## Q. 12.8

The pressure head and velocity at entrance to the runner of a Francis turbine are 250 m and 10 m/s. The elevation of the turbine above the tail race is 5 m. The turbine runs at 400 rpm discharging 14 $m^3/s$ of water. The diameter and width of runner at inlet is 1.8 m and 0.275 m respectively. The hydraulic and overall efficiencies are 0.975 and 0.91 respectively. Find out the guide vane angle and the inlet vane angle of the runner. Also find the power output and total head across the turbine.

## Verified Solution

Given: Refer Figure 12.17. $p_1 / W=250 m ; V_1=10 m / s ; h=5 m ; N=400 rpm; Q=14 m ^3 / s ; D_1=1.8 m ; B_1=0.275 m ; \eta_h=0.975 ; \eta_o=0.91$

$H=\frac{p_1}{w}+\frac{V_1^2}{2 g}+h=250+\frac{10^2}{2 \times 9.81}+5=260.1 m$

Power output $(S P)=\eta_0 \times w Q H$

\begin{aligned}&=0.91 \times 9810 \times 14 \times 260.1 \\&=32506.75 W=32.507 kW\end{aligned}

The tangential velocity of runner at inlet

\begin{aligned}u_1 &=\frac{\pi D_1 N}{60}\\&=\pi \times \frac{1.8 \times 400}{60}=37.7 m / s\end{aligned}

From the equation for discharge, $Q=\pi D_1 B_1 V_{f 1}$, the flow velocity at inlet is:

$V_{f 1}=\frac{Q}{\pi D_1 B_1}=\frac{14}{\pi \times 1.8 \times 0.275}=9 m / s$

The tangential velocity $u_1$ is greater than $V_1$, and hence the inlet velocity triangle is as shown in Figure 12.17. From the velocity triangle,

\begin{aligned}\sin \alpha &=\frac{V_{f 1}}{V_1}=\frac{9}{10}=0.9 \\\alpha &=64.16^{\circ}\\\tan (180-\theta) &=\frac{V_{f 1}}{\left(u_1-V_1 \cos \alpha\right)}=\frac{9}{\left(37.7-10 \cos 64.16^{\circ}\right)}=0.2699\end{aligned}

∴            \begin{aligned}180-\theta &=15.1^{\circ}\\\theta &=164.9^{\circ}.\end{aligned} 