Low Voltage burns out electric motor - How ?

[ZenBeam]

How about low voltage and heating elements? I bought a heated dog dish in January so the dog’s water wouldn’t freeze outside. By the end of February, it was no longer working, so I took it back, and the clerk flipped it over and showed me the sticker that said “Do not use with extension cords.”, which of course is what I had done. Why would using a 12’ heavy duty extension cord cause the heating element to fail?

I doubt it did. Googling on “Do not use with extension cords” brings up a bunch of hits, so it seems to be a common warning. None of the hits looked like they gave any explanation why you’re not supposed to use one*. I suspect it’s more to prevent tripping over the cord than because it will cause any damge to the item.

• there was one link to a … plant … growing site where the questioner began “i just bought a timer for my grow on the back of the box it says do not use with extension cords…”, but I’m blocked from getting to the site.

[David_Simmons]

@ZenBeam

I doubt it did. Googling on “Do not use with extension cords” brings up a bunch of hits, so it seems to be a common warning. None of the hits looked like they gave any explanation why you’re not supposed to use one*. I suspect it’s more to prevent tripping over the cord than because it will cause any damge to the item.

• there was one link to a … plant … growing site where the questioner began “i just bought a timer for my grow on the back of the box it says do not use with extension cords…”, but I’m blocked from getting to the site.

It’s also possible becaise a heater draws a lot of current. Heaters run unattended especially at night. People tend to put carpets and things over extension cords to hide them and to prevent tripping. An extension cord with lots of current in the it under a carpet is asking for trouble.

[BubbaDog]

@David Simmons

It’s also possible becaise a heater draws a lot of current. Heaters run unattended especially at night. People tend to put carpets and things over extension cords to hide them and to prevent tripping. An extension cord with lots of current in the it under a carpet is asking for trouble.

And the warning is a universal application which the most likely targets are those flimsy “lampcord” extension cords which aren’t designed to carry the currents drawn by motors and heating elements.

[ZenBeam]

One of the items that showed up in the Google search was Glade plug-ins. I don’t think those are big current draws.

[Fear_Itself]

@David Simmons

It’s also possible becaise a heater draws a lot of current. Heaters run unattended especially at night. People tend to put carpets and things over extension cords to hide them and to prevent tripping. An extension cord with lots of current in the it under a carpet is asking for trouble.

This was a heated water dish, not a space heater. It wasn’t designed to boil the water, just keep it just above freezing. It never threw off any noticeable heat, so it didn’t draw a lot of current.

[empty_filing_cabinet]

@Fear Itself

This was a heated water dish, not a space heater. It wasn’t designed to boil the water, just keep it just above freezing. It never threw off any noticeable heat, so it didn’t draw a lot of current.

Hm. It was probably just a liability CYA thing for them, as the others have basically suggested.

If all it is, is basically a resistor that gets slightly hot, then I don’t think the extension cord caused it to break. [But yeah, for other things that draw a lot of current, a flimsy extension cord could overheat. Of course, you had a heavy duty one, so that would most likely have been OK anyways.]

In other words, it was defective or damaged somehow.

[empty_filing_cabinet]

DC motors, universal motors, induction motors … I think I have some reading cut out for me. Thanks for all your responses so far.

The real world application that spurred my question is motors in airplanes that turn those rotating beacon lights. An aircraft mechanic told me that if you’re working on a plane that runs on 28 volts, but accidentally plug it in to a 14 V source to run things (such as the rotating beacon light) while you work on it, it is possible to burn out motors. Like the rotating beacon will turn slower, and it takes a while to fry.

That and jammed electric motors being a fire hazard.

[David_Simmons]

@Fear Itself

This was a heated water dish, not a space heater. It wasn’t designed to boil the water, just keep it just above freezing. It never threw off any noticeable heat, so it didn’t draw a lot of current.

Well, then my guess can’t be it.

[David_Simmons]

@empty filing cabinet

DC motors, universal motors, induction motors … I think I have some reading cut out for me. Thanks for all your responses so far.

The real world application that spurred my question is motors in airplanes that turn those rotating beacon lights. An aircraft mechanic told me that if you’re working on a plane that runs on 28 volts, but accidentally plug it in to a 14 V source to run things (such as the rotating beacon light) while you work on it, it is possible to burn out motors. Like the rotating beacon will turn slower, and it takes a while to fry.

That and jammed electric motors being a fire hazard.

Motors are interesting gadgets.

An induction motor is one having a stator (the stationary part) coil with lots of turns. This coil is wound around an iron core and establishes a stationary magnetic field. The rotor contains a sqirrel-cage assembly of conductors (imagine a squirrel-cage made of heavy copper or aluminum conductors) also contained in an iron armature and electrically insulated from it. The rotor is mounted on bearings in the air-gap of the stator core.

