transformer impedance base conversions
transformer impedance base conversions
(OP)
Good day,
I am new to the field of power engineering and this is my first post to eng-tips. If I have not provided enough information, then please let me know for future posts.
I have been given one line data and field information to enter into a model for a short circuit study. I came across a transformer that has the following nameplate data:
Manufacturer: Ferranti-Packard
KVA: 1500/1680
Type: LNAN 55/65 rise
Primary: 6.9/13.8kV Delta
Secondary: 600Y/347
Impedance: 5.6% at 1500KVA, 13.8kV
My line side base voltage is 6.9kV, so what is a valid impedance for the transformer if impedance is only given for 13.8kV?I figured to use the standard p.u. impedance conversion of
Znew = Zold x (Vll_old^2)/(Vll_new^2)
--> .056 x (13.8^2)/(6.9^2)
This yields: Z% = 22.4%
Have I performed this calculation correctly or is there a different method to apply to transformers with multiple line side voltage options?
Thanks for the help,
Tim
I am new to the field of power engineering and this is my first post to eng-tips. If I have not provided enough information, then please let me know for future posts.
I have been given one line data and field information to enter into a model for a short circuit study. I came across a transformer that has the following nameplate data:
Manufacturer: Ferranti-Packard
KVA: 1500/1680
Type: LNAN 55/65 rise
Primary: 6.9/13.8kV Delta
Secondary: 600Y/347
Impedance: 5.6% at 1500KVA, 13.8kV
My line side base voltage is 6.9kV, so what is a valid impedance for the transformer if impedance is only given for 13.8kV?I figured to use the standard p.u. impedance conversion of
Znew = Zold x (Vll_old^2)/(Vll_new^2)
--> .056 x (13.8^2)/(6.9^2)
This yields: Z% = 22.4%
Have I performed this calculation correctly or is there a different method to apply to transformers with multiple line side voltage options?
Thanks for the help,
Tim






RE: transformer impedance base conversions
Zbase at 13.8kV = kV^2/MVA = 127 ohms
Z = 5.6% * 127 = 7.11 ohms
@ 6.9kV, you will have (6.9/13.8)^2 * 7.11 = 1.78 ohms
Zbase at 6.9kV = 31.7 ohms
1.78 / 31.7 = 5.6%
RE: transformer impedance base conversions
One other question on the same topic, since my equipment experience is even less than my practical knowledge at this point
How are the xfmrs physically constructed? Is there essentially a centre tap that drops the voltage by 1/2 on the primary or are they dual wound (or is it something completely different)? I am guessing this may be a very arbitrary question given that there is likely more than one way to construct one.
Thanks for the help
Tim
RE: transformer impedance base conversions
The machine who has one winding split it by half it's called Autotransformer, and It works in a symilar way.
Some books you can read about it: Hindmarsh, Kostenko, Wikipedia
RE: transformer impedance base conversions
RE: transformer impedance base conversions
You might take a look at some of the basic power system references in FAQ238-1287: What are good references for a Power Engineer?.
RE: transformer impedance base conversions
Primary: 6.9/13.8kV Delta
Secondary: 600Y/347
sorry
RE: transformer impedance base conversions
I would say that tricard's calculation was correct.
Short circuit test will give the impedance voltage. Normally, the secondary windings are shorted and a voltage apply to the primary until the secondary current reaches its full load current at its base rating (1500kVA). If the primary voltage is 13.8kV and the impedance is 5.6%, which means 5.6% of the primary voltage was applied on the 13.8kV tap to get full load current on the shorted secondary windings.
When the primary tap changed to 6.9kV, the voltage and current ratio has been changed. In order to push the secondary current to its full load current during the short circuit test, you need to give more voltage on the primary side to get there.
So, I believe that tricard's result:
"Znew = Zold x (Vll_old^2)/(Vll_new^2)
--> .056 x (13.8^2)/(6.9^2)
This yields: Z% = 22.4%"
stays valid.
BTW the complete equation is
Znew = Zold x (Vll_old^2)*Snew/(Vll_new^2)Sold
In the above case Snew=Sold
RE: transformer impedance base conversions
Watch what happens when you change from 13.8 to 6.9kV. It is likely that you now have two parallel primary windings where you used to have only one, and each of these two windings has only half as many turns.
So you have less ABSOLUTE applied primary voltage required to deliver full load current on the secondary side (but these volts are on a different Vbase, so the %V is larger).
RE: transformer impedance base conversions
Either way, the percent impedance is the same so long as the base voltage of your calculations is the same as the winding voltage. In other words, if you use the 6.9kV connection and your base voltage on that side of the transformer is 6.9kV, the impedance is 5.6%.
pwrtran,
The equation you reference is only for changing per-unit values from one base to another.
Zpu(new) = Zpu(old) * [Vbase(old)/Vbase(new)]^2 * [kVAbase(new)/kVAbase(old)]
RE: transformer impedance base conversions
Select your per unit bases properly and you do not need to do conversions unless your individual transformer size is different from your base MVA....as a general rule.
Ie: 2X voltage= 1/2 current....series windings
Or 1/2 voltage = 2X current....parallel windings
That is for the same kVA rating.
Alan
Democracy is two wolves and a sheep deciding what to have for dinner. Liberty is a well armed sheep!
Ben Franklin
RE: transformer impedance base conversions
You are all correct! I missed the parallelled primary windings.
Thanks for the correction
RE: transformer impedance base conversions
Forgot to say, a star to each of you!