Pelton wheel power with penstock losses

Given: $H_g=450$ m, $L=2000$ m, $D=0.6$ m, $f=0.016$, $Q=0.5$ m³/s, $C_v=0.98$, $\phi=0.46$, $D_w=1.4$ m, deflection $165^\circ$, $k=0.95$ ($V_{r2}=kV_{r1}$), $\eta_m=0.97$.

(a) Net head

Velocity in penstock: $V_p=\dfrac{Q}{\frac{\pi}{4}D^2}=\dfrac{0.5}{0.2827}=1.768$ m/s.

$$h_f=\frac{fLV_p^2}{2gD}=\frac{0.016\times2000\times1.768^2}{2\times9.81\times0.6}=8.50\ \text{m}$$ $$H = H_g - h_f = 450 - 8.50 = 441.50\ \text{m}$$

(b) Jet diameter

$$V_1 = C_v\sqrt{2gH} = 0.98\times\sqrt{2\times9.81\times441.50} = 0.98\times93.07 = 91.21\ \text{m/s}$$ $$a=\frac{Q}{V_1}=5.482\times10^{-3}\ \text{m}^2,\qquad d=\sqrt{\frac{4a}{\pi}}=83.5\ \text{mm}$$

(c) Speed of wheel

$$u=\phi\sqrt{2gH}=0.46\times93.07=42.81\ \text{m/s},\qquad N=\frac{60u}{\pi D_w}=\frac{60\times42.81}{\pi\times1.4}=584\ \text{rpm}$$

(d) Power developed and hydraulic efficiency

Velocity triangles: $V_{w1}=V_1=91.21$ m/s; $V_{r1}=V_1-u=48.40$ m/s; $V_{r2}=0.95V_{r1}=45.98$ m/s.

The bucket turns the jet by $165^\circ$, so the relative exit velocity makes $180-165=15^\circ$ with the direction opposite to the bucket motion.

$$V_{w2}=u-V_{r2}\cos15^\circ=42.81-44.41=-1.60\ \text{m/s (backwards)}$$ $$V_{w1}+|V_{w2}|=91.21+1.60=92.81\ \text{m/s}$$ $$\begin{aligned} P&=\rho Q\,u\,(V_{w1}+|V_{w2}|)=1000\times0.5\times42.81\times92.81=1986.7\ \text{kW}\\ \text{Water power}&=\rho g Q H=2165.6\ \text{kW}\\ \eta_h&=\frac{1986.7}{2165.6}=91.7\% \end{aligned}$$

(e) Shaft power

$$P_s=\eta_m P=0.97\times1986.7=1927.1\ \text{kW}$$

Answer: (a) 441.50 m; (b) 83.5 mm; (c) 584 rpm; (d) 1986.7 kW, $\eta_h=91.7\%$; (e) 1927.1 kW.