Symmetrical Components in Power Systems
Demonstrative Video
Problem-1
\[V_a = 300 \angle 20^{\circ}~V, ~V_b = 360\angle
90^{\circ}~V, ~\text{and}~V_c = 500 \angle -140^{\circ}~V\]
The voltages across a 3-phase unbalanced load are
Phase sequence is abc
Solution-1
Symmetrical component voltages:
\[\left[\begin{array}{c}
V_{a 0} \\
V_{a 1} \\
V_{a 2}
\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & a & a^{2} \\
1 & a^{2} & a
\end{array}\right]\left[\begin{array}{l}
V_{a} \\
V_{b} \\
V_{c}
\end{array}\right]\]
\[\begin{aligned}
\mathrm{V}_{\mathrm{a}
0}&=\frac{1}{3}\left[\mathrm{~V}_{\mathrm{a}}+\mathrm{V}_{\mathrm{b}}+\mathrm{V}_{\mathrm{c}}\right]
\\
\mathrm{V}_{\mathrm{a}
1}&=\frac{1}{3}\left[\mathrm{~V}_{\mathrm{a}}+\mathrm{aV}_{\mathrm{b}}+\mathrm{a}^{2}
\mathrm{~V}_{\mathrm{c}}\right] \\
\mathrm{V}_{\mathrm{a}
2}&=\frac{1}{3}\left[\mathrm{~V}_{\mathrm{a}}+\mathrm{a}^{2}
\mathrm{~V}_{\mathrm{b}}+\mathrm{a} \mathrm{V}_{\mathrm{c}}\right]
\end{aligned}\]
On expansion:
Given data:
\[\begin{aligned}
\mathrm{V}_{\mathrm{a}} &=300 \angle 20^{\circ}
\mathrm{V}=281.91+\mathrm{j} 102.61 \mathrm{~V} \\
\mathrm{~V}_{\mathrm{b}} &=360 \angle 90^{\circ}
\mathrm{V}=0+\mathrm{j} 360 \mathrm{~V} \\
\mathrm{~V}_{\mathrm{c}} &=500 \angle-140^{\circ}
\mathrm{V}=-383.02-\mathrm{j} 321.39 \mathrm{~V}
\end{aligned}\]
Find values:
\[
\begin{aligned}
a V_{b}&=1 \angle 120^{\circ} \times 360 \angle 90^{\circ}=360
\angle 210^{\circ}=-311.77-j 180 \mathrm{~V} \\
a^{2} V_{b}&=1 \angle 240^{\circ} \times 360 \angle
90^{\circ}=360 \angle 330^{\circ}=311.77-j 180 \mathrm{~V} \\
a V_{c}&=1 \angle 120^{\circ} \times 500 \angle-140^{\circ}=500
\angle-20^{\circ}=469.85-j 171.01 \mathrm{~V} \\
a^{2} \mathrm{~V}_{c}&=1 \angle 240^{\circ} \times 500
\angle-140^{\circ}=500 \angle 100^{\circ}=-86.82+\mathrm{j} 492.40
\mathrm{~V}
\end{aligned}
\]
We know that \(V_{a 0}=V_{b 0}=V_{c
0}\)
\[\begin{array}{l}
\mathrm{V}_{\mathrm{a} 0}=57.89 \angle 126^{\circ} \mathrm{V} \\
\mathrm{V}_{\mathrm{b} 0}=57.89 \angle 126^{\circ} \mathrm{V} \\
\mathrm{V}_{\mathrm{co}}=57.89 \angle 126^{\circ} \mathrm{V}
\end{array}\]
The zero sequence components are We know that, \(\mathrm{V}_{\mathrm{b}
1}=\mathrm{a}^{2} \mathrm{~V}_{\mathrm{al}} \quad ; \quad
\mathrm{V}_{\mathrm{cl}}=\mathrm{a}
\mathrm{V}_{\mathrm{al}}\)
\[\begin{array}{l}
\mathrm{V}_{\mathrm{a} 1}=143.70 \angle 106^{\circ} \mathrm{V}
\\
\mathrm{V}_{\mathrm{b} 1}=\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a}
1}=1 \angle 240^{\circ} \times 143.70 \angle 106^{\circ}=143.70 \angle
346^{\circ} \mathrm{V} \\
\mathrm{V}_{\mathrm{c} 1}=\mathrm{aV}_{\mathrm{al}}=1 \angle
120^{\circ} \times 143.70 \angle 106^{\circ}=143.70 \angle 226^{\circ}
\mathrm{V}
\end{array}\]
The positive sequence components are We know that \(, \mathrm{V}_{\mathrm{b}
2}=\mathrm{a} \mathrm{V}_{\mathrm{a}} ; \quad \mathrm{V}_{\mathrm{c}
2}=\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 2}\)
