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I'm trying to solve some problems and I'm confusing one thing . We have , let's say a 30MVA, 24.5/345 KV 3-phase transformer connected in wye-delta. I try to use one-phase analysis and want to find the voltage ratio and make use of the following equation: $$S=\sqrt 3VI$$.

The values V and I are phase or line voltage/current? And when I find the voltage ratio which of the following is right? $$a=\frac{24.5}{345}$$ $$a=\frac{\frac{24.5}{\sqrt3}}{345}$$

I divide by the root of 3 because of the Y connection.

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  • \$\begingroup\$ The equations of you second question can be misleading. The letter \$a\$ is usually used for the turns ratio, which is equal to the ratio of phase voltages. However, you wanted to compute the voltage ratio, which I think you meant transformer ratio, which is equal to the ratio of line voltages. \$\endgroup\$ – Alejandro Nava Jan 13 at 12:53
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The nominal voltages given to describe a 3-phase transformer are always the line-to-line voltages, regardless of the connection. The voltage used for the single-phase equivalent circuit are always line-to neutral voltages. The transformer ratio is 24.5/345 but the single phase transformer voltages are those voltages each divided by root 3.

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  • \$\begingroup\$ Thank you for answering to so many of my electrical machine questions sir. I'd like to ask one more thing which I forgot to mention. If in this case I'm given the voltage across the load and the impedence of the transformer can I use the voltage ratio and from the given load voltage find the primary voltage or do I need to take impedence into account? In fewer words , does the ratio include the voltage drop because of the impedence or not? \$\endgroup\$ – John Katsantas Jun 8 '17 at 14:30
  • \$\begingroup\$ If you are given the impedance of the transformer you should probably assume that the voltages given are the open-circuit voltages and you must take the impedance into account. \$\endgroup\$ – Charles Cowie Jun 8 '17 at 15:03
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The three-phase apparent power or total apparent power of a device/machine/load with balanced voltages and currents can be computed as:

\$ S_{3 \phi} = 3 \, V_\phi \, I_\phi = \sqrt{3} \, V_\text{L} \, I_\text{L} \tag*{} \$

regardless of the connection.

In the previous equation, remember the following:

  • In a wye connection, line current is equal to phase current; and in a balanced linear system, line (or line-to-line) voltage is in magnitude \$\sqrt{3}\$ times the phase (or line-to-neutral) voltage:

\$ I_\text{L} = I_\phi \tag*{}\$

\$ V_\text{L} = V_\text{LL} = \sqrt{3} \, V_\phi = \sqrt{3} \, V_\text{LN} \tag*{}\$

  • In a delta connection, line (or line-to-line) voltage is equal to phase voltage; and in a balanced linear system, line current is in magnitude \$\sqrt{3}\$ times the phase current. A delta connection has no neutral point, so it has no physical line-to-neutral voltage, however, we still define a fictitious line-to-neutral voltage because per-phase equivalent circuits of three-phase systems use line-to-neutral voltages (whether the neutral is real or fictitious):

\$ V_\text{L} = V_\text{LL} = V_\phi = \sqrt{3} \, V_\text{LN} \tag*{}\$

\$ I_\text{L} = \sqrt{3} \, I_\phi \tag*{}\$


The values V and I are phase or line voltage/current?

As I explained above, they are line voltages and line currents.


Now to address your second question. Below, \$a\$ is defined as the turns ratio. The following equations only relate the magnitude (usually RMS values) of the phasors, and not their phase shift (angle). The following equations are valid for three-phase two-winding transformers in balanced linear systems.

