Induction Motor Question

But, if by overload, you're talking 'overvoltage' the power factor will decrease.

And, since we're on the topic, the power factor will increase for under-voltage.

Well an overloaded motor heats up, that should be your first clue...

Is the motor to be supplied from this?

http://cr4.globalspec.com/thread/76700

Hire a consultant and then resign.

Mechanical overload to the point of a stalled rotor. Electrical overload in the form of excessive voltage causing stator winding arcing. Thermal overloading to the point of burning insulation. Which these types of an overload are you talking about?

As mechanical load increases, slip increases, motor rpm decreases. The graph below shows the resulting power factor. http://what-when-how.com/induction-motor/steady-state-characteristics-induction-motor/

Your curve is true for only one of the many different rotor designs. (Looks like a NEMA A or B type) There are many different rotor designs.

An induction motor runs at very low power factor when unloaded, . . . it power factor improves - increases - as the motor is loaded.

It will have the highest power factor at its rated load.

Do some reading:

ELECTRICITY FUNDAMENTALS

Fundamentals of Electricity - Eaton

Learning 101-Basics of Electricity - Newark

In my opinion, from 0.5 to 1.25 p.u. the power factor will rise a bit[1-2%]-the resistance grows up with temperature and the reactance will stay constant [if the voltage will be -approx.-constant].

As a rule-of-thumb PF gradually improves as it speeds up to rated output then reduces when speed drops due to excess load (on the shaft) and then to zero effect when it trips out.

If it does not trip then it will heat up. I'm not sure what actually happens to the PF during the minutes it takes to burn out when it catches fire.

Dear Mr.magdumshirish,

The three phase Induction Motor - the load current will follow the CIRCLE DIAGRAM.

The DC Motor, where the load current is a linear function of the load - it will foppow the pattern of y = m.x + c.

Induction Motor load current will follow the locii of a circle in which the Power Factor will be poor at no-load say 0.35 to 0.4 (approximately 35 t0 40 percent of full load current) and at short circuit current level (say 4.5 to 6 times of the full load current) the Power Factor will be still poorer.

The sort circuit current is calculated, by locking the rotor and inject low voltage so that the full load current is circulated in the winding, and the ratio of the rated voltage to this low voltage is to be multiplied by the full load current.

These two current value points lie on the circle, join these two points and draw a bisector and draw a circle. This circle is the locii for the load current.

Refer the link below and know more about Circle Diagram. You can also refer Standard Text Book.

http://kbreee.blogspot.in/2013/08/circle-diagram-of-induction-motor.html

Circle Diagram of Induction motor

By using the data obtained from the no load test and the blocked rotor test, the circle diagram can be drawn using the following steps :

Step 1 : Take reference phasor V as vertical (Y-axis).

Step 2 : Select suitable current scale such that diameter of circle is about 20 to 30 cm.

Step3 : From no load test, I_o and are Φ_o obtained. Draw vector I_o, lagging V by angle Φ_o. This is the line OO' as shown in the Fig. 1.

Step 4 : Draw horizontal line through extremity of I_o i.e. O', parallel to horizontal axis.

Step 5 : Draw the current I_SN calculated from I_sc with the same scale, lagging V by angle Φ_sc, from the origin O. This is phasor OA as shown in the Fig. 1.

Step 6 : Join O'A is called output line.

Step 7 : Draw a perpendicular bisector of O'A. Extend it to meet line O'B at point C. This is the centre of the circle.

Step 8 : Draw the circle, with C as a center and radius equal to O'C. This meets the horizontal line drawn from O' at B as shown in the Fig. 1.

Step 9 : Draw the perpendicular from point A on the horizontal axis, to meet O'B line at F and meet horizontal axis at D.

Step 10 : Torque line.

The torque line separates stator and rotor copper losses.

Note that as voltage axis is vertical, all the vertical distances are proportional to active components of currents or power inputs, if measured at appropriate scale.

Thus the vertical distance AD represents power input at short circuit i.e. W_SN, now which consists of core loss and stator, rotor copper losses.

Now FD = O'G = Fixed loss

Where O'G is drawn perpendicular from O' on horizontal axis. This represents power input on no load i.e. fixed loss.

Hence AF α Sum of stator and rotor copper losses

Then point E can be located as,

AE/EF = Rotor copper loss / Stator copper loss

The line O'E under this condition is called torque line.

Power scale = W_SN/l(AD) W/cm

where l(AD) = Distance AD in cm

Location of Point E : In a slip ring induction motor, the stator resistance per phase R₁ and rotor resistance per phase R₂ can be easily measured. Similarly by introducing ammeters in stator and rotor circuit, the currents I₁ and I₂ also can be measured.

K = I₁/I₂ = Transformation ratio

Now AF/EF = Rotor copper loss / Stator copper loss = (I₂²R₂)/(I₁²R₁) = (R₂/R₂)(I₂²/I₁²) = (R₂/R₂).(1/K²)

But R₂'= R₂/K² = Rotor resistance referred to stator

AE/EF = R₂'/R₁

Thus point E can be obtained by dividing line AF in the ratio R₂' to R₁.

In a squirrel cage motor, the stator resistance can be measured by conducting resistance tset.

Stator copper loss = 3I_SN² R₁ where I_SN is phase value.

Neglecting core loss, W_SN = Stator Cu loss + Rotor Cu loss

Rotor copper loss = W_SN - 3I_SN² R₁

AE/EF = (W_SN - 3I_SN² R₁)/(3I_SN² R₁)

Dividing line AF in this ratio, the point E can be obtained and hence O'E represents torque line.

Note: I tried to put the diagram but not depicting. Pl. open the link above and you can find the diagram.

DHAYANANDHAN.S

A great deal of information and good of you to take the trouble to present it.

