Question

Direct solar energy is incident on the horizontal surface at an average rate of 200 W/m². If 20% of this energy can be converted to useful electrical energy, how much area is needed to supply 8 kW?

• A. \(1000 \, {m}^2 \)
• B. \(100 \, {m}^2 \)
• C. \(200 \, {m}^2 \)
• D. \(2000 \, {m}^2 \)

Hint:

When calculating the area required for solar energy conversion, remember to account for the efficiency of the system. The total energy required will be higher than the output power due to losses.

Answer 1

The Correct Option is C

Given:
- Solar energy incident on the surface: \( I = 200 \, \text{W/m}^2 \)
- Efficiency: \( \eta = 0.20 \)
- Required electrical power: \( P = 8 \, \text{kW} = 8000 \, \text{W} \)

Step 1: Calculate the total energy that needs to be absorbed
The total energy required to be absorbed is:
\[ P_{\text{absorbed}} = \frac{P}{\eta} = \frac{8000}{0.20} = 40000 \, \text{W} \]

Step 2: Calculate the area required to absorb this energy
The area \( A \) required to absorb the energy is:
\[ A = \frac{P_{\text{absorbed}}}{I} = \frac{40000}{200} = 200 \, \text{m}^2 \]