7.2.1 Bus Admittance Matrix

Table 7.1 presents a review of circuit-component definitions and laws. Ohm’s and Kirchhoff’s laws can be used for matrix formulations that are useful for load flow analyses. This approach yields two matrices:

• Kirchhoff’s current law (KCL) generates bus matrices, which result from the fact that the sum of currents entering a bus is zero.

• Kirchhoff’s voltage law (KVL) results in loop matrices, which describe voltages in a circuit loop.

The bus matrix written in terms of admittance originates directly from the KCL applied to each circuit bus other than the reference (ground) bus.

Solution to Application Example 7.1

The single-phase representation of the power system is shown in Fig. E7.1.1.

Let j be an arbitrary circuit bus other than the reference bus (which is denoted as bus 0), as shown in Fig. 7.2. Then

${\displaystyle \sum _{i=1}^n{\tilde{I}}_{ji}={\tilde{I}}_j}\text{,}$

where ${\tilde{I}}_{ji}\text{≡}\mathrm{line}$ line current, which leaves bus j and enters bus i via a line and ${\tilde{I}}_j\text{≡}\mathrm{current}$ current injected at bus j from an external source or bus injection current. The tilde (~) above Ĩ_ji and Ĩ_j denotes a complex line current and a complex bus injection current, respectively.

Application of Ohm’s law (Fig. 7.3) leads to

${\tilde{I}}_{ji}={\tilde{y}}_{ji}\left({\tilde{V}}_j-{\tilde{V}}_i\right)\text{,}$

where ỹ_ji is the complex line admittance of a line from bus j to bus i.

Certain elements of ${\displaystyle \sum _{i=1}^n{\tilde{I}}_{ji}={\tilde{I}}_j}$ are zero:

• The current Ĩ_jj is zero since no line joins bus j with bus j and this is reflected in ${\tilde{I}}_{jj}={\tilde{y}}_{jj}\left({\tilde{V}}_j-{\tilde{V}}_j\right)$ because it gives zero for Ĩ_jj due to $\left({\tilde{V}}_j-{\tilde{V}}_j\right)=0\text{;}$ and

• Ĩ_ji will be zero if no line exists between j and i, the “line impedance” is infinite, and the “line admittance” ỹ_ji is zero. One can write now (introducing Eq. 7-2 into Eq. 7-1)

${\displaystyle \sum _{i=1}^n{\tilde{y}}_{ji}\left({\tilde{V}}_j-{\tilde{V}}_i\right)={\tilde{I}}_j}\text{.}$

Note that ${\tilde{V}}_j$ and ${\tilde{V}}_i$ are bus voltages with respect to bus 0, the reference (ground) bus. Equation 7-3 is written for i = 1, 2, …, n and the coefficient of voltage ${\tilde{V}}_j$ in the Kirchhoff equation at bus j is ${\displaystyle \sum _{i=1}^n{\tilde{y}}_{ji}}$ whereas the coefficient of ${\tilde{V}}_i\left(i\ne j\right)$ in the jth equation is $\left(-{\tilde{y}}_{ji}\right)\text{.}$

All equations now read (n × n equation system) as follows,

for j = 1:

${\tilde{I}}_1=\left({\displaystyle \sum _{\begin{array}{l}i=1\\ i\ne 1\end{array}}^n{\tilde{y}}_{1i}}\right){\tilde{V}}_1+\left(-{\tilde{y}}_{12}\right){\tilde{V}}_2+\dots +\left(-{\tilde{y}}_{1n}\right){\tilde{V}}_n\text{,}$

for j = 2:

${\tilde{I}}_2=\left(-{\tilde{y}}_{21}\right){\tilde{V}}_1+\left({\displaystyle \sum _{\begin{array}{l}i=1\\ i\ne 2\end{array}}^n{\tilde{y}}_{2i}}\right){\tilde{V}}_2+\dots +\left(-{\tilde{y}}_{2n}\right){\tilde{V}}_n\text{,}$

for j = n:

${\tilde{I}}_n=\left(-{\tilde{y}}_{n1}\right){\tilde{V}}_1+\left(-{\tilde{y}}_{n2}\right){\tilde{V}}_2+\dots +\left({\displaystyle \sum _{\begin{array}{l}i=1\\ i\ne n\end{array}}^n{\tilde{y}}_{ni}}\right){\tilde{V}}_n\text{.}$

