Power factor correction experiment

[Besoeker]

I tried my level best.I am not able to see the wrong in the calculation.Can't you do any thing about it?

The values for efficiency and power factor are not the same at high values of slip as at operation at rated duty.
I thought my comment "a slip up on your part" might have been a clue.
In round figures, your example would give half the volts but three times the current.
The power factor simply cannot be the same. With the higher current, the losses will be greater so the power factor simply cannot be the same.

If you do not mind,please give a description how an auto-transformer starter works and compare the current it delivers to motor with those you mentioned.You will then see those are starting currents of motor and not its running currents.

Simple. The auto-transformer is a reduced voltage starter. It reduces the starting and run up current before switching (preferably with closed transition) to full voltage.
We don't do many of those. The last one of those we did was a couple of years ago and rated at 6600kW.
But what does that have to do with running currents?

That rotor resistor is an optional item.

Install resistors. Exercise caution when wiring as the
resistor element operates at 375 ?C (707?F).

 

[Besoeker]

111205-0919 EST

Besoeker:

The Kill-A-Watt is a meter with a three prong plug and a three prong socket. This meter is for individual plug-in loads. In between is the meter and I suspect a resistor shunt for current measurement.

Maximum current is 15 A and it complains above this. For the EZ model the voltage range is spec'd at 85 to 125 V with 0.1 V resolution. It doesn't complain on any overvoltage to 135 V (I have not taken it higher). It is RMS reading and measures or calculates --- voltage, current, power, VA, frequency, and power factor.

Below about 1 A it resolves 0.1 W in the newer EZ units, doesn't work too well below about 2 W on some units. Above something around 1 A it switches to 1 W resolution.

For about $30 at Home Depot it is a rather good value and useful with knowledge of its limitations.

Handles reactive loads very well as indicated by data from measurements that I have made. A limited useful device for the average consumer. No data collection except by the eyeball.


The TED Systems consist of two different designs, but similar. TED stands for "The Energy Detective". These systems are basically power and energy monitors with data collection possible. Voltage and power are directly measured, albeit, with substantial error on power of non-resistive loads. Time is measured with a Dallas real time clock chip, a rather good clock. Current is measured but only available as VA. On the 5000 series power factor is obtained somehow. For low power factor they apparently fudge VA to equal about 15 possibly to avoid a divide by zero problem.

Spec'd for 100 to 130 V with 0.1 V resolution. Maximum current per current transformer is 200 A. Two current transformers are supplied, one for each phase of the center tapped single phase system. Thus, maximum power is about 200 * 240 = 48,000 watts.

Power Line Communication is used and is a major problem. Software and circuit design are also major problems. There is much to be disliked about the TED systems, but for the money it can provide useful information.

This is also a useful system for the average consumer at $165 to $240 for a minimum system. It is intended for whole house monitoring. But it is worthless for evaluating a PFC capacitor.

Will $250 plus spent of the TED system save the average consumer that amount. Maybe, if it helps them keep things turned off that do not need to be on. In other words change their life style.

Gar

Thank you for the very informative post.
I haven't seen such devices marketed in UK. I've checked the websites for a couple of our distributors and got no hits.
My measurements were made with the instruments I routinely use.

I think your comment about life style changes is spot on. It needs good habits and a bit of common sense for the average consumer to save rather than waste energy.

 

[mivey]

Simple. The auto-transformer is a reduced voltage starter...

Bes,

T.M. must be thinking along the lines of slide 8 from here:

http://www.me.ua.edu/me360/spring05/Notes/Topic17-AC_Motor_Speed_and_Other_Motors_sv.pdf

and "high slip" usually means around 12% or so for running slip instead of the usual 2-4%. But, like has been said, the efficiency gets a lot worse as you lower the voltage.

 

[dkarst]

That was put forward just to show you that an induction motor speed can be varied greatly by change in stator voltage only.

I just know I'm going to wake up in the morning and say "What was I thinking hopping in on this thread 300+ posts down the line" but here is a more heuristic type explanation considering the generally accepted motor circuit model has failed us. This is also the way my simple mind works.
The statement in the box above is generally false and here's why...

1. I'm going to make the base assumption that we are talking about cage-type motors so there will be no discussion of external rotor resistance etc.

