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worty [1.4K]
3 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]3 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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3 years ago
Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
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Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

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h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

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The power that pump adds to the fluid is:

\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

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