Observe the pattern of the coefficients of the bus voltages on the right-hand side of the equation system. This pattern will be generalized as the rule for the formation of a matrix of coefficients of ${\tilde{V}}_j$. The above n equations can be written in matrix form

${\overline{I}}_{\mathrm{bus}}={\overline{Y}}_{\mathrm{bus}}{\overline{V}}_{\mathrm{bus}}\text{,}$

where Ī_bus is the bus injection vector representing the n phasor injection currents, ${\overline{Y}}_{\mathrm{bus}}$ is the bus admittance matrix (it is an n × n matrix of coefficients formed as indicated above), and ${\overline{V}}_{\mathrm{bus}}$ is the bus voltage vector of n phasor voltages referenced to bus 0:

${\tilde{V}}_n={\tilde{\mathrm{\Phi }}}_n-{\tilde{\mathrm{\Phi }}}_0={\tilde{\mathrm{\Phi }}}_n\text{,}$

where ${\tilde{\mathit{\Phi }}}_n$ and ${\tilde{\mathit{\Phi }}}_0$ are the potentials at bus n and bus 0, respectively, and ${\tilde{\mathit{\Phi }}}_0$ = 0.

If ${\overline{Y}}_{\mathrm{bus}}$ is a nonsingular matrix then its inverse ${\left({\overline{Y}}_{\mathrm{bus}}\right)}^{-1}$ can be found and there exists a solution to the matrix equation. If ${\overline{Y}}_{\mathrm{bus}}$ is singular then ${\left({\overline{Y}}_{\mathrm{bus}}\right)}^{-1}$ cannot be found and there exists no solution to the matrix equation.

$\begin{array}{cccc}\hfill {\overline{I}}_{\mathrm{bus}}\hfill & \hfill =\hfill & \hfill {\overline{Y}}_{\mathrm{bus}}\hfill & \hfill {\overline{V}}_{\mathrm{bus}}.\hfill \\ \hfill \uparrow \hfill & \hfill \hfill & \hfill \uparrow \hfill & \hfill \uparrow \hfill \\ \hfill n\times 1\hfill & \hfill \hfill & \hfill \underbrace{n\times nn\times 1}\hfill & \hfill \hfill \\ \uparrow \hfill & \uparrow \hfill & \hfill n\times 1\hfill & \hfill \hfill \\ \mathrm{row}\hfill & \mathrm{column}\hfill & \hfill \hfill & \hfill \hfill \end{array}$

Note that the injection currents find a return path through bus 0 (ground). Also note that the Kirchhoff current law is not applied to bus 0 in these equations; however, the total current entering bus 0 is the negative sum of the currents Ĩ₁, Ĩ₂, … Ĩ_n. Therefore, the Kirchhoff current law at bus 0 is

$\begin{array}{l}-\sum _{i=1}^n{\tilde{I}}_j=-\left[\left(-{\tilde{y}}_{21}-{\tilde{y}}_{31}-\dots -{\tilde{y}}_{n1}+{\displaystyle \sum _{\begin{array}{l}i=1\\ i\ne 1\end{array}}^n{\tilde{y}}_{1i}}\right)\right.{\tilde{V}}_1+\\ \left.\left(-{\tilde{y}}_{12}-{\tilde{y}}_{32}-\dots -{\tilde{y}}_{n2}+{\displaystyle \sum _{\begin{array}{l}i=1\\ i\ne 2\end{array}}^n{\tilde{y}}_{2i}}\right){\tilde{V}}_2+\dots \right]\text{,}\end{array}$

which can be readily simplified as

${\tilde{I}}_0=-{\tilde{y}}_{01}{\tilde{V}}_1-{\tilde{y}}_{02}{\tilde{V}}_2-\dots -{\tilde{y}}_{0n}{\tilde{V}}_n\text{,}$

where ỹ_0k is the primitive admittance joining buses 0 and k.

Because the relation in Eq. 7-9 is a linear combination of the previous n equations, it cannot be used to obtain additional information beyond that given in the n × n equation system. Inspection of the coefficients of the $\tilde{V}$ terms in the n × n equation system reveals a rule for the formation of ${\overline{Y}}_{\mathrm{bus}}$. The following rules are used to “build” the bus admittance matrix referenced to bus 0.

The Admittance Matrix Building Algorithm

• y_jj: the diagonal entries of ${\overline{Y}}_{\mathrm{bus}}$ are formed by summing the admittances of lines terminating at bus j. Note there is no tilde on y_jj.

