Question

A hydroelectric power station takes its water from a lake whose water level is 50 m above turbine . assuming an overall efficiency of 40%,calculate mass of water which must flow through turbine each second to produce power output of 1 MW

Answer 1

since effieciency is 40 %
useful energy = 40% mgh
= ( 40 / 100) * m * 10 *50
1MW = 200m
1000000 = 200m
m =1000000/200= 5000kg
mark branliest if it was

Answer 2

Given:

Height = 50m above the turbine

Efficiency = 40%

Power output = 1MW

g = 10 ms^-2

Solution:

Potential energy = mgh

Efficiency = 40%

Let's find:

Useful work done = 40% of potential energy

= 40 / 100 (mgh)

= 0.4 (m × 10 × 50)

= 200 m

Power = work done per second

1MW = 200 × mass of water per second

1 × 106W = 200 × mass of water flowing per second

Therefore:

Mass of water flowing per second = (1 × 106) / 200

= 5000 kg