\[\begin{array}{l}
\mathrm{V}_{a2}=364.05 \angle-13^{\circ} \mathrm{V} \\
\mathrm{V}_{\mathrm{b} 2}=\mathrm{aV}_{\mathrm{a} 2}=1 \angle
120^{\circ} \times 364.05 \angle-13^{\circ}=364.05 \angle 107^{\circ}
\mathrm{V} \\
\mathrm{V}_{\mathrm{c} 2}=\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 2}=1
\angle 240^{\circ} \times 364.05 \angle-13^{\circ}=364.05 \angle
227^{\circ} \mathrm{V}
\end{array}\]
\(\therefore\)
Problem-2
\[\mathrm{V}_{\mathrm{a} 0}=10
\angle 180^{\circ} \mathrm{V}, \mathrm{V}_{\mathrm{al}}=50 \angle
0^{\circ} \mathrm{V} \text { and } \mathrm{V}_{\mathrm{a} 2}=20 \angle
90^{\circ} \mathrm{V}\]
The symmetrical components of phase-a voltage in a 3-phase unbalanced
system are
Solution-2
The phase voltages of \(\mathrm{V}_{a},
\mathrm{V}_{\mathrm{b}}\) and \(\mathrm{V}_{\mathrm{c}}\) are given by
\[\left[\begin{array}{c}
\mathrm{V}_{\mathrm{a}} \\
\mathrm{V}_{\mathrm{b}} \\
\mathrm{V}_{\mathrm{c}}
\end{array}\right]
=
\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & \mathrm{a}^{2} & \mathrm{a} \\
1 & \mathrm{a} & \mathrm{a}^{2}
\end{array}\right]\left[\begin{array}{l}
\mathrm{V}_{\mathrm{a0}} \\
\mathrm{~V}_{\mathrm{a1}} \\
\mathrm{~V}_{\mathrm{a2}}
\end{array}\right]\]
On expansion
\[\begin{array}{l}
V_{a}=V_{a0}+V_{a1}+V_{a2} \\
V_{b}=V_{a0}+a^{2} V_{a1}+a V_{a2} \\
V_{c}=V_{a0}+a V_{a1}+a^{2} V_{a2}
\end{array}\]
Given data
\[\begin{aligned}
\mathrm{V}_{\mathrm{a} 0} &=10 \angle 180^{\circ}
\mathrm{V}=-10+\mathrm{j} 0 \\
\mathrm{V}_{\mathrm{a} 1} &=50 \angle 0^{\circ}
\mathrm{V}=50+\mathrm{j} 0 \\
\mathrm{V}_{22} &=20 \angle 90^{\circ}
\mathrm{V}=0+\mathrm{j} 20
\end{aligned}\]
Find values
\[\begin{array}{l}
\mathrm{aV}_{\mathrm{a} 1}=1 \angle 120^{\circ} \times 50 \angle
0^{\circ}=50 \angle 120^{\circ}=-25+\mathrm{j} 43.30 \\
\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 1}=1 \angle 240^{\circ}
\times 50 \angle 0^{\circ}=50 \angle 240^{\circ}=-25 .-\mathrm{j} 43.30
\\
\mathrm{aV}_{\mathrm{a} 2}=1 \angle 120^{\circ} \times 20 \angle
90^{\circ}=20 \angle 210^{\circ}=-17.32-\mathrm{j} 10 \\
\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 2}=1 \angle 240^{\circ}
\times 20 \angle 90^{\circ}=20 \angle 330^{\circ}=17.32-\mathrm{j} 10
\end{array}\]
Phase voltages are:
\[\begin{aligned}
\mathrm{V}_{\mathrm{a}} &=\mathrm{V}_{\mathrm{a}
0}+\mathrm{V}_{\mathrm{a} 1}+\mathrm{V}_{\mathrm{a} 2}=40+\mathrm{j}
20=44.72 \angle 27^{\circ} \mathrm{V} \\
\mathrm{V}_{\mathrm{b}} &=\mathrm{V}_{\mathrm{a}
0}+\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 1}+\mathrm{a}
\mathrm{V}_{\mathrm{a} 2}=-10-25-\mathrm{j} 43.30-17.32-\mathrm{j} 10 \\
&=-52.32-\mathrm{j} 53.30=74.69 \angle-134^{\circ}
\mathrm{V} \\
\mathrm{V}_{\mathrm{c}} &=\mathrm{V}_{\mathrm{a}
0}+\mathrm{a} \mathrm{V}_{\mathrm{a} 1}+\mathrm{a}^{2}
\mathrm{~V}_{\mathrm{a} 2}=-10-25+\mathrm{j} 43.30+17.32-\mathrm{j} 10
\\
&=-17.68+\mathrm{j} 33.3=37.70 \angle 118^{\circ} \mathrm{V}
\end{aligned}\]