• y_ij: the off-diagonal entries are the negatives of the admittances of lines between buses i and j. If there is no line between i and j, this term is zero. Note there is no tilde on y_ij.

Usually, bus 0 is the ground bus, but other buses may be chosen as the reference (for specialized applications). Since bus 0 is commonly chosen as the reference, the notation

${\overline{Y}}_{\mathrm{bus}}={\overline{Y}}_{\mathrm{bus}}^{\left(0\right)}$

is employed to denote the bus admittance matrix referenced to ground. If some other bus is used as the reference, for example bus k, the following notation is used:

${\overline{Y}}_{\mathrm{bus}}^{\left(k\right)}\text{.}$

7.2.1.2 Application Example 7.2: Construction of Bus Admittance Matrix

Construct the bus admittance matrix for the simple 4 bus power system of Fig. E7.2.1.

Solution to Application Example 7.2

The ${\overline{Y}}_{\mathrm{bus}}$ admittance matrix is for Figure E7.2.1

$\left[{\overline{Y}}_{\mathit{bus}}\right]=\left[\begin{array}{l}{y}_{11}{y}_{12}{y}_{13}{y}_{14}\\ y_{21}{y}_{22}{y}_{23}{y}_{24}\\ y_{31}{y}_{32}{y}_{33}{y}_{34}\\ y_{41}{y}_{42}{y}_{43}{y}_{44}\end{array}\right]$

where the entries y_ij of Eq. E7.2-1 are as a function of the admittances ỹ_ij for i = 1…4 and j = 1…4:

y₁₁= ỹ₁₂+ ỹ₁₄, y₁₂=-ỹ₁₂, y₁₃= 0, y₁₄=-ỹ₁₄,

y₂₁=-ỹ₁₂, y₂₂= ỹ₁₂+ ỹ₂₃, y₂₃=-ỹ₂₃, y₂₄= 0,

y₃₁= 0, y₃₂=-ỹ₂₃, y₃₃= ỹ₂₃+ ỹ₃₀, y₃₄= 0,

y₄₁=-ỹ₁₄, y₄₂= 0, y₄₃= 0, y₄₄= ỹ₁₄, and

$-{\displaystyle \sum _{j=1}^4{\tilde{I}}_j=-\left({\tilde{I}}_1+{\tilde{I}}_2+{\tilde{I}}_3+{\tilde{I}}_4\right)={\tilde{I}}_0}\text{.}$

Inspection of the ${\overline{Y}}_{\mathrm{bus}}$ building algorithm reveals several properties of the bus matrix:

• ${\overline{Y}}_{\mathrm{bus}}$ is complex and symmetric;

• ${\overline{Y}}_{\mathrm{bus}}$ is sparse since each bus is connected to only a few nearby buses, and this causes many y_ji = 0 entries; and

• Provided that there is a net nonzero admittance tie to the reference bus (e.g., ground), ${\overline{Y}}_{\mathrm{bus}}$ is nonsingular and may be inverted to find ${\overline{V}}_{\mathrm{bus}}$ from

${\overline{I}}_{\mathrm{bus}}={\overline{Y}}_{\mathrm{bus}}{\overline{V}}_{\mathrm{bus}}\text{,}$

or

${\overline{V}}_{\mathrm{bus}}={\left({\overline{Y}}_{\mathrm{bus}}\right)}^{-1}{\overline{I}}_{\mathrm{bus}}\text{.}$

If there are no ties to the reference bus, ${\overline{Y}}_{\mathrm{bus}}$ is singular and cannot be inverted because no solution to the n × n equation system exists.

7.2.1.3 Application Example 7.3: Building of Nonsingular Bus Admittance Matrix

Construct the bus abmittance matrix for the 3 bus power system of Fig. E7.3.1.

