Question

At a hydroelectric power plant the water pressure head is at a height of 300 m and the water flow available is $100m^{3}s^{- 1}$ If the turbine generator efficiency is 60%, then the electric power available from the plant is

$(\text{Take }g = 10\text{ m }\text{s}^{- 2},\ \rho = 10^{3}\text{kg }\text{m}^{3})$

• A. 1.8 MW
• B. 18 MW
• C. 180 MW
• D. 1800 MW

Answer 1

Answer: (C)

180 MW

Answer 2

: Here, height , h = 300 m

Rate of water flow, $V = 100m^{3}s^{- 1}$

Efficiency ,

$\eta = 60\%,g = 10ms^{- 2},\rho = 10^{3}kgm^{- 3}$

Hydroelectric power

$= \frac{work}{time} = \frac{force \times dis\tan ce}{time} = \frac{force}{area} \times \frac{dis\tan ce \times area}{time}$

= pressure × area × velocity

But area ×velocity = volume/second = V

$\therefore$ Hydroelectric power = P×V

Now, Efficiency $= \frac{Poweravailable}{Hydroelectricpower}$

$\therefore\frac{60}{100} = \frac{Poweravailable(miyC/k\ 'kfDr)}{P \times V}$

or Power available ,$= \frac{60}{100} \times P \times V$

$= \frac{3}{5} \times (h\rho g)V$ $\lbrack\because P = h\rho g\rbrack = \frac{3}{5} \times 300 \times 10^{3} \times 10 \times 100 = 1.8 \times 10^{8}W = 180MW$