A solar power park is planned over an area of 10 km². The average solar insolation is 5 kWh m⁻² day⁻¹, the photovoltaic panels have an efficiency of 15 %, and the plant is expected to operate 300 days a year. Calculate the total electricity that can be generated in a year in gigawatt‑hours (GWh). (1 km² = 10⁶ m²)
Answer
2250 GWh per year
The correct answer is 2250 GWh per year — Convert kWh to GWh (1 GWh = 10⁶ kWh): 2.25×10⁹ kWh ÷ 10⁶ = 2250 GWh/year.
Step-by-step solution
- Convert the area to square metres: 10 km² × 10⁶ = 1.0×10⁷ m².
- Compute daily solar energy incident on the area: 5 kWh/m² × 1.0×10⁷ m² = 5.0×10⁷ kWh/day.
- Apply panel efficiency to obtain useful energy: 15 % of 5.0×10⁷ kWh = 0.15 × 5.0×10⁷ = 7.5×10⁶ kWh/day.
- Multiply by the number of operating days: 7.5×10⁶ kWh/day × 300 days = 2.25×10⁹ kWh/year.
- Convert kWh to GWh (1 GWh = 10⁶ kWh): 2.25×10⁹ kWh ÷ 10⁶ = 2250 GWh/year.
Common mistake: Omitting the conversion of km² to m² or forgetting to apply the 15 % efficiency factor, leading to a much higher estimate.
Concept tested: Solar energy generation calculation
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