Portable Generator Internal Circuit Breaker vs Inrush Current

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Tainted

Tainted

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I'm designing a backup system for (2) 2HP motors. Both motors have an inrush current of 54 amps.

I looked into one of generac's portable 17500W generator which says it could handle up to 26250 starting watts and 17500 running watts. The generator has a 50A internal circuit breaker which feeds a 50A outlet. I want to utilize this outlet to feed the (2) 2hp motors but not sure if the internal 50A circuit breaker could handle it due to inrush. There's no information on the time characteristic curve for that 50A circuit breaker for me to refer to. I am not sure if the starting watts applies to the internal 50A circuit breaker because it has many other addition outputs.

I want to ask the manufacturer but it's hard to get a hold of them, see cut sheet below.

https://www.generac.com/generaccorp...onal-power/gp series/0196640-a-gp17500e_1.pdf

thank you.
 
Will both motors start at the same time. If not, when the 1st motor starts and is running it can help the 2nd motor start.
 
220126-1348 EST

Motors have starting current, and do not have inrush current like transformers and incandescent bulbs.

Induction motors will have a relatively constant starting current from probably 6 cycles unloaded to more cycles loaded.

On a 1/3 HP motor I have measured at 120 V 50 A peak, and thus 35 A RMS starting current. If you have a low impedance source I might expect that at 120 V and 2 HP you might see 180 A RMS starting current for one 2 HP motor. Go to 240 and this might be 45 A RMS.

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gar said:
220126-1348 EST

Motors have starting current, and do not have inrush current like transformers and incandescent bulbs.

Induction motors will have a relatively constant starting current from probably 6 cycles unloaded to more cycles loaded.

On a 1/3 HP motor I have measured at 120 V 50 A peak, and thus 35 A RMS starting current. If you have a low impedance source I might expect that at 120 V and 2 HP you might see 180 A RMS starting current for one 2 HP motor. Go to 240 and this might be 45 A RMS.

.
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Sorry I used wrong vocabulary, the motor is 240V. I wanted to know if the internal breaker has the ability to go past 50A for a short period of time without tripping when the motors start. The starting watts of the generator is 26250 which is good but wasn't sure if it applies to the 50A internal breaker

tom baker said:
Will both motors start at the same time. If not, when the 1st motor starts and is running it can help the 2nd motor start.
Click to expand...

I would like to know both scenarios. I think they will start at the same time.
 
The starting watts given in the cutsheet doesn't mean anything without knowing the corresponding voltage drop and if that value is acceptable to your installation.
 
Usually with portable generators, the generator motor will stall down before the breaker trips on an overload. If it’s an inverter type, the inverter will trip out quicker. If it is a push button type breaker, it is probably thermal trip only, not magnetic, so the trip time will most likely be increased.
 
hillbilly1 said:
Usually with portable generators, the generator motor will stall down before the breaker trips on an overload. If it’s an inverter type, the inverter will trip out quicker. If it is a push button type breaker, it is probably thermal trip only, not magnetic, so the trip time will most likely be increased.
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I found a closeup picture of the circuit breaker
1643232406592.png

Could this handle the starting of the motors?
 
The sKVA value is determined by the alternator. It is a curve with sKVA on one axis and % voltage dip on the other. A typical acceptable value would be around 30% voltage dip. If your voltage dip is greater, you risk dropping loads due to things like the coil voltage of your motor starter dropping so low that the relay opens. This can lead to chatter and premature failure of your equipment. If you have sensitive electronics, even 30% voltage dip may be too great. You might need to design to 20%, 10%, 5%, etc., whatever your particular situation requires.

In your specific example 54A*2*240V = 25920VA. That is pretty close to the 26250W stated by your cutsheet. We also are ignoring power factor in these calculations. If the 26250W is at 30%, you are right on the edge of not being able to start both motors simultaneously. If that starting watts is at a higher voltage dip, you probably can't start both motors at the same time. If it is at a lower voltage dip, then you are probably fine.
 
You will definitely need to stage them, if possible. Do they have a control that starts them automatically? Or do they have a manual control, and they run continuously once started?
 
d0nut said:
The sKVA value is determined by the alternator. It is a curve with sKVA on one axis and % voltage dip on the other. A typical acceptable value would be around 30% voltage dip. If your voltage dip is greater, you risk dropping loads due to things like the coil voltage of your motor starter dropping so low that the relay opens. This can lead to chatter and premature failure of your equipment. If you have sensitive electronics, even 30% voltage dip may be too great. You might need to design to 20%, 10%, 5%, etc., whatever your particular situation requires.

In your specific example 54A*2*240V = 25920VA. That is pretty close to the 26250W stated by your cutsheet. We also are ignoring power factor in these calculations. If the 26250W is at 30%, you are right on the edge of not being able to start both motors simultaneously. If that starting watts is at a higher voltage dip, you probably can't start both motors at the same time. If it is at a lower voltage dip, then you are probably fine.
Click to expand...
How do I find the voltage dip based on my situation?
 
hillbilly1 said:
You will definitely need to stage them, if possible. Do they have a control that starts them automatically? Or do they have a manual control, and they run continuously once started?
Click to expand...
It’s powered via HOA controller has internal ladder logic components. Sends a signal to start motors under certain conditions
 
Tainted said:
How do I find the voltage dip based on my situation?
Click to expand...
It would have to be provided by the manufacturer. I get nervous when i see sKVA numbers without an associated voltage dip because that is only telling half the story and I worry that the manufacturer is trying to hide something or that the marketing department got their hands on the literature and wanted a flashy number to publish. That being said, if you can stagger your motor starts it is probably a moot point.
 
d0nut said:
It would have to be provided by the manufacturer. I get nervous when i see sKVA numbers without an associated voltage dip because that is only telling half the story and I worry that the manufacturer is trying to hide something or that the marketing department got their hands on the literature and wanted a flashy number to publish. That being said, if you can stagger your motor starts it is probably a moot point.
Click to expand...
You have a good point. Would the dip even matter if I’m only dedicating the generator to the 2 motors? The HOA control panel needs a 120V from neutral too. Can’t stagger, the motors are prewired to the unit by manufacturer
 
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Tainted said:
I'm designing a backup system for (2) 2HP motors. Both motors have an inrush current of 54 amps.

I looked into one of generac's portable 17500W generator which says it could handle up to 26250 starting watts and 17500 running watts. The generator has a 50A internal circuit breaker which feeds a 50A outlet. I want to utilize this outlet to feed the (2) 2hp motors but not sure if the internal 50A circuit breaker could handle it due to inrush. There's no information on the time characteristic curve for that 50A circuit breaker for me to refer to. I am not sure if the starting watts applies to the internal 50A circuit breaker because it has many other addition outputs.

I want to ask the manufacturer but it's hard to get a hold of them, see cut sheet below.

https://www.generac.com/generaccorporate/media/library/content/all-products/portable-recreational-power/gp series/0196640-a-gp17500e_1.pdf

thank you.
Click to expand...
One thing I will point out you mentioned inrush (or whatever you want to call the peak current at starting) may be 54 amps assuming voltage input remains stable. Reduced voltage soft starting methods intentionally reduce that current surge. If the source has high enough impedance, there is undersized conductors, etc. you also get some that effect that will reduce the surge current. Sounds like the starting ability of your generator may be high enough it isn't going to be too significant, but that is something to consider. a generator often has more severe voltage drop than utility power when you suddenly put a larger load on it, particularly if large enough load that prime mover temporarily slows down during that surge.
 
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