FORMATION OF BUS IMPEDANCE MATRIX USING BUILDING ALGORITHM OR STEP BY STEP PROCEDURE OF FINDING BUS IMPEDANCE MATRIX Z BUS

 ZBUS building

FORMATION OF BUS IMPEDANCE MATRIX

The bus impedance matrix is the inverse of the bus admittance matrix. An alternative method is possible, based on an algorithm to form the bus impedance matrix directly from system parameters and the coded bus numbers. The bus impedance matrix is formed adding one element at a time to a partial network of the given system. The performance equation of the network in bus frame of reference in impedance form using the currents as independent variables is given in matrix form by

Now assume that the bus impedance matrix Zbus is known for a partial network of m buses and a known reference bus. Thus, Zbus of the partial network is of dimension mxm. If now a new element is added between buses p and q we have the following two possibilities:

(i) p is an existing bus in the partial network and q is a new bus; in this case p-q is a

branch added to the p-network as shown in Fig 1a, and

 

(ii) both p and q are buses existing in the partial network; in this case p-q is a link

added to the p-network as shown in Fig 1b.

If the added element ia a branch, p-q, then the new bus impedance matrix would be of order m+1, and the analysis is confined to finding only the elements of the new row and column (corresponding to bus-q) introduced into the original matrix. If the added element ia a link, p-q, then the new bus impedance matrix will remain unaltered with regard to its order. However, all the elements of the original matrix are updated to take account of the effect of the link added.

FORMATION OF BUS IMPEDANCE MATRIX ZBUS USING NODE ELIMINATION METHOD

 FORMATION OF BUS IMPEDANCE MATRIX

NODE ELIMINATION BY MATRIX ALGEBRA

Nodes can be eliminated by the matrix manipulation of the standard node equations. However, only those nodes at which current does not enter or leave the network can be considered for such elimination. Such nodes can be eliminated either in one group or by taking the eligible nodes one after the other for elimination, as discussed next.

CASE-A: Simultaneous Elimination of Nodes:

Consider the performance equation of the given network in bus frame of reference in admittance form for a n-bus system, given by:

IBUS = YBUS EBUS (1)

Where IBUS and EBUS are n-vectors of injected bus current and bus voltages and YBUS is the square, symmetric, coefficient bus admittance matrix of order n. Now, of the n buses present in the system, let p buses be considered for node elimination so that the reduced system after elimination of p nodes would be retained with m (= n-p) nodes only. Hence the corresponding performance equation would be similar to (1) except that the coefficient matrix would be of order m now, i.e.,

IBUS = YBUSnew EBUS (2)

Where YBUSnew is the bus admittance matrix of the reduced network and the vectors

 

IBUS and EBUS are of order m. It is assumed in (1) that IBUS and EBUS are obtained with their elements arranged such that the elements associated with p nodes to be eliminated are in the lower portion of the vectors. Then the elements of YBUS also get located accordingly so that (1) after matrix partitioning yields,

Where the self and mutual values of YA and YD are those identified only with the nodes to be retained and removed respectively and YC=YBt is composed of only the corresponding mutual admittance values, that are common to the nodes m and p.

Now, for the p nodes to be eliminated, it is necessary that, each element of the vector IBUS-p should be zero. Thus we have from (3):

IBUS-m = YA EBUS-m + YB EBUS-p IBUS-p = YC EBUS-m + YD EBUS-p = 0

(4)

Solving,

EBUS-p = - YD-1YC EBUS-m (5)

Thus, by simplification, we obtain an expression similar to (2) as,

IBUS-m = {YA - YBYD-1YC} EBUS-m (6)

Thus by comparing (2) and (6), we get an expression for the new bus admittance matrix in terms of the sub-matrices of the original bus admittance matrix as:

YBUSnew = {YA – YBYD -1YC} (7)

This expression enables us to construct the given network with only the necessary nodes retained and all the unwanted nodes/buses eliminated. However, it can be observed from

(7) that the expression involves finding the inverse of the sub-matrix YD (of order p). This would be computationally very tedious if p, the nodes to be eliminated is very large, especially for real practical systems. In such cases, it is more advantageous to eliminate the unwanted nodes from the given network by considering one node only at a time for elimination, as discussed next.

CASE-B: Separate Elimination of Nodes:

Here again, the system buses are to be renumbered, if necessary, such that the node to be removed always happens to be the last numbered one. The sub-matrix YD then would be a single element matrix and hence it inverse would be just equal to its own reciprocal value. Thus the generalized algorithmic equation for finding the elements of the new bus admittance matrix can be obtained from (6) as,

Yij new = Yij old – Yin Ynj / Ynn " i,j = 1,2,…… n. (8)

 

Each element of the original matrix must therefore be modified as per (7). Further, this procedure of eliminating the last numbered node from the given system of n nodes is to be iteratively repeated p times, so as to eliminate all the unnecessary p nodes from the original system.