Question

1. In an impulse turbine of the Pelton type, the jet is turned by the bucket by 165°. The head available at the nozzle is 750 m and blade speed ratio is 0.46. Cv = 0.98. Relative velocity is reduced by 12% due to friction. Determine the hydraulic efficiency. If the flow available is 20 m3/s. What is the power potential. Assuming 5 units of equal power, determine the jet diameter and wheel diameter if d/D = 12.

Answer 1

o type - Impulse – Pellon type. ☺_Headchyle 750 m _ _blade ratio (0)= 0.16 © Co-efficient of Velocity (cv) = 0.98 ☺ bucket deflochon angle = 1650 ® flow dole 00) discharge (Q): 20m3/s What going to be find is © bydraulic elficiency efficiency n = 2 [V with wo] x U - v² VW, = Velocity of whirl at Inler (VW, = V1 (velocity of jet) TV W2 = Vr₂ - Coso-u, u=4, = U2 = speed of bucket Vr = Vria Ur2 = relative velocing.

due to gravity O firstly, we calculate. Velocity of_Jot () - Vi = Cragh ::g=9.81 - = 0.98 V 269.81*750_ acceleration V = 118.8TH m/s [v=Vwil Calculation of speed of bucket = ublade speed ratio x V 2014 = 0.46 x V 2x9.8) X750 - 4 = 55.800 m/s :.(U=U,=U2) ③ relative velocity (uri) at Inlet of turbine Vorg = 4 - =(118.874 -55.800) Vri= 63.0TH BIS 4 exit blade angle (8) 0 = 180 - angle of backed deflechon '=180 - 1650 Q = 150

due to gravity O firstly, we calculate. Velocity of_Jot () - Vi = Cragh ::g=9.81 - = 0.98 V 269.81*750_ acceleration V = 118.8TH m/s [v=Vwil Calculation of speed of bucket = ublade speed ratio x V 2014 = 0.46 x V 2x9.8) X750 - 4 = 55.800 m/s :.(U=U,=U2) ③ relative velocity (uri) at Inlet of turbine Vorg = 4 - =(118.874 -55.800) Vri= 63.0TH BIS 4 exit blade angle (8) 0 = 180 - angle of backed deflechon '=180 - 1650 Q = 150

80 Reduchon relative velocty (uri) at Inlet due to poichon Vr (friction) = 88 % vr 124. friction 1 5.88Vors 3,88 x 63.074 m/s Nr(trice) = 55.505 m/s ③ whirling velocity at outlet (WW2) VW2 = Vraffiction) Cosp U2 <( 55.505 x 603 15° - 55.800) Vw z = -2.186 m/s. by draulic efficiency: nh: 2 [V, + VWzJXU 12 3 2 118.874 + (-2.186) * 55.800 (118.874)2 2 x 116.688 x 55-800 (118.87472 X 56.800

in = 13,022.38 --(H18-87412 n = 13,022.38 14731-02 hh - 0.9215 (24) Inn = 92.15% Answor. Pelton wheel hydraulic efficiany = 92.15%. Power Potential of deurbine Powor = Workdone by set 1000 i woredone by jet on sunnor por second - Waga [ Vw, + Vw ₂J xu. Q = flow rate 100 discharge m3/3 = 20 - density of water 1000 kg/m3 W z 1000 x 20 [118-874 + (-2.186) JX 516.800 W = 180, 223, 808 Nm/s

power = 130, 223,808 4000 = 130,223.80 kw Lanswer. P Jet dia meher given jet ra discharge Q = Area ok jer x Velocity or jet . Q - "/4 xd² x vi 20 = 1/2 (a) x 118-814 02- 20 XH Ox_118.874. d =V20x4_ t go - 3.14 H8-874 373.26 d = 10. 214327 diameter of het = 0.462m = 462 mm Alsoming 5 Unit of equal pocoor. Id=5x462 d 2310 mm on 231 m.