Power spent to keep a spring compressed

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Discussion Overview

The discussion revolves around calculating the power required to keep a spring compressed using an engine or solenoid. It explores the relationship between force, work, and power in the context of maintaining a spring's compression, as well as practical considerations for solenoid operation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how to calculate the power necessary to keep a spring compressed, noting that without movement, it seems there would be no work done and thus no power required.
  • Another participant asserts that if the motor is turned off and locked in place, it can hold the spring without consuming power.
  • A participant argues that if a coil is used to keep the spring compressed, electrical power must flow to maintain the constant force, implying that energy is still required.
  • It is suggested that using a motor to hold the spring in place could be inefficient, as power may be converted to heat rather than doing useful work.
  • One participant expresses difficulty in calculating the necessary current for a solenoid valve and suggests consulting the data sheet for power requirements instead.
  • A later reply indicates that the electrical power needed for a solenoid depends on the force required to compress the spring, which is related to the spring's rate constant and the amount of compression.
  • There is a comparison made to how peak-n-hold fuel injectors operate, requiring a burst of power to initiate movement before reducing power to maintain position.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of power to maintain a spring's compression, with some asserting that power is required while others suggest it is not necessary when the motor is locked in place. The discussion remains unresolved regarding the calculation of power in practical applications.

Contextual Notes

Participants note the complexity of calculating power requirements and suggest reliance on manufacturer data sheets for solenoids, indicating limitations in deriving these values theoretically.

MatheusMazur
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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?
 
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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.
 
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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?
 
MatheusMazur said:
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).
 
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MatheusMazur said:
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.
 
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anorlunda said:
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!
 
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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