NCEES 535

[colo_elec]

Ok, so the answer for this problem requires dividing the transformer power by 3, not root 3. It's 3 phase, wye connected. Why do we divide power by 3 to find output per phase in order to find current?

 

[phowardtx]

Ok, so the answer for this problem requires dividing the transformer power by 3, not root 3. It's 3 phase, wye connected. Why do we divide power by 3 to find output per phase in order to find current?

In this problem we are asked to solve per-phase and in order to do that we have to convert the Voltage to per-phase. That's where the root 3 comes in... |Vp|=|VL|/√3. In 3-phase systems line-line voltage is always assumed given unless specified otherwise. Once the voltages are in per-phase quantities you can treat it as 3 separate systems where the voltage and current magnitudes are identical and the power is 1/3 of the total system. Thus the 3-phase power equation using per-phase quantities is |S|=3*|Vp|*|Ip|

using per phase voltage:

|S|=3*|Vp|*|Ip|

using line-line voltage:

|S|=√3*|VL|*|IL|

Wye-systems

|Vp|=|VL|/√3

|Ip|=|IL|

Δ-systems

|Vp|=|VL|

|Ip|=|IL|/√3

PS: The power equations are identical for both wye and Δ systems, the one you use depends on which quantities (line or phase) are most convenient to work with.

 

[BamaBino]

The 180 MVA rating is for the autotransformer and not the rating for the original isolation transformer that was used to form the autotransformer, right?

 

[BamaBino]

The answer 16.5 MVA is also the rating of transformer (for each phase) before it was connected as an autotransformer, right?