How much power can an open delta transformer deliver?
How much power can an open delta deliver compared to a standard three phase delta transformer?
Where does the power formula for the open delta transformer connection come from?
What are the phasor diagrams for an open delta transformer?
What about the current and voltages for an open delta transformer?
What are the advantages of an open delta transformer?
If you’ve ever tried finding this information on the open delta transformer connection in a book, or on the internet, you’ve no doubt come across over simplified diagrams and empty explanations that left you grasping for more information and still feeling uncertain.
To fix that, we made a complete step by step break down analysis on the open delta transformer connection starting with a standard three phase delta connection and finishing with a two-winding open delta connection.
This is by far the most complete and in-depth information you will find on the internet or elsewhere on the subject of open delta transformer connections.
- Three Phase Delta Transformer Connection Diagram
- Three Phase Delta Transformer Phase Voltages
- Three Phase Delta Transformer Phase and Line Current
- Three Phase Delta Maximum Power Formulas
- Three Phase Sample PE Exam Practice Problem
- Open Delta Transformer Connection Diagram
- Open Delta Transformer Phase Voltages
- Open Delta Transformer Line Current
- Open Delta Transformer Phase Current
- Open Delta KCL Relationship For Phase and Line Current
- Open Delta Phase and Line Current Phasor Diagram
- Open Delta Maximum Power Formula
- Open Delta Max Power vs Standard 3ø Delta Max Power
- Three Open Delta Sample Exam PE Practice Problem Examples
Let’s start by looking at the standard three phase delta connection.
Below is a diagram of three single phase transformers connected in delta to form a three phase delta transformer. For the sake of this article let’s assume we are working with a balanced, positive ABC sequence system.
For our phasor diagrams, let’s use a reference of zero degrees for the A phase line to neutral voltage of the system, meaning that the A-phase voltage of the delta, which is equal to the A line voltage of the system, will be at a reference of 30 degrees.
With the chosen reference angle and a balanced, positive sequence system we can determine the delta phase voltages:
Here is the resulting phasor diagram for the delta phase voltages:
Next, let’s assign the three phase delta phase and line currents that will occur during maximum power conditions.
Maximum power conditions for a transformer exist when the power factor is unity.
When this is true, the delta phase current will be exactly in phase with the delta phase voltage, and the active power output in watts (P) will be equal to the apparent power volt-amp rating (S) of the transformer:
Since we know that each delta phase current is now in phase with each delta phase voltage, the phase currents will have the same angle as their respective phase voltage. This lets us determine the complex value of each delta phase current.
From there, we can determine each system line current since by definition they are larger by a factor of √3 and lag by 30º compared to their respective delta phase current:
Notice each phase current has the same angle as the previously determined delta phase voltage since power factor is equal to unity.
Here is the resulting delta phase current and system line current phasor diagram:
If we graph the delta phase current and delta phase voltage on the same phasor diagram we can verify that power factor is unity since the angles are equal for each pair of phase current and phase voltage.
Recall that this is the correct relationship that will result in maximum power output for the delta transformer:
Now, let’s solve for the total three phase power of the delta transformer during these conditions by adding up the power delivered by each of the three windings.
Theta is equal to the difference between the phase voltage angle and the phase current angle. Since these angles are equal, theta is equal to zero for each pair:
The maximum power delivered by a three phase delta transformer is equal to three times the delta phase voltage magnitude times the delta phase current magnitude.
Example – Determine the maximum power of a three phase transformer if three individual single-phase transformers rated for 150kVA and 4,160V/480V are connected in delta.
Let’s use the previous formula we derived to solve for maximum power.
Note that we can use either the secondary phase voltage with the secondary phase current, or the primary phase voltage with the primary phase current and still get the same answer.
In our calculations we are using the secondary values:
The maximum output power of this transformer configuration is 450kW.
Since the active power output in watts is equals the apparent power rating of a transformer in volt-amps during maximum power conditions, we can check our work by adding up the power ratings of each individual single-phase transformer used to make the three phase transformer:
So far so good, this is the easy stuff.
Now that we are familiar with the maximum power relationships for the three phase delta transformer, let’s look at what happens when one winding is removed and we are left with the open delta transformer connection.
