Transformer Z value

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mbrooke

mbrooke

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How do I turn this 3.13% Z transformer into an R and X value so I can accurately compute the fault current in this following circuit?

1597725384507.png



Nameplate:

1597725567924.png
 
Checking the transformer load losses at different manufacturers I found that the losses are between 2350 and 2900 watts for transformers in 150 kVA in oil. If the rated current it is 150/sqrt(3)/1.2=72.17 A then R=Loadlosses/3/I^2 [neglecting stray currents]=0.147 up to 0.186 ohm/phase.
If Z=1.2^2/0.15*3.13%=0.3 ohm then:
Xmax=sqrt(0.3^2-0.147^2)=0.26 ohm R=0.147ohm
Xmin=sqrt(0.3^2-0.186^2)=0.235 min and R=0.186
 
Julius Right said:
Checking the transformer load losses at different manufacturers I found that the losses are between 2350 and 2900 watts for transformers in 150 kVA in oil. If the rated current it is 150/sqrt(3)/1.2=72.17 A then R=Loadlosses/3/I^2 [neglecting stray currents]=0.147 up to 0.186 ohm/phase.
If Z=1.2^2/0.15*3.13%=0.3 ohm then:
Xmax=sqrt(0.3^2-0.147^2)=0.26 ohm R=0.147ohm
Xmin=sqrt(0.3^2-0.186^2)=0.235 min and R=0.186
Click to expand...


Any idea of the X value or that can't be determined?

I had someone explain to me that I could take 150,000/3=50,000/480=104x1.73=180/0.0313=5,764 amps of fault current.

5,764/1.73=3,328 amps per coil.

3,328amps at 277 volts gives a coil impedance of 0.08323 ohms or 12.0148984741 Siemens.

Of course that doesn't give me an R or X value.
 
The previous answer listed approx. values for Xmax and Xmin.

But do I understand the diagram right? You have 500' of #12 wire? That alone is pretty much going to determine the fault current at the end. The transformer %Z is going to make very little difference. And dividing the %Z into X and R components is going to make even less difference.

If its purely an academic exercise, I understand, but otherwise why make so much extra work for a fraction of a percent accuracy?
 
steve66 said:
The previous answer listed approx. values for Xmax and Xmin.
Click to expand...

Im missing the "Xs" so I assumed both values were in regards to resistance only.

But do I understand the diagram right? You have 500' of #12 wire? That alone is pretty much going to determine the fault current at the end. The transformer %Z is going to make very little difference. And dividing the %Z into X and R components is going to make even less difference.
Click to expand...


The 500 feet of number 12 is just for the simplicity of discussion, as are the full size equipment grounds. IMO its easier to discuss math and technical subjects with fewer, evenly proportional variables.


If its purely an academic exercise, I understand, but otherwise why make so much extra work for a fraction of a percent accuracy?
Click to expand...

In the real world transformer, service drop and feeder impedance can make or break anyone single scenario by a small factor. Thus Ze must be taken into account and can not be ignored in the whole of the circuit:

1597781660060.png
 
And, in some cases, by a large factor such as large buildings like hospitals or feeders to ball fields, barns, trailers, or remote buildings.
 
I have to apologize for a serious mistake - I quickly introduced the primary voltage 1.2 kV instead of 12 kV. Now I want to start all over again.
Transformer data check:
12/0.48 kV 150 kVA 3.13% short-circuit impedance p.u.
You need only R and X viewed from low voltage terminals. As I said, the average load losses [from 2350 to 2900 W] it is [2350+2900]/2=2625 W.
Let's revise the theory:
S=sqrt(3)*I*VL_L apparent power
VL_L*vsc=sqrt(3)*I*Zph vsc=short-circuit impedance p.u.
Then:
Zph=VL_L*vsc/(SQRT(3)*I). By multiply up and down by VL_L:
Zph=VL_L^2*vsc/(SQRT(3)*I*VL_L)
Zph=vsc*VL_L^2/S
Then the transformer impedance viewed from low voltage it is:
Zph=0.48^2/0.15*3.13% [voltage in kV, apparent power in MVA]=0.048077 ohm/phase
If ploss=2625 W and Ilv=150000/sqrt(3)/480=180.42 A then:

