Question 12.16: A Kaplan turbine operating under a net head of 6 m develops ......

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.

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

$d_2=2.8 m ; \eta_d=0.78 ;\left(p_a-p_2\right) / w=4.5 m$

From the equation for overall efficiency, $\eta_o=S P / w Q H$,

Discharge

$Q=\frac{S P}{W H \eta_o}=\frac{2350 \times 10^3}{9810 \times 6 \times 0.88}=45.37 m ^3 / s$

Velocity 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

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