Performance of a Hydraulic Turbine–Generator The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine–generator at a location where the depth of the water is 50 m (Fig. 2–60). Water is to be supplied at a rate of 5000 kg/s. If the electric power

Question

Performance of a Hydraulic Turbine–Generator The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine–generator at a location where the depth of the water is 50 m (Fig. 2–60). Water is to be supplied at a rate of 5000 kg/s. If the electric power generated is measured to be 1862 kW and the generator efficiency is 95 percent, determine (a) the overall efficiency of the turbine– generator, (b) the mechanical efficiency of the turbine, and (c) the shaft power supplied by the turbine to the generator.

Accepted Answer

A hydraulic turbine–generator is to generate electricity from the water of a lake. The overall efficiency, the turbine efficiency, and the turbine shaft power are to be determined.

Assumptions 1 The elevation of the lake remains constant. 2 The mechanical energy of water at the turbine exit is negligible.
Properties The density of water can be taken to be ρ = 1000 kg/m³.
Analysis (a) We take the bottom of the lake as the reference level for convenience. Then kinetic and potential energies of water are zero, and the change in its mechanical energy per unit mass becomes

[latex]\begin{aligned}e_{ ext {mech,in }}-e_{ ext {mech,out }} &=\frac{P}{ ho}-0=g h=\left(9.81 m / s ^{2} ight)(50 m )\left(\frac{1 kJ / kg }{1000 m ^{2} / s ^{2}} ight) \&=0.491 kJ / kg\end{aligned}[/latex]

Then the rate at which mechanical energy is supplied to the turbine by the fluid and the overall efficiency become

[latex]\begin{aligned}\left|\Delta \dot{E}_{ ext {mech,fluid }} ight| &=\dot{m}\left(e_{ ext {mech,in }}-e_{ ext {mech,out }} ight)=(5000 kg / s )(0.491 kJ / kg )=2455 kW \\eta_{ ext {overall }} &=\eta_{ ext {turbine-gen }}=\frac{\dot{W}_{ ext {elect,out }}}{\left|\Delta \dot{E}_{ ext {mech,fluid }} ight|}=\frac{1862 kW }{2455 kW }=0.76\end{aligned}[/latex]

(b) Knowing the overall and generator efficiencies, the mechanical efficiency of the turbine is determined from

[latex] \eta_{ ext {turbine-gen }}=\eta_{ ext {turbine }} \eta_{ ext {generator }} ightarrow \eta_{ ext {turbine }}=\frac{\eta_{ ext {turbine }- ext { gen }}}{\eta_{ ext {generator }}}=\frac{0.76}{0.95}= 0 . 8 0[/latex]

(c) The shaft power output is determined from the definition of mechanical efficiency,

[latex] \dot{W}_{ ext {shaft,out }}=\eta_{ ext {turbine }}\left|\Delta \dot{E}_{ ext {mech,fluid }} ight|=(0.80)(2455 kW )=1964 kW [/latex]

Discussion Note that the lake supplies 2455 kW of mechanical energy to the turbine, which converts 1964 kW of it to shaft work that drives the generator, which generates 1862 kW of electric power. There are losses associated with each component.

Question 2.16: Performance of a Hydraulic Turbine–Generator The water in a ...

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