motor calculations

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fisherelectric

Senior Member
Location
Northern Va
I have a potential customer who needs to hook-up 2 pumps for a water recirculating system for doing testing on automotive engines (dynomometer). Not having much experience with motors I need help interpreting Article 430 as to ratings on over-current, branch circuit, disconnect, and motor controller ratings. Main pump is 10HP, single phase, 48 rla, code c, class F. Return pump is 2 HP, 115/230v, 21/10 amp. Both pumps are to run simultaneously off same start-stop switch. I am not sure how to size the OCP, branch circuits, and magnetic starters properly for these 2 motors. This whole exercise may be moot as the service for this space is 100 amp 3 phase,full, with a 100 a sub-panel with some minor loads. Main loads are other small automotive machines, lifts, welder, lighting and a small office AC unit (gas heat). Owner says the dynomometer will only be used after hours with no other equipment running and that the dynomometer will run for an hour to an hour and a half doing tests that take about 10 minutes each.
 
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fisherelectric

Senior Member
Location
Northern Va
The customer has purchased a used system which has single phase motors. I did suggest he go to 3phase and save himself some bucks but he wants to wait until he moves his shop to a new location in January. He needs to use what he has in the meantime.
 

bsh

Senior Member
OK
for cable size is 125% of FLC - 430.22(A)
overloads are 125% of FLC
OC protection per table 430.52
Does this help?
 

fisherelectric

Senior Member
Location
Northern Va
OC protection according to table 430.52 would be 250% of 48 amps, or 120 amps max, (with inverse time breaker) for the large motor. This is what I am not sure about. Would that be the correct size breaker for this application?
So, I'm thinking 100 or 125 amp breaker with #6 thhn to disconnect and magnetic starter w/ 60 amp heaters and I'm good for the large (48 amp fla) motor. Is this correct? It seems as though the start-up current for this motor is more than the ampacity of the !00 amp service its being fed from. I'm dubious about the whole set-up and wonder if I need to tell this customer he's got a problem.
I believe, according to Table 430.7(B), that the locked rotor amperage of this motor is ~173 amps.
 
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bsh

Senior Member
The motor overloads (430.32) will provide overload protection. The inverse time breaker you are using will provide the short circuit protection. If this panel is fed with a 100 amp breaker located in the upstream panel then that breaker may trip during motor starting. If the owner is willing to risk the installation cost you can try it and see if it works.
 

haskindm

Senior Member
Location
Maryland
Heaters should be sized at 115% of FLA per nameplate. I believe that your service is undersized for this level of motor load. It may work or it may not. It certainly is borderline. Not a safety problem, but certainly a poor design.
 

fisherelectric

Senior Member
Location
Northern Va
I agree with you all and have told the customer that with both motors starting simultaneously plus existing loads he stands a good chance of tripping his main, and that he needs to go 3 phase with the motors to insure he'll be able to use this system. Thanks for all the input from everyone. I've learned a lot from this situation.



One thing I was wondering is if there is a way to calculate the inrush current of a motor ( other than direct measurement). The large motor mentioned above, according to Table 430.7(B) has a locked rotor current of ~173 amps and I understand that the inrush current can be much higher than the locked rotor current.
 
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jtester

Senior Member
Location
Las Cruces N.M.
fisherelectric said:
One thing I was wondering is if there is a way to calculate the inrush current of a motor ( other than direct measurement). The large motor mentioned above, according to Table 430.7(B) has a locked rotor current of ~173 amps and I understand that the inrush current can be much higher than the locked rotor current.

I believe that inrush and locked rotor currents are the same.

Jim T
 

haskindm

Senior Member
Location
Maryland
Locked rotor is the amperage that the motor will pull when starting, which some people refer to as inrush. If you think about it, when current is first applied to a motor the rotor is "locked". Thus it will draw locked rotor amperage momentarily. As the rotor starts to rotate, it is no longer locked and the amperage will drop as the motor comes up to speed.
Hope that helps.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
It's not going to work on several fronts.

