transformer impedance base conversions 3

[tricard]

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

 

[alehman]

The percent impedance of a transformer is the same, regardless of the voltage. If it's rated 1500kVA at 6.9kV (which it appears to be), the impedance would still be 5.6%.

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%

 

[tricard]

Excellent. Thank you very much for your help.

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

 

[tonyjony]

Most transformers have two windings over a ferromagnetic material core. So the magnetic flux created by the first winding induce voltage over the second one.
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

 

[jghrist]

A dual voltage transformer would have two primary windings which would be connected in series for the larger voltage and in parallel for the lower voltage. The proper designation should be 6.9x13.8 kV, not 6.9/13.8 kV.

 

[alehman]

jghrist is right. If you simply tap the winding, you have only 1/2 the full-voltage kVA capacity available because the current is limited to the full-winding value. By paralleling two separate windings, you double the current capacity at the same time you reduce the voltage by half.

You might take a look at some of the basic power system references in faq238-1287.

 

[tonyjony]

I just read this:

Primary: 6.9/13.8kV Delta
Secondary: 600Y/347

sorry

 

[pwrtran]

Hi Guys

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

 

[tinfoil]

I disagree with pwrtran.

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).

 

[alehman]

I agree with tinfoil. It seems more likely that the 6.9kV connection would use paralleled connection of two windings for each leg of the delta than taps.

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)]

 

[racobb]

Tinfoil amd Alehman are correct. Ampere turns are the same, it is just the difference in primary windings being in parrallel or series....But the %z for the transformer does not change! The iron and copper of the unit are the same from when it was built either way.

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

 

[pwrtran]

tinfoil, alehman, racobb

You are all correct! I missed the parallelled primary windings.
Thanks for the correction

 

[pwrtran]

tinfoil, alehman, racobb

Forgot to say, a star to each of you!