Power Generation from a Hydroelectric Plant
Electric power is to be generated by installing a hydraulic turbine–generator at a site 70 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kg/s steadily (Fig. 2–60). If the mechanical power output of the turbine is 800 kW and the electric power generation is 750 kW, determine the turbine efficiency and the combined turbine–generator efficiency of this plant. Neglect losses in the pipes.
A hydraulic turbine-generator installed at a large reservoir is to generate electricity. The combined turbine-generator efficiency and the turbine efficiency are to be determined.
Assumptions 1 The water elevation in the reservoir remains constant. 2 The mechanical energy of water at the turbine exit is negligible.
Analysis We take the free surface of water in the reservoir to be point 1 and the turbine exit to be point 2 . We also take the turbine exit as the reference level \left(z_{2}=0\right) so that the potential energies at 1 and 2 are \mathrm{pe}_{1}=g z_{1} and \mathrm{pe}_{2}=0. The flow energy P / \rho at both points is zero since both 1 and 2 are open to the atmosphere \left(P_{1}=P_{2}=P_{\text {atm }}\right). Further, the kinetic energy at both points is zero \left(\mathrm{ke}_{1}=\mathrm{ke}_{2}=0\right) since the water at point 1 is essentially motionless, and the kinetic energy of water at turbine exit is assumed to be negligible. The potential energy of water at point 1 is
\mathrm{pe}_{1}=g z_{1}=\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(70 \mathrm{~m})\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^{2} / \mathrm{s}^{2}}\right)=0.687 \mathrm{~kJ} / \mathrm{kg}
Then the rate at which the mechanical energy of water is supplied to the turbine becomes
\begin{aligned}\mid \Delta \dot{E}_{\text {mech,fluid }}\mid &=\dot{m}\left(e_{\text {mech,in }}-e_{\text {mech,out }}\right)=\dot{m}\left(\mathrm{pe}_{1}-0\right)=\dot{m} \mathrm{pe}_{1} \\&=(1500 \mathrm{~kg} / \mathrm{s})(0.687 \mathrm{~kJ} / \mathrm{kg}) \\&=1031 \mathrm{~kW}\end{aligned}
The combined turbine-generator and the turbine efficiency are determined from their definitions to be
\begin{array}{c}\eta_{\text {turbine }-\text { gen }}=\frac{\dot{W}_{\text {elect,out }}}{\left|\Delta \dot{E}_{\text {mech,fluid }}\right|}=\frac{750 \mathrm{~kW}}{1031 \mathrm{~kW}}=0.727 \text { or } 72.7 \% \\\eta_{\text {turbine }}=\frac{\dot{W}_{\text {shaft,out }}}{\mid \dot{E}_{\text {mech,fluid }}\mid}=\frac{800 \mathrm{~kW}}{1031 \mathrm{~kW}}=0.776 \text { or } 77.6 \%\end{array}
Therefore, the reservoir supplies 1031 \mathrm{~kW} of mechanical energy to the turbine, which converts 800 \mathrm{~kW} of it to shaft work that drives the generator, which then generates 750 \mathrm{~kW} of electric power.
Discussion This problem can also be solved by taking point 1 to be at the turbine inlet and using flow energy instead of potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir.