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Solution Manual for Photovoltaic Systems Engineering, 4th Edition

Solution Manual for Photovoltaic Systems Engineering, 4th Edition makes solving textbook exercises easier with step-by-step solutions and helpful tips.

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    1 Photovoltaic Systems Engineering, 4 th Ed. Chapter 1 Problem Solutions 1.1 Prove equation 1.5. Using (1.5) with N(n) = 2N o , results in ) i 1 ln( n o o e N N 2   . Solving for n yields the result i D i n 693 . ) 1 ln( 2 ln     since ln(1+i)  i for small i. 1.2 Calculate the approximate and exact doubling times for annual percentage increases of 5%, 10%, 15% and 20%. % Increase 5 10 15 20 Exact D 14.2 yr 7.27 yr 4.96 yr 3.80 yr Approx D 14 yr 7.00 yr 4.67 yr 3.50 yr Approx D/Exact D 0.99 0.96 0.94 0.92 1.3 The human population of the earth is approximately 7 billion and is increasing at approximately 2% per year. The diameter of the earth is approximately 8000 miles and the surface of the earth is approximately two-thirds water. Calculate the population doubling time, then set up a spreadsheet that will show a) the population, b) the number of square feet of land area per person, and c) the length of the side of a square that will prod uce the required area per person. Carry out the spreadsheet for 15 doubling times, assuming that the rate of population increase remains constant. What conclusions can you draw from this exercise? For 2% annual growth rate, the doubling time is 35 years. The total area of the earth is 4  (4000 mi  5280 ft/mi) 2 = 5.61  10 15 ft 2 . Hence, the land area of the earth is approximately 1.87  10 15 ft 2 . The following table results. # Doubling Times Year Population Land Area/ Person (ft^2) Length of side of square/Person (ft) 0 2017 7.00E+09 2.67E+05 516.9 1 2052 1.40E+10 1.34E+05 365.5 2 2087 2.80E+10 6.68E+04 258.4 3 2122 5.60E+10 3.34E+04 182.7 4 2157 1.12E+11 1.67E+04 129.2 5 2192 2.24E+11 8.35E+03 91.4 6 2227 4.48E+11 4.17E+03 64.6 7 2262 8.96E+11 2.09E+03 45.7 8 2297 1.79E+12 1.04E+03 32.3 9 2332 3.58E+12 5.22E+02 22.8 10 2367 7.17E+12 2.61E+02 16.2 11 2402 1.43E+13 1.30E+02 11.4 12 2437 2.87E+13 6.52E+01 8.1 13 2472 5.73E+13 3.26E+01 5.7 14 2507 1.15E+14 1.63E+01 4.0 15 2542 2.29E+14 8.15E+00 2.9 Conclusions: These are left up to the creative thought processes of the problem solver. 1.4 Show that in an exponential growth scenario, the amount accumulated in a doubling time equals the amount accumulated in all previous history.

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    2 Let T represent the doubling time, r the rate, P 0 the initial population, and let t>0 represent any arbitrary time. The problem is to show that t r) (1 P r) (1 P r) (1 P 0 t 0 t T 0       . This is equivalent to showing 0 T 0 2P r) (1 P   , which is true by assumption (this equation is what it means for T to be the doubling time). 1.5 Assume there is enough coal left to last for another 300 years at current consumption rates. a. Determine how long the coal will last if its use is increased at a rate of 5% per year. b. If there is enough coal to last for 10,000 years at current consumption rates, then how long will it last if its use increases by 5% per year? c. Can you predict any other possible consequences if coal burning increases at 5% per year for the short or long term? d. Determine the annual percent reduction in coal consumption to ensure that coal will last forever, assuming the 300-year lifetime at present consumption rates. Using (1.9) a) 36 . 56 ) 05 . 0 1 ln( ] 1 ) 05 . 0 1 ln( 300 ln[ m      yr b) 91 . 126 ) 05 . 0 1 ln( ] 1 ) 05 . 0 1 ln( 10000 ln[ m      yr c) To mention a few: significant increases in CO, CO 2 , NO 2 , SO 2 and particulates in the air, affecting acid rain, global warming and respiratory ailments. d) Set       ) n 1 ln( ] 1 ) n 1 ln( 300 ln[ m and solve for n. Note that if n < 0, then ln(1+n) < 0. Also, if [300ln(1+n) + 1] = 0, then ln[300ln(1+n) + 1] =  . So ln(1+n) =  1/300 and, hence, 3 300 1 10 328 . 3 1 e n        , or, approximately 0.333% per year decrease in consumption to ensure the supply will last forever. 1.6 The half-life is the time it takes to decay exponentially to half the original amount. If the half-life of a radioactive isotope is 500 years, how many years will it take for an amount of the isotope to decay to 1% of its original value? Assume the isotope decays exponentially. The time constant is given by 5 . 721 693 . 500 2 ln t 2 / 1     yr. Hence, setting 01 . 0 e t    and solving for t gives the result 3323 100 ln 5 . 721 100 ln t     yr. 1.7 If a colony of bacteria lives in a jar and doubles in number every day, and it takes 30 days to fill the jar with bacteria, a. How long does it take for the jar to be half full? b. How long before the bacteria notice they have a problem? (You may want to pretend you are a bacterium.) c. If on the 30th day, 3 more jars are found, how much longer will the colony be able to continue to multiply at its present rate? a) 29 days b) Probably about when the jar is about 1/8 full. This is 3 days before the jar will. be full. It sort of depends whether bacteria are as smart as humans. c) 2 more days 1.8 An enterprising young engineer enters an interesting salary agreement with an employer. She agrees to work for a penny the first day, 2 cents the second, 4 cents the third, and so on, each day doubling the amount of the previous day. Set up a spreadsheet that will show her daily and cumulative earnings for her first 30 days of employment.

