Vd at locked rotor

[coulter]

Question(2) of the week:

Again, I looking for opinions (and the science/industry practices behind them).

The installation
MCC is 480V, 3ph, Z grounded. Motor is 31A, 225LRA, high starting torque load. Feeder is 3 - #6CU, Okonite CLX, MCHL (Al sheath), ~1000ft.

The quandry:
I am looking at the voltage drop under starting current. So I go to Ch 9, Table 9 and use the equation in note 2:
Ze = R cos(θ) + XL sin(θ)

For R, I used the value in the Al conduit column R = .49 ohms. For XL, I used the value in the AL conduit column XL = .051 ohms. For power factor, I considered the load at locked rotor to to be pure inductive, cos(θ) = 0, and sin(θ) = 1.

Which gives Vd@LR = (1.732)(225)(.051) = 20V

For a check, calculating the Vd at FLA. using pf = .85, and the effective Z for CU in Al conduit, Ze = .45 ohms
Vd@FLA = (1.732)(31)(.45) = 24V

The Question: Is this right? It seems strange to me the Vd at LR is less than the Vd at FLA. What am I missing?

Amazingly in the last 35 years this hasn't come up and I really haven't paid much attention to it. Usually I have used I = FLA, pf = .85, Ze from the .85pf column. Probably the first time I have used pf = 0 and Ze = XL. Today it is bugging me. I'm going to dig out my red book and maybe Electric Machinery Fundamentals.

Any thoughts are graciously appreciated.

carl (no code to hide in and not yet understanding the science :-? )

 

[kingpb]

I believe the current you are looking at in the equation is for the cable voltage drop. Therefore the current through the cable is the locked rotor current. You are correct in that the load during starting is mostly reactance, hence the low power factor on start, e.g. 0.14, 0.17, etc. But it is not 0.00.
Assuming a starting pf = 0.17

Z = 0.49(.17) + 0.051(sin(acos(.17))
Z = .134

Vd = 1.732 x 0.134 x 225
Vd = 52.22V

running @ pf = 0.85
Vd = 1.732 x 0.45 x 31
Vd = 24 V

 

[coulter]

king -

Again, thanks for taking time to look at this. I have the vector math down now. It was pretty easy once I sketched it out and put some numbers to it - that and the picture in the red book:roll:

One part eludes me. You suggested to use an LR pf of .14 - .17, which looks reasonable. Other than doing the IEEE tests - suggested in Chapman - where does this range come from?

I ran a couple of calcs assuming the inductance didn't change from running at FLA to LR - that didn't pan out, oboviously it does. Chapman agrees. Surely someone out there has a paper on this. (I know, "Don't call you Shirley".)

carl

 

[winnie]

You are correct in that the load during starting is mostly reactance, hence the low power factor on start, e.g. 0.14, 0.17, etc. But it is not 0.00.
Assuming a starting pf = 0.17

The above number doesn't sit well with me, and seems far too low. Could you point me to the reference that you are using which suggests such low power factors?

I dug out a model of a 4 pole 15 Hp three phase induction motor, simulated using the 'SPEED PC-IMD' software (http://www.speedlab.co.uk/software/imd.html) , and specifically looked at power factor as speed went to zero. Full load power factor was about 83%, locked rotor power factor was about 60%. This could, of course be simulation error. I no longer have direct lab access, so cannot set up the experiment to test the locked rotor power factor.

I did find one other online reference: http://www.bauergear.info/DanWeb/EP3000E.PDF?Cat_Ident=47 see page 3/2, figure 3.3...typical locked rotor power factors are shown to range from 0.4 to 0.95!!

I wonder if this is a scaling law issue; I mostly work with small (<50 kW) motors. Rather than thinking about a motor, think about an iron core inductor. The inductance depends upon the magnetic flux encircled by the coil (among other factors); the resistance will depend upon the _length_ of the conductor (among other factors). In other words, inductance scales as the area, resistance scales as the circumference. A quick 'back of the hand' evaluation suggests to me that as pole area in a motor gets larger, the inductance will increase faster than the resistance; or looking at things from the other direction, small motors will have relatively higher winding resistance for their inductance, and relatively higher power factor at locked rotor.

The 0.95 locked rotor power factor referenced above was for 100 watt motors; the figure shows a clear trend to lower locked rotor power factor as the motor power factor increases. Do you perhaps work with much larger motors than either Carl or myself?

-Jon

 

[kingpb]

My experience has been that it is dependent on the motor design itself, e.g. voltage (LV or MV) # of poles, torque versus speed, etc. Typically, I see lower starting pf on MV motors, the larger the motor the lower the starting pf.

Unfortunately, motor manufacturers do not readily offer this information, it usually has to be requested. The typical published data you see is usually only applicable for Design B motors, which is probably where the 0.4 to 0.95 mentioned comes from, just a guess. NEMA MG-1 is also not much help.

So, yes, I have seen them this low (0.14), but I have also seen them in the 50's as well, which means it is probably best to get the info from the manufacturer. Request it early as it is like pulling teeth out of a lion while he is awake.