3 Terminal Shaded Pole Inductance Question

Are the secondary coils open or shorted. If open, they should not make any difference other that the larger one perhaps being larger and affecting the cooling.

We are considering the thermal insulation of having extra coils but it is the inductance that I am trying to get my head around.

http://www.electrical4u.com/what-is-inductor-and-inductance-theory-of-inductor/

Drew K

Is it correct to assume that the stators are mechanically identical, and that the "primary" windings are the same gauge and physically distributed in both machines with the same pitch, slot, direction, etc.? What is the purpose of the secondary winding, some sort of start/run winding, with capacitors, switches, etc.?

The secondary winding is the same wire as the primary. The purpose is to allow the same motor to work with 50 hz and 60 hz 220 voltage. We look at amp draw and torque out of the gearbox to ensure the motor will perform. The coil has 3 connections, one on the left and 2 on the right; directly across is 60 hz and offset is 50 hz.

Yesterday I took Ω and Α readings I used 60 hz on the 50 and 60 hz connections to keep things the same. A is the 2375/265 and B is 2375/425.

##: 50 hz: 60 hz: 50 hz Α: 60 hz Α

1(A) : 135.4: 120.3: .157 : .198

2(A) : 135.2: 120.5: .156 : .196

3(B) : 138.2: 112.6: .147 : .222

My understanding is that because the coils are in effect similar to a straight wire next to another straight wire that the magnetic field of one will affect the other and induce a current. When the 60 hz connection is used the extra coils for the 50 hz side are an open circuit and acting like a capacitor as the fields change and the current alternates in the extra coils.

Thoughts?

Drew K

Be clear about one thing, an open circuit coil has voltage induced in it, but there is no current or extra watts input unless you load the coil or short it (which just means you add the resistance of the coil as a load).

The resitance for B, 50 Hz at 138 is only a little more than A. Why worry?

I am confused when you say you used 60 Hz on 50 Hz & 60 Hz connections. Do you mean you applied 220V 60 Hz supply to 2375 turns & then to 2375 + secondary turns?

This does not seem a true test since surely the shaft speed will be lower on 50 Hz (less load torque??), also its shaft mounted cooling fan will move less air?

The other basic consideration is the iron core flux, assuming the same peak flux you need 20% more turns at 50Hz and the same 220V (winding B). The peak flux must be limited to avoid magnetic saturation (although the motor might be designed for well below saturation, for all we know) - Winding A might be expected to have more iron loss at 50 Hz but maybe the extra airspace with less turns improves cooling.

The motor was in a locked rotor condition during the testing so I could remove that variable from the problem. Yes, I did apply 230V 60 Hz to all connections to eliminate another variable.

I understand the open circuit has no load but is it acting like a capacitor and due to the induction there should be a current from the cycling of the potential. My thought is that is generating heat and drawing some power. I don't understand the concepts well enough and keep seeing too many variables to narrow it down.

We are pretty confident that there is sufficient iron to prevent saturation (an EE in past determined our laminate count).

Our goal is to increase torque without increasing the heat. We have to keep the motor cool enough to pass testing but have been asked to increase torque. We are looking to make a more efficient motor but most of the motor development so far has been trial and error. I am attempting to understand the physics behind what is going on so we can optimize our design.

Drew K

The mutual coupling between conductors will be almost purely inductive, the only measurable capacitance is to ground, and it is so miniscule as to be ignored except in the most rigorous analyses. An additional coil laid parallel to another coil will have a measurable open circuit potential whose value will be close to the turns ratio, but as soon as you connect a load to it it just acts like the secondary of a transformer and will present an additional load to the input of your motor.

The "...goal is to increase torque without increasing the heat..." is difficult since increasing the torque results in additonal power consumption, and that comes at the price of additional current, all other things being equal.

Cooler more efficient motors usually have more iron and/or more copper to reduce the no-load (iron) and load (copper) losses, and that means a more expensive motor. Something else to look at is the rotor, the shape, form factor, construction, and material of the rotor bars.

Here are some pictures. These are test windings that were made trying to find the sweet spot.
There is not any money for changing the rotor or stater until we optimize the windings...and probably not even after that.
My intuition says that if we reduce the number of secondary coils to a minimum for the required torque it will reduce the inductive load and therefore heat.
We already know that the best solution for increasing torque is to change the gearbox but the overlords have already forbidden that for some reason unbeknownst to the minions in Engineering.

Drew K

Which torque are you trying to increase; starting, running, pullout, minimum, maximum, at what point on the speed-torque curve, and what "sweet spot" are you trying to hit?

The gearbox is driving a 2 in or 3 in plastic valve that drains commercial laundry machines. One concern is that the motor be strong enough to break the 'sticking' of the closed valve and the other that it have sufficient torque to hold the normally open valve closed through the entire cycle. The motors can be in a locked rotor condition holding the valve closed for long periods of time.

