Question

Suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 45.0 MWMW of electricity? This is a typical value for a small hydroelectric dam.

Answer 1

Answer:

$\dot m = 163877.114\,\frac{kg}{s}$

Explanation:

Let assume that the water dam has a fall of 35 meters. The required instantaneous potential energy is:

$\dot E_{g} = \frac{\dot E_{el}}{\mu_{el}}$

$\dot E_{g}= \frac{45\times 10^{6}\,W}{0.8}$

$\dot E_{g} = 56.25\cdot 10^{6}\,W$

The instantaneous potential energy has the following model:

$\dot E_{g} = \dot m \cdot g \cdot \Delta h$

The mass flow rate needed to produced the required energy production:

$\dot m = \frac{56.25\times 10^{6}\,W}{(9.807\,\frac{m}{s^{2}} )\cdot (35\,m)}$

$\dot m = 163877.114\,\frac{kg}{s}$