Calculation of Unbalanced Transformer Loads 1

[7JLAman4]

I'm attempting to calculate the Line currents (with no sucess) of a 120V Delta transformer secondary with 3,4 & 5 kVA on phase windings A, B & C respectively with a phase sequence ABC. I originally started out with a 15kVA 3Ph load (5kVA per phase) and knowing how to calculate for a balanced system, tried to achieve the currents using nodal analysis which I was then going to apply to the problem above. I can't seem to calculate the line currents from the phase currents using vectors. I had started out with:
Ia=3kVA/120V=25A@0deg, Ib=4kVA/120V=33.3A@-120deg, Ic=5kVA/120V=41.7A@120deg. IA=Ia-Ic, IB=Ib-Ia, IC=Ic-Ib. IA=(25cos(0)+i25sin(0)) - (41.7cos(120)+i41.7sin(120))
I'm not sure if this is the correct method. However using a sample that I know what the outcome should be, Ican't seem to get near close enough to the answer.

 

[jghrist]

Your method seems correct assuming that Ia is the current in winding A, Ib is the current in winding B, etc.; the line currents are IA, IB, and IC; and the transformer connection is D_AC (top of winding A connected to bottom of winding C).

If the connection were D_AB (top of winding A connected to bottom of winding B), then IA=Ia-Ib, IB=Ib-Ic, and IC=Ic-Ia.

 

[cflatters]

Sounds like a college exercise to me !

 

[7JLAman4]

College didn't explain unbalanced systems. Work is driving this goal of mine to better understand certain theories and formula shortcuts. When using the aboe calculations on a "balanced" system, Line current IC always somehow comes out much higher than line currents IA or IB.

 

[jghrist]

I get IA=58.3A, IB=50.6A, IC=65.0A using your equations for a D_AC winding. IC doesn't seem inordinately high considering the unbalance in the winding kVA.

Using 5 kVA in each phase gives 72.1A in each line, the same as winding current time sqrt(3).

 

[stevenal]

Why are you assuming currents are evenly displaced by 120 degrees?

 

[7JLAman4]

If only given kVA of the loads connected to the transformer, how would the phase angles be calculated then? and is it necessary?

 

[stevenal]

You can't. Necessary? Not if you have a meter.

 

[waross]

Further to stevenal's comments:
If the power factors of the different loads are different, you add another level of complexity to the problem.
I suggest that you draw some vectors to scale and double check your calculations.
Another suggestions to help you.
On a balanced system:
50 amps phase current will result in 50 x 1.73 = 86.5 amps line amps.
65 amps phase current will give 65 x 1.73 = 112.4 amps.
When you combine 50 amps on one phase with 65 amps on another phase, the result will be more than 86.5 amps and less than 112.4 amps.
I suggest that you use scale vector drawings and the high and low limits to self check your calculations, until you develop confidence in your solutions.
respectfully

 

[7JLAman4]

Thankyou for the replies.

 

[stevenal]

Even without differing pf loads, this is not an easy problem. A balanced delta can be converted to an equivalent wye, and solved on a per phase basis. Unbalanced systems are generally handled by breaking the problem down to its balanced sequence components, each of which can be solved on a per phase basis and recombined.

Note that on a delta system, all the line currents must add to zero since there is no neutral return path. Even if the loads are 100% resistive, unequal magnitude line currents cannot be evenly displaced by 120 degrees and still sum to zero.

If this is really necessary for work, I suggest you look for a course on sequence components.

 

[jghrist]

Solving for the line currents with the winding currents known is an easy problem. It is just nodal analysis with two currents known going into each node. The equations are as jlamann gave in the original post.

The more difficult problem is solving for winding currents with the line currents known. In this case, you know one current going into each node and need to find the other two.

 

[stevenal]

Solving for the line currents with the winding currents known is an easy problem.

Agreed, but in this case all that is known is the KVA per winding. From this jlamann got a set of winding currents that don't close the delta graphically. The equations are good, but the inputs into them don't work. GIGO.

 

[stevenal]

Another thought, though. The winding current magnitudes calculated above can only describe one triangle. You should be able to use the laws of cosines and sines to find the relative angles of the three phasors. Then use the formulas to find the line currents. Pf and sequence components avoided.

 

[jghrist]

The winding currents do not have to add to zero. The line currents do. In the original example:

IA = Ia - Ic = 58.33@-38.21°
IB = Ib - Ia = 50.69@-145.28°
IC = Ic - Ib = 65.09@93.67°

IA + IB + IC = 0

 

[tinfoil]

Further to jghrist:

The line currencts must sum to zero

The nodal currents (i.e. the sum of currents at each corner of the delta) must sum to zero

The winding currents are a function only of the load. Imagine a delta service with only single phase load on one leg. Assuming ideal transformers, only one winding will have non-zero current.

 

[jghrist]

Or consider a delta-wye transformer with a single phase-to-ground fault on the wye secondary and no load on the unfaulted phases. There will be current in only one of the windings (primary and secondary). Obviously, with current in only one winding, the sum of the winding currents will not add to zero.

The problem of calculating winding currents from the line currents arises because the set of equations

IA = Ia - Ic
IB = Ib - Ia
IC = Ic - Ib

cannot be solved for winding currents Ia, Ib, or Ic. If a constant is added to each winding current, the equations still hold. The constant adder (magnitude and phase) might be recognized as a zero-sequence current circulating in the delta.

 

[waross]

Hi tinfoil;
I don't undestand what you mean by an ideal transformer.

Imagine a delta service with only single phase load on one leg. Assuming ideal transformers, only one winding will have non-zero current.

What about two transformers in parallel with a single phase load. Won't ideal transformers share the load?
When a single phase load is applied to "A" phase of a delta transformer bank, "B" phase and "C" act as an open delta transformer. The open delta may be resolved to a single phase transformer in parallel with "A" phase. The "B" phase and "C" phase transformers share 50% of the load.
respectfully

 

[jghrist]

Let's say you have a wye-delta transformer with A phase of the primary open. You could still serve load between the 'a' and 'b' terminals (ends of the delta A winding). There would be zero current in the A winding and the currents in the B and C windings would be equal. This current would flow out of the 'b' terminal (top of B winding, bottom of A winding), through the load, and back into the 'a' terminal, which is connected to the bottom of the C winding. This would be equivalent to an open-wye, open-delta connection.

Then again, if B and C of the primary were open, the currents in the delta B and C windings would be zero and all of the load current would flow in the A winding. This would be equivalent to a single-phase transformer.

It just shows that if the delta winding currents are known, you can calculate the line currents, but not vice versa.

 

[stevenal]

I agree that in the general case, winding currents do not necessarily sum to zero. However in this case, with a delta secondary feeding the load, winding currents will sum to zero. There is no source given for the circulating zero sequence current.

For jghrists example to occur; there must be a source on the delta side and a wye winding with unbalanced loading or fault. Possibly an unbalanced source voltage on the wye side could also create circulating current, but the problem assumed the transformed voltage was 120V across each winding.

Tinfoil,

You cannot have a single phase load on a delta that involves only one leg unless you are speaking of that 4 wire delta that center taps one of the windings. Given only the three nodes of the original post, I assume this is a 3 wire delta. Single phase loads can only be connected l-l and therefore involve all three windings. Current consists of + and - sequence only. Winding current sums to zero to give no 0 sequence current.

With eng-tips MVPs challenging me on this, I've double checked my memory using Aspen. Grounded wye source with l to l fault on the delta side. No zero sequence current on either side. Memory confirmed.