Chiller Waveform Question 2

[testtech]

I obtained voltage and current waveforms from a centrifugal chiller of about 200 HP. It was running at about 70% full load. The voltage waveform looks fine. RMS voltage for A phase is 476.4V and the peak is 664V. The current waveform for each phase is flat topped at the positive and negative peaks. The A phase RMS is 161.5 amps. The peak is 230 amps. The duration of the positive and negative flat peaks is about 2.9 miliseconds each. Total harmonic distortion is 1.53% for voltage (A phase) and 7.21% for current (A phase). Here is the question: What characteristic of the motor produces the flat topped current waveform?

 

[jschwei314]

The problem is one's current sensor has gone into satuation.

 

[testtech]

No chance--I used two different types of sensors to ensure there was no clamp problem. One type of clamp was rated for 1000 amps. The other was rated for 2000 amps. The load was only about 150 amps. I do not believe saturation could occur here. Also, I read the current waveforms on a CSI 2120 spectrum analyzer and a Hioki Power Analyzer. Both provided consistent results.

 

[jraef]

2.9ms? Is that a problem for you? It could be a lot of things. Is it only on Phase A or is it present on all 3 phases?

Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[testtech]

Jraef--
Except for the flat current waveform, I have no evidence of a problem. The vibration data I took on the machine looks great. The three phase current and voltage data is well balanced. The flat topping does occur on all three phases. I suspect there is no problem here but I am curious why the flat topping occurs in the current waveform.

 

[electricpete]

Is it possible you are using a digital instrument near the bottom of it's range where the step due to bit size is significant (maybe 1% or more of your peak value). The software draws a straight line between samples (linear interpolation) so where the slope is high you don't see it (no stairstepping due to linear interpolation), but where the slope is lowest (at the peaks), the current does not change enough to toggle the last bit.

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[testtech]

Pete
As I mentioned, I have taken this test on two instruments using two different clamps. The CSI 2120 spectrum analyzer autoranges to prvide the best signal resolution. Here it is seeing an RMS voltage of .15 This is certainly nowhere near the bottom of its range. The second instrument offers a 500 amp range for a 1mv/amp clamp. A 150 amp load will be accurately resolved.

 

[itsmoked]

Is this chiller cooling the motor with the refrigerant? If so I could imagine the V/f to be slightly too high so the motor is just barely saturating but with large cooling available not showing up otherwise.

Just a guess though.

Give the maker a call and ask them.

Keith Cress
Flamin Systems, Inc.- <

http://www.flaminsystems.com>

 

[testtech]

itsmoked

The chiller is hermetic; it does cool the motor with refrigerent. I do a lot of chiller testing and see this issue somewhat regularly, although not usually to the extent exhibited by this machine. Could you possibly explain in a little more detail how this would produce the appearance of saturation. Thanks for your insights.

 

[waross]

Help am I figuring this right? I get a flat spot of about 60 electrical degrees. That's about one third of a peak.
However an RMS of 161.5 times root 2 = 228.4
Compared to an indicated 230 peak volts, that is an error of 0.7%
Your RMSeak ratio is consistent with an almost perfect wave form. Try measuring the current with an old amprobe with a D'arsonval meter movement. If an old amprobe indicates any where near 161.5 amps, assume a good waveform and check the settings to see how you have caused two different instruments to clip the displays the same way. The old D'arsonval meter movements responded to average current or voltage rather than RMS. The scale was adjusted to take into account the ratio between the average value of a sin wave and the RMS value of a sin wave. (1.1 as I remember). This was called the form factor (from wave form). The indicated value only equalled the RMS value with a sign wave. It didn't take much wave-form distortion to throw the reading off.
respectfully

 

[electricpete]

Although THD is lower on the current, your crest factor is more out of whack on the voltage than on the current:

CF(V)=664/476 = 1.39
CF(I)=230/161.5=1.42
For a sinusoid we expect crest factor sqrt(2)=1.41 so the current is closer to sinusoid.

In fact, one would have thought a current sinusoid with top chopped off would have crest factor < 1.41. The deviation from sinusoidal is maybe a little more complex than just chopping off the top... I'm imagining the slope of this waveform as it crosses 0 must be less than we expect for a perfect sinusoid, lingering at lower values bringing the rms down. The slope coming off of that flat peak higher rate of change than a sinusoid to get down closer to 0 faster. I don't exactly know what to make of that but just trying to reconcile the high crest factor with the chopped-off-peak description.