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    3 Day Daily Earnings Cumulative Earnings Day Daily Earnings Cumulative Earnings 1 $0.01 $0.01 16 327.68 655.35 2 $0.02 $0.03 17 $655.36 $1,310.71 3 $0.04 $0.07 18 $1,310.72 $2,621.43 4 $0.08 $0.15 19 $2,621.44 $5,242.87 5 $0.16 $0.31 20 $5,242.88 $10,485.75 6 $0.32 $0.63 21 $10,485.76 $20,971.51 7 $0.64 $1.27 22 $20,971.52 $41,943.03 8 $1.28 $2.55 23 $41,943.04 $83,886.07 9 $2.56 $5.11 24 $83,886.08 $167,772.15 10 $5.12 $10.23 25 $167,772.16 $335,544.31 11 $10.24 $20.47 26 $335,544.32 $671,088.63 12 $20.48 $40.95 27 $671,088.64 $1,342,177.27 13 $40.96 $81.91 28 $1,342,177.28 $2,684,354.55 14 $81.92 $163.83 29 $2,684,354.56 $5,368,709.11 15 $163.84 $327.67 30 $5,368,709.12 $10,737,418.23 1.9 Burning a gallon of petroleum produces approximately 25 pounds of carbon dioxide and burning a ton of coal produces approximately 7000 pounds of carbon dioxide. a. If a barrel of petroleum contains 42 gallons, if the world consumes 80 million barrels of petroleum per day and if the atmosphere weighs 14.7 pounds per square inch of earth surface area, calculate the weight of carbon dioxide generated each year from burning petroleum and compare this amount with the weight of the atmosphere. b. If a total of 16 million tons of coal are burned every day on the earth, calculate the weight of carbon dioxide generated each year from coal burning and compare it with the weight of the atmosphere. a) (25 lb CO 2 /gal)  (42 gal/bbl)  (80  10 6 bbl/day)  (365 days/yr) = 3.066  10 13 lb/yr. The atmosphere weighs (14.7 lb/in 2 )  (5.61  10 15 ft 2 )  (144 in 2 /ft 2 ) = 1.188  10 19 lb. Hence, annual CO 2 production is 2.584  10 -4 % of the total weight of the atmosphere under the assumptions given. b) (16  10 6 ton/day)  (7000 lb CO 2 /ton)  (365 days/yr) = 4.088  10 13 lb/yr, or 3.445  10 -4 % of the weight of the atmosphere under the assumptions given. 1.10 Assume a world population of 7.3 billion and a U.S. population of 323 million. a. Look up the present total annual U.S. primary energy consumption. Then determine the total world energy consumption in quads if the rest of the world were to use the same per capita energy as in the U.S. b. If the energy source mix were to remain the same as the present mix in achieving the scenario of part a, what would be the percentage increase in CO 2 emissions? a) The US EIA shows 97 quads of U.S. Primary Energy Consumption in 2013 and total world primary energy consumption of 543 quads. The U. S. fraction of total world primary consumption thus was 18% in 2013. Thus,on a per person basis, U.S. energy consumption was 97x10 15 Btu/323x10 6 people = 3x10 8 Btu/pers and worldwide energy consumption is 543x10 15 Btu/7.3x10 9 people = 7.44x10 7 Btu/pers. Thus, if everyone used 3x108 Btu/yr, total world consumption would have been 7.3x10 9 x3x10 8 = 21.9x10 17 Btu = 2,190 quads, which is 4.03 times the actual consumption. b) For the same energy mix, the CO 2 production rate would be 4.03 times as much, or 403% of the actual 2013 rate.

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    4 1.11 Obtain data on worldwide energy consumption by sector from the United States Department of Energy, Energy Information Administration website. Plot the data and estimate annual percentage growth rates for the seven regions reported and then for the world. Note that the sectors listed on the EIA website include residential, industrial, commercial and transportation. The seven regions of the world, as called for in the problem, are: North America, Central and South America, Europe, Eurasia, Middle East, Africa, Asia and Oceania. New data becomes available frequently, so the reader is invited to access this data and plot historic trends. The end product might be a set of graphs with energy use by sector for each region, or vice versa. There is a lot of data that can be massaged in the various databases, with projections of energy use. These projections can then be compared with projections made by the EIA. 1.12 The following measurements of x(t) are made: t 0 1 2.3 3.0 4.5 5.2 6.5 8.0 x 2.1 8.4 31 65 360 850 3700 20,000 Construct a semilog plot of x(t) either manually or with a computer, and determine whether the function appears to have an exponential dependence. If so, determine x(t). The semi-log plot as an Excel plot is as follows: This is clearly a pretty good straight line on semilogarithmic coordinates. The function can thus be represented fairly closely as bt ) 10 ( A x   , where A = 2.1 and 497 . 0 8 979 . 3 0 8 1 . 2 log 20000 log t t x log x log b 1 2 1 2         . Alternatively, using the Excel trendline option yields x = 2.2995e 1.1334t with an R 2 value of 0.9994. Since e t = 10 tloge , x can also be expressed as x = 2.2995  (10) 0.4922t , which is reasonably close to the graphical estimate. 1.13 a. What does the area under the Gaussian curve represent? b. Show that 68% of the area under the Gaussian curve lies within one standard deviation, s, of maximum value of the function. c. What percentage of the area lies within 2s? a) The area under the Gaussian curve represents the total sample, or total supply. b) This is probably most easily proven by a numerical integration or else by looking up the function in a book of tables, such as the CRC Standard Mathematical Tables . The 20th Edition of the CRC Standard Mathematical Tables tabulates the Normal probability function and related functions, beginning on p. 582. 1 10 100 1000 10000 100000 0 1 2 3 4 5 6 7 8

