Question 2.65: Electric power is to be generated by installing a hydraulic ......
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. 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 is generating electricity from the water of a large reservoir. The combined turbine-generator efficiency and the turbine efficiency are to be determined.
Assumptions 1 The elevation of the reservoir remains constant. 2 The mechanical energy of water at the turbine exit is negligible
Analysis We take the free surface of 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), and thus the potential energy at points 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
Then the rate at which the mechanical energy of the fluid is supplied to the turbine become
The combined turbine-generator and the turbine efficiency are determined from their definitions,
\displaystyle \begingroup \begin{aligned} & \eta_{\text {turbine-gen }}=\frac{\dot{W}_{\text {elect,out }}}{\left\vert \Delta \dot{E}_{\text {mech,fluid}}\right\vert }=\frac{750 \mathrm{~kW}}{1031 \mathrm{~kW}}=0.727 \text{~or } \mathbf{7 2 . 7 \%} \\[15pt] & \eta_{\text {turbine }}=\frac{\dot{W}_{\text {shaft,out }}}{\left\vert \Delta \dot{E}_{\text {mech,fluid}}\right\vert }=\frac{800 \mathrm{~kW}}{1031 \mathrm{~kW}}=0.776 \text{~or } \mathbf{7 7 . 6 \%} \end{aligned} \endgroupTherefore, the reservoir supplies 1031 kW of mechanical energy to the turbine, which converts 800 kW of it to shaft work that drives the generator, which generates 750 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
