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steezmo

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Hey All,

Sorry if this was covered elsewhere, couldn't find a specific answer. Want to know the rule of thumb for choosing an infinite source (on the PRIMARY) based on XFMR size. Secondary side I know the calc. For example, if I have a 750 kVA XFMR according to a SKM model I need a 15000 MVA (3PH) and 7500 MVA (SLG) primary source contribution to get the theoretical max fault current out of that transformer on the secondary. 

In this example the source needed was 20,000X the rating of the XFMR. Anyone know the calculation to prove this or if there is a rule of thumb or even IEEE doc covering this?

Thanks for the help.

 
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This is how we size transformers.  Assuming a 2-winding transformer, your primary would be: Amps= (MVA*10^6)/((kV*10^6)Sqrt(3)).

For a 3 winding transformer, you would need to look at the manufacturers drawing to figure out the MVA on the tertiary then use the same formula.

 
Last edited by a moderator:
In your example, assuming infinite primary fault current

the secondary fault MVA would be .750 MVA/% Z, let's say 5% transformer impedance = .750/0,05 = 15 MVA

You can use MVA method too. let's say the Fault MVA on primary side is infinity(99999999 MVA) and transformer fault mva is .750/.05 = 15MVA

(1/99999999 + 1/15)^-1 = 14.9999 MVA

If the Primary fault level is 20,000 MVA then

(1/20,000+ 1/15)^-1 = 15 MVA

I think this is where you may have got 20,000, anything over 20,000 MVA in this case will not change the fault level on the secondary side of the transformer. 

 
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