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Topic Title: Three Phase Motor
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Created On: 03 October 2013 09:55 PM
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 03 October 2013 09:55 PM
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scottseedell

Posts: 58
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All,

Interesting one (not sure if it's in the right section):

I'm having an ongoing discussion with a colleague who insists that a three phase motor will always consume the same power once it is up to speed and who insists that once in this condition, if the voltage dropped then the current would rise in order to achieve this constant power. However, I am of the opinion that if the voltage (or current for that matter) decreased then this would result in a proportional decrease in consumed power. The basis for his point of view seems to be that once accelerated and up to synchronous speed, the motor would always turn at the same speed (n = f/p) so he seems to think that the work done will always be the same and power must be constant.

The basis for my point of view is that despite turning at a constant speed, the motor will have a reduced torque should the voltage drop and this will therefore reduce power also. The example I gave was a star/delta motor: the motor will run at the same speed in both configurations (a small percentage slip in star) but as the voltage across each winding increases by root 3 and the current also increases by root 3, the power increases by a factor of 3 despite still turning at the same speed. The additional power surely must be an increased torque.

The more I think about it, the more I'm sure that I am correct but if there are any electrical engineers out there it would be nice to hear any opinions. Accept my apologies if this is as simple as I feel it is!

-------------------------
Scott Seedell - BSc(Hons) IEng MIET
 03 October 2013 10:56 PM
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statter

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I don't agree power used will depend on the load on the motor it's behaviour under reduced voltage will depend on the type of motor. Few motors are used at full load and often larger motors are fitted when a spare is needed. A motor won't run at its rated power unless it's attached to a load that will allow the power to be developed at the motor speed.... The easiest way to think about this is a no load condition when the motor will consume very little power ...... Then consider progressive loading .... The power input must rise ..... Until the motor rating is reached. Hope this helps
 03 October 2013 11:03 PM
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mabx

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as the voltage drops the current would increase to keep the motor running at (almost) the same speed under load - until the voltage drops to the point where the motor stalls.

so your colleague is right.
 03 October 2013 11:24 PM
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Zuiko

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for a sync-motor running at constant speed, the change in supply voltage is proportional to the change in torque!

you can't get something for nothing!
 04 October 2013 07:33 AM
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statter

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.... so were all right..... as normal it depends on the details!
 04 October 2013 08:36 AM
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bartonp

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But what about the supply impedance?
LOL.
 04 October 2013 10:14 AM
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Jaymack

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Originally posted by: scottseedell
I'm having an ongoing discussion with a colleague who insists that a three phase motor will always consume the same power once it is up to speed and who insists that once in this condition, if the voltage dropped then the current would rise in order to achieve this constant power.

Considering the expression for the nameplate power of an A.C. asynchronous motor at 50 Hz. say, P (Output in Watts) = Volts (Line) × Amps (Line) × P.F. × 1.73 × Efficiency. The power output is dependent upon the connected load at the shaft. On no load, the power requirements are just sufficient to meet the copper, iron and windage losses, but with a low P.F.; the power requirement in kW of the motor increases to meet the increase in the connected torque requirement for the mechanical load, .......... if the torque requirement is too much, it will cause a motor overload. ............ Circle diagram anyone?

Regards

Edited to clarify.
In the case of a synchronous motor the Amperage and hence the kW requirement will increase to meet the connected mechanical load at the virtual synchronous speed, large motors are sometimes used for power factor correction, but require a separate means of starting and are relatively more expensive.

Edited: 04 October 2013 at 10:28 AM by Jaymack
 04 October 2013 08:54 PM
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scottseedell

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Originally posted by: mabx

as the voltage drops the current would increase to keep the motor running at (almost) the same speed under load - until the voltage drops to the point where the motor stalls.



so your colleague is right.


Hi,

I'm not saying you're wrong but can you explain how this is possible? Are we saying that voltage is not a factor in the power consumed by the motor? By thinking of a star/delta arrangement we can prove that an increased voltage results in greater available power (should the mechanical load demand it) so if this was reversed (i.e. the voltage across each winding was reduced), how would this not affect consumed power? Surely if we ran a pump up to speed in delta and then changed it to star, despite no change in rotor speed the amount of liquid being pumped would reduce (as a result of decreased torque available) and as the mechanical load (the pumped liquid) has effectively decreased, the consumed power would also drop?

