Help needed on Electrical PE Sample Question...

[rr]

Hey ya'll.

I'm taking my Electrical PE in October and I need some assistance on a sample question that I'm trying to solve.

I don't have the question in front of me, but this is what I remember from memory:

Two transformers are in parallel. What is the maximum load (kVA) that the transformer bank can withstand?

Transformer A: 1000kVA 4% Impedance
Transformer B: 2000kVA 6% Impedance

I've never come across this before in a design. So any insight from the wise ones on this board would be appreciated.

Thanks in advance!

 

[sceepe]

Re: Help needed on Electrical PE Sample Question...

Come on now, aren't you going to give us the choices????

Here is idea. Represent the transformer as a circuit element. For distribution xfrm (over 500 kva) you can represented as a reactance only. got this from EERM section 36.9 Now you have two reactances in parallel.
1. Use given info to calc the reactance.
2. Simplify the circuit from there to get a total reactance
3. Do the opposite of what you did in step one to get an equivalent single transformer with a KVA rating and an inpedance.
4. See if this is one of the choices. If not ignore all that I have said..... My free advice is worth what you paid for it.

 

[bob]

Re: Help needed on Electrical PE Sample Question...

The total load sharing is inversley proportional
to the impedance. With the impedance you have given the 1000 kva will share 4%/6% = 0.67 of the total. Therefore the max load on both transformers must be such that load should not exceed 1000 kva on the smaller transformer.
1500 x .67 = 1005 on the 1000 kva and 1500 x .33 = 495 on the 2000 kva

 

[rr]

Re: Help needed on Electrical PE Sample Question...

Originally posted by rr:

Two transformers are in parallel. What is the maximum load (kVA) that the transformer bank can withstand?

Transformer A: 1000kVA 4% Impedance
Transformer B: 2000kVA 6% Impedance

The Impedance in Transformer A is 4.5% NOT 4%. The answer is 2500 kVA.

But I'm not sure how to work the problem in order to get this answer. Could someone lead me in the right direction?

Thanks in advance!

 

[steve66]

Re: Help needed on Electrical PE Sample Question...

For the origonal impedences of 4% and 6%, I think Bob meant the 4% transformer will get 4/(4+6) or 4/10 or 0.4 of the load while the larger xformer gets 0.6 of the load.

So the total load would be calculated as the smaller of 1000KVA/0.4 or 2000KVA/0.6. The smaller answer would be 1000KVA/0.4 = 2500KVA.

But you said the answer is 2500KVA for a impedence of 4.5% on the smaller transformer. That would give an answer of 2222KVA So I'm confused.

Steve

 

[rattus]

Re: Help needed on Electrical PE Sample Question...

Can anyone explain the relationship between these parameters and the parameters of the "exact" transformer equivalent circuit? Someone like rbalex maybe?

 

[jsteed]

Re: Help needed on Electrical PE Sample Question...

I had to remember things that I haven't used since college but I think I've got it now.

The impedance values are "per unit" values and expressed as a percentage. The percentage is based on the kVA rating of the transformer. So the first step is to get the impedances on a common power base. You can choose either the 1000 kVA or the 2000 kVA as the base. If you choose the 2000 kVA then the impecances become 9% for TA and 6% for TB, using 1000 kVA as the common base the impedances would be 4.5% for TA and 3% for TB. Doesn't matter as long as the numbers have the same reference.

The voltages involved are the same for both transformers, so the division of power will be proportional to the division of current. The 1000 kVA will see 6/(9+6) or 6/15 of the power.

Ptotal = Pta + Ptb = Ptotal*(6/15) + Ptotal*(9/15)
Assume the 1000 kVA will reach its rating before the 2000 kVA unit would give
Pta = Ptotal*(6/15) or
Ptotal = Pta/(6/15) = 1000 /(6/15)
Ptotal = 2500 kVA
Ptb = 2500 * (9/15) = 1500 kVA

jim

[ August 11, 2005, 12:51 PM: Message edited by: jsteed ]

 

[rattus]

Re: Help needed on Electrical PE Sample Question...

Just found in my ancient text that per unit impedance is defined as:

1) Ir x Z/Vr

where Ir is the rated current, Z is the effective impedance, and Vr is the rated voltage.

Multiply by 100 to obtain percent impedance, then for Ta,

2) 100 x Ira x Za/Vra = 4.5

for Tb,

3) 100 x Irb x Zb/Vrb = 6.0

Since Vra = Vrb,

4) 100 x 2Ira x Zb/Vra = 6.0
5) 100 x Ira x Zb/Vra = 3.0

Dividing eqn 2 by eqn 5 we obtain,

6) Za/Zb = 4.5/3 = 1.5, and

7) Ib/Ia = 1.5

This does not change the result, but perhaps it provides a deeper understanding of the mechanisims involved.

[ August 11, 2005, 10:20 PM: Message edited by: rattus ]

 

[bob]

Re: Help needed on Electrical PE Sample Question...

Can anyone explain the relationship between these parameters and the parameters of the "exact" transformer equivalent circuit?

Rattus
I do not think these values can be related to the values in the transformer equivalent circuit
because of the way they are obtained.
The %Z of a transformer is obtained in the following manner:
The mfg short circuits the transformer secondary.
Then a variable voltage is applied to the primary terminals. The voltage is increased until the secondary amps equal the transformers
full load amps. The voltage is measured and indicated as a percent of the primary voltage.
As an example, if the applied voltage is say 750 volts and the normal primary voltage is 12460 volts the the percent is 750/12460 = 0.06 or 6%.
I think the value is only used when caculation
fault current.

 

[rattus]

Re: Help needed on Electrical PE Sample Question...

Bob, that is an interesting bit of info on the testing, and it is quite logical. If we force maximum current into the primary with a shorted secondary, the current and voltage are related by the sum of primary impedance and reflected secondary impedance. The current however is not greater than the rated maximum, so we are not testing fault current.

My ancient text also has problems similar to the one posted in this thread which tells me that percent impedance is used to compute regulation.