How Much Power Does a Pump Need to Move Water from a Deep Hole?

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SUMMARY

The discussion centers on calculating the minimum power required for an electric pump to move 1200 kg of water from a depth of 40 meters to the surface in 2 minutes. The key formulas involved include gravitational potential energy (U = mgh) and the relationship between work and power (P = W/t). The water is released at a velocity of 12 m/s at the surface, indicating that both potential and kinetic energy must be considered in the calculations. The final conclusion confirms that the initial confusion regarding the force of the pump and the constant speed of water was resolved, leading to a correct solution.

PREREQUISITES
  • Understanding of gravitational potential energy (U = mgh)
  • Knowledge of kinetic energy and its calculation
  • Familiarity with power and work concepts (P = W/t)
  • Basic principles of fluid dynamics, particularly regarding pump operation
NEXT STEPS
  • Study the principles of energy conservation in fluid systems
  • Learn about the efficiency of electric pumps and how to calculate it
  • Explore the impact of flow rate on pump power requirements
  • Investigate the effects of varying water velocities on energy calculations
USEFUL FOR

This discussion is beneficial for high school physics students, engineering students, and anyone interested in understanding the mechanics of pumps and fluid dynamics in practical applications.

quark001
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This is a high school work/power problem that I'm struggling with:

An electric pump is used to pump water from a 40m deep hole to the surface. If 1200kg of water is needed every 2 minutes and the water is released at 12 ms-1 at the surface, what should the minimum power of the pump be?

I'm not sure whether I can use Wnet=ΔK in this example. Can I assume that the water moves at constant speed? Is the force of the pump present at all times during the movement of the water, or does it just accelerate the water initially? I know that the water gains potential energy, so I would simply use U=m g h to calculate the work. But then, why is the speed of the water at the surface given?

Once I have the work, the power is easy (just divide the work by 120s, obviously).
 
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Focus on the change in energy of the 1200 kg "parcel" of water that is transported in 2 min. How much energy does it have at the start, before it is pumped to the surface? How much more energy (in all relevant forms) does it have after it reaches the surface? So, if that amount of energy was gained in 2 min, what is rate of change of energy due to the pump?
 
Thanks, I sorted it out and my solution was correct :)
 

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