Power in = Power outbut what about for solenoids?

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When operating a solenoid at 12V DC and 1 amp, the input power is 12W, but efficiency varies based on core materials and design. Different core materials, such as iron versus air, affect the power required to operate the solenoid, with iron cores generally being more efficient. The solenoid only performs useful work when the plunger is in motion; once it stops, the power supplied is primarily converted to heat, which can be seen as wasted energy. However, from a mechanical perspective, maintaining the plunger's position may not be considered wasted power, as it serves a functional purpose. Understanding these dynamics is crucial for optimizing solenoid performance and energy use.
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Ok. so assume that you have a solenoid running about 12V DC and 1 amp. that's 12W. So that would suggest that this solenoid would have 12W of power when it performed tasks or needs to use that much power. But, different core materials make the solenoid more POWERful correct? (i may be misinterpreting this completely...) but how does it do this, and to what degree? (since power in = power out)
 
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It depends how efficient the solenoid is in converting the incoming power into something useful. A certain amount of the "power in" will be lost to electrically resistive loading for example, and dissipated as heat. Or the mechanics of the solenoid may use up the power by friction in the movement.

Different core materials have different reluctances which may effect how much power is needed to drive each type of material. For example it will take more power to persuade an air-cored solenoid to move than an iron-cored solenoid.
 
so then using iron as a core just technically makes it more power efficient?
 
pretty much
 
Power in = waisted power . . .
A solenoid does "usefull work" only during the fraction of a second when the plunger moves. Once the plunger stops moving - all the power you are supplying is turned into heat - waisted.

Oleh
 
Oleh Iwanusiw said:
Power in = waisted power . . .
A solenoid does "usefull work" only during the fraction of a second when the plunger moves. Once the plunger stops moving - all the power you are supplying is turned into heat - waisted.

Oleh
It may be "wasted" from an electrical point of view, from a mechanical view to keep the plunger/valve it is connected to in the desired position then that power is not really wasted.
 
FredGarvin said:
It may be "wasted" from an electrical point of view, from a mechanical view to keep the plunger/valve it is connected to in the desired position then that power is not really wasted.

Alot of people make this misconception. I can hold the plunger in place just as easily by running a pin through it to hold it in place and expend absolutely NO power whatsoever. The advantage is simply one of convenience. The power IS in fact wasted.
 
You can't always do that when the solenoid is unaccessible for whatever reason.
 
True, but the point still is that no actual work is being done.
 
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That is indeed true. I was looking at it from a broader sense, not the absolute definition. If I have a solenoid that is a power to open, so as to supply fuel to an engine that is doing work, can I really say that the power to hold the solenoid is being wasted? From the system involving the solenoid alone then I would say yes. It's just something I always get hung up on.
 
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