Motor Starting Current Interpretation 2

[electricpete]

Here is starting current for a 800HP, 4KV vertical motor driving induction motor. FLA is 100A. Current is monitored at secondary of 40:1 CT.

http://www.reliability-magazine.com/pub/EP_MotorStart.jpg

The waveform appears to include two components.

First component is sinusoidal component of approx 25A peak =>17.7A rms => 707A rms on primary => approx 7x FLA as expected.

Second component is decaying DC component of approx +10A (ct secondary) on phase A and -15A on phase C. As discussed in recent posts this is expected when applying voltage to an inductive load.

Lastly I notice that there is a perturbation in the waveform approx 90 degrees after the positive peaks on phase A and negative peaks on phase C. It appears obvious these are in some way related to the dc component. I would tend to think they are somehow related to saturation which occurs once per cycle (on same polraity as dc offset). However, I cannot reconcile why the effects of saturation appear at 90 degrees beyond the peak. My model of induction motor would include predominantly inductive magnetizing branch in parallel with predominatly inductive (during starting) rotor/stator branch. I would expect saturation to appear as a drastic increase in current near the peak of the excitatio curve.

Can anyone explain this curve?

 

[electricpete]

Let me reword my next-to-last sentence:
I would expect saturation to appear as a drastic increase in current near the peak of the current curve (comprised of excitation plus stator/rotor current in phase with each other), NOT at 90 degrees away.

 

[jbartos]

Suggestions:
1) If you intend to pursue industry standards and associated contexts, then the following may be a good start:
1a) IEEE Std 399-1997 IEEE Recommended Practice for Industrial and Commercial Power Systems Analysis, Sections:
1a1) 7.3.1.2 DC Decrement and system impedances, where dc decrement is also characterized by the fact that because the prefault system current cannot change instantaneously, a significant unidirectional component may be present in the fault current depending on the exact instant of the occurrence of the short circuit. This unidirectional current component, often referred to as dc offset, decays with time exponentially. ......
1a2) 7.4.1.1 Accounting for ac and dc decrement. Table 7-2 gives Reactance values for first cycle and interrupting duty calculations for different system components.
1a3) 7.4.3 Differences between the ANSI and IEEE 37 standards and IEC 60909 (1988) explains differences. IEC concentrates on DC decrement modeling which is fault location dependent and it quantifies the rotating machinery's proximity to the fault. The IEEE Std recommends universal, system-wide ac decrement modeling. Plus, there are additional differences.
1b) IEEE Std C37.013-1997 "IEEE Standard for AC High-Voltage Generator Circuit Breakers Rated on a Symmetrical Current Basis" gives the generator version (i.e. opposite version) of your motor modeling and measurements.
2) If you intend to pursue textbooks then the following book includes a smooth start to the topic:
Stevenson W. D. "Elements of Power System Analysis," McGraw-Hill, Third or Fourth edition
(A derivation of equations is available.)

 

[electricpete]

jbartos - thanks. I believe I understand the dc component. I don't understand the "glitch" in the waveform and why IF it is caused by saturation does it occur at 90 degrees from the main current peak.

 

[rhatcher]

Hey electricpete,

Very cool picture. I agree that the glitch is related to the DC offset. This is based on the fact that it is limited to the half of the waveform in the direction of the DC offset for a given phase and the fact that the degree of peterbation appears somewhat proportional to the degree of offset. Note that the 'C' (blue) phase has a greater DC offset and is more affected by the glitch than the 'A' phase. 'B' phase appears to have been 'in phase' at the time of contactor (breaker) closure and is unaffected by the offset and the 'glitch'.

However, it appears that the glitch is not in fact 90 degrees out...that is an optical illusion based on the zero crossing. When you look at all of the cycles it appears that the glitch location is unrelated to the phase relation to the peak but is directly related to the zero crossing. I agree from the shape of the glitch that you are seeing a saturation affect, but my gut feeling is that it is in the CT's and not the motor. How are they connected? Some CT's are delta connected and some I have seen w/ delta connection (LV systems) are grounded at the neutral. Perhaps they are saturating on peak current....

