transformer Z% in parallel

[arc101]

Say you have two step down transformers (240v/120v, Z=5%) wired in series. Will this make total Z=10%? Then actual Z would vary by a factor of 4 (due to the different voltages)?

 

[JensenDrive]

I should say that I am assuming in my thoughts above that the two primaries are connected in parallel, so this becomes a 240:240V isolation xfmr.

 

[jghrist]

JensenDrive,

I assume differently. I assume that the primaries are in series and the secondaries are in series. The object is to construct a 480-240 volt transformer from two 240-120 volt transformers.

The total %Z would be twice the individual %Z if you keep the base kVA and base voltage equal to that of one transformer. If you use a base kVA equal to the total of the two transformers, and a base voltage equal to the total voltage, then the %Z would be equal to the original.

Total Znew (each transformer) = Zgiven x (base kVgiven/base kVnew)² x (base kVAnew/base kVAgiven)

Total Z = 2 x Znew

Example: two 10 kVA 240-120 volt tranformers, 10% impedance.

Znew = 2 x 10% x (0.24/0.48)² x (20/10)
= 10%

 

[Sideswiper]

Well since the title says Parallel, I'll use this thread to ask about it.

What if you had two transformers in parallel with slightly different Z values (say 7.94% and 7.90%), and you are trying to calculate high side voltages and currents using low side PTs and CTs... what would you use for your Z value? Split the difference with 7.92%?

 

[BJC]

I think the impedance of the combined bank is 5%.
If you use the diagram Dave Beach has drawn on 21 Nov and calculate the secondary short circuit current and the calculate the normal full load as a percentage of the short circuit current you get 5%.

 

[dpc]

%Z means nothing unless you specify a base kVA. But basically you're on the right track provided your base kVA for the 7.92% is twice the kVA of a single transformer (and the transformers have the same kVA rating. You're going to have twice the fault current, so either the base kVA doubles or the impedance goes down by 50%.

Of course, the primary current can be computed with knowing the impedance, just using the turns ratio. Ignoring the exciting current, which the impedance tells you nothing about anyway.

The difference in impedance when you average these is about 0.2% which is probably smaller than the error bar on the PT accuracy anyway. And probably less than the change in impedance if you ever change taps.

 

[BJC]

dpc
You can assume any KVA( for both xfmrs) and calculate the %Z of the combination.
It's the same as the % Z of one of the transformers.

 

[dpc]

On what base kVA?

 

[BJC]

"On what base kVA? "
What ever you want.
If you assume the transformer were 60 KVA ( no size was given initially) the short circuit current for that transformer connected to a 240 volt primary would be 10,000 amps. 500 amps/5%= 10,000amps. The 5% would be for 60kva.
If you connect the transformer as shown in hte 27 Nov post
the full load current is still 500 ampas and the short circuit current is still 10,000 amps.
The % Z is still 5% but now on a 120 KVA base.

 

[dpc]

The % Z is still 5% but now on a 120 KVA base.

Exactly. The only way the impedance is still 5% is if you change the base kVA from 60 to 120 kVA.

 

[BJC]

Exactly
That was inherent in the original question. He had two transformers.
The way I looked at was the two transformers go into a "black box" with 4 terminals. What label do you put on the box?
Ans 480/240, 120KVA, X=5%