Solution to Application Example 7.3

Note: Off-diagonal entries = sum of negatives of the line admittances

$\begin{array}{c}\begin{array}{c}\hfill y_{21}=y_{12}=–\left(1+j\right)1/\mathrm{\Omega },\hfill \\ \hfill y_{31}=y_{13}=–\left(1+j\right)1/\mathrm{\Omega },\hfill \\ \hfill y_{32}=y_{23}=–\left(1+j\right)1/\mathrm{\Omega },\hfill \end{array}\\ \mathrm{Main}-\mathrm{diagonal}\mathrm{entry}\text{for}\mathrm{bus}1:y_{11}=\left(2+2j\right)1/\mathrm{\Omega },\hfill \\ \mathrm{Main}\mathrm{diagonal}\mathrm{entry}\text{for}\mathrm{bus}2:y_{22}=\left(2+2j\right)1/\mathrm{\Omega },\hfill \\ \mathrm{Main}-\mathrm{diagonal}\mathrm{entry}\text{for}\mathrm{bus}3:y_{33}=\left(3+3j\right)1/\mathrm{\Omega }\text{,}\hfill \end{array}$

Therefore,

${\overline{Y}}_{\mathrm{bus}}=\left[\begin{array}{ccc}\hfill y_{11}\hfill & \hfill y_{12}\hfill & \hfill y_{13}\hfill \\ \hfill y_{21}\hfill & \hfill y_{22}\hfill & \hfill y_{23}\hfill \\ \hfill y_{31}\hfill & \hfill y_{32}\hfill & \hfill y_{33}\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill \left(2+2j\right)\hfill & \hfill -\left(1+j\right)\hfill & \hfill -\left(1+j\right)\hfill \\ \hfill -\left(1+j\right)\hfill & \hfill \left(2+2j\right)\hfill & \hfill -\left(1+j\right)\hfill \\ \hfill -\left(1+j\right)\hfill & \hfill -\left(1+j\right)\hfill & \hfill \left(3+3j\right)\hfill \end{array}\right]\text{.}$

${\overline{Y}}_{\mathrm{bus}}$ is complex and symmetric, but not sparse.

7.2.1.4 Application Example 7.4: Building of Singular Bus Admittance Matrix

Construct the bus abmittance matrix for the simple 4 bus power system of Fig. E7.4.1.

Solution to Application Example 7.4

${\overline{I}}_{\mathrm{bus}}={\overline{Y}}_{\mathrm{bus}}{\overline{V}}_{\mathrm{bus}}=\left[\begin{array}{ccc}\hfill \left(2+2j\right)\hfill & \hfill -\left(1+j\right)\hfill & \hfill -\left(1+j\right)\hfill \\ \hfill -\left(1+j\right)\hfill & \hfill \left(2+2j\right)\hfill & \hfill -\left(1+j\right)\hfill \\ \hfill -\left(1+j\right)\hfill & \hfill -\left(1+j\right)\hfill & \hfill \left(2+2j\right)\hfill \end{array}\right]\left[\begin{array}{c}\hfill {\tilde{V}}_1\hfill \\ \hfill {\tilde{V}}_2\hfill \\ \hfill {\tilde{V}}_3\hfill \end{array}\right]\text{,}$

${\tilde{I}}_1=\left(2+2j\right){\tilde{V}}_1-\left(1+j\right){\tilde{V}}_2-\left(1+j\right){\tilde{V}}_3$

${\tilde{I}}_2=-\left(1+j\right){\tilde{V}}_1+\left(2+2j\right){\tilde{V}}_2-\left(1+j\right){\tilde{V}}_3$

${\tilde{I}}_3=-\left(1+\mathrm{j}\right){\tilde{V}}_1-\left(1+j\right){\tilde{V}}_2+\left(2+2j\right){\tilde{V}}_3$

Add Eqs. E7.4-2 and E7.4-3:

$\left({\tilde{I}}_1+{\tilde{I}}_2\right)=\left(1+j\right){\tilde{V}}_1+\left(1+j\right){\tilde{V}}_2-\left(2+2j\right){\tilde{V}}_3$

or

$\underset{I_3}{\underbrace{\left({\tilde{I}}_1+{\tilde{I}}_2\right)}}=-\left(1+j\right){\tilde{V}}_1-\left(1+j\right){\tilde{V}}_2+\left(2+2j\right){\tilde{V}}_3$

Note that Eq. E7.4-2 + Eq. E7.4-3 = Eq. E7.4-5, which is the same as Eq. E7.4-4. That is, these equations are dependent. One concludes that the matrix is singular and no solution exists for Eqs. E7.4-2 to E7.4-4.

7.2.2 Triangular Factorization

In the United States there exist three power grids: the Eastern, Western, and Texan power systems. Because of stability problems these grids are linked with small (100 MW range) DC tie lines. Nevertheless if even one of the grids is analyzed, the number of buses will be very large: a few thousand buses or more, depending on the detail of modeling. Fortunately, not all buses are mutually linked through a transmission line and for this reason the resulting bus admittance matrix is very sparse, that is, only about 1% of the entire matrix entries are nonzero. However, because of the large dimensions (few thousands) a direct solution (e.g., Gaussian elimination) of the equation system is not feasible and one uses the Newton–Raphson method together with the Jacobian matrix to arrive at the solution via iterations.