 

[jbartos]

Suggestion: Rather than to keep on guessing without any mathematical descriptions and normal waveforms corresponding to the mathematical models, it is better to develop those first, and then see the discrepancy and explain or troubleshoot them. The simple model is
Vmax x sin(wt+alpha)=R x i + L x di/dt
with its solution
(Vmax/|Z|) x [sin(wt+alpha-theta)-e**(-Rxt/L) x sin(alpha-theta)]
where |Z|=[R**2+(wL)**2]**0.5 and theta=PowerFactor=arctan(wL/R)
Now, each phase has a different alpha and potentially different theta. Now, what is needed to post more of those oscilloscope snapshots to see how waveforms appear for different alphas and potentially thetas.

 

[electricpete]

rhatcher - thanks for the comments. I think we're looking at this the same way.

jbartos - the form you describe is a sinusoid plus a decaying exponential. I agree 100% that is what is expected in theory. I am trying to mentally fit the observed curves to the theoretical prediction. The part that doesn't fit are the glitches mentioned. I believe these might be atributed to saturation caused by decaying dc component since: #1 - the linear model does not account for saturation and #2 -the glitch occurs on only one of two half-cycles per cycle, which appear related to the polarity of the dc based on comparison between phases A and C.

My first choice is to try to understand the waveform by eye-balling the solution. I've got excel file which gives phase-to-phase voltages captures with same time scale as the currents. When I get time I will play with it to try to establish the power factor angle after dc component decays when current reaches pseudo-steady state at LRA. Also will extrapolate voltages backwards to determine phase relationships at time of closing. In the meantime, if someone can give the insight based on what has already been posted that would be appreciated.

 

[electricpete]

I went thru excercize assuming the motor was ungrounded wye (not positive). Notes peaks of Vab, followeed by Vca followed by Vbc. By my computation that corresponds to a similar rotation of Line-TO-Neutral voltage peaks in Van, Vcn, Vbn.

Recorded time of zero-crossing in sec (during steady-state portion) as noted below. Convert to phase theta =(T-mod(t,T)*(360/T) where t is zero crossing and T is 1/60. This corresponds to theta in the equation X=Xmax*sin(2Pi60t+theta).

A pleasing result was that the sequence of current peaks IA, IC, IB match the sequence of volt peaks. All three-phase sets are very close to 120 degrees apart. Looking at 10-cycle space I compute frequency varied less than 1/2% from 60hz. IB appears in-phase with Tclose - which makes sense... breaker was closed at the point when Ib would have been zero anyway, so no dc offset.

One unpleasant result which makes me question my whole analysis. Power factor is computed at 0.9 during starting. This seems way too high. IEEE3999-1997 section 9.5.2.c indicates should be between 0.15-0.2. Suggests big error in one of the above calcualtions. I'll double check whether this might be delta winding (I'm not sure offhand if that changes the conclusions, since it should behave similarly to an equivalent wye system).


Closing and zero-crossing times*
Value(sec)
** Phase=(T-mod(t,T)*(360/T)
t_close -0.0105 226.8
tVab0 * 0.060008496 143.8164826
tVbc0 0.054442688 264.0379447
tVca0 0.065558391 23.93875051
tVa0=tVab0+T/6 0.062786274 83.81648258
tVb0=tVbc0+T/6 0.057220466 204.0379447
tVc0=tVca0+T/6 0.068336169 323.9387505
tIa0 0.211595225 109.5431394
tIb0 0.206018462 230.0012308
tIc0 0.217044023 351.8491115
pfA 0.900875164
pfB 0.89907476
pfC 0.883680999
* Note - above times are times of zero crossing in the increasing direction
** - Conventions for defining phase - phasor rotates ccw, X=Xmax*sin(2Pi60t+Phase)
T = 1/F = 1/60
Currents and voltages follow ACB sequence.
Phase of B indicates it would have been zero at closing => no dc offset of phase B

 

[electricpete]

Not only do my phase angles predict 0.9 pf during starting, but Ia is leading Van. More proof that I've done something wrong.

