A solar power park of 15 km² is planned. The average solar insolation is 6 kWh m⁻² day⁻¹. Photovoltaic panels have an efficiency of 18 %, and the plant operates 320 days a year. Calculate the total electricity that can be generated in a year in gigawatt‑hours (GWh). (1 km² = 10⁶ m²).

Step-by-step solution

1. Compute the area in square metres: $15\times 10^6=1.5\times 10^7{\text{m}}^2$.
2. Calculate energy incident per day: $6{\text{kWh/m}}^2\times 1.5\times 10^7{\text{m}}^2=9\times 10^7\text{kWh/day}$.
3. Find total incident energy per year: $9\times 10^7\text{kWh/day}\times 320\text{days}=2.88\times 10^{10}\text{kWh}$.
4. Apply panel efficiency: $0.18\times 2.88\times 10^{10}=5.184\times 10^9\text{kWh}$.
5. Convert to GWh: $\frac{5.184\times 10^9}{10^6}=5,184\text{GWh}$.

Concept tested: Solar power generation calculations

Published 7 October 2026 · How this solution was prepared and checked