To reduce computer memory requirements, sparsity programming is used: it is a digital programming technique whereby sparse matrices are stored in compact forms. The usual basis of sparsity programming techniques is the storage of only nonzero entries. Two broad classes of sparsity programming techniques are

• the entry-row-column storage method, and

• the chained-data-structure method.

Applications of sparsity programming methods are primarily in ${\overline{Y}}_{\mathrm{bus}}$ methods. In this chapter the Jacobian will be used; this matrix is as sparse as the ${\overline{Y}}_{\mathrm{bus}}$ matrix and lends itself to sparsity programming.

In power engineering, as well as in many other branches of engineering, it is necessary to solve a simultaneous set of algebraic linear equations. The general form is the solution of

$\overline{A}\cdot \overline{x}=\overline{b}$

Given are the n by n matrix Ā and the n-dimensional vector $\overline{b}$. Such is the case in the solution of

${\overline{I}}_{\mathrm{bus}}={\overline{Y}}_{\mathrm{bus}}{\overline{V}}_{\mathrm{bus}}$

or

${\overline{V}}_{\mathrm{bus}}={\left[{\overline{Y}}_{\mathrm{bus}}\right]}^{-1}{\overline{I}}_{\mathrm{bus}}$

for ${\overline{V}}_{\mathrm{bus}}$ where the injection currents and the admittance matrix are given. Using the notation of $\overline{A}\cdot \overline{x}=\overline{b}$, let Ā be factored into two matrices called triangular factorization:

$\overline{A}=\overline{L}\cdot \overline{U}$

where $\overline{L}$ is lower left triangular (e.g., l_rc is zero for c > r) and Ū is upper right triangular (e.g., u_rc is zero for r > c). Then

$\overline{A}\cdot \overline{x}=\left[\overline{L}\cdot \overline{U}\right]\overline{x}=\overline{b}$

Let $\left(\overline{U}\cdot \overline{x}\right)$ be a new variable $\overline{w}$, which is an n-dimensional vector; then

$\overline{w}=\overline{U}\cdot \overline{x}$

$\overline{L}\cdot \overline{w}=\overline{b}$

where

$\overline{w}=\left[\begin{array}{c}\hfill w_1\hfill \\ \hfill w_2\hfill \\ \hfill ⋮\hfill \\ \hfill w_n\hfill \end{array}\right],\overline{b}=\left[\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill ⋮\hfill \\ \hfill b_n\hfill \end{array}\right]\text{,}$

$\overline{L}=\left[\begin{array}{cccc}\hfill l_{11}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill .\hfill \\ \hfill l_{21}\hfill & \hfill l_{22}\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill l_{31}\hfill & \hfill l_{32}\hfill & \hfill l_{33}\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \end{array}\right],\overline{U}=\left[\begin{array}{cccc}\hfill u_{11}\hfill & \hfill u_{12}\hfill & \hfill u_{13}\hfill & \hfill .\hfill \\ \hfill 0\hfill & \hfill u_{22}\hfill & \hfill u_{23}\hfill & \hfill .\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill u_{33}\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \end{array}\right]\text{.}$

Note that $\overline{x}$ and $\overline{b}$ correspond to the ${\overline{V}}_{\mathrm{bus}}$ vector and the current vector Ī_bus, respectively.

Solution to Application Example 7.5

$\overline{L}\cdot \overline{U}=\left[\begin{array}{cc}\hfill l_{11}\hfill & \hfill 0\hfill \\ \hfill l_{21}\hfill & \hfill l_{22}\hfill \end{array}\right]\cdot \left[\begin{array}{cc}\hfill u_{11}\hfill & \hfill u_{12}\hfill \\ \hfill 0\hfill & \hfill u_{22}\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill \left(l_{11}{u}_{11}\right)\hfill & \hfill \left(l_{11}{u}_{12}\right)\hfill \\ \hfill \left(l_{21}{u}_{11}\right)\hfill & \hfill \left(l_{21}{u}_{12}+l_{22}{u}_{22}\right)\hfill \end{array}\right]$

