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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photovoltaic Panels
Photovoltaic panels, often simply known as solar panels, are devices designed to convert sunlight directly into electricity. This magic happens due to the photovoltaic effect, where sunlight strikes the panel's surface and excites electrons, creating an electric current.

These panels are generally made up of many solar cells linked together, usually constructed from silicon, a semiconductor material that effectively captures solar energy.
  • Photovoltaic panels are a popular choice for generating clean, renewable energy right at your home or business location.
  • They are typically positioned on rooftops, facing the sun to capture maximum solar energy.
Investing in photovoltaic panels is often seen as a long-term step towards reducing electric bills and minimizing carbon footprints. Their performance largely depends on their efficiency and the intensity of available sunlight.
Solar Radiation
Solar radiation is the key to unlocking the power stored in our sun. It's the energy we receive from the sun's rays and encompasses a range of wavelengths, including visible, ultraviolet, and infrared light.

For photovoltaic panels, understanding solar radiation helps in assessing how much energy panels can realistically harness. Factors affecting solar radiation include the time of day, weather conditions, and geographical location.
  • The intensity of solar radiation is crucial for the amount of electricity generated by the panels.
  • It is measured in watts per square meter (W/m²), providing a quantitative measure of energy that can be theoretically captured.
Efficient use of solar radiation means capturing as much sunlight as possible throughout the year, making it essential for sites with high sunlight exposure to optimize energy output from solar systems.
Energy Conversion
Energy conversion is the core process that solar panels rely on. It involves converting solar radiation into electrical energy that can be used to power homes and devices.

The efficiency of a solar panel indicates how well it can convert the solar energy it receives into electric power. For example, a 10% efficient solar panel means that 10% of the solar energy is converted into usable electricity.
  • Higher efficiency means more electricity is produced for the same amount of solar radiation, resulting in more cost-effective energy generation.
  • Energy conversion efficiency is a vital consideration when choosing solar panels, influencing both the space needed and the overall energy yield.
It is worth noting that various factors, such as temperature and shading, can impact a panel's conversion efficiency, thus affecting the total energy output.
Kilowatt-Hour Calculation
Understanding kilowatt-hour (kWh) calculations is crucial for evaluating the impact and benefits of installing solar panels. Kilowatt-hour is a unit of energy that represents the power consumption of 1000 watts over one hour.

Calculating the kWh output of your solar panels involves multiplying the system's power output in watts by the number of hours the panels are effective, and then further considering the days in question.
  • In our example, with a panel generating 720 watts over 24 hours for 30 days, we convert this into kWh by dividing by 1000.
  • This results in a total generation of 518.4 kWh for the month, representing substantial energy savings and reduced reliance on nonrenewable sources.
Regularly monitoring kWh output helps in tracking efficiency and optimizing panel placement and performance. It's a practical way to assess the financial benefits of solar panels too.

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Most popular questions from this chapter

Silica aerogel, an extremely porous, thermally insulating material made of silica, has a density of \(1.00 \mathrm{mg} / \mathrm{cm}^{3}\). A thin circular slice of aerogel has a diameter of \(2.00 \mathrm{~mm}\) and a thickness of \(0.10 \mathrm{~mm}\). a) What is the weight of the aerogel slice (in newtons)? b) What is the intensity and radiation pressure of a \(5.00-\mathrm{mW}\) laser beam of diameter \(2.00 \mathrm{~mm}\) on the sample? c) How many \(5.00-\mathrm{mW}\) lasers with a beam diameter of \(2.00 \mathrm{~mm}\) would be needed to make the slice float in the Earth's gravitational field? Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

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Scientists have proposed using the radiation pressure of sunlight for travel to other planets in the Solar System. If the intensity of the electromagnetic radiation produced by the Sun is about \(1.40 \mathrm{~kW} / \mathrm{m}^{2}\) near the Earth, what size would a sail have to be to accelerate a spaceship with a mass of 10.0 metric tons at \(1.00 \mathrm{~m} / \mathrm{s}^{2} ?\) a) Assume that the sail absorbs all the incident radiation. b) Assume that the sail perfectly reflects all the incident radiation.

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