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    5 For one standard deviation from the mean, x = 1.00. F(1) represents the area under the curve between x =  and x = 1.00, which is tabulated as 0.8413. Note that f(x) is normalized so that F(  ) = 1.00. Since f(x) is symmetrical about x = 0, this means that the area under the curve between x = 0 and x = 1 is 0.8413  0.500 = 0.3413. Thus, the area between x =  1 and x = 0 is also 0.3413, so the area between x =  1 and x = 1 is 0.6826 . Hence, 68.26% of the distribution lies within 1 standard deviation of the mean. c) In a similar fashion, F(2) = 0.9772. Hence, 2  (0.9772-0.500) = 0.954, or, 95.4% of the distribution lies within 2 standard deviations of the mean. 1.14 Determine R m , t o and s for the worldwide graph of Figure 1.8. The worldwide graph shows R m  36.5  10 9 bbl/yr consumption rate occurring somewhere near the year 2005. The Gaussian function has the property that it has decreased to 60.7% of its peak value at one standard deviation from the peak (mean). Thus, .607  R m = 22.2  10 9 bbl/yr, occurs at the years 1980 and 2030. So s  25 years and hence this implies that 68% of the world’s petroleum will be used between the years 1980 and 2030. 1.15 Look up actual U. S. and W orld petroleum production figures and plot them on Hubbert’s curves to compare the actual production with the theoretical production. Data is available from the U.S. Energy Information Agency. 2016 World production 96 million barrels/day. 2015 US production 9.25 million barrels/day. These figures change monthly, but the idea is to see where they lie on the Hubbert curve, so they need to be converted to barrels/yr. The results: U.S. = 3.4x10 9 and World = 35x10 9 . 2016 Hubbert curves show U.S.  2x10 9 barrels/yr and World  35x10 9 barrels/yr. Is this interesting, or what? 1.16 Based on the data of Figure 1.9, a. Estimate the year when PV shipments will reach 100 GW. b. Estimate the year when PV shipments will reach 1,000 GW. c. Estimate the year when PV shipments will reach 2,700 GW. The annual percent increase needs to be estimated from Figure 1.9. Since there is some uncertainty involved in predicting the future, one needs to decide upon how much of the curve of Figure 1.9 to use to estimate the annual growth rate, since the slope of the curve is time dependent. Noting the recent trend, since about 2000, estimates can be made of shipments between 2000 and 2015 from the data on Figure 1.9 as follows: Year 2000 2003 2006 2009 2012 2015 Normalized year (t) 0 3 6 9 12 15 GW shipped (GW) 0.22 0.55 1.7 7.5 30.0 50.0 Plotting this data in Excel and obtaining the best exponential curve fit yields GW = 0.200e 0.3868t with R 2 = 0.9875, a reasonable fit. Thus, to determine t to reach a particular value of GW, solve the equation for t to get 200 MW(t) ln 0.3868 1 t  . Thus GW(t) = 100 GW when t = 16 yr (i.e., in 2016) GW(t) = 1,000 GW when t = 22 yr (i.e., in 2022) GW(t) = 2,700 GW when t = 24.6 yr (i.e., in 2025) 1.17 The Earth Policy Institute [31] reports the following worldwide PV production figures. Plot the data on an Excel graph, establish an equation to represent the data, and then answer the three questions posed in Problem 1.16. Compare the results of the two problems.

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    6 Year 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 MW 77.6 88.6 126 155 201 277 386 547 748 1194 1786 2521 Remember to set 1995 as year 0. An Excel plot will give MW = 62.5570e 0.3226t with R 2 = 0.9897. Hence, solving for t gives 62.5570 MW(t) ln 0.3226 1 t  . Thus MW(t) = 10,000 MW when t = 15.7yr (i.e., in 2010) MW(t) = 50,000 MW when t = 20.7 yr (i.e., in 2015) MW(t) = 2700 GW = 2,700,000 MW when t = 33.1 yr (i.e., in 2028) An unnamed philosopher once claimed that the future will be different from the past. Which, if either, scenario from the last 2 problems do you think is more realistic? What might happen to change the trend in either direction?

