Figure out this transformer connection

[dwverzwy]

On a 13.09 kV customer distribution substation, two PT's are used for a watt hour meter. They are supposedly connected open Delta to give 110 V phase to phase and phase to ground for a watt hour meter. Measuring secondary voltage, however, I read 200V A-C, 109V A-B, and 109V C-B. The B is grounded. To me, it is not hooked up correctly but I can't take a closer look. Anyone got any clues about how it might actually be hooked up?
Thx

 

[BJ Conner]

Two wattmeter power measurement.

Two wattmeter power measurement.

Google up the "Two wattmeter power measurement method"
I don't think I can explaine here, but it's covered in most power books. Basically it says you need one less wattmeter than lines in your system. Two wire system - one wattmeter. three wire, three phase system- two wattmeter, three phase, four wire system 3 wattmeters. The proof requires some phasor diagrams- good luck.

 

[dwverzwy]

Thanks I checked that out. This watthour meter is microprocessor based. I'm measuring before the meters, actually on the secondary of the xfmr's, which on the drawing show to be open delta..

I did some figuring and found the angle between A ph and C ph to be 90 degrees... Still at a loss as to what connection will give these secondary volts and angles..

 

[mivey]

You must have 120:1 PTs. You should be getting something more like 189 across A&C.

It sounds like the rotation is wrong (one of the windings is reversed). Swap B & C or swap A & B secondary leads and you should get 109 volts across the open phase.

 

[dwverzwy]

Cool that's good to hear.. Something definitely didn't seem right. I just wish I had the transformer intuition to describe what you're saying with a phasor diagram because it will take some convincing because this is to a critical customer and it would take an outage to check the connections because they didn't run all of the taps down to the junction box (you would have to climb up near the pt and look at the connections, which are near HV Energized conductors..)

Thanks again

Also, you were correct, they are 120:1

 

[mivey]

For example, using:
AB = 109<30?
BC = 109<270?
CA = 109<150?

then
AB-CB = 109<30? - 109<90? = 109<330?

also
AB-BC = 109<30? - 109<270? = 188.8<60?

the only difference is B & C were swapped (the winding was reversed).

 

[mivey]

I just wish I had the transformer intuition to describe what you're saying with a phasor diagram

How about a combination diagram that shows the windings oriented like a closed phasor diagram?

 

[dwverzwy]

For example, using:
AB = 109<30?
BC = 109<270?
CA = 109<150?

then
AB-CB = 109<30? - 109<90? = 109<330?

also
AB-BC = 109<30? - 109<270? = 188.8<60?

the only difference is B & C were swapped (the winding was reversed).

I worked your calculations above and I agree with the results, after converting to rectangular and back.. I'm not sure what the quantity "AB-CB" is though. The measured voltage is just CA.. Are you saying that, if the connection was correct, that (AB-CB) = CA = 109<330

This makes sense I'm just not sure how to get from the phasor diagram to that conclusion .. Need to read more on xfmrs which I'll be working on.

 

[mivey]

I worked your calculations above and I agree with the results, after converting to rectangular and back.. I'm not sure what the quantity "AB-CB" is though. The measured voltage is just CA.. Are you saying that, if the connection was correct, that (AB-CB) = CA = 109<330

This makes sense I'm just not sure how to get from the phasor diagram to that conclusion .. Need to read more on xfmrs which I'll be working on.

BC = -CB

You get the CB by reversing the leads on the winding taken as BC, and vice-versa.

In other words, if "C" has been mis-labeled and is actually "B", then you can see the results in the vector math and the winding diagram. In the diagram, winding 1 has b1 correctly labeled. Then if we brought c2 from the second winding and thought it was b2, we would be expecting to get the upper diagram but in reality we got the lower diagram for our meter.

 

[mivey]

I'm not sure what the quantity "AB-CB" is though. The measured voltage is just CA.. Are you saying that, if the connection was correct, that (AB-CB) = CA = 109<330

It was just a calculation to get the voltage across the missing winding: AB - CB = AC = -CA

also
AB - BC /= AC
and
AB - BC /= CA

so that shows that swapping B & C gives you a result that is sqrt(3) times larger in magnitude than what we expect.

