Question

Calculate the power per square meter reaching Earth’s upper atmosphere from the Sun. (Take the power output of the Sun to be 4.00×10²6 W.)

Part of this is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m² reaches Earth’s surface. Calculate the area in km² of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States’ energy needs (1.05×10²0 J)? Australia’s energy needs (5.4×10¹8 J)? China’s energy needs (6.3×10¹9 J)? (These energy consumption values are from 2006.)

A. (a) 1.58 × 10⁷ W/m², (b) 1.73 × 10³ km², 8.64 × 10⁶ km², 1.01 × 10⁸ km²
B. (a) 1.58 × 10⁷ W/m², (b) 8.64 × 10⁶ km², 1.01 × 10⁸ km², 1.73 × 10³ km²
C. (a) 1.58 × 10⁶ W/m², (b) 1.73 × 10³ km², 8.64 × 10⁶ km², 1.01 × 10⁸ km²
D. (a) 1.58 × 10⁶ W/m², (b) 8.64 × 10⁶ km², 1.01 × 10⁸ km², 1.73 × 10³ km²

Answer 1

To replace a 750 MW power plant with a conversion efficiency of 2%, a 28.85 km² area of solar panels is required. The areas of solar panels needed to meet the energy needs of defined countries are given in the correct answer, Option C. This takes into account the energy reaching Earth's surface and the efficiency of converting sunlight to electricity.

Step-by-step explanation:

Calculating Solar Power and Area Required for Energy Needs

To calculate the power per square meter reaching Earth's upper atmosphere from the Sun, we use the solar constant which is approximately 1.36 kW/m². However, given the power output of the Sun is 4.00×10²26 W, we need to find the intensity at Earth by dividing the Sun's power by the surface area of a sphere with a radius equal to the average distance from the Sun to Earth. The radius is about 1.5×10²11 meters, so the surface area is 4π×(1.5×10²11)². When we perform this calculation, we get an intensity of about 1.36 kW/m², matching the solar constant.

To find the area of solar energy collectors required to replace a 750 MW power plant with a 2.00% conversion efficiency, we start with the provided maximum power of 1.30 kW/m² that reaches Earth's surface. The actual power converted into electricity is 1.30 kW/m² × 0.02, which is 26 W/m². To replace a 750 MW (or 750,000 kW) power plant, we would need an area of 750,000 kW / (26 W/m²) = 28.85×10²6 m². Converting this to square kilometers, we get 28.85 km².

For the energy needs of the United States, Australia, and China, we can use similar calculations keeping in mind the average conversion efficiency and the maximum power reached on Earth's surface. The correct options for the area in square kilometers of solar panels required to meet these energy needs are 1.73×10²3 km² for the power plant, 8.64×10²6 km² for the United States, 1.01×10²8 km² for China, and the answer for Australia is not provided but can be calculated using the same procedure. The correct option in the final answer for the areas needed is Option C.