Well certainly there is some core loss, but from the spec values it would appear that all but a couple watts are due to that, which seems rather excessive.

Hi,

Yes i agree something may not be right here. We need more measurements, for example phase angle and actual running temperature after say 5 hours run time.

 

No, you can't use this information to solve for the power. There is substantial core loss and the given information doesn't allow you to calculate what the core loss would be. To do so, you would need to know what grade of iron is used and what the maximum flux density is.

Why do you need to be able to calculate the power consumption rather than just accepting the specification?

Hi,

I would guess that it is because the power spec seems too high.
More measurements would help here, such as the phase angle current to voltage.
An actual temperature measurement too after a long time running, along with the physical dimensions. We could get a much better picture of what the power is and if it matches the spec.

25 watts seems high for a little relay, but then again i dont know the actual size. I have heat sinks that run much less than 25 watts and they get pretty hot already.

 

No, you can't use this information to solve for the power. There is substantial core loss and the given information doesn't allow you to calculate what the core loss would be. To do so, you would need to know what grade of iron is used and what the maximum flux density is.

Why do you need to be able to calculate the power consumption rather than just accepting the specification?

I am accepting the specification. I just wanted to understand the concept behind it. I'm sure most people in this forum understand that.

I appreciate your direct answer.

 

I am accepting the specification. I just wanted to understand the concept behind it. I'm sure most people in this forum understand that.

I appreciate your direct answer.

Hi,

Yes but we need the phase shift measurement to be able to show you how to calculate this actual power, if it is actually correct.

That's because the actual power is:
P=V*I*cos(TH)

where TH is the angle between current and voltage.

 

What's not in the spec is whether or not the measured values are at the rated pull. I would THINK that they are. This MIGHT explain the discrepancy. The plunger is pretty big too.

 

Hi,

Yes but we need the phase shift measurement to be able to show you how to calculate this actual power, if it is actually correct.

That's because the actual power is:
P=V*I*cos(TH)

where TH is the angle between current and voltage.

In post #1, the TS said "I need to do it on paper, so no tools". I take that to mean that he doesn't want to make electrical measurements of things like current and phase angle. He wants to calculate the power just from the info given in the specification.

In a post to the Projects forum last December, the TS mentioned that he is a mechanical engineering student, presumably not as well versed in electrical things as some of us who are responding to him.

What's not in the spec is whether or not the measured values are at the rated pull. I would THINK that they are. This MIGHT explain the discrepancy. The plunger is pretty big too.

The 23.5 watt and .33 amp spec is for the solenoid in the "seated" condition. As long as the pull is not enough to unseat the solenoid, the power consumption will be independent of the pull the solenoid is providing.

 

Never seen a AC solenoid quite like this with a variable stroke range from 1/8" to 1 1/4"!
A typical solenoid will often burn out if the armature is not seated.
Also I would have guessed the pull force to be greater not less at the minimum stroke due to a higher current?
This is usually the only time a AC solenoid has an advantage over its DC counter part at pull in due to the higher current causing faster pull in.
Max.

 

In post #1, the TS said "I need to do it on paper, so no tools". I take that to mean that he doesn't want to make electrical measurements of things like current and phase angle. He wants to calculate the power just from the info given in the specification.

In a post to the Projects forum last December, the TS mentioned that he is a mechanical engineering student, presumably not as well versed in electrical things as some of us who are responding to him.



The 23.5 watt and .33 amp spec is for the solenoid in the "seated" condition. As long as the pull is not enough to unseat the solenoid, the power consumption will be independent of the pull the solenoid is providing.

Hi,

Yes good point. My post was partly to try to get him to measure it (ha ha) and partly to show that the calculation can not be done without at least one more measurement.

I had a feeling that all along he was using a data sheet for all his work and might not actually have a unit on hand. I can say why i think this but then he'll know too

 

The thing about a magnetic component is the initial charge current surge required to pull in the contact. once that surge passes, the holding current required will be much less. Don't confuse momentary or DC calculations with AC. AC calculations are based on peak values, not RMS. P = I*I*R = (Ipeak * .707)ac * (Ipeak * .707)ac * R = (.33 *.707) * (.33 *.707) * R = 0.14 W (average power loss). I think the 23W spec value is way too high & wrong or you've misinterpreted the pull-in power at the instant power is applied which is a very short term power requirement. Only resistive losses determine power consumed. careful reading of the spec sheet's test conditions may clear things up.