Problem 64 - Chemistry: The Central Science

Question

The solar power striking Earth every day averages 169 watts per square meter. The peak electrical power usage in New York City is 12,000 megawatts. Considering that present technology for solar energy conversion is only about \(10 \%\) efficient, from how many square meters of land must sunlight be collected in order to provide this peak power? (For comparison, the total area of the city is \(\left.830 \mathrm{~km}^{2} .\right)\)

Step-by-Step Solution

Answer

Sunlight must be collected from an area of approximately \(710.65 \text{ km}^2\) to provide the peak power usage of New York City with solar energy at \(10\%\) efficiency.

Step 1: Convert peak power usage to watts

First, we need to convert the peak power usage of New York City from megawatts (MW) to watts (W). There are 1,000,000 W in 1 MW, so we can easily calculate this by multiplying: \[12000 \text{ MW} \times 1000000 = 12000000000 \text{ W}\]

Step 2: Calculate the required solar power

Now, we need to find how much solar power is required to provide this peak power. Since the solar energy conversion is only about \(10\%\) efficient, we need to calculate the actual solar power required by dividing the peak power usage by the efficiency: \[\frac{12000000000 \text{ W}}{0.10} = 120000000000 \text{ W}\]

Step 3: Determine the necessary land area

To determine the area needed to collect enough sunlight, we can divide the required solar power by the average solar power striking Earth per square meter: \[\frac{120000000000 \text{ W}}{169 \text{ W/m}^2} = 710650887.57 \text{ m}^2\]

Step 4: Compare the land area to the total area of New York City

Finally, we can compare the necessary land area to the total area of New York City, which is \(830 \text{ km}^2\). To do this, let's convert the land area we found to square kilometers by dividing by \(1,000,000 \text{ m}^2/\text{km}^2\): \[\frac{710650887.57 \text{ m}^2}{1000000} = 710.65 \text{ km}^2\] Therefore, sunlight must be collected from an area of approximately \(710.65 \text{ km}^2\) to provide the peak power usage of New York City with solar energy at \(10\%\) efficiency.

Key Concepts

Solar Power Calculation

Understanding solar power calculations is key to harnessing the sun's energy effectively. To determine how much solar power is needed for any particular application, start by identifying the total power requirement. For instance, in our exercise, New York City's peak power usage is given as 12,000 megawatts (MW).

To work with this value in subsequent calculations, it should be converted to watts (W), the standard unit of power. Knowing that 1 MW is equal to 1,000,000 W, simply multiply the megawatt value by this conversion factor to find the number of watts:

• New York's peak power usage in watts: \[12,000 ext{ MW} \times 1,000,000 = 12,000,000,000 ext{ W}\]

This number will serve as the basis for determining how much solar energy must be collected. By accurately converting and calculating these initial values, we lay the groundwork for a precise assessment of solar power needs.

Energy Conversion Efficiency

Energy conversion efficiency is an essential concept to understand when discussing solar power systems. It explains how much of the incoming solar energy is actually converted into usable electrical energy. In the exercise, we see that the current technology allows for only about a \(10\%\) conversion efficiency.

Why is this important? Because it directly affects how much solar energy must be captured to meet energy demands. Only a fraction of the solar energy collected can be transformed into electrical energy. Thus, when calculating the required solar power to meet a certain energy demand, you need to consider this efficiency:

• First, divide the desired electrical power output by the efficiency percentage expressed as a decimal to find the total solar power needed.
  For example, \[\frac{12,000,000,000 ext{ W}}{0.10} = 120,000,000,000 ext{ W}\]
• This tells us the actual amount of solar energy that must be captured, taking into account the losses due to inefficiency.

Understanding energy conversion efficiency ensures accurate planning and implementation of solar power solutions.

Land Area for Solar Power

Calculating the land area required for solar power involves understanding the average solar energy received per unit of land and the total solar power needed. In our example, the average solar radiation hitting Earth is 169 watts per square meter (W/m²).

With the total required solar power calculated earlier due to efficiency losses, the next step is to derive how much land area is needed. This is done by dividing the total required solar power by the solar power per square meter:

• \[\frac{120,000,000,000 ext{ W}}{169 ext{ W/m}^2} \approx 710,650,887.57 ext{ m}^2\]

This result represents the amount of land required in square meters. For practical understanding and comparison, converting to square kilometers is beneficial. Divide by 1,000,000 (the number of square meters in a square kilometer):

• \[710,650,887.57 ext{ m}^2 \div 1,000,000 = 710.65 ext{ km}^2\]

This process shows not just the extent of land necessary for new installations but also highlights the substantial space requirements of solar energy, which is crucial for urban planning and environmental considerations.