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    7 Chapter 2 Problem Solutions 2.1 Show that, for a surface temperature of 5800 K, the sun will deliver 1367 W/m 2 to the Earth. This requires integration of the blackbody radiation formula over all wavelengths to determine the total available ene rgy at the surface of the sun in W/m 2 . Numerical integration is recommended. Be careful to note the range of wavelengths that contribute the most to the spectrum. Then note that the energy density decreases as the square of the distance from the source, similar to the behavior of an electric field emanating from a point source. The diameter of the sun is 1.393  10 9 m, and the mean distance from sun to Earth is 1.5  10 11 m. Start by plugging the values of the various constants along with T = 5800 K into (2.1). This gives 1 e 10 749 . 3 w 6 10 4845 . 2 5 16           W/m 2 /m . This expression must now be integrated over the interval 0 <  <  . Since most of the spectral intensity appears in the wavelength range between 0.1 and 5  m, values for evaluating the expression must be carefully chosen if numerical integration is chosen. The result of the integration is 7 0 10 44 . 6 d w       W/m 2 . Since this is the power density at the surface of the sun, the total power of the sun can be found by multiplying this power density by the surface area of the sun. The result is P = (6.44  10 7 W/m 2 )  (4  )  (0.5  1.393  10 9 m) 2 = 3.926  10 26 W. This power is radiated uniformly in all directions. The power density at the earth is thus found from the total power of the sun divided by the area of the sphere surrounding the sun with radius equal to the distance from the sun to the earth. The result is 1389 m 10 827 . 2 W 10 926 . 3 A P S 2 23 26      W/m 2 , which compares nicely with the value 1367 W/m 2 that results from the not-quite-perfect blackbody spectrum received by the earth. 2.2 If the diameter of the sun is 1.393  10 9 m, and if the average density of the sun is approximately 1.4  the density of water, and if 2  10 19 kg/yr of hydrogen is consumed by fusion, how long will it take for the sun to consume 25% of its mass in the fusion process? The density of 1.4 means the density of the sun is 1.4  103 kg/m3 . The mass of the sun thus computes to be 1.981  10 30 kg. Henc e, 25% of the sun’s mass is 4.953  10 29 kg. Hence, (4.953  10 29 kg)  (2  10 19 kg/yr) = 24.76 billion years. 2.3 Calculate the zenith angles needed to produce AM 1.5 and AM 2.0 if AM 1.0 occurs at zero degrees. AM = AM(90 o )csc  . Thus, if AM(90 o ) = 1.0, then for AM = 1.5, csc  = 1.5 and  = 41.81 o . If  = 41.81 o , then  Z = 90   = 48.19 o . For AM = 2, sin  = 0.5 so that  = 30 o and  Z = 90   = 60 o . 2.4 Calculate the zenith angle at solar noon at a latitude of 40 o north on May 1. At solar noon,  Z =    = 40 o         365 ) 80 121 ( 360 sin 45 . 23 o = 24.79 o . 2.5 Calculate the number of hours the sun was above the horizon on your birthday at your birthplace.

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    8 To find n to find  , the date is needed. Then the latitude of the birthplace is needed. Finally, solving (2.8) gives 5 . 7 ) tan tan ( cos DH 1      hr . 2.6 Calculate the irradiance of sunlight for AM 1.5 and for AM 2.0, assuming no cloud cover, using (2.2) and (2.3). Then write a computer program that will plot irradiance vs. AM for 1  AM  10 for each equation. Equation (2.2) results in irradiance = 801 W/m 2 for AM1.5 and irradiance = 670 W/m 2 for AM2.0. Equation (2.3) results in irradiance = 855 W/m 2 for AM1.5 and irradiance = 772.5 W/m 2 for AM2.0. Irradiance vs. Air Mass using (2.2) Irradiance vs. Air Mass using (2.3) 2.7 Assume no cloud cover, and, hence, that the solar irradiance is predominantly a beam component. If a nontracking collector is perpendicular to the incident radiation at solar noon, estimate the irradiance on the collector at 1, 2, 3, 4, 5 and 6 hours past solar noon, taking into account air mass and collector orientation. Then estimate the total daily irradiation on the collector in kWh. Assume AM 1.0 at solar noon and a latitude of 20  N. AM1.0 at solar noon means the sun is directly overhead, or  = 90 o and  = 0 o . This means the collector is horizontal. Thus, to calculate the direct beam irradiation on the collector at 2, 3 and 4 hours past solar noon, it is necessary to calculate  at these times. 0 200 400 600 800 1000 0 2 4 6 8 10 0 100 200 300 400 500 600 700 800 900 1000 0 2 4 6 8 10