I'm not being finnicky here, I just have trouble getting my head around why switching to delta results in an increase in voltage, greater applied torque and more available power to be consumed by the load, yet the reverse is somehow not true??

-------------------------
Scott Seedell - BSc(Hons) IEng MIET
 04 October 2013 08:58 PM
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scottseedell

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Originally posted by: statter

I don't agree power used will depend on the load on the motor it's behaviour under reduced voltage will depend on the type of motor. Few motors are used at full load and often larger motors are fitted when a spare is needed. A motor won't run at its rated power unless it's attached to a load that will allow the power to be developed at the motor speed.... The easiest way to think about this is a no load condition when the motor will consume very little power ...... Then consider progressive loading .... The power input must rise ..... Until the motor rating is reached. Hope this helps


Thanks but I knew this already. Maybe my original post did not explain this clearly

-------------------------
Scott Seedell - BSc(Hons) IEng MIET
 04 October 2013 09:42 PM
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mabx

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Originally posted by: scottseedell

... Surely if we ran a pump up to speed in delta and then changed it to star, despite no change in rotor speed the amount of liquid being pumped would reduce (as a result of decreased torque available) and as the mechanical load (the pumped liquid) has effectively decreased, the consumed power would also drop?


If there's no change in speed, then there's no change in the amount of liquid being pumped, or torque, or consumed power.


switching the pump from delta to star (reducing voltage) would reduce the maximum torque that the pump could produce, but if it was very lightly loaded (i.e. producing minimal torque; using minimal power) it may keep going at it's rated speed as it can still produce enough torque to maintain rpm and will still use the same amount of power.

If it's fully loaded when you switch to star, then it would stall - i.e. it couldn't maintain the torque needed to maintain rated rpm so it would fall to a low speed and pump a lot less liquid.
 05 October 2013 08:24 AM
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statter

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To summarise

I'm having an ongoing discussion with a colleague who insists that a three phase motor

Answer depends on the type of motor: synchronous will always turn at the same speed until it stalls; asynchronous e.g. induction, will not get back to exactly the same speed if voltage is reduced. the rest of this bit applies to induction motor. In this case the power will be down slightly but for small reductions in voltage we tend to neglect this. Its hard to consider any of this in the absence of any specification of the load on the motor. As has been said if the speed of the motor is exactly the same most loads will absorb the same power. If the speed drops slightly most loads will require slightly less power. How far the speed drops for a given reduction in voltage will depend how close the motor is in the first place to its maximum output. If it is very close then even a small reduction in voltage may lead to stalling. The sort of change associated with star : delta starting will be much larger and there will be a much greater prospect of the motor stalling. Think about something like a large fan where the load increases with the speed, star delta starting was often used and worked well as power absorption characteristic of the load was behind the production characteristic of the motor in star delta if well set up.


will always consume the same power once it is up to speed and who insists that once in this condition, if the voltage dropped then the current would rise in order to achieve this constant power.

.... up to a point but for e.g. induction motors it wont quite get back to where it started. so if you friend is talking exactly he is mistaken for induction motors - if he is talking practically then he is probably correct providing the motor isn't too tightly sized.

However, I am of the opinion that if the voltage (or current for that matter) decreased then this would result in a proportional decrease in consumed power.

..... not proportional but there would be small decrease for an induction motor or a larger decrease if it was closer to stalling.

The basis for his point of view seems to be that once accelerated and up to synchronous speed, the motor would always turn at the same speed (n = f/p) so he seems to think that the work done will always be the same and power must be constant.

....... synchronous motors are different. There will be no drop in power provided the motor isn't close to its rating (neglecting the small change in copper losses in the motor).

..... induction motors don't run at synchronous speed they run very close to it but speed drops slightly with load.

The basis for my point of view is that despite turning at a constant speed, the motor will have a reduced torque should the voltage drop and this will therefore reduce power also.

