Current in 3-phase Secondary side of transformer

I think you have it right IF your are talking the same winding configuration on primary and secondary. That is Y-Y or delta-delta. Remember that the power going into a leg of the transformer is the same as that going out. Your Es*Is=Ep*Ip.

Bill

-- Ferme le Bush

---------- Salmon egg has it right but it would be easier if you gave all the data at the beginning. it appears that the transformer may be 415/208 line to line and 240/120 line to neutral Y-Y possibly an autotransformer.

Hi there Guys, I'll give you a bit more info.

I have recently started a job in a large manufacturing plant in the UK.

This plant has alot of American equipment/machines so voltages are converted to 110 and 208v.

The plant has been having problems as previously there was no standards or conditions for electrical loading on Distribution boards, eg. many board were overloaded and many had very little loading... Therefore leading to some transformers being close to their load limit.

I have recently carried out a survey on the distribution system and have recomended changes, eg cables moved from heavey loaded DB's to light loaded DB's.. this has been actioned and all seems ok... the loads are now shared.

Within the plant there is a computer system monitioring the kW in various boards, these are fed to a PC where loading can be checked etc.

5 of these readings come from 208v circuits.. eg to find the loading on the transformers on the 208 circuits.

The problem I have is that when being installed in 1993 the contractor then installed the CT's on the primary side of the transformer, eg 415v rater than the secondary (i think they were cutting corners). So now... when I get the reading I am getting the reading (kW) from the primary side 415v and not the secondary 208v. This is not giving me the correct reading as this is taking into consideration the transformer power... I just want to see the power/current in my 208 side.

Here's the data I have from the plate on the transformer: The transformer is: ABB Trafo-BB, Type DZV

500kVA V= 433/216 I=66.7/1336.5 Vector=Yy0 Impedance 3.7%

Here's the reading i've takin for one of the transformers: The line current is: 762A The voltage is: 208v so therfore the kW is: 247kW

These are manual readings... but the reading on the PC which is coming from the Primary 415v side is 324kW.

*Note... these are approx values since its such a big num and from different sources so any calculations will not be exact*

Its nearly impossible for me to install 110v or 208v CT's as the equipment must be shut down. Does anyone have any ideas how Is there any way I can get around this... all I'm looking for is the current or power on the secondary side. Thanks loads, Shay

I guess the decimal points are not at correct places - should be "I=667/1336.5A"

762A*208V*sqrt(3)=274kW

The difference between 324 and 274 is 15%

The current at low voltage side in your case should be 2 times less than on primary:

P(high voltage)=433V*sqrt(3)*Current(high voltage)*(power factor) P(low voltage)=216V*sqrt(3)*Current(low voltage)*(power factor)

If we neglect the losses in the transformer - then P(hv)=P(lv), so Current(lv)/Current(hv)=433/216=2 => Current(lv)=2*current(hv)

I hate such expressions. What would "one time less" mean?

-- Ferme le Bush

OK: 433/216 is effectively 2 to 1 and the rated currents are in the inverse ratio 667/1337A. These are line currents on either side, not phase currents and I believe that the transformer isY-Y (Yy0 ). For your load case, you have not given any power factor figures but the measured output is 1.732*208*762 =274KVA (not KW) For the impedance given, at 0.8 pf, and estimating 3% magnetising current, the input KVA should about 285-290KVA, depending on losses, not 324KW on the basis of your measured values. However a 10% under reading error (see 1 below) in current would bring things closer to your PC values. Transformer losses should be low- say 1% so measuring on the input isn't going to introduce a serious error unless it is imperative that an accurate output measurement was needed and this doesn't appear to be the case when it was installed.

1)you apparently used a clip on ammeter - good for indication but accuracy is not that great. Did you check all 3 phases and neutral? Did you take input current measurements with this instrument at the same load to compare with the information sent to the PC? 2)Check your instrumentation setup feeding the PC and the algorithms involved.--

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

Salmon Egg =D0=BD=D0=B0=D0=BF=D0=B8=D1=81=D0=B0:

I guess it should mean "exactly the same" :-) However, English is not my native language, so maybe I used the phrase improperly - sorry for that.

I think it was misused EXACTLY as it is misused by many Americans. Sometimes people learning English should not emulate the people they learn from.

Bill

-- Ferme le Bush