Question 9.1.3: In the Hirakund power station, there are four Kaplan and two......
In the Hirakund power station, there are four Kaplan and two Francis turbines, as in 1985, operating under an average head of 26.5 m. The overall efficiency of the Kaplan turbine is 90% and of the Francis turbine is 88%. Both types have a mechanical efficiency of 94.4% between the turbine and the alternator. (a) The design flow rate for the Kaplan turbine is 170 m³/s and the operating speed 150 rpm. Calculate the shaft power, the electric power output, and the specific speed. (b) For Francis turbine, the specific speed is 2.25, and an electrical power rating of P_e = 24 MW.
Determine the flow rate and the rotational speed. Take g = 9.8 m/s².
(a) For Kaplan turbine:
Water power = P_w = wQH,
Shaft power = P_M = η_o × P_w = 0.9 × 9.81 × 170 × 26.5 × 10^{–3} MW = 39.734 MW
Input to alternator = P_e = η_m × P_M = 0.944 × 39.734 = 37.5 MW
ω = 2πN/60 = π × 150/30 = 15.7 rad/s
From Eq. 9.1.15
\eta ,P=f(\rho ,Q,N,D,gH) (9.1.15)
\pi_1=\omega _s=\frac{\omega \sqrt{P_M/\rho } }{(gH)^{5/4}}=\frac{15.7\times \sqrt{39.7 \times 10^6/10^3} }{(9.8 \times 26.5)^{5/4}} =3
Note Definition of specific speed is different from what is used in practice. The difference lies in that it is (i) non dimensional (ii) Watt is used instead of kW, and (iii) angular speed is used instead of revolutions per second.
(b) For Francis turbine
P_M=\frac{P_e}{\eta _m}=\frac{24}{0944} =25.42=\eta _owQH
\rightarrow Q=\frac{P_M}{\eta _owH}=\frac{25.42\times 10^6}{0.88 \times 9810 \times 26.5} =111.2 m^3/s
\pi_1=\omega _s=\sqrt{\eta _o}\times \frac{w\sqrt{Q} }{(gH)^{3/4}}
\rightarrow \omega =\frac{\omega _s(gH)^{3/4}}{\sqrt{\eta _oQ} }=\frac{2.25 \times (9.81 \times 26.45)^{3/4}}{\sqrt{0.88 \times 111.2} }=14.71 \mathrm{rad/s} =\frac{30\times 14.71 }{\pi} =140.5 \mathrm{rpm}