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    9 To find  , (2.9) is used:         cos cos cos sin sin sin At solar noon, the hour angle,  = 0. Also if the sun is directly overhead,  =  so  = 90 o . Since -23.45 o <  < 23.45 o , it is necessary to choose a location to determine  . If  = 20 o N, it is possible to evaluate  for times when   0. So, at 2 hours past solar noon,  =  30 o and 8817 . 0 ) 30 cos( 20 cos 20 sin sin o o 2 o 2      so  = 61.85 o . Similarly, at 3 hours past solar noon,  = 47.85 o , and at 4 hours past solar noon,  = 33.95 o . Next, for the calculated solar altitudes, the air mass and then the irradiance need to be calculated. Using (2.3), the following table can be generated for solar altitude, air mass, irradiance and the component of irradiance on the horizontal array, given by (Irradiance)  (sin  ): Time sin   , degrees AM Irradiance, W/m 2 Direct array component, W/m 2 Solar noon 1.0000 90 1.000 957 957 noon + 1 0.9699 75.9 1.020 950 921 noon + 2 0.8817 61.9 1.134 927 817 noon + 3 0.7414 47.9 1.349 883 655 noon + 4 0.5585 34.0 1.791 805 450 noon + 5 0.3455 20.2 2.894 657 227 noon + 6 0.1170 6.7 8.55 297 35 An estimate of the total daily irradiation on the collector can then be made by summing the direct array components multiplied by appropriate time intervals. Assuming the solar noon component from 11:30 a.m. to 12:30 p.m., the noon + 1 component from 12:30 to 1:30 p.m. and 10:30 to 11:30 a.m., the noon + 2 value from 1:30 to 2:30 p.m. and 9:30 to 10:30 a.m., etc., the approximate total output of the array due to the direct component of the incident sunlight over the 12-hour period becomes kWh/day = 0.957 + 2  (0.921 + 0.817 + 0.655 + 0.450 + 0.227 + 0.035) = 7.167 . 2.8 Write a computer program using Matlab, Excel or something similar that will plot solar altitude vs. azimuth. Plot sets of curves similar to Figure 2.8 for the months of March, June, September and December for Denver, CO, Mexico City, Mexico, and Fairbanks, AK. The program needs to plot  and  as determined by (2.9) and (2.10), with appropriate values of  and  for the months and the locations. Note that since  reaches its maximum and minimum values on or about the 21 st of June and December, rather than at the 15 th of the months, the curves for May and July, April and August, etc., will be slightly different. 2.9 Using eq. (2.9 and 2.10), show that at solar noon,  Z =  and  = 0. Start with equation (2.9) since it does not depend on  . Using some trigonometric identities, and  =0 at solar noon gives: δ) cos( φ cos φ cos δ sin sin δ sin α ) θ sin(90 cos θ Z Z        φ . This gives  Z = ±(  ). Explain why it isn’t the negative . Now for equation (2.10). Using the result from above, and a trigonometric identity, we have: cos α cos φ sin α sin φ α) cos( φ 90) α sin( φ ) θ sin( φ sin δ Z           . Plugging this value into equation 2.10 gives the answer.

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    10 2.10 Write a computer program that will generate a plot of solar altitude vs. azimuth for the 12 months of the year for your home town. It would be nice if each curve would have time-of-day indicators. Appropriate choices need to be made for  and  for use of (2.9) and (2.10) again. 2.11 Neglecting the analemma effect, calculate the time of day at which solar noon would occur at your longitude. Then compare the north indicated by a compass, corrected for compass declination, with the north indicated by a shadow at calculated solar noon and estimate the error in the shadow direction (i.e., solar noon) due to the analemma effect. Estimate actual solar noon time on the basis of the difference between compass north and the shadow “north.” True north is in the direction of the shadow at solar noon in the northern hemisphere, provided that  >  . If  <  , then the shadow points directly south. As noted in the text, neglecting the analemma effect, solar noon occurs at clock noon at longitudes that are multiples of 15 o from Greenwich, England (with a few noted exceptions). For other longitudes, solar noon is found from interpolation, where 15 o = 60 minutes (i.e., 1 hour), so 1 o = 4 minutes = 1/15 hr. For example, at 80 o W, solar noon occurs at (80  75)  60  15 = 60  3 = 20 minutes past noon, clock time. The difference between true north and compass north is the compass declination, not to be confused with the declination of the sun, δ . This difference is relatively small in eastern United States, but is in the range of 20  in northwestern Canada. Thus, north indicated by a compass must first be corrected to true north from a compass declination table. Next, the direction of the shadow at calculated solar noon, neglecting the analemma effect, is compared with true north determined from the corrected compass reading. Then, note that if the sun travels 15 ° /hr, if the shadow “north” is west of true (corrected compass) north, then solar noon has not yet occurred. If the shadow “north” is east of true north, then the s un is west of south and solar noon has already happened. Using the angular spacing between true north and shadow “north,” one can now estimate either the time until solar noon or the time elapsed since solar noon occurred. 2.12 Noting the dependence of air mass on  z , a. Generate a table that shows  ,  and I (using 2.3) vs. time for a latitude of your choice on a day of your choice. b. Assuming a tracking collector and no cloud cover, estimate the total energy available for collection during your chosen day. The selection of latitude is to be made by the student. This solution will use 30 o N on March 21. For this case,  = 30 o and  = 0 o . Hence,       cos 866 . 0 cos 30 cos 0 cos 30 sin 0 sin sin o o o o and           tan 5774 . 0 cos cos sin sin sin cos . Now, using     sin 1 sin ) 90 ( AM AM o and 678 . 0 ) AM ( ) 7 . 0 ( 1367 I  with  = 15(12  T) for times when the sun is above the horizon enables the construction of a table of I vs. T, which, when summed as I  T over daylight hours, gives the total irradiation per day incident on a tracking collector. The results are tabulated in the following table.

