I read and derived formulas for balanced delta source to Wye source; it is given by Vy = Vd / sqrt(3)<30 but in some places I see people only dividing by sqrt(3) for conversions are keeping angle 0, is it the correct way for conversion ?
Electrical – Balanced delta source to Wye source by considering magnitude only is ok
acpowerthree phasey-to-delta
Related Solutions
Now, my main question is how do I work out the phase angle of each component?
Draw a phasor diagram: -
In delta, the loads are connected across three line voltage namely Vab, Vbc and Vca. These are shown above on the diagram - they are 120 degrees apart. It's simpler because the three delta loads are all equal and mainly reactive (hence the angle close to 90 degrees). So, for each of the individual line voltages, the current will lag by 76 degrees.
Here's another diagram that is for a delta load of just resistors: -
Important to note is that Ia = Iab - Ica and this tells you how the individual load currents sum to give you the phase current.
The zero sequence has no phase shift between voltages. Therefore on the delta side of the transformer, voltages will remain equal at all times between terminals, which means no current can flow through these terminals.
In general; no current can flow into a delta when fed by a zero sequence source.
Best Answer
If one doesn't care about the angle but only the magnitude then
\$ V_Y = \frac {1}{\sqrt 3} V_{\Delta} \$ is all you need.
For example, if I have a delta-star transformer feeding an induction motor nobody cares if the phases are 30° out because it is completely isolated from the delta side. Only the voltage magnitude will matter.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Phasor representation of star (wye) and delta voltages in a three-phase system.
It is a fairly trivial exercise to prove the √3 relationship between the phase to phase and phase to neutral phasors.