As I said before, in the engineering dept we just want to use a stronger gearbox but the benevolent overlords have crushed that idea outright without considering logic.

Using the archaic measuring notation we only need around 10 to 14 inch pounds of torque.

Drew K

That's going to depend upon the gear ratio and the amount of "wind-up" before the resistance from the valve gets transmitted back to the motor. If the motor is "tight" to the valve then starting torque is probably your main concern. Adding turns to the running winding will have limited effect once you start saturating the iron, but the real problem may lie with the single-turn shading winding.

That's there to provide the phase shift that gets a single phase motor moving from standstill and plays a fundamental role in the amount of starting torque available, you may have to find another way to increase the starting torque perhaps by adding a starting capacitor to your auxiliary winding. It may be time for you to hire a consultant to help you, it shouldn't take an experienced one more than a day to get you moving in the right direction.

I do not have a good grasp of the saturation effect you guys speak of. Can you explain it or point me to something I can read to get an understanding?

How would a starting capacitor help, I thought i read you could start a motor with permanent magnets, shaded pole, a capacitor and something else I can't recall.

The problem with hiring a consultant is that they have been making this motor for 30 years or so and don't see the need, in engineering we would love to contract an expert and learn how it ticks...

Drew K

This is a saturation curve [Thanks to "Electrical Engineers Reference Book" by M.G. Say]. Vertical is the strength of the magnetic field you get in an "iron" core due to passing a current through a coil around it. Horizontal is the number of winding turns per metre of iron path x amps current in the winding you apply to make the field.

Look at the Stalloy curve, this is representative of laminations used for transformers and motors. Note that the flux you can get stops increasing with A-t/m and flattens out above about 1.2 webers/square metre. This is called saturation. Motors & transformers would not have more than about 1.4 Wb/m² PEAK flux because both iron core and winding resistance losses would make for excessive heating and low efficiency.

Your motor stator is roughly a ring with a a core length of a few tenths of a meter, with the winding only around part of the path, that does not matter because the field flows easily round the iron.

Remember Faraday and Henry's discovery that the voltage induced in a coil depends on the rate of change of magnetic flux linking the coil.

When you apply 230V AC 60 Hz line voltage to any coil of wire, a current flows and this is limited by both the winding resistance and the voltage induced in the coil by the magnetic field caused by the current which is always changing, swinging positive then negative 60 times a second.

In a motor or transformer most of the voltage is across the coil and the induced "back e.m.f." voltage in the coil is almost equal to the 115V applied.

Think of the positive half cycle of line voltage, the induced emf must be always positive, so the magnetic flux must always increase to produce an induced voltage equal to the line voltage at the instant. The flux will be a maximum at the end of the half cycle, then it will fall - this is changing in the opposite direction and produces a negative induced voltage opposing the supply voltage which has swung negative. During one half cycle, flux increases to a positive maximum, on the next half cycle it changes direction and reaches a negative maximum.

The important point is that increasing the line voltage requires a proportional increase in peak flux in the iron.

Referring back to the saturation curve, however, there is a limit to the peak flux at which the ampere-turns increase needed to get that flux increase causes overheating of the coil wire by the current. Note that with a fixed number of turns, the current must increase to give more ampere turns.

I hope you can see why it was always drummed into me as a student that "the voltage sets up the flux".

If losses and overheating are not to become excessive a magnetising winding must be designed for a maximum AC voltage which matches the line voltage maximum.

Because at 50 Hz line frequency, the magnetising current changes at 50/60 of the rate, for the same peak magnetising current ( and peak magnetic flux) to get the same 230V line voltage there have to be 60/50 times the number of turns.

Note that the magnetising current waveform develops a noticeably "peaky" current as the line voltage is increased to saturation levels.

The following curve is the current taken (mauve) and line voltage (blue) for a 220V 50 Hz nameplate "plugtop" transfo (unloaded) with 244V 50 Hz applied. Notice the current is not sinusoid but distinctly "peaked". The magnetising current is not maximum at voltage zero because of the voltage drop across the primary coil resistance, caused by the magnetising current (which was ignored in the above explanation). Where the voltage is near maximum and almost steady, the current rate of change is almost constant and the flux (current) is low, well away from saturation.

67model

My reply is under yours because this is way over my head! I am going to have to read this a few times to get a good grasp of it. I asked if anyone ever did a study on our windings and stators before, I will look for the answer and learn from there too.

Drew K

Well, you did ask? Anyhow, understanding what is going on helps.

Getting back to your problem. As I understand, you have two demands to meet a) more torque at standstill b) 230V 50 Hz operation.

The torque running increases much with speed from the standstill & your motor sits continuously at standstill - so your standstill test is the essence.

A current I in a wire in a magnetic field of strength M has a force on it which depends on M x I. In your motor M is made by current in the stator coil Is and the current in the rotor I is induced in it by transformer action - so M x I is really (Is x k) x I, a function of Is x I.

