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Analytic Loss Minimization: A Proof

Abstract

Loss minimizing generator dispatch profiles for power systems are usually derived using optimization techniques. However, by partitioning and manipulating the \(Y_{bus}\) matrix, it is possible to derive the \(K_{GL}\) matrix, which can be used to analytically determine a loss minimizing dispatch for generators. This letter draws on recent research on the characterization of transmission system losses to demonstrate how the \(K_{GL}\) matrix achieves this impressive function. A new proof of the observed zero row summation property of the \(Y_{GGM}\) matrix is provided to this end.

Various works, such as (Visakha 2004, Thukaram 2009, Thukaram 2009a) have noted that the \(K_{GL}\) matrix allows the direct calculation of a generator dispatch profile that appears to minimize active power losses. The literature offers scant explanation for how this impressive result is achieved. The recent work of Abdelkader et al. (Abdelkader 2014, Abdelkader 2011, Abdelkader 2008), which draws on the \(Y_{bus}\) partitioning approach introduced in (Kessel 1986), offers useful insight here. Crucially, (Abdelkader 2011) shows how transmission losses can be separated into three distinct components. One of these loss components arises solely because of mismatched generator voltages, which cause circulating currents to flow through the system (Abdelkader 2014). This letter demonstrates how the \(K_{GL}\) matrix can give a generator dispatch profile which equalizes generator complex voltages, so no circulating currents will flow and this loss component is nullified.

The \(Y_{bus}\) is reordered, per (Kessel 1986), such that the \(m\) generator buses (\(G\) subscript) and \(n\) load buses (\(L\)) are grouped together: \[\label{eq:YLL} \begin{bmatrix}I_{G} \\ I_{L} \end{bmatrix} = \begin{bmatrix}Y_{GG} & Y_{GL} \\ Y_{LG} & Y_{LL} \end{bmatrix}\begin{bmatrix}V_{G} \\ V_{L} \end{bmatrix}\] \(I_{G}\) and \(I_{L}\) are complex-valued vectors representing the nodal currents at generator and load buses, respectively, while \(V_{G}\) and \(V_{L}\) are corresponding complex nodal voltages. Manipulation of (\ref{eq:YLL}) gives: \[\label{eq:YMod} \begin{bmatrix}V_{L} \\ I_{G} \end{bmatrix} = \begin{bmatrix}Z_{LL} & F_{LG} \\ K_{GL} & Y_{GGM} \end{bmatrix}\begin{bmatrix}I_{L} \\ V_{G} \end{bmatrix}\] Where: \[\label{eq:Yggm} Y_{GGM} = Y_{GG}-Y_{GL}Z_{LL}Y_{LG}\] \[\label{eq:Flg} F_{LG} = -Z_{LL}Y_{LG}=K^T_{GL}\]

The treatment of (Abdelkader 2011) gives an expression for the network loss using the difference between aggregate powers on the generation and load sides of the network. To find the total power generated, use the \(I_{G}\) expression from (\ref{eq:YMod}), left-multiplied by \(V_{G}\) to give a single complex value:

\[\label{eq:SGTOT} S^{Tot}_{G} = V^T_{G}I_{G} = V^T_{G}K_{GL}I^*_{L} + V^T_{G}Y_{GGM}V^*_{G}\]

Equivalently, multiplying the \(V_{L}\) expression from (\ref{eq:YMod}) by \(I_{L}\) expresses the total load power consumed as a complex number: \[\label{eq:SLTOT} S^{Tot}_{L} = I^T_{L}V_{L} = I^T_{L}Z_{LL}I_{L} + I^T_{L}F_{LG}V_{G}\] Given that generator powers will be positive, and loads negative, summing these powers gives the total system loss: \[\label{eq:TotalLoss} S^{Tot}_{Loss} = S^{Tot}_{G} + S^{Tot}_{L}\] Which equals \[\label{eq:TotalLossExpand} V^T_{G}Y_{GGM}V_{G} + V^T_{G}(Y^*_{GL}Z^*_{LL} - Y^T_{LG}Z^T_{LL})I^*_{L} + I^T_{L}Z^T_{LL}I^*_{L}\] The system losses are seen to naturally separate themselves into three distinct components. The last component is the load current loss, which cannot be affected by the system operator, being a pure consequence of system topology and load currents. The middle component, the mismatch loss, will only arise when branch \(X/R\) ratios are heterogeneous, and will in any case generally be small and imaginary (Abdelkader 2011). The first term in the circulating current loss, and it depends on generator power outputs. As will be shown, the \(Y_{GGM}\) matrix has the property that its rows sum to zero, and so its product can be brought to zero by right-multiplying it by a vector whose elements are homogeneous. Therefore, if the system is operated in such a way that generator voltages are homogeneous, there will be no excess circulating current loss, and the active power losses will reduce to the irreducible component which arises from serving load currents.

