# NCEES problem# 525 - parallel transformers ### Help Support Engineer Boards: P

#### Platinum

525.) Two single-phase transformers with identical voltage ratings are proposed to be connected in parallel to serve a load. The following rating and impedance data are provided:

Transformer 1: 1,000kVA, Z= 4.5%

Transformer 2: 2,000kVA, Z= 6.0%

The maximum total load (kVA) that can be served by the bank without overloading either transformer is most nearly:

A.) 2,333

B.) 2,500

C.) 2,667

D.) 3,000

Okay, I don't know much about paralleling transformers. Could someone give me an explanation on this one.....particularly on the solution's statement: "The transformers will load in inverse proportion to their impedances."

Thanks guys

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#### Dark Knight

##### Silent Guardian
Plat,

I would try to explain but I am not good explaining things. For this one the best way I had to understand was looking at the relationships between per units quantities and transformers turn ratios. You can get the final formula combining these. I had red notes on this one so I got it worng practicing but then have references in blue that took me to these topics. I am going to take a look and see if I can explain it in a coherent way.

#### Art

##### Veteran
answer B 2500 kva, my 'guess'

look at it this way...

you know since the 1000 has lower impedance it will load up to it's rating fast...draws more current at the same voltage...

and we know kva ~ Z, ie, I^2 Z or V^2/Z

so 4.5/6 x 2000 + 1000 = 2500

that's my story and I'm sticking to it...

G

#### grover

525.) Two single-phase transformers with identical voltage ratings are proposed to be connected in parallel to serve a load. The following rating and impedance data are provided:
Transformer 1: 1,000kVA, Z= 4.5%

Transformer 2: 2,000kVA, Z= 6.0%

The maximum total load (kVA) that can be served by the bank without overloading either transformer is most nearly:

A.) 2,333

B.) 2,500

C.) 2,667

D.) 3,000
per-unit impedances have to be relative to the same number. So, Transformer 1 (1000kVA/Z=4.5%) has a per-unit impedance of Z=9% relative to 2000kVA.
Resisters in parallel, The 1000kVA transformer is getting 6/(6+9) of the total apparent power. Logic dictates transformer 1 will be saturated first. So, since we're maximizing this, it means the TOTAL power is 1000kVA / (6/15) = 2500kVA. B.

Is that right? I don't have my book handy to check...

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#### singlespeed

##### Tormented
Art and Grover show that this problem can be solved in two ways, both using the same logic. The pu calculation by Grover is the way you'll find in the EERM and the NCEES solution.

#### eng.dork

##### Member
Great discussion so far. Why is it not good to parallel transformers? Is it because it doubles fault current? Or if I parallel transformers to the same bus and a fault occurs then both transformers are gone and there is no redundancy what so ever? If anyone has a good explaination for this please help!

P

#### Platinum

B. 2500kVA is the correct answer

Art- I didn't follow your method.

Grover- I understand converting the two transformer impedances to the same base. Then treating them like parallel resistors and perform a current division, ie your branch power= (total power)*(.06/.15)

So total power = 1000/(.06/.15)......gotcha.

I don't understand the solution's equation: (Load T1/Load T2) = (Z1/Z2)

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#### Dark Knight

##### Silent Guardian
Great discussion so far. Why is it not good to parallel transformers? Is it because it doubles fault current? Or if I parallel transformers to the same bus and a fault occurs then both transformers are gone and there is no redundancy what so ever? If anyone has a good explaination for this please help!
The thing when paralleling transformers is that you have to be aware of the impedances. They have to be the same. If not you are going to have circulating current flowing between the transformers. Not good.

But the main cause of why utilities don't like to do that is the available fault current. It is going to be higher, way much higher, because now you will have two sources instead of one.

G

#### grover

I like paralleling transformers, though, for redundancy. Power distribution can never have too much redundancy! Hrm, of course, usually we'll split the bus with an open tie breaker when we do....