But where does the OP find the applicable overload conditions - and the accompanying power factor.

<...applicable overload conditions...>

Overload is not applicable, because if the installation is correctly installed and the protection arrangements are correctly set up, overload simply will not happen.

If overload were applicable, then the original poster risks being unpopular with the facility's fire insurance company.

Yes....but during 'overload' what effect does it have on the power factor - is what the OP asked. Although the circle diagram provides lots of information - it is not at all clear where overload occurs and what the PF is.

Overload needs to be defined. In this context here, overload to me is the motor not being able to deliver the required mechanical shaft power and thus slows down under load.

To my mind, motor rating is a matter of opinion, backed up by appropriate facts, and boils down to safe dissipation of heat

I learn't this lesson many years ago and was chatting to an aero engineer when he showed me a small transformer. It was only about one inch cube and rated at 2 kw he said. I said I did not believe it because at that load it would surely burn out in a few minutes. He said it does. So I asked what use it was. He said it was used in an air-to-air missile - it didn't have to last long - it would be destroyed when it hit the target - keeping the weight down was most important.

Pure anecdote of course - no supporting facts - but I like the story.

<...during 'overload' what effect does it have on the power factor...>

It's like asking what is further north than the North Pole. The question is nonsense.

A correctly-set overload protection device will disconnect the <...overload...> from the supply, at which point the concept of <...power factor...> is irrelevant.

<unsubscribes>

I've now glanced at the blog location link you posted and have some apprehensions in this analysis technique. There are several entries that bother me :

• Slip s = Rotor Cu loss = QR/PR
• Slip is defined as a percentage difference between synchronous speed and running speed divided by synchronous speed. This page implies that slip is constant. It is not constant.
• The torque is the rotor input in synchronous watts.
• Wrong units. Also to get an induction motor to be perfectly synchronous ( 0% slip) an external torque must be applied.
• Location of Point E : In a slip ring induction motor, the stator resistance per phase R₁ and rotor resistance per phase R₂ can be easily measured.
• Induction motors do not have slip rings. The current in the rotor gets there by induction coupling (like a transformer). Synchronous motors have slip rings.

This might be a valid technique for some application but it does not explain that application well.

The circle diagram is well known for many years-I learned it 50 years ago in the college. It is interesting of course. But are some limits.

The circle diagram is based on Steinmetz induction motor equivalent circuit. The motor parameters are considered constant. Actually, the rotor resistance and leakage reactance are variable as the slip changes-the rotor frequency changes.Even in the normal bar case-not considered dip-bars-the resistance presents skin effect and the reactance changes all the time.

In order to overcome this phenomenon some authors purpose two circle diagram as in the double-cage case. See- for instance:

http://machineryequipmentonline.com/electric-equipment/computations-and-circle-diagramsdouble-squirrel-cage-motor/

By the way it is difficult to see from this diagram how the power factor behaves in an overload case.

...back in those days I recall hours of study on the subject where magnetic flux distribution around the teeth of the slots in the rotor and stator and across the air gap was covered - boring, boring, boring - no doubt important at the time (to pass exams) but something never ever used by me since leaving school - until this OP popped up.

I have forgotten most of it, but I do remember the circle diagram was constructed using basic principles (Steinmetz you say - I've forgotten) - so you end up with results extracted from the arc that are no more accurate than your basic assumptions about Steinmetz.

Better than nothing I suppose. But as you say it is not clear how the power factor performs - where you have to make assumptions to work it out.

As I thought the text from the earlier blog using the circle diagram is flawed. For those who wish to make a circle diagram the instructions and methodology to use for analysis can be found here. This graphical analysis technique is based on Steinmetz model but is attributed to A. Heyland and B. A. Behrend.

Thank you redfred!

Thanks Redfred

The link was a trip down Memory Lane .... brought back distant memories of burning the midnight oil drawing circle diagrams with the aid of a ruler, protractor, compass, slide-rule, trig and log tables, drawn on graph paper using pen and pencil....and definitely a rubber...

...and in the lab measuring LRA and torque etc to construct the circle - and then later to measure brake hp and speed under load etc to check out homework results....and then to explain why sometimes there big variation in theory and practice.

It's surprising how much 'forgotten' detail came back after 50 years....but don't ask me to sit an exam.

But back to the OP - your link tells how to work out the power factor --- Great.

But without all this theory, I know from my work with air compressors over the years that PF is very low when compressors run off-load - thus getting better on full load - but as to what the PF actually is in overload conditions I can't say, I've used my supply of midnight oil already.....so I leave the OP to play with the circle diagram.....

Perhaps the OP can come back and tell us.

Respectd sir,

Thanking you for guiding me in understanding induction

motor.

s s magdum

The only people who want motors to run in overload are the motor manufacturing companies and the motor rewinding companies. It's so good for business! Motors should never be run in overload!

Seriously, who gives a damn about power factor other than those trying to shave a few quid from their payments to the utilíty company?

If you think you've got a problem, hire a consultant to investigate, and to recommend the solution.

Here is a typical chart that includes the effect of overloading the motor on the power factor.

It only goes to 120% load, but as you can see, the slip increases, efficiency drops slightly, the the current rises rapidly, and the PF stays relatively flat. I have an old book version that takes this out beyond 120% to 150%, and the curves continue on those trends, with current increasing more rapidly and efficiency dropping more rapidly, but PF remaining relatively flat.

No idea why this is important to anyone, given that overloading a motor is dangerous to begin with, so what the power factor does is somewhat moot. It's kind of like asking about what effects being submerged had on the furniture polish used on the Titanic. But here you go.

....what effects being submerged had on the furniture polish used on the Titanic....

...or the toxic risk of using lead bullets

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Power factor of the motor reduces with overload or underload. It is maximum when the motor is loaded to rated capacity.