# Question 12.7: 0.25 m^3/s of water under a net head of 10 m is flowing thro......

0.25 $m^3/s$ of water under a net head of 10 m is flowing through the runner of an inward flow reaction turbine. The diameter at inlet is twice that at outlet. The vane is radial at inlet and the discharge is radial at outlet. The velocity of flow is constant at inlet and outlet and is 1.8 m/s. If the wheel makes 300 rpm, find the dimensions of the runner and the guide vane angles. Take the speed ratio as 0.72.

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Given: Refer Figure 12.16 $Q=0.25 m ^3 / s ; H=10 m ; D_1=2 D_2 ; \theta=90^{\circ} \beta=90^{\circ} ; V_{f 1}=V_{f 2}=1.8 m / s ; N=300 rpm ; \rho=0.72$

Then, Absolute velocity at inlet  $V_1=\sqrt{2 g H}=\sqrt{2 \times 9.81 \times 10}=14 m / s$

The tangential velocity of blade at inlet, $u_1=\rho V_1=0.72 \times 14=10.085 m / s$

From the inlet velocity triangle,

\begin{aligned}\tan \alpha &=\frac{V_{f 1}}{u_1}=\frac{1.8}{10.085}=0.17848 \\∴ \alpha &=10.12^{\circ}\\V_{r 1}&=V_f=1.8 m / s\end{aligned}

The diameter of runner at inlet and outlet may be obtained from

\begin{aligned}u &=\frac{\pi D N}{60}\\D_1 &=u_1 \times \frac{60}{\pi N}=\frac{10.085 \times 60}{\pi \times 300}=0.642 m \\D_2 &=\frac{D_1}{2}=0.321 m\end{aligned}

Discharge through runner (assuming negligible vane thickness) is given by

$Q=\pi D_1 B_1 V_{f 1}$

∴             The width of wheel at inlet $B_1$ is:

$B_1=\frac{Q}{\pi D_1 V_{f 1}}=\frac{0.25}{\pi}\times 0.642 \times 1.8=0.689 m$

Width of wheel at outlet

$B_2=\frac{Q}{\pi D_1 V_{f 2}}=\frac{0.25}{\pi \times 0.321 \times 1.8}=0.1378 m$

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