Work Energy Theorem Homework: Water Pump Power Output

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SUMMARY

The discussion focuses on calculating the power output of a pump required to lift 800 kg of water from a depth of 14.0 m and eject it at a speed of 8 m/s. The work done per minute in lifting the water is calculated using the formula W = mgh, where m is mass, g is gravitational acceleration, and h is height. Additionally, the kinetic energy imparted to the water is considered, emphasizing the application of the Work Energy Theorem. The power output of the pump is determined by combining the work done against gravity and the kinetic energy of the ejected water.

PREREQUISITES
  • Understanding of the Work Energy Theorem
  • Knowledge of gravitational force calculations (mgh)
  • Familiarity with kinetic energy formulas (KE = 0.5mv²)
  • Basic integration concepts for calculating work done
NEXT STEPS
  • Calculate the total work done using both gravitational and kinetic energy methods
  • Explore the implications of varying pump speeds on power output
  • Investigate real-world applications of the Work Energy Theorem in fluid dynamics
  • Learn about efficiency calculations for pumps in hydraulic systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy concepts, as well as engineers involved in fluid dynamics and pump design.

ehabmozart
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Homework Statement



A pump is required to lift 800 kg of water (about 210 gallons)
per minute from a well 14.0 m deep and eject it with a speed
of 8ms-1 (a) How much work is done per minute in lifting the
water? (b) How much work is done in giving the water the kinetic
energy it has when ejected? (c) What must be the power output of
the pump?

Homework Equations



Work Energy theorem

The Attempt at a Solution



The book attempt was that work done = mgh... Why don't we add tension as well... Or force applied by pump?
 
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hi ehabmozart! :smile:
ehabmozart said:
The book attempt was that work done = mgh... Why don't we add tension as well... Or force applied by pump?

you can calculate the work done in two ways …

by integrating the tension T times the displacement, total ∫ T dh

or (applying the work energy theorem) by simply finding the difference in energy! :-p

(if the acceleration is 0, so that T is constant, then T = mg, and ∫ T dh = mgh :wink:)

the whole point of these work-energy exercises is to show you how to avoid using the usual work done formula …

which would be very difficult if the acceleration wasn't 0!​
 
Last edited:

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