Symmetrical Components in Power Systems

Demonstrative Video

Symmetrical Components & Sequence Networks

Symmetrical Components

Three symmetrical components:

A balanced three-phase system with the same phase sequence as the original sequence

A balanced three-phase system with the opposite phase sequence as the original sequence

Three phasors that are equal in magnitude and phase

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Symmetrical Component Transformation

let us first define the \(\alpha\)-operator for which the following relations holds \[\begin{aligned} \alpha & =e^{j120^{0}}=-\dfrac{1}{2}+j\dfrac{\sqrt{3}}{2}\\ \alpha^{2} & =e^{j240^{0}}=-\dfrac{1}{2}-j\dfrac{\sqrt{3}}{2}=\alpha^{*}\\ \alpha^{3} & =1\\ \alpha^{4} & =\alpha\\ \alpha^{5} & =\alpha^{2}~\text{and so on} \end{aligned}\] Note that we have \[\boxed{1+\alpha+\alpha^{2}=0}\]

Using the \(\alpha\)-operator we can write \[\begin{aligned} V_{b1} & =\alpha^{2}V_{a1} \quad V_{c1}=\alpha V_{a1}\\ V_{b2} & =\alpha V_{a2} \quad V_{c2}=\alpha^{2}V_{a2} \end{aligned}\] Finally, we get \[\begin{aligned} V_{a0} & =V_{b0}=V_{c0}\\ V_{a} & =V_{a0}+V_{a1}+V_{a2}\\ V_{b} & =V_{a0}+\alpha^{2}V_{a1}+\alpha V_{a2}=V_{b0}+V_{b1}+V_{b2}\\ V_{c} & =V_{a0}+\alpha V_{a1}+\alpha^{2}V_{a2}=V_{c0}+V_{c1}+V_{c2} \end{aligned}\]

Therefore, \[\begin{aligned} \left[\begin{array}{c} V_{a}\\ V_{b}\\ V_{c} \end{array}\right] & =\left[\begin{array}{ccc} 1 & 1 & 1\\ 1 & \alpha^{2} & \alpha\\ 1 & \alpha & \alpha^{2} \end{array}\right]\left[\begin{array}{c} V_{a0}\\ V_{a1}\\ V_{a2} \end{array}\right] \end{aligned}\] The symmetrical component transformation matrix is then given by \[\begin{aligned} \left[\begin{array}{c} V_{a0}\\ V_{a1}\\ V_{a2} \end{array}\right] & =\dfrac{1}{3}\left[\begin{array}{ccc} 1 & 1 & 1\\ 1 & \alpha & \alpha^{2}\\ 1 & \alpha^{2} & \alpha \end{array}\right]\left[\begin{array}{c} V_{a}\\ V_{b}\\ V_{c} \end{array}\right]\\ \Longrightarrow & V_{a012}=C\cdot V_{abc}\\ \Longrightarrow & I_{a012}=C\cdot I_{abc} \end{aligned}\] where \(C\) is the symmetrical component transformation matrix

The original phasor components can be obtained from the inverse symmetrical component transformation, i.e., \[\begin{aligned} V_{abc} & =C^{-1}V_{a012}\\ I_{abc} & =C^{-1}I_{a012} \end{aligned}\]

Real and reactive power

The three-phase power in the original unbalanced system is given by \[\begin{aligned} P_{abc}+jQ_{abc} & =V_{a}I_{a}^{*}+V_{b}I_{b}^{*}+V_{c}I_{c}^{*}=V_{abc}^{T}I_{abc}^{*}\\ & =\left(C^{-1}V_{a012}\right)^{T}\left(C^{-1}I_{a012}\right)^{*}\\ & =V_{a012}^{T}C^{-T}\left(C^{-1}\right)^{*}I_{a012}^{*} \end{aligned}\] we can deduce, \[C^{-T}\left(C^{-1}\right)^{*}=3\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]\]

Therefore, \[P_{abc}+jQ_{abc}=3\left(V_{a0}I_{a0}^{*}+V_{a1}I_{a1}^{*}+V_{a2}I_{a2}^{*}\right)\] Thus, complex power is three times the summation of the complex power of the three phase sequences.

Some facts about Sequence Current