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Jet on series of curved vanes: power and efficiency

Short answer

Inlet: Vw1=40 m/s, Vr1=25 m/s. Outlet: Vr2=25 m/s, Vw2 = u - Vr2 cos20° = -8.49 m/s. Force F = ρaV(Vw1 - Vw2) = 3809 N. Power = Fu = 57.13 kW. Efficiency = 90.9%.

A jet of water of 50 mm diameter and velocity 40 m/s strikes a series of curved vanes mounted on a wheel and moving at 15 m/s in the direction of the jet. The jet enters tangentially at the inlet tip of the vanes, and the vane tip angle at outlet is 20 degrees to the direction of motion. Assuming no friction loss, calculate the force on the vanes, the power developed and the efficiency of the wheel.

Given: d=50d=50 mm, V=V1=40V=V_1=40 m/s, u=15u=15 m/s (same at inlet and outlet), outlet vane angle ϕ=20∘\phi=20^\circ, no friction so Vr2=Vr1V_{r2}=V_{r1}.

Jet area a=π4(0.05)2=1.9635×10−3a=\frac{\pi}{4}(0.05)^2=1.9635\times10^{-3} m². Because the vanes form a series on a wheel, the entire jet flow is used: m=ρaV=78.54m=\rho a V=78.54 kg/s.

Inlet triangle

The jet enters tangentially, so Vw1=V=40V_{w1}=V=40 m/s and Vr1=V−u=25V_{r1}=V-u=25 m/s.

Outlet triangle

 Outlet (vane moves right, water leaves backwards)

   Vr2 = 25 m/s
        \
         \ 20 deg
   <------*-------------> u = 15 m/s
   Vw2 = u - Vr2 cos20  (negative = backwards)

Vr2=Vr1=25V_{r2}=V_{r1}=25 m/s. Component of Vr2V_{r2} opposite to motion: Vr2cos⁡20∘=23.49V_{r2}\cos20^\circ=23.49 m/s.

Vw2=u−Vr2cos⁡ϕ=15−23.49=−8.49 m/sV_{w2} = u - V_{r2}\cos\phi = 15 - 23.49 = -8.49\ \text{m/s}

The negative sign means the exit whirl is opposite to the direction of the vane. The change in whirl velocity is Vw1−Vw2=40+8.49=48.49V_{w1} - V_{w2} = 40 + 8.49 = 48.49 m/s.

Force, power, efficiency

F=ρaV (Vw1−Vw2)=78.54×48.49=3809 NP=Fu=3809×15=57129 W=57.13 kWJet power=12ρaV3=62.83 kWη=Pjet power=57.1362.83=90.9%\begin{aligned} F &= \rho a V\,(V_{w1}-V_{w2}) = 78.54\times48.49 = 3809\ \text{N}\\ P &= F u = 3809\times15 = 57129\ \text{W} = 57.13\ \text{kW}\\ \text{Jet power} &= \tfrac12\rho a V^3 = 62.83\ \text{kW}\\ \eta &= \frac{P}{\text{jet power}} = \frac{57.13}{62.83} = 90.9\% \end{aligned}

Check with the formula η=2u(V−u)(1+cos⁡ϕ)V2=2×15×25×(1+0.9397)1600=90.9%\eta=\dfrac{2u(V-u)(1+\cos\phi)}{V^2}=\dfrac{2\times15\times25\times(1+0.9397)}{1600}=90.9\%, which agrees.

The remaining energy is lost as the kinetic energy of the water leaving the vanes, with V2=Vw22+Vf22=12.05V_2=\sqrt{V_{w2}^2+V_{f2}^2}=12.05 m/s.

Answer: Force =3809=3809 N, power =57.13=57.13 kW, efficiency =90.9%=90.9\%.

✓ Answer checked by Claude Sonnet 5.5

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