Homework 5 solution

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Apr 3, 2024

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© 2023 Kenneth Bryden 1 ME 433 Homework 5 - Thermal solar energy 1. You have been asked to size a water heating system for camping. The system is a simple black polyethylene bag filled with 1 gallon of water laid flat in the midday sun. The solar absorptivity of polyethylene black plastic is 0.94 and the emissivity to the sky is 0.92. The water is initially at 50°F and needs to be heated to 110°F in 4 hours. For design of this bag, we will use the midpoint of the temperature band to determine the required energy to heat the water, specifically, at 80°F the density of water and specific heat are 62.215 lbm/ft 3 and 1.00 Btu/lbm-°F, respectively. The radiative and convective losses will be higher at higher temperatures, as a result you will need to radiant energy capture of the water heating bag based on the highest temperature. The bag is well insulated from the ground and there will no losses due to conduction to the ground (it will be placed on folded sleeping bag). At the design conditions the convective heat transfer coefficient is 0.215 Btu/ft 2 -h-°F, the air temperature is 60°F, the temperature of the sky is -15°F, the solar irradiation is 150 Btu/ft 2 - h. a. How much energy is required to heat the water (Btu)? ࠵? !"#$% = ࠵? !"#$% ࠵?࠵? !"#$% ∆࠵? !"#$% ࠵? !"#$% = 1 gal 1 × 1.0 Btu lbm ∙ °F × (110 − 50)°F 1 × 62.215 lbm ft & × ft & 7.481 gal = 499.0 Btu = 499 ± Btu b. What is the solar heat flux absorbed by the bag at design conditions (Btu/h-ft 2 )? ࠵?" ’()*+ = ࠵? ’()*+ ࠵?" ’()*+ ࠵?" ’()*+ = 0.94 1 × 150 Btu h ∙ ft , = 141.0 Btu h ∙ ft , = 141 ± 1 Btu/h ∙ ft , c. What are the convective heat flux losses of the bag at design conditions (Btu/h-ft 2 )? ࠵?" -./0 = ℎ J K࠵? 1"2 − ࠵? "3% L ࠵?" 4(56 = 0.215 Btu h ∙ ft , ∙ °F × (110 − 60)°F 1 = 10.75 Btu h ∙ ft , = 10.8 ± 0.1 Btu/h ∙ ft , d. What are the radiative heat flux losses of the bag at design conditions (Btu/h-ft 2 )? ࠵?” +*7 = ࠵?࠵?K࠵? s 9 − ࠵? sky 9 L ࠵?” +*7 = 0.92 1 × 1.714 × 10 <= Btu h ∙ ft , ∙ R 9 × [(110 + 459.67) 9 — (15 + 459.67) 9 ]R 9 1 ࠵?” +*7 = 104.42 Btu h ∙ ft , = 104 ± 1 Btu/h ∙ ft ,

© 2023 Kenneth Bryden 2 e. How much usable heat flux is provided to the bag at design conditions (Btu/ft 2 -h)? ࠵?" >’*?)@ = ࠵?" ’()*+ − ࠵?" 4(56 − ࠵?" +*7 ࠵?” >’*?)@ = 141.0 Btu h ∙ ft , − 10.8 Btu h ∙ ft , − 104.4 Btu h ∙ ft , = 25.8 Btu h ∙ ft , = 25.8 ± 0.1 Btu/h ∙ ft , f. What should the exposed surface area of the bag be (ft 2 )? ࠵? A*B@+ = ࠵?" >’*?)@ ࠵? ?*C ࠵? rearranging and solving ࠵? ?*C = ࠵? A*B@+ ࠵?" >’*?)@ ࠵? = 499 Btu 1 × h ∙ ft , 25.83 Btu × 1 4 h = 4.830 ft , = 4.83 ± 0.02 ft , g. How long would it take to heat the bag if the convective heat transfer coefficient were 3 times larger (h)? ࠵? A*B@+ = ࠵?" >’*?)@ ࠵? ?*C ࠵? rearranging ࠵? = ࠵? A*B@+ ࠵? ?*C ࠵?" >’*?)@ ࠵?" >’*?)@ = ࠵?" ’()*+ − ࠵?" 4(56 − ࠵?" +*7 ࠵?" >’*?)@ = 141.0 Btu h ∙ ft , − 3(10.75 Btu) h ∙ ft , − 104.42 Btu h ∙ ft , = 4.33 Btu h ∙ ft , ࠵? = 499 Btu 1 × 1 4.830 ft , × h ∙ ft , 4.33 Btu = 23.9 h = 24 ± 0.5 h

