Question

A hydroelectric power plant uses spinning turbines to transform the kinetic energy of moving water into electric energy with 84 % efficiency. That is, 84 % of the kinetic energy becomes electric energy. A small hydroelectric plant at the base of a dam generates 45 MW of electric power when the falling water has a speed of 18 m/s.

Answer 1

Answer:

m_{flow} = 330754.35[kg/s]

Explanation:

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Description

A hydroelectric power plant uses spinning turbines to transform the kinetic energy of moving water into electric energy with 84 % efficiency. That is, 84 % of the kinetic energy becomes electric energy. A small hydroelectric plant at the base of a dam generates 45 MW of electric power when the falling water has a speed of 18 m/s.

Question

What is the water flow rate −kilograms of water per second − through the turbines?Express your answer with the appropriate units.

We know that the efficiency is 84%, that is 16% of the remaining energy is lost because the mechanism of transformation from Kinetic to rotational mechanical energy, is not perfect. We must calculate the energy of the water stream taking into account this efficiency.

45 = 0.84 * Pf

Pf = 45 / 0.84

Pf = 53.57 [MW] "Kinetic energy of the water stream when impacting the turbine"

Now using the energy conservation law that tells us that the potential energy is transformed into kinetic energy, we can find the height H, with this value we can find later the volumetric flow.

$E_{k}=E_{p}\\m*g*h=0.5*m*v^2\\g*h=0.5*v^2\\9.81*h=0.5*(18)^2\\h = 16.51[m]$

The power of a turbine is calculated with the following expression:

$P_{tur}=m_{flow}*g*h\\therefore:\\m_{flow}=P_{tur}/g*h\\m_{flow}=(53.57*10^6)/(9.81*16.51)\\m_{flow}=330754.35[kg/s]$