About electrical motors and their energy consumption - Page 1
Forum section: electronics / beginners

author: axero
author_rank: Regular Contributor
timestamp: October 27, 2016, 02:20:20 pm
content: I have these questions since I never got around to experimenting with electrical motors and measuring their amperages in different applications.Let's say that I have a 3-phase motor that is rated for 1 kW. If I turn it on and let it run without applying any load, what would the typical power consumption be for such an operation?If I put it to crank a certain load, say that the resulting torque from that load would be constant at say 200 Nm, during operation, what would then be the power consumption of the motor?Is there an overload condition, let's say that the load is so tough that the motor fails to crank, what would happen to the power consumption and can it damage the motor or surrounding components in any way?Let's say that the motor gets hot, e.g. friction during operation heats up the shaft. How would that affect the power consumption?Can we say anything about how these above mentioned conditions affect the amperage differences between phases?Is there a difference in this regard when looking at brushless motors DC, AC or any other types of electricle motors? If there are some white papers or scientific articles about this, I'd be happy to read them.

author: IanB
author_rank: Super Contributor
timestamp: October 27, 2016, 02:31:58 pm
content: I don't have knowledge of all the details, but some simple rules apply with any motor. Firstly the input power equals the output mechanical power divided by some efficiency. There is typically an optimum operating condition where the motor has the best efficiency. A motor will have a performance map showing speed, torque, power and efficiency so you can select the best motor for a job. Different kinds of motor will have different performance maps.If a motor is free spinning without load it will consume relatively little power. The power consumption will go up as the load increases.If a motor is overloaded it will tend to overheat. Large motors may have a thermal overload cutout to prevent damage if this happens.

author: axero
author_rank: Regular Contributor
timestamp: October 27, 2016, 02:53:04 pm
content: Ok, interesting. I'm asking about the power consumption and amperages during overload because motor protection circuits generally don't measure temperature, so it sounds kind of strange to call it overheat protection. What they do is that they break the circuit should the amperage reach a certain level. I also think it trips on differences in amperage between phases.Usually different amperage over the phases is apparently commonly used as an indication that something is wrong. What kind of wrong that would be I don't know.

author: dmills
author_rank: Super Contributor
timestamp: October 27, 2016, 03:28:19 pm
content: One of the loss mechanisms in a motor is simple I^2R heating of the windings resistance, so more current equals more heat (The other major one is magnetic hysteresis, I^2R is referred to as 'copper loss', hysteresis is 'Iron loss' .A motor can be thought of as a relatively small resistance in series with a back emf produced by the interaction of the various fields, as you load the motor one of several mechanisms (depending on motor type) will cause the back EMF to fall, leaving more voltage across the winding resistance and causing the supply current to increase to deliver the required power.A locked rotor (Or equivalently a motor at the moment of startup) can pull a LOT of current (30 times normal running is not uncommon for a small machine, and it can go much higher then that).The subtle point here is that a better motor will have lower copper losses and so will have a higher locked rotor or starting current.Big motors are never started by switching them directly across the line, because currents can get really silly, as can peak torque. Either star delta or starting resistors are used, or in modern systems a VFD that brings speed and voltage up together smoothly.Phase current imbalance would typically signal a short circuit in one winding, or possibly the loss of a phase at the input (A motor already turning will sort of run under these conditions, but it is not good).HTH.Regards, Dan.

author: woodchips
author_rank: Frequent Contributor
timestamp: October 27, 2016, 05:42:59 pm
content: Presumably your 3ph motor is an induction type? An induction motor is a strange beast, far more interesting than modern digital electronics. A 1kW motor will take about 1kW power, running light. Why? What you have to take into consideration is the power factor. At low load this will be poor, possibly only 0.1. What happens as you increase the load on the motor is that the power factor increases so that the current stays pretty constant but more of it is 'real' and contributes to the mechanical output power.Motor start currents are high, but probably not up to 30x due to the leakage inductance limiting the current. Direct on line can only be used to a certain motor size before the magnetising forces within the stator physically distort and ruin the winding.Power is torque x speed, 200nm x ? rad/sec = ? W.

