Puzzled by a tripping breaker

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rai

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Hi,

I installed a 20amp 12-2awg ckt for an air compressor for a friend of mine. (dedicated ckt)

He is telling me that when he turns on the garage lights, the breaker for the compressor trips.( only if he is using the compressor)The lights are on a completely different ckt????

Is it possible there is an unbalance in the panel?

Thanks,
Ryan
 
Re: Puzzled by a tripping breaker

My only thought is that, perhaps, the two are fed from the same panel, and that panel is operating at a reduced voltage (for reasons that you would have to discover). With the compressor running on its own, the available voltage is just enough. When you turn on that one more load, the startup current for the light (by the way, is it a fluorescent?) brings the panel voltage just a little bit too low. The lower voltage causes the current in the compressor to increase, and it?s just enough to trip the compressor breaker. The situation would be even more likely if the two are fed by the same phase, and if there is a significant imbalance (as you had suggested).

This is, of course, pure theory, and I don?t think it is likely to be the cause. But it is all I can offer. Perhaps someone else can take it to the next step?
 
Re: Puzzled by a tripping breaker

Ryan, What is the startup and full load current for the air compressor? I am unable to come up with a theory different from what Charlie mentioned :confused:
 
Re: Puzzled by a tripping breaker

Charlie B, I know you, if anybody, knows that V/I=R or V/R=I. or this:

The lower voltage causes the current in the compressor to increase
Click to expand...
doesn't comply with Ohm's law. :confused:
 
Re: Puzzled by a tripping breaker

Induction motor loads don't behave the same as a pure resistance load.
The motor's impedance consists of a mix of resistance and inductive reactance, and the current drawn depends partly on the mechanical (shaft) load, which affects the slip.
With a constant load, a reduction in supply voltage will result in an increase in current draw.

Ed

[ December 03, 2004, 06:43 AM: Message edited by: Ed MacLaren ]
 
Re: Puzzled by a tripping breaker

Originally posted by physis:
Charlie B, I know you, if anybody, knows that V/I=R or V/R=I. or this:

The lower voltage causes the current in the compressor to increase
Click to expand...
doesn't comply with Ohm's law. :D

It gets a little hard to explain but the short answer is you can not get the same 'work' from a motor with less electrical power.

Lets say that the mechanical load on the motor is consistent. A positive displacement pump like a hydraulic pump is one example of that.

We will also say that this pump requires 5 HP to spin it any less and it stops.

The motor is a 208 volt 1 phase motor drawing 31 amps to produce 5 HP

Now reduce the voltage, one of two things has to happen.

</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">The motor produces less than 5 HP and therefore stops turning as we have a mechanical load of 5 HP

    Or</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">The motor draws more current to 'make up' for the low voltage.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
You can not get something for nothing. :D

If the amount of work produced remains the same and the voltage drops the current must rise.

Other mechanical loads like a fan react differently. Less voltage, the fan spins slower (less work produced) the current may even drop.

Bob

[ December 03, 2004, 06:47 AM: Message edited by: iwire ]
Click to expand...
 
Re: Puzzled by a tripping breaker

When I was learning about motors, the instructor drove home several major points, one of which is "The key to understanding inductive loads is back EMF!"

The back EMF produced by a motor opposes the current from the source. Upon start up, the motor is getting up to speed and there is very little back EMF produced.

Consequently, the inrush current is only limited by the resistance of the windings, which is practically zero, so the current surges until the motor comes up to speed.

That is why, in a nutshell, motors have very high inrush currents.

I concur with the others, this sounds like a voltage drop problem.
 
Re: Puzzled by a tripping breaker

Originally posted by physis: ?The lower voltage causes the current in the compressor to increase? doesn't comply with Ohm's law.
Click to expand...
It is a good and reasonable observation. But not correct, alas. I like the way the others have addressed your point. Let me add one minor thing. Ohm?s Law does hold within a motor, and Bob?s statement that ?motor loads do not know Ohm?s Law? should be taken as a jest.

What you have to realize, however, is that the three variables, V, I, and R, are all, in fact, variables! V can change; I can change; R can change. Your suggestion that "a drop in V should result in a drop in I" would be true if, and only if, R is constant. In a motor, R is not constant.

An easy (though not completely accurate) way to think of it is to split the world into two classes of loads: ?Constant Impedance Loads,? and ?Constant Power Loads.? A light fits into the former category. A motor fits into the later category.

Recall that Power equals V times I. So if power is constant in a motor (until it stalls, as Bob has described), and if you reduce V, you will observe that I must go up.
 
Re: Puzzled by a tripping breaker

Here's another way of looking at it.

An induction motor actually functions much like a transformer, with the stator winding serving as the primary, and the rotor as the secondary.

Basically, as has been stated, the power in has to be equal to the power out plus the losses, like any other transformer.

Ed
 
Re: Puzzled by a tripping breaker

Originally posted by charlie b:
Ohm?s Law does hold within a motor, and Bob?s statement that ?motor loads do not know Ohm?s Law? should be taken as a jest.
Click to expand...
Thank you Charlie that is an important point. :)

I should leave the comedy to professionals. :eek:
 
Re: Puzzled by a tripping breaker

I didn't figure you would actually be wrong Charlie B. :)

Motors certainly aren't resistive.

Let me see if this sounds close to why a motor would draw more current at lower voltage.

The RPM is directly related to the AC frequency. A reduction in voltage causes the power delivered to the armature to decrease. The reduction in power causes the armeture to lag in phase in relation to the AC frequency. Depending on what type of motor, this may cause the stator coils to be energized for a larger rms under the power curve. :)

Edit: I added an extra smiley guy :)

[ December 03, 2004, 04:30 PM: Message edited by: physis ]
 
Re: Puzzled by a tripping breaker

If the amount of work produced remains the same and the voltage drops the current must rise.
Click to expand...
That's well put Bob, if the motor can do that. It agrees with Ohm's law too though. :)
 
Re: Puzzled by a tripping breaker

Motors are very interesting. I've never really looked at them as closely as I have other things. I'm sure I'll get it one day, probably soon. There was another thread recently about finding the value for a missing capacitor and it really bugs me that I have no idea how to do that. :)
 
Re: Puzzled by a tripping breaker

By: Peter D:

The back EMF produced by a motor opposes the current from the source
Click to expand...
This is slightly backwards. The inductance of the motor opposes the current not the back EMF. The opposition comes from the inflation of the magnetic field around the coils. Coils don't like the idea of current unless they have a field first. The back EMF comes from the collapse of the field if the current is removed. When the field is established it is 90? to the right of the direction of the current. When the field colapses it induces current back into the coil's conductors at an additional 90? to the right. The resulting current is 180? out of phase with the original current and hense is called back EMF.
 
Re: Puzzled by a tripping breaker

I'm sorry to keep posting things here but I keep seeing new ways to look at this.

By Charlie B.:

Recall that Power equals V times I. So if power is constant in a motor (until it stalls, as Bob has described), and if you reduce V, you will observe that I must go up.
Click to expand...
Do motors really attempt to maintane constant power?
 
Re: Puzzled by a tripping breaker

To add to what I just said. Here is how it was explained to me:
Electric induction motors have a built in governor, When the speed of the motor is reduced the the motor starts to slip and causes the poles in the armature to be out of sync with the frequency of the supplying field. this causes more power to be applied to the field to cause the rotor to catch up sort to speak. When there is a lower voltage the motor is slipping out of sync and catching up all at the same time and this is why you will see a higher current. :eek:
 
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