/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Interactive Solution \(\underlin... [FREE SOLUTION] | 91Ó°ÊÓ

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Interactive Solution \(\underline{24.53}\) at provides one model for problems like this one. The drawing shows an edge-on view of the solar panels on a communications satellite. The dashed line specifies the normal to the panels. Sunlight strikes the panels at an angle \(\theta\) with respect to the normal. If the solar power impinging on the panels is \(2600 \mathrm{~W}\) when \(\theta=65^{\circ}\), what is it when \(\theta=25^{\circ} ?\)

Short Answer

Expert verified
When \( \theta=25^\circ \), the power on the panels is approximately 5577 W.

Step by step solution

01

Understanding the Conceptual Setup

The power received by the panels is determined by the component of sunlight that is perpendicular to the panel. This component is given by multiplying the total solar power density (incident sunlight power) by the cosine of the angle of incidence \( \theta \). This is based on the formula \( P = P_0 \cos(\theta) \), where \( P_0 \) is the total power and \( P \) is the power received.
02

Given Parameters and Formula

We are given that when \( \theta = 65^\circ \), the power \( P = 2600 \, \text{W} \). The formula to find the power received at a new angle is \( P_1 = P_0 \cos(\theta_1) \cos(\theta_2) \). Here, we need to calculate the power when \( \theta = 25^\circ \).
03

Finding Power at \( \theta = 25^\circ \)

Using the previous step's equation, we interpose the new angles. Calculate:\[P_1 = 2600 \, \text{W} \times \frac{\cos(25^\circ)}{\cos(65^\circ)}\]Calculate the cosine values: \( \cos(25^\circ) \approx 0.9063 \) and \( \cos(65^\circ) \approx 0.4226 \). Substitute back into the equation:\[P_1 = 2600 \, \text{W} \times \frac{0.9063}{0.4226} \approx 2600 \, \text{W} \times 2.145 = 5577 \, \text{W}.\]
04

Conclusion of Calculation

The calculated power at \( \theta = 25^\circ \) is approximately \( 5577 \, \text{W} \). This completes the calculation, affirming the inverse relationship between angle \( \theta \) and the power received due to the cosine factor.

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Key Concepts

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

Angle of Incidence
In the realm of solar power calculations, the angle of incidence plays a pivotal role in determining how much sunlight effectively reaches the solar panels. The angle of incidence is essentially the angle between the incoming sunlight and a line perpendicular to the surface of the panel. This perpendicular line is often referred to as the "normal." When sunlight hits the panel directly (at a 0° angle of incidence), the maximum amount of energy is absorbed. As this angle increases, the effective energy received reduces.
Understanding the angle of incidence is crucial because as the angle grows, the sunlight covers a larger area and must travel further through the atmosphere, reducing energy concentration. In practical terms, the smaller the angle, the more efficient the panel becomes at absorbing sunlight. Therefore, optimizing the angle of the solar panels relative to the sun is key for maximizing energy collection.
Cosine Function
The cosine function is fundamental in calculating the true amount of power received by solar panels. Recall from trigonometry, the cosine of an angle in a right triangle gives the ratio of the adjacent side to the hypotenuse.
For solar power calculations, using the formula \(P = P_0 \cos(\theta)\), we apply the cosine of the angle of incidence \(\theta\) to determine the effective power contribution. The cosine function decreases as the angle increases, reflecting the reduction in power received as the sunlight's approach becomes more oblique.
  • At 0°, \(\cos(0°) = 1\), meaning full power is received.
  • At 90°, \(\cos(90°) = 0\), resulting in no power.
This relationship emphasizes the sensitivity of power received to changes in the angle of incidence, making it crucial for the optimization of solar panel orientation.
Solar Panels
Solar panels are devices that convert sunlight into electricity. They are composed of many solar cells, which are typically made of silicon. When photons from sunlight hit these cells, they knock electrons loose, creating a flow of electricity. The efficiency of solar panels is heavily dependent on how they are positioned in relation to the sun to capture maximum sunlight.
Since solar panels perform optimally when the sunlight strikes them perpendicularly, understanding the angle of incidence is essential for their efficient operation. Engineers often design solar tracking systems that adjust the position of the panels throughout the day to maintain optimal angles with respect to the sun's position.
By leveraging insights from the cosine function, solar panel systems can be fine-tuned to optimize energy yield, thus making the investment in solar technology more worthwhile.
Power Received Calculation
Calculating the power received by solar panels involves understanding how the angle of sunlight affects the solar energy installation. The power received is calculated by finding the component of sunlight perpendicular to the solar panel surface.
Using the formula \(P = P_0 \cos(\theta)\), where \(P_0\) is the initial power or total solar radiation, and \(\theta\) is the angle of incidence, we determine the effective power \(P\) actually absorbed by the panel. This calculation involves:
  • Determining the angle of incidence.
  • Using cosine trigonometric calculations to assess how much energy is effectively captured.

For instance, if the power at an angle \(\theta = 65°\) is given as 2600 W, adjusting the angle to 25° and recalculating power using cosine ratios allows us to find the new power received, which equates to more power input due to a reduced angle. Such calculations are crucial for understanding and maximizing the efficiency and output of solar power systems.

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Most popular questions from this chapter

Obtain the wavelengths in vacuum for (a) blue light whose frequency is \(6.34 \times 10^{14} \mathrm{~Hz}\) and (b) orange light whose frequency is \(4.95 \times 10^{14} \mathrm{~Hz}\). Express your answers in nanometers \(\left(1 \mathrm{nm}=10^{-9} \mathrm{~m}\right)\)

A laser emits a narrow beam of light. The radius of the beam is \(1.0 \times 10^{-3} \mathrm{~m}\), and the power is \(1.2 \times 10^{-3} \mathrm{~W}\). What is the intensity of the laser beam?

The human eye is most sensitive to light having a frequency of about \(5.5 \times 10^{14} \mathrm{~Hz}\), which is in the yellow-green region of the electromagnetic spectrum. How many wavelengths of this light can fit across the width of your thumb, a distance of about \(2.0 \mathrm{~cm} ?\)

The average intensity of sunlight reaching the earth is \(1390 \mathrm{~W} / \mathrm{m}^{2}\). A charge of \(2.6 \times 10^{-8} \mathrm{C}\) is placed in the path of this electromagnetic wave. (a) What is the magnitude of the maximum electric force that the charge experiences? (b) If the charge is moving at a speed of \(3.7 \times 10^{4} \mathrm{~m} / \mathrm{s}\), what is the magnitude of the maximum magnetic force that the charge could experience?

A heat lamp emits infrared radiation whose rms electric field is \(E_{\mathrm{rms}}=2800 \mathrm{~N} / \mathrm{C} .\) (a) What is the average intensity of the radiation? (b) The radiation is focused on a person's leg over a circular area of radius \(4.0 \mathrm{~cm}\). What is the average power delivered to the leg? (c) The portion of the leg being radiated has a mass of \(0.28 \mathrm{~kg}\) and a specific heat capacity of \(3500 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) How long does it take to raise its temperature by \(2.0 \mathrm{C}^{\circ}\) ? Assume that there is no other heat transfer into or out of the portion of the leg being heated.

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