Hello all,

I need to find the power consumption of an AC Solenoid, but can't seem to locate the correct way to do this. (I need to do it on paper, so no tools)
The solenoid is 240VAC @ 60Hz with a coil resistance of 26 Ohms.
When "seated," it draws 0.33A of current.

Can I use this information to solve for the power?
If so, how?

Thanks!

 

The power is a product of the current and voltage, you can also roughly find out the inductive reactance the 26ohms coil resistance will be in series with this.
Max.

 

Is this a Homework/Exam question?

 

Is this a Homework/Exam question?

No. I am trying to understand a datasheet for a Laminate solenoid.

 

The power is a product of the current and voltage, you can also roughly find out the inductive reactance the 26ohms coil resistance will be in series with this.
Max.

This is in regards to a Dormeyer 3001-M-1 solenoid.
At "seated," the datasheet says it consumes 23.5 Watts. That number isn't the product of 240 and 0.33.
Although I do have the answer, I need to understand the process behind the answer.

 

Hello,

There is a phase shift in the circuit.
The solenoid can be seen as a resistor and an inductor in series.
This page of the eBook will tell you more:
http://www.allaboutcircuits.com/tex...ent/chpt-3/series-resistor-inductor-circuits/

Bertus

 

Hello,

There is a phase shift in the circuit.
The solenoid can be seen as a resistor and an inductor in series.
This page of the eBook will tell you more:
http://www.allaboutcircuits.com/tex...ent/chpt-3/series-resistor-inductor-circuits/

Bertus

I've actually already read that page while trying to come up with the answer myself. My problem is that I don't know the inductance and cannot therefore determine the phase angle. Correct?

I'm trying to treat it as a resistor and an inductor, but am not having any luck trying to match the datasheet's power.

 

Hello,

Can you post the datasheet?

Bertus

 

Again, for reference, I'm looking at the 3001-M-1

 

Hello,

You have to deal with the power factor:
http://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/calculating-power-factor/

Bertus

 

You don't need to know or calculate the inductance, you just need the current through the solenoid resistance.
Just multiply the square of the current times the solenoid resistance.
This gives a solenoid power of 0.33A^2 * 26Ω = 2.83W.

Since it states a power 23.5W there would seem to be a very high amount of eddy current loss in the core..

 

You don't need to know or calculate the inductance, you just need the current through the solenoid resistance.
Just multiply the square of the current times the solenoid resistance.
This gives a solenoid power of 0.33A^2 * 26Ω = 2.83W.

Since it states a power 23.5W there would seem to be a very high amount of eddy current loss in the core..

With regards to an AC solenoid, you're telling me that I can ignore the reactance?

 

This is in regards to a Dormeyer 3001-M-1 solenoid.
At "seated," the datasheet says it consumes 23.5 Watts. That number isn't the product of 240 and 0.33.
Although I do have the answer, I need to understand the process behind the answer.

Hello there,

One question: Does it get hot when it is 'seated' ??

The power is calculated from the inductance, the DC resistance, and an equivalent resistance in parallel with the coil. We dont know what the equivalent resistance is without another measurement. For example, if you could measure the phase angle between the current and voltage, we could calculate the equivalent resistance and then the inductance also. That is to determine if they are right or not.
But if we assume they are right, then we might be able to calculate the missing items knowing the power is 23.5 watts like the data sheet says, but a phase angle measurement would provide more information we could use to verify or disprove this.

 

Hello there,

One question: Does it get hot when it is 'seated' ??

The power is calculated from the inductance, the DC resistance, and an equivalent resistance in parallel with the coil. We dont know what the equivalent resistance is without another measurement. For example, if you could measure the phase angle between the current and voltage, we could calculate the equivalent resistance and then the inductance also. That is to determine if they are right or not.
But if we assume they are right, then we might be able to calculate the missing items knowing the power is 23.5 watts like the data sheet says, but a phase angle measurement would provide more information we could use to verify or disprove this.

As the solenoid would be dissipating some amount of heat, I'd imagine it gets "hot" over time.
These solenoid datasheets are based on an 80 C rise in coil temp. from ambient.

 

As the solenoid would be dissipating some amount of heat, I'd imagine it gets "hot" over time.
These solenoid datasheets are based on an 80 C rise in coil temp. from ambient.

Hi,

Yes i could imagine too, but i was asking if you could feel it or better yet measure the temperature. That might help figure out if it was really dissipating 25 watts.
A single measurement is worth 1000 data sheets

 

With regards to an AC solenoid, you're telling me that I can ignore the reactance?

The reactance is only needed to determine the AC current.
Since that is given in the data sheet, than all you need is the coil resistance (also in the data sheet) to determine the power dissipated in the resistance.
But the power seems to be much higher than that calculation gives, which I don't understand.

 

The reactance is only needed to determine the AC current.
Since that is given in the data sheet, than all you need is the coil resistance (also in the data sheet) to determine the power dissipated in the resistance.
But the power seems to be much higher than that calculation gives, which I don't understand.

Hi,

Yes you do...core loss
They are probably banging the core with high flux.
That's unless they misquoted the spec of course

 

Hi,

Yes you do...core loss
They are probably banging the core with high flux.
That's unless they misquoted the spec of course

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.

 

Could the high value relate to the peak initial power when the solenoid is fully extended and has minimal inductance? Current is then 6.5A, rather than the 0.33A when seated, so the figures still don't seem consistent.

 

Hello all,

I need to find the power consumption of an AC Solenoid, but can't seem to locate the correct way to do this. (I need to do it on paper, so no tools)
The solenoid is 240VAC @ 60Hz with a coil resistance of 26 Ohms.
When "seated," it draws 0.33A of current.

Can I use this information to solve for the power?
If so, how?

Thanks!

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?