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    11 T  sin    AM I, kW/m 2 cum I, kWh/m 2 6 90 0.0000 0.00 90.01 870319 0 0 6.5 82.5 0.1130 6.49 86.24 8.847 286 0.07 7 75 0.2241 12.95 82.37 4.462 511 0.27 7.5 67.5 0.3314 19.36 78.30 3.017 643 0.56 8 60 0.4330 25.66 73.90 2.309 729 0.90 8.5 52.5 0.5272 31.82 69.01 1.897 788 1.28 9 45 0.6124 37.76 63.44 1.633 831 1.69 9.5 37.5 0.6870 43.40 56.91 1.456 863 2.11 10 30 0.7500 48.59 49.10 1.333 886 2.55 10.5 22.5 0.8001 53.14 39.63 1.250 903 2.99 11 15 0.8365 56.78 28.17 1.195 914 3.45 11.5 7.5 0.8586 59.16 14.72 1.165 920 3.91 12 0 0.8660 60.00 0.00 1.155 923 4.37 12.5 -7.5 0.8586 59.16 14.72 1.165 920 4.83 13 -15 0.8365 56.78 28.17 1.195 914 5.29 13.5 -22.5 0.8001 53.14 39.63 1.250 903 5.74 14 -30 0.7500 48.59 49.10 1.333 886 6.19 14.5 -37.5 0.6870 43.40 56.91 1.456 863 6.63 15 -45 0.6124 37.76 63.44 1.633 831 7.05 15.5 -52.5 0.5272 31.82 69.01 1.897 788 7.45 16 -60 0.4330 25.66 73.90 2.309 729 7.83 16.5 -67.5 0.3314 19.36 78.30 3.017 643 8.18 17 -75 0.2241 12.95 82.37 4.462 511 8.47 17.5 -82.5 0.1130 6.49 86.24 8.847 286 8.66 18 -90 0.0000 0.00 90.01 870318.7 0 8.74 2.13 Plot I vs. AM and then plot I vs.  . The plots can be generated from the data from Problem 2.11 with the following results: 0 100 200 300 400 500 600 700 800 900 1000 0 2 4 6 8 10 Irradiance, W/m^2 Air Mass Irradiance vs. Air Mass

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    12 2.14 Calculate the collector orientation that will produce maximum summer output between 2 p.m. and 5 p.m. in Tucson, AZ. Use July 21 as the assumed midpoint of summer and correct for the longitude of the site. The collector should face the sun at 3:30 p.m. (assume MDT) on July 21. Hence, the sun position in Tucson, AZ is needed at 3:30 p.m. on July 21. Note that on July 21, 3:30 p.m. is daylight savings time, or 2:30 p.m. standard time. The longitude of Tucson is 110.9  W, so solar noon occurs at 12:24, sun time. Thus, 2:30 p.m. standard time is equivalent to 2:06 p.m. sun time. For Tucson,  = 32.2 o N and for July 21, o 24 . 20 365 ) 80 202 ( 360 sin 45 . 23     and  = 15(12  T) = 15(12  14.1) =  31.5 o . Hence,         cos cos cos sin sin sin = 0.861 and thus  = 59.5 o . Next,         cos cos sin sin sin cos = .188 so that  =  79.2 o . (Note negative since west of south.) Hence, the collector is pointed close to west at an altitude of 59.5 o . As an interesting follow-up exercise, one might want to calculate the position without the longitude correction. Another interesting follow-up is to note that Arizona is on pacific time instead of mountain time. Thus, the problem could also be worked using 3:30 p.m. PDT, which is based on longitude of 120  , rather than 105  for mountain time solar noon. This means solar noon is at 11:24 a.m. rather than 12:24 p.m. PST. Obviously the same procedures will be followed. 2.15 Calculate the collector orientation that will produce maximum summer output at 9 a.m. Daylight Savings Time in Minneapolis, MN, using July 21 as the assumed midpoint of summer and using a longitude correction. Minneapolis, MN has  = 45 o N and longitude = 93.3 o W. For July 21,  = 20.24 o (See Problem 2.13). To find how 9 a.m. CDT relates to solar noon, note that at 90 o W, solar noon is at 1 p.m. CDT. Hence, at 93.3 o W, solar noon is 3.3  60  15 = 13.2 minutes past the hour, or at 1:13 p.m. This means that 9 a.m. CDT corresponds to 4 hr 13 min before solar noon. (4 hr 13 min = 4.2167 hr) Hence T  12 =  4.2167 hr and  = 15(12  T) = 63.25 o . Solving for  and  gives 5432 . 0 cos cos cos sin sin sin          and thus  = 32.9 o and 0643 . cos cos sin sin sin cos          so that o 316 . 86   East of South. 0 200 400 600 800 1000 0 20 40 60 80 Irradiance, W/m^2 Altitude, degrees Irradiance vs. Solar Altitude

