Francis turbine velocity triangles and power

Given: $H=45$ m, $N=400$ rpm, $D_1=1.0$ m, $D_2=0.5$ m, $B_1=0.12$ m, $\eta_h=0.93$, $V_{w2}=0$, $V_{f1}=V_{f2}=7$ m/s.

Blade speeds

$$u_1=\frac{\pi D_1N}{60}=\frac{\pi\times1.0\times400}{60}=20.94\ \text{m/s},\qquad u_2=\frac{\pi\times0.5\times400}{60}=10.47\ \text{m/s}$$

Inlet whirl velocity

For $V_{w2}=0$: $\eta_h=\dfrac{V_{w1}u_1}{gH}$

$$V_{w1}=\frac{0.93\times9.81\times45}{20.94}=19.60\ \text{m/s}$$

(a) Guide vane angle

$$\tan\alpha=\frac{V_{f1}}{V_{w1}}=\frac{7}{19.60}\ \Rightarrow\ \alpha=19.7^\circ$$

(b) Runner vane angles

Since $V_{w1}<u_1$, the whirl component of the relative velocity ($V_{w1}-u_1$) is opposite to $u_1$, so the blade angle $\theta$ (measured from the direction of $u_1$) is obtuse, greater than 90 degrees.

$$\tan(180^\circ-\theta)=\frac{V_{f1}}{u_1-V_{w1}}=\frac{7}{1.34}\ \Rightarrow\ 180^\circ-\theta=79.1^\circ,\ \ \theta=100.9^\circ$$

At outlet ($V_{w2}=0$):

$$\tan\phi=\frac{V_{f2}}{u_2}=\frac{7}{10.47}\ \Rightarrow\ \phi=33.8^\circ$$

(c) Discharge

$$Q=\pi D_1B_1V_{f1}=\pi\times1.0\times0.12\times7=2.639\ \text{m}^3/\text{s}$$

(d) Width at outlet

Continuity: $\pi D_1B_1V_{f1}=\pi D_2B_2V_{f2}$, with $V_{f1}=V_{f2}$:

$$B_2=\frac{D_1B_1}{D_2}=\frac{1.0\times0.12}{0.5}=0.24\ \text{m}=240\ \text{mm}$$

(e) Power developed by runner

$$P=\rho Q\,V_{w1}u_1=1000\times2.639\times19.60\times20.94=1083.4\ \text{kW}$$

Check: water power $\rho gQH=1165.0$ kW, and $1083.4/1165.0=93.0\%=\eta_h$ (satisfied).

Answer: (a) $\alpha=19.7^\circ$; (b) $\theta=100.9^\circ$, $\phi=33.8^\circ$; (c) $2.639$ m³/s; (d) 0.24 m; (e) 1083.4 kW.