Question 12.22: A Kaplan turbine with specific speed 600 is suggested for a ......

Given: N_s=600 ; P=5 \times 10^3  kW ;  f=50  cps ;  H=15  m ; \eta_o=9 ; \Psi=0.62 ; D_{h t} / D_o=0.4 ; H_ν=p_ν / w=0.25  m ; \sigma_c=0.7 ;  \text { Scale }=1: 10 ;  H_m=8  m

(i) From the equation for specific speed,

N_s=\frac{N \sqrt{P}}{H^{5 / 4}}

Speed of turbine N=\frac{N_s H^{5 / 4}}{\sqrt{P}}=\frac{600 \times 15^{5 / 4}}{\sqrt{50 \times 10^3}}=79.21  rpm

For speed to be synchronous, N=\frac{60 f }{P}

∴          Number of poles  P=\frac{60 f }{N}=\frac{6 \times 50}{79.21}=37.87

Taking the number of poles as 38, the required speed is:

N=\frac{60 \times 50}{38}=78.95  rpm

(ii) From the equation for  \frac{\eta_1}{\eta_o}=\frac{S P}{W P}=\frac{S P}{w Q H}

Discharge  Q=\frac{S P}{w H \times \eta_o}=\frac{50 \times 10^6}{9810 \times 15 \times 0.6}=377.54  m ^3 / s

Flow velocity  V_f=\psi \sqrt{2 g H}=0.62 \times \sqrt{2 \times 9.81 \times 15} = 10.64  m/s

From the equation for discharge

Q=\frac{\pi}{4}\left(D_0^2-D_h^2\right) V_f=\frac{\pi}{4} D_0^2\left(1-0.4^2\right) V_f

Runner diameter  D_o \sqrt{\frac{4 Q}{\pi\left(1-0.4^2\right) V_f}}=7.334  m

Hub diameter  D_h=0.4 \times D_o=2.933   m

(iii) From the definition of critical Thoma factor,

\sigma_c=\frac{H_a-H_ν-H_s}{H}

0.7=\frac{10.3-0.25-H_s}{15}

Suction head H_s=-0.45  m

The negative sign suggests that the turbine must be installed below the tail race level.

(iv) For similar turbines, the three coefficients, viz; head, flow and power must be the same.

\left(\frac{H}{N^2 D^2}\right)_m=\left(\frac{H}{N^2 D^2}\right)_p

Speed of model  N_m=N_p \times\left(\frac{D_p}{D_m}\right)\left(\frac{H_m}{H_p}\right)^{1 / 2}=78.95(10)\left(\frac{8}{15}\right)^{1 / 2}=576.57    rpm

\left(\frac{Q}{N D^3}\right)_m=\left(\frac{Q}{N D^3}\right)_p

Discharge for model                          Q_m=Q_p\left(\frac{N_m}{N_p}\right)\left(\frac{D_m}{D_p}\right)^3

=377.54 \times\left(\frac{576.57}{78.95}\right) \times\left(\frac{1}{10}\right)^3=2.76  m ^3 / s

\left(\frac{P}{N^3 D^5}\right)_m=\left(\frac{P}{N^3 D^5}\right)_p

Power developed by model  P_m=P_p\left(\frac{N_m}{N_p}\right)^3\left(\frac{D_m}{D_p}\right)^5

=50 \times 10^3 \times\left(\frac{576.57}{78.95}\right)^3\left(\frac{1}{10}\right)^5=194.75   kW

For similarity in cavitation,

\begin{aligned}& \sigma_m=\sigma_p \\& \sigma_m=0.7=\frac{10.3-0.26-H_s}{8}\\&∴                  H_s=4.44  m\end{aligned}

i.e., the model runner must be tested by keeping it 4.44 m above tail race.