Question

A hydroelectric power plant uses spinning turbines to transform the kinetic energy of moving water into
electric energy with 80 % efficiency. That is, 80 % of the kinetic energy becomes electric energy. A small
hydroelectric plant at the base of a dam generates 50 MW of electric power when the falling water has a
speed of 18 m/s.
What is the water flow rate -kilograms of water per second-through turbines?

Answer 1

Solution.

$P=50\sdot10^6W;$

$\upsilon=18m/s;$

$\eta=80$ %;

$P=\dfrac{W}{t};\implies W=Pt;$

Since 80% of kinetic energy becomes electrical, we have:

$\eta=\dfrac{W}{E_k}\sdot100$ %; $\implies E_k=\dfrac{W}{\eta}\sdot100$ %;

Kinetic energy is calculated by the formula:

$E_k=\dfrac{m\upsilon^2}{2}; \dfrac{m\upsilon^2}{2}=\dfrac{Pt}{ \eta}\sdot100$ %;$\implies$

Water consumption per second is equal to:

$\dfrac{m}{t}=\dfrac{2P}{\eta\upsilon^2}\sdot100$ %;

$\dfrac{m}{t}=\dfrac{2\sdot50\sdot10^6W}{80\%\sdot(18m/s)^2}\sdot100\%=3.86\sdot10^5kg/s;$

Answer:$\dfrac{m}{t}=3.86\sdot10^5kg/s;$