Over correction

Status
Not open for further replies.
B

Bugman1400

Senior Member
Location
Charlotte, NC
I'd like to see the theoretical when you get a few mins. I am still having a tough time understanding how an induction motor can be turned into a generator because of excess PFC. Since the motor is driving the load and not vice versa, the motor cannot supply any kWs. So, you must be referring to kVars. However, seems like caps would do that anyway without the need for a motor or generator.

Thanks in advance for the lesson.
 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
@ Bugman...

Once again, I strongly disagree. The motor can deliver kW's! But, I suggest it be discussed in a separate thread!


@ PtonSparky...

My response to your original query was, "How does PFC impact an induction motor's Overload setting?" I gave a 3-part answer:

1) "If the PFC is switched with the motor, problems could occur!"

2) I said a motor's PF is not materially changed, but instead the PFC, offsets the motor's kVAr!

3) Analysis of your data revealed an increase in kW flowing into the motor's feeder! I said that although small, it could be explained.


To answer the 3rd question let's start with several facts!

A) When PFC is applied, at the motor starter, the VPP, and VLN may rise!

B) The line-current, IL, should drop!

C) Although a motor's usually represented as a constant kVA load, i.e., when voltage increases, amps decrease, but not in all cases!

D) A capacitor's load-factor is 100%!


Stay tuned! Regard, Phil Corso
 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Gentlemen... finally, THE END!

This discussion has one goal: to make clear(er) why PFC, with its promise of loss reduction, did not materially alter power consumption as measured on the surface.

Consider an example:
A motor having a 100 kW power draw, and a 90% efficiency! The power-draw must satisfy three conditions: a) conversion to mechanical output (90% x 100 => 90kW); b) winding-loss (say 8% x 100 => 8kW): and c) core-loss (say 2% x 100 =>2kW)! Now include submersible-cable losses (say 8kW)! Thus, surface-measurement should show a 108kW power-draw from the power source, while down-hole the motor's power-draw is 100kW!

How are losses affected by voltage and current?
Theoretically, the power-conversion value is fixed; winding and cable-losses vary with the square of current; and core-losses vary with terminal voltage! Assume the PFC reduces current by 50%, resulting in a reduced cable V-drop! Down-hole, the motor's terminal voltage-rises 5%!

How are losses in the example affected?
Winding-loss is reduced (0.5² or 0.25 x 8 = 2kW), cable-loss is reduced (0.5² or 0.25 x 8 = 2kW), and core-loss is increased (1.05 x 2kW = 2.1kW). Theoretically, then, the surface-measurement should be (90kW+2kW+2kW+2.1kW) ~ 96kW !

Why, then (in PtonSparky’s case) did power-draw not decrease?
Instead, it increased a bit! The resultant improvement in the motor’s performance was a decrease in motor-slip, hence an increase in motor speed. Flow-rate, per The Affinity Law, increased in proportion to speed raised to some exponent! That exponent is between 2 and 3, depending on pump design. It follows then, that there is a corresponding increase in mechanical power requirement.

Regards, Phil
 
Last edited:
ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
:thumbsup:
Phil Corso said:
Gentlemen... finally, THE END!

This discussion has one goal: to make clear(er) why PFC, with its promise of loss reduction, did not materially alter power consumption as measured on the surface.

Consider an example:
A motor having a 100 kW power draw, and a 90% efficiency! The power-draw must satisfy three conditions: a) conversion to mechanical output (90% x 100 => 90kW); b) winding-loss (say 8% x 100 => 8kW): and c) core-loss (say 2% x 100 =>2kW)! Now include submersible-cable losses (say 8kW)! Thus, surface-measurement should show a 108kW power-draw from the power source, while down-hole the motor's power-draw is 100kW!

How are losses affected by voltage and current?
Theoretically, the power-conversion value is fixed; winding and cable-losses vary with the square of current; and core-losses vary with terminal voltage! Assume the PFC reduces current by 50%, resulting in a reduced cable V-drop! Down-hole, the motor's terminal voltage-rises 5%!

How are losses in the example affected?
Winding-loss is reduced (0.5² or 0.25 x 8 = 2kW), cable-loss is reduced (0.5² or 0.25 x 8 = 2kW), and core-loss is increased (1.05 x 2kW = 2.1kW). Theoretically, then, the surface-measurement should be (90kW+2kW+2kW+2.1kW) ~ 96kW !

Why, then (in PtonSparky’s case) did power-draw not decrease?
Instead, it increased a bit! The resultant improvement in the motor’s performance was a decrease in motor-slip, hence an increase in motor speed. Flow-rate, per The Affinity Law, increased in proportion to speed raised to some exponent! That exponent is between 2 and 3, depending on pump design. It follows then, that there is a corresponding increase in mechanical power requirement.

