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Anestetic [448]
3 years ago
10

A pump increases the pressure in the flow of water from 120 kPa to 400 kPa. The inlet and outlet diameters are 9 cm and 3 cm res

pectively. If the flow rate is 57 m3 /hr (neglecting losses) what is power delivered by the pump to the water? Neglect elevation changes.
Physics
1 answer:
Alina [70]3 years ago
8 0

Answer:

4.433 kW

Explanation:

Suppose the flow rate is the same at the inlet and outlet at 57 m3/hr.

Since 1 hour = 60 minutes = 3600 seconds we can calculate the flowrate in m3/s

\dot{V} = 57 / 3600 = 0.0158 m^3/s

As we are neglecting loss from friction and elevations changed, the power delivered by the pump is the product of the change in pressure and the flow rate

P = \Delta P \dot{V} = (400 - 120)*0.0158 = 4.433 kW

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Answer:

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the formula for voltage (if power is involved)

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28.6666666667

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