Does anyone know how to calculate this at 120VAC?

- Thread starter Isaiah
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Does anyone know how to calculate this at 120VAC?

Your supplier should be able to tell you what the pull in current of that coil is. I would think 2.5 amps at 120v should be adequate.

Does anyone know how to calculate this at 120VAC?

Correct. The pull-in VA varies greatly from one mfr to the next, it depends on the magnet design, stroke length of the armature, spring strength etc. etc. If you know the make and model of the starter, that is data that can usually be looked up on-line now. So for example, on an A-B Bulletin 509 starter the coil inrush is 1490VA, on a Square D 8536 starter it is 2970VA for the same NEMA size 5. That makes a big difference in the size of CPT you need.Your supplier should be able to tell you what the pull in current of that coil is. I would think 2.5 amps at 120v should be adequate.

This is exactly what I am considering. Thanks

My notes tell me (re NEMA Heavy Duty Contactors):

Does anyone know how to calculate this at 120VAC?

1) Sealed VA = 6.7 VA

2) Inrush VA = 590 VA

3) Drop out = 60% of rated voltage

4) Rated V = 110 - 127V

5) Frequency = 50/60 Hz.

Depending on your CPT, if you have a 300 VA, that thing can possibly supply about 2,000 VA on inrush with secondary output voltage dropping to 95% of rated.

That's only valid for whatever brand of contactor you are looking at. They are all different, i.e. what I posted above.My notes tell me (re NEMA Heavy Duty Contactors):

1) Sealed VA = 6.7 VA

2) Inrush VA = 590 VA

3) Drop out = 60% of rated voltage

4) Rated V = 110 - 127V

5) Frequency = 50/60 Hz.

Depending on your CPT, if you have a 300 VA, that thing can possibly supply about 2,000 VA on inrush with secondary output voltage dropping to 95% of rated.

My notes tell me (re NEMA Heavy Duty Contactors):

1) Sealed VA = 6.7 VA

2) Inrush VA = 590 VA

3) Drop out = 60% of rated voltage

4) Rated V = 110 - 127V

5) Frequency = 50/60 Hz.

Depending on your CPT, if you have a 300 VA, that thing can possibly supply about 2,000 VA on inrush with secondary output voltage dropping to 95% of rated.

I've contacted the vendor and he has stated M coil draws 1158VA on pickup and the size 5 combo-starter comes with a 300VA CPT.

The control circuit runs from the MCC to a 'Start-Stop' push-button hand-station located nearly 1000 feet away at the motor. Standard wiring for this circuitry is #14AWG,...using an online calculator, (1158VA/120V = 9.65A) amounts to 49VD, which leaves only 71V. I feel this is excessive and believe I need an auxiliary relay as mentioned above and energize the M coil on the primary side of the CPT at 480V....

Your thoughts?

Very important concern, however, it hasn’t been addressed despite the responses.

Does anyone know how to calculate this at 120VAC?

Since you are asking how to calculate the current draw, perhaps you can go down to basics.

If doubts still persist the following will at least something to work on.

Ohm's Law is your friend.

You know the voltage. Put the VA rating aside for now, since that is what you are not sure (ie.doubtful) of its size.

Take the resistance of the magnet coil.

Ohm's Law:

E=IR

I= E/R

Value of I based on nameplate is 300 VA divided by 120 vac. = 300/120= 2.5A

Your coil resistance based on reading from your meter is X

From what are known:

R=120/I

2.5=120/R

Therefore:

R=120/2.5=

Any value lower than 48 ohms (from your meter) would require a larger transformer VA rating. You can increase the VA as a safety factor.

Easy peazzy Physics can sometimes save you from falling into a deep precipice. How’s that for a metaphor. .

Enlarging the control transformer is easier than adding those relays. . . and it also keeps the original concept.

Purists will argue:

But. we are talking about AC.

Well, you won’t be far off if you use the example above.

Great post - I will definitely keep this in mind. thanks.Very important concern, however, it hasn’t been addressed despite the responses.

Since you are asking how to calculate the current draw, perhaps you can go down to basics.

If doubts still persist the following will at least something to work on.

Ohm's Law is your friend.

You know the voltage. Put the VA rating aside for now, since that is what you are not sure (ie.doubtful) of its size.

Take the resistance of the magnet coil.

Ohm's Law:

E=IR

I= E/R

Value of I based on nameplate is 300 VA divided by 120 vac. = 300/120= 2.5A

Your coil resistance based on reading from your meter is X

From what are known:

R=120/I

2.5=120/R

Therefore:

R=120/2.5=48 Ohms

Any value lower than 48 ohms (from your meter) would require a larger transformer VA rating. You can increase the VA as a safety factor.

Easy peazzy Physics can sometimes save you from falling into a deep precipice. How’s that for a metaphor. .

