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Question 14.6: The Pelton wheel in Fig. 14–11 has a diameter of 3 m and buc......

The Pelton wheel in Fig. 14–11 has a diameter of 3 m and bucket deflection angles of 160°. If the diameter of the water jet striking the wheel is 150 mm, and the velocity of the jet is 8 m/s, determine the power developed by the wheel when it is rotating at 3 rad/s.

Fig. 14-11
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Fluid Description. We have steady flow onto the blades, and we will assume water to be an ideal fluid for which \rho_w = 1000 kg/m³.
Kinematics. The average speed of the buckets is

U = ωr = 3 rad/s(1.5 m) = 4.50 m/s

The kinematic diagram in Fig. 14–11b shows the flow of the water onto and then off each bucket. Here the relative speed of the flow onto the bucket is V_{f/cs} = 8 m/s – 4.50 m/s = 3.50 m/s.

Power. We will use Eq. 14–22, with θ = 20° and Q = VA. Thus,

\dot{W}_{turb} = T ω = \rho QV_{f/cs}U(1 + \cos θ)     (14-22)

\dot{W}_{\mathrm{turb}}\,=\,\rho_{w}Q V_{f/cs}U(1\,+\,\cos\theta)\\=(1000\;\mathrm{kg/m^{3}})[(\mathrm{8\;m/s})\pi(0.075\;\mathrm{m})^{2}](3.50\;\mathrm{m/s})(4.50\;\mathrm{m/s})(1+\cos20^{\circ})\\=4.32\,{\mathrm{kW}}

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