As we know, a three phase Transformer bank connected in Delta-Delta when supplies a three phases balanced load then individual Transformers share 1/3 of the total load. But if there is some fault in any one of the Transformer and due this one Transformer is taken out of service, then also three phase power can be supplied though at a reduced power level. The resulting connection obtained after the removal of one of the Transformer from a three phase Transformer bank connected in Delta-Delta is known as an Open Delta or V-connection of Transformer. Figure below depicts this open Delta or V-connection.

*Open Delta or V-connection is missing then how can we deliver three phase power?*Since only two phase voltages are available i.e. V_{ac} and V_{bc} so we need to calculate the remaining voltage V_{ab} and then only we can judge whether we can supply three phase power or not. Note that for supplying three phase power, the three voltages i.e. V_{ca}, V_{bc} and V_{ab} must be balanced in the sense that all of them are equal in magnitude and 120° displaced mutually in time domain.

Thus to find, V_{ab}we need to draw voltage triangle first.

From the above voltage triangle, it can be written that

**V _{ab}**+

**V**+

_{bc}**V**= 0, bold letters mean phasor form.

_{ca}So, voltage across the open Delta terminal can be found as

V_{ab} = Voltage across open delta terminal

= – (**V _{bc}**+

**V**)

_{ca}Here **V _{bc}** and

**V**are equal in magnitude, say V but 120° displaced. Therefore their resultant will be V and 120° apart from the reference voltage i.e.

_{ca}**V**&

_{bc}**V**.

_{ca}Thus we see that Open Delta connection, secondary line voltages form a balanced system and balanced three phase load can be supplied.

Now, we will calculate the amount of VA that can be supplied by an Open Delta connected Transformer.

**VA Delivered by Open Delta:**

**Case1: When all the three transformers of three phase Transformer bank are in service.**

Suppose,

I_{ph} = Phase current of each of Secondary

V_{ph} = Phase voltage of each of Secondary

Therefore,

Line Voltage V_{L} = V_{ph} (because of Delta connection)

Line Current I_{L} = 1.732I_{ph}

Thus,

VA Rating of Bank of three Transformers in Delta

= 1.732V_{L}I_{L}

= 1.732xV_{ph}x1.732xI_{ph}

= 3V_{ph}I_{ph}

**Case2: Open Delta Connection**

As in Open Delta connection, only two Transformers are there in service so,

VA Rating of Open Delta

= 2V_{ph}I_{ph}

*But this is not correct. Why?*

Because as can be seen from the first figure,

Line Voltage in Open Delta V_{L}= V_{ph}

Line Current in Open Delta I_{L}= I_{ph} as there is no path to bifurcate the line current. Same current is flowing in line as well as in phases.

Therefore,

VA Rating of Open Delta

= 1.732xV_{L}xI_{L}

= 1.732 V_{ph}I_{ph}

Thus,

**VA Rating of Open Delta / VA Rating of Close Delta**

= 1.732V_{ph}I_{ph}/ 3V_{ph}I_{ph}

=1 / 1.732

=**0.577**

Thus the VA delivering capacity of Open Delta becomes 57% of that of the full capacity when all the three Transformers are in service. It shall also be noted that, though the total capacity of Transformers in Open Delta is 2V_{ph}I_{ph}but still Open Delta can only deliver 1.732V_{ph}I_{ph}.

**The Ratio of actual available kVA rating to the sum of the kVA rating of installed Transformer is called Utilization Factor and given by**

**U.F = Ratio of actual available kVA / Sum of the kVA rating of installed Transformer**

For Open Delta connection,

U.F = 1.732V_{ph}I_{ph}/ 2V_{ph}I_{ph}

= 0.866

Thus it is beneficial to operate the bank of Transformer in Open Delta at 86% of rated capacity while the faulty Transformer is under maintenance.

If you deliver…A phasor diagram….It would be much better…

If you deliver a phasor diagram ,students will be more helped.

Thank you very much for your suggestion. I will definitely look into it to make topic more easy to understand.

Hi,

I don´t quite get it. If line currents are the same as phase currents, then phase b on the right does not have current at all. No three phase power is achieved then