Enlarging the control transformer is easier than adding those relays. . . and it also keeps the original concept.

Purists will argue:

But. we are talking about AC.

Well, you won’t be far off if you use the example above.

That is typical on irrigation wells around here. Pump panel with 480 volt coil. "Hand" operation directly switched in the 480 volt control circuit, "Auto" operation driven by the 120 volt center pivot controls to an interposing relay to switch the 480 volt pump control circuit. Sizes 3 and 4 contactors are pretty common on these wells. Maybe an occasional size 5, but mostly on older installs. New installs larger than 100 HP POCO's typically want soft start, reduced voltage starting, etc. I have only run into a few that were over 100HP.Yes, an interposing relay is the answer. If it were me though, I would change the contactor coil to 480V and use a relay with 600V rated contacts, that way the CPT is never burdened by the contactor inrush at all.

Since you are asking how to calculate the current draw, perhaps you can go down to basics.

If doubts still persist the following will at least something to work on.

Ohm's Law is your friend.

You know the voltage. Put the VA rating aside for now, since that is what you are not sure (ie.doubtful) of its size.

Take the resistance of the magnet coil.

Ohm's Law:

E=IR

I= E/R

Value of I based on nameplate is 300 VA divided by 120 vac. = 300/120= 2.5A

Your coil resistance based on reading from your meter is X

From what are known:

R=120/I

2.5=120/R

Therefore:

R=120/2.5=48 Ohms

Any valuelowerthan 48 ohms (from your meter) would require a larger transformer VA rating. You can increase the VA as a safety factor.

Easy peazzy Physics can sometimes save you from falling into a deep precipice. How’s that for a metaphor. .

Enlarging the control transformer is easier than adding those relays. . . and it also keeps the original concept.

Purists will argue:

But. we are talking about AC.

Well, you won’t be far off if you use the example above.

[/QUOTE

Myspark, did you mean to say "Any valueHIGHERthan 48 ohms (from your meter) would require a larger transformer VA rating"?

Thanks, Isaiah

Let me answer for that, if I may?

Since you are asking how to calculate the current draw, perhaps you can go down to basics.

If doubts still persist the following will at least something to work on.

Ohm's Law is your friend.

You know the voltage. Put the VA rating aside for now, since that is what you are not sure (ie.doubtful) of its size.

Take the resistance of the magnet coil.

Ohm's Law:

E=IR

I= E/R

Value of I based on nameplate is 300 VA divided by 120 vac. = 300/120= 2.5A

Your coil resistance based on reading from your meter is X

From what are known:

R=120/I

2.5=120/R

Therefore:

R=120/2.5=48 Ohms

Any valuelowerthan 48 ohms (from your meter) would require a larger transformer VA rating. You can increase the VA as a safety factor.

Easy peazzy Physics can sometimes save you from falling into a deep precipice. How’s that for a metaphor. .

Enlarging the control transformer is easier than adding those relays. . . and it also keeps the original concept.

Purists will argue:

But. we are talking about AC.

Well, you won’t be far off if you use the example above.

[/QUOTE

Myspark, did you mean to say "Any valueHIGHERthan 48 ohms (from your meter) would require a larger transformer VA rating"?

Thanks, Isaiah

myspark used the 300VA CPT you have given as the basis: Irated = 300/120V = 2.5 amps! The resulting calculations yielded a limiting resistance of 48 ohms (120V/2.5A).

.Let me answer for that, if I may?

myspark used the 300VA CPT you have given as the basis: Irated = 300/120V = 2.5 amps! The resulting calculations yielded a limiting resistance of 48 ohms (120V/2.5A).

Not entirely sure how that works, if I increase the VA the amps aregreater.400/120V = 3.3 amps Therefore the VD is greater.

Explain what you are looking for..

Not entirely sure how that works, if I increase the VA the amps aregreater.400/120V = 3.3 amps Therefore the VD is greater.

Do you want to size the control transformer adequately for the size 5 coil?

Do you want to know what the VD will be for a size 5 coil if the transformer is located xxx feet away?

I am trying to size the control circuity for a 125Hp motor with size 5 starter, (i.e. #14's, #12's OR #10's?). The pickup on the M coil is 1158VA; the circuit contains 2 LEDs and a simple start-stop control (no load basically). The space heater is fed from an external source so we're only looking at the M coil pickup and seal-in.Explain what you are looking for.

Do you want to size the control transformer adequately for the size 5 coil?

Do you want to know what the VD will be for a size 5 coil if the transformer is located xxx feet away?

Normally I would simple use Table 8 NEC resistances/calc to find the VD or an on-line calculator for ball-park estimate. The std CPT for a size 5 starter from Eaton is 300VA and I am not sure how the size of the CPT figures in the V-Drop on the control circuit.