Continuing on the same theme as above, you are required to size a grid-connected
ID: 2080575 • Letter: C
Question
Continuing on the same theme as above, you are required to size a grid-connected south-facing PV system for a customer in Miami to deliver 5000 kWh/yr. Use a tilt equal to latitude minus 15 degree. a. What should be the ac-rated power of the PV/inverter system? b. What temperature de-rating should you use if the modules have a nominal operating cell temperature (NOCT) of 43 degree C, the maximum power drops by 0.39% per degree C, and for ambient the average annual daily high can be depicted from the Table below for Miami. c. For 3% dirt, 3% mismatched modules, and a 92% efficient inverter, what should the DC, STC rated power of the modules? d. If the modules are 13% efficient, what area of PVs would be required? Estimate the total cost of the systems making practical assumptions.Explanation / Answer
Given that Energy consumed per year by a Customer is 5000 kWh / year
The tilt angle of solar modules is Latitude - 150 = 25.800 - 150 = 10.800
Operating Temperature of Module is Tm = Tab + (NOCT - 200) *S/800 where
Tab is ambient temperature, NOCT = 430 , S is Radiation of given location in W/m2 depends on hours of Sun available at a considered location. Consider Sun availability at location on an average 7 hours a day .
Overall efficiency = 0.97*0.97*0.92*0.13 = 0.1125
Average Energy required per day is = 5000/365 = 13.7 kWh/day
Average system size required is = 13.7/24 = 0.5708 kW under ideal condition
a ) Actual System size P = 0.5708/0.1125 = 5 kW
Average Dc rating PV system is P = 5.5 kW
d) Area of PV system required is
Consider module rating of 260 W peak, No.of Modules Required = 5500 / 260 = 22 modules
Consider area of each module is = 1.66 m2
Total are of solar PV system is 22*1.66 = 36.52 m2
Considering Temperature deratring factor
(kW)
=24 + (43 - 20)(4100/(7*800)) = 40.830
= 24.7 + (23/5600)*4700 = 44.00
= 0.0039*(44 - 24.7)
= 0.07527
= 31.7 + (23/5600)*5600 = 54.70
Month Irradiance kWh/m2 per day Temperature (oC) Operating Temperature Derating Factor of PV system due to temperature effectConsidering Temperature deratring factor
(kW)
Energy Output From Installed System Jan 4.1 24=24 + (43 - 20)(4100/(7*800)) = 40.830
= 0.0039 * (40.83 - 25) = 0.0617 = 5/(1-0.0617) = 5.4 Feb 4.7 24.7= 24.7 + (23/5600)*4700 = 44.00
= 0.0039*(44 - 24.7)
= 0.07527
= 5/(1-0.07527) = 5.4 March 5.5 26.2 = 26.2 + (23/5600)*5500 = 48.780 = 0.0039 * (48.78 - 25) = 0.09274 = 5/(1-0.09274) = 5.5 April 6.2 28 = 28 + (23/5600) *6200 = 53.460 = 0.0039 * (53.46 - 25) = 0.11 = 5/(1-0.11) = 5.6 May 5.9 29.6 = 29.6 + (23/5600)*5900 = 53.830 = 0.0039 *(53.83 - 25) = 0.11 = 5.6 June 5.5 30.9 = 30.9 + (23/5600)*5500 = 53.480 = 0.0039 (53.48 - 25) = 0.111 5.6 July 5.7 31.7 = 31.7 + (23/5600)*5700 = 55.110 = 0.0039(55.11 - 25) = 0.117 = 5/(1-0.117) = 5.7 Aug 5.6 31.7= 31.7 + (23/5600)*5600 = 54.70
= 0.0039(54.7 - 25) = 0.115 = 5/(1-0.115)= 5.6 Sep 5.1 31 = 31 + (23/5600)*5100 = 51.940 = 0.0039(51.94 - 25)= 0.105 = 5/(1-0.105) = 5.6 Oct 4.7 29.2 = 29.2 + (23/5600)*4700 = 48.500 = 0.0039(48.5 - 25) = 0.091 =5/(1-0.091) = 5.5 Nov 4.2 26.9 = 26.9 + (23/5600)*4200 = 44.150 = 0.0039 (44.15 - 25)= 0.074 = 5/(1-0.074) = 5.4 Dec 3.9 24.8 24.8 + (23/5600)*3900 = 40.810 = 0.0039*(40.81-25) = 0.0616 = 5/(1-0.0616) = 5.3Related Questions
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