CTs in 3-phase transformer with swapped leads to meter

[AHarb]

Hello,

I'm new to the forums, but I've used this forum to find the neutral current in a 3-phase unbalanced system. I figured this would be a good place to ask the question. I hope that I can make sense of it in text because it's kind of hard to explain. I've attached a poorly drawn diagram trying to graphically show what I'm asking as well.

Basically, I work at an electric utility company and we use meters rated for 200 amps. If the current through the meter is greater, we use CTs and calculate the demand based on the CT ratio. Well we had an issue the other day where the C-phase CT lead was going to the B phase current lead on the meter. The B phase CT lead was going to the C phase current lead on the meter. Basically the two current leads going to the meter were swapped for B and C phase. The correct voltage leads from the transformer all went to the correct voltage leads on the meter and the A phase current lead was correctly wired as well.

My manager came in to ask me a question about this. I've scanned my three engineering books from college and can't figure out if what I'm doing is correct or not. 3 phase power (real) is define as:

P = V*I*cos(theta) for each phase (where theta is the angle between the voltage and the current) and the total power is defined as:

Ptotal = Va*Ia*cos(thetaA)+Vb*Ib*cos(thetaB)+Vc*Ic*cos(thetaC)

Lets assume Va=Vb=Vc, Ia=Ib=Ic, and our power factore is 1.0. The voltage phase angle between phases A, B, and C are 120 degrees. The same is true for the current.

In my attached example, I've setup an example problem with some random numbers. I defined the voltage phase angles as 0 degrees, -120 degrees, and +120 degrees for A, B, and C respectively. I've assumed that we calculated the current phase angle on A phase is 40 degrees. Therefore, the current phase angles for B and C have to be -80 degrees, and 160 degrees respectively (120 degrees apart).

Now, if we use the Ptotal formula above we get (this is with the B and C current leads swapped):

Ptotal = V*I*cos(0 - 40 degrees)+V*I*cos(-120 - 160 degrees)+V*I*cos(120 +80 degrees)

**Notice I used the normal calculation for A phase power since the voltage and current leads are correctly wired. However, I used B phase voltage phase angle and C phase current phase angle for the calculation of B phase theta since the connections are swapped going to the meter.

--> Ptotal = V*I*cos(-40 degrees)+V*I*cos(-280 degrees)+V*I*cos(200 degrees)

--> Ptotal = V*I*(0.766044443 + 0.173648178 - 0.939692621)

--> Ptotal = V*I*(0) = 0 W

Is this correct? Will the meter read 0 W for the demand since the B and C phase current leads are swapped going to the meter?

I hope this makes sense. I would appreciate any help on the matter.

Thank you,
Adam

 

[rcwilson]

Check your math.

If A current is 40 degrees then it is leading phase A by 40 degrees. You calculated it as lagging by 40 degrees. I think your calculations for B & C phase are for leading current.

If the currents are balanced and lagging by 40 degrees, wouldn't the actual B current phase angle be =-120-40 = -160 degrees and C phase = +120-40 = 80 degrees. With the swapped phases, the meter B phase element angle woudl be =-120-80 = -200 and C element angle = 120 -80 = 40

Assuming balanced voltages and currents, the final equation would be:
Ptotal = VxI x[ cos(-40) +cos(-200) + cos(40)] =.766 -0.940 +0.766 = 0.592

 

[AHarb]

I see what you're saying about the leading lagging errors. I'm still not following where you're getting the angle between Vc and Ib to be +40 degrees. I'm getting 80 degrees when I draw out the phase diagram. If you see the attached document, you can see that Ib lead Vc by 80 degrees or am I missing something.

Sorry for the bad drawing. Obviously, the magnitudes are supposed to be be equal.

Thanks you for the help btw.

 

[AHarb]

I've been thinking about this throughout the day and I think I have an idea of what may be happening. The phase current is based upon the impedance and the phase voltage. If the load is a balanced three phase load then the impedances are equal for each phase.

Therefore, Ia = V (0 degrees) / Za (some angle)
and Ib = V (120 degrees) / Zb (same angle as Za)
and the same for C.

Therefore, does it make sense that since the phase voltage connections going to the meter were not swapped, the phase angles would still be 120 degrees apart and the meter would see everything as normal (basically you could wire the CTs from the transformer to any marked phase on the meter and still get correct kWh/kW readings)?

I'm not sure if this is correct, but just throwing an idea out there.

 

[rcwilson]

Good catch on my math error and spelling mistakes. I should have checked mine also.

Phase B current lags phase b-n voltage by 40 degrees so its angle is -120-40 =-160 degrees. The C phase voltage is at -240 degrees so B phase current leads C voltage by 80 degrees. The C meter element will have a V to I phase angle of + 80 degrees.

Ptotal = V X I x [cos(-40) + cos(-200) + cos(80)=.766 -0.940 + 0.174 = 0.0

At 200 degrees difference, the Vbn and Ic phasors are almost exactly opposite. Think of it as 0.94 power factor (cos20) load in the opposite direction. That element subtracts kW from the total. Phase C only adds kw equivalent to a 0.17 power factor load. Only phase A reads the correct 0.77 power factor load. Result is 0 kw recorded.

 

[AHarb]

I think that's the conclusion that my manager and I have come to. We debunked my theory about the voltages since the CTs are not connected to the load and therefore do not determine the phase angle of the current. The meter just reads the angle.

My manager told me that the meter was reading close to zero. I think he was wanting to make sure that the meter and CT were functioning properly.

 

[AdrianWint]

I'm not sure that I agree with your conclusion.

If we step back a little here to think about definitions for moment & forget the heavy math happyyes.....

What is 'real power' (Watts)? It is the product of the voltage & the component of the current in that line which is in phase with that voltage. This is what the meter is calculating for each line & then adding together to get the total 3 phase power.

So, if we mis-connect the L1 current with the L2 voltage the answer that we get will have no meaning - what is the product of the L2 voltage & that component of L1 which is in phase with it? It isnt defined - it certainly isnt real power or reactive power or even apparent power.

Also bear in mind that power is a vector quantity - it has direction. By misconnecting the meter, it is entirely possible that the meter will calculate that the power flow is in the opposite direction to the actual direction & thus subtract from the 'total' power.

So, my answer is that, although the scalar quantities (Volts & Amps) may well have numbers which are correct (but in the 'wrong place') the vector & product quantities (powers, pf etc) will be rubbish.

Adrian