Question

A hydroelectric generating plant is supplied from a reservoir of capacity 3.6 × 10⁶ m³ at a head of 100 m. Find the total energy available in kWh if the overall efficiency is 75%
1. 63675
2. 35750
3. 735750
4. 536750

Answer 1

Correct Answer - Option 3 : 735750

Concept:

A hydroelectric generating station is a plant that produces electricity by using water to plant the turbine, which, in turn, drives the generators.

Potential energy, E = mgh

Total output energy, E'  = Potential energy x overall efficiency

\(\left[ {Density = \frac{{Mass}}{{Volume}}} \right]\)

Where,

M = Mass

g = gravitational acceleration

h = height

1 J = 1/ 3600 Wh

Calculation:

Given that

overall efficiency = 75%

Reservoir capacity (V) = 3.6 × 106 m3

Water Head (h) = 100 m

Now to calculate the potential energy, we know that

Density of water = 997 kg/m3

g = 9.8 m/s2

Put all these in formula of potential energy, we get

E = (D × V) × 9.8 × 100     

E = 997 × 3.6 × 106 × 9.8 × 100

E = 3517416 × 106 J

E =  (3517416 / 3600) × 103 kWh

E = 977060 kWh

Hence, total output energy can be calculated as

E' = 977060 × 0.75 

∴ E' ≈ 732795 Wh ≈ 732.7 kWh