Question 12.9: The runner of a reaction turbine running at 200 rpm has inte......

The runner of a reaction turbine running at 200 rpm has internal and external diameters 0.5 m and 1 m respectively. The flow is inward with an inlet guide angle of 10° to the tangent to the wheel. The discharge at outlet is radial. The flow velocity at inlet and outlet are the same and is 2 m/s. The width of the turbine at inlet is 0.25 m. Find the runner blade angles and width of the runner at outlet. Also estimate the power developed by the runner and its hydraulic efficiency.

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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}

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