At a hydroelectric power plant,the water pres | vedclass

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(C) The potential energy of the water per unit time is the power input to the turbine.Power input $P_{in} = \frac{mgh}{t} = \rho V g h$,where $\rho$ i

Vedclass · Accepted Answer

$176.4$

Vedclass · Answer

(C) The potential energy of the water per unit time is the power input to the turbine.Power input $P_{in} = \frac{mgh}{t} = ho V g h$,where $ho$ is the density of water $(10^{3}\; kg/m^{3})$,$V$ is the volume flow rate $(100\; m^{3}/s)$,$g$ is the acceleration due to gravity $(9.8\; m/s^{2})$,and $h$ is the height $(300\; m)$.$P_{in} = 10^{3} 	imes 100 	imes 9.8 	imes 300 = 294 	imes 10^{6}\; W = 294\; MW$.The electric power available is the output power,which is the product of efficiency and input power.$P_{out} = \eta 	imes P_{in} = 0.60 	imes 294\; MW$.$P_{out} = 176.4\; MW$.

At a hydroelectric power plant,the water pressure head is at a height of $300\; m$ and the water flow available is $100\; m^{3} s^{-1}.$ If the turbine generator efficiency is $60\%,$ estimate the electric power available from the plant (in $MW$) $\left(g=9.8\; m s^{-2}\right).$

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