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    13 2.16 A collector in Boca Raton, FL (  = 26.4 o N, longitude = 80.1  W) is mounted on a roof with a 5:12 pitch, facing 30 o S of W. Determine the time of day and days of the year that the direct beam radiation component is normal to the array. A roof with a 5:12 pitch has o 1 62 . 22 12 5 tan     . Thus,  = 90   = 67.38 o and 30 o S of W corresponds to  =  60 o . For Boca Raton,  = 26.36 o N with longitude of 80.08 o W. From this information it is necessary to determine  and  . Equation (2.12) can be solved for sin  . The result is 2375 . 0 cos cos cos sin sin sin          and thus  = 13.74 o . Next, (2.11) can be used to determine  , with the result 9394 . 0 cos cos sin sin sin cos          and thus  =  20.05 o (i.e., W of S). To find the days of the year that the sun shines directly onto the array, solve for the days for which  = 13.74 o . This is done using (2.5), which gives the result o 1 87 . 35 45 . 23 74 . 13 sin 365 ) 80 n ( 360     or 144.13 o . For 35.87 o , n = 116, corresponding to April 26. For 144.13 o , n = 226, corresponding to August 14. The time of day relative to solar noon for direct sun is found from (2.8), resulting in 3367 . 13 15 05 . 20 12 T    , which corresponds to 13 hr 20 min solar time. To find when solar noon occurs at 80.08 o W, note that solar noon occurs at 60 15 75 08 . 80   = 20 minutes past the hour, i.e., at 12:20 p.m. EST, or at 1:20 p.m. EDT. Since direct sun occurs 1 hr 20 min later, it occurs at 2:40 p.m. on April 26 and August 14. 2.17 Determine the location (latitude and longitude) where, on May 29, sunrise is at 6:30 a.m. and sunset is at 8:08 p.m. EDT. At what time does solar noon occur at this location? Solar noon is halfway between sunrise and sunset. Thus, solar noon is at         60 8 20 5 . 6 2 1 = 13.317 hr , which corresponds to 1:19 p.m. EDT. From this information, the longitude can be determined to be 75 o + . W 75 . 79 15 60 19 o o   The declination on May 29 is found to be o 75 . 21 365 ) 80 149 ( 360 sin 45 . 23     . From sunrise to sunset, DH = 20.1333  6.500 = 13.633 hours on May 29. Finally, since DH is known to be 13.633, and  is known, (2.8) can be used to solve for the latitude. 633 . 13 5 . 7 ) tan tan ( cos DH 1       , so that 531 . 0 399 . 0 212 . 0 tan    and thus  = 27.985 o N. 2.18 Using Figure 2.13, make a table of unshaded collector times for each month of the year. The line is drawn on the face of the Solar Pathfinder such that above the line represents shaded and below the line is unshaded. Shaded or unshaded is determined by the reflection of shading objects from the face of the instrument. Hence, the following table can be generated based on the position of the chalk/crayon line on the instrument face.

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    14 Month Unshaded times Month Unshaded times Jan 12:45 – 1:30 p.m. Jul 5:15 a.m. – 6:45 p.m. Feb noon – 2:00 p.m. Aug 5:30 a.m. – 6:30 p.m. Mar 9:30 a.m. – 2:45 p.m. Sep 9:15 a.m. – 3:15 p.m. Apr 5:30 – 6:30 a.m. & 8:30 a.m. – 6:30 p.m. Oct 10:00 a.m. – 2:15 p.m. May 5:15 a.m. – 6:45 p.m. Nov 12:30 – 1:30 p.m. Jun 5:00 a.m. – 7:00 p.m. Dec 12:45 – 1:15 p.m. 2.19 An array of collectors consists of three rows of south-facing collectors as shown in Figure P2.1. If the array is located at 40  N latitude, determine the spacing, d, between the rows needed to prevent shading of one row by another row. The smallest value of  over the year is needed for this problem. If sunrise to sunset is considered, then  = 0 and the spacing must be infinite. Normally it is considered to be adequate if the collector is unshaded between 9 a.m. and 3 p.m. So this means that  =  45  . For either value of  ,  will be the same. The smallest value of  will occur when  =  23.45  . So for  = 40  , sin  = sin(  23.45  )sin40  + cos(  23.45  )cos40  cos45  = 0.2411, so  = 13.95  . Then, since tan  = 3  d  , where d  is the shadow length, d  = 3  tan  = 12.07 ft. But this shadow distance is measured from the projection of the high side of the module onto the roof toward the northwest, since the azimuth of the sun in the morning is toward the southeast. So the value of  is also needed. Using 2.10,  =  42 ° (i.e., east of south) . The distance between rows, d, is thus found from d = d  cos  = 8.97 ft. 2.20 The top of a tree is found to be at an angle of 15  between the horizontal and the corner of a collector and at an azimuth angle of  30  . If the site is at a latitude of 30  N, determine the months (if any) when the tree will shade the array at the point from which the measurements are taken. This problem would most easily be solved with an instrument to calculate sun path. But, of course, for an engineer, this would be too simple. So the mathematical approach is used instead. The problem is to determine whether the sun altitude dips below 15  when the azimuth is  30  during any month of the year. So this means calculating the declination when the altitude, azimuth and latitude are known. Solving (2.10) for the declination gives sin  = sin  sin   cos  cos  cos  . Plugging in  = 15  ,  =  30  and  = 30  gives sin  =  0.60 and thus  =  36.52  . Since  23.45  <  < 23.45  , this means that the tree will never be a problem unless it grows taller. For the curious person, it is interesting to calculate how tall the tree can grow before it becomes a problem, i.e., for what value of  is  =  23.45  . 2.21 A residence has a 5:12 roof pitch and is located at latitude 27  N and longitude 83  W, facing 15  west of south. If two rows of 66 cm x 142 cm collectors are to be mounted on the back roof so they will have a tilt of latitude, as shown in Figure P2.2, a. How far apart will the rows need to be if collectors are to remain unshaded for 6 hours on December 21? b. Assuming Eastern Standard Time, over what time period on December 21 will the collectors be unshaded, assuming the collectors are facing 15  west of south. c. Assuming no other shading objects, over what time period will the collectors be unshaded on March 21? a. On December 21,  =  23.45  . For 6 hours of unshaded operation, the collector will remain unshaded from 9 a.m. to 3 p.m. sun time. Thus, the altitude must be computed when  = 45  at the latitude of 27  . The fact that the collector is facing slightly west of south means that if the collector is not shaded at 9 a.m., then it will not be shaded at 3 p.m. so, in fact, the collector will receive a bit more than 6 hours of sun. Using (2.9) to find  gives  = 23.4  . Referring to Figure P2.2, note that  The 5:12 roof slope determines  = tan  1 (5/12) = 22.62  . This means that  = 0.78  .

