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
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"Why not the best?"
Jimmy Carter