Let’s pretend that while in service, the “B” phase of a standard three phase delta transformer failed, this is the winding located between the B and C connection. Let’s also pretend that the other windings were left intact and unaffected by the failure.
This results in an open delta transformer configuration that looks like this:
Just as before, we will use a reference of zero degrees for the A phase line to neutral voltage of the system, meaning that the A-phase voltage of the delta, which is equal to the A line voltage of the system, will be at a reference of 30º.
So far, nothing has changed for the delta A-phase and C-phase voltage, but we are not so sure about the delta B phase voltage just yet:
Let’s pretend we can connect a voltage meter across the B to C connection where the missing winding and measure what the new open delta voltage is across B to C.
We will use a positive polarity reference at B and a negative polarity reference at C, just like what the line would see prior to the winding failure.
Looking at this as a KVL loop, we can see that we have two voltages in series which we can add by summing the values.
Starting from the B to C connection we have negative A-phase delta voltage followed by the negative C-phase delta voltage from the A to C connection.
Let’s use a delta phase voltage magnitude of 1 so that we can plug these values into our calculator and see if the magnitude and angle increases or decrease in value:
The magnitude of the voltage across the missing winding has neither increased or decreased in value, it is equal to the previous magnitude of the same balanced phase voltage.
The angle of the voltage across the missing winding has also neither increased or decreased in value. It is still equal to the angle of the voltage that was there prior to the winding failure.
The surprising finding is that the connected three phase system sees no difference in the supply voltage with the missing winding!
The open delta phase voltages are identical to the three phase delta voltages prior to the winding being removed:
The resulting open delta phase voltage phasor diagram is identical to the previous delta phase voltage phasor diagram with all three windings in service:
Now let’s inspect the open delta phase and line current. This is where it gets a little bit more difficult.
We will start with the line current.
First, recognize that there is no change in the system line voltage since the open delta phase voltage is identical to the three phase delta phase voltages. The result of this is that the angles of each system line current will be the same as before.
The magnitudes of the system line currents will still be balanced and equal, but will be different in value since the previous KCL formulas for a three phase delta no longer apply with the missing winding:
Now let’s look at the open delta phase current.
Recognize that when we look at each of the single-phase transformers individually, even though the B phase transformer is no longer connected, it does not affect the full load amps of each of the two remaining single phase transformers that make up the two remaining windings of the open delta transformer.
The result is that each phase current in the two remaining open delta windings will have the same magnitude as before.
However, since the KCL relationship has been changed by removing a connection between the B an C phase, the angles will be different than before:
Let’s look at the open delta connection again and see if we can set up KCL formulas to relate the open delta phase and line currents:
Let’s start with the C-line and C-phase current since they have the easiest relationship:
Looking at the open delta digram, the C line current is now equal to the C phase current for the open delta connection since the B winding is missing.
Since we know the magnitude of the C line current, and the angle of the C phase current, we can determine the complex value for both:
The C-line current and the C-phase current both have a magnitude equal to the previous phase current magnitude with an angle of 120 degrees.
Next, let’s move on to the B line and phase current since they have the next easiest relationship:
Looking at the open delta diagram, the B-phase current is equal to the negative B-line current since they are flowing in opposite directions. This is the same as multiplying the complex value by negative one to change the direction.
Multiplying a vector by negative one does not change the magnitude, it only rotates the vector by positive or negative 180 degrees.
Since the B-phase current equals the negative B-line current, we know that their magnitudes are equal.
We already know the B-phase current magnitude and the B-line current angle.
We can find the missing B-phase angle by rotating it 180 degrees and solving for the complex values for both:
The B phase and line current are equal in magnitude, and 180 degrees apart.
Last, let’s move on to the A line current since it requires summing up the two previous open delta phase currents that we just solved for.
Looking at the open delta diagram and using KCL node equations, the A-line current is now equal to B-phase current minus the C-phase current.
We will use a phase current magnitude of 1 so that we can plug these values into our calculator and solve to see if the magnitude or angle increases or decreases with this new KCL relationship:
The A-line current has a magnitude equal to the phase current magnitude with an angle of 0º degrees.
