Closed Delta Transformer Bank Analysis

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fgreco2

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Hi,

I?m looking for some help/advice here. I work for an electric utility company in Arizona. I recently pulled some actual loads from a closed delta OH transformer bank. The transformer sizes are 50-100-50. I had a lineman perform a snap test for me to measure the amps on the secondary side. I took the high amps on each leg to calculate the load (KVA) on each transformer. I am having difficulty determining what portion of this load is 3-phase and what portion is 1-phase. The lighter transformer (commonly called pot in the utility industry) is the 100kVA transformer, and the power pots (2) are the 50kVAs. I used P=IE on each transformer using the AMPS from the snap test to see what the actual load is seen by each transformer (individually). For T1 (50kVA) I get 19kVA (by multiplying 240V by 78A); for T2 (100kVA) I get 40kVA (by multiplying 240V by 167A); for T3 (50kVA) I get 28kVA (by multiplying 240V by 117A).

My overall goal is to determine what portions of the loads seen by the transformer bank is single phase and what portion is three phase (where my difficulty is). With a Delta system (secondary side) the lighting transformers must be sized at 67% of the single phase load, and 33% of the three phase load (seen by T1 and T3). The power transformer is sized at 33% of the single phase load and 33% of the three phase load (see by T2). I?m trying to see if I can convert this bank to an OPEN DELTA system (one power transformer and one lighter transformer), but I?m not so sure what I need to do with the data collected to verify this. Please e-mail me your response at frank.greco@aps.com.

Thanks a million for the help,

Frank Greco
 
For an open delta-delta connection I use:
Lighter kVA = .58(T)+S
Power kVA = .58(T)

For a wye-delta connection I use:
Lighter kVA = (T+2.5(S))/3 = .33(T)+.83(S) or for simplification .33(T)+S
Power kVA = T/3 = .33(T)

where T = balanced three phase load and S = balanced single phase load.
 
Looking at the diagram, how can A1 not equal B1, and B2 not equal C2?

Presuming T2 is the center-tapped 100kva transformer, I can understand A2 not equalling C1.
 
Larry,

I agree .... there is quite an imbalance here. These are the results we obtained from the snap test.

Thanks,

Frank Greco
 
Jim,

Sounds good! So to solve for T I can divide 40kVA (actual load on the power pot) by .33, then the result (121kVA) I can use to solve for S:

S = 47kVA (lighter pot actual loads) - (.33 * 121kVA) Is this the correct approach????

Thanks,

Frank Greco
 
fgreco2 said:
Jim,

Sounds good! So to solve for T I can divide 40kVA (actual load on the power pot) by .33, then the result (121kVA) I can use to solve for S:

S = 47kVA (lighter pot actual loads) - (.33 * 121kVA) Is this the correct approach????
Click to expand...

If you assume that your balanced three phase loading is 19kVA per transformer, then your formula is S= 40-19 = 21kVA, but your worst case is actually T=28*3=84kVA.

So in your situation I would size an open delta as:
Lighter = .58(84)+21 = 70kVa
Power = .58(84) = 49kVA
 
Last edited:
080423-139 EST USA

There is a disparity in labeling if each row in the table corresponds to the transformer in the row.

Assuming T2 is the center tapped single phase transformer, then A2 and C1 could have unbalanced readings since these are secondary readings. However, A1 must equal B1 and B2 must also equal C2. A1 and B1 are close enough to be called equal depending upon instrumentation and how the load might change between the readings.

.
 
Confirm whatever analysis you use by measuring the V-I phase angle for each load, unless they really are pure resistances.
 
080423-2011 EST

Assuming T2 is the center tapped transformer, and all transformers have no internal shorts, then no matter what current measuring instrument is used (oscilloscope, peak, RMS, average, or phase sensitive) so long as the same meter in the same mode is used to measure A1 or B1 the readings must be the same. Also B2 must equal C2.

Only for T2 might you want to know the phase angles, and also the neutral current.

.
 
Charlie, one thing about that doc: the top line in the second example data box, the single-phase load. The figures add to 40kw instead of the 30kw it seems they should.

Maybe the 1/3, the 10kw, should have been split between A-B and B-C, and have been 5kw each. Anyone?
 
LarryFine said:
. . . the top line in the second example data box, the single-phase load. The figures add to 40 kW instead of the 30 kW it seems they should.
Click to expand...
I know, it is counter intuitive but it is correct. Perhaps one of our engineers would like to opine since I don't know enough about the theory of putting a single phase load on a closed delta bank. :smile:
 
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