5.2: Bus Admittance

Now, if the network itself is linear, interconnections between buses and between buses and ground can all be summarized in a multiport bus impedance matrix or its inverse, the bus admittance matrix. As it turns out, the admittance matrix is easy to formulate.

The network consists of a number \(\ N_{b}\) of buses and another number \(\ N_{\ell}\) of lines. Each of the lines will have some (generally complex) impedance \(\ Z\). We form the line admittance matrix by placing the admittance (reciprocal of impedance) of each line on the appropriate spot on the main diagonal of an \(\ N_{\ell} \times N_{\ell} \) matrix:

\[\ \mathbf{Y}_{\ell}=\left[\begin{array}{rrrr}
\frac{1}{\mathbf{Z}_{1}} & 0 & 0 & \cdots \\
0 & \frac{1}{\mathbf{Z}_{2}} & 0 & \cdots \\
0 & 0 & \frac{1}{\mathbf{Z}_{3}} & \cdots \\
\vdots & \vdots & & \ddots
\end{array}\right]\label{2} \]

Interconnections between buses is described by the bus incidence matrix. This matrix, which has \(\ N_{\ell}\) columns and \(\ N_{b}\) rows, has two entries for each line, corresponding to the buses at each end. A “direction” should be established for each line, and the entry for that line, at location \(\ \left(n_{b}, n_{\ell}\right)\) in the node incidencd matrix, is a 1 for the “sending” end and a −1 at the “receiving” end. Actually, it is not important which end is which. The bus incidence matrix for the network described by Figure 1 below is:

\(\ \underline{\underline{N I}}=\left[\begin{array}{llll}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
-1 & 1 & 1 & 0 \\
0 & 0 & -1 & -1 \\
0 & -1 & 0 & 0
\end{array}\right]\)

It is not difficult to show that the bus admittance matrix is given by the easily computed expression:

\[\ \underline{\underline{\mathbf{Y}}}=\underline{\underline{\mathbf{N I}}} \quad \underline{\mathbf{Y}_{\ell}} \quad \underline{\underline{\mathbf{N I}}^{\prime}}\label{3} \]

The elements of the bus admittance matrix, the self– and mutual– admittances, are all of the following form:

\[\ \mathbf{Y}_{j k}=\frac{\mathbf{I}_{k}}{\mathbf{V}_{j}}\label{4} \]

with all other voltages equal to zero.

Thus an alternative way to estimate the bus admittance matrix is to:

• Assume that all nodes (buses) are shorted to ground,
• Assume that one node is unshorted and connected to a voltage source,
• Calculate all node currents resulting from that one source.
• Do this for each node.

We may observe:

• Reciprocity holds:

  \[\ \mathbf{Y}_{j k}=Y_{k j}\label{5} \]

• Driving point admittance \(\ \mathbf{Y}_{k k}\) is just the sum of all admittances of lines connected to bus \(\ k\), including any fixed impedances connected from that bus to ground.
• Mutual admittance \(\ \mathbf{Y}_{j k}\) is minus the sum of the admittances of all lines connected directly between buses \(\ j\) and \(\ k\). Usually there is only one such line.

Network currents are then given by:

\[\ \underline{\mathbf{I}}=\underline{\underline{\mathbf{Y}}} \quad \underline{\mathbf{V}}\label{6} \]

Where \(\ \underline{\mathbf{I}}\) is the vector of bus currents (that is, those currents entering the network at its buses. \(\ \underline{\mathbf{V}}\) represents the bus voltages and \(\ \underline{\underline{Y}}\) is the bus admittance matrix. We will have more to say about estimating the bus admittance matrix in another section. For the moment, note that an individual bus current is given by:

\[\ \mathbf{I}_{k}=\sum_{j=1}^{N} \mathbf{Y}_{j k} \mathbf{V}_{j}\label{7} \]

where \(\ N\) is the number of buses in the network. Then complex power flow at a node is:

\[\ \mathbf{S}_{k}=\mathbf{V}_{k} \sum_{j=1}^{N} \mathbf{Y}_{j k}^{*} \mathbf{V}_{j}^{*}\label{8} \]

Now, the typical load flow problem involves buses with different constraints. It is possible to specify six quantities at each bus: voltage magnitude and angle, current magnitude and angle, real and reactive power. These are, of course, inter–related so that any two of these are specified by the other four, and the network itself provides two more constraints. Thus it is necessary to, in setting up a load flow problem, specify two of these six quantities. Typical combinations are:

• Generator Bus: Real power and terminal voltage magnitude are specified.
• Load Bus: Real and reactive power are specified.
• Fixed Impedance: A fixed, linear impedance connected to a bus constrains the relationship between voltage and current. Because it constrains both magnitude and angle, such an impedance constitutes two constraints.
• Infinite Bus: This is a voltage source, of constant magnitude and phase angle.

The load flow problem consists of solving Equation \ref{8} as constrained by the terminal relationships.

One bus in a load flow problem is assigned to be the “slack bus” or “swing bus”. This bus, which is taken to be an “infinite bus”, since it does not have real nor reactive power constrained, accommodates real power dissipated and reactive power stored in network lines. This bus is necessary because these losses are not known a priori. Further, one phase angle needs to be specified, to serve as an origin for all of the others.