2. At start up, I think of the rotor at standstill (very high slip), the stator field is zooming around and flux lines are cutting the rotor at a high rate, now think of the motor model as similar to transformer... end result --> high starting current

3. At rated speed/load/voltage/current... the slip is reasonably low (say 2 - 3%), resulting in flux lines cutting rotor conductors at rational rate and current therefore the motor draws rated current.

4. Under the conditions described in previous posts with the slip at ~40% and applied voltage at 55%, even when the load is a fan-type centrifugal load and therefore the required torque drops significantly, the flux lines are cutting the rotor much too quickly and the resultant current is higher than at rated load (estimated to be ~3pu).

Sorry to hop in and back out so quickly but thought if there was a chance this helps anyone is was worth my effort typing. Bye as this will be my only post on topic.

 

[Besoeker]

Bes,

T.M. must be thinking along the lines of slide 8 from here:

Not numbed on my download.
The last maybe?

 

[mivey]

Not numbed on my download.
The last maybe?

Page #8. They are not numbered so just step through or look at the page count in the reader you use.

Labeled "Variable Voltage Transformer"

 

[Besoeker]

Page #8. They are not numbered so just step through or look at the page count in the reader you use.

Labeled "Variable Voltage Transformer"

TY.
Appreciated.

 

[T.M.Haja Sahib]

But, like has been said, the efficiency gets a lot worse as you lower the voltage.

This is not the case.Please refer to any motor soft starter with energy saving feature.It reduces voltage to the motor at part load to reduce power consumption i.e efficiency does not decrease when voltage is reduced to partly loaded motor.See also post #151:The power factor also increases at reduced 'star' voltage.

Please note that I am not responsible for any confusion.

See post #151 again.The load torque at 'delta' speed of water pump motor is less than that at 'star' speed of speed, as can be seen from the motor current:Its 'star'current is less than its 'delta' current.

If the load torque remains the same at 'star' connection also,the motor current would have increased much to the delight of Besoeker and others.Alas,it does not,because it is a fan type load.Its torque requirement at 'star; speed is less than that at 'delta' speed and so its current is correspondingly less.

The critical question Besoeker and others did not ask,while objecting to my calculation in post #244 is this:Can the load torque curve of the subject centrifugal load intersect the unstable region of the torque-speed curve of o.55 p.u voltage of motor of post # 170? The answer is NO.

The result?

Even though the induction motor is capable of operating in the unstable region for suitable load torque profiles,the centrifugal load curve does not intersect the torque-speed curve in the unstable region of the motor of post #170 at 0.55 p.u voltage.It could intersect the above motor curve only at the stable region and hence the motor speed can not be reduced greatly for the motor of post#170 . And generally for induction motors with very low rotor resistance.

 

[T.M.Haja Sahib]

the motor speed can not be reduced greatly for the motor of post#170 . And generally for induction motors with very low rotor resistance.

But if the fan type load torque curve is such that it intersects the motor curve at the unstable region for still more reduced supply voltage,operation of the motor at greatly reduced speed is possible anyway.

 

[Besoeker]

This is not the case.

If you are operating at high slip, which is what your are suggesting, it is.

See also post #151

Yes. The one where significant voltage reduction gives insignificant speed reduction. And, as far as I recall, the only real world example you have given to support your assertions that the speed of an induction motor can be varied greatly with reduction in supply voltage. Let me remind you again that the speed change resulting from that significant voltage reduction was just 1%.

The power factor also increases at reduced 'star' voltage.

Again rendering your assumption "same efficiency and power factor values" from post #244 invalid.
And maybe this too:
Please note that I am not responsible for any confusion.

See post #151 again.The load torque at 'delta' speed of water pump motor is less than that at 'star' speed of speed, as can be seen from the motor current:Its 'star'current is less than its 'delta' current.

Not really. There is no direct relationship between torque and current for an induction motor. As I'm sure you, as an expert would know, typically an induction motor will take 30% of rated current at no load. But yes, the torque would be less given that 1% speed reduction assuming it is a centrifugal type load.

The critical question Besoeker and others did not ask,while objecting to my calculation in post #244 is this:

I objected to it because it had invalid and incorrect assumptions. No question about it.

