inductive vs resistive loads

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Re: inductive vs resistive loads

I do have The American Electrician's Handbook and the Rosenburg Electric Motor Repair.

I also looked at several sites on the net... and a typical comment from the books and the sites is this:

"Low voltage will result in the motor drawing higher current to deliver the same horsepower " (italics are mine)

Okay, well and good, I agree. But what kind of driven load would keep a constant horsepower load on a motor as it slows down due to low voltage?

Would a fan? Seems to me that lower voltage would result in more slip = lower RPM and less horsepower output, not a higher current draw.

Now perhaps I am being asinine, but what about the case where the voltage = zero? Hmm... lower voltage causes a higher current?

Obviously there is more to the situation than a blanket statement that says "if you lower the voltage, the current will increase."

This is sort of like saying "electricity always takes the least resistive path to ground."
 
Re: inductive vs resistive loads

Yes, it is almost always true, emphasis on almost. There are exceptions to everything. A motor is a mechanical device. Whatever load it is carrying, Note: not necessarily full or rated load, even unloaded (mechanically). That device wants to do work. Use power, at whatever it is drawing at it's designed parameter's. Power basically is I X E (I know, I know, throw in a square root of 3 or other crazy engineering constant, but those are constant.). Again the motor wants to draw power, whatever it need's. Voltage goes down, current goes up.

Now we can get into core saturation and all that other good stuff also. You will find if you put a motor on a test bench and you raise the voltage from 0 then as the voltage goes up and once the motor is running, the current will go down. Until the core is saturated, then the current will take off and increase at an amazing fast rate. This is due to magnetic saturation of the core and basically, again, no more extra generated CEMF.

You used fan's as an example before. Must be careful with fan's. Something else that happens as the voltage goes down and current goes up, you will notice that the motor will slow down just a bit (yes it does for everyone that just looks at the formulas and a perfect world). A fan loading changes dramatically with speed changes, just look at most fan curves.

Now again what the motor draws depends on what it needs. Just like a little bitty motor trying to turn a big load. It's going to draw the current it needs to try and turn that bad boy until the wires let the smoke out as they say, unless of course the electrician has properly installed overload protection. But if you check all that everything is rated to be able to handle starting current which could be 7 to 10 times rated FLA. So the motor will get to give a good shot at trying to run.
 
Re: inductive vs resistive loads

I have heard all my career that with a motor, "if the voltage starts decreasing, the current will start increasing." I am certain this is not always true
Click to expand...
You are correct, it is not true with all types of motors. It happens only with induction and synchronous motors.

I tried to explain this earlier today, in my 07:17 AM post, but apparently didn't do a very good job. :)

Ed
 
Re: inductive vs resistive loads

Also crossman, electricity does not always take the least resistive path. It prefers it and is used as a parallel path. Just most of the current will go that way, there is still a high potential for current to travel in other path's that's why people get zapped even when things are grounded.
 
Re: inductive vs resistive loads

The windings of a 120V motor read a resistance of 3 Ohms by ohms law the current should have been 40A but the name plate said FLA 2.4A.
Click to expand...
I'm gonna use this motor to calculate it's inductance at two different voltages and constant power.

V/I=Z
120?2.4=50

The impedance at FLA is Z=50 Ohms
The DC resistance is R=3 Ohms

50?=2500
2500-3?=2491
√2491=49.91

The inductive reactance at FLA is XL=49.91 Ohms

L=XL?2Pi?
L=49.91?(2*3.14*60)
.132=49.91?376.8

At 120 volts and 288 watts the inductance of the motor at FLA is L=.132 Henrys
_________________________________________________

Now at 100 volts and 288 watts.

V/I=Z
100?2.88=34.72

The impedance is Z=34.72 Ohms
The DC resistance is R=3 Ohms

34.72?=1205.5
1205.5-3?=1196.5
√1196.5=34.6

The inductive reactance is XL=34.6 Ohms

L=XL?2Pi?
L=34.6?(2*3.14*60)
.092=34.6?376.8

At 100 volts and 288 watts the inductance of the motor is L=.092 Henrys


So if there's nothing wrong with the way I did this, then in order for a motor to maintain constant power with lower voltage it's inductance will have to decrease.
 