1) Your circuit is probably overloaded, even if "everything else is off". If you have a 100A Main, you cannot really load it to 100A (unless someone specifically ordered a 100% rated breaker, very rare). Standard breakers can only be loaded to 80% continuously. Assuming the service is 230V 3 phase, you already have 48A from the big motor, 10A from the smaller one, which leaves you only 22A for lights, outlets, air conditioning etc. I seriously doubt you can work in the dark and heat, plus I would assume a dyno has other associated loads, i.e. a computer of some sort, some instrumentation, a printer etc. Maybe you'll get lucky though.

2) The starting current is more likely to be 6 times the FLC, so in this case 288A for the large motor, but that is somewhat irrelevant anyway. Although a typical 100A 3 phase thermal-mag main breaker is going to have magnetic trips factory set for 400% of rated current (400A in this case) and therefore probably capable of holding in during a fast startup, keep in mind that when you go to energize that big motor there will almost surely be a severe voltage drop, even with nothing else on. That will cause a severe drop in starting torque and unless your control system has some method of unloading that motor, it may not accelerate to full speed. The typical "rule-of-thumb" for circuit capacity sizing to avoid a voltage drop of more than 5% when starting a motor across the line is 2.5 - 3 times the motor HP as kVA, so for that 10HP motor that means 25kVA minimum. On a 100A 3 phase 230V service you theoretically have 40kVA available, but in this case the motor is 1 phase so not all of the 3 phase kVA is available to you. You really only have 23kVA available for starting power. Were it a 3 phase motor I might say go for it, or put on a soft starter. But 1 phase motors can't have soft starters applied to them (some can, but not cap-start as this almost surely is), plus if you don't accelerate to 90% speed, the start switch never opens to take the caps off-line, so the motor never can finish accelerating. Even if you get lucky again and it does manage to accelerate, it will undoubtedly take longer and probably trip the thermal trip elements of the breaker anyway.

3) As to breaker sizing, you are correct about the MAXIMUM breaker size being 120A, but breakers don't come in that size, plus don't forget you have a 100A main ahead of it, so you will have to settle for a 100A anyway. It might still hold in for starting though, with the provisos I laid out above.

It could possibly work if he changes to 3 phase motors however. What I would do is change them now because I am convinced it won't work, but if he wants to be cheap, try it with 1 phase and see if he gets lucky. The worst case; he smokes the motors and has to replace them, then replace them with 3 phase. To keep the installation cost risk lower, size a 3 phase motor starter for the 1 phase current and use 2 of the 3 poles for the 1 phase trial. If it doesn't work and you have to go to 3 phase, you don't have to toss the motor starter and re-install, all you will need to do is change the overloads. If you buy some types of solid state OL relays, you won't even have to do that, just dial it down. Do the same for the motor leads, run all 3 but only use 2, sized for the 1 phase service, then they are just a little over sized if you go for 3 phase but you don't have to re-pull them. You will need to swap out the breaker though. You will need a 60A 3P breaker for the 3 phase starter.

Just keep the fire extinguisher handy at startup for those motors!
 

fisherelectric

Senior Member
Location
Northern Va
Thanks Jraef, thats exacatly what I was looking for. I just wanted to be able to quantify to this customer precisely why his system wouldn't work without being vague, and you've explained it plainly. I knew inrush current could be many times what locked-rotor current is and wasn't sure if that "instantaneous" current would add to the problem or not. I appreciate the clarification. I have told the customer to get in touch with the tech people at the manufacturer of this equipment and run the 3-phase option by them with this service before he purchased 3-phase pump assemblies. Also appreciate the rule of thumb info on the 2.5 to 3 times HP of motor for circuit capacity to avoid damaging voltage drop. Good to know.
Thanks again.
 
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Jraef said:
It's not going to work on several fronts.