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    15 Next, note that  =  +  = 49.62  . If the modules are mounted in landscape position, then sin  = y/66 and cos  = x/66 (x and y in cm). So x = 42.8 cm and y = 50.3 cm. Finally, tan  = y/(d-x), so d = 3737 cm = 37.37 m. Hence, it is unlikely that the roof will be large enough to meet the spacing constraint. b. As a first approximation, the collectors will be unshaded from 9 a.m. to 3 p.m. sun time if the spacing of part a can be achieved. At a longitude of 83  W, solar noon will occur at 12:32 p.m. EST, so the collectors will be unshaded from 9:32 a.m. to 3:32 p.m. c. On March 21,  = 0, so if  = 23.4  and  = 27  , then, using (2.11),  =  63.5  . This corresponds to unshaded hours 4.24 hr either side of solar noon, or from 7:46 a.m. to 4:14 p.m. sun time. Correcting time for longitude, this results in unshaded hours from 8:18 a.m. to 4:46 p.m. 2.22 A commercial building has a flat roof as shown in Figure p2.3. The roof has a parapet around it that is 3 ft high, and an air conditioner located as shown that is 4 ft above the roof. The building is located at 40 degrees north. A PV system is to be installed using modules that measure 3 ft by 5 ft, in rows that will remain unshaded for at least 6 hours every day of the year. The modules are to be mounted at a tilt of 25 degrees, in portrait mode, facing south. The bottom edges of the modules will be 6 inches above the roof. Determine the maximum number of modules that can be installed on the roof so that no module will be shaded during the 6- hour period. Pay attention to module rows shading other rows, shading from the air conditioner and shading from the parapet. Draw a diagram that shows your result. Since the modules are raised 6 inches above the roof, this problem can be solved by treating the modules as though their bottom edges are at ground (roof) level, the parapet is 2.5 ft tall, and the AC is 3.5 ft tall. It may be wise to first check that at least 6 hours of sunlight are available every day of the year. The least amount of daylight hours occur when δ =  23.45°. Using equation (2.8) it is found that the shortest day has about 9.15 hours of sunlight. Hence the problem is reasonable. It should also be clear that the desirable unshaded hours are from 9am – 3 pm solar time. To calculate the shadow length (and direction) for a given day and time it is necessary to find  and  . In the following table, using equations (2.6), (2.9), and (2.10) the required angles for the first day of winter, spring/fall, and summer can be found. Also, since cosine is an even function, one must be careful to make sure the sign of the azimuth angles calculated are correct. Rather than calculate the shadow length of each side of the parapet, the modules, and the AC for each of the 3 declinations, the shadow length of a pole 1 foot high will be found. Then it is only necessary to multiply any of these shadow lengths by the actual height of our object to get the actual shadow length. In the table, Length corresponds to the shadow length, N Length corresponds to the length of the shadow in the northern direction, and similarly for W Length ((N Length) 2 + (W Length) 2 = (Length) 2 ). Length was found using (1ft)/tan  , and for N Length and W Length, Length is multiplied by cos  and sin  respectively. | ------ Winter ------ | | ---- Fall/Spring ---- | | ----- Summer ------ | TIME: ω α ψ α ψ α ψ 9:00 45.00 13.95 41.95 32.80 57.27 48.83 80.19 9:30 37.50 17.56 35.86 37.43 50.05 54.42 73.73 10:00 30.00 20.66 29.36 41.56 41.93 59.82 65.83 10:30 22.50 23.17 22.45 45.05 32.80 64.83 55.65 11:00 15.00 25.03 15.19 47.73 22.63 69.17 41.89 11:30 7.50 26.17 7.67 49.42 11.57 72.28 23.17 12:00 0.00 26.55 0.00 50.00 0.00 73.45 0.00 12:30 -7.50 26.17 -7.67 49.42 -11.57 72.28 -23.17 1:00 -15.00 25.03 -15.19 47.73 -22.63 69.17 -41.89 1:30 -22.50 23.17 -22.45 45.05 -32.80 64.83 -55.65 2:00 -30.00 20.66 -29.36 41.56 -41.93 59.82 -65.83 2:30 -37.50 17.56 -35.86 37.43 -50.05 54.42 -73.73 3:00 -45.00 13.95 -41.95 32.80 -57.27 48.83 -80.19

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