Question

In a hydroelectric power plant, 65 m3/s of water flows from an elevation of 90 m to a turbine, where electric poweris generated. The overall efficiency of the turbine–generator is84 percent. Disregarding frictional losses in piping, estimate the electric power output of this plant.

Answer 1

Electric power output of this plant is
`48.192* 10^(6)\,W`.

Step-by-step 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 (
`\dot W`), measured in watts, is determined by this expression, which is derived from definition of mechanical energy and energy efficiency:

`\dot W = \eta \cdot \rho\cdot g\cdot H \cdot \dot V` (1)

Where:

`\eta` - Energy efficiency, no unit.

`\rho` - Density, measured in kilograms per cubic meter.

`g` - Gravitational acceleration, measured in meters per square second.

`H` - Fluid column, measured in meters.

`\dot V` - Volume flow, measured in cubic meters per second.

If we know that
`\eta = 0.84`,
`\rho = 1000\,(kg)/(m^(3))`,
`g = 9.807\,(m)/(s^(2))`,
`H = 90\,m` and
`\dot V = 65\,(m^(3))/(s)`, then the estimated electric power output of this plant is:

`\dot W = (0.84)\cdot \left(1000\,(kg)/(m^(2)) \right)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot \left(90\,m\right)\cdot \left(65\,(m^(3))/(s) \right)`

`\dot W = 48.192* 10^(6)\,W`

Electric power output of this plant is
`48.192* 10^(6)\,W`.