[TD valign="bottom"]ploss=3*Rlv*Ilv^2[/TD]
[TD valign="bottom"]Rlv=ploss/3/Ilv^2[/TD]
[TD valign="bottom"][/TD]
Rlv=2625/3/180.42^2=0.02688 ohm/phase
Then Xph=sqrt(Zph^2-Rph^2)=sqrt(0.048077^2-0.02688^2)=0.03986 ohm/ph
The ratio X/R=1.483 [it is still odd since usually it has to be 2-7].
 
The phase-to-ground short-circuit current I”k1=sqrt(3)*VL_L/(Z1tot+Z2tot+Zotot) in complex.
Ztot= (Z1tot+Z2tot+Zotot)
Ztot={[(Rxfr+2*R(4/0)+2*R(6)+2*R(12)]+j[3*Xxfr+2*X1(4/0)+Xo(4/0)+2*X1(6)+Xo(6)+2*X1(12)+Xo(12)]}
Rxfr=0.0480768 ;X1xfr=X2xfr=Xoxfr=0.03986 ohm
500 ft=0.1524 km
R4/0=0.207 ohm/km;X1(4/0)=X2(4/0)=0.135 ohm/km ;Xo(4/o)=0.135
R6=1.61 ohm/km;X1(6)=X2(6)=0.167 ohm/km; Xo(6)=0.167
R12=6.6 ohm/km;X1(12)=X2(12)=0.223 ohm/km; Xo(12)=0.223
If we take all the resistance and all reactances in series including grounding wire and parallel the earth we get 314.84 A short-circuit current at the end
of 12 awg wire. If we neglect the earth 314.55 A. If we neglect the transformer we get 322.44 A[2.51% more]
 
steve66 said:
The previous answer listed approx. values for Xmax and Xmin.

But do I understand the diagram right? You have 500' of #12 wire? That alone is pretty much going to determine the fault current at the end. The transformer %Z is going to make very little difference. And dividing the %Z into X and R components is going to make even less difference.

If its purely an academic exercise, I understand, but otherwise why make so much extra work for a fraction of a percent accuracy?
Click to expand...

To clarify my inference and my equations:

1597866574007.png

Zs= Total impedance of the faulted circuit. Or System impedance.

Ze= External Impedance. The impedance of the utility. This would be the sum of the transformer and the service drop.

R1+R2 = the impedance of the phase and ground wires from the main disconnect to the fault point.

Zero ohms is assumed for the fault itself.

Any parallel paths like conduit, pipes or grounding electrodes are ignored.

Wire impedance values are derived from chapter 9, table 9.

Drawing shows 230 volts, however this holds true for any voltage.
 
JoeStillman said:
SKM Lists a "typical" X/R ratio for this size transformer as 3.5855 which works out to 0.8409 %R and 3.0149 %X.
Click to expand...


Question- If I simply add the trafo impedance to the wire impedance like this:

R= 0.8409+0.062+0.49+2.0=3.3929

X=3.0149+0.041+0.051+0.054=3.1609

Z= 4.637139120190379

I=E/R ---- 277/4.63714=59.73511

Loop current = 59.73511

400% of 15 amps = 60 amperes.

Technically I'm just 300 milliamps from having a ground fault current path.


My question is- how does my way vary from actual numbers?

For example, the transformer is mostly reactive. It is my understanding that the short circuit current will lag substantially, thus a current of other than 60 amperes will be "seen" by the TEY breaker?
 
Interestingly, using an infinite source I'm getting 2.55617 ohms:

1598085015803.png


So- 2.55617 vs 4.63714... Thats a big difference 😲
 
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