Autotransformer phase shift?

[oldsparky52]

I'm way out of my education level here. I'm trying to conceptualize what's happening in an autotransformer. For this discussion let's say we are boosting a voltage from 208 to 240. So a 32 volt boost is what we desire. My question involves the way the boost voltage (line, vector, ??) is drawn.

1) Is this drawn correctly in that the boost voltage stays in a straight line with the 2 phases that make up the 208?

2) Should the line show some kind of phase shift for the 32 volts? If so, how many degrees would the phase shift be and which direction? Any explanation (high school level if possible) would be greatly appreciated.

 

[kwired]

I'm way out of my education level here. I'm trying to conceptualize what's happening in an autotransformer. For this discussion let's say we are boosting a voltage from 208 to 240. So a 32 volt boost is what we desire. My question involves the way the boost voltage (line, vector, ??) is drawn.

1) Is this drawn correctly in that the boost voltage stays in a straight line with the 2 phases that make up the 208?

2) Should the line show some kind of phase shift for the 32 volts? If so, how many degrees would the phase shift be and which direction? Any explanation (high school level if possible) would be greatly appreciated.




View attachment 2560549

I think you drew that correctly.

The single phase volts between the two points of 240 would be at a 180 degree angle, the 120 volt legs of the wye would all be 120 degrees apart, so seems there be no other choice of how to draw it.

The boosted end there would have voltage to the neutral that is same as the yellow line you drew I would think, a little trigonometry skills needed to calculate it out.

 

[jim dungar]

You may find the 32V boost may be split on both sides of the 208V, so each 120V leg is lengthened in your sketch. It depends on if your transformer is single phase, delta, or Wye.

 

[gar]

220511-0951 EDT

oldsparky52:

Consider a three terminal ordinary autotransformer. It is best to describe that device as a single phase transformer no matter how you configure the three leads for input and output. This means that any combination of leads used for input and output that the phase relationship between input and output is either a 0 or 180 degree phase shift ( other than for some very small phase shifts ).

Thus, your 208-32 line is drawn correctly.

When you put this autotransformer in the three phase wye circuit as shown, then the phasor drawing you show is correct, and your ??? vector is calculated using trig.

.

 

[kwired]

You may find the 32V boost may be split on both sides of the 208V, so each 120V leg is lengthened in your sketch. It depends on if your transformer is single phase, delta, or Wye.

Most commonly used buck boost transformers would be like what was drawn, but yes one could have one long coil with selectable taps within and have other possible outcomes

 

[synchro]

View attachment 2560549

The 32V phasor is 150° from the 120V phasor in red, since 180° - 30° = 150°.
Therefore to find the voltage on the yellow phasor we can use the law of cosines to get:
√ [ 32² + 120² - 2 x 32 x 120 x cos(150°) ] = 148.6 volts

 

[oldsparky52]

Thanks everyone. I didn't know if that 32 V section had a slight phase shift or not, you know like a little kick up or down.

 

[gar]

220511-1111 EDT

oldsparky52:

Any real world single phase transformer has some very minor phase shift between input an output even under no load conditions. This results from a transformer having series leakage inductance and resistance, and further some shunt inductance and losses.

Shunt inductance and resistance is generally ignored. Series inductance and resistance is not usually ignored when a transformer is loaded.

.

 

[winnie]

To amplify what gar said:

A real world transformer has a bit of inductance and resistance that will cause some phase shift (a 'kick' such as mentioned in post #7).

For buck-boost applications, I think this can safely be ignored. The magnitude of this 'kick' is the transformer impedance rating * output voltage, so in the original example (208V to 240V) might be on the order of 5% * 32V at full load.

-Jon

 

[oldsparky52]

Thanks again gentlemen.