Solution to Application Example 7.6

$\begin{array}{ll}\overline{w}=\overline{U}\cdot \overline{x}& =\underset{2\times 2}{\underbrace{\left[\begin{array}{cc}\hfill u_{11}\hfill & \hfill u_{12}\hfill \\ \hfill 0\hfill & \hfill u_{22}\hfill \end{array}\right]}}\underset{2\times 1}{\underbrace{\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]}}\\ & \uparrow \uparrow \uparrow \uparrow \\ & {}_{}^{}\mathit{row}\mathit{column}\mathit{row}\mathit{column}\end{array}$

$\overline{w}=\left[\begin{array}{c}\hfill u_{11}{x}_1+u_{12}{x}_2\hfill \\ \hfill u_{22}{x}_2\hfill \end{array}\right]\text{.}$

Therefore, $\overline{w}$ is a 2 × 1 matrix or a column vector.

7.2.3 Jacobian Matrix

The proposal to use for a partial derivative “$\partial$” instead of “d” stems from Carl Gustav Jacob Jacobi, 1804–1851, a German mathematician. At about the same time as Jacobi, the Russian mathematician Michail Vasilevich Ostrogradski (1801–1862) also introduced the Jacobian of n functions y₁, y₂, …, y_n and n unknowns x₁, x₂, …, x_n as follows:

$\overline{J}=\left[\begin{array}{cccc}\hfill \frac{\partial y_1}{\partial x_1}\hfill & \hfill \frac{\partial y_1}{\partial x_2}\hfill & \hfill \dots \hfill & \hfill \frac{\partial y_1}{\partial x_n}\hfill \\ \hfill \frac{\partial y_2}{\partial x_1}\hfill & \hfill \frac{\partial y_2}{\partial x_2}\hfill & \hfill \dots \hfill & \hfill \frac{\partial y_2}{\partial x_n}\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill \\ \hfill \frac{\partial y_n}{\partial x_1}\hfill & \hfill \frac{\partial y_n}{\partial x_2}\hfill & \hfill \dots \hfill & \hfill \frac{\partial y_n}{\partial x_n}\hfill \end{array}\right]\text{.}$

It is possible to factorize arbitrary nonsingular matrices $\overline{J}$ into $\overline{L}$ and Ū without the need for new computer storage areas: in other words, $\overline{J}$ is destroyed and replaced by the table of factors. The table of factors is approximately as sparse as $\overline{J}$ and, therefore, the table of factors does not require additional storage.

Techniques that result in the destruction of the original data and the replacement of those data by the solution are known as in situ methods. Matrix $\overline{J}$ is factored into $\overline{L}$and Ū in situ and $\overline{J}$ is destroyed and replaced by the table of factors.

The formulas required for the calculation of the table of factors are readily found, noting that

$\overline{L}\cdot \overline{U}=\overline{J}\left(=\overline{A}\right)$

Write $\overline{L}$ and Ū, and find the entries of $\overline{L}$ and Ū, that is, the table of factors:

$\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill .\hfill \\ \hfill l_{21}\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill .\hfill \\ \hfill l_{31}\hfill & \hfill l_{32}\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \end{array}\right]\left[\begin{array}{cccc}\hfill u_{11}\hfill & \hfill u_{12}\hfill & \hfill u_{13}\hfill & \hfill .\hfill \\ \hfill 0\hfill & \hfill u_{22}\hfill & \hfill u_{23}\hfill & \hfill .\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill u_{23}\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \end{array}\right]=\overline{J}=\left[\begin{array}{cccc}\hfill J_{11}\hfill & \hfill J_{12}\hfill & \hfill J_{13}\hfill & \hfill .\hfill \\ \hfill J_{21}\hfill & \hfill J_{22}\hfill & \hfill J_{23}\hfill & \hfill .\hfill \\ \hfill J_{31}\hfill & \hfill J_{32}\hfill & \hfill J_{33}\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \end{array}\right]\text{.}$

Solution to Application Example 7.7

$\begin{array}{c}{J}_{11}=u_{11}\mathrm{or}{u}_{11}=J_{11}\\ J_{12}=u_{12}\mathrm{or}{u}_{12}=J_{12}\\ J_{21}=l_{21}{u}_{11}\mathrm{or}{l}_{21}=J_{21}/u_{11}\\ J_{22}=l_{21}{u}_{12}+u_{22}\mathrm{or}{u}_{22}=J_{22}-l_{21}{u}_{12}\end{array}$

In general,

$\begin{array}{ll}{u}_{11}=& J_{11}\\ u_{12}=& J_{12}\\ ⋮& ⋮\\ u_{1c}=& J_{1c}\\ ⋮& ⋮\\ u_{1n}=& J_{1n}\end{array}$

Since the first row of the table of factors is row 1 of Ū, the first row of the table of factors is simply row 1 of $\overline{J}$.

