   # If a typical home requires an average power of 1 kW, calculate how many square meters of PV cells would be required to supply that power. Assume that solar power is 1 .0 × 10 3 W/m 2 and that the cells are 15% efficient in converting sunlight to electrical energy. Could such an array meet the peak power demands? Could such an array fit on a rooftop? ### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692

#### Solutions

Chapter
Section ### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 10, Problem 30E
Textbook Problem
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## If a typical home requires an average power of 1 kW, calculate how many square meters of PV cells would be required to supply that power. Assume that solar power is 1 .0 × 10 3 W/m2 and that the cells are 15% efficient in converting sunlight to electrical energy. Could such an array meet the peak power demands? Could such an array fit on a rooftop?

Interpretation Introduction

Interpretation:

The surface area of the PV cells is to be calculated and whether the given array would meet the peak power demands and fit on a rooftop or not is to be explained.

Concept Introduction:

Power input on a PV cell is the total active area of the PV cell multiplied by the solar power.

The percent of the input power that is converted into the output power by a PV cell is called its efficiency.

Percent efficiency of a PV cell is 100 multiplied by the ratio of power out and power in.

The expression is as follows:

Percent efficiency=Power outPower in×100 …… (1)

### Explanation of Solution

Given information:

The average power output required is 1 kW or 1000 W.

The solar power is 1.0 ×103 W/m2or 1000 W/m2.

The efficiency of the PV cell is 15 %.

Consider the area of the PV cell to be A m2.

Calculate the power input by multiplying the area with solar power as:

Power input=(A m2×1000 Wm2)=1000A W

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