Question 12.6: The electric generator coupled to a Pelton turbine develops ......
The electric generator coupled to a Pelton turbine develops an output power of 12 MW. The generation efficiency of the generator is 94% and the efficiency of the Pelton turbine is 85%. The coefficient of velocity of the nozzle is 0.98. The bucket velocity to jet velocity ratio is 0.45 and the blade velocity coefficient is 0.85. If the effective head of water is 800 m and the jet deflection angle is 165°, find the force exerted on the buckets by the jet. If the jet ratio should not be less than 10, find the best synchronous speed and the corresponding mean rotor diameter for power generation at 50 Hz.
Given: Refer Figure 12.12. P=12 \times 10^6 W ; \eta_g=0.94 ; \eta_o=0.85 ; C_ν=0.98; \rho=0.45 ; K=0.85 ; H=800 m ; \phi=180-165=15^{\circ} ; D / d=10 ; f=50 Hz
Generator efficiency \eta_g=\frac{\text { Output of generator }}{\text { Shaft power of turbine }}
∴ Shaft power of turbine =\frac{12 \times 10^6}{0.94}=12765957 W =12765.96 kW
Also, Overall efficiency
\eta_o=\frac{\text{Power generated by turbine (SP)}}{\text{Power available from water jet}( WP )}=\frac{12765.96 \times 10^3}{9810 \times Q \times 800}
∴ Flow rate through turbine Q=\frac{12765.96 \times 10^3}{9810 \times 800 \times 0.85}=1.9137 m ^3 / s
Velocity of the jet V_1=C_ν \sqrt{2 g H}=0.98 \times \sqrt{2 \times 9.81 \times 800}=122.78 m / s
Flow rate Q=a \times V_1=\frac{\pi d^2 V_1}{4}
∴ Diameter of jet d=\sqrt{\frac{4 Q}{\pi V}}=\sqrt{\frac{4 \times 1.937}{\pi \times 122.78}}=0.142 m
Peripheral velocity of bucket u=\rho V_1=0.45 \times 122.78=55.25 m / s
From the inlet velocity triangle,
\begin{aligned}V_{r 1}&=V_1-u=122.78-55.25=67.53 m / s \\V_{w 1}&=V_1=122.78 m / s\end{aligned}
From the outlet velocity triangle,
\begin{aligned}& V_{r 2}=K V_{r 1}=0.85 \times 67.53=57.4 m / s \\& V_{w 2}=V_{r 2}\cos \phi-u=57.4 \cos 15-55.25=0.194 m / s\end{aligned}
Force exerted by water jet F=\rho Q\left(V_{w 1}+V_{w 2}\right)=1000 \times 1.9137 \times(122.78+0.194)= 235335.3 N = 235.335 kN
For a jet ratio of 10, the mean bucket circle diameter is:
D=10 \times d=10 \times 0.142=1.42 m
Peripheral velocity u=\frac{\pi D N}{60}=\frac{\pi \times 1.42 N}{60}
∴ Speed of the turbine N=\frac{60 \times 55.25}{\pi \times 1.42}=743.18 rpm
Frequency of generation f=\frac{N_{s y} n_p}{120}
where np is the number of poles and N_{sy} is the synchronous speed. If we take number of poles as 8, then the synchronous speed is:
N_{s y}=\frac{f \times 120}{\eta_p}=\frac{50 \times 120}{8}=750 rpm
The value is near to 743.1 rpm. So, 750 rpm may be taken as the synchronous speed. The mean rotor diameter D, corresponding to this synchronous speed is:
D=\frac{1.42 \times 743.1}{750}=1.4069 m
