At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m³s^–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms^–2).

Solution

Here given,
Height of water pressure head, h = 300 m
Volume of the water flowing per second, V = 100 m³Mass of water flowing per second, m = 100 × 10³ kg
= 10⁵ kg
acceleration due to gravity, g = 9.8 ms^–2Potential energy of water fall during one second
$= mgh = 10^{5} \times 9.8 \times 300 = 29.4 \times 10^{7} J .$

Therefore, input power = $294 \times 10^{6} {Js}^{- 1}$

$Efficiency, η = \frac{output power}{input power}$
$∴$

$Output power = η \times input power = 0.6 \times 294 \times 10^{6} = 176.4 \times 10^{6} watt = 176.4 MW .$

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Electrostatic Potential And Capacitance

At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m³s^–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms^–2).

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Electrostatic Potential And Capacitance

At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3s–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms–2).

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At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m³s^–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms^–2).