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Chapter 31: Problem 59

As shown in the figure, sunlight is coming straight down (negative \(z\) -direction) on a solar panel (of length \(L=1.40 \mathrm{~m}\) and width \(W=0.900 \mathrm{~m}\) ) on the Mars rover Spir- it. The amplitude of the electric field in the solar radiation is \(673 \mathrm{~V} / \mathrm{m}\) and is uniform (the radiation has the same amplitude everywhere). If the solar panel has an efficiency of \(18.0 \%\) in converting solar radiation into electrical power, how much average power can the panel generate?

Short Answer

Expert verified
Solution: After performing the calculations, the average power generated by the solar panel is: \(P = \left(\frac{1}{2} \cdot 8.85 \times 10^{-12} \, \text{F/m} \cdot 3.00 \times 10^8 \, \text{m/s} \cdot (673 \, \text{V/m})^2 \right) \cdot 0.180 \cdot (1.40\,\text{m} \cdot 0.900\,\text{m})\) Calculate the value of \(P\), and provide the result in watts.

Step by step solution

01

Calculate the intensity of solar radiation

Using the given amplitude of the electric field, \(E = 673 \,\text{V/m}\), we can calculate the intensity of the solar radiation using the formula: \(I = \frac{1}{2} \epsilon_0 c E^2\), where \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\) is the permittivity of free space, and \(c = 3.00 \times 10^8 \, \text{m/s}\) is the speed of light. \(I = \frac{1}{2} \cdot 8.85 \times 10^{-12} \, \text{F/m} \cdot 3.00 \times 10^8 \, \text{m/s} \cdot (673 \, \text{V/m})^2\)
02

Calculate the active surface area of the solar panel

Given the dimensions of the solar panel, we can find the active surface area by multiplying its length and width: \(A = L \cdot W = 1.40\,\text{m} \cdot 0.900\,\text{m}\)
03

Calculate the average power generated by the solar panel

To find the average power generated, we have to multiply the intensity of solar radiation, the efficiency of the solar panel, and the active surface area: \(P = I \cdot \eta \cdot A\), where \(\eta = 18.0\% = 0.180\) \(P = \left(\frac{1}{2} \cdot 8.85 \times 10^{-12} \, \text{F/m} \cdot 3.00 \times 10^8 \, \text{m/s} \cdot (673 \, \text{V/m})^2 \right) \cdot 0.180 \cdot (1.40\,\text{m} \cdot 0.900\,\text{m})\) Now, calculate the value of \(P\) to get the average power generated by the solar panel.

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Key Concepts

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

Understanding Solar Radiation Intensity
Solar radiation intensity refers to the power per unit area received from the Sun in the form of electromagnetic radiation. This intensity can significantly affect how much power a solar panel can generate. To find it, we need the formula:
  • \( I = \frac{1}{2} \epsilon_0 c E^2 \)
Here:
  • \( \epsilon_0 \) is the permittivity of free space, \( 8.85 \times 10^{-12} \, \text{F/m} \)
  • \( c \) is the speed of light, \( 3.00 \times 10^8 \, \text{m/s} \)
  • \( E \) is the electric field amplitude
This expression tells us how solar energy is spread across an area at a point in space, influencing how efficiently solar panels on devices like the Mars rover can capture this energy.
As your calculations reveal intensity, it becomes easier to connect these insights with practical applications.
Role of Electric Field Amplitude in Solar Panels
Electric field amplitude (\( E \)) is vital for calculating solar radiation intensity. It represents the maximum strength of the electric field component of solar radiation. In our scenario, the amplitude is \( 673 \, \text{V/m} \). This value is crucial because solar panels convert the electromagnetic energy of sunlight into electrical energy, and the electric field plays a central role in this conversion process.
  • Higher electric field amplitude means higher potential energy transformations.
  • It's directly involved in the equations that help us determine the potential power output of a solar panel.
Thus, knowing the electric field amplitude helps predict the efficiency and potential power output of solar modules, which is essential for planning energy needs and solutions for exploration missions like those on Mars.
Mars Rover Power Output: Meeting Energy Demands
Mars rovers, like Spirit, rely heavily on solar panels for their energy. This power output is pivotal, as it determines the rover's ability to perform scientific tasks and maintain essential functions. The steps for assessing power output include:
  • Calculating solar radiation intensity, which provides the baseline solar power available to the panel.
  • Determining the solar panel鈥檚 efficiency to ascertain how much of the received solar energy can be converted to electrical energy.
Given the values, a solar panel with an efficiency of \( 18\% \) means only a fraction of the received solar radiation converts into usable power.
Knowing the power output helps engineers and scientists ensure that the rover has enough energy to complete its mission tasks and survive in the harsh Martian climate.
Solar Power Calculation: Harnessing Energy Efficiently
To calculate the solar power generated by a panel, we need to understand both the solar radiation intensity that hits the panel and the panel's efficiency. The expression used is:
  • \( P = I \cdot \eta \cdot A \)
Where:
  • \( I \) is the intensity of solar radiation
  • \( \eta \) is the efficiency of the solar panel (expressed in decimals)
  • \( A \) is the active surface area of the solar panel
The active surface area is simply the panel's length times its width, allowing us to calculate the total area exposed to sunlight.
Efficiency consideration ensures we understand how much of the sunlight energy is converted into usable electricity.
Proper calculation assists in optimizing energy usage to meet the specific power requirements of devices or missions, like the Mars rover, and can be critical in planning explorative or sustainable energy systems.

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Most popular questions from this chapter

The most intense beam of light that can propagate through dry air must have an electric field whose maximum amplitude is no greater than the breakdown value for air: \(E_{\max }^{\operatorname{air}}=3.0 \cdot 10^{6} \mathrm{~V} / \mathrm{m},\) assuming that this value is unaffected by the frequency of the wave. a) Calculate the maximum amplitude the magnetic field of this wave can have. b) Calculate the intensity of this wave. c) What happens to a wave more intense than this?

.An industrial carbon dioxide laser produces a beam of radiation with average power of \(6.00 \mathrm{~kW}\) at a wavelength of \(10.6 \mu \mathrm{m}\). Such a laser can be used to cut steel up to \(25 \mathrm{~mm}\) thick. The laser light is polarized in the \(x\) -direction, travels in the positive \(z\) -direction, and is collimated (neither diverging or converging) at a constant diameter of \(100.0 \mu \mathrm{m} .\) Write the equations for the laser light's electric and magnetic fields as a function of time and of position \(z\) along the beam. Recall that \(\vec{E}\) and \(\vec{B}\) are vectors. Leave the overall phase unspecified, but be sure to check the relative phase between \(\vec{E}\) and \(\vec{B}\) .

Two polarizing filters are crossed at \(90^{\circ}\), so when light is shined from behind the pair of filters, no light passes through. A third filter is inserted between the two, initially aligned with one of them. Describe what happens as the intermediate filter is rotated through an angle of \(360^{\circ} .\)

A tiny particle of density \(2000 . \mathrm{kg} / \mathrm{m}^{3}\) is at the same distance from the Sun as the Earth is \(\left(1.50 \cdot 10^{11} \mathrm{~m}\right)\). Assume that the particle is spherical and perfectly reflecting. What would its radius have to be for the outward radiation pressure on it to be \(1.00 \%\) of the inward gravitational attraction of the Sun? (Take the Sun's mass to be \(\left.2.00 \cdot 10^{30} \mathrm{~kg} .\right)\)

What is the radiation pressure due to Betelgeuse (which has a luminosity, or power output, 10,000 times that of the Sun) at a distance equal to that of Uranus's orbit from it?
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