Homework 10 solution

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Apr 3, 2024

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© 2023 Kenneth Bryden 1 ME 433 Homework 10 - Solar Chimney Power Plants There are two parts of Homework 10 (10a and 10b). Part 10a is worth 10 points and Part 10b is 15 points. • Homework 10a consists of 10 questions. You have only two attempts to work Homework 10a. The highest score will be kept. • Homework 10b consists of 10 numerical questions focused on solar chimney power plan ts. You have unlimited attempts to work Homework 10b . The highest score will be kept. Note that due to the estimates that we are creating in this homework, we will be reporting 2 places of significance for all the Homework 10b problems.

© 2023 Kenneth Bryden 2 Homework 10a Questions 1-7 in HW 10a are based on the journal article in the reading: A. Babkir, “Techno- economic optimization for the design of solar chimney power plants,” Energy Conversion and Management 138:461-473 (2017). 1. In the article “Techno-economic optimization for the design of solar chimney power plants,” how many design cases are examined? 12 2. In the article “Techno-economic optimization for the design of solar chimney power plants,” what are the assumed collector, turbine, and electric generator efficiencies? 42%, 85%, 80% respectively 3. The article “Techno-economic optimization for the design of solar chimney power plants” examines three types of solar chimney power plants. Which type of solar chimney power plant is found to be the optimum? The floating solar chimney 4. What assumed electricity price was used in the optimization algorithm discussed in the article “Techno-economic optimization for the design of solar chimney power plants”? $0.10/kWh 5. In the article “Techno-economic optimization for the design of solar chimney power plants,” how does the construction cost of a floating solar chimney compare with the construction cost of a corresponding reinforced concrete chimney? The floating solar chimney is 5-6 times less in cost 6. Compared to conventional electricity generation sources what are the challenges of solar chimney power plants discussed in the article “Techno-economic optimization for the design of solar chimney power plants”? Choose all correct answers. • huge initial costs • long construction periods • large land needs • low efficiency • uncertainty about materials and construction techniques 7. In the article “Techno-economic optimization for the design of solar chimney power plants” what is the advantage of a solar chimney power plant? low environmental impact on air and water Questions 8-10 in HW 10a are based on the videos. 8. Based on the second video, what is the height of the EnviroMission solar chimney power plant proposed for construction in Australia (ft)? 3281 feet 9. What is the air temperature at the base of the proposed solar chimney power plant (i.e., the temperature of the air entering the solar chimney)? 158 F

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© 2023 Kenneth Bryden 4 Homework 10b 1. Noting that the solar chimney power plant is a Brayton cycle heat engine and using the information available in the second video for this homework, estimate the Carnot cycle efficiency for the proposed EnviroMission solar chimney power plant (%). ࠵? ! = 68℉ = 68 + 460 R = 528 R ࠵? " = 158℉ = 158 + 460 R = 618 R ࠵? #$%&’( = 1 − ࠵? # ࠵? ) = 1 − 528 618 = 14.6% = 15 ± 1% 2. Using the data and equations from the article “Techno-economic optimization for the design of solar chimney power plants,” estimate the overall efficiency of the proposed EnviroMission solar chimney power plant (%). Hint: use the assumed collector, turbine, and electric generator efficiencies and equations 1 and 3 from the article. ࠵? $*%(*&) = 86℉ = 303.15 K ࠵? #) = ࠵?(࠵?) ࠵? - ࠵? $ !" = 9.807 m s . × 3281 ft 1 × m 3.281 ft × kg ∙ K 1.005 kJ × 1 303.15 K × J ∙ s . kg ∙ m . × kJ 1000 J ࠵? #) = 3.129% ࠵? ’/0%$11 = ࠵? # ࠵? #) ࠵? 2 ࠵? 3 = 0.03219(0.42)(0.85)(0.80) = 0.009193 = 0.92% ± 0.01