 

[dwverzwy]

Great! That helps me a lot. I never quite knew how to determine the phase angle of the missing winding, but now I see how it can be done. Thanks a lot I'll repost when I have an update...

 

[mivey]

I did some figuring and found the angle between A ph and C ph to be 90 degrees.

Could you clarify that? 90? relative to what? Do you mean AB relative to CB or do you mean AC relative to some other reference? If you mean AB relative to CB I hope your figuring is wrong. If not there are limited explanations (a two-winding secondary like used with residual schemes comes to mind but I doubt that).

because it will take some convincing because this is to a critical customer and it would take an outage to check the connections because they didn't run all of the taps down to the junction box (you would have to climb up near the pt and look at the connections, which are near HV Energized conductors..)

You could verify what is going on without having to climb up if you used a phase-angle meter. I would do that before scheduling an outage on a critical load.

Any chance the POCO would help since they might could get you the terminal markings without an outage?

 

[dwverzwy]

Could you clarify that? 90? relative to what? Do you mean AB relative to CB or do you mean AC relative to some other reference? If you mean AB relative to CB I hope your figuring is wrong. If not there are limited explanations (a two-winding secondary like used with residual schemes comes to mind but I doubt that).

Hey Good morning,
When I took the readings I scribbled down 200V a-c. Using the law of Cosines with the 109 volts to ground on a and c, I came up with 90⁰ between a and c. Looking back though I see my mistake.
with C being the angle in question (the angle between a and c)the formula is
Cos C = (a²+b²-c²)/(2ab)
I subtraced a² and b² making the numerator 200² with makes cos C = 0.
Looking back and correcting the mistake I come up with Cos C = -14008/25992 = -.5389 so the angle between A and C would be Cos^-1(-.5389) = 120⁰

 

[dwverzwy]

You could verify what is going on without having to climb up if you used a phase-angle meter. I would do that before scheduling an outage on a critical load.

This would be a good thing to do, hopefully there will be an opportunity to do so. The meter tech has expressed that the PT's will need to be changed either way because the watt meter is requiring a true secondary with three PT's... Not sure why, but he said this even before he knew the secondary voltages were wrong.. Not a meter tech so not sure about this reasoning..

Any chance the POCO would help since they might could get you the terminal markings without an outage

Polarity drwgs? Were not provided, I do have the xfmr connection dwgs which is what i am going by that show it hooked up open delta..

 

[mivey]

The meter tech has expressed that the PT's will need to be changed either way because the watt meter is requiring a true secondary with three PT's... Not sure why, but he said this even before he knew the secondary voltages were wrong.. Not a meter tech so not sure about this reasoning..

It is dependant on the form type required by the meter (the form type tells how many voltages and currents are needed and how they are used). That should have been specified up front.

Any chance the POCO would help since they might could get you the terminal markings without an outage?

Polarity drwgs? Were not provided, I do have the xfmr connection dwgs which is what i am going by that show it hooked up open delta..

POCO = POwer COmpany. They know how to work safely near energized MV & HV equipment.

 

[brad9m]

Hey Good morning,
When I took the readings I scribbled down 200V a-c. Using the law of Cosines with the 109 volts to ground on a and c, I came up with 90⁰ between a and c. Looking back though I see my mistake.
with C being the angle in question (the angle between a and c)the formula is
Cos C = (a²+b²-c²)/(2ab)
I subtraced a² and b² making the numerator 200² with makes cos C = 0.
Looking back and correcting the mistake I come up with Cos C = -14008/25992 = -.5389 so the angle between A and C would be Cos^-1(-.5389) = 120⁰

I think your math is still a little off. Let's call the voltage from A-N, a. Let's call the voltage from C-N, b. Then that makes the voltage from A-C, c, where the angle C represents the angle between a and b.

cos C = (109^2 + 109^2 - 200^2)/(2*109*109)
C=133 degrees the angle between the A and C phase