author: IanB
author_rank: Super Contributor
timestamp: October 27, 2016, 07:32:37 pm
content: Quote from: woodchips on October 27, 2016, 05:42:59 pmIf a 1 kW motor consumed 1 kW when running idle then it would become a 1 kW heater and would rapidly let out the magic smoke. I rather think this doesn't happen.It may be true that the power factor at idle is low and there is a significant reactive current, but this is still lower than the full load current and the real power consumed by the motor is not that much. It is important to distinguish between real power (kW) and reactive power (VAR).

author: Ammar
author_rank: Regular Contributor
timestamp: October 27, 2016, 08:23:15 pm
content: All really good questions. We did a whole subject on this at Uni. I can recommendhttp://www.abebooks.com/9780131776913/Electrical-Machines-Drives-Power-Systems-0131776916/plpif you are interested. It is a fantastic subject!

author: Zero999
author_rank: Super Contributor
timestamp: October 27, 2016, 08:32:52 pm
content: A motor and generator are really the same thing.As the motor spins, it generates a voltage (back EMF (Electro Motive Force)) which opposes the power supply voltage. When the power is first applied, the motor will be stationary and the back EMF will be zero, causing the current to be high, as the motor reaches full speed, the back EMF will be near the power supply voltage and the current drops very low.If the shaft is locked, then the situation will be the same as the motor starting, except the shaft will never spin and generate an EMF. With such a high continuous current, the motor will overheat.One interesting thing is if the motor is made to spin faster than it would when connected to the power supply (say an engine is connected to its shaft) the back EMF will become greater than the supply voltage and current would flow back into the power supply: now you have a generator.

author: woodchips
author_rank: Frequent Contributor
timestamp: October 27, 2016, 08:49:50 pm
content: You have just explained why the motor will take 1kW, as seen by the thing providing this power. It will be volts x amps, the fact that most the amps are reactive doesn't make their number any the less.The snag with an induction motor is that it is vastly more complicated, and interesting, than a collection of copper wire, iron laminations and an AC source has any right to be. The induction generator is a case in point, but only if the load is leading, not lagging.

author: dmills
author_rank: Super Contributor
timestamp: October 27, 2016, 10:14:16 pm
content: Nope, it might take 1kVA (I doubt the magnetising current is anything like that high unless you are running a 20kw motor with only a 1kW load, but whatever), the units matter here.The supply source has to supply that much current, but NOT that much real power.Regards, Dan.

author: AlfBaz
author_rank: Super Contributor
timestamp: October 27, 2016, 11:57:18 pm
content: I think this talk about back emf limiting current is a bit misplaced when talking about ac induction motors. It's predominant in DC motors where there is a stationary magnetic field in which the armature is rotating. In an AC motor, a rotor spinning with a small percentage of slip relative to the rotating magnetic field in the stator will produce bugger all back emf.An ac induction motor is best viewed as a transformer.On start up this "transformer" has the secondary shorted by virtue of the rotor bars in a squirrel cage rotor being shorted and the fact the rotor is stationary. As the rotor gets up to speed the rotating stator field cuts the rotor bars less causing less current to be induced in the secondary (rotor bars). This in turn causes less current to be drawn from the "primary"

author: dmills
author_rank: Super Contributor
timestamp: October 28, 2016, 01:03:07 am
content: You can sort of view a transformer as a back emf driven device as well, series resistance, shunt magnetising inductance and a shunt voltage source that is the back EMF (Proportional to the core flux), loading the secondary lowers the core flux this lowering the back EMF and raising the real power consumed.As ever in this game there are multiple levels of abstraction, and multiple models used to simplify the reality, you pick the one that works to answer the question.73 Dan.

author: AlfBaz
author_rank: Super Contributor
timestamp: October 28, 2016, 01:44:18 am
content: Quote from: dmills on October 28, 2016, 01:03:07 amI don't understand. Loading the secondary increases current causing a stronger magnetic field (aka flux) this is why AC induction motors are designed to have their greatest torque at full load, if flux is reduced, how are we developing a greater torque under load?

author: dmills
author_rank: Super Contributor
timestamp: October 28, 2016, 07:08:07 am
content: That would be intuitive, but incorrect for a transformer, and I suspect also for the induction motor (But that case is more complex due to the variable slip and I cannot be bothered to draw the phasors and do the math).In a transformer, load current in the secondary produces a field in the core that opposes that produced by the primary, reducing flux and with it back EMF induced in the primary and thus increasing primary current. Good transformers will have sufficiently low copper loss and leakage inductance that the increased primary current will put the flux almost back where it was and thus maintain output voltage under load.I have to admit that realising this was something of a lightbulb moment for me.Regards, Dan.