When current flows in the field coil an alternating field is established in the rotor and this induces current in the squirrel cage (that’s why is called an induction motor). This current produces a magnetic field that interacts with the stationary field so as to turn the rotor. When the rotor is standing still at start the squirrel-cage current is large. As the rotor speeds up the squirrel cage currents fall off to a lower and lower value. At no-load very little induced current is required to turn the thing and the rotor runs at almost the synchronous speed. For 60 Hz current and a two pole motor this is nearly 3600 rpm. When a load is applied the rotor slows down and this increases the induced current in the rotor winding. At full load the rotor might be running at 3500-3525 rpm. When the rotor current increases this produces a magnetic field that reacts with the stator field so as to increase the input current which provides the torque on the rotor that turns the load.

The fly in the ointment with ordinary induction motors that you use around the house is getting it started. The stationary magnetic field merely builds up to a maxumum in one polarity (direction), falls back to zero and builds up to maximum in the opposide polarity. There are three ways to get it started.

The first is the so-called shaded pole motor. A small slot is cut in the iron core of the stator. A shorted turn of copper wire is placed in this slot so that it encircles a part of the stator core. The current induced in this turn by the main field current produces a small magnetic field that is phased (timed) relative to the main field so as to produce a net rotating field and this gets the rotor moving. The starting torque on this type motor is small and it is used only for things like electric clocks that don’t require much to start them.

The second type of starting arrangement is the capacitor motor. This one has a second winding on the stator core. A capacitor connects this winding to the input voltage and the phase (timing) shift resulting in the current from the series capacitor-inductor (the auxilliary field winding) combination produces a rotating magnetic field with lots of starting torque.

The third type is the capacitor-start motor and it is just the same as the second type with a centrifugal switch added that disconnects the auxilliary winding when the rotor gets up to some predetermined speed.

Motors like this are commonly used in things like shop tools, drill presses, table saws, refrigerators, ceiling fans and the like. Their advantage is that there aren’t any things like brushes to need replacing. The disadvantage is that they run pretty much at a constant speed.

[Desmostylus]

@David Simmons

I did not write about universal dc motors but rather a parallel connected dc motor.

Doesn’t matter whether it’s series or shunt wound, the model’s the same.

@David Simmons

I’m not so sure that you are right about about a parallel connected dc motor. At no load if you reduce the voltage in such a motor you also reduce the field strength which results in less back emf and the armature current increases temporarily until the speed increases to bring it back down to that current sufficient to overcome losses. However you also reduce the torque so that under load the motor will slow down and the current increase. It does seem to me that if you reduce the voltage in such a motor the current could increase enough to damage it.

It’s not that difficult conceptually.

A d.c. motor will have a locked rotor torque:

T[sub]0[/sub] = K V[sub]s[/sub] / R, where

K = a constant for a particular motor,
V[sub]s[/sub] = supply voltage,
R = locked rotor resistance.

It will have a no-load speed of:

w[sub]0[/sub] = V[sub]s[/sub] / K

and in between locked rotor speed (0) and no-load speed, the torque will be:

T = T[sub]0[/sub] (1 - w / w[sub]0[/sub])

It will draw a current of:

I = T / K

Let’s say we have a 12 V motor, which draws 2 A with a locked rotor, and has a no-load speed of 3,000 rads[sup]-1[/sup]. Then:

R = 12 / 2 = 6 ohms,
K = 12 / 3000 = 0.004 NmA[sup]-1[/sup],
T[sub]0[/sub] = 0.004 x 12 / 6 = 0.008 Nm.

What will happen if we reduce the supply voltage to 8 V? No-load speed will drop to 8 / 0.004 = 2,000 rads[sup]-1[/sup]. This doesn’t tell us much about current, though, because in this model, no-load current is zero (i.e., frictional losses are treated as a load).

So let’s have the thing drive a fan. We’ll assume that with a 12 V supply, it’ll drive the fan at 2,700 rads[sup]-1[/sup]. We’ll further assume that the torque required to drive the fan is of the form:

T[sub]f[/sub] = a w

where a is a constant for the fan.

So for our fan,

a x 2,700 = 0.008 (1 - 2,700 / 3,000)
a = 2.963 x 10[sup]-7[/sup],

and the motor will draw a current of:

0.008 (1 - 2,700 / 3,000) / 0.004 = 0.2 A.

What happens now when we reduce the supply voltage to 8 V?

2.963 x 10[sup]-7[/sup] x w = 0.004 x 8 / 6 (1 - w / 2,000)
w = 1,800 rads[sup]-1[/sup]
I = 0.004 x 8 / 6 (1 - 1,800 / 2000) / 0.004 = 0.133 A

So speed dropped and current dropped as the supply voltage dropped.

The above result will hold for any monotonically increasing load torque function, i.e., provided it always takes more torque to make the thing go faster. This will generally hold for any type of load, once the initial static friction has been overcome.

That’s the end of the story for d.c. motors.

@David Simmons

DC motor speed control is either with armature current control or field current control but the armature and field are not in parallel but are separately excited. In that case you can either vary the field current to change the stationary magnetic field or the armature current so as to vary the rotating magnetic field.

