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worty [1.4K]
2 years ago
15

a pump takes water from the bottom of a large tank where the pressure is 50 psi and delivers it through a hose to a nozzle that

is 50ft above the bottom of the tank at a rate of 100 ibm/s. the water exits the nozzle into the atmosphere at a velocity of 70ft/s.if a 10hp motor is required to drive the pump which is 75% efficient, find:the friction loss in the pump
Engineering
1 answer:
Digiron [165]2 years ago
4 0

Answer:

The friction loss in the pump is 442.12 lb*ft/slug

Explanation:

the efficiency is

n=\frac{W_{shaft}-W_{loss}}{W_{shaft} } =0.75\\W_{shaft}=\frac{pump-power}{flow-rate}

pump power = 10 hp = 5500 lb*ft*s⁻¹

flow rate = 100 lbm/s = 3.11 slug*s⁻¹

W_{shaft} =\frac{5500}{3.11} =1768.48 lbft/slug

Clearing Wloss in equation of efficiency

0.75=\frac{1768.48-W_{loss} }{1768.48} \\W_{loss} =442.12lbft/slug

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