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The solar power striking Earth every day averages 168 watts per square meter. The peak electrical power usage in New York City is 12,000 MW. Considering that present technology for solar energy conversion is about \(10 \%\) efficient, from how many square meters of land must sunlight be collected in order to provide this peak power? (For comparisen, the total area of New York city is \(830 \mathrm{~km}^{2}\).)

Short Answer

Expert verified
To provide the peak power usage of 12,000 MW in New York City with solar panels that are 10% efficient, approximately 714,285,714 m² (or 86% of the city's total area) would be needed.

Step by step solution

01

Convert the peak power usage to watts

First, we must convert the peak power usage, which is given in megawatts (MW), into watts (W) using the conversion factor: 1 MW = 1,000,000 W So, 12,000 MW = 12,000 * 1,000,000 W
02

Adjust the required power according to the solar panel efficiency

The given solar panel efficiency is 10%, which means that only 10% of the sunlight can be converted into electricity. To calculate the effective required power, we must divide the peak power usage by the efficiency factor: Effective required power = total required power / efficiency Efficiency = 10% = 0.1 Thus, Effective required power = (12,000 * 1,000,000) / 0.1 = 120,000,000,000 W
03

Calculate the required area for solar panels

Now, to find the required area for solar panels, we can divide the effective required power by the average solar power striking Earth per square meter: Required area = Effective required power / average solar power per square meter Required area = 120,000,000,000 W / 168 W/m² Required area ≈ 714,285,714 m²
04

Compare the required area with the total area of New York City

The total area of New York City is given as 830 km². To compare the required area for solar panels to the total area of the city, we need to convert the city's area into square meters: 1 km² = 1,000,000 m² So, 830 km² = 830 * 1,000,000 = 830,000,000 m² Now, we can compare the required area for solar panels (714,285,714 m²) to the total area of New York City (830,000,000 m²): Percentage of New York City area needed for solar panels = (required area / total city area) * 100 Percentage = (714,285,714 m² / 830,000,000 m²) * 100 ≈ 86 % Hence, approximately 86% of New York City's total area would need to be covered with solar panels to meet its peak power usage with the given solar panel efficiency.

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Key Concepts

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

Solar Power Calculation
Understanding the calculation of solar power is crucial for anyone looking to harness the sun's energy. Solar power, measured in watts per square meter (\( W/m^2 \)), denotes the rate at which energy is radiated onto a given area of the Earth's surface by the sun. To begin with, the average solar power incident on Earth is approximately 168 watts per square meter. However, not all of this energy can be converted to electricity due to the efficiency limitations of solar panels.

In essence, solar power calculation involves assessing the total power output that can be obtained from the sunlight hitting an area. This is pivotal in planning for the energy yield especially in large electricity grids like New York City. For instance, with a conversion efficiency of about 10%, a mere fraction of the 168 watts is useable. Consequently, much larger areas are needed to collect sufficient energy to meet power demands. The calculation, therefore, requires an understanding of both the incoming solar power and the efficiency of the solar energy system at hand.
Electrical Power Usage
Electrical power usage denotes the amount of electrical energy consumed by a household, business, city, or region over a certain period. It is measured in watts (W) when referring to the rate of consumption at any instant or watt-hours (\( Wh \) or kilowatt-hours (\( kWh \) for the energy used over time. Peak power usage, such as the 12,000 megawatts (MW) figure for New York City, represents the maximum load or the highest level of electricity consumption at a particular instant, often occurring during peak hours of the day or season.

For our calculations, it's essential to convert megawatts to watts since solar irradiance (solar power) is usually calculated in watts per square meter. This approach helps in pinpointing the exact requirement in terms of area and energy production necessary for solar panels to meet the peak electricity demand. When considering such a significant figure like New York City's peak power usage, it becomes clear that proficient energy management and solar panel installation is necessary to meet the demands sustainably.
Solar Panels Area Requirement
The area requirement for solar panels is determined by dividing the total power demands by the product of the solar irradiance and the conversion efficiency of the solar panels. In the context of New York City's peak power needs, and with the solar efficiency provided, a substantial area is required to capture enough sunlight to convert into 12,000 MW of electricity.

Using the calculation from the step-by-step solution, we find that the required area for solar panels would be around 714 million square meters, which is nearly 86% of New York City's land area. This figure accentuates the importance of maximizing the efficiency of solar panels to reduce the total area needed. It also prompts exploration into multi-layered approaches, combining solar panels with other renewable sources or integrating them into buildings and infrastructure to minimize land use. With advancements in technology and more efficient solar cells, the area requirement for solar panel installations could decrease, making solar energy a more viable option for large urban spaces.

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Most popular questions from this chapter

The enthalpy of evaporation of water is \(40.67 \mathrm{~kJ} / \mathrm{mol}\). Sunlight striking Earthis surface supplies 168 W per square meter \((1 \mathrm{~W}=1 \mathrm{watt}=1 \mathrm{~J} / \mathrm{s})\). (a) Assuming that evaporation of water is due only to energy input from the Sun, calculate how many grams of water could be evaporated from a \(1.00\) square meter patch of ocean over a 12 -h day. (b) The specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\). If the initial surface temperature of a \(1.00\) square meter patch of ocean is \(26{ }^{\circ} \mathrm{C}\), what is its final temperature after being in sunlight for \(12 \mathrm{~h}\), assuming no phase changes and assuming that sunlight penetrates uniformly to depth of \(10.0 \mathrm{~cm}\) ?

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