## Q. 12.14

The outer diameter and hub diameter of the runner of a Kaplan turbine are 3.4 m and 1.7 m respectively. It develops 12 MW power while working under a net head of 21 m. The guide blade angle at the extreme edge of the runner is 35°. At outlet the whirl is zero. The hydraulic efficiency of the turbine is 90% and overall efficiency is 85%. Find out the inlet and outlet vane angles of the runner near the outer periphery.

## Verified Solution

Given: Refer Figure 12.25. $D_o=3.4 m ; D_h=1.7 m ; S P=12 \times 10^6 W$;$H=21 m ; \alpha=35^{\circ} ; \eta_h=0.9 ; \eta_o=0.85$

From the equation for overall efficiency,

$\eta_o=\frac{S P}{W P}=\frac{S P}{W Q H}=\frac{12 \times 10^6}{9810 \times Q \times 21}=0.85$

∴      Discharge $Q=68.53 m ^3 / s$

Also, discharge is given by

$Q=\frac{\pi}{4}\left(D_0^2-D_h^2\right) \times V_{f 1}$

$68.53=\frac{\pi \times\left(3.4^2-1.7^2\right) \times V_{f 1}}{4}$

∴            Flow velocity $V_f$ = 10.06  m/s

From the inlet velocity triangle (at the extreme edge)

$V_{w 1}=\frac{V_{f 1}}{\tan \alpha}=\frac{10.06}{\tan 35^{\circ}}=14.37 m / s$

From the equation for hydraulic efficiency,

\begin{aligned}\eta_h &=\frac{V_{w 1}u_1}{g H}\\u_1 &=\frac{\eta_h \times g H}{V_{w 1}}=\frac{0.9 \times 9.81 \times 21}{14.37}=12.9 m / s\end{aligned}

Also,  $\tan \theta=\frac{V_{f 1}}{V_{w 1}-u_1}=\frac{10.06}{14.37-12.9}=6.844$

or      $\theta=81.7^{\circ}$

In the outlet velocity triangle,  $V_{f 2}=V_{f 1} \text { and } u_2=u_1$

$\tan \phi=\frac{V_{f 2}}{u_2}=\frac{10.06}{12.9}=0.7798$

or                                    $\phi=37.95^{\circ}$.