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

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.

Question Data is a breakdown of the data given in the question above.
  • Depth of the water: 50 m
  • Water supply rate: 5000 kg/s
  • Electric power generated: 1862 kW
  • Generator efficiency: 95%
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Step 1:
In this problem, we are considering a hydraulic turbine-generator system that generates electricity from the water of a lake. We need to determine the overall efficiency, turbine efficiency, and turbine shaft power.
Step 2:
We make some assumptions for our analysis. First, we assume that the elevation of the lake remains constant. This assumption helps us simplify the analysis. Second, we neglect the mechanical energy of water at the turbine exit. This assumption allows us to focus on the energy changes within the system.
Step 3:
We start our analysis by considering the mechanical energy change of water per unit mass. Since we are taking the bottom of the lake as the reference level, the kinetic and potential energies of water are zero. The change in mechanical energy per unit mass becomes gh, where g is the acceleration due to gravity and h is the height of the water column.
Step 4:
We can calculate the change in mechanical energy per unit mass using the given values. The density of water is given as ρ = 1000 kg/m³. Plugging in the values, we find that the change in mechanical energy per unit mass is 0.491 kJ/kg.
Step 5:
Next, we calculate the rate at which mechanical energy is supplied to the turbine by the fluid. This can be determined by multiplying the mass flow rate (given as 5000 kg/s) by the change in mechanical energy per unit mass. The result is 2455 kW.
Step 6:
The overall efficiency of the system is equal to the turbine-generator efficiency. We can calculate it by dividing the electrical power output (given as 1862 kW) by the rate at which mechanical energy is supplied to the turbine by the fluid (2455 kW). The overall efficiency is found to be 0.76 or 76%.
Step 7:
We can also determine the mechanical efficiency of the turbine using the overall and generator efficiencies. The mechanical efficiency of the turbine is given by the product of the turbine efficiency and the generator efficiency. Dividing the overall efficiency (0.76) by the generator efficiency (0.95), we find that the turbine efficiency is 0.80 or 80%.
Step 8:
Finally, we calculate the shaft power output of the turbine. This can be done by multiplying the turbine efficiency (0.80) by the rate at which mechanical energy is supplied to the turbine by the fluid (2455 kW). The shaft power output is determined to be 1964 kW.
Step 9:
In summary, the lake supplies 2455 kW of mechanical energy to the turbine. The turbine converts 1964 kW of this energy to shaft work, which drives the generator. The generator then generates 1862 kW of electric power. It is important to note that there are losses associated with each component of the system.

Final 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

\begin{aligned}e_{\text {mech,in }}-e_{\text {mech,out }} &=\frac{P}{\rho}-0=g h=\left(9.81 m / s ^{2}\right)(50 m )\left(\frac{1 kJ / kg }{1000 m ^{2} / s ^{2}}\right) \\&=0.491 kJ / kg\end{aligned}

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

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

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

\eta_{\text {turbine-gen }}=\eta_{\text {turbine }} \eta_{\text {generator }} \rightarrow \eta_{\text {turbine }}=\frac{\eta_{\text {turbine }-\text { gen }}}{\eta_{\text {generator }}}=\frac{0.76}{0.95}= 0 . 8 0

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

\dot{W}_{\text {shaft,out }}=\eta_{\text {turbine }}\left|\Delta \dot{E}_{\text {mech,fluid }}\right|=(0.80)(2455 kW )=1964 kW

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.

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