Transformer Calculation

[larslars]

I have extended our HV installation with one additional transformer 4,160/208 DY

A subcontractor is asking for some data concerning the LV part of the transformer – I think it is for the panels and other gear he wants to install.

I suppose he needs to know the maximal prospective short circuit current level Ik3ph. and a belonging X/R value – Do any of you now if this could be correct??

The only usable information I have found on the rating plate on the (used) transformer, is the value IZ 4.7% - 150KW

I think I’m getting old. I really can’t remember a lot of this stuff, maybe some of you could help me out.

1: The available data on the supply terminal is: The AC-Comp (7,779A) and the Asym (8,100A)
Which of this information will I use in my calculations (I think that it is the AC-Comp, the X/R value informs about the size of Asym, but I’m not sure about this)

2: The same data says that the X/R = 1.98, how can I translate this to useful X values and R values.

3: The 2-0-5KV cable is 120 ft. ( R=0.0764ohm/1000ft. and X=0.0414ohm/1000ft.). All right, I got that.

3: IZ 4.7% - 150KW on the transformers rating plate. Is that really enough information to calculate anything??

Thanks,
Lars

 

[wareagle]

Lars
I think you are correct. The contractor is trying to size
the equipment for the fault current. What size is the transformer?

 

[larslars]

Hi

It's a 150KVA transformer

 

[waross]

150KVA /3 Phase = 50 KVA per phase
50KVA / 120 volt = 417 amps
417 amps / 4.7% = 8865 Amps.
At this point I would use breakers and equipment with an interupting rating of 10,000 amps. symetrical and not bother with any more calculations. If the available fault current was very close to or slightly over 10,000 amps I would check the impedance of the feeder to see if 10,000 amp equipment would be safe.
The X/R ratio is used for calculating the transformer voltage regulation at various power factors.
respectfully

 

[sitl]

If your subcontractor only wants to size the equipment in switchboard immediately after the transformer, then you need to give him:
a) value of initial short circuit current Ik'' (which is in distribution networks equal to a steady state short circuit current) and
b) peak value of the short circuit current Ip,
both for three phase faults.
First one then should be less than breaking capacity of the equipment in the switchboard and second one less than making capacity of the fore mentioned equipment. Hopefully the breakers are of category A for which no short-circuit
trip delay is provided, therefore you don’t need to calculate the short-time withstand current Icw.

In order to compute Ik'' you need to know what is the three phase short circuit power Ssc (usually given in MVA) at the point of coupling of your transformer to the network. This should be given to you by your utility. I am not sure about those markings ''AC comp'' and ''Asym'' (is that Isc), so you should check with them. Since you don't have parameters of the transformer (there is always an option of calling transformer manufacturer), you should calculate Isc from Ssc and then transfer it directly on the secondary side of the transformer (you are neglecting the dampening effect of transformer on the short circuit current, therefore being on the safe side).
Since you don't have enough parameters to calculate the exact value X/R on low voltage side, you should calculate Ip as 2*square root (2)*Ik'', again being very much on the safe side.
These two values should be enough for him.

If your subcontractor wants to determine short circuit currents on all levels of installation downstream of the transformer, then in order to calculate them you need to give him Ssc (or Isc), transformer parameters (R and X, or Pcu and uk from which he can compute R and X) and cable parameters from transformer to the switchboard.

To waross: X/R ratio is used to calculate Ip as Ip=K*square root (2)*Ik''.

 

[larslars]

My subcontractor has accepted all of my calculations; again I can smile to the sun

Thanks to both of you

Lars