For a wye-wye or a delta-delta transformer:

\$ a = \dfrac{N_\text{p}}{N_\text{s}} = \dfrac{V_\text{LL,p}}{V_\text{LL,s}} = \dfrac{V_{\phi \text{,p}}}{V_{\phi \text{,s}}} = \dfrac{V_\text{LN,p}}{V_\text{LN,s}} \tag*{} \$

\$ a = \dfrac{N_\text{p}}{N_\text{s}} = \dfrac{I_\text{L,s}}{I_\text{L,p}} = \dfrac{I_{\phi \text{,s}}}{I_{\phi \text{,p}}} \tag*{} \$

For a delta (primary)-wye (secondary) transformer:

\$ a = \dfrac{N_\text{p}}{N_\text{s}} = \sqrt{3} \, \dfrac{V_\text{LL,p}}{V_\text{LL,s}} = \dfrac{V_{\phi \text{,p}}}{V_{\phi \text{,s}}} = \sqrt{3} \, \dfrac{V_\text{LN,p}}{V_\text{LN,s}} \tag*{} \$

\$ a = \dfrac{N_\text{p}}{N_\text{s}} = \sqrt{3} \, \dfrac{I_\text{L,s}}{I_\text{L,p}} = \dfrac{I_{\phi \text{,s}}}{I_{\phi \text{,p}}} \tag*{} \$

For a wye (primary)-delta (secondary) transformer:

\$ a = \dfrac{N_\text{p}}{N_\text{s}} = \dfrac{1}{\sqrt{3}} \, \dfrac{V_\text{LL,p}}{V_\text{LL,s}} = \dfrac{V_{\phi \text{,p}}}{V_{\phi \text{,s}}} = \dfrac{1}{\sqrt{3}} \, \dfrac{V_\text{LN,p}}{V_\text{LN,s}} \tag 1 \$

\$ a = \dfrac{N_\text{p}}{N_\text{s}} = \dfrac{1}{\sqrt{3}} \, \dfrac{I_\text{L,s}}{I_\text{L,p}} = \dfrac{I_{\phi \text{,s}}}{I_{\phi \text{,p}}} \tag*{} \$

The turns ratio is defined as the ratio of individual turns, which is equal to the ratio of phase voltages and phase currents. On the other hand, the transformer ratio, as Charles Cowie said, is defined as the ratio of line (or line-to-line) voltages.

Note: which winding you consider as the primary and which as the secondary is in general arbitrary. What is not arbitrary, though, is which is the high-voltage winding and which is the low-voltage winding.

For the reader who wonders whether the previous equations are correct, he/she can check equations (2-88), (2-89), (2-90), (2-91) and table A-1 (or equations (A-10), (A-11), (A-14) and (A-15)) of the textbook Electric Machinery Fundamentals (5th edition) by Stephen Chapman. He/she can also check table 2.1 of the textbook Power Systems Analysis by John Grainger and William Stevenson.


And when I find the voltage ratio which of the following is right?

Remember the following:

  • As Charles Cowie said, the rated voltages of transformers are always line (or line-to-line) voltages.

  • As far as I know, the format when indicating nominal voltages of a transformer is "primary/secondary", regardless of which is the high-voltage and which is the low-voltage. So in your example, since it is stated as "24.5 kV/345 kV", the 24.5 kV winding (which is the LV side) is considered as the primary, and the 345 kV winding (which is the HV side) is considered as the secondary.

  • As far as I know, the format when indicating the connection of a three-phase two-winding transformer is "primary-secondary", regardless of which is the high-voltage and which is the low-voltage. So in your example, since it is stated as "wye-delta", the primary side is in wye, and the secondary is in delta. Therefore we must use the equation I labeled as (1) above.

In summary, in your transformer the wye side is the primary at 24.5 kV line-to-line, and the delta side is the secondary at 345 kV line-to-line. So the turns ratio (ratio of phase voltages) is:

\$ a = \dfrac{N_\text{p}}{N_\text{s}} = \dfrac{1}{\sqrt{3}} \, \dfrac{V_\text{LL,p}}{V_\text{LL,s}}= \dfrac{1}{\sqrt{3}} \, \dfrac{24.5 \text{ kV}}{345 \text{ kV}} \approx 0.041 \tag*{} \$

If you wanted to compute the transformer ratio (ratio of line voltages), then:

\$ \dfrac{V_\text{LL,p}}{V_\text{LL,s}}= \dfrac{24.5 \text{ kV}}{345 \text{ kV}} \approx 0.071 \tag*{} \$

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