transformer calculations and sq mil versus cir mil formu

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champion

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Assistance requested to clerify transformer calculations (open delta) and square mils versus circular mils.
I am reviewing some books of transformer calculations and formula's. In my review, I have discovered that the author may have made a mistake. I would like some assistance in reassuring myself that I may be correct in my assumptions.
********************************************** 1st topic, transformer calculations----
Two question are listed within this book in regard to open delta kva ratings.
1st question-------If you were operating three 25kva 1ph transformers in a delta connected 3ph system and removed one transformer, what is the kva capacity of the two remaing transformers connected delta three-phase?
The answer given in the book indicates 75kva x .577 = 43.275kva.
2nd question-------Three 10kva transformers are delta connected, if one phase opens up, what will be the kva rating of the open delta?
The answer given in the same book indicates 10kva x .577 = 5.77kva.

Open winding or removed winding, would not the electrical impact of the kva remain the same.
Do you agree that the answer to question 2 above should really by 30kva x .577 = 17.3kva.--or is there something I'am not grasping here?
***********************************************
Square mils vs circular mils-----
Book indicates the following:
1 circular mil = .7854 square mils
1 square mil = 1.2732 circular mils
circular mils = square mils x 1.2732
square mils = circular mils x .7854

It seems to me that, if one circular mil = .7854 square mils then circular mil would = square mils x .7854 not 1.2732.
Am I correct in assuming the author may have made an error?
Confused?
***********************************************

[ August 04, 2003, 02:32 PM: Message edited by: champion ]
 
Re: transformer calculations and sq mil versus cir mil formu

it looks as if the 10kva question is a in error and was caught when proofed which would be confusing to the student.
 
Re: transformer calculations and sq mil versus cir mil formu

Champ
It appears that you are correct regarding the transformers. While the two transformers provide
66.6% capacity of the origional bank( 3-10 kva)
they can only provide 57.7 % when one transformer is out of service. Therefore 57.7% x 30 kva = 17.4

Regarding the circuclar mill question
If 1 circular mil = .7854 square mils then
1 square mil = 1 circular mils/ 0.7854 or
1 square mil = 1.2732 circular mills
 
Re: transformer calculations and sq mil versus cir mil formu

Could someone guide me to the first principle equations, how to arrive at the kVA capacity of an open delta/wye system, compared to a closed one? In other words, how to get there, starting at the very bottom?

Thanks
 
Re: transformer calculations and sq mil versus cir mil formu

Could someone guide me to the first principle equations, how to arrive at the kVA capacity of an open delta/wye system, compared to a closed one? In other words, how to get there, starting at the very bottom?
Click to expand...
Vaguely remembering open delta from trade school, I thought that you would simply find the Kva of three transformers as a closed Delta and then multiply by 57% if only two transformers were operating as an open delta. Am I close?
 
Re: transformer calculations and sq mil versus cir mil formu

Yes, open-delta bank would be 57.7% of the capacity of a closed-delta bank.
57.7% is the reciprocal of the square root of 3.
 
Re: transformer calculations and sq mil versus cir mil formu

Here is a quick calculation using the example of the three 25 kVA transformers. The capacity of the closed delta connected bank is 75kVA.

In an open delta connection, the phase current is limited to the full load current of one transformer, since the two transformers are the only source of current for two of the three phases.

25*1000/240 = 104.167 amperes (FLA)

104.167*240*1.732/1000 = 43.3 kVA

43.3/75*100 = 57.7%

MKB
 
Re: transformer calculations and sq mil versus cir mil formu

Could someone guide me to the first principle equations, how to arrive at the kVA capacity of an open delta/wye system, compared to a closed one? In other words, how to get there, starting at the very bottom?
Click to expand...
Yes, the load capacity of an open-delta bank is indeed 57.7% of the capacity of a closed-delta bank of the same transformers.
And 57.7% is indeed the reciprocal of the square root of 3.

But, the reason that the the load capacity of an open-delta bank is reduced to 57.7%, and not 66.6%, as one would expect, is a function of the current-carrying capacity of the individual transformer winding.

In the example below, each transformer is rated 4800 va, with secondary winding current rating of 10 amps.

When converted to open-delta, the 10 amp secondary winding rating cannot be exceeded, which limits the new maximum load to 8314 va.

Ed
P.S. MKB's example is correct too, and he types faster than me. :D

Trans13.gif


[ June 16, 2005, 09:49 PM: Message edited by: Ed MacLaren ]
 
Re: transformer calculations and sq mil versus cir mil formu

Referring to Ed's diagram, let me add that there is significant phase shift in the windings. This is necessary in order for 10A + 10A = 10A.
 
Re: transformer calculations and sq mil versus cir mil formu

Can someone explain that phase shift a little more to answer how the top line in drawing B remains 10 A?
 
Re: transformer calculations and sq mil versus cir mil formu

lauraj
The load does not remain 10 amps. The load was 17.3 amp. Look at Eds dwg again. He notes that the
"New Maximum Load Capacity is 8413 va". The load has been reduced because of the loss of the one transformer.
 
Re: transformer calculations and sq mil versus cir mil formu

I was refering to Rattus's comment 10A+10A = 10A. I assumed he is talking about I(C) + I(A) = I(B). It is I(B) I'm asking about.
 
Re: transformer calculations and sq mil versus cir mil formu

Lauraj

This is difficult to explain without diagrams, but here goes:

First assume balanced resistive loads.

Now let the voltage on the open side be V@0 deg and let the other voltages be V@120 deg and V@240 deg. The load currents will also be a 0, 120, and 240 degs.

The open side current must be supplied by the remaining windings, and when we compute the phase currents vectorially we obtain

IphaseA = 1.732*Iload@30
IphaseC + 1.732*Iload@150

IlineAC= 1.732*Iload*2*cos(60) = 1.732*Iload@90

The key is the cos(60) factor.
 
Re: transformer calculations and sq mil versus cir mil formu

Three 25 kva transformers connected in a Delta configuration provides 75 kva of capacity. Removing one of the transformers reduces the capacity to .577 x 75kva = 43.27kva. OR 75kva divided by 1.73 = 43.27kva.

Three 10kva transformers connected in Delta configuration provides 30kva of capacity. The question asks what the result is when you loose one phase of the supply, like a blown fuse on one leg. The result is a reduction in capacity of .577 x 30kva = 17.3kva.
 
Re: transformer calculations and sq mil versus cir mil formu

rattus,

Are you sure you're not confusing the results of your 10A + 10A = 10A with the calculation (120 degree vector subtraction)for the neutral current of a 4-wire, wye-connected system?

The combining of transformer currents in a closed delta-connected system is 120 degree vector addition. 10A + 10A = 17.3A

MKB
 
Re: transformer calculations and sq mil versus cir mil formu

MKB,

No, I am talking about a 3-wire open delta with a balanced resistive load. The line and phase currents are equal in magnitude as indicated in Ed's diagram.

In order for this to be true, Ia and Ic are defined as flowing into the junction of windings A and C, and the phase separation of Ia and Ic must be 120 degrees.

For example,

Iline = 10@150 + 10@30
= [10*cos(60) + 10*cos(60)]@90
= 10@90

In the full delta, the separation would be 60 deg., and the magnitude of I line would be 17.3.

You have to go off in a corner and draw about 50 loop current diagrams and vector diagrams before this makes any sense. It is near impossible to do this with standard formulas.
 
Re: transformer calculations and sq mil versus cir mil formu

John,

I think we all accept the 0.577 factor

The question now is,

How do the two 10A phase currents combine to yield a 10A line current?

I hope I have answered that in my previous posts.

Rattus
 
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