We’re not talking about negative resistance devices, here, so reducing current through the thing necessarily implies reducing the voltage across it.

[Desmostylus]

@Desmostylus

We’re not talking about negative resistance devices, here, so reducing current through the thing necessarily implies reducing the voltage across it.

Note that this statement doesn’t apply to induction motors. The situation there is more complicated. Reducing the voltage to an induction motor will reduce the in-phase current through it, and reduce the real power delivered to the load. However, the reduced voltage will increase the out-of-phase current through it. So an induction motor isn’t a negative resistance device, but over a certain range of speeds, it’s approximately a constant VA device.

[David_Simmons]

@Desmostylus

The above result will hold for any monotonically increasing load torque function, i.e., provided it always takes more torque to make the thing go faster. This will generally hold for any type of load, once the initial static friction has been overcome.

That’s the end of the story for d.c. motors.We’re not talking about negative resistance devices, here, so reducing current through the thing necessarily implies reducing the voltage across it.

I’ve spent some time going over all my old stuff on motors and for a motor that is loaded and if the armature voltage is reduced along with the field voltage you’re right.

There is a pathological case that we were always warned about when playing with large motors in the lab. With the motor unloaded so that the only resistance to greater speed is viscous friction (i.e. ball bearings) and with the armature having full voltage or nearly so, if the field voltage is reduced to zero or the field current is interrupted there is still the residual magnetic field of the field core. In case of large machines with low friction the torque from the large armature current that results when the back emf in the armature drops suddenly is sufficient to drive the motor to destructive rotational speeds.

But you are correct that if both armature and field current voltages and currents are reduced and if the motor is loaded it will slow down. By far the great majority of machines that we have anything to do with have sufficient bearing friction that they will not run away in any case, except possibly series field, universal motors without a load of some kind…

[empty_filing_cabinet]

OK, so is there some authoritative work on motors that you all are consulting? If so, what is it? (I looked at the public library and couldn’t find anything in sufficient depth.)
So is the back-EMF relevant, or not?

And how about eddy currents?

[David_Simmons]

@empty filing cabinet

OK, so is there some authoritative work on motors that you all are consulting? If so, what is it? (I looked at the public library and couldn’t find anything in sufficient depth.)
So is the back-EMF relevant, or not?

Absolutely. The armature current, or what makes the motor go depends upon the net voltage applied to the armature and the total resistance in the armature. That net voltage is the applied voltage minus the back emf. The armature resistance can be represented by the actual winding resistance and a fictitious impedance representing the load that is in parallel with the winding resistance. When the motor load increases the rotor, which is usually the armature, slows down decreasing the back emf which results in more armature current in order to drive the increased load. The current increase flows through the impedance representing the load.

And how about eddy currents?

Eddy currents are currents that are generated because of the fluctuating magnetic fields in the magnetic core. They circulate in the core and are part of the losses of the motor. These currents are made as small as possible by building up the core from thin sheets of iron that are electrically insulated from each other.

[David_Simmons]

In practice, the armature resistance used in the equivalent circuit is also a fictititious resistance composed of the actual armature resistance along with a resistance representing all other motor losses, eddy current, hysteresis, friction, windage. The loss resulting from the field resistance can either be treated separately or lumped in with the other losses in the armature.

[Desmostylus]

@empty filing cabinet

So is the back-EMF relevant, or not?

The “back emf” concept is another way of considering the equations I used above.

We can write:

I = V[sub]s[/sub] (1 - w/w[sub]0[/sub]) / R = (V[sub]s[/sub] - V[sub]s[/sub] w/w[sub]0[/sub]) / R

The model has two voltage sources, the supply voltage V[sub]s[/sub], and a “back-EMF” V[sub]s[/sub] x w/w[sub]0[/sub]. They’re connected to each other via resistance R.

It makes no difference whether you introduce the “back-EMF” concept or not, as long as you use the equations properly.

[Desmostylus]

@David Simmons

if the motor is loaded it will slow down. By far the great majority of machines that we have anything to do with have sufficient bearing friction that they will not run away in any case, except possibly series field, universal motors without a load of some kind…

No. I already covered this.

@Desmostylus

What will happen if we reduce the supply voltage to 8 V? No-load speed will drop to 8 / 0.004 = 2,000 rads-1. This doesn’t tell us much about current, though, because in this model, no-load current is zero (i.e., frictional losses are treated as a load).

No-load speed, with no frictional losses, is directly proportional to supply voltage. Cut the supply voltage by, say, 10%, and you cut the frictionless no-load speed by 10%.

[David_Simmons]

From Series Motors

“The coils in the series field are made of a few turns of large gauge wire, to facilitate large current flow. This provides high starting torque, approximately 2 ¼ times the rated load torque. Series motors have very poor speed control, running slowly with heavy loads and quickly with light loads. A series motor should never drive machines with a belt. If the belt breaks, the load would be removed and cause the motor to overspeed and destroy itself in a matter of seconds”