In row 2 of $\overline{J}$:

$\begin{array}{c}\hfill l_{21}{u}_{11}=J_{21}\hfill \\ \hfill l_{21}{u}_{12}+u_{22}=J_{22}\hfill \\ \hfill l_{21}{u}_{13}+u_{23}=J_{23}\hfill \\ \hfill ......\hfill \end{array}\text{.}$

These equations yield l₂₁, u₂₂, u₂₃, u₂₄, …

$\begin{array}{c}\hfill l_{21}=J_{21}/u_{11}\hfill \\ \hfill u_{22}=J_{22}-l_{21}{u}_{12}\hfill \\ \hfill u_{23}=J_{23}-l_{21}{u}_{13}\hfill \\ \hfill ......\hfill \end{array}\text{.}$

Thus in the expansion of row 2 of $\overline{J}$, one l and (n – 1) u’s are formed. The generalization for the expansion of row r of $\overline{J}$ is straightforward:

$l_{rc}=\frac{J_{rc}-{\displaystyle \sum _{q=1}^{c-1}{l}_{rq}{u}_{qc}}}{u_{\mathit{cc}}},\mathrm{f}\mathrm{o}\mathrm{r}c=1,2,\dots ,\left(r-1\right)$

$u_{rc}=J_{rc}-{\displaystyle \sum _{q=1}^{r-1}{l}_{rq}{u}_{qc}},\mathrm{f}\mathrm{o}\mathrm{r}c=r,r+1,\dots ,n\text{.}$

The rules for the formation of the table of factors in situ are as follows:

1) Row 1 of $\overline{J}$ is not modified.

2) Row 2 of $\overline{J}$ is modified to yield one l and (n – 1) u’s. These entries are found using the above two equations for l_rc and u_rc. For these equations, when the summation upper index is zero, there are no terms in the sum (e.g., ignore the summation term). When the upper and lower indices of summation are the same, there is a single term in the sum.

3) Rule 2 is repeated for each row and $\overline{J}$ is replaced by the table of factors. In row r, there will be (r – 1) l-terms calculated and (n – r + 1) u-terms. The above equations for l_rc and u_rc are used.

7.3.1 Fundamental Bus Admittance Matrix

As discussed and formulated in the previous section, for an n bus system (not including the ground bus, which is the reference bus), the fundamental bus admittance matrix is an n × n matrix:

${\overline{Y}}_{\mathrm{bus}}=\left[\begin{array}{cccc}\hfill y_{11}\hfill & \hfill y_{12}\hfill & \hfill \dots \hfill & \hfill y_{1n}\hfill \\ \hfill y_{21}\hfill & \hfill y_{22}\hfill & \hfill \dots \hfill & \hfill y_{2n}\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill \\ \hfill y_{n1}\hfill & \hfill y_{n2}\hfill & \hfill \dots \hfill & \hfill y_{nn}\hfill \end{array}\right]\text{,}$

where f₁ = 60 Hz and ω₁ = 2πf₁ = 377 rad/s.

The entries of ${\overline{Y}}_{\mathrm{bus}}$ are found as follows:

Main-diagonal entries:

y_kk = sum of line admittances connected to bus k (k = 1, 2, …, n).

Off-diagonal entries:

y_kj = –(line admittance connected between buses k and j).

Since y_kj = y_jk, the matrix ${\overline{Y}}_{\mathrm{bus}}$ is symmetric.

7.3.2 Newton–Raphson Power Flow Formulation

The power (load) flow problem consists of calculations of power flows (real, reactive) and voltages (magnitude, phase) of a network for specified (demand and generation powers) terminal or bus conditions. All voltages and currents are assumed to have fundamental (h = 1, f₁ = 60 Hz) frequency components. A single-phase representation is considered to be adequate because power systems are approximately balanced within all three phases. Associated with each bus (e.g., bus j) are four quantities:

• real power P,

• reactive power Q,

• root-mean-squared (rms) voltage value $\left|\tilde{V}\right|$, and

• phase angle of voltage δ.

