Open Delta Calculations

[abdenham]

Please help me understand how on a 240v open delta system how the voltage between phase A on the open end of one transformer 1 and phase B on the open end of transformer 2 is 240v. I have studied the wiring diagrams and the vector diagrams but I do not uderstand the mathmatics behind how this works after it is between the coils of two transformers instead of one.

Thank You.

 

[mivey]

Please help me understand how on a 240v open delta system how the voltage between phase A on the open end of one transformer 1 and phase B on the open end of transformer 2 is 240v. I have studied the wiring diagrams and the vector diagrams but I do not uderstand the mathmatics behind how this works after it is between the coils of two transformers instead of one.

Thank You.

If the phase angle of the voltage on the first coil is "fixed" relative to a point, and the phase angle of the voltage on the second coil is "fixed" relative to the same point, then the relationship is fixed.

Fold a piece of paper into a triangle. Pick any two sides to define the coils for transformer 1 & 2. No matter how you orient the paper, the third side is fixed relative to the other two.

 

[Smart $]

Welcome to the forum, abdenham.

Did mivey's explanation satisfy your knowledge quest, or do you require a more in depth explanation?

 

[mivey]

Yes, welcome. I did not notice this was your first post.

As for the math:

Let's look at a one-dimensional example (x-axis). Given that there are three numbers that add up to 10. One of the numbers is 2. Another of the numbers is 5. The unknown number is constrained to be 3.

The same thing happens with the delta voltages except we now have numbers that use the x & Y axis. We have a 3-sided triangle that we know must close and is located in a 2-dimensional plane. Given any two sides, the third side is constrained.

For vectors we have:
Side one with endpoints (x1, y1) & (x2, y2).
Side two with endpoints (x2, y2) & (x3, y3).
Side three must have the endpoints (x3, y3) & (x1, y1).

For phasors and voltage A@a?, B@b?, C@c? we have:
A@a? + B@b? + C@c? = 0
which gives us one equation and one unknown so we can solve for the missing voltage.

The key is that the delta is a closed triangle. There is no missing voltage, i.e. no smoking gun. If the voltages don't add to zero, you have another path somewhere that is not being accounted for.

 

[LarryFine]

Abby, welcome to the forum!

The reason there is 240v across the open side of the Delta is the same reason that (theoretcially) no current flows through the secondaries with a full Delta when unloaded.

If you were to add a third secondary to an open Delta, its 240v matches the 240v between the two phases it connects to. Anything else would cause elevated no-load current flow.

By the same mechanism, if you opened one corner of a full Delta, there would be (again, theoretically) no voltage between the two secondary ends that make up that corner.

 

[Marc L]

One way to visualize it is to recognize that at any two of the three lines there is the equivalent of a parallel connection: One secondary in parallel with the series connection consisting of the other two secondaries. If you try it, you'll see that regardless of which two lines you choose to get your single phase 240 volts, you'll always get this parallel circuit. This means that you can eliminate either parallel connection and your voltage will remain the same. You can eliminate either the single secondary or the other two secondaries in series, voltage remains the same.

Not a kosher explanation for an engineering classroom, but it helps to get a beginner started on the concept.

Sherlock

 

[Smart $]

Another way to visualize the physics of an open delta secondary is to relate it to the nowadays more common wye secondary. By the latter's very nature we utilize the "open" windings connections to get a usable voltage, commonly 208 or 480 volts with winding voltages of 120 and 277 respectively. The difference is that we have a 120? out-of-phase relationship between the voltage of each winding on a wye secondary. The open-delta configuration has a 60? out-of-phase relationship between windings. Therefrom the question arises as to how we get a 60? out-of-phase relationship when the system has a 120? shift between phases. Easy... we just reverse the polarity of one of the secondary windings (e.g. make the other end the common connection without changing the primary configuration).