Answer:
[tex]\dot m = 163877.114\,\frac{kg}{s}[/tex]
Explanation:
Let assume that the water dam has a fall of 35 meters. The required instantaneous potential energy is:
[tex]\dot E_{g} = \frac{\dot E_{el}}{\mu_{el}}[/tex]
[tex]\dot E_{g}= \frac{45\times 10^{6}\,W}{0.8}[/tex]
[tex]\dot E_{g} = 56.25\cdot 10^{6}\,W[/tex]
The instantaneous potential energy has the following model:
[tex]\dot E_{g} = \dot m \cdot g \cdot \Delta h[/tex]
The mass flow rate needed to produced the required energy production:
[tex]\dot m = \frac{56.25\times 10^{6}\,W}{(9.807\,\frac{m}{s^{2}} )\cdot (35\,m)}[/tex]
[tex]\dot m = 163877.114\,\frac{kg}{s}[/tex]
