Symmetrical Components in Power Systems

Demonstrative Video

Symmetrical Components & Sequence Networks

  • An unbalanced \(3\phi\) system can be resolved into three balanced systems in the sinusoidal steady state.

  • This method of resolving an unbalanced system into three balanced phasor system has been proposed by C. L. Fortescue.

  • This method is called resolving symmetrical components of the original phasors or simply .

  • Symmetrical components transformation will present how unbalanced components like Y-or-\(\varDelta\) -connected loads, transformers, generators and transmission lines can be resolved into symmetrical components.

  • We can then combine all these components together to form what are called .

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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  • The easiest method to analyse unbalanced system operation due to faults is through the use of symmetrical components

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Symmetrical Component Transformation

\[\begin{aligned} \alpha & =e^{j120^{0}}=-\dfrac{1}{2}+j\dfrac{\sqrt{3}}{2}\\ \alpha^{2} & =e^{j240^{0}}=-\dfrac{1}{2}-j\dfrac{\sqrt{3}}{2}=\alpha^{*}\\ \alpha^{3} & =1\\ \alpha^{4} & =\alpha\\ \alpha^{5} & =\alpha^{2}~\text{and so on} \end{aligned}\]
\[\boxed{1+\alpha+\alpha^{2}=0}\]
-operator for which the following relations holds let us first define the
\[\begin{aligned} V_{b1} & =\alpha^{2}V_{a1} \quad V_{c1}=\alpha V_{a1}\\ V_{b2} & =\alpha V_{a2} \quad V_{c2}=\alpha^{2}V_{a2} \end{aligned}\]
\[\begin{aligned} V_{a0} & =V_{b0}=V_{c0}\\ V_{a} & =V_{a0}+V_{a1}+V_{a2}\\ V_{b} & =V_{a0}+\alpha^{2}V_{a1}+\alpha V_{a2}=V_{b0}+V_{b1}+V_{b2}\\ V_{c} & =V_{a0}+\alpha V_{a1}+\alpha^{2}V_{a2}=V_{c0}+V_{c1}+V_{c2} \end{aligned}\]
-operator we can write Using the
\[\begin{aligned} \left[\begin{array}{c} V_{a}\\ V_{b}\\ V_{c} \end{array}\right] & =\left[\begin{array}{ccc} 1 & 1 & 1\\ 1 & \alpha^{2} & \alpha\\ 1 & \alpha & \alpha^{2} \end{array}\right]\left[\begin{array}{c} V_{a0}\\ V_{a1}\\ V_{a2} \end{array}\right] \end{aligned}\]
\(C\)
\[\begin{aligned} \left[\begin{array}{c} V_{a0}\\ V_{a1}\\ V_{a2} \end{array}\right] & =\dfrac{1}{3}\left[\begin{array}{ccc} 1 & 1 & 1\\ 1 & \alpha & \alpha^{2}\\ 1 & \alpha^{2} & \alpha \end{array}\right]\left[\begin{array}{c} V_{a}\\ V_{b}\\ V_{c} \end{array}\right]\\ \Longrightarrow & V_{a012}=C\cdot V_{abc}\\ \Longrightarrow & I_{a012}=C\cdot I_{abc} \end{aligned}\]
Therefore,
\[\begin{aligned} V_{abc} & =C^{-1}V_{a012}\\ I_{abc} & =C^{-1}I_{a012} \end{aligned}\]
The original phasor components can be obtained from the inverse symmetrical component transformation, i.e.,

Real and reactive power

\[\begin{aligned} P_{abc}+jQ_{abc} & =V_{a}I_{a}^{*}+V_{b}I_{b}^{*}+V_{c}I_{c}^{*}=V_{abc}^{T}I_{abc}^{*}\\ & =\left(C^{-1}V_{a012}\right)^{T}\left(C^{-1}I_{a012}\right)^{*}\\ & =V_{a012}^{T}C^{-T}\left(C^{-1}\right)^{*}I_{a012}^{*} \end{aligned}\]
\[C^{-T}\left(C^{-1}\right)^{*}=3\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]\]
The three-phase power in the original unbalanced system is given by
\[P_{abc}+jQ_{abc}=3\left(V_{a0}I_{a0}^{*}+V_{a1}I_{a1}^{*}+V_{a2}I_{a2}^{*}\right)\]
Therefore,

Some facts about Sequence Current

  • Balanced \(3\phi\) system consists of \(I_1\) only; \(I_2\) & \(I_0\) being zero

  • Presence of \(I_2\) or \(I_0\) in a \(3\phi\) system introduces un-symmetry and is indicative of an abnormal condition of the circuit

  • In an unbalanced 3-phase system \(\overrightarrow{I_{1}}+\overrightarrow{I_{2}} = 0\).

  • \[\begin{aligned} = \vec{I_{R0}} + \vec{I_{Y0}} + \vec{I_{B0}} \end{aligned}\]
    i.e. Vector sum of all sequence currents in 3-phase unbalanced system The resultant solely consists of
  • \[\begin{aligned} \text{Zero sequence current} = \dfrac{1}{3} \cdot \left[\text{Current in neutral wire}\right] \end{aligned}\]
    In a 3-phase, 4 wire unbalanced system, the magnitude of

  • In the absence of path through the neutral of a 3-phase system, \(I_n =0\) and \(I_L\) contain no \(I_0\).

  • \(\Delta\) load \(\Rightarrow\) no path to \(I_n\) & \(I_L\), and can contain no \(\overrightarrow{I_{0}}\)

  • In a 3-phase unbalanced system, \(|\overrightarrow{I_{2}}|\) cannot exceed that of \(|\overrightarrow{I_{1}}|\)

  • If \(|\overrightarrow{I_{2}}|\) were the greater, the phase sequence of the resultant system would be reversed.

  • \[I_1=I_2=I_0\]
    The current of a single phase load drawn from a 3-phase system comprises