.... this depends on the load. If the speed is the same then the torque should be the same for a given load. if the voltage falls then what happens depends on the factors above

The example I gave was a star/delta motor: the motor will run at the same speed in both configurations (a small percentage slip in star) but as the voltage across each winding increases by root 3 and the current also increases by root 3, the power increases by a factor of 3 despite still turning at the same speed. The additional power surely must be an increased torque.


.... the power factor changes too. The eventual power factor will depend on the type of motor and how close to rated capacity it is working. The change in power factor will result in changes to voltage and current not being reflected in the power delivered.


Overall I think its important to understand that there will be a difference between the effects of small changes and larger changes especially when the characteristics of the load are taken into account. The size of the change in the end result will also critically depend on how close the motor is to its maximum output.
 05 October 2013 09:48 AM
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Jaymack

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Originally posted by: statter
I'm having an ongoing discussion with a colleague

With due respect, I suggest that this bullet maker, is the one who should be contributing to this post.

Regards
 05 October 2013 10:11 AM
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scottseedell

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Originally posted by: mabx

Originally posted by: scottseedell



... Surely if we ran a pump up to speed in delta and then changed it to star, despite no change in rotor speed the amount of liquid being pumped would reduce (as a result of decreased torque available) and as the mechanical load (the pumped liquid) has effectively decreased, the consumed power would also drop?




If there's no change in speed, then there's no change in the amount of liquid being pumped, or torque, or consumed power.


Respectfully, I'm not sure I agree with that statement. Are we saying that a star-connected centrifugal pump running at 50 Hz would deliver the same torque and pump the same volume of liquid as an identical centrifugal pump also running at 50 Hz but connected in delta? If that was the case, besides the obvious advantages in limitation of start-up current, why would they even bother changing from star to delta? Instead they could run continuously in star and save themselves 2/3 power consumption? Surely the increase of voltage across each winding provides an increased torque which can be used to sustain an increased mechanical load? By my reckoning, unless there are constant mechanical load and torque requirements, a variable load (such as a pump) would surely work harder and convert more electrical energy into kinetic energy during transition from star to delta, irrespective of rotor speed? Think of a car travelling 20mph in 1st gear and compare to an identical car travelling at 20mph in 2nd gear. The speed is the same but the torque is greatly increased in 1st gear and as a result, could sustain a greater mechanical load than 2nd gear. I'm certain we could look at pump characteristic graphs and see a rise in torque that is proportional to voltage.

-------------------------
Scott Seedell - BSc(Hons) IEng MIET
 05 October 2013 09:04 PM
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mabx

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if you re-read the bit of your text I quoted in bold:-
...and then changed it to star, despite no change in rotor speed the amount of liquid being pumped would reduce


...you state that switching to star results in no change in rotor speed. If that were the case then what I said:-

If there's no change in speed, then there's no change in the amount of liquid being pumped, or torque, or consumed power.


...holds true.

In practice however, when you switch to star there's a good chance the motor will stall as it is unable to sustain the pump at full speed and will pump less liquid.
 06 October 2013 08:46 AM
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Jaymack

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Originally posted by: mabx
In practice however, when you switch to star there's a good chance the motor will stall as it is unable to sustain the pump at full speed and will pump less liquid.

An increased torque requirement on a S.C. motor shaft will cause an increased "slip", with a consequent increase in electrical power requirement for an asynchronous motor. The key here is the slip; this is a comparatively small percentage for full load, e.g. 5%, and is the key to such motor operation. Most S.C. motors have 4 poles for economical design, and never reach the synchronous speed of 1500 RPM for 50 Hz., dependent upon the slip which in turn, directly corresponds to the load torque requirement, as you probably know. Whether a motor will trip, is dependent on the designed allowable heat balance for that motor, and hence the allowable thermal heating current etc. Star motor starting is used to reduce the load on the supply and the system downstream, with the penalty of a reduction to 33% of the comparable torque in a delta connection.

