Given: Refer Figure 12.18. N = 200 rpm; D_1=1  m ; D_2=0.5  m ; \alpha=10^{\circ} ; \beta=90^{\circ} (radial discharge); V_{f 1}=V_{f 2}=2  m / s ; B_1=0.25  m

Peripheral velocity at inlet and outlet of the turbine blade are:

\begin{aligned}& u_1=\frac{\pi D_1 N}{60}=\frac{\pi \times 1 \times 200}{60}=10.472  m / s \\& u_2=\frac{\pi D_2 N}{60}=\frac{\pi \times 0.5 \times 200}{60}=5.234  m / s\end{aligned}

From the inlet velocity triangle, Absolute velocity at inlet

V_1=\frac{V_{f 1}}{\sin \alpha}=\frac{2}{\sin 10^{\circ}}=11.52  m / s

Velocity of whirl at inlet

V_{w l }=V_1 \cos \alpha=11.52 \times \cos 10^{\circ}=11.34  m / s

Relative velocity at inlet

\begin{aligned}V_{r 1}&=\sqrt{\left(V_{w 1}-u_1\right)^2+V_{f 1}^2}\\&=\sqrt{(11.34-10.472)^2+2^2}=2.18  m / s\end{aligned}

Also,  \tan \theta=\frac{V_{f 1}}{V_{w 1}-u_1}=\frac{2}{11.34-10.472}=2.304

or                          \theta=66.54^{\circ}

From the outlet velocity triangle,

\begin{aligned}\tan \phi &=\frac{V_{f 2}}{u_2}=\frac{2}{5.234-0.3821}\\\phi &=20.9^{\circ}\\V_2 &=V_{f 2}=2  m / s \\V_{w 2}&=0\end{aligned}

Flow rate through the turbine (neglecting blade thickness),

Q=\pi D_1 B_1 V_A=\pi \times 1 \times 0.25 \times 2=1.571  m ^3 / s

Width of runner at outlet  B_2=\frac{Q}{\pi D_2 V_{f 2}}=\frac{1.571}{\pi} \times 0.5 \times 2=0.5  m

Head of water at inlet to the turbine H

= Head converted into work + Kinetic head at exit = \frac{V_{w 1} u_1+V_{w 2} u_2}{g}+\frac{V_2^2}{2 g}

=\frac{(11.34 \times 10.472+0)}{9.81}+\frac{2^2}{2 \times 9.81}=12.31  m

Power developed

\begin{aligned}&=\rho Q\left(V_{w 1}u_1 \pm V_{w 2}u_2\right)=1000 \times 1.571 \times(11.34 \times 10.442+0) \\&=186560 W=186.56  kW\end{aligned}

Hydraulic efficiency

\begin{aligned}\eta_h &=\frac{\text{Power developed by runner}}{\text{Water power}}\\&=\frac{186.56}{9810 \times 1.571 \times 12.13}=0.983=98.3 \% .\end{aligned}