Here's a thought (might be way off base). Even thought the voltage THD is lower than current THD, the problem might still be associated with the voltage (power supply) and somehow magnified in the current? (I would think this is more common in capacitive loads). I have heard that during power quality surveys, it is not uncommon for a small Vthd to cause a larger Ithd (although I would tend to associate this with capacitive loads which amplify the higher frequency harmonics).

I assume you would have told us if vfd. I really can't put my finger on what about the power system would cause it so just throwing out some off-the-wall stuff... Anything weird in the power system? Attached capacitors? Unusual upstream transformer configuration? Nearby electronic power supplies? Or maybe not.

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[electricpete]

Hey waross - I wasn't ignoring you, just typing at the same time (I must be a lower typer than you).

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[waross]

Hey electricpete
I'm not a very fast typist. I think I started a lot sooner.
I was thinking along the lines of generators with non sinusoidal voltages depending on the distribution of the windings. Motors act as induction generators. I thought that this may be an explanation until I checked the peak ratio.
Remember, 2.9 mili-seconds is almost one third of a half cycle, and with a perfect wave-form ratio?

Did I ever tell you about the time I taugh a theory lesson right out of the text book on the ratio of line to line voltage to line to neutral voltages to a class? Then I sent them out to the shop to connect a bunch of small dry type transformers in Wye-Wye and verify the lesson. I got in trouble more than once that way. It wasn't long before the quicker students were all over me, because the voltage ratios were obviously too far off to be explained by acceptable experimental errors. We ended up putting a scope on the transformer outputs. It's so long ago that I don't remember which was which. That is the wave form from line to line versus the wave form from line to neutral. I clearly remember one wave form being normal and the other waveform having a flat spot on the zero line. The distorted wave-form was a fairly good looking half wave and then a few electrical degrees of zero, and then a good looking negative half wave followed by a few electrical degrees of zero. The disparity in the voltage ratios as indicated by the D'arsonval meters was readily apparent.
I later ran into the same effect on a pumping station start-up, amidst a very puzzeled group of engineers.
The effect disappears as soon as a load is applied.
respectfully

 

[itsmoked]

Sorry testtech this isn't saturation.. I just looked again at your original post. In saturation the current wouldn't flat top it would spike!

A saturated supply transformer would result in your motor's voltage flat topping.

I'm still thinking probe.

How about this.
The probe sees a large flux change during all of the current cycle except at just the peaks where essentially over a short period there is NO flux variation. The output (a voltage) flattens out briefly during this point in the waveform. Essentially DC very briefly. (Not to be mistaken with saturation)

You could check this if there is any resistance in the power circuit. Wouldn't need much. The contactor's contacts for one phase would be plenty. Watch the voltage drop across the contacts. This will give the current without any magnetics in the picture. (Note: it might get ugly if the contactor opens while you're looking at the 0.5 Volt scale)

Keith Cress
Flamin Systems, Inc.- <

http://www.flaminsystems.com>

 

[testtech]

HI All:

Thanks for your thoughts.

Electicpete-I wonder if you are not right about the small distorion in the voltage being the culprit. I have lots of saved chiller voltage/current signatures. I will go through some tonight and see if this makes sense. I will let you know.

 

[waross]

Hello testtech
I have a few questions.
Are there any capacitors in the system and if so are you looking at just the motor current or the combined motor-capacitor current?
Do I understand correctly that you have a flat line across the top of your wave-form for 2.9 Milliseconds. Almost a third of the width of the half cycle?
Can you describe the voltage wave-form please?
Thanks.
respectfully

 

[testtech]

Hi Waross:

There are no capacitors. If there were a way to post images, I could attach the waveform. However, if you cut off the positive and negative peak and drew a straight line to connect the shoulders, you'd have this waveform.

I reviewed lots of data that I have here with chiller waveforms. There is no obvious relationship between voltage waveform crestfactor and THD and current crestfactor and THD. Certainly, I found quite a few that had the same voltage crestfactor or lower than the case here but had normal looking or just slightly distorted current waveforms. The same applies to THD.

So, at this point, I cannot explain the flat topping and high THD for this machine.

 

[bklauba]

The problem might be one purely of instrumentation. I wonder whether the preamp of the analyzer is being overloaded, similar to when a mic overloads a preamp on a tape deck or mixing board. Does the amp probe give you the options for the ratio of V / A, such as my Fluke 80i-110s does: 100 mV/A and 10 mV/A? If it does, does the waveform look more "complete", less clipped, when using lower settings? If it does, you are clipping the inputs of your analyzer.

BK

 

[itsmoked]

Picture posting:

faq238-1161

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[testtech]

Thanks itsmoked.

Here are the voltage and current waveforms.

http://i1.tinypic.com/s1t5l1.jpg