(i) What is the available energy in kWh if the hydraulic efficiency be 0.8 and the electrical efficiency 0.9?

Question

An hydroelectric generating station is supplied form a reservoir of capacity 6 million m³ at a head of 170 m. 

(i) What is the available energy in kWh if the hydraulic efficiency be 0.8 and the electrical efficiency 0.9? 

(ii) Find the fall in reservoir level after a load of 12,000 kW has been supplied for 3 hours, the area of the reservoir is 2.5 km². 

(iii) If the reservoir is supplied by a river at the rate of 1.2 m³ /s, what does this flow represent in kW and kWh/day ? Assume constant head and efficiency. Water weighs 1 tonne/m³ .

Answer 1

(i) Wt. of water W = 6 × 10⁶ × 1000 kg wt = 6 × 10⁹ × 9.81 N 

Water head = 170 m 

Potential energy stored in this much water 

= Wh = 6 × 10⁹ × 9.81 × 170 J = 10¹² J 

Overall efficiency of the station = 0.8 × 0.9 = 0.71 

∴ energy available = 0.72 × 10¹³ J = 72 × 10¹¹/36 × 10⁵ 

= 2 × 10⁶ kWh 

(ii) Energy supplied = 12,000 × 3 = 36,000 kWh 

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

= 36,000/0.72 = 5 × 10⁴ kWh 

= 5 × 10⁴ × 36 × 10⁵ = 18 × 10¹⁰ J

If m kg is the mass of water used in two hours, then, since water head is 170 m 

mgh = 18 × 10¹⁰ or m × 9.81 × 170 = 18 × 10¹⁰ 

∴ m = 1.08 × 10⁸kg 

If h metre is the fall in water level, then 

h × area × density = mass of water 

∴ h × (2.5 × 10⁶) × 1000 = 1.08 × 10⁸ 

∴ h = 0.0432 m = 4.32 cm 

(iii) Mass of water stored per second = 1.2 × 1000 = 1200 kg 

Wt. of water stored per second = 1200 × 9.81 N 

Power stored = 1200 × 9.81 × 170 J/s = 2,000 kW 

Power actually available = 2,000 × 0.72 = 1440 kW 

Energy delivered /day = 1440 × 24 = 34,560 kWh