 

[jbartos]

Suggestion: Please, could you post manufacturer's parameters (nameplate data) of that motor rather than to present some models that may have many shorcomings hard or impossible to decipher.
The motor could have the following (nameplate) data:
Manufacturer name
Model Number
Type
Serial Number
LRA
Xd"
Frequency
Design letter
Nema Code letter
X/R ratio
starting power factor
etc.

 

[redtrumpet]

Could the perturbation in the waveform be due to some type of resonance with nearby capacitors when the motor is first switched on, due to the change in inductance on the system? Just guessing.

 

[electricpete]

jbartos - Data for the motors is
Reliance, 900RPM 800HP Vertical,Induction Motor 4kv, 60hz 3-phase, Mfr Date - ~1985, FLA=111A, LRA=555A, NLA=38A, Nema Design B, I have effeciencies and power factors at 1/4,1/2,3/4 if you really need 'em. Oddly enough, this is particular induction motor does not have an Xd" stamped on its nameplate ;-),

My model for the time period of interest (first 10 cycles or so) is a simple wye-connected series L-R load with voltage applied to the line terminals at t=0. It's simple and it seems to work based on field observation.

 

[electricpete]

#1 - We repeated the test for total of two tests on each of two motors and found very similar results (allowing for random closing angle).

# 2 - I believe I have the explanation for the "glitch" which occurs 90 degrees after the positive peaks for positive dc offset currents and 90 degrees after the negative peaks for negative dc offset currents.

It is CT saturation.

Remember that the current that you see plotted is CT secondary current which goes to approx 38 amps peak at first half-cycle (ct ratio is 40:1). Obviously we should be able to go a lot higher without significant loss of accuracy, but some waveform distortion is expected at this level.

Here's my CT model, all quantities referenced to the secondary.
I1 V I2
==>=========>====
| |
|Im |I2
L R
| |


I assume the relay load is purely resistve.
V is therefore in phase with I2.
The magnetizing branch can be modeled as a current Im or a flux Phi_m, BOTH of which will be related to the integral of the voltage (integral of I2), and both will therefore lag I2 by approx 90 degrees.

90 degrees after an offset peak in I2 I hit a zero in I2. This will correspond to the highest possible value of Phi since we have just completed integrated an offset-high half-cycle of voltage. At this point we are furthest into saturation.

There are two alternative ways to visualize the effects of the saturation which occurs 90-degrees after an offset-high peak:
#1 - CT becomes incapable of generating any secondary voltage because Phi flatlines at a max value and there is no more dPhi/dt to produce current. Accordingly secondary voltage goes to zero and resistive secondary current goes to zero at the same time.
#2 - Im will greatly increase to try to match the voltage proportional to dPhi/dt. Remember dPhi/dt = dPhi/dI * dI/dt. (When Phi maxes out then dPhi/dI approaches zero and dI/dt must get very high). Im will increase greatly but is limited to no more than I1. Net result is I2=I1-Im is forced to zero during the saturation period.

Both of the above results predict I2 forced to zero while the CT remains in saturation at 90-degrees after the offset peak. This is of course the natural point for I2 to be zero but the CT will try to keep I2 at zero for a longer period of time while the CT remains in saturation... resulting in a "flat spot" in the waveform as it crosses zero.

It's a little bit of a hokey explanation from the standpoint that I had to start with an assumption of sinusoidal secondary current and voltage to arrive at my conclusion of non-sinusoidal secondary current waveform. But I still think it gives a pretty good approximation which can be visualized without solving implicit equation or numerical method.

 

[jbartos]

Suggestion to the previous posting marked ///\\\:
This is of course the natural point for I2 to be zero but the CT will try to keep I2 at zero for a longer period of time while the CT remains in saturation...
///Generally, if any transformer saturates, the Im is peaking or excessive. Notice, that Im appears to be a part of I2 in your posting. It needs some clarification.\\ resulting in a "flat spot" in the waveform as it crosses zero.

 

[electricpete]

jbartos

Sorry that my circuit diagram got a little garbled.
I was trying to convey that I1 enters the CT and then divides into IM and I2. i.e. I1=IM+I2

Rearrange that as: I2 = I1 - Im.

At saturation, Im peaks (as you point out) and I2 decreases (toward zero).