Question

helppp

Nozzle Problem 3) In a hydraulic Blade Front view turbine, water exits from a jet of the blade and strikes the blades smoothly, as shown in the figure. Here, only two blades are shown. The angles of velocity vectors, before and after the blades, are shown. Determine (a) the force Wheel of blade exerted by the water on the blades (5 points), and (b) the power output of the turbine (1 point). Neglect friction on the blades. Assume V = 15 m/s, nozzle diameter = 10 cm, B. = 35°, B. =75°, R = 0.6 m, o = 10 rad/s. Top view

Answer 1

UNG u tvog Given that, V = 15 mly d = locm ß = 35° B2 = 75° R = 0.6m w = 10 rad If tri blode velocity (W) = PE DN gut U= Rw -0.6 x 10 To * Vw u = 6 m/s Vr = √ (Vw,-4) ²+ 4 - Viv, como Ag - u)+ CV, sinß )? - J(1500x35°- 6)2 + (155in 35)2 Nasy - 10,65 m/s There is no friction loss, so Vr Vr = 10,65 m 18 Now, Exit velocity triangle Star Vr₂ sing2 = V₂ Sinßz Vro CxO2 = u - V₂ Cup B2 Vyz = (v2 Sin B2)2 + (4 - V2C%B2) 2 Nr.2: Vz? + u? - Qu V, Capßz 10.652 - 2 + 6² – 2x6x cox 75 x V₂ 1, ² – 3.105 "₂ - 77.42 = 0 By solving quadratic equation VY2

V = 10.48 and ~7.38 non acceptable because velocity can't be negative 10.48 m/s ₂ = @ force Exented by water an the blade & change in momentum - m [ vw, - vw2] = PAV, [VW, Voz Mere Vi should be taken not wr, because there is series of vanes. F - Ped?) Vi (vicone, - V. Con B2] - loºx I x 0.12 x 15 [15 Guy 35° – 10.48 cup 75°] If = 1.128 kN] 6 Power output of the turbine P= fxu 1:128 X 6 1P = 6.768 kW I Am
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