Transformer Nameplate

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OCPD

Member
Location
Tempe, AZ
Hello Everyone,

I came across this transformer nameplate which shows 4 impedance values. My interpretation is a combination of voltage and mVA has a corresponding %Z. Has anyone seen this before and have any idea why four impedance values Capture1.JPG Capture1.JPG Capture2.JPG ?

 

publicgood

Senior Member
Location
WI, USA
Transformer Nameplate

Remember how %Z is determined...short secondary and raise voltage on primary until rated FLA flows on secondary. Take said voltage and divide by rated primary voltage for Z.

If there are (4) MVAs, there will be (4) different FLAs; thus, (4) different %Z.
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
Remember how %Z is determined...short secondary and raise voltage on primary until rated FLA flows on secondary. Take said voltage and divide by rated primary voltage for Z.

If there are (4) MVAs, there will be (4) different FLAs; thus, (4) different %Z.
The short-circuit impedance-except for 10 MVA-meet even IEC 60076-5.For 10 MVA 5.04% it seems to me very low.[8% it is minimum according to IEC 60076-5-and ANSI C57.12.00 also].
 

Ingenieur

Senior Member
Location
Earth
The short-circuit impedance-except for 10 MVA-meet even IEC 60076-5.For 10 MVA 5.04% it seems to me very low.[8% it is minimum according to IEC 60076-5-and ANSI C57.12.00 also].

here's a 5 mva
I took the pic yesterday during a commissioning <5%
most of the 5 to 10 mva I see are in the 5% range
 

Ingenieur

Senior Member
Location
Earth
Remember how %Z is determined...short secondary and raise voltage on primary until rated FLA flows on secondary. Take said voltage and divide by rated primary voltage for Z.

If there are (4) MVAs, there will be (4) different FLAs; thus, (4) different %Z.

Yep
if you compare the 10 and 20 it is doubled
pu z = applied v to achieve rated fla / rated v to to get twice the current (or va) need twice the V (it is also (xfmr ratio / applied V) / rated current in Ohms)

actual Z in Ohm is still ~ the same though
the pu Z has doubled but so has the base va
base Zxx = v base^2 / Sxx va
Therefore Zbase10 = 2 Zbase20
and since pu Z20 is x2 that of pu Z10 the act Z is the same
Z act = Zbase10 x pu Z10 = Zbase20 x pu Z20
 
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