Power spent to keep a spring compressed

  • #1

Summary:

How to calculate the power necessary to keep a spring compressed?

Main Question or Discussion Point

I was wondering, I have an engine that should keep a spring compressed. How can I calculate the power necessary for this?
The work is Force x Distance, as there is no distance, there is no work, so no power... But obviously to keep the spring compressed the engine will have to produce a constant force and spend energy... So what am I missing?
 

Answers and Replies

  • #2
.Scott
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You are not missing anything.
If you turn the motor off and lock it into place, it will hold the spring and not consume any power.
 
  • #3
But, for example, if the spring is kept compressed by a coil, electricity must be flowing through the coil to keep the pressure, won't it require electrical power to keep producing this constant force?
 
  • #4
.Scott
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But, for example, if the spring is kept compressed by a coil, electricity must be flowing through the coil to keep the pressure, won't it require electrical power to keep producing this constant force?
That's just an inefficient was of holding the motor in place. The motor will not move, and the power will go into heating up the motor. If the motor was using superconducting windings and other ideal components, it would hold so long as the power input was kept open (disconnected).
 
  • #5
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Got it. This question came to me after observing a spring return solenoid valve. It works with 24V and I wondered how could I calculate the necessary current to keep the spring compressed... Is there a way to calculate it?
It is extremely difficult to calculate. Much simpler to look up the data sheet for the solenoid. The data sheet should say how much power is needed to keep it energized.
 
  • #6
It is extremely difficult to calculate. Much simpler to look up the data sheet for the solenoid. The data sheet should say how much power is needed to keep it energized.
Thanks!
 
  • #8
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Electrical power going into solenoid is dependent upon force that it is required to push.
This is then dependent upon the spring's rate constant multiplied by compression amount (displacement) needed.

This is similar to how peak-n-hold fuel-injectors work. A larger burst of power is needed to initiate the opening of the injector and open the solenoid fully against spring. Then power is backed down to just enough to keep it in that position for duration of injector pulse.
 

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