Even though the induction motor is capable of operating in the unstable region for suitable load torque profiles,the centrifugal load curve does not intersect the torque-speed curve in the unstable region of the motor of post #170 at 0.55 p.u voltage.It could intersect the above motor curve only at the stable region and hence the motor speed can not be reduced greatly for the motor of post#170 . And generally for induction motors with very low rotor resistance.

Ya mean REAL motors?!!!

 

[T.M.Haja Sahib]

Besoeker sir,

How is it possible that any time I log in,you also come in,even though we are separated by time zones.Here time now is 3.10 p.m.Are you taking any rest or not.

 

[Besoeker]

Besoeker sir,

How is it possible that any time I log in,you also come in,even though we are separated by time zones.Here time now is 3.10 p.m.Are you taking any rest or not.

I'm psychic.....obviously.

 

[T.M.Haja Sahib]

If you are operating at high slip, which is what your are suggesting, it is.

But tell me if you agree to the following and, if not,please state the reason.

Torque developed by a motor is proportional to the square of the motor voltage.At steady state operation, the motor torque is equal to the load torque.So load torque is proportional to the square of supply voltage at steady state.In other words the supply voltage is approximately proportional to the square root of load torque.So the voltage controller such as an auto-transformer can reduce the supply voltage to the motor approximately proportional to the square root of load torque.

 

[Besoeker]

.So load torque is proportional to the square of supply voltage at steady state.

Not really.
It could be full rated voltage at any load including zero if, for example, the motor was running uncoupled.
It could even be a negative load if the motor is being driven.

 

[T.M.Haja Sahib]

Not really.
It could be full rated voltage at any load including zero if, for example, the motor was running uncoupled.
It could even be a negative load if the motor is being driven.

I said at equilibrium between motor torque and applied load torque.Do you now agree?

 

[Besoeker]

I said at equilibrium between motor torque and applied load torque.Do you now agree?

Most induction motors run at supply voltage regardless of loading.

 

[T.M.Haja Sahib]

Most induction motors run at supply voltage regardless of loading.

I do not know why you do not seem to get what I am saying.

The torque developed by the induction motor is approximately proportional to the square of applied voltage and hence the applied voltage is directly proportional to the square root of the load torque at steady state.So voltage of the variable voltage controller applied to the motor is approximately proportional to the square root of load torque.An IEEE paper states in its abstract
(see http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=4111787 ) the following

''Efficiency is shown to be independent of output power when a variable voltage controller reduces the voltage approximately as the square root of the load torque to maintain the required slip during part load operation.''

 

[GeorgeB]

Not really. There is no direct relationship between torque and current for an induction motor. As I'm sure you, as an expert would know, typically an induction motor will take 30% of rated current at no load.

Sticking my nose in where I was not invited ... Semantics ... no LINEAR relationship, but yes, a direct versus an inverse relationship. AFAIK, with an operating motor at a constant voltage, there is no case where an increase in current does not reflect an increase in torque load. Baldor has published very good empirically determined APPROXIMATIONS of approximate load data given unloaded current, actual current, and nameplate FLA. I'll post from their downloadable documents if desired.

 

[dkarst]

Sticking my nose in where I was not invited ... Semantics ... no LINEAR relationship, but yes, a direct versus an inverse relationship. AFAIK, with an operating motor at a constant voltage, there is no case where an increase in current does not reflect an increase in torque load. Baldor has published very good empirically determined APPROXIMATIONS of approximate load data given unloaded current, actual current, and nameplate FLA. I'll post from their downloadable documents if desired.

I believe what you stated is true for points near the rated operating point and think that is shown by the references from the "Cowern papers" from Baldor. You do have to recall that here it is being proposed to operate far away from this point. In fact, just to the "left" of the peak of the Torque-speed curve, the torque is increasing and current is decreasing (due to slip decreasing). I would agree that this is not a normal operating mode.

 

[Besoeker]

hence the applied voltage is directly proportional to the square root of the load torque at steady state.

The applied voltage is the applied voltage. It doesn't depend on torque.

''Efficiency is shown to be independent of output power when a variable voltage controller reduces the voltage approximately as the square root of the load torque to maintain the required slip during part load operation.''

I have no disagreement with that at all. But please note the "maintain the required slip" part of this. Same slip, same speed.