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Answer:
Electric power output of this plant is [tex]48.192\times 10^{6}\,W[/tex].
Explanation:
From First Law of Thermodynamics we understand that hydroelectric power plant transforms mechanical energy from fluid into electric energy. The power output of this plant ([tex]\dot W[/tex]), measured in watts, is determined by this expression, which is derived from definition of mechanical energy and energy efficiency:
[tex]\dot W = \eta \cdot \rho\cdot g\cdot H \cdot \dot V[/tex] (1)
Where:
[tex]\eta[/tex] - Energy efficiency, no unit.
[tex]\rho[/tex] - Density, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]H[/tex] - Fluid column, measured in meters.
[tex]\dot V[/tex] - Volume flow, measured in cubic meters per second.
If we know that [tex]\eta = 0.84[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]H = 90\,m[/tex] and [tex]\dot V = 65\,\frac{m^{3}}{s}[/tex], then the estimated electric power output of this plant is:
[tex]\dot W = (0.84)\cdot \left(1000\,\frac{kg}{m^{2}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \left(90\,m\right)\cdot \left(65\,\frac{m^{3}}{s} \right)[/tex]
[tex]\dot W = 48.192\times 10^{6}\,W[/tex]
Electric power output of this plant is [tex]48.192\times 10^{6}\,W[/tex].