Determine the surface (m2) occupied by 120 kWpk of CIS modules that have an efficiency of 8.8% at 1000 W / m2. if it is known that c-Si modules with an efficiency of 15% (in STC conditions) require 20 m2 / kWpk when arranged on the ground with the appropriate row spacings.
To determine the surface area occupied by the CIS modules, we can use the information given and calculate it step by step.
1. First, let's find the total power output of the CIS modules. We are given that the power output is 120 kWpk (kilowatt peak).
2. Next, we need to determine the power density of the CIS modules. The efficiency of the CIS modules is given as 8.8% at 1000 W/m2. This means that for every 1000 W/m2 of sunlight, the CIS modules can produce 8.8% of that power. Therefore, the power density is 1000 W/m2 multiplied by 0.088 (8.8%).
Power density = 1000 W/m2 * 0.088 = 88 W/m2
3. Now, we can calculate the area required for the CIS modules. We know that c-Si modules with an efficiency of 15% (in STC conditions) require 20 m2/kWpk.
Area required for CIS modules = 120 kWpk * (20 m2/kWpk) = 2400 m2
Therefore, the surface area occupied by 120 kWpk of CIS modules is 2400 m2.