Impedance Transformer Calculation Based on Short Circuit Tests.
Impedance Transformer Calculation Based on Short Circuit Tests.
(OP)
Hi, all.
We are testing a new three winding transformer (4,5 MVA 13.8 kV Delta / 2,25 MVA 0,48 kV Wye / 2,25 MVA 0,48 kV Wye) to obtain the percentual impedance of windings. We obtained the following results:
-Voltage applied in 13.89 kV winding with secondary short-circuited and tertiary open: 1259 V / 94.86 A.
-Voltage applied in 13.89 kV winding with secondary open and tertiary short-circuited: 1238 V / 94.24 A.
-Voltage applied in secundary with tertiary short-circuited and primary open: 35,67 V / 1276,7 A.
How to calculate the impedances Zps, Zpt and Zst? Is there any considerations to do abou de Delta winding, like an Delta-Wy transformation?
Best Regards.
We are testing a new three winding transformer (4,5 MVA 13.8 kV Delta / 2,25 MVA 0,48 kV Wye / 2,25 MVA 0,48 kV Wye) to obtain the percentual impedance of windings. We obtained the following results:
-Voltage applied in 13.89 kV winding with secondary short-circuited and tertiary open: 1259 V / 94.86 A.
-Voltage applied in 13.89 kV winding with secondary open and tertiary short-circuited: 1238 V / 94.24 A.
-Voltage applied in secundary with tertiary short-circuited and primary open: 35,67 V / 1276,7 A.
How to calculate the impedances Zps, Zpt and Zst? Is there any considerations to do abou de Delta winding, like an Delta-Wy transformation?
Best Regards.






RE: Impedance Transformer Calculation Based on Short Circuit Tests.
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
Zps=(1259/13800)*(94.13/94.86)*100%=9.05% tested on 4.5MVA base
Zpt=(1238/13800)*(94.24/94.86)*100%=9.04% tested on 4.5MVA base
Zst=(35.67/480)*(2706/1276.7)*100%=15.75% tested on 2.25MVA base
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
Zps=(1259/13800)*(188.3/94.86)*100%=18.1%
Zpt=(1238/13800)*(188.3/94.24)*100%=17.9%
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
13.89 kV was a typing error. The correct is 13.8 kV.
We was calculating Zps and Zpt like in your first answer, founding 9.05% and 9.04% respectively, assuming that power base por HV winding was 2.25 MVA (half total), for a full load in secondary OR tertiary winding.
My doubt is the presence of delta winding in HV side and if it requires a delta-wye transformation before calculate de bases of voltage and current.
In your calculus, isn't it necessary to consider 1.732 for calculate the current bases?
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
You don't need Delta-wye transformation because this is the pair winding equivalent impedance
Sbase=4.5MVA, Vbase=13800
Zbase=V^2/S=42.32 Ω
Zps=1259/1.732/94.86/42.32=0.181 pu
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
This was another pointing of discussion here, during de transformer testing.
I'm not sure what base use in primary winding, 4,5MVA or 2,25MVA. Until the moment no one coul answer this for me.
First I was thinking that correct is use 4,5MVA, but another guy here said do use 2,25MVA as base in sense that it corresponds to a full load on secondary winding. I'm confused...
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
Normally, the rated (apparent) power and nominal voltage are arbitrarily selected as base quantities. So, is not important what is the base value, just one have to use it on all three windings.
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
See this explanation make a better sense -
Let us treat a transformer as a black box and we could careless what are inside including the winding configurations. When doing a short circuit test a voltage is applied to one side(normally primary) and the secondary is shorted and tertiary is open. So, if look in to the winding that the voltage is applied, it is a two terminal network (a black box with tertiary open), now one thing we need to do is to reach the rated short circuit current and then calculated the equivalent impedance. It is obviously that this winding is the base voltage and MVA that the test is performed, and therefore the real impedance value is calculated as a ratio of the base value, so to use based values from other windings does not make sense. When doing the pair windings S-t, the base then becomes the secondary or tertiary whichever the voltage is applied. If both secondary and tertiary are shorted and voltage applied on primary, it more like paralleling transformer and the impedance is about cut by half around 9%
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
Impedance test. Short the secondary and increase the primary current until rated current flows.
If this is rated current in the secondary and the base must be the 2.25 MVA winding.
However, if this is rated current in the primary, the current and hence the voltage must be twice as much. The base will be 4.5 MVA and twice the voltage on twice the base will give the same PU value.
However, the secondary will be carrying 200% of rated current if the transformer is tested with this base (4.5 MVA).
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
This was what we were doing. Full load at secondary 2.25 MVA (half of primary winding) and use as base 2.25 MVA at primary. We was in doubt in what MVA base to use (2.25 or 4.5 MVA), but considering what 7anoter4 said, now, I really think that I should use 2.25 MVA in all three windings. Thinking in a two winding transformer, power base is the same for both windings.
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
RE: Impedance Transformer Calculation Based on Short Circuit Tests.
Base MVA is not important.Generally these are expressed for the rated secondary MVA ie 2.25 MVA. But in this case there is one more %impedance ie the impedance between primary and with both secondaries shorted.For this transformer it will be roughly 10.2% on 4.5 MVA base.Note for this impedance usually the base will be the rated MVA of primary.
This is an arrangement that is called with less coupled secondaries. ie when you load one secondary the terminal voltage of the other secondary will not be affected due to the regulation in the loaded secondary. This is because of the almost double percentage impedance between secondraies compared to primary to secondary impedance.In this case in three limbed impedance circuit( relevant for multi circuit transformer) the impedance of primary will be almost zero and hence the laoding onn one secondary will not create voltage drop in the terminal voltage of the unloaded secondray.This is physically achieved by placing the secondaries axially one above the other with the primary also axially split equally.