A Kaplan turbine operating under a net head of 6 m develops 2350 kW power with an overall efficiency of 88%. The draft tube of inlet diameter 2.8 m having an efficiency of 78% is to be attached to the turbine unit. From cavitation consideration, the pressure head at entry to the draft tube must not drop more than 4.5 m below atmosphere. Calculate the maximum height at which the runner may be set above the tail race.

Step-by-Step

Learn more on how do we answer questions.

Given: Refer Figure 12.34.H=6 m ; S P=2350 \times 10^3 W ; \eta_o=0.88

d_2=2.8 m ; \eta_d=0.78 ;\left(p_a-p_2\right) / w=4.5 mFrom the equation for overall efficiency, \eta_o=S P / w Q H,

Discharge

Q=\frac{S P}{W H \eta_o}=\frac{2350 \times 10^3}{9810 \times 6 \times 0.88}=45.37 m ^3 / sVelocity at the inlet of the draft tube can be calculated using, Q=\frac{\pi}{4} d_2^2 \times V_2

V_2=\frac{4 Q}{\pi d_2^2}=\frac{4 \times 45.37}{\pi \times 2.8^2}=7.37 m / s

For the draft tube,

\frac{p_2}{W}=\frac{p_a}{W}-H_s-\left[\frac{V_2^2-V_3^2}{2 g}-h_f\right]

But, \frac{p_a-p_2}{w}=4.5

\frac{V_2^2-V_3^2}{2 g}-h_f=4.5-H_s

From the equation for efficiency of draft tube,

\begin{aligned}& \eta_d=\left(\frac{V_2^2-V_3^2}{2 g}-h_f\right) / \frac{V_2^2}{2 g}=\frac{4.5-H_s}{7.37^2 / 2 \times 9.81}=0.78 \\& H_s=2.34 m\end{aligned}

i.e., Maximum height above tail race = 2.34 m.

Question: 12.22

Given: N_s=600 ; P=5 \times 10^3 kW ; f=5...

Question: 12.21

Given: P_m=5 kW ; N_m=280 rpm ; H_m=2 m ...

Question: 12.20

Given: \text { Scale ratio }=1: 10 ; P_p=1 ...

Question: 12.19

H_1=600 m ; P_1=6 \times 10^6 W ; N_1=200...

Question: 12.18

Given:Q=250 m ^3 / s ; H=18 m ; N=180 rp...

Question: 12.17

Given: d_2=2.75 m ; H=7 m ; S P=1700 \tim...

Question: 12.15

Given: Refer Figure 12.26. D_o=5 m...

Question: 12.14

Given: Refer Figure 12.25. D_o=3.4 m ; D_h...

Question: 12.13

Given: H=6 m ; S P=10 \times 10^6 W ; D_h...

Question: 12.12

Data. Refer Figure 12.21. N=275 rmm ; H=2...