TX sanity check plea?!? 2

[pavement]

Could somebody assist me in a quick sanity check please, following simulations results of a 13.8kV marine system, it was found that existed normal operating currents of around 15kA?!?!? on the bus following the secondary side, obviously to a guy who has limited experience even this seems exceesive.So the first instinct is that either the cable quantities or the transformer data is wrongly entered, this is where somebody can confirm im either hopeless or I still remeber my uni education.


For a 13.8/0.44kV 3MVA Ddo TX which has R=1% and X=12% what are the ohmic values of R and X for both sides? Im i right in what i was doing was using the Vrating on either side as Vbase, and the rated MVA as the VA base to get Zbase, and then assuming that the 12% is split 50:50 between each side i.e 0.06 (6%) * Zbase(pri) to get the ohmic value of Xl for the primary side.


Sorry its been a bit long winded but less isnt more incase of sanity checks.


Cheers

 

[cuky2000]

Here are some basic calcs.:

a-Full Load Current: IFL=1000*MVA/(1.73*kV)
IFL_sec=3000kVA/(1.73*0.44kV)~3,941 A

IFL_prim=3000kVA/(1.73*13.8kV) ~125 A.

b- Impedance(Z%) & Angle (A):
Z%=SQRT(X%^2+R%^2) =SQRT(12%^2+1%^2)=SQRT(145) ~ 12.04%
A=atan(X%/R%) = atan(12%/1%) = 85.24 Degree.

c- Transformer Impedance: Z=(Z%/100)*(kV^2)/MVA
Z_sec= 0.1204*(0.44kV)^2/ 3MVA~0.00777 Ohm
Z_prim= 0.1204*(13.8kV)^2/ 3MVA~7.64299 Ohm

d- X & R Ohmic Values: X=Z*sin(A); R=Z*cos(A).
X_prim= 7.616591Ohm & R_sec= 0.634716 Ohm
X_sec= 0.007743 Ohm & R_sec= 0.000645 Ohm

e- Approx. Sec. Short Circuit (for infinite source)
*SC At Transformer bushings without Motor Contribution:
Isc1= IFLsec/10*Z%~3941/10*12.04%~ 32 kA.

*ASSUMING 50% Motor Contribution:
Isc2 = Isc~1.5*32 kA.= 48 kA

_{NOTE: for more accurate result include cable SC damping and actual load.}

 

[pavement]

Well what can I say cuky2000, a big thank you to you, throughly desereved the star!

 

[pavement]

Where i went wrong was dividing the p.u % by 2, for each side, I dont know where Ive picked that reengineering trick up, but thanks again!

 

[jghrist]

cuky2000's calculations of Z_sec and Z_pri do not refer to the impedance of the secondary and primary windings, but to the entire transformer impedance referenced to the secondary side and the primary side respectively.

 

[rbulsara]

Isc= Ifla/%Z in p.u.

I am not sure why cuky's formula show "10" as a dividing factor, although his answer is correct

3941 A/0.12 Z = 32,841 A or 32kA

 

[jghrist]

rbulsara,

cuky is dividing by 1000 to get the answer in kA instead of A.
Z_pu=Z%/100
1000·Z_pu=10·Z%

 

[rbulsara]

jghrist:

Thanks....I always think and use % in decimals only)