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A residential PV system needs to supply 2 kW.

Assume that:
- The solar panel has an efficiency of 18%.
- Solar irradiance on the surface of the solar panel is 1 kW/m².
- A 60-cell solar panel is used with an area of 1.66 m².

How many 60-cell panels will be needed to generate 2 kW for the home?

Answer :

To solve the problem it is necessary to find the total power generated by each of the panels and then through a proportional relationship find the number of total panels.

There is an efficiency of 18% of which the area of [tex]1.66m ^ 2[/tex] receives a surface radiation of [tex]1kW / m ^ 2[/tex]

Therefore the power is subject to

[tex]P_u = 1*1.66*(18\%)[/tex]

[tex]P_u = 0.2988kW[/tex]

Each panel will have the capacity to produce 0.2998kW

The number of panels required is determined by the total power given by each of the panels, that is

[tex]P_T = \#Panels * P_u[/tex]

[tex]\#Panels = \frac{P_T}{P_u}[/tex]

[tex]\#Panels = \frac{2kW}{0.2988kW}[/tex]

[tex]\#Panels = 6.69 \approx 7[/tex]

Therefore the number of panels required are 7

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