Question

Question 2 (5 points) A Pelton wheel has a head of 100 m and head lost due to friction in the penstock is 20 m. The main bucket speed is 15 m/s and the nozzle discharge is 1 ms. If the deflection angle is 160 and - 0.98, find the power of Pelton wheel and hydraulic efficiency

Answer 1

Solution - 2 Given as: - 100m ht = 2om. CV = 0.98 u=4,=42= 15 misec. Iz 1 m/sec. $ = 180–160 = 20° 1 p = ? nh= ? Angle of Od Deflection (160) lub VAL A vwi -0, 0-0 figca) (velocity triangles of Pelton wheel twiline) : Net head (H) = Glasshead (Hg) - Head lest due to friction (ht) .. H = loo -20 = 90 m. [Hz 80 m. | velocity of jet, (v) = CV J 2GH Y= 0.98 J 2x 9.8 1x 80 Vi = 38.83 m)s.

from velocity triangle- LV, - VW, = 38.83 m)s.] V = 4, +ve 38-83 = 15 + vsk, =) Vj = 38.83-15 JVA, = 23.83 mis. from vri= ve = 23.83 misec. outlet velocity triangle - - cwq = "₂ tuwa 1 Vwz= le cose - 42 Xw2 = 23.83 CuS 20 - 15 1 YW₂ = 7. 39 mis. Hydraulic efficiency (24) = 24 [vw, twa] yz 24 = 2x15 [ 38-83+7:39] (3.0.83)2 24 = 0.92 [RH = 92-4. Ang.

- Hydraulic efficiency (NH) Power developed by the tuuline (S.P) water Power (W.P) weater power (W.P) = WQH = Pg OH w.p = nu = put in es. 1000 x 9.81 x 1 x 80 0.92 - I s.p. = nHX w.p. | S.p. = = 0.92 x 1000 X 9.% 1 1x 80 72 2016 Nm) sec oz J/s. Or watia 72 2016 1000 | S.P = 722.016 kW. Ty