Question 14.12: Effect of Component Efficiencies on Plant Efficiency A hydro...

We are to estimate the power production from a hydroelectric plant.
Properties   The density of water at T = 20°C is 998 kg/m³.
Analysis   The ideal power produced by one hydroturbine is

$\dot{W}_{\text {ideal }}=\rho g \dot{V} H_{\text {gross }}$

$=\left(998  kg / m ^3\right)\left(9.81  m / s ^2\right)\left(12.8  m ^3 / s \right)(325  m )$

$\times\left(\frac{1  N }{1  kg \cdot m / s ^2}\right)\left(\frac{1  W }{1  N \cdot m / s }\right)\left(\frac{1  MW }{10^6  W }\right)$
= 40.73 MW

But inefficiencies in the turbine, the generator, and the rest of the system reduce the actual electrical power output. For each turbine,

$\dot{W}_{\text {electrical }}=\dot{W}_{\text {ideal }} \eta_{\text {turbine }} \eta_{\text {generator }} \eta_{\text {other }}=(40.73  MW )(0.952)(0.945)(1  –  0.035)$
= 35.4 MW

Finally, since there are 12 turbines in parallel, the total power produced is

$\dot{W}_{\text {total  electrical }}=12  \dot{W}_{\text {electrical }}=12(35.4  MW )= 4 2 5  MW$

Discussion   A small improvement in any of the efficiencies ends up increasing the power output and it thus increases the power company’s profitability.