Re: inductive vs resistive loads

Sorry physis but there lies the problem. Formulas don't work it all out. The inductance of the coil in the motor is going to change, as the speed changes and as the current changes. When you turn on the motor and apply voltage you have a coil, actually several coils. The motor as a unit wants to turn. If it does not or has to large of a mechanical load that keeps it from turning within it's design, the current will go up due to the reduced CEMF in the coils. Said CEMF is what effects the inductance of the coil. Yes you can figure out the inductance at a point in time but the change in inductance does not affect the current the changing current affects the inductance.
 
Re: inductive vs resistive loads

change in inductance does not affect the current the changing current affects the inductance.
Click to expand...
Well, both of those things happen interdependently. If you just look at the values in the circuit and ignore the fact that it's a motor.

I see:

Resistance. [I'm assuming that to be constant]
Inductance.
Current.
Voltage.
Power. [I'm assuming that to be constant]

So the only way to lower voltage and maintain power is by increasing current. And the only way to increase current is to reduce impedance. I'm making the assumption that the resistance isn't capable of dropping so that only leaves inductance as a possibility for decreasing impedance.
 
Re: inductive vs resistive loads

Yes they are happening at the same time. Just thing that is hard to realize is that the motor as a unit wants to work and will draw actually whatever it needs to meet it's design. (sorry but it's not like some people that just draw and don't do anything). Just like a text book short circuit. Current goes to infinite (but lets not start that discussion again) and voltage goes to zero. (but how can you have infinite current with no volts :p )
 
Re: inductive vs resistive loads

Originally posted by 69boss302:
Also crossman, electricity does not always take the least resistive path. It prefers it and is used as a parallel path.
Click to expand...
Boss, I realize that, and am using it as an "old wive's tale" that is wrong, yet is believed by many electricians and is handed down through the years. (I will have to throw in that electricity doesn't want to "go to ground" it actually wants to return to the other side of its source)

When a blanket statement of "lower voltage = higher current" is thrown out there, well, it is my opinion that it is another "old wive's tale". It may be true sometimes, but not all of the time.

I will do some experimentation tomorrow... and report back.
 
Re: inductive vs resistive loads

And I will state this again:

Today, I took an induction motor that only had a frictional load, reduced the voltage from 120V all the way down to 50V, and the current went down. Obviously the horsepower output went down too.

So a statement of "a motor always wants to produce its rated output horsepower" is another statement that is wrong.

I will perform some more experiments tomorrow and get back with the results.
 
Re: inductive vs resistive loads

Sam, quit talking about changing resistance, and changing inductance. The counter emf changes with load. If we lock the rotor, there is no CEMF. Then R & L limit the current, but not enough to keep it from smoking. As RPM increases, the CEMF increases, but this is due to generator action in the motor, not R & L. And to first order, R & L of the motor are constants, as they are in the RL circuit we discussed a few days ago. That is, L is constant until the core saturates, then it drops drastically.

I repeat, trying to explain complex machinery with changing R, L, & C is meaningless!
 
Re: inductive vs resistive loads

Today, I took an induction motor that only had a frictional load, reduced the voltage from 120V all the way down to 50V, and the current went down.
Click to expand...
That's a very drastic reduction in voltage. If the motor was under normal load it would have stalled before the voltage got that low.

Try a 15% to 25% reduction in voltage on a motor with a normal load on it and let us know what happens.

Ed
 
Re: inductive vs resistive loads

There are messy formulas which predict motor behavior, and there are performance curves. That is what you have to do to predict performance.

Dropping the voltage on a lightly loaded motor doesn't tell us much. Drop the voltage on a heavily loaded motor and see what happens. Have the fire extinguisher handy.

The old ceiling fans and buzz fans changed speed by inserting inductance in series with the line, but the load dropped non-linearly as the speed dropped. No problem in this case.
 
Re: inductive vs resistive loads

Crossman,

You raise a good point that "voltage down, current up, constant HP" is not always true. Obviously an induction motor is much more complex than that across all the possible ways it can be forced to run.

The statement is true, however, in the normal load range of an operating induction motor, a motor that is working into a load it is rated for.

Just last week I was working with an editor of a woodworking magazine, demonstrating for him the effects of voltage, voltage drop, and useable torque. We went out to the shop, plugged in a jointer (1? HP, 120 Volt) and watched the voltage at the motor as a 6" wide board was pushed into the cutting blade.

We put the jointer on the end of 200' of #12, and then on the end of 3' of #12.

As the wood was fed at a faster rate, the current would climb, leading to more voltage drop along the branch circuit conductors and less voltage at the motor. The motor would respond with greater current draw as the wood was fed faster, and the cutting blade would work with very little loss of speed (torque) up to a point. . .after that, the characteristics of the motor would move into breakdown mode, rather quickly stalling.