© 2023 Kenneth Bryden 5 3. On their website (http://www.enviromission.com.au/IRM/content/solar-tower-faq- s.aspx?RID=305) EnviroMission states that approximately 5 km diameter area of land would be used for the proposed 200 MW solar chimney power plant. Assuming that rated power of 200 MW is based on a solar insolation of 1000 W/m 2 , estimate the overall efficiency of the proposed solar chimney power plant (%). ࠵? ̇ *& = ࠵? × ࠵?" 4’1$% = ࠵?(5 km) . 4 × 10 5 m . km . × 1000 W m . × MW 10 5 W = 19,630 MW ࠵? ’/0%$11 = ࠵? ̇ ’6( ࠵? *& = 200 MW 19,630 MW = 1.019% = 1.02 ± 0.05% 4. In their video EnviroMission states that the proposed 200 MW solar chimney power plant will provide power for 100,000 households. If each household uses 800 kWh monthly (this is a high estimate of household energy use) estimate the capacity factor for the solar chimney power plant (%). ࠵? 789 = 200 MW 1 × 8760 h y = 1,752,000 MWh y ࠵? :01*/0%0: = 100,000 homes 1 × 800 kWh mo × 12 mo y × MWh 1000 kWh = 960,000 MWh y ࠵?࠵? = ࠵? :01*/0%0: ࠵? 789 = 960,000 MWh y × y 1,752,000 MWh = 54.79% = 54.8 ± 0.5% 5. The solar wind energy tower proposed for construction in Arizona is a downdraft chimney in which the air at the top of the chimney is cooled by evaporation using a mist water. That is the solar wind energy tower is heat engine similar the drinking bird in which the temperature difference between dry bulb and wet temperature provides the energy input for the engine. Assuming that air temperature at the top and bottom of the solar wind energy tower are equal and 25.0°C, that the relative humidity of the air is 10.0%, and that the pressure is 1 bar estimate the Carnot cycle efficiency for this proposed solar wind energy tower (%). Hint: I suggest the use of https://www.psychrometric-calculator.com/HumidAirWeb.aspx to determine the needed psychrometric values for problems 5 through 9. For a dry bulb 25°C and relative humidity =10% the wet bulb temperature is 10.36°C. ࠵? #$%&’( = 1 − ࠵? # ࠵? ) = 1 − (10.36 + 273.15)K (25 + 273.15)K = 4.91% = 4.9 ± 0.1%

© 2023 Kenneth Bryden 6 6. Water usage (specifically the kg of water used to generate 1 kWh of electricity) is a critical concern in developing the solar wind energy tower proposed for construction in Arizona described in Question 5. It is tempting to assume that heat input to the solar wind energy tower is heat removed from the air by means evaporation. However, we need to remember that the cold sink is the temperature at which the energy is removed from not input to the system. That is, in this system the hot ambient air is cooled and saturated at the top of the tower, creating heavier more dense air that sinks to the base of the chimney. We can determine the work done by a cubic meter of air through the system by considering the gross potential energy due to the evaporative cooling. This is ࠵?′′′ $*% = ∫ ; ) # V࠵? $*%((’<0%) (࠵?) − ࠵? $*%($=>) (࠵?)Y ࠵?dx ≈ ࠵? $*%((’<0%) ࠵?࠵? # ࠵? $*%($=>) − ࠵? $*%((’<0%) ࠵? $*%($=>) Where ࠵? $*%($=>) − ࠵? $*%((’<0%) is the mean difference between the temperatures outside the tower ]࠵? $*%($=>) ^ and the inside of the tower ]࠵? $*%((’<0%) ^ . And ࠵? $*%($=>) is the mean temperature of the air outside the tower. Noting that the ambient air temperature varies by approximately 10°C from the bottom to the top of a typical 600 - 1000 m high tower, we can further simplify the analysis as follows. ࠵?′′′ $*% ≈ ࠵? $*%((’<0%) ࠵?࠵? # ࠵? :%?_>61> − ࠵? <0(_>61> ࠵? :%?_>61> _ AB; where all quantities are determined at the ground ( ࠵? = 0 ). Using this equation and estimate the energy contained in one cubic meter of air traveling through the solar wind energy tower described in Question 5 (Wh avail /m 3 ). ࠵?′′′ $*% ≈ ࠵? $*%((’<0%) ࠵?࠵? # ࠵? :%?_>61> − ࠵? <0(_>61> ࠵? :%?_>61> _ AB; ࠵? $*%((’<0%) = 1.214 kg CDE 8GD m $*%((’<0%) H × kg 8GD 0.9921 kg CDE 8GD = 1.224 kg 8GD m $*%((’<0%) H ࠵? III $*% ≈ 1.224 kg m $*%((’<0%) H × 9.807 m s . × 2250 ft 1 × m 3.281 ft × W ∙ s H kg ∙ m . × h 3600 s × (25 − 10.36)K (25 + 273.15)K ࠵? III $*% ≈ 0.1123 Wh/m $*%((’<0%) H = 0.11 ± 0.01 Wh $/$*1 /m $*%((’<0%) H

