Guillermo_Morales3.3

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1 Module 3 Chapter Assignment Guillermo Morales Embry-Riddle Aeronautical University ENGR 400: Fundamentals of Energy Systems Dr. Abdulhakim Agll January 27, 2024

CHAPTER ASSIGNMENT MODULE 1 1 Exercise 6.1 A solar panel consists of four parallel columns of PV cells. Each column has 10 PV cells in series. Each cell produces 2 W at 0.5 V. Compute the voltage and current of the panel. Each parallel column is 5 A ∗ 10 with 4 cells where 4 ∗ 50 = 200 A . Each parallel series contains 10 where, TotalVoltage = 10 ∗ 0.5 = 5 V Each cell produces 2W where, Power = VI 0.5 I = 2 W I = 4 A Therefore, Voltage = 5 V Current = 200 A

3 Exercise 6.5 A solar power density for a given area has a standard deviation of 3 h, and a maximum power of 200 W at noon. Compute the solar energy in 1 day. E s = P ma √ 2 π σ E s = 200 √ 2 ( π ) ( 3 ) E s = 1.5 kWh Exercise 6.14 A solar cell with a reverse saturation current of 1nA is operating at 35°C. The solar current at 35°C is 1.1A. The cell is connected to a 5 Ω resistive load. Compute the output power of the cell. We may state that I s is the solar current, I o is the reverse saturation current, and V T is the thermal voltage. I = [ I s − I o ( e V V T − 1 ) ] = V R V T = T 11600 V T = 35 + 273.15 11600 = 0.02656 V Then we can solve for the above formula where, [ 1.1 − 1 ∗ 10 − 9 ( e V 0.02656 − 1 ) ] = V 5 V = [ 5.5 − 5 ∗ 10 − 9 ( e V 0.2656 − 1 ) ] V − 5.5 − 5 ∗ 10 − 9 + 5 ∗ 10 − 9 e V 0.2656 V − 5.00000005 + 5 ∗ 10 − 9 e 37.59 V = 0 V + 5 ∗ 10 − 9 e 37.59 − 5.500000005 = 0 V = 0.5509 V

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4 Therefore, P = V 2 R = 0.5509 2 5 = 0.06069 W P = 0.06069 W Exercise 6.16 For the solar cell in the previous problem, compute the load resistance at the maximum output power. R load = V t 2 ( ReverseSaturationCurrent ) From the previous exercise we can plug in the values where, R load = 25.85 mV ( 2 ) 1 ∗ 10 − 9 A = 25.85 ∗ 10 − 3 V ( 2 ) 1 ∗ 10 − 9 A = 25.85 ∗ 10 − 3 V 2 ∗ 10 − 9 A = ( 25.85 2 ) ∗ 10 6 ohms R load = 12.925 ∗ 10 6 ohms Exercise 6.19 A solar cell is operating at 30°C where the output current is 1.1 A and the load voltage is 0.5 V. The series resistance of the cell is 20 mΩ and the parallel resistance is 2 kΩ. Compute the electrical efficiency of the PV cell. Assume that the irradiance efficiency is 22%, compute the overall efficiency of the PV cell. P out = IV = ( 1.1 ) ( 0.5 ) = 0.55 W η irradiance = 22% = 0.22 P ¿ = P out η irradiance = 0.55 W 2.5 W = 0.22

5 Electrical efficiency of the PV cell is 22%. η overall = η electrical η series η ¿ R series = 20 mΩ = 0.02 Ω R ¿ = 2 k Ω = 2000 Ω η series = 1 1 + R series R prallel = 1 1 + 0.02 Ω 0.5 Ω = 1 1 + 0.04 = 0.9615 η ¿ = 1 1 + R ¿ R l oad = 1 1 + 2000 Ω 0.5 Ω = 1 1 + 4 000 = 0.000249875 η overall = η electrical η series η ¿ = ( 0.22 ) ( 0.9615 ) ( 0.000249875 ) = 0.0000539 The overall efficiency of the PV cell would be approximately 0.00539%.

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