Guess how many Amps this 220 VAC HVAC motor draws at 110 VAC?

Guess how many Amps this 220 VAC HVAC motor draws at 110 VAC?
After running a GE 5KCP392G M730BS fan/motor (salvaged from a
condenser) for a week at 110 VAC, and after all the naysayers said it
could not be done, I finally meaured the current.
There was no large insurge of current as seen on an analog 0-30 Amp
Simpson meter.
The spec on the motor reads 208-230 VAC 1.9 Amp
It has a run cap of 5 mfd/370 V
BTW The motor case was hot but I could still hold my hand on the metal
case.
.
The motor starts easily and runs smoothly and quietly at 110 VAC
What is your estimate of the Amp draw at 110 VAC?
Reply to
stu
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My guess would be between 3 and 4 amps.
These motors are routinely slowed down by effectively decreasing the voltage ("effectively" means by putting more turns on the windings.)
A "high slip" motor slows down with decreased voltage.
A "low slip" motor doesn't slow down as much but, rather, "sucks" more current from the supply.
Single speed motors tend to be "low slip" for efficiency reasons. When you reduce the voltage they tend to maintain speed (slight "droop" of course but not much). With nearly the same speed they have nearly the same load and must draw nearly the same power. With reduced voltage the motor draws more current to keep the VAs constant. That would be "OK" except that the heating of the wirings is proportional to the square of the current (I^2*R).
Single speed induction motors run at reduced voltage tend to overheat and have a much shorter life.
Reply to
John Gilmer
I won't try to estimate the current as it depends on the load. In general:
Lower no-load current at 110V -due to a lower magnetising current. Full load current will be higher. There is some in between load where there is the cross-over. Note that available torque will be lower so that the motor won't drive a fan as fast as it would at 220V and with some loads might not start.
I note that you have not given any load measurements.
Note that Ohm's Law doesn't apply to motors except at standstill (where the inrush occurs) .
Reply to
Don Kelly
is there a prize involved?
Reply to
TimPerry
This is a motor turning fan blades 1075 rpm at 220 VAC.
The prize is knowing the parameters when running a split capacitor 220 VAC HVAC motor at 110 VAC.
Reply to
stu
| Single speed induction motors run at reduced voltage tend to overheat and | have a much shorter life.
Even when a motor normally expecting 240 volts get connected to 208. Even with this little of a voltage drop, there's 33% more heating due to I^2*R. In extreme usage (continuous operation in high ambient heat) they can burn out with even a 15.47% voltage decrease. I know this from experience.
This is why I believe that switching from 240CTD/120 to 208Y/120 is a bad idea. It means we have yet another voltage to worry about and many things have to be made for both 240 volts and 208 volts. We need to reduce the number of voltages available, not increase them. And that can be done while eliminating delta services.
Power companies do want to reduce the costs by eliminating 240CTD/120. But they should not do so by passing the costs on to the customers.
Reply to
phil-news-nospam
I would say about 4 amperes. You can do strange things with motors. In 1975 I ran a 120 volt motor on single phase 208 using a variac. It was used by painters painting a fuel line that ran about a mile out across the tundra at the Pow 2 Dewline site in Alaska. I ran a 1500 foot No. 14 extension cord to the paint motor and set the input voltage at about 180 volts from a 208 volt volt supply. The motor ran fine for about 4 hours then burn up!
Reply to
electrician
If the motor runs at the same speed the electrical power would stay the same - resulting in high current - would saturate the iron?
But if the RPM went down the power could go down a lot. If the motor is still running a fan - IIRC the mechanical power is proportional to the 4th power of the RPM. A relatively small change in RPM could produce a large change in mechanical power used, and electrical power, and current. Could run below 1.9A? The temperature would indicate that happened.
bud--
Reply to
Bud--
-------- No saturation- flux density proportional to voltage for AC machines.
Reply to
Don Kelly
Magnetics is not one of my storng points. Isn't flux density proportional to Ampere turns [until saturation]?
bud--
Reply to
Bud--
Yep! The key word is "net."
For an example, you might have a transformer where with no load it would only take a fraction of an amp to saturate the iron. Yet the transformer can easily draw many amps in service. The secondary amps cancel the primary amps.
In an induction motor there are currents "induced" in the rotor. For "saturation" considerations, these cancel the current in the windings.
Reply to
John Gilmer
|> "Bud--" wrote in message | |>>If the motor runs at the same speed the electrical power would stay the |>>same - resulting in high current - would saturate the iron? |> |> -------- |> No saturation- flux density proportional to voltage for AC machines. | | Magnetics is not one of my storng points. Isn't flux density | proportional to Ampere turns [until saturation]?
Yes, it is proportional to current until the saturation is reached. With an unloaded transformer, that is also proportional to voltage. And it is inversely proportional to frequency because the impedance goes down and current goes up at a lower frquency. But as soon as you put a load on the transformer, the secondary current flows in the opposite direction and cancels out the magnetic field. Then more current flows on the primary and brings it back, along with a wee bit more that keeps the core magnetized.
