Kaplan turbine discharge and blade angles

Given: $P=9000$ kW, $H=20$ m, $\eta_o=0.88$, $\eta_h=0.93$, $D_o=4.0$ m, $D_b=1.8$ m, $N=120$ rpm, $V_{w2}=0$, $u_1=u_2$ (axial flow).

Discharge

$$Q=\frac{P}{\rho g H\eta_o}=\frac{9\times10^{6}}{1000\times9.81\times20\times0.88}=52.13\ \text{m}^3/\text{s}$$

Flow velocity

Annular flow area:

$$A=\frac{\pi}{4}(D_o^2-D_b^2)=\frac{\pi}{4}(4.0^2-1.8^2)=10.022\ \text{m}^2$$ $$V_{f1}=V_{f2}=\frac{Q}{A}=\frac{52.13}{10.022}=5.201\ \text{m/s}$$

Blade speed at mean diameter

$D_m=\dfrac{4.0+1.8}{2}=2.90$ m

$$u_1=u_2=\frac{\pi D_mN}{60}=\frac{\pi\times2.90\times120}{60}=18.22\ \text{m/s}$$

Whirl velocity at inlet

$$\eta_h=\frac{V_{w1}u_1}{gH}\ \Rightarrow\ V_{w1}=\frac{0.93\times9.81\times20}{18.22}=10.014\ \text{m/s}$$

Angles

Guide vane angle:

$$\tan\alpha=\frac{V_{f1}}{V_{w1}}=\frac{5.201}{10.014}\ \Rightarrow\ \alpha=27.4^\circ$$

Runner blade angle at inlet. Here $V_{w1}<u_1$, so the blade angle is obtuse:

$$\tan(180^\circ-\theta)=\frac{V_{f1}}{u_1-V_{w1}}=\frac{5.201}{8.21}\ \Rightarrow\ 180^\circ-\theta=32.4^\circ,\ \ \theta=147.6^\circ$$

Runner blade angle at outlet:

$$\tan\phi=\frac{V_{f2}}{u_2}=\frac{5.201}{18.22}\ \Rightarrow\ \phi=15.9^\circ$$

(Angles of blades measured from the direction of blade motion. Because $u$ changes from hub to tip, the blade is twisted; the angles at the other radii are found in the same way with the local $u$.)

Answer: $Q=52.13$ m³/s; $V_f=5.201$ m/s; guide vane angle $27.4^\circ$; blade angle $147.6^\circ$ at inlet and $15.9^\circ$ at outlet.