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Question 58 – In the design of a solar power system steps of calculations below are followed : (a) The power output of the inverter of the of the solar panel is 100 watts. What is the power input, Pin to the inverter when the efficiency of the inverter is 50 %? (b) If the rated power of the inverter is 300 watts, how many inverter is needed for the solar panel? (c) Charge controller of V = 12 volts is used to supply power to the inverter. What is the input current I to the inverter? (d) If the charge controller capacity is 10 A, how many charge controllers are needed? (e) If a biochemical mixer consumes 100 watts, running for 2 hours per day, what is the energy consumption in kilowatt hour per day? (f) What is the input energy needed when the efficiency of the inverter is 50 %? (g) If your area receives 2.88 hours of full sunlight per day, how much energy, in kilowatt hour can be produced per day when one solar panel can produce 20 watts of power? (h) If you know that you have to produce total energy as the answer for (f), how many solar panels are needed? (i) Each V = 12 V battery has 5 ampere hours. If the total energy needed is in answer (f), then how many batteries are needed to run the biochemical mixer if without sunlight for 3 days?
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Answer Posted / kang chuen tat (malaysia - pen
Answer 58 – (a) Input = Output / Efficiency = 100 watts / 0.5 = 200 watts. (b) Number of inverter = Input / Rated power = 200 / 300 = 2 / 3 < 1. Practically only 1 inverter is needed. (c) I = Pin / V = 200 / 12 = 16.667 A. (d) Number of charge controllers = Input / Rated current = 16.667 / 10 = 1.6667 < 2. Practically only 2 are needed. (e) Energy consumption = 0.1 kilowatt x 2 hours / day = 0.2 kilowatt hour per day. (f) Input energy = Energy consumption / Efficiency = 0.2 / 0.5 = 0.4 kilowatt hour per day. (g) Energy generation per day per solar panel = 2.88 hours x 0.02 kilowatt = 0.0576 kilowatt hour. (h) Number of solar panels needed = Answer for (f) / Answer for (g) = 0.4 / 0.0576 = 6.944 < 7. Practically only 7 are needed. (i) Ampere hour per day to run the load = 0.4 kilowatt hour per day / 12 V = 1 / 30 k. Total ampere hour needed for 3 days = 1 / 30 k x 3 = 0.1 k. Number of batteries needed = 0.1 k / 5 = 100 / 5 = 20. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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