Regards, Phil
Click to expand...
Thank you.
 
kwired

kwired

Electron manager
Location
NE Nebraska
so does that mean we did more work?

With a well pump, that would have to mean we pumped more water. If not where did that increase in energy go? If it went anywhere besides pumping more water that would be an inefficiency wouldn't it? I guess the motor maybe wouldn't have more losses but maybe the pump could have increased mechanical losses if it is "overdriven".
 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
kW...

I guess it's not THE END!

BTW, it's not increased energy, which is power x time. The Power increase is explained in the example, and I repeat, it's due to the reduction in cable and motor losses! Perhaps you have another theory why power-draw did not decrease with the PFC connected?

I'm more than happy to send you the Excel calc using generic data, just contact me!

Phil
 
kwired

kwired

Electron manager
Location
NE Nebraska
Phil Corso said:
kW...

I guess it's not THE END!

BTW, it's not increased energy, which is power x time. The Power increase is explained in the example, and I repeat, it's due to the reduction in cable and motor losses! Perhaps you have another theory why power-draw did not decrease with the PFC connected?

I'm more than happy to send you the Excel calc using generic data, just contact me!

Phil
Click to expand...
I will assume the motor output remained fairly constant, maybe by reducing some losses we have less slip and more potential output - but how the driven load responds will have an impact as well. If you speed up a variable torque load because you have less slip in the driving motor then you do increase the load on the output shaft. So did Tom pump more water as a result of the power increase? If he had fewer losses but still increased power then more work must have been done, or losses were increased someplace else.
 
Besoeker

Besoeker

Senior Member
Location
UK
ptonsparky said:
With less slip I would expect a bit more work being done.
Click to expand...
More slip is more torque. The motor will slip enough to produce the torque required by the load.
Power output is torque times speed. Slip is quite small, typically not more than 1% for the motors I've dealt with so we have close to constant speed. Thus more slip is more power.

TorqueandCurrentvs_zpse1150a90.jpg
 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Besoeker,

I said slip decreases, hence speed increases! If speed increases, then the pump's flow-rate increases, thus requiring more power to the motor! But the power increase is not proportional to the speed increase! Instead it increases exponentially.

The phenomenon, referred to as The Law of Affinity, is expressed mathematically as:

Pi = Po x (Ni / No)X , where the subscripts i and o, represents the increased, and original conditions, respectively. And as I said in my discussion X can be between 2 and 3... but in most cases it's = 3!

Thus, the reduction of power input because of loss reduction is replaced with the increase of power required because of the Affinity Law!

Phil
 
Last edited:
W

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Besoeker is describing the power output law of the motor. As slip _decreases_ motor torque _decreases_, and thus power output goes _down_. This is only true if all other parameters are held constant, eg. terminal voltage. For slip to decrease, speed _increases_ slightly.

Phil Corso is describing the power _input_ law for the pump. As speed increases the power consumed by the pump increases.

When the motor drives the pump, the power output of the motor must equal the power input to the pump.

If a change to the supply circuit causes an increase in system speed, then I'd have to agree that system power has increased. The pump is consuming more mechanical power and thus the motor must be producing more mechanical power. The fact that slip has decreased means that something else must have changed to increase the torque/slip characteristic.

For example, if the supply circuit change causes an increase in terminal voltage, then this will increase the torque per slip, and increase the power output of the motor even with reduced slip.

-Jon
 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Winnie...

Thanks! Perhaps a practical exercise might be useful!

Consider a speed change from 1,750 rpm to 1760 rpm. The percent speed change is ~ 1/2%. But the pump's brake-Hp would increase by ~2%!

This problem arose shortly after the introduction of NEMA's Premium Efficiency Motor! The expected efficiency never materialized!


Phil
 
Last edited:
Besoeker

Besoeker

Senior Member
Location
UK
Phil Corso said:
Besoeker,

I said slip decreases, hence speed increases!
Click to expand...
You seem to have missed the vital point.

The slip decreases if the load torque decreases. Since the slip on a typical cage motor is very small as I pointed out and, as my diagram shows, that can only happen with reduced load power requirements.

Your attempt at a fancy mathematical explanation does not change the basic motor characteristics.
 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Besoeker...

Hmmm, so for a 2-pole, 50 Hz, with a slip = 0.05, then its rpm =1,425! But if its slip is lower, say 0.04, then the motor speed is not 1,440 rpm, but should be less than 1,425! Thus, you believe believe the formula, Nr = Ns (1.0-slip) is incorrect!

Wow, I can't imagine what you will come up for the case when slip = 1.0!

Phil





 
P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Sorry, I have a rather slow keyboard and I hit the wrong key!

Besoeker, I believe that you overlooked the fact the at a higher terminal-voltage, the torque-rpm curve shifts higher!

Phil
 
Status
Not open for further replies.
Top