SINGLE-PHASE LOAD ON A 3-PHASE DELTA SECONDARY TRAFO

[ver]

Please tell me where I got wrong with my math below:

P(3-phase) = sqrt(3) X Vrated X Irated                                 ---> (1)
P(1-phase) = Vload X Iload                                                   ---> (2)
In this case, the rated phase current will limit the capacity:
Iphase = 2/3 X Iload                                                             ---> (3)

But we know:
Iphase = Irated/sqrt(3)                                                        ---> (4)

Equating (3) and (4):

2/3 X Iload  =  Irated/sqrt(3); or:
Iload  = sqrt(3) X Irated/2                                                    ---> (5) or 0.866 Irated

Substituting (5) in (2):
P(1-phase)  = Vload X sqrt(3) X Irated/2 or if Vload = Vrated then:

P(1-phase)  = sqrt(3)/2 X (Vrated X Irated)                          ---> (6)

Then, P(1-phase)/P(3-phase):

Code: [Select]

                   sqrt(3)/2 X Vrated X Irated
          =      ---------------------------------------------           = 1/2   = 0.50  = 50%
                   sqrt(3)   X Vrated X I rated
For your info, this question has been hanging in my mind for sometime and I hope to find closure on this bugging question. Hope you could help me on this.
Respectfully.

Truly you are right. This problem should be considered as a 1-line open fault. Not as a normal system as I have previously considered. My bad .

This makes the 3 transformers acting as single phase, two of which are in series and in parallel with the other phase. That was where your equation (3) came about.

Let's say the transformers are A, B, C where B & C are in series then in parallel with A.

Transformer impedance, assuming they are identical, will distribute transformer currents according to where they are located into the circuit. If transformer A will carry 2/3 of the load current while the other two will on have 1/3.

That leads to a maximum load size equal to  50% x transformer rating.

[Parchie]

. . . . . .
or

VA_{transformer per phase} = VA_load / 1.732 = 0.577 x VA_load

Thank you for responding. That figure is what I got from one colleague also! I must admit I'm a little bit puzzled.
I was also trying to relate that same percentage which came out from a transformer configuration commonly known as the V-configuration (Open-delta)! I was wondering why the two arrangements yield the same percentage when in this case you have the three phases intact (not just two! Besides, on the V-configuration, the capacity of 58% is supposed to be the total 3-phase load of the trafo and not a single-block, single-phase load!

I happened to look at a copy of the J&P transformer handbook (p.633), it says:"the loaded phase carries two-thirds of the total current, while the remainder flows through the other two phases, which are in series with each other and in parallel with the loaded phase."
If this will be the case, that will make the maximum load current to be equal to 0.866 of the rated trafo current, and the maximum single-phase, single-block load will be just 50% of the 3-phase capacity (very low)!

Please tell me where I got wrong with my math below:

P(3-phase) = sqrt(3) X Vrated X Irated                                 ---> (1)
P(1-phase) = Vload X Iload                                                   ---> (2)
In this case, the rated phase current will limit the capacity:
Iphase = 2/3 X Iload                                                             ---> (3)

But we know:
Iphase = Irated/sqrt(3)                                                        ---> (4)

Equating (3) and (4):

2/3 X Iload  =  Irated/sqrt(3); or:
Iload  = sqrt(3) X Irated/2                                                    ---> (5) or 0.866 Irated

Substituting (5) in (2):
P(1-phase)  = Vload X sqrt(3) X Irated/2 or if Vload = Vrated then:

P(1-phase)  = sqrt(3)/2 X (Vrated X Irated)                          ---> (6)

Then, P(1-phase)/P(3-phase):

Code: [Select]

                   sqrt(3)/2 X Vrated X Irated
          =      ---------------------------------------------           = 1/2   = 0.50  = 50%
                   sqrt(3)   X Vrated X I rated
For your info, this question has been hanging in my mind for sometime and I hope to find closure on this bugging question. Hope you could help me on this.
Respectfully.

[ver]

What is the biggest single-block, 1-phase load can a 3-phase, delta-secondary transformer  carry safely without overloading?
Any answer accepted, but please explain.

This could be explained mathematically.

For single phase load:

VA_load = I_line x V_phase -->(1)

Expressing Iline in terms of the secondary current of the transformer:
I_line = 1.732 x I_phase  -->(2)

Substituting (2) in (1):

VA_load = 1.732 x I_phase x V_phase -->(3)

On a per phase basis at the transformer:
I_phase x V_phase = VA_{transformer per phase} -->(4)

Substituting (4) in (3):

VA_load = 1.732 x VA_{transformer per phase}

or

VA_{transformer per phase} = VA_load / 1.732 = 0.577 x VA_load

[Parchie]

I hate to ask this question that looks like a simple one, but I got different answers from friends. Here it goes:

What is the biggest single-block, 1-phase load can a 3-phase, delta-secondary transformer  carry safely without overloading?
Any answer accepted, but please explain.