In breakdown mode, the torque will decline very quickly as the current rises sharply and the voltage at motor declines due to IR drop in the branch circuit conductors.

Then our circuit breaker would trip. We couldn't practically explore further into the behaviour of the motor.

Here's my point. The average electrician deals with motors working under load, most of the time. The breakdown mode I mention above is avoided. In this "under load" state the "voltage down, current up, constant HP" statement is true within enough accuracy to be a general statement.
 
Re: inductive vs resistive loads

Originally posted by al hildenbrand:
Here's my point. The average electrician deals with motors working under load, most of the time. The breakdown mode I mention above is avoided. In this "under load" state the "voltage down, current up, constant HP" statement is true within enough accuracy to be a general statement.
Click to expand...
I consider myself an average electrician :p but I feel that most of the motors I run into are not under an unchanging load.

Fans, centrifugal pumps, power equipment etc. When voltage the drops to these types of loads the output RPM drops and current does too, until we reach that 'break down point'
 
Re: inductive vs resistive loads

Sam, quit talking about changing resistance, and changing inductance. The counter emf changes with load. If we lock the rotor, there is no CEMF. Then R & L limit the current, but not enough to keep it from smoking. As RPM increases, the CEMF increases, but this is due to generator action in the motor, not R & L. And to first order, R & L of the motor are constants, as they are in the RL circuit we discussed a few days ago. That is, L is constant until the core saturates, then it drops drastically.
Click to expand...
Ok Rattus, I've been reading up on induction motors and I have a somewhat better idea of how they work. Of course I'm going to have to address your post here, :p in the order they appear.

I assumed resistance to be constant.

I disagree that the inductance doesn't change.

I don't see where CEMF is particularly relevant. (I'm not saying it isn't, I just don't see it.)

R and L will always limit current.

And
I repeat, trying to explain complex machinery with changing R, L, & C is meaningless!
Click to expand...
I wasn't defining the motor, I was looking for what has to happen for I to get bigger when V gets smaller.

But I'm just learning about these induction motor thingies. So I might not have it right yet.

_____________________________________________________________________________________________

I think I leaned a couple things.
I don't think that you actually get constant power when voltage drops. You get whatever that motor does with what it's doing.

dol2.jpg


This curve doesn't describe all motors, just this one.

The Red line shows current at full supply voltage from zero to 100% rotor speed. What will happen if you drop the supply voltage is the red current line will have a lower amplitude on the graph. But when it gets to around 75% rotor speed the curve is going mostly vertical, so that end of the curve will hardly change.

Well, it's true, lower voltage does give you lower current. I guess that sett.....Wait a minute, look at that, the rotor's slowing down! And now the rotor speed is under the big meaty part of the current curve.

Here's an equation that I think might give you the stator impedance.

The stator circuit equivalent impedance Zrf for a rotor / stator frequency ratio s is:
Zrf = Rrs / s + jXrs
Click to expand...
And here's a circuit that's supposed to be equivalent to a motor.

motor-eq-cct.jpg


Edit: That peak on the blue line is the breakdown point.

[ March 02, 2005, 01:42 PM: Message edited by: physis ]
 
Re: inductive vs resistive loads

I dunno Bob. The data is pretty hard on this.
MotorVoltageVariation.jpg
 
Re: inductive vs resistive loads

My broad generalization that a motor draws more amps as voltage is reduced comes from the amperage on the nameplate on dual voltage 208-230/460 3 phase motors. The 208 rating doesn't involve a reconnection, just an amperage increase when load stays at rated load.

I see now it is an inappropriate generalization across the whole spectrum of motors.

Jim
 
Re: inductive vs resistive loads

How can you try and explain that current goes up when voltage goes down with a motor torque speed curve. That curve does not even look at voltage.

I have one question for you.

Why do they put UV (Under voltage relay) in motor starters and who know's how many other pieces of electrical equipment? And which by the way the "engineers" that put these UV's together came up with about 75 to 80% of rated voltage to drop out. (I'll give the "engineers" credit here, they did know what they are doing.

Give up? I'll tell you why. Because as voltage decreases, current increases and the load needs to be protected. Standard NEC OCPD are not designed to protect this kind of situation. They protect short circuit and overload conditions but with the under voltage condition they OCPD is not designed to function correctly.
 
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