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  $B_1=\frac{Q}{\pi D_1 V_{f 1}}=\frac{2.3524}{\pi \times 0.371 \times 13.31}=0.1516  m$