Question 12.14: The outer diameter and hub diameter of the runner of a Kapla......

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}.