author: woodchips
author_rank: Frequent Contributor
timestamp: October 28, 2016, 03:55:09 pm
content: I was trying to keep the terminology within the OP question, power can be real or reactive, as I said. Measure the phase current off load and on load and there won't be much difference, phase angle will change though.No, the transformer, and motor, are constant flux devices, this is, after all, what magnetising flux is. There is a complex dance of phase angles, slip, power factor etc as a load comes on a motor or transformer, but the core flux doesn't change.

author: axero
author_rank: Regular Contributor
timestamp: October 28, 2016, 09:42:18 pm
content: Not that I know that much about electrical motors, but I'd venture to guess that the rating of a motor is just a label for what it is designed for, not necessarily what it draws at any given point in time. For short periods, it may be able to draw perhaps 4 times as much or more and at idle, the power draw may be considerably lower. The "1kW" label merely gives a hint of what continuous operation one could get from it while staying within safe limits and what ampere ratings should be used on power supply cables, fuses and other switching and protection circuits. If the motor is part of an operation where for example a fluid is being pumped and that fluid can act as a coolant, perhaps one could squeeze a few more Watts out of it than it is rated for, given that the electric circuitry allows for it. That's my two cents anyway.I skimmed through the book by Wildi, thanks for the suggestion! I jumped directly to the chapter on 3-phase motors and cursively read the chapter on "abnormal conditions". It says that if voltage drops on one of the phases while the other stay the same, the motor will basically run as a one-phase motor and the amperage will double on the other two phases, giving rise to overheat. Even small voltage differences can lead to a serious load imbalance over the phases and a difference as small as 2.5% can lead to a temperature increase of 15°C, it says.But a scenario where the voltages are the same over all three phases while there is a difference in currents is not described in the bookBtw, this forum supports MathJax so I assume that the wire load is denoted by \({{I^2R}}\)

author: Zero999
author_rank: Super Contributor
timestamp: October 28, 2016, 10:35:05 pm
content: Quote from: axero on October 28, 2016, 09:42:18 pmIt's cooling which limits how much power a motor can safely output. If the motor is specified at an ambient temperature of 30oC, then it should be able to output more power at 10oC. There's a limit on how low the temperature can go of course: too cold and the grease in the bearings will solidify, causing more friction and higher power dissipation.QuoteWhat's worse, is if there are other loads connected to the phase which has been disconnected: the motor will then start to try to power them.QuoteBecause, if the voltages are the same over all three phase, then there won't be any differences in currents drawn by each phase of the motor.

author: axero
author_rank: Regular Contributor
timestamp: October 29, 2016, 09:33:52 am
content: Quote from: Hero999 on October 28, 2016, 10:35:05 pmI actually think that there is more to the power rating of a motor than just the cooling. Let's say that I could put infinite cooling onto the motor while operating it. Then I would say that a big motor will be stronger than a small one. Of course, if the bigger motor is poorly designed, then this won't apply. I guess that internal resistance plays a role here. So either internal resistance or maximum torque @ a given RMS voltage level would give what's missing in the equation. I guess.Quote from: Hero999 on October 28, 2016, 10:35:05 pmThat I don't entirely understand. I can understand that the disconnected winding now will act as a power generator. But for that winding to be able to supply voltage to other loads in that circuit, one of the ends of that winding must somehow be connected to Neutral. As I understand it, the windings of a 3-phase motor are only connected between the phases, not against neutral.Quote from: Hero999 on October 28, 2016, 10:35:05 pmI don't understand this. If something happened to one of the windings so that it draws more current/power than the other windings, for one reason or another. Then why should that lead to any significant voltage drop? I'm talking about voltage between the phases here. If I measured the voltage over a 2kW heater and then a 20W light bulb, there wouldn't be any significant voltage difference. Amperage difference yes, voltage difference no. Not if the cabling is good anyway.

