Question

A proposed hydro-electric station has an available head of 30 m, catchment area of 50 × 10⁶ sq.m, the rainfall for which is 120 cm per annum. If 70% of the total rainfall can be collected, calculate the power that could be generated. Assume the following efficiencies : Pen stock 95%, Turbine 80% and Generator 85. 

Answer 1

Volume of water available = 0.7(50 × 10⁶ × 1.2) = 4.2 × 10⁷m³ 

Mass of water available = 4.2 × 10⁷ × 1000 = 4.2 × 10¹⁰ kg 

This quantity of water is available for a period of one year. 

Hence, quantity available per second 

= 4.2 × 10¹⁰/365 × 24 × 3600 

= 1.33 × 10³ . 

Available head = 30 m 

Potential energy available = mgh = 1.33 × 10³ × 9.8 × 30 = 391 × 10³ J 

Since this energy is available per second, hence power available is = 391 × 10³ J/s = 391× 10³ W = 391 kW 

Overall efficiency = 0.95 × 0.80 × 0.85 = 0.646 

The power that could be generated = 391 × 0.646 = 253 kW