An industrial facility decided to install a PV system to supply load with averag
ID: 2080574 • Letter: A
Question
An industrial facility decided to install a PV system to supply load with average 2MWh. The system will be grid connected. The average power consumed from the utility over 24h should remain close to zero. Design a PV system to meet the industrial facility requirements. For Irradiance pattern use real pattern for Miami, FL. Your design should shows (the PV array size, converters rating and estimated cost) Verify your design using the PV model given to you earlier. For verification you should use irradiance pattern for a winter and summer day. Compare the peak output power, average power in winter and summer. Observe the impact of the variable PV generation on the grid voltage. If the system will be isolated from the grid, calculate the energy storage size required for standalone operation. Prepare a detailed report and an oral presentation. A presentation in class will be required. Due March 10, 2017.Explanation / Answer
Given to design grid connected PV system size to meet required energy of 2 MWh of an industry
The designed system should meet the demand and should offset the energy consumed from the grid.
Average Energy irradiance per m2 at given location Miami, F.L is obtained using NREL solar maps and PVwatts
Here temperature effect is considered constant throught out the year. It will Vary throught out the year.
Considering Solar PV modules selected of 260 W and Area of Each Panel is = 1.65*0.992 = 1.6368 m2 and Efficiency of each panel is 16.16%.
Energy generated per day on an average is
5.25 kWh ------------------------------- 1 m2
? ---------------------------------------- 1.6368 m2
Energy to be generated per unit are of each panel is = (5.25*1.6368)* efficiency of panel = 8.5932*0.1616 = 1.3886 kWh per day .
Considering ideal conditions for Inverters , cables, temperature effect etc
If these loss are considered then
Actual energy output per panel in one day is = (0.98*0.99*0.85)*1.3886 = 1.1451 kWh / day .
Total energy need to meet by the designed PV system is on an average 2 MWh / day
Total Panels Required is N = 2*1000 / 1.1451 = 1746.439 = 1750
Total PV system DC rating is P = N*260 = 1750*260 = 455000 W = 455 kW
Conver rating required is S = 455/0.9 = 505 kVA , near to 500 kVA
Cost of the PV system designed is
Let each panel cost = $ 190
Total Cost of panels is Cp = 1750*190 = $ 332500
Total Cost of inverter Cinv = $159545.45
Balance of system Cost CBOS = $ 99750
Total PV system Cost = Cp + Cinv + CBOS = $ (332500 + 159545.45 + 99750) = $ 591795.45
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NOW Design storage system if PV system is not connected to Grid
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Energy required to store which should be used by Industry during NIght time is on an average 1 MWh / day
Battery rating is given by Ah .
Let Battery Voltage be 12 V each, efficiency of the battery is 50 %
Total required Ah = 1000000/(12*0.5) = 166666.6667 Ah
Total Energy need to be stored is = 2 MWh
Now the PV system should be designed to generate 3 MWh per day P = 3000 / (1.1451) * 260 = 682 kW
Now Cost of the system is
Panels = $ 2620 * 190 = $ 497,800
Inverter = $ 319 * 750 = $ 239,250
Bos cost = $ 149,340
Battery Cost = $ 2 * 166666.6667 = $ 333,333.333
Total Cost of system = $ ( 497,800 + 239,250 + 149,340 + 333,333.333) = $ 121972.333
Month Irradiance (kWh/m2/day) Actual Energy Output per day of Month (kWh) Jan 4.42 =4.42 * 1750 * 1.6368 * 0.1616 * 0.8246 = 1687.09 Feb 5.27 = 5.27 * 381.69 = 2011.54 March 5.65 = 5.65 * 381.69 = 2156.54 April 6.28 = 6.28 * 381.69 = 2397 May 5.80 = 5.8*381.69 = 2213.80 June 5.44 = 5.44 * 381.69 = 2076.39 July 5.69 = 5.69 * 381.69 = 2171.18 August 5.64 = 5.64 * 381.69 = 2152.73 Sep 5.17 = 5.17 *381.69 = 1973.33 Oct 4.96 = 4.96 * 381.69 = 1893.18 Nov 4.38 =4.38 * 381.69 = 1671.80 Dec 4.33 = 4.33 * 381.69 = 1652.71