Finding the Phase Voltages

Time for me to create some practice problems.

Objective: Find the Line Voltages (Va, Vb, Vc) for the given voltages at the terminals of a balanced load connected in wye.

Vab = 100 ∟ 0° V
Vbc = 141.4 ∟ 225° V
Vca = 100 ∟ 90° V

Strategy:
1) Find the symmetrical components (positive sequence, negative sequence, and zero sequence)
2) Convert symmetrical components of Line-to-Line to symmetrical components of Line-to-Neutral.
3) Superposition the new symmetrical components back to regular voltages for Va.
4) Find the voltages for Vb and Vc using the symmetrical components from Va.


Execution:
1)
The standard equation for the Positive Sequence Components is:
V(1)ab = (Vab + Vbc*∟120° + Vca*∟240° ) /3
so, V(1)ab = (100 ∟ 0° + 141.4 ∟225+120° + 100∟240+90° ) /3
V(1)ab = (100 ∟ 0° + 141.4 ∟345° + 100∟330° ) /3
V(1)ab = [100 + (136 -j36.6) + (86.6-j50) ]/3

V(1)ab = (322.6 - j86.6)/3

V(1)ab = 107.53-j 28.86 = 111.3∟-15°

The standard equation for the Negative Sequence Components is:
V(2)ab = (Vab + Vbc∟+240° + Vca∟120° ) /3
so, V(2)ab=(100 ∟ 0° + 141.4 ∟105° + 100∟330° ) /3
V(2)ab= (327.95-j46.3)/3

V(2)ab=109.3-j15.4 = 110.4∟-8°

The standard equation for the Zero Sequence Components is:
V(0)ab = (Vab + Vbc + Vca ) /3
so, V(0)ab = (100 ∟ 0° + 141.4 ∟225° + 100∟90° ) /3
V(0)ab= 0 V


2) Convert line-to-line components to line-to-neutral with the standard equations:

V(1)ab = V(1)an*sqrt(3) *∟-30° V(2)ab = V(2)an*sqrt(3) *∟+30°

Zero sequence voltage components have no line-to-neutral components V(0)a=0 V

so, V(1)an= V(1)a /sqrt(3) *∟-30°
V(1)an = (111.3/1.73) ∟ (-15+30)°
V(1)an = 64.3 ∟15°

V(2)an= V(2)a /sqrt(3) *∟+30°
V(2)an = 110.4/1.73 ∟-8+30°

V(2)an= 63.8∟22°

3) Superposition the new symmetrical components back to regular voltages.
Van = V(0)an + V(1)an + V(2)an Van

Van = 0 + 64.3 ∟15° + 63.8∟22°
Van = 62.1 + j16.6 + 59.1 +j 23.9

Van = 121.25 + j 40.5 = 127.8 ∟18.5° V

Vbn = V(0)a + V(1)an*∟240 + V(2)an*∟120
Vbn = 0 + 64.3 ∟(15+240)° + 63.8∟(22+120)°
Vbn = 64.3 ∟(255)° + 63.8∟(142)°

Vbn = -16.6 -j62.1 +50.3 + j39.28

Vbn = -66.9 -j 22.8= 70.7 ∟(-161)°

Vcn = V(0)a + V(1)an*∟120 + V(2)an*∟240
Vcn = 0 + 64.3 ∟(15+120)° + 63.8∟(22+240)°

Vcn = 64.3 ∟(135)° + 63.8∟(262)°

Vcn = -54.3 -j 17.7 = 57.16 ∟(-162)°

Done! Some check my math.

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