Star Delta (YNd11) Transformer - Single-Phase Loading

[Elec81]

Hi,

Please can someone give me some guidance or point me in the direction of some literature relating to single phase loading of a three-phase Star Delta (YNd11) transformer.

My understanding is that when the primary neutral is connected, a sole single phase load connected between two phases of the delta secondary would result in a primary current flowing in one phase and neutral only.

May thanks in advance.

 

[jghrist]

It gets a little tricky to calculate the currents in the primary of a Star Delta transformer from the known currents in the secondary. This is because there can be current circulating in the Delta secondary windings if there is an unbalanced voltage on the primary.

 

[waross]

Assume that there is no voltage drop in the supply conductors and the voltages are balanced at the transformer primary terminals.
Consider a virtual single phase transformer comprised of the open side of an open delta. Under load, the terminal voltage and voltage drop of this virtual transformer will be described by the same vectors as the real transformer that it is replacing.
In a perfect world, under single phase loading, the real transformer and the virtual transformer formed by the open delta will share the load equally.
Of the two transformers considered in open delta that form the virtual transformer, one will have a leading power factor and one will have a lagging power factor.
A further comment in support of jqhrist: When there is a voltage unbalance on the primary so that the voltages of the virtual transformer formed by two transformers considered in open delta and the real transformer that completes the full delta are not equal, there will be a current in the delta driven by the voltage difference and limited by three times the transformer impedance(s).
In the real world this is complicated by the voltage drop on the primary neutral conductor.
Simply stated, when a single phase load is connected to a delta transformer the current in all three phases is equal.
The transformer in phase with the load will supply half of the current, half of the KVAs and half of the KW.
The transformers not in phase with the load will each supply half of the current, KVAs equal to the first transformer and 1/4 of the KW.
From the statement that the KVAs of each transformer are equal and also one half of the load KVAs, it follows that the maximum loading on one phase of a delta transformer or bank is limited to 2/3 of the KVA rating of the transformer or bank.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Elec81]

Thanks for your replies.
However my original post was probably a bit misleading, I should have said that the load was connected between two lines (not phases) of the secondary.

So to summarise, a YNd11 transformer just feeding a sole single-phase load

 

[waross]

That is what my post was based on.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ijl]

You are discussing a situation like the attached picture: An YD transformer with grounded neutral on the primary (Y) side. The primary side is connected to a source with grounded neutral. (In the picture, the source is drawn as three single-phase generators, for clarity). On the secondary (D) side there is a single phase load between (say) phases a and b. (Did I get this right?)

Your question is (isn't it?): Is there current only in the A-N-winding on the primary side, and only on the a-b-winding on the secondary side?

I think that this is not correct. There are currents in all windings. Consider the current coming to point b from the load. What should prevent it from continuing to both windnigs a-b and b-c?

I calculated the case with the following parameters: Phase-phase voltage 1000V, an 1:1 YD transformer, and a 10 Ohm load. The currents on the primary side are given in the second picture

 

[ijl]

Some problems with the uploadig of pictures. Here is a link

http://pp.kpnet.fi/ijl/YDtransfomer.htm

 

[waross]

If you connect a single phase load to A phase of a single phase bank, one half of the load in KW will be supplied by the A phase transformer and one half of the load will be supplied by the B and C phase transformers. The KVA loading will be equal in all three transformers.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jghrist]

Use symmetrical components for a line-to-line fault (b-c) with a fault impedance equal to the load impedance.
I₁ = -I₂
I₀ = 0
I_a = I₁ + I₂
= I₁ - I₁ = 0
I_b = a²·I₁ + a·I₂
= (-0.5-j0.866)·I₁ - (-0.5+j0.866)·I₁
= -j1.732·I₁
I_c = a·I₁ + a²·I₂
= (-0.5+j0.866)·I₁ - (-0.5-j0.866)·I₁
= j1.732·I₁

 

[pwrtran]

one phase is 0.67 pu
the other two phases 0.33 pu

 

[waross]

Consider one half the load connected to the open side of an open delta transformer bank. The KVA in each transformer will be equal to the KVA of the load due to phase shifts.
Calculate the load voltage and the voltage drop vectors across the load.
Consider one half of the load connected to a single phase transformer of the same phase angle as the open side of the delta.
Calculate the load voltage and the voltage drop vectors across the second load.
The vectors should be equal so the single transformer may be paralleled with the open side of the delta. 50% of the load will still be supplied by the two out of phase transformers and 50% of the load will be supplied by the in phase transformer.

Bill
--------------------
"Why not the best?"
Jimmy Carter