Calculating Zero Sequence Impedance for Power Transformers 2

[tgott]

How is zero sequence impedance calculated from transformer test reports? And how is the test performed? The test reports give volts, amps, & kW for H-X, X-Y, and H-Y. This is a gnd-Y to gnd-Y with delta tertiary (Y-winding). Does connection matter for calculations? I thought I had the calculations figured out using short circuit test equations, but when the test data was given on different taps my equations no longer worked.

 

[electricpete]

I'm not sure exactly what your question is (and I'm sure I wouldnt know the answer if I did understand the question).

The transformer connections play a huge role in the zero sequence impedance. For example grounded-wye/grounded wye will conduct zero sequence and ungrounded-wye/grounded wye will not.

The impedance data that you have is likely for balanced positive sequence operation. I do see in a power textbook (Bergen Power Systems Analysis page 432) where it says Z0~Z1, but I am skeptical. Obviously does not cover 3-winding configuration. Also I'm sure that core construction (shell form or core form and number of legs) would have dramatic effect on zero sequence impedance, not accounted for in that simple thumbrule.

 

[tgott]

It is rather confusing. Let me try to clarify. I have a test report for a 161/69 kV gnd Y-gnd Y with 13.2 kV delta tertiary that gives me this data:
Zero Phase Sequence Impedance
HV Tap 1&N at 30 MVA
H-Y H-X X-Y
Volts = 7063.00 1734.00 810.10
Amps = 151.20 117.80 132.10
% Imp.= 14.71 4.64 11.59
KWATTS= 108.80 17.98 8.729

My question is, assuming the volts, amps and kWatts were measured, how is the % impedance calculated? And then also, how is the transformer connected to take those measurements?

 

[peterb]

The measurement is done by short circuiting one winding (the lower voltage) and applying a lower than rated 3-phase voltage to the higher voltage winding (third winding open circuited).

The percent impedance is calculated as follows -

Base impedance = (kV)^2/(Base MVA) Ohms
Test impedance = [(test volts)/(sqrt3)]/(test amps) Ohms
Percent impedance = 100*(test impedance)/(base impedance)

The kW measurement allows you to calculate the equivalent resistance and then to resolve the tranformer impedance into (R+jX).

 

[electricpete]

peterb - I couldn't get the numbers to work. Using the H-Y values.
> v:=7063;
v := 7063
> i:=151.20;
i := 151.20
> z:=v/SQRT3/i;
z := 26.97
> zb:=(161E3)^2/10e6;
zb := 2592.
> zpu:=z/zb;
zpu := .01041
> zpu*SQRT3;
.01803

Where I used Sb=S_3PHASE divided by 3. My reasoning was: there are two ways of working the bases. One way uses per-phase quantities Vb=VLL/sqrt(3) along with Sb=S3phase/3. The other uses three-phase quantities Vb=VLL along with S=S3phase. Computing Zb=Vb^2/Sb should come out the same either way. But even if I change result above by factor of sqrt(3) or 3 it still doesn't work. What did I do wrong?

 

[stevenal]

Peterb,

Your procedure sounds like it's for the positive sequence impedance. Zero sequence impedance will require a unidirectional current (same single phase source applied to all three phases in parallel). IEEE C57.12.90 gives a little information. The formula given is: Z0(%)=300((E*Ir)/(Er*I))

E=measured excitation voltage
Er=rated phase to neutral voltage of excited winding
I=measured total input current flowing in the three parallel connected windings
Ir=rated current per phase of excited winding

Trouble is, I get 5.11% H to X with the given information, assuming H was the excited winding.

 

[tgott]

Thanks Stevenal. That's it. The HV tap is on 1 which corresponds to 169.05 kV. Using that gives you 4.64 for the H-X. Apparently the 69 kV winding was excited for the X-Y test. Also thanks to all who weighed in on the topic.