© 2023 Kenneth Bryden 3 2. A student group is building a small concentrating solar power system with a parabolic dish and a Stirling engine. You have been asked to help on the design. In the rated design case, the direct beam solar flux is 250 W/m 2 , the air temperature is 30.0°C, and the receiver temperature is 325°C. The emissivity of the receiver to the sky is 0.100 and the temperature of the sky is -20.0°C. The parabolic dish will be 2.00 m in diameter and will be focused on the final 2.00 cm of the Stirling engine receiver/piston which has an outside diameter of 10.00 cm. The parabolic mirror has a reflectivity of 92.5%, a cleanliness factor of 90.0%, and a field efficiency of 93.0%. The absorptivity of the receiver to solar radiation is 0.960. The convective heat transfer coefficient is 1.50 W/m 2 · K. The Stirling engine high temperature is 300°C and the cooling fin temperature is 85.0°C. Well-designed commercially available Stirling engines operate at approximately 70.0% of Carnot efficiency. As this Stirling engine will be locally manufactured and will be relatively simple, assume that the engine efficiency will be 40.0% of Carnot efficiency. The generator efficiency is 87.3%. The heat transfer efficiency from the receiver to engine is 82.5%. a. How much solar energy is received by the parabolic mirror (kW)? ࠵? DE++(+ = ࠵?" ’()*+ ࠵? DE++(+ ࠵? DE++(+ = ࠵?(2 m) , 4 = 3.142 m , ࠵? DE++(+ = 250 W m , × 3.142 m , 1 × kW 1000 W = 0.7854 kW = 0.785 ± 0.002 kW ⟸ b. What is the concentration ratio of this system? ࠵?࠵? = ࠵? DE++(+ ࠵? +@4@E6@+ ࠵? +@4@E6@+ = ࠵?࠵? +@4@E6@+ ࠵? +@4@E6@+ 1 = ࠵? 1 × 10 cm 1 × 2 cm 1 = 62.83 cm , ࠵?࠵? = 3.142 m , 62.83 cm , × 10 9 cm , m , = 500.1 = 500 ± 1 c. What is the concentrator efficiency of this system? ࠵? -./-$/#%"#$ = ࠵? ∙ ࠵? -F$"/ ∙ ࠵? G3$FH = 0.925(0.90)(0.93) = 77.42% = 77.4 ± 0.1%

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© 2023 Kenneth Bryden 4 d. What is the energy flux absorbed by the receiver (kW/m 2 receiver )? ࠵?" %$-$30$% = [࠵?࠵? × ࠵? × ࠵? -F$"/ × ࠵? G3$FH × ࠵?" I.F"% ]࠵? %$-$30$% = [࠵?࠵? × ࠵? -.FF$-#.% × ࠵?" I.F"% ]࠵? %$-$30$% ࠵?" %$-$30$% = c 500.1 1 × 0.7742 1 × 250 W m , × kW 1000 W d 0.96 1 = 92.92 kW m , = 92.9 ± 0.1 kW/m , e. What are the convective energy losses from the receiver (kW/m 2 receiver )? ࠵?" 4(56 = ℎ J (࠵? %$-$30$% − ࠵? "3% ) ࠵?" 4(56 = 1.50 W m , ∙ K × (325 − 30) K 1 × kW 1000 W = 0.4425 kW m , = 0.442 ± 0.001 kW/m , f. What are the radiative energy losses from the receiver (kW/m 2 receiver )? ࠵?" %"H = ࠵?࠵?K࠵? %$-$30$% 9 − ࠵? IJK 9 L ࠵?” +*7 = 0.1 1 × 5.670 × 10 <L W m , ∙ K 9 × [(325 + 273) 9 — 20 + 273) K 9 1 × kW 1000 W = 0.7019 kW m , = 0.702 ± 0.001 kw/m , g. What is the usable energy flux delivered by the receiver (kW/m 2 receiver )? ࠵?" >’*?)@ = ࠵?" +@4@E6@+ − ࠵?" 4(56 − ࠵?" +*7 ࠵?" MI"1F$ = (92.92 − 0.442 − 0.702) kW/m , = 91.78 kW/m , = 91.8 ± 0.1 kW/m , h. What is the capture efficiency of the system (%)? ࠵? -"N#M%$ = ࠵?" MI"1F$ ࠵?࠵?(࠵?" I.F"% ) ࠵? -"N#M%$ = 91.78 kW/m , 500.1(0.250 kW/m , ) = 0.7341 = 73.4 ± 0.2%

© 2023 Kenneth Bryden 5 i. How much usable energy is delivered by the receiver (kW)? ࠵? MI"1F$ = ࠵?" MI"1F$ ࠵? %$-$30$% ࠵? MI"1F$ = 91.78 kW m , × 62.83 cm , 1 × m , 10 9 cm , = 0.5767 kW = 0.577 ± 0.001 kW Alternately ࠵? MI"1F$ = ࠵? %$-$30$% ࠵? -"N#M%$ ࠵? MI"1F$ = 0.7854 kW I.F"% 1 × 0.7341 kW MI"1F$ kW I.F"% = 0.5766 kW = 0.577 ± 0.001 kW j. How much usable energy is delivered to the Stirling engine (kW)? ࠵? $/23/$ = ࠵? MI"1F$ ࠵? OP ࠵? $/23/$ = 0.5767 kW MI"1F$ 1 × 0.825 kW $/23/$ kW MI"1F$ = 0.4758 kW = 0.476 ± 0.003 kW k. What is the efficiency of the Stirling engine (%)? ࠵? $/23/$ = 0.40࠵? Q"%/.# ࠵? Q"%/.# = 1 − (85 + 273.15) K (300 + 273.15) K = 0.3751 ࠵? $/23/$ = 0.40(0.3751) = 0.1500 = 15.0 ± 0.1% l. What is the overall efficiency (i.e., system efficiency) of the power system (%)? ࠵? .0$%"FF = ࠵? -"N#M%$ ∙ ࠵? OP ∙ ࠵? $/23/$ ∙ ࠵? 2$/ ࠵? .0$%"FF = 0.7341(0.825)(0.1500)(0.873) = 0.07931 = 7.93 ± 0.01% m. How much electrical power will this system deliver at rated conditions (W)? ࠵? ̇ $F$- = ࠵? ̇ R3%%.% ࠵? .0$%"FF ࠵? ̇ $F$- = 0.7854 kW(0.07931) = 0.06229 kW = 62.3 ± 0.2 W