Question Oil with a specific gravity of 0.87 is being pumped from a lower reservoir to an elevated tank, as shown. The pump in the system is 78 percent efficient and is rated 185 kW. Determine the flow rate (L/s) of the oil in the pipe if the total head loss from point 1 to point 2 is 12 m of oil. Hint: Pout = Eff% * Pin (2) -Elev. 200 m TE! 200-mm-diameter pipe Pump -160-mm-diameter pipe -Elev. 150 m Oil
2JFVNC The Asker · Civil Engineering
Oil with a specific gravity of 0.87 is being pumped from a lower reservoir to an elevated tank, as shown. The pump in the system is 78 percent efficient and is rated 185 kW. Determine the flow rate (L/s) of the oil in the pipe if the total head loss from point 1 to point 2 is 12 m of oil. Hint: Pout = Eff% * Pin
Transcribed Image Text: (2) -Elev. 200 m TE! 200-mm-diameter pipe Pump -160-mm-diameter pipe -Elev. 150 m Oil
More Transcribed Image Text: (2) -Elev. 200 m TE! 200-mm-diameter pipe Pump -160-mm-diameter pipe -Elev. 150 m Oil
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Given:-{:[rho_(alpha_(1))=0.87,],[z_(1)=150m,],[d=0.2m,eta=78%],[z_(2)=200m,p=185km],[h_(L)=12m,]:}To Find:- varphisol?.{:[" Using Bemoullies eqn at (1) \& (2) "],[((P_(1))/(gamma))+((V_(1)^(2))/(2g))+z_(1)+E_(p)=((P_(2))/(gamma))+((V_(2)^(2))/(2g))+z_(2)+h_(L)],[(0)+(0)+150+E_(p)=0+(V_(2)^(2))/((2xx9.81))+200+12]:}Here, V_(2)=(varphi )/(A)=(varphi)/((pi) ... See the full answer