2) The starting current is more likely to be 6 times the FLC, so in this case 288A for the large motor, but that is somewhat irrelevant anyway. Although a typical 100A 3 phase thermal-mag main breaker is going to have magnetic trips factory set for 400% of rated current (400A in this case) and therefore probably capable of holding in during a fast startup, keep in mind that when you go to energize that big motor there will almost surely be a severe voltage drop, even with nothing else on. That will cause a severe drop in starting torque and unless your control system has some method of unloading that motor, it may not accelerate to full speed. The typical "rule-of-thumb" for circuit capacity sizing to avoid a voltage drop of more than 5% when starting a motor across the line is 2.5 - 3 times the motor HP as kVA, so for that 10HP motor that means 25kVA minimum. On a 100A 3 phase 230V service you theoretically have 40kVA available, but in this case the motor is 1 phase so not all of the 3 phase kVA is available to you. You really only have 23kVA available for starting power. Were it a 3 phase motor I might say go for it, or put on a soft starter. But 1 phase motors can't have soft starters applied to them (some can, but not cap-start as this almost surely is), plus if you don't accelerate to 90% speed, the start switch never opens to take the caps off-line, so the motor never can finish accelerating. Even if you get lucky again and it does manage to accelerate, it will undoubtedly take longer and probably trip the thermal trip elements of the breaker anyway.

Assuming it's a delta service, you really only have 13.3KVA (40K/3) available for starting power, right? I know it probably doesn't matter at this point but I just thought I'd point that out.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
georgeshall said:
Assuming it's a delta service, you really only have 13.3KVA (40K/3) available for starting power, right? I know it probably doesn't matter at this point but I just thought I'd point that out.

No, it is 40K divided by the square root of 3, which is 1.732, not divided by 3.
 

eric stromberg

Senior Member
Location
Texas
Jraef said:
No, it is 40K divided by the square root of 3, which is 1.732, not divided by 3.

Another way of looking at it is that a three-phase transformer is really just three single phase transformers. A 45kVA three phase is three 15kVA transformers.

So, let's look at the current of a 45kVA transformer. Let's say that it is 208V.
You can either do it this way:
45000 / 208 / sqrt(3) = 125 Amps
...or you can do it this way:
15000 / 120 = 125 Amps

In keeping with all of this, you can do the same thing with Power. If all three legs are loaded, you can get 45kVA out of the transformer. But let's say you only have one leg loaded. The maximum you can get out of the transformer is 15kVA. If you connect up a 'single phase' 208 load across two of the legs and leave the other unloaded, you can get 30kVA out of the transformer.

So, using the number of 40kVA in the original post, connecting up two legs would come out to 27kVA.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
Well, I maybe if you were sizing transformers, but I was recalculating the motor power absorption. kVA in a single phase circuit is V x A, in a 3 phase circuit it is V x A x 1.732. So in comparison, power required in a 1 phase circuit for the same load as a 3 phase circuit would be the 3 phase power divided by 1.732. I was assuming the pf and eff. would be the same either way.
 

eric stromberg

Senior Member
Location
Texas
Jraef said:
Well, I maybe if you were sizing transformers, but I was recalculating the motor power absorption. kVA in a single phase circuit is V x A, in a 3 phase circuit it is V x A x 1.732. So in comparison, power required in a 1 phase circuit for the same load as a 3 phase circuit would be the 3 phase power divided by 1.732. I was assuming the pf and eff. would be the same either way.
You're right. I was definately looking at transformers there. Ok, let's look at some motors.

Let's compare a 5 HP motor at three phase and at single phase.
430.250 (230 Volts 3 phase) gives us 15.2 Amps
430.248 (230 Volts 1 phase) gives us 28 Amps

3 phase: Power = 230 * 15.2 * sqrt(3) = 6055 VA
1 phase: Power = 230 * 28 = 6440 VA

Three phase motors will have better performance than the same size single phase motor.
 
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