author: Zero999
author_rank: Super Contributor
timestamp: October 29, 2016, 01:42:13 pm
content: Quote from: axero on October 29, 2016, 09:33:52 amWhat limits the power rating is the maximum temperature rating of the motor windings. At lower temperatures, the windings will be able to dissipate more energy before overheating. Another thing is the resistance of copper and other metals is lower at lower temperatures, so the power dissipation will also fall, given the same load (the efficiency will increase): running at lower temperatures is a win-win.QuoteWell it will be connected to neutral, via the other windings and loads on the other phases, which will be connected to neutral.QuoteLike a shorted turn or open circuit? Then yes, that would lead to excessive current draw and power dissipation, even though the voltage will be the same.

author: AlfBaz
author_rank: Super Contributor
timestamp: October 30, 2016, 12:51:48 am
content: Quote from: woodchips on October 28, 2016, 03:55:09 pmHow do you reconcile those 2 statements? Given how and why a constant flux machine works, namely that for the flux to remain constant the primary current must proportionally match the secondary current so that there phasor sum of the constituent magnetic fields remain constantAs an industrial electrician with 30+ years in the field, I can tell you that the current difference between no load and full load can be several orders of magnitude in difference. If this weren't the case the majority of overload protection devices would not work as they monitor current amplitude only. The older types relied on the heating effect of a given current (I^2R nothing else), the newer ones model that same heating effect in softwareQuote from: Hero999 on October 29, 2016, 01:42:13 pmAssuming we are talking about the power a motor can output and not just the power it consumes, then mechanical constraints come into play as well such as shaft sizes, the bearings, construction of the rotor bars. A common failing is cracked rotor bars due to mechanical over loading and shock loading. Electrically, iron core saturation comes into play as wellQuote from: axero on October 29, 2016, 09:33:52 amIf the currents are imbalanced for what ever reason, imaging the following.The rotating magnetic field can be represented by a vector with an origin fixed at zero, spinning around in some direction with a fixed amplitude if all phase currents are balanced.If we take into account only one phase current and look at what that produces as a magnetic field it would be stationary and its amplitude would grow, diminish, changed direction, grow, diminish, following the currents sinusoidal amplitude. If we take all three phases' magnetic fields and sum them we get our original spinning vector.This is easier to visualise if you take just 2 phases 90 degrees out of phase with each of their poles physically displaced by 90 degrees also.Back to our original scenario with one phase having different current. Aside from now having an imbalanced 3 phase system with phase currents all over the place the magnetic field created by these currents is no longer of fixed amplitude and spinning in a nice synchronous manner. This causes the motor to sound angry as the rotor hunts and slips all over the place as it locks and unlocks from the now ruined rotating magnetic field.

author: IanB
author_rank: Super Contributor
timestamp: October 30, 2016, 01:10:43 am
content: Since a picture is worth a thousand words, here is an interesting (moving) picture:https://youtu.be/PI-OT6-TZRA

author: woodchips
author_rank: Frequent Contributor
timestamp: October 30, 2016, 10:17:44 am
content: You are making two mistakes.Firstly, that the current supplied will vary in phase angle to the voltage as the load changes, but as measured by an ammeter it will not vary much. Real and reactive power will vary considerably.Secondly, that the core flux is constant, but the source of that flux will change as the load changes.An overload works on excessive current, and if the transformer or motor is overloaded then the balancing core flux from the secondary or rotor can't do this, primary or stater current rises anyway and trips the overload.As I said, there is a complex dance between currents, flux, slip etc going on, you can't just pick one and explain everything with it.

author: IanB
author_rank: Super Contributor
timestamp: October 30, 2016, 10:40:14 am
content: Here's an example of a performance curve showing the supply current in relation to load. For this motor the ratio between no load current and full load current is about 3:1.Look also at the video linked above: the peak current of 500 A fell to just 26 A at idle (the normal full load current is 10x that at 233 A).

author: axero
author_rank: Regular Contributor
timestamp: October 30, 2016, 12:58:53 pm
content: QuoteI think I need to see something more tangible before I can say that I understand this.I'm still wondering about amperage differences between phases while the voltages are the same: Let's assume that the motor itself is working properly, but that the load it operates is unbalanced. If we visualize each revolution of the motor shaft in terms of degrees, let's say that between 0-30° the load is higher than the rest of each revolution. What then would happen to the current draw on each phase? I would suspect that one of the phases will draw more current than the other two, merely due to the uneven load. I could imagine that the uneven load may be due to a bearing that is torn on a certain spot which gives rise to slightly higher friction than on the rest of the bearing. Something that makes it "pinch" the shaft at that particular spot.