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Chapter 31: Problem 55

A house with a south-facing roof has photovoltaic panels on the roof. The photovoltaic panels have an efficiency of \(10.0 \%\) and occupy an area with dimensions \(3.00 \mathrm{~m}\) by \(8.00 \mathrm{~m} .\) The average solar radiation incident on the panels is \(300 . \mathrm{W} / \mathrm{m}^{2}\), averaged over all conditions for a year. How many kilowatt hours of electricity will the solar panels generate in a 30 -day month?

Short Answer

Expert verified
Answer: The solar panels will generate 518.4 kilowatt-hours of electricity in a 30-day month.

Step by step solution

01

Understanding the solar panel's efficiency

The solar panel's efficiency is given as \(10.0 \%\). This means that it converts \(10.0 \%\) of the incident solar radiation into usable electricity.
02

Find the area of the solar panels

The dimensions of the solar panels are \(3.00 \mathrm{~m}\) by \(8.00 \mathrm{~m}\). To find the area of the solar panels, we will multiply their length and width: Area \(= 3.00 \mathrm{~m} \times 8.00 \mathrm{~m} = 24.0 \mathrm{~m^2}\)
03

Calculate the solar radiation energy incident on the panels

The average solar radiation incident on the panels is given as \(300 . \mathrm{W} / \mathrm{m}^{2}\). We will multiply the average solar radiation per square meter with the total area of the panels to find the total solar radiation incident on the panels. Total solar radiation energy \(= 24.0 \mathrm{~m^2} \times 300 . \mathrm{W} / \mathrm{m}^{2} = 7200 \mathrm{~W}\)
04

Calculate the electricity generated by the panels

Now, we'll find how much electricity the solar panels generate by multiplying the total solar radiation energy incident on the panels with their efficiency. Electricity generated \(= 7200 \mathrm{~W} \times 10.0 \% = 720 \mathrm{~W}\)
05

Calculate the total electricity generated in a 30-day month

We have the electricity generated by the solar panels in watts. Now, we need to find the total electricity generated in a 30-day month (assuming average solar radiation remains constant). Total electricity generated in a 30-day month \(= 720 \mathrm{~W} \times 24 \mathrm{~hours/day} \times 30 \mathrm{~days}\)
06

Convert the electricity generated into kilowatt-hours

Finally, we need to convert the electricity generated from watts to kilowatt-hours. Since \(1 \mathrm{~kW} = 1000 \mathrm{~W}\) and \(1 \mathrm{~kWh}=1 \mathrm{~kW} \cdot 1 \mathrm{~hour}\): Total electricity generated \(= \frac{720 \mathrm{~W} \times 24 \mathrm{~hours/day} \times 30 \mathrm{~days}}{1000 \mathrm{~W/kW}}= 518.4 \mathrm{~kWh}\) The solar panels will generate 518.4 kilowatt-hours of electricity in a 30-day month.

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