Question

(1) Hydraulic turbines convert the potential energy of water into shaft work, which, in turn, rotates the electric generator coupled to it in producing electric power. Pelton turbine named after Lester Allen Pelton of the USA, an impulse turbine used for high head and low discharge. In an impulse turbine all the available head of water is converted into kinetic energy in a nozzle. The water shoots out of the nozzle in a free jet into a bucket which revolves around a shaft. The free jet is at atmospheric pressure before and after striking the bucket. A Pelton wheel has to be designed to develop 6000 kW under a net head of 300 m at a speed of 500 rpm. If the diameter of the jet is 1/10 of the wheel diameter, and the speed ratio of bucket to jet = 0.45, estimate (a) the flow of water required, (b) diameter of the wheel, (c) diameter of each jet and (d) the number of jets.

Answer 1

1 Wap Power - 6000 kW - Gooo xid Rspeed = 0.45 Hnet = 300 m N = 500 rpm djet + (to) (a = 0.45 Dwheel u= bucket speed Reservoir 3oom Applying Energy equation in terms of head (einen in question) A PA = P = Atmospherie {A=0 - free surface 300 (24-76) = (NA) 300 - Vit VB = √ 29x300 VB - 269.81 x300 [VB = 76.72 m/s

B = Vjet a velocity of jet through Nozzles in velocity of bucket (coneel) u = 0.45 vjet n. ce = 0.45 * vjet ce = 0.45*76.72 u= 340524 m/s speed of wheel = u = 34.524 w formula diameter of wheel a- u= 34.524 = (TON) & - Diameter of wheel DO 34.524 x 60 -TXN D wheel - (34.524 x 60 3.14 x 500 wheel = 1.319 m / en diameter of jet = Dwheel

djet = (Owenee BE 1-319 0.1319 m 16 power No of jet Required in (Assuming No losses during Свилядіви): - Power = (eg 84) CH-) to Net Head 0 flow rate en density of water = 1000 km gh Granity & 9.81 m2 2 1000X9.81 X 300 6000 12 = 2.038 m's Each Nozzle Q = - No of jets & flow rate through Nyx (7 der ) yet

2.038 N = N; * * (0.1312) ? x 76.72 = 1.945 ^2 7 of jet Required = 2