Phase rotation meter 4

[bdf5526]

Hi all, I need to make a 3phase rotation meter (to measure rotation for 400V,50hz,incoming supply from diesel) since we don't have one (we will get one soon). The test equipments which we suppost to receive are missing and its a long story. To check and confirm the incoming phase rotation is a requirement for the client.

I have 1 capacitor 3uF (+-5%)/400Vac/40-60-Hz, 2 bulbs (filamen type)220volt, 50w. All these are cnnected in star with capacitor in L3, bulb no.2 in L2 and bulb no.1 in L1.

First question, what is the capative reactance at 50Hz, since the power frequency available is 50Hz? I am confused with the rating 40-60hz.

Say the cap. is same 3uF,so the -jXc is 1/(2*pi*50*3uF) = 1.06kohm. Each bulp is 220v*220v/50w = 0.968kohm.

Second question, the cap. reactance and bulp ohmic value are bit different is this acceptable?

3rd question, two bulps will have 400v accross their terminal and this should be allrite, correct?

4th question, Neutral Voltage formula is =Ya*Van+Yb*Vbn+Yc*Vcn/Ya+Yb+Yc

Guys I need your input before I make one. I shall draw and attach a file in the next post.

Thanks in advance.

 

[electricpete]

Attached is a "brute force" numerical solution for the specific values mentioned, which confirms indeed the behavior will be as stated by skogsgurra.

I'm sure there are a number of more direct approaches. One mentioned by skogsgurra.

Below is another attempt at a simpler approach that I started, but don't have the time to finish (have to do my taxes):
Simply consider that the system is a superposition of three voltage sources (A,B,C) applied between the phase terminal and the system neutral or ground (diffrerent than the floating ground we have created).

Solve each phase source by itself (with the other two phase sources shorted out to ground) to find the contribution to Vn from that source.
Add together the contributions of the three sources to deterimine the final Vn. If you align Vb along the real axis as I have done in the attached, then it is very easy to see which source vector (Va or Vc) the Vn will be closer to by observing whether the imaginary portion is + or -.


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[electricpete]

Here was an attempt to use the superposition approach. The math is simpler, but the conclusion came out backwards from before. Must have been some problem... have to look some more.

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[electricpete]

There was just an error at the very end - forgot to negate I bringing it from the denominator to the numerator - will fix that and re-post.

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[electricpete]

Here is the corrected version. I have also added some explanation and symbols. I think it is a pretty straightforward proof of the conclusion stated by skogsgurra. It relies on the assumtpion R*w*C <<1 (I don't think that assumption is required for the conclusion, but it makes the math easier and it should be easy to meet for practical values of C).

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[electricpete]

One more revision (attached) corrects an obvious (?) typographical error in the very last line.

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[electricpete]

One thing was not explicitly stated in the writeup but should be obvious. With angle B assumed 0, then B is a real number and the sum Va + Vc is also a real number. Those observations allows us to isolate the imaginary component and determine that it is positive.

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[electricpete]

And of course Vb is not only real but also positive.

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[electricpete]

Attached is my last revision - it incorporates the clarifying observations of my last two posts.

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[waross]

Just as important as knowing what the phase rotation is is knowing what phase rotation the pump wants. I would bite the bullet and connect the pump motor and check the motor rotation. If need be, disconnect the motor from the pump before checking rotation.
Skogs, I like your cheap and simple rotation meter. It answers the original question. However, the original question didn't ask everything that needed to be asked.
Next question: I know the incoming phase rotation, how do I connect the pump so it will run the right way.
I still remember when it was common in a large plant for a mechanical guy with an electrical guy to check the rotation of every motor in the plant and correct it if needed. Later, a couple of electrical guys would take a phase rotation meter and a motor rotation meter and check every motor. If there was a special motor that could not be turned by hand for the motor rotation meter, the mechanical guy was called for assistance.
The point;
You have to check both the supply rotation and the motor rotation.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Looks like the sequence of execution within the worksheet resulted in some premature results that might be confusing to the reader. Here is the real last version which may be less confusing.

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[Skogsgurra]

Thanks Pete!

You saved me some work. Still, I will try and do it "old fashion" with pencil and paper. But it will not be within the next hours. Stand by!

Gunnar Englund

http://www.gke.org

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[bdf5526]

Thanks for the post guys and I need some time to understand the formulas, at least to tell client and the principle behind it.