If a motor starter didn't change over to delta within a reasonable time period, the motor will still continue to run but would run with the comparative increased slip, since the torque is largely the same as with a delta connection, but could be overloaded/overheated, dependent on the load requirement at the motor shaft.

Regards
 06 October 2013 09:03 AM
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davezawadi

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I am slightly surprised at some of the comments so far. I will attempt to clarify the thinking needed in this case, as Scott is somewhat correct and his colleague is somewhat wrong, but no one seems to know exactly why!

So lets forget the pump for a moment, and just look at motors, and I will start with the induction motor.
The winding connection is not important in this case, the reasoning is exactly the same for star or delta. I have a motor powered and rotating with no external load, what real (the important bit) power does it take and what can I measure? I can measure the supply voltage and current but this does not tell me much, because under this condition the power factor is poor. I can only tell what the power loss in the motor is with a Wattmeter, which works by integrating (adding up if you like) the instantaneous voltage and current across the mains cycle. With a low power factor there will be instants where the supply voltage is high but little current is drawn (so little real power) and points where the voltage is low but the current is higher (again little real power). Our free running motor absorbs just enough real power to overcome its internal losses, and as a result does not get hot as is implied by it taking a constant power as suggested by the colleague!

Now we need to understand why the motor takes a load dependent current, and not some value controlled by its internal resistance. Motors have an interesting property which is best explained by saying that the operate as generators at the same time, producing a voltage in opposition to the supply (known as a back EMF) and the motor current is proportional to the difference between the supply and the back EMF. As the motor mechanical load is increased the back EMF falls and so the current increases, and the real power input to the motor increases as a consequence. A side effect of this is that the power factor also improves, but this is not very important to the discussion. Inside an induction motor there are some other changes as well, the slip angle and speed change slightly but again we need not consider these unless we wish to design the motor.

Lets look at the pump now to understand what load this places on the motor. The only thing that matters is that the pump will provide a fixed mechanical load at a given speed, as long as the head etc. is constant.

Back to the OP. If I have a motor and run it in star I measure a current which depends on the mechanical load (plus a few losses but these are small). If I take the same motor and connect it delta and apply the same mechanical load I should expect it to take the same real power (plus the losses which will be slightly different). Does the supply voltage actually matter? It matters in only one way, and that is that the motor can develop sufficient power output to drive the load at its proper running speed without exceeding its winding current rating (resulting in overheating), and it is not so high that the motor is damaged (excessive losses, insulation etc). An overloaded motor will run too slowly, reducing the cooling from its fan and will at some point fail.

The error is that the current will increase between star and delta connection, whereas the answer is that it will be less in delta, to keep the absorbed power identical (ignoring losses)! That may seem very counter intuitive, and voltage and current readings may be misleading because of the phase differences, so I see the root of the discussion between colleagues! The important bit is that the motor is essentially self regulating, it takes real power input to exactly match its load (ignoring losses) however you connect it up. If it cannot develop the necessary power output it will slow or stall, and get very hot in the process!

The synchronous motor is exactly the same except that its load has a much greater effect on the power factor, and it either runs at synchronous speed or stalls. Unloaded it has a very high leading power factor which is why it may be used to correct system power factors under some conditions.

A more complete explanation may be found in Hughes "Electrical Technology" (there is mathematical treatment!)

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David
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 06 October 2013 09:09 AM
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Jaymack

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Originally posted by: davezawadi
A more complete explanation may be found in Hughes "Electrical Technology" (there is mathematical treatment!)

Read it?, I wrote it!

Just a poor country boy, trying to earn a crust.

Regards
 06 October 2013 09:20 AM
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Jaymack

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Originally posted by: davezawadi

I am slightly surprised at some of the comments so far. I will attempt to clarify the thinking needed in this case, as Scott is somewhat correct and his colleague is somewhat wrong, but no one seems to know exactly why!

Pontificating DC rides again!

Regards
 06 October 2013 04:25 PM
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davezawadi

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Jaymack, if you wrote it then why did you not explain correctly?

Never mind, just keep mocking those that really try to help, fun isn't it?

-------------------------
David
CEng etc, don't ask, its a result not a question!
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