Question

A hydraulic turbine has 50m of head available at a flow rate of 1.30 m^3/s, and its overall turbine. generator efficiency is 78 percent. Determine the electric power output of this turbine. Given that ρ=1000 kg/m^3

Answer 1

P = 4.96 *10^5 watts

Step-by-step explanation:

we know that, power is given as

`power =(\dot m gh)/(t)`

`power = (\dot m)/(dt) gh`

we know that

`1L = 10^(-3) m^3`

also 1L = 1kg

`(\dot m)/(dt) = 1.30 m^3/s`

therefore

`P = 50 * ((1.30)/(10^(-3))* 9.8`

P = 6.37*10^5 W

78% of power = 6.37 *0.78

P = 4.96 *10^5 watts