Thanks electricpete for the valuable attachement.

 

[electricpete]

Attached is the phasor drawing which shows the orientation I had in mind (angle B = 0)

If you compare my very first and very last of my previous attachments, I reached the same expression for Vn using two different approaches: KCL (first attachment) and superposition (last attachment).

The KCL approach is simpler, but when I did it in my first attachment I substituted in the values of V too early which made the expression complicated.

The bottom line quickest way to get to that same result is simply KCL as follows:

KCL at the midpoint (floating neutral):
(Va-Vn)/R + (Vc-Vn)/R + (Vb-Vn)*j * w * C=0

Move Vn to RHS
Va/R + Vc/R + Vb * j*w*C = Vn * (2/R + j*w*C)

Solve for Vn by dividing by (2/R + j*w*C)
Vn = [Va/R + Vc/R + Vb * j*w*C] / (2/R + j*w*C)

Multiply by (-j*R)/-j*R)
Vn = [-j*Va -j* Vc + Vb w*C*R] / (-2*j + R*w*C)

Above is the same result as the last attachment (from superposition) and the analysis can proceed as in the last attachment under the assumption R*w*C << 2.

Even if we don't make that assumption, we can use this expression for Vn to compute
|Va-Vn| / |Vc - Vn| for a given value R*w*C under assumption ABC phase sequence. We know that swapping the phase sequence will simply swap the roles of A and C and the ratio would be inverted.

If I get a chance tonight I will try to make a plot of that function ( |Va-Vn| / |Vc - Vn| vs R*w*C )

Still maybe there is a simpler and more direct path to the conclusion. I will be interested to see what Gunnar comes up with.

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[electricpete]

Here is a plot of the function
|Va-Vn| / |Vc - Vn| vs (R*w*C)
under assumption of ABC phase sequence.

(note it uses the complex functions in the excel "analysis toolpak" add-in)

If you swap phase sequence you can just relabel the plot as |Vc-Vn| / |Va - Vn| (or else keep the same label and invert the ratio).

You can see the minimum ratio (max difference in brightness) occurs for R*w*C ~ 1 which is where resistance is approximately equal to capacitive reactance. But there is still noticeable difference as we move away from that ideal point.

Of course as we all know there are temperature effects on resistance which would tend to push the two currents closer together, but that's a complication for another day.


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[Skogsgurra]

OK, Pete.

That's a challenge. Using Indian ink, paper and a Faber Castell 1/54 Slide Rule ;-) , I got the attached result.

I have plotted the locus for the star point voltage in the phasor diagram (using the Real and Imaginary values for x and y) so one can see the distance between star point and phase voltages.

It is not at all necessary to use the optimal capacitor value (when Xc = R). Already a five percent difference in lamp voltage makes a huge difference in lamp brightness. Using larger capacitor values will burn out the brighter lamps in no time at all if lamp voltage is chosen to be equal to phase voltage, which is the natural thing to do.


Gunnar Englund

http://www.gke.org

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[bdf5526]

Thanks for the input guys, I really appreciate the post,this is very valuable for me.

From the understanding after all this posts, as advised by Skogsgurra and electricpete to prevent the light from burn I can connect 2 x 50w/220v bulbs in series on each leg A and C.

Looks like you guys have given me enough bulletproof vest to face the client.

 

[Skogsgurra]

Yes, two in series is, as you say, bullet proof. And so will you be, too.

Gunnar Englund

http://www.gke.org

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[sibeen]

bdf5526, if you have two 3uf capacitors, putting them in parallel will give you a better voltage match between the two globes whilst still letting you tell the difference in brightness.

With a 3uf capacitor the voltages will be (approximately) 344 volts across one lamp and 92 volts across the other.

With a 1.5uf capacitor the voltages will be (approximately) 280 volts across one lamp and 119 volts across the other.

I like these types of threads as they force you to go back to the basics. I did a branch analysis to get the above figures. I used Gunnar's method of using a pen and paper to get the equations, but I admit I shamelessly and outrageously used a computer program to solve the complex simaltaneous equations.

See attached PDF.

 

[Skogsgurra]

sibeen, shall I send you a Slide Rule?

So you don't have to be ashamed using a computer?


Gunnar Englund

http://www.gke.org

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[bdf5526]

Hi guys, sorry for the late reply. It worked as it should. Thanks sibeen, for bring back those old memories with your method. This thread is my lesson of the week.

Thanks again guys.