. For the efficiency of the combined cycle unit,
need to first convert TH in °F to K. The result is (1500 − 32)(5/9) = 815.55 °C, which becomes 815.55 +
273.15 = 1088.7 K. Thus, the Carnot efficiency is 1 − 300/1088.7 = 72.4%.
b. 80% of 8,760 = 7008 hr of annual operation at 1 GW. Thus, the total annual energy output is 7,008 × 1
GW = 7,008 GW for each plant. Thus, the total input energy for the PWR plant is 7008/0.536 = 13,065
GWh and the total energy input for the combined cycle plant is 7008/0.724 = 9680 GWh.
c. The direct impact on the environment for each plant is the “waste heat” from each. For the PWR plant,
this is 13,065 − 7,008 = 6057 GWh/yr, or 30 × 6057 = 181,710 GWh over the 30-year life of the plant. For
the combined cycle plant, this is 9,680 − 7,008 = 2672 GWh/yr, or 80,160 over the 30-yr lifetime. Since 1
kWh = 3600 kJ, it is left as an exercise for the reader to convert these GWh values into kJ or any other
more familiar units of energy.