In particular, the standstill torque is due to the single copper turns wrapped around part of the poles (without these the motor will not turn). They get their current by transformer action from the main winding.

Unless you change the laminations to a higher permeability material, which gives more field strength per amp-turn of the 230V winding or reduce the air gap around the rotor, it is all down to increasing the amp-turns.

Increasing the supply voltage would increase all the currents, giving a more than proportional increase in torque - on the other hand a low supply voltage gives torque reduced in a similar ratio and must meet your minimum torque requirement.

Reducing the number of turns on your main winding will increase the magnetic field strength - there must be more field to get the supply voltage induced with less turns.Doing this is limited by the heating of the increased currents.

Hopefully, your materials list will tell you what the make/type of the winding wire & enquiry to maker gives you the maximum working temperature (he may be able to help on how working temperature affects life) usual values are 120-180 Celsius. You should also be able to find the lamination size/number/material & get magnetic data.

Your design file may give you the location of "hot spots" and their temperature and their relation to the winding resistance. Many standards give the permitted temperature rise in terms of the cold to hot change of the measured winding resistance ( the temperature vs resistance of electrical copper is very exact & is used as RTD, resistance at 100 'C is ~34% higher than 20'C) - you may find test values in the design file.

My hope would be that the original designer was conservative, so the turns can be reduced.

I would start by measuring rms current at 80% rated voltage, then increase it in steps recording current at each voltage. It is best to have a "true rms" ammeter because common meters read mean current but are scaled to rms, assuming sinusoidal current - winding heating depends on rms current. This will give a rough "saturation" curve.

Motors are commonly designed to withstand 30s start at 6 times full load current, so you do not have to be delicate about it. If you do it quickly, winding temperature will be close to ambient & not a variable (as I wrote, you can measure the winding resistance to gauge its temperature).

Maybe you could measure torque at each voltage too.

My thought is that if 120% voltage is no problem, then 10% less turns will be OK at rated voltage + 10%. Do not forget figures will ultimately be sensitive to tolerances on air gaps between stator & rotor.

67model

There is always this motor design software.....

http://bldcmotordesign.com/pdf/Shaded_Pole_AC.pdf

They also do design for customers.

67model

Have you priced that software! I contacted them and in it's current format the software will not work with our motors. He offered to add a module that would at no extra cost. I thought wow, that seems really nice...then I found the price tag on the quote...$9k U.S.!!!

I think for now we will continue the trial and error method, until I understand everything you guys have given me to learn.

Drew K

Further to RAMCONSULT'S comments, the other factor in the torque is the current induced in the rotor.

The strength of the rotating magnetic field depends on the currents in the main winding and the shorted-turn on the stator, but its effect depends on the current induced in the rotor and its phase.

I remember a short-term rated induction motor in which the rotor bars were deliberately made high resistance, because it increased the torque.

So what is the rotor like?

Um...the rotor is a cylinder on a shaft?

I am much more mechanical than electrical but I am looking to learn some programming. I have a handy picture of a rotor from one of our 80 lam motors and have played with a sample that was not cast. It appears to be stacked plates of steel in a spiral designed to be grabbed by the alternating magnetic field. This one is from a returned device and stopped working because the bushing wore away enough to allow the rotor to contact the stator.

Drew K

OK, Drew, some remarks on your measurements in post #3 and other things:-

a) Motor B has lower resistance at 60 Hz tap, not higher resistance compared to A & so temperature is not, as you suggest in original post, higher .

b) I suppose your ohm readings are "hot". But we do not know temperature rise unless there is a cold temperature and ohms. Also, without the winding insulation temperature limits, it is not clear if a temperature is excessive. If temperature rise increases, it may be you could use a higher temperature class coil wire.

c) I ask now, seeing your test method, do you want "breakaway" torque at standstill as the major requirement?

d) I note that the standardised 50 Hz voltages in Europe are now 230V +/- 10%, 207 to 253V (at the consumer's incoming meter), plus a further volt loss of 5% maximum (due to building cable voltage drop) to the wall outlet.

e) You seem now to have two requirements, 50 or 60 Hz operation at 230V nominal and "more torque".

Regarding e), I think you are doing the right thing with 20% more turns for 50 Hz (motor B). However, more torque requires more current and less turns which may hit saturation problems. Also, if reduced voltage below nominal is considered, this will reduce torque a lot, if I remember right, 10% volt reduction is 20% torque reduction.

Finally, if you are to be sure 50 Hz operation is OK, you need to test at 50Hz and reduced [and increased]voltage. Since you only need about 60 volt-amps, 50Hz could be supplied from an 50 Hz inverter/variable frequency motor drive (VFD) or a mobile generator with the governor "modified", but you have to be careful you have a good sine wave - cheap inverters/VFDs can be square waveform (which will cause higher motor losses) .

Higher inductance at 50 hz to have same maximum current, I would suppose.