Chapter 12

Q. 12.22

A Kaplan turbine with specific speed 600 is suggested for a power station to generate 50 MW at 50 cycles frequency. Available head is 15 m. (i) Find out the speed and discharge through the turbine assuming that it has an overall efficiency of 90%. (ii) For a flow ratio of 0.62 and the hub to outer diameter ratio of 0.4, find the main dimensions of the turbine. (iii) For a vapour pressure of 0.25 m of water and critical Thoma factor of 0.7, find the maximum elevation of runner to avoid cavitation. (iv) Test runs of a geometrically similar model of 1/10 size were conducted in a laboratory where head available is 8 m. Find the discharge through the model, power developed and elevation of model runner above sump in the laboratory.


Verified Solution

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,



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.