Reply to
phil-news-nospam
--------- Not for AC excitation. For a sine wave Erms =4.44FNABmax so at a given area A, frequency f turns N, and applied voltage E, Bmax is determined. If the voltage alone is halved, then so is Bmax. This is based on Faraday's Law. Note that the core material and magnetic path don't enter into this. What they do is determine the magnetising ampere turns required at a given Bmax. A wood core excited at 60Hz, 120V will have an exciting current that is about 5000 times that of a transformer steel core of the same size but the same Bmax.
What you say is true for DC but Faraday's law doesnt' apply there.
Reply to
Don Kelly
| A wood core excited at 60Hz, 120V will have an exciting current that is | about 5000 times that of a transformer steel core of the same size but the | same Bmax.
I bet that would be exciting, if the source has the available current and the wiring leading to it is short enough and big enough, and none of those pesky fuses or circuit breakers is in the way.
What if the steel core is actually laminated or insulated steel wire wound continuously in the shape of the core, with the ends connected?
Reply to
phil-news-nospam
Thanks to all 3. What I missed was the (now) obvious rotary transformer action, with Faraday's law as a further detail.
bud--
Reply to
Bud--
Hi Don,
So, a question for you. All the large generators I've worked with have a 'saturation curve' listed in the documentation. Open circuit voltage (P.U.) versus field current. Nice and linear, up to about AFNL (Amps, Field, No-Load), then from there the line curves off as the field iron saturates.
(they also have a line for short-circuit operation (short-circuit current P.U. versus field current) and AFFL (Amps, Field Full-Load) that is always quite a straight line. Between these, I can get a pretty good idea of the syncronous impedance of the machine. )
But like you say, the iron doesn't saturate during operation. I've been modeling steady-state generator operation as an ideal voltage source, and an inductance (equal to the synchronous impedance) in series. But I've had some 'discussions' with others trying to model these machines about what to use for the 'ideal voltage source'. I've been using a linear calculation of the field current times the slope of the linear portion of the 'saturation line'. Another bloke as been trying to argue that his results are 'better' because he uses a curve fit of the saturation line that 'flattens' as field current rises.
I maintain that the MMF of the field current is being countered by the MMF of the armature current and the iron doesn't saturate. What you've said here seems to support that idea.
Thoughts, comments??
daestrom
Reply to
daestrom
----------------------------
----------- What I said with respect to saturation was a decrease in operating voltage results in lower flux in a motor excited from the mains. The synchronous machine has a DC excitation as well as a possible AC excitation. Both are involved. The mmf due to the stator winding currents will not be directly co-linear with the rotor mmf so x amp turns from the stator will not counter x amp turns from the rotor. In fact a large part of the synchronous reactance is actually due to armature reactance mmf, rather than conventional leakage reactance. Park came up with a model about 1928 which is still used today. (See any text such as those by Krause) The conventional synchronous machine model is based on his equations However, strictly speaking the model, in steady state is a DC referred to the rotor model which has the advantage of being expressible in terms of equivalent phasor quantities and the mmf components replaced by direct and quadrature reactances. If saturation exists, there are ways to approximate this by reducing the direct axis reactance accordingly and keeping the linear voltage -field excitation curve that you use. If you have a given no load voltage and apply a lagging pf load, the terminal voltage will fall as the reactive component of the armature mmf reduces the direct axis mmf. If the field is not changed, then there will, in fact be less saturation. However, if you restore the terminal voltage- an increase in field will be needed to counter this component of armature reactance plus a bit more to account for the effect of leakage reactance. You will then return, in terms of direct axis mmf and saturation to nearly where you were before. Your use of the air gap line and the short circuit current curve does give a good measure of the synchronous reactance in unsaturated conditions. Your ignoring saturation is quite reasonable (note that if you are 5% above rated voltage- there will be some saturation occuring but at rated voltage there shouldn't be. Using the actual open circuit curve and the short circuit curve can be done to determine the saturated reactance but not directly- it is a bit messier than that. I'm not sure that any added accuracy(?) is worth the hassle of trying to work with a variable synchronous reactance. For fault studies, and approximate voltage regulation calculations, the air gap line that you use is satisfactory for finding the synchronous reactance of a round rotor machine- inductive load currents demagnetise. For more careful (as required of students 60 years ago and then there were several different approaches, all not quite agreeing, to be forgotten and never used again) voltage regulation calculations-then saturation should be taken into account- but not by using the no load saturation curve directly if I recall correctly. I note that most texts stick with the linear model and mention a saturation correction if needed. It shouldn't be needed if you are within the rated operating region.
In the case of a synchronous motor- then there is a double excitation- from the stator and this dictates the flux seen by the stator and which may not be the same as the flux seen by the stator which is due to the DC rotating field. Exciting current from the stator makes up the difference so armature reaction may boost or buck the field.
What I tried to say earlier was that for AC there is a direct relationship between flux and voltage- independent of the magnetic path. The flux- mmf relationship depends on the path and also exists. For DC only the latter exists. A synchronous machine is only partly AC in that, in steady state, all mmfs as seen from the rotor are DC. This is where any saturation may occur and this is where the bulk of synchronous reactance lives. The rest comes from leakage reactance of the stator windings. --
Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer
Reply to
Don Kelly
The magnetising current would still depend on the material - whatever its B-H curve is. I should hope that the core is laminated. The steel wire core would work and I think that the original transformer may have had such a core. The connection at the ends really wouldn't matter. I note that continuous strip wound cores are being used and it is a hell of a job to connect the ends other than making a multiturn Moebus strip:).
