# A power plant has a total peak capacity of 826MW from its seven generators. The water intake is 25m below the surface level of the lake. There are seven 7.4m diameter tunnels at the intake for each...

A power plant has a total peak capacity of 826MW from its seven generators. The water intake is 25m below the surface level of the lake. There are seven 7.4m diameter tunnels at the intake for each turbine. Calculate the speed of water flowing in the tunnel assuming the efficiency of generators are 50% and density of water 1000kg/m^3.

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The kinetic energy of the turbines will convert to the electric energy of generators. But the potential energy of the water at lake level will give this kinetic energy to water.

Power generated by plant `= 826xx10^6W`

Since the efficiency is 50% actual amount of potential energy in water is twice the energy generated.

Potential energy of water `= 826xx2xx10^6W`

Potential energy = mgh

`mxx9.81xx25 = 826xx2xx10^6`

`m = 6.736xx10^6(kg)/s`

The mass flow rate of the water is `6.736xx10^6(kg)/s.`

Volume flow = (mass flow)/(density)

Volume flow `= (6.736xx10^6)/10^3`

Volume flow `= 6736m^3/s`

Volume flow of one tunnel `= 6736/7m^3/s = 962.29m^3/s`

We know that;

Volume flow = Cross sectionxVelocity

`Q = AV`

`V = Q/A`

` A = pixx(7.4/2)^2 = 43m^2`

`V = 962.29/43 = 22.38m/s`

*So the velocity of the water of each tunnel is 22.38m/s.*

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