**Theorem** If every row of the \(Y_{bus}\) matrix sums to zero and \(detY_{LL}\neq 0\), then every row of \(Y_{GGM}\) sums to zero. If every row of the \(Y_{bus}\) matrix sums approximately to zero, or if it sums to zero but \(detY_{LL}\)=0, then every row of the matrix \(Y_{GGM}\) sums approximately to zero.

**Proof** The matrix \(Y_{GGM}\) can be written as \[\label{eq:eq1}
Y_{GGM}=Y_{GG}+Y_{GL}F_{LG},\] since \(F_{LG}=-Z_{LL}Y_{LG}\). Where \[\label{eq:eqzll}
Z_{LL}=\left\{\begin{array}{c}Y_{LL}^{-1},\quad det(Y_{LL})\neq 0\\Y_{LL}^\dagger,\quad det(Y_{LL})=0\end{array}\right\}\] For ease of notation set \(Y_{GG}=[a_{ij}]_{i=1,2,...,m}^{j=1,2,...,m}\), and similarly:

\[\label{eqn:notation} \begin{tabular}{l l} $Y_{GL}=[b_{ij}]_{m \times n}$ & $F_{LG}=[c_{ij}]_{n \times m}$ \\ $Y_{GL}F_{LG}=[d_{ij}]_{m \times m}$ & $Y_{GGM}=[g_{ij}]_{m \times m}$ \\ \end{tabular}\]

By substituting the previous expressions into \eqref{eq:eq1}, for every row \(i=1,2,...,m\) we have \[\label{eqn:eq2} \sum_{j=1}^mg_{ij}=\sum_{j=1}^ma_{ij}+\sum_{j=1}^md_{ij}.\]

In addition \[\begin{array}{c} d_{i1}=\sum_{k=1}^n b_{ik}c_{k1} \\ \vdots\\ d_{im}=\sum_{k=1}^n b_{ik}c_{km}. \end{array}\] By taking the sum of the above equalities, \(\forall i=1,2,...,n\) we arrive at

\[\label{eq:eq3} \sum_{j=1}^md_{ij}=\sum_{j=1}^m\sum_{k=1}^n b_{ik}c_{kj}.\]

Disregarding shunt elements, the \(Y_{bus}\) is a weighted Laplacian matrix (Edström 2014), and so each of its rows will sum to zero. Accordingly:

\[\label{eq:eq4} \sum^m_{j=1}a_{ij}+\sum^n_{j=1}b_{ij}=0,\quad \forall i=1,2,...,m.\]

By replacing \eqref{eq:eq3} into \eqref{eqn:eq2} we get:

\[\label{eq:eq4.1} \sum_{j=1}^mg_{ij}=\sum_{j=1}^ma_{ij}+\sum_{j=1}^m\sum_{k=1}^n b_{ik}c_{kj}\]

and using \eqref{eq:eq4} \[\label{eq:eq4.2} \sum_{j=1}^mg_{ij}=-\sum_{k=1}^nb_{ik}+\sum_{j=1}^m\sum_{k=1}^n b_{ik}c_{kj}\]

which equals \[\label{eq:eq4.3} -\sum_{k=1}^nb_{ik}+\sum_{k=1}^nb_{ik}(\sum_{j=1}^mc_{kj})\]

which equals

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