Question 12.10: A Francis turbine working under a head of 10 m develops 180 ......

Given: Refer Figure 12.19.H=10  m ; S P=18 \times 10^3  W ; \eta_0=0.78;N=180  rpm ; h_L=0.2  H ; u=0.25 \sqrt{2 g H} ; \quad V_f=0.95 \sqrt{2 g H} ; \quad \beta=90^{\circ}

From the given data,

Tangential velocity of runner at inlet u_1=0.25 \sqrt{2 g \times 10}=3.5   m / s

Flow velocity at inlet

V_{f 1}=0.95 \sqrt{2 g \times 10}=13.31  m / s

The hydraulic losses are 20% of the total head. Therefore,

Hydraulic efficiency  \eta_h=\frac{H-0.2 H}{H}=0.8

From the equation for hydraulic efficiency (with radial discharge),

\eta_h=\frac{V_{w 1} u_1}{g H}

Whirl velocity at inlet

V_{w 1}=\frac{0.8 \times 9.81 \times 10}{3.5}=22.42  m / s

From the inlet velocity triangle,

\tan \alpha=\frac{V_{f 1}}{V_{w 1}}=\frac{13.31}{22.42}=0.5937

Guide blade angle α = 30.7°

Also  \tan \theta=\frac{V_{f 1}}{V_{w 1}-u_1}=\frac{13.31}{(22.42-3.5)}=0.7035

Vane angle at inlet \theta=35.13^{\circ}

From the equation for peripheral velocity at inlet, u_1=\frac{\pi D_1 N}{60}

Diameter of wheel at inlet  D_1=\frac{3.5 \times 60}{\pi \times 180}=0.371  m

From the equation for overall efficiency,

\eta_o=\frac{\text { Shaft power }}{\text { Water power }}=\frac{S P}{w Q H}=\frac{180 \times 10^3}{9810 \times Q \times 10}

∴                        Discharge  Q=\frac{180 \times 10^3}{9810 \times 10 \times 0.78}=2.3524  m ^3 / s

Also from the equation,  Q=\pi D_1 B_1 \times V_f

Width of runner at inlet  B_1=\frac{Q}{\pi D_1 V_{f 1}}=\frac{2.3524}{\pi \times 0.371 \times 13.31}=0.1516  m