A family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to to fulfill the requirement of family?

FindTutor 08/01/2023 36

Answer

200 m²

Solution

Required Power = 8 kW = 8,000 W
Power production per square meter = 20 % of Solar Energy Incident = (20*200)/100 = 40 W
Hence, Required Area = 8,000 W/40 W = 200 m²

A family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to to fulfill the requirement of family?

Question 1

A typical family in Delhi needs 10 kW of power supply. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 15% of this energy can be converted to useful electrical energy then how much area is required to fulfill the requirement of the family?

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A typical family in Delhi needs 10 kW of power supply. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 15% of this energy can be converted to useful electrical energy then how much area is required to fulfill the requirement of the family?

Question 1