Question

The reservoir area of a hydro-electric generating plant is spread over an area of 4 sq km with a storage capacity of 8 million cubic-metres. The net head of water available to the turbine is 70 metres. Assuming an efficiency of 0.87 and 0.93 for water turbine and generator respectively, calculate the electrical energy generated by the plant. 

Estimate the difference in water level if a load of 30 MW is continuously supplied by the generator for 6 hours.

Answer 1

Since 1 cubic metre of water weighs 1000 kg., the reservoir capacity 

= 8 × 10⁶ m³ = 8 × 10⁶ × 1000 kg. = 8 × 10⁹ kg. 

Wt. of water, W = 8 × 10⁹ kg. 

Wt. 8 × 10⁹ × 9.81 = 78.48 × 10⁹ N. 

Net water head = 70 m. 

Potential energy stored in this much water = Wh = 78.48 × 10⁹ × 70 = 549.36 × 10¹⁰J 

Overall efficiency of the generating plant = 0.87 × 0.93 = 0.809 

Energy available = 0.809 × 549.36 × 10¹⁰ J = 444.4 × 10¹⁰ J 

= 444.4 × 10¹⁰/36 × 105 = 12.34 × 105 kWh 

Energy supplied in 6 hours = 30 MW × 6 h = 180 MWh 

= 180,000 kWh 

Energy drawn from the reservoir after taking into consideration, the overall efficiency of the station 

= 180,000/0.809 = 224,500 kWh = 224,500 × 36 × 10⁵ = 80.8 × 10¹⁰ J 

If m kg. is the mass of water used in 6 hours, then since water head is 70 m, 

mgh = 80.8 × 10¹⁰ or m × 9.81 × 70 = 80.8 × 10¹⁰ 

∴ m = 1.176 × 10⁹ kg. 

If h is the fall in water level, then h × area × density = mass of water 

∴ h × (4 × 10⁶ ) × 1000 = 1.176 × 10⁹ 

∴ h = 0.294 m = 29.4 cm.