## Questions & Answers > 5S 277/480/ 4wire /3cts

Thanks for your question Dan.

First, the easiest way to figure out the affect of the customer’s billing is to correct the installation wiring. Then, after the meter has been read for a couple of months, compare those to the reads prior [when the meter was mis-wired]. You should be able to determine the differences and show these comparisons to the customer (if necessary).

In addition, I am working on the actual proof and explanation of the metering error for this mis-wired installation and will post that information here as soon as I have completed it.

Brad

Thank you for your reply I also recently corrected a very similar issue at a resturaunts metering but it was 120/208 , 6S with 3 CTs .... it also had a failed transformer 6 months earlier were the CTs were put back on BAC the customer called us because of 6 months of a low Bills ..........after correcting it......it worked out to be a 13 fold incresase / decrease error This was only a 6 month error .. .. the 277 /480 install appears to of occurred back in1995 on transformer failure ... if the error factor is similar we are talking a lot of lost revenue ...............Dan

In answering your question based on a more thorough mathematical and direct approach, I would first need to begin with a few assumptions:

ONE: that these figures and calculations are going to be based on unity power factor and a balanced 3-Phase load.

TWO: that “a” and “b” [lowercase] are referencing the secondary labels and that B A C [uppercase] are also referring to secondary labels. [I say this because generally lowercase a,b,c refers to secondary labels and uppercase A,B,C refer to primary labels.]

THREE: that the ‘b’ and ‘a’ CTs were swapped or mis-wired and that the voltages to the meter potential coils have remained in their correct positions.

1.) A correctly wired 5S meter installed on a 4 wire 277/480 circuit with 3 CTs would have the following wiring connection with Va-n and Vc-n being the two voltages connected to the meter potential coils.

2.) The ‘a’ phase CT would be wired to the first meter stator current coil with Va-n voltage on the stator potential coil (polarity of ‘a’ phase CT current matched with the polarity voltage of ‘a’ to non-polarity ‘n’ on the potential coil.)

3.) The ‘c’ phase CT would be wired to the second meter stator current coil with Vc-n voltage on the stator potential coil (polarity of ‘c’ phase CT current matched with the polarity voltage of ‘c’ to non-polarity ‘n’ on the potential coil.)

4.) The ‘b’ phase CT is connected to both meter stator current coils:

4a.) One would be wired to the first meter stator current coil with Va-n voltage on the stator potential coil (the non-polarity of ‘b’ phase CT current matched with the polarity voltage of ‘a’ phase to non-polarity ‘n’ on the potential coil).

4b.) The second would be wired to the other meter stator current coil with Vc-n voltage on the stator potential coil (the non-polarity of ‘b’ phase CT current matched with the polarity voltage of ‘c’ phase to non-polarity ‘n’ on the potential coil). [Note: ‘b’ phase CT current is wired so that it goes backwards through each of the two meter current coils].

This correctly wired 5S meter on a 4 wire WYE service will measure all connected loads to an accuracy that is viewed correct by industry standards.

To see the vector diagram for reference you will need to locate the image with the title: Correctly wired 5S Image which can be found by going to this link (you can copy and paste into your browser or highlight the link and right click to "go to") http://metergod.com/images-for-q-a-page/

Now, to address the mis-wired installation you described and what that metering error was:

1.) Va-n and Vc-n are still the two voltages connected to the meter potential coils.

2.) The ‘b’ phase CT would have been wired to the first meter stator current coil with Va-n voltage on the stator potential coil (polarity of ‘b’ phase CT current matched with the polarity voltage of ‘a’ phase to non-polarity ‘n’ on the potential coil).

3.) The ‘c’ phase CT would have been wired to the second meter stator current coil with Vc-n voltage on the stator potential coil (polarity of ‘c’ phase CT current matched with the polarity voltage of ‘c’ phase to non-polarity ‘n’ on the potential coil.)

4a.) The ‘a’ phase CT would have been wired to the first meter stator current coil with Va-n voltage on the stator potential coil (the non-polarity of ‘a’ phase CT current matched with the polarity voltage of ‘a’ phase to non-polarity ‘n’ on the potential coil).

4b.) The ‘a’ phase CT would have been wired to the second meter stator current coil with Vc-n voltage on the stator potential coil (the non-polarity of ‘a’ phase CT current matched with the polarity voltage of ‘c’ to non-polarity ‘n’ on the potential coil). Now ‘a’ phase CT current is wired so that it goes backwards through each of the two meter current coils.

To see the vector diagram for reference you will need to locate the image with the title: Mis-wired 5S Image which can be found by going to this link (you can copy and paste into your browser or highlight the link and right click to "go to") http://metergod.com/images-for-q-a-page/

So, if each of the 3 CT currents were 5 amps along with Va-n and Vc-n voltages being 120 volts each, the following proof would be given:

The first meter element or stator would have Va-n on the meter potential coil and the sum of ‘b’ phase CT current and ‘-a’ phase CT current. This is not direct addition but vector addition, where ‘b’ CT current would be 5 amps at 240 degrees + ‘-a’ CT current being 5 amps at 180 degrees = 8.66 amps at an angle of 210 degrees. Multiplying 120 volts at zero degrees by 8.66 amps at 210 degrees = 120 x 8.66 x -.866 (Cos 210 degrees) = -900 watts for meter element #1.

The second meter element or stator would have Vc-n on the meter potential coil and the sum of ‘c’ phase CT current and ‘-a’ phase CT current. Again, this is not direct addition but vector addition where ‘c’ CT current would be 5 amps at 120 degrees + ‘-a’ CT current being 5 amps at 180 degrees = 8.66 amps at an angle of 150 degrees. Multiplying 120 volts at 120 degrees by 8.66 amps at 150 degrees = 120 x 8.66 x .866 (Cos -30 degrees) = 900 watts for meter element #2.

Therefore: -900 watts (element #1) + 900 watts (element #2) = 0 watts

The correct amount of watts should have been 5 amps and 120 volts for the three phases making a total of 1800 watts. The meter was measuring 0 watts, so the billing error was 100% or none of the load is being metered correctly. These values offer proof that the meter would be stopped at unity power factor, if my three listed assumptions are correct.

If there is anything that I need to correct in my noted assumptions; please let me know what those would be and I will fix those part(s) including the calculations.

Take care and Be Safe All Ways, Brad

I have recently corrected the above metering at the start of the billing cycle . . after my last post and during rewire it came to my attention that A phase potential was on the pad mounts B phase lug along with the errors listed above . I am sure this would put new light on your reply to billing error ..., more info to consider in this is its a very busy heavy duty fabrication /weld shop . they weld and cut there steel with the 3 arc welders 3 men steady 8 to 10 hrs a day 6 days a week. I will post the billing error correction after a 3 month cycle ............thanks Dan

Recently I Discovered an old installation .. 3 PHASE ...277/480 4 WIRE w/3 CT's wired to a 5S METER the problem is some time ago the pad mount failed and during that work "a" and "b" CTs were accidentally installed B A C ..is there a way to figure the effect on customers billing thanks Dan