#### Dark Knight

##### Silent Guardian
I like paralleling transformers, though, for redundancy. Power distribution can never have too much redundancy! Hrm, of course, usually we'll split the bus with an open tie breaker when we do....
This is what we do here. A bus tie breaker to tie the buses in case a transformer failure. In that case the failing transformer is left out. In case we have to put, for some reason, two transformers in parallel we are sure they same impedances. The biggest design I worked, at the the distribution level, was a 3 transformers station. I have done testing jobs when I put two transformers in parallel for a short period of time. The risks are there though.

Way to go Grover.

#### kris7o2

##### Member
I put Z,new = Z,old * (S,new/ S,old)*(V,old/V,new)^2 and since Z,new = .045*(2000/1000) = .09

However, I can't understand the formula Load, T1/ Load, T2 = Z, T2 / Z, T1

Does anyone know where this formula came from?

I also tried to understand Art and Grover's solutions, but I just can't understand.

If anyone can help it would be really appreciated, thanks

And, does any of this apply, below? And, by using this formula, would I use

S1=1000kVA = (1000/.045) / (1000/.045 + 2000/.06) * S, sum
= 1000 kVA = .4 * S, sum
Then S,sum = 1000 kVA /.4 = 2500 kVA

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#### akyip

##### Member
When it comes to problems with 2 parallel transformers, I like to use Zach Stone's limiting transformers method from Electrical PE Review. I included my work for this problem based on Zach's limiting transformers method.

The limiting transformer is the transformer that is loaded to its rated KVA value, from which the other parallel transformer cannot be simultaneously loaded to its rated KVA value because it would then be overloaded. You can determine which transformer is the limiting transformer and from there the max KVA value allowed from both parallel transformers without overloading either transformer.

Checking if transformer 1 is limiting:

S1 = S1 rated * (Z2 pu / Z1pu)
If S1 > S1 rated, then transformer 1 is the limiting transformer
With transformer 1 being limiting: S max = S1 rated + S2 rated (Z1 pu / Z2 pu)

Checking if transformer 2 is limiting:

S2 = S2 rated * (Z1 pu / Z2 pu)
If S2 > S2 rated, then transformer 2 is the limiting transformer
With transformer 2 being limiting: S max = S2 rated + S1 rated (Z2 pu / Z1 pu)

So problem 125 from the NCEES practice exam, transformer 1 (1000 KVA, Z = 4.5%) is the limiting transformer because:

S1 rated = 1000 KVA
S1 = S1 rated * (Z2 pu / Z1 pu) = 1000 KVA * (6.0%/4.5%) = 1333.333 KVA > S1 rated

Then the max KVA allowed from both parallel transformers without overloading either transformer becomes:

S max = S1 rated + S2 rated (Z1 pu / Z2 pu)
S max = 1000 KVA + (2000 KVA)(4.5%/6.0%) = 2500 KVA

(When I first studied for the PE Power exam, I used the method from the A.S. Graffeo book and NCEES practice exam solutions. But I found this method to be harder to follow and understand. In contrast, I have a much easier time using and understanding Electrical PE Review's limiting transformers method to deal with problems involving 2 parallel transformers.)

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• Zach Stone P.E.

#### kris7o2

##### Member
Thanks for the help, and I understand your solution, but I am uncertain when I am supposed to use the other formula... or when do I use that kind of logic to solve a problem? I hope I am making sense, please let me know.

#### akyip

##### Member
Thanks for the help, and I understand your solution, but I am uncertain when I am supposed to use the other formula... or when do I use that kind of logic to solve a problem? I hope I am making sense, please let me know.
You mean the other formulas you posted earlier from Engineering Pro Guides?

IMO, I think you use these formulas for when you have the total load KVA being supplied by both transformers, the rated KVA of each transformer, and the p.u. impedance of each transformer. This is when you are trying to find how much KVA each parallel transformer contributes to the total load.

I have seen this being used in some parallel transformer problems from the Complex Imaginary practice exams set.

#### Zach Stone P.E.

##### Learn how to Pass the Power PE Exam at electricalp
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When it comes to problems with 2 parallel transformers, I like to use Zach Stone's limiting transformers method from Electrical PE Review. I included my work for this problem based on Zach's limiting transformers method.

Thanks for the mention!

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