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© 2023 Kenneth Bryden 7 7. Utilizing the Wh of available energy in one cubic meter of air at the top of the solar wind energy tower from problem 6 and determining the difference in the weight fraction of water in the dry bulb air and the wet bulb air, determine the quantity of water needed to generate one kWh of available energy (kg w /kWh avail ). ࠵?̇ < ࠵? ̇ $/$*1 = m $*%((’<0%) H 0.1123 Wh $/$*1 × 1000 Wh kWh × 1.224 kg $*% m $*%((’<0%) H × c 0.007892 kg < kg 8GD − 0.001981 kg < kg 8GD d ࠵?̇ < ࠵? ̇ $/$*1 = 64.4 kg < /kWh $/$*1 = 64 ± 1 kg < /kWh $/$*1 8. Assuming that the pump used to raise the water to the top of the proposed solar wind energy tower is 40% efficient (this includes piping losses and energy to create the fine water mist needed), how much energy is needed to raise the water to the top of the tower (kWh w /kWh avail )? ࠵? < ࠵? $/$*1 = 64.4 kg < kWh $/$*1 × 9.807 m s . × 2250 ft 1 × m 3.281 ft × W ∙ s H kg ∙ m . × kW 1000 W × h 3600 s × kWh < 0.4 kWh 7GJ ࠵? < ࠵? $/$*1 = 0.301 kWh < kWh $/$*1 = 0.30 ± 0.01 kWh < /kWh $/$*1 9. Estimate the overall efficiency of the proposed solar wind energy tower discussed in questions 5 - 8, including the energy required to pump the water, the turbine efficiency, and the generation efficiency (%). Hint: use the turbine and generation efficiencies given in the article “Techno-economic optimization for the design of solar chimney power plants,” as estimates for this system. ࠵? ’/0%$11 = ࠵? # (࠵? 2 ࠵? 3 − 0.301) = 0.0491[(0.85)(0.80) − 0.301] = 0.0186 = 1.9% ± 0.1

© 2023 Kenneth Bryden 8 10. Consider the development of a solar chimney power system for your home in the desert. You estimate that your home will require 1000 kWh per month and you estimate that the system will have a capacity factor of 30% and hence you will need a 5 kW power system. In reviewing the project you believe that you can build a 200 ft high floating chimney. Assume the collector inlet temperature is 25°C and a solar insolation of 1000 W/m 2 . What is the expected efficiency of your proposed solar chimney power system(%)? Hint: Given that maximum air temperature is unknown without additional analysis, use the specific heat of air at 300 K. @300 K ࠵? - = 1.005 kJ/kg ∙ K ࠵? #) = ࠵?࠵? ࠵? - ࠵? $_*& = 9.807 m s . × 200 ft 1 × m 3.281 ft × kg ∙ K 1.005 kJ × 1 298.15 K × J ∙ s . kg ∙ m . × kJ 1000 J ࠵? #) = 0.1995% ࠵? ’/0%$11 = ࠵? # ࠵? #) ࠵? 2 ࠵? 3 = 0.001995(0.42)(0.85)(0.80) = 0.0570% = 0.057% ± 0.001 11. Using the information in the article “Techno-economic optimization for the design of solar chimney power plants,” what diameter of solar canopy will be required for the proposed solar chimney power system for your desert home (ft)? Eq. 7 in “Techno-economic optimization for the design of solar chimney power plants,” ࠵? ̇ 010K = ࠵? ’/0%$11 ࠵? K’110K(’% ࠵?" rearranging ࠵? K’110K(’% = ࠵? ̇ ’6( h D8LMC ࠵? ’/0%$11 ࠵?"| D8LMC = 5 kW 1 × kW *& 0.000563 kW 010K × m . 1 kW = 8772 m . ࠵? K’110K(’% = k 4(8772 m . ) ࠵? l N/. = 106 m = 349 ft = 350 ± 10 ft