Reply to
Don Kelly
|> |> |> | A wood core excited at 60Hz, 120V will have an exciting current that is |> | about 5000 times that of a transformer steel core of the same size but |> the |> | same Bmax. |> |> I bet that would be exciting, if the source has the available current |> and the wiring leading to it is short enough and big enough, and none |> of those pesky fuses or circuit breakers is in the way. |> |> What if the steel core is actually laminated or insulated steel wire |> wound continuously in the shape of the core, with the ends connected? | ---------- | The magnetising current would still depend on the material - whatever its | B-H curve is. I should hope that the core is laminated. The steel wire core | would work and I think that the original transformer may have had such a | core. The connection at the ends really wouldn't matter. I note that | continuous strip wound cores are being used and it is a hell of a job to | connect the ends other than making a multiturn Moebus strip:).
What I was wondering is the physic of the flux loop ... if it had to be connected back to itself directly in the same material, or if there would be any difference in the fact that it went to a "different" wire the next pass around. I would think wire would help cut down hysteresis loss.
Reply to
phil-news-nospam
Yes, this was my thinking. Reactive load currents (lagging) create an mmf that is exactly along the direct axis, but displaced 180. There is a 90 shift between the field winding and the maximum induced voltage in the stator, and another 90 shift between the voltage in the stator and the lagging pf load current. This puts the mmf associated with lagging load current in direct opposition to the field mmf, resulting in poor voltage regulation.
The text I've been using ("Standard Handbook for Electrical Engineers", Knowlton) gives the vector diagram for regulation and field excitation, taken from an old ASA Standard. That, along with a 'Blondel diagram' ('two-reaction diagram')have been my primary tools/sources of data. Doesn't mention 'Park' by name though...
As I said, I've been using the linear, no-saturation model to calculate the steady-state excitation requirements at various load levels, and they seem to agree with the actual operating unit data. But some of it is supposition on my part. I felt that the air-gap flux *must* be the net result of these two mmf's (field and armature), and therefore saturation couldn't be occurring under normal operation. Another clue is the much larger field currents that are normally seen when operating near rated conditions.
Interestingly, I found one of our units operating with the field current showing *below* the AFNL value, yet it was carrying enough MVAR to be running about 0.95 lagging. So I concluded that the field current transducer input to the logging computer must be defective. The computer tech swore that his computer was right, and the I&C tech swore that it really was reading the number of milliamps that corresponded to the field current shown. But when we checked the field-current to mA transducer, lo and behold it was way off, reading low. I told them that my 'calculations' predict the field current should be 'X', and the 'as-found' testing of the transducer showed I was within 3%. They were suitably impressed :-)
I haven't tried to tackle transient or sub-transient features in my model yet though. Still working on the steady-state. I am working on the transient rise in voltage with a load-reject sceanrio. It *seems* like I could just drop the load current to zero while maintaining rated field current. But that gives some pretty incredible voltages (as in three times rated). So
Yes, I've found that the synchronous reactance is really the major player in voltage regulation of the unit. In my calculations of generator, main transformer and 'infinite bus', I've found that from the 'ideal voltage' source viewpoint, it carries quite a large 'reactive load', most of it right inside the machine in the 'synchronous reactance' component. And yes, when I compare the synchronous reactance with the winding inductance, there is a large difference.
Haven't had to model one, but was thinking I could use the exact same components, just swap the torque angle to the opposite side. Sort of just what happens in the machine. This reverses the real component of the current. And what was a 'lagging' reactive current is now 'leading' (the reactive current vector doesn't change, just its relationship to the real current vector that is now 180 from where it was in a generator). But as I say, haven't tried it yet, so.....
Not sure what you mean about "the bulk of synchronous reactance lives" on the rotor.
Your previous statements also help explain why a constant V/Hz is desirable in variable speed applications. As frequency is reduced, the voltage must be reduced concurrently to avoid saturation. I've also seen some small isolation transformers used for attaching instruments to AC servo-motor machinery. Along with other ratings, the label listed frequency as "> V/1.5". As long as you connected to a system with a frequency higher than that, no problem. When it was inadvertently hooked up to a servo-motor incorrectly, the output was nasty 'spikes' every half-cycle as the iron within went from saturated in one direction, to saturated in the opposite (fortunately, the thing wasn't damaged, must have had enough impedance to limit the primary current)..
daestrom P.S. Not to worry, these models aren't being used for design/analysis, I'm not a PE. Just what you might call, 'professional curiosity'. If I can calculate/predict behavior, then I figure I've got a better understanding. (getting too old to actually crawl inside the things, besides carbon dust is hard to wash off :-) P.P.S. Although we have some salient pole units, MG-sets (~5000 hp) and diesel generators (nice EMD 4500kW units), I'm still just working through round-rotor machines (tubine driven).
Reply to
daestrom

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