We can check our work by verifying that we already determined that the A-line current had an angle of 0º in our previous steps.
The open delta phase and line currents we determined are:
Let’s draw all three line currents and both phase currents on the same phasor diagram.
Note that the C-line current and the C-phase current are equal and share the same vector:
Last step, it’s time to finally understand where the formula for open delta power comes from.
Even though the open delta can still serve a three phase load provided it has enough capacity, to find the maximum power we will only be looking at the two windings that remain.
Let’s draw a phasor diagram that shows both pairs of the remaining phase values, the A-phase voltage and current, and the C-phase voltage and current.
Note that we are not using line values because we want to find the complex sum of both remaining individual single phase transformers that each make up the two remaining phase windings of the open delta transformer:
Even though the open delta configured transformer is operating at maximum power conditions the phase currents and phase voltages are no longer perfectly in phase and have a displacement between them that is no longer equal to zero degrees.
Let’s take the complex sum of both remaining windings to determine the total maximum power that the open delta transformer can deliver:
The maximum three phase power that the open delta configured transformer can deliver is √3 times the phase voltage magnitude times the phase current magnitude.
Intuitively we know that an open delta transformer cannot supply as much power as a standard three phase delta since one of the windings is missing.
Since we already derived the formulas for both maximum open delta power and maximum standard delta power, let’s see if we can determine the percent difference.
This only works we are comparing an open delta three phase transformer with the same single phase transformer ratings. For example, if a three phase delta transformer loses one of its windings, or, if the same rated single phase transformers are used to make both a three phase delta transformer and an open delta transformer.
In both cases, the magnitudes of the phase voltage and phase current will be equal between the two types of delta transfomers:
An open delta transformer can only supply 57.7% of the power that a three phase delta transformer with equal single phase transformer ratings can deliver.
This is quite a large decrease in power when only losing a third of the overall three phase transformer windings.
Ex 1. Three single phase transformers rated for 10MVA 13.8kV/4,160V are connected in delta to form a three phase transformer. Determine the maximum power that this three phase transformer can deliver.
We can solve this two ways.
The first method:
The second method:
Note that both methods yield the same maximum value of 30MW of power delivery by the three phase delta transformer.
The first method we used the secondary voltage and the secondary FLA for the phase voltage and phase current magnitudes.
We could have just as easily used the primary voltage and the primary FLA instead (try it and verify the results).
Ex 2. Two single-phase transformers rated for 10MVA 13.8kV/4,160V are connected in open delta to form a three phase transformer. Determine the maximum power that this three phase transformer can deliver.
Time to use the max power formula for an open delta transformer:
Notice that the three phase open delta transformer with two windings will only deliver 17.3MW compared to the previous 30MW with three windings.
The trick here is that with open delta, we can’t just add the power ratings of both windings (10MW + 10MW ≠ 17.3MW).
Again we used the secondary voltage and the secondary FLA for the phase voltage and phase current magnitudes but we could have just as easily used the primary voltage and primary FLA values instead to get the same results (try it and verify).
Ex 3. Three single phase transformers rated for 10MVA 13.8kV/4,160V are connected in delta to form a three phase transformer. While in operation, one winding fails without damaging the other two. The transformer continues to provide power in an open delta configuration. Determine the percentage of power that the open delta configuration is able to supply compared to when all three windings were in service.
We can solve this two different ways.
Since we already did the math in the previous two examples, let’s use the previous values we calculated to find the percentage of power that the open delta can supply compared to when all three windings were in service.
We can do this by dividing the max power we determined for the open delta transformer by the max power we determined for the delta transformer.
The first method:
The 17.3MW of power that the open delta can deliver is only 57.7% of the previous 30MW of power that the transformer was able to deliver before the winding failure.
The second way to calculate this is to compare the formulas for open delta maximum power to standard delta maximum power.
Since the phase voltage magnitudes and phase current magnitudes are equal, they will cancel.
The second method:
Notice that we still get the same answer without having to know the actual values of the phase voltage or phase current magnitudes!
Have you ever come across a real open delta during your career in the industry?
Is this the first time you’ve seen a full-length explanation on the open delta configuration?
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