Induction motor power factor during run-up

[JPetersen]

Hi (again),

We are installing a 150kW induction motor, for a lifting device. Motor has to start on full load torque.

The terminal voltage must not drop too much during start, to ensure enough torque. My problem is how to calculate max. voltage drop. Voltage drop will of course depend on current and power factor during run-up.

Is voltage drop maximum in the excact moment of start? Or can it be expected to drop further when runing up, as power factor is changing towards full load PF?

Of course current is reduced as the motor gains speed, and thus voltage drop is reduced. But I can't figure which effect (PF change VS current change) is strongest!

When energized PF is 0.2 - full load PF approx. 0,89.

Any thoughts?

Brgds/JP

 

[rbulsara]

Motor starting voltage drop is normally calculated by assuming (which are very reasonable) the following:

Pre-starting voltage: Nominal voltage, 1 p.u.

Staring current = rated locked rotor current. You can get it from its Code letter or assume 5-6 times the FLA (5 or 6 pu.)

Impedance of the motor used: Transient reactance, usually 20-25% or 0.2-0.25 p.u. or get it from the mfr.

Power factor really have no bearing on the starting voltage dip.

Draw simple one line showing this and calculatin is very simple.

If you want to get too fancy and detailed and know the impedances of the upstream system and you can covnert them to right base, feel free to use them too.

 

[JPetersen]

rbulsara - very fast reply, thanks.

But:

Power factor has, as far as I can calculate, a large impact on voltage drop.

If assuming PF=1, then voltage drop will be in phase with supply voltage, ie a large drop.

If assuming PF=0, then voltage drop will be displaced 90 deg from supply voltage, ie absolute terminal voltage on motor is almost unchanged (cable reactance neglected)! Right??

Brgds/JP

 

[JPetersen]

formula:

dU= sqrt(3) x current x (cable_resistance x cos_phi + cable_reactance x sin_phi)

phi = phase angle of current

JP

 

[rbulsara]

The above method, in fact ignores the resistive element or assumes power factor of 0 as it will give you the most conservative result. It is viewed as simple voltage drop across a reactance. Once the motor picks up load, actual current drops as you said, but at that point you are already past the point of maximum voltage sag.

Your assumption of 0.2 pf is very reasonable, not sure how it helps determine max. votage dip which occurs at the very instant of starting when motor is drawing LRA or very close to it.

Or look at it this way when the motor is locked up, it is not rotating and no work is being done, all power is reactive at that time and that is the instant you want to know the max. voltage dip.

 

[Marke]

Hello JPetersen

The power factor of an inductrion motor at zero speed is in the order of 0.15 - 0.25 This varies from motor to motor depending on the motor design, but it is always in that order. The power factor gradually increases as the motor speed increases peaking at about the speed of maximum torque capacity.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[JPetersen]

Hi Mark,

Yes, the power factor increases - and peaking at max T seems fair.
But still: the increasing PF will increase voltage drop, according to previous formula. At the same time current is reduced because rotor-slip is reduced. So my original question is still open.

rbulsara: I think we are not talking about the same - please note I am talking about voltage drop experienced on motor terminals due to the impedance of supply cable!

Next I do not understand your statement

"The above method, in fact ignores the resistive element or assumes power factor of 0"

The resistive element is NOT ignored - power factor is not assumed zero!

Hope we can clarify what I have started ;-)

JP

 

[rbulsara]

JP:

Your questions are:

"The terminal voltage must not drop too much during start, to ensure enough torque. My problem is how to calculate max. voltage drop."

Well, what I said before still holds true. You can not ignore motors reactance when trying to determine the voltage drop during a start. You can add cable elements to the calcs. See more at end of this reply.

Your other questions are:

"Is voltage drop maximum in the excact moment of start? Or can it be expected to drop further when runing up, as power factor is changing towards full load PF?"

The answer to the first part is Yes.

The answer to the second part: No, because voltage at the motor will improve as motor comes to speed and the current "through the cable" drops.

As I said in the my first post, if you know any more data about the upstream elements, place them in the circuit (calculation).

Maximum drop thru the cable will also occur when the maximum current is passing throuh it which is when motor is drawing close to LRA when the pf is lowest, and then the VD across the cable will fall until current reaches a steady state.

I think you are confusing the power factor of the motor with the Voltage trangle you will make for the VD acorss cable using its R and X.

But in short the cable only sees the 'magnitude' of the curernt going throgh it. You take that magnitude I (max. I will be the LRA) and draw a IR and IX vector at 90 degrees to each other, add them up to get total V which is the VD across the cable. So if you know the I , the pf of the motor does not matter.

Again, the magnitude of the current will change based on the load and pf of the motor, but it will only decrease after the initial start.

 

[rbulsara]

By the way, I will look at my own write up with more clearer head when I get back home, and post any correction clarifications.

For one thing, beside the cable, you must know source impedance to figure out the Voltage at the motor terminal ( or in otherwords know the available SCC at that point).

Then you place all reacntances/impedances in series, and calculate the VD using I (LRA) through it. You only need to know the magnitude of I.

 

[polyphase]

I hope that I can add a little to this discussion without causing more confusion.

The starting PF does matter.

dV = I(R cos phi + X sin phi), where:

I = LRA of the Motor
R = The System Resistance (Cable + Source)
X = The System Reactance (Cable + Source)
cos phi = Starting PF of the Motor

If you do a series of calculations, while varying the PF from 0 through 1.0, you will find that the voltage drop goes from a lower value up to a peak and then goes back down.

The highest voltage drop will occur at the PF that is equal to the equivalent PF of the System Impedance (System R / System Z). (System Z = Sq Rt(R Sq'd + X Sq'd)).

So if you want a conservative approach, just use the System Impedance and PF to calculate the maixmum voltage drop, regardless of the motor starting PF.

Regards,

polyphase