The velocity of steam, leaving the nozzle of an impulse turbine, is 1200 m/s and the nozzle angle is 20º. The blade velocity is 375m/s and the blade velocity coefficient is 0.75. Assuming no loss due to shock at inlet, calculate for a mass flow of 0.5 kg/s and symmetrical blading: (a) blade inlet angle; (b) driving force on the wheel; (c) axial thrust on the wheel; and (d) power developed by the turbine.
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Explanation:
The velocity of steam, leaving the nozzle of an impulse turbine, is 1200 m/s and the nozzle angle is 20º. The blade velocity is 375m/s and the blade velocity coefficient is 0.75. Assuming no loss due to shock at inlet, calculate for a mass flow of 0.5 kg/s and symmetrical blading: (a) blade inlet angle; (b) driving force on the wheel; (c) axial thrust on the wheel; and (d) power developed by the turbine.
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Answer:
Vw=1127.68ms
Vf=410.42ms
theeta=28.60,VR=857.30ms
Vr1=642.975ms
because aur blade is symmitrical then theeta =phy
phy =28.60
Vw1=189.52ms
vf1=307.78ms
beeta=25.57<90
power = 0.5(1127.68+189.52)×375=247.066kw
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