Three types of buses are represented and two of the four quantities are specified at each bus:

• Swing bus. It is necessary to select one reference bus (called the swing or slack bus) providing the transmission losses because these are unknown until the final solution is available. The voltage amplitude and its phase angle are given at the swing bus (usually 1.0 per unit (pu) and zero radians, respectively), and all other bus voltage angles are measured with respect to the phase angle of this reference bus. The remaining buses are either PV or PQ buses.

• Voltage controlled (PV) buses. The real power P and the voltage magnitude (amplitude) $\left|\tilde{V}\right|$ are specified for PV buses. These are usually the generation buses at which the injected real power is specified and held fixed by turbine settings (e.g., steam entry valve). A voltage regulator holds $\left|\tilde{V}\right|$ fixed at PV buses by automatically varying the generator field excitation.

• Load (PQ) buses. These are mostly system loads at which real and reactive powers (P and Q, respectively) are specified.

The following notations will be used:

• For a given bus (e.g., the jth bus):

• P_j = real power

• Q_j = reactive power

• $\left|{\tilde{V}}_j\right|$ = rms value of voltage, and

• δ_j = angle of voltage.

• For the swing or slack bus (j = 1):

• $\left|{\tilde{V}}_1\right|$ = 1.0 pu

• δ₁ = 0 radians, and

• It will cover all transmission line losses of the system (“slack”).

• For a PV bus (e.g., the jth bus):

• It is a generation bus

• P_j is given by turbine setting, and

• $\left|{\tilde{V}}_j\right|$ is given by generator excitation.

• For a PQ bus (e.g., the jth bus):

• It is a load bus,

• P_j and Q_j are given by the load demand,

• because power values are given (p = v · i) one obtains an algebraic, nonlinear system of equations.

Fundamental Bus Voltage Vector

For power system modeling at rated frequency, the bus voltage vector is defined as (note: the bar above the x means vector, and t means transpose):

$\overline{x}={\left[{\delta }_2,\left|{\tilde{V}}_2\right|,{\delta }_3,\left|{\tilde{V}}_3\right|,\dots {\delta }_n,\left|{\tilde{V}}_n\right|\right]}^t$

  (7-42)

where δ_j is the phase shift of voltage at bus j with respect to the swing bus voltage phase angle and $\left|{\tilde{V}}_j\right|$ is the voltage rms magnitude at bus j. Note that if bus 1 is the swing bus, $\left|{\tilde{V}}_1\right|$ and δ₁ are known (usually 1.0 per unit and zero radians).

Although power system components are assumed to be linear in the conventional power flow formulation, the problem is a nonlinear one because power p is the product of voltage v and current i. The load flow problem requires the solution of algebraic but nonlinear equations, which are solved using the Newton–Raphson method as will be discussed next.

i) Linear approximation, one-dimensional analysis (Fig. 7.6):

$F\left(x_1\right)=\left(x_1-x_0\right)\frac{dF\left(x_0\right)}{dx}\text{,}$

  (7-43)

where h = (x₁ – x₀), or

$F\left(x_1\right)=h\frac{dF\left(x_0\right)}{dx}\text{,}$

  (7-44)

${\left[\frac{dF\left(x_0\right)}{dx}\right]}^{-1}F\left(x_1\right)=\left(x_1-x_0\right)\text{,}$

  (7-45)

$x_1=x_0+{\left[\frac{dF\left(x_0\right)}{dx}\right]}^{-1}F\left(x_1\right)\text{,}$

  (7-46a)

or

$\begin{array}{cccc}\hfill {\displaystyle \underset{\begin{array}{c}\hfill \uparrow \hfill \\ \hfill \begin{array}{l}\mathit{updated}\mathit{guess}\mathit{after}\\ \mathit{first}\mathit{iteration}\end{array}\hfill \end{array}}{x_1}}\hfill & \hfill {\displaystyle \underset{\begin{array}{c}\hfill \hfill \\ \hfill \hfill \end{array}}{=}}\hfill & \hfill {\displaystyle \underset{\begin{array}{c}\hfill \uparrow \hfill \\ \hfill \mathit{first}\mathit{guess}\hfill \end{array}}{x_0}}\hfill & \hfill +\frac{F\left(x_1\right)}{\left[\frac{dF\left(x_0\right)}{dx}\right]}\hfill \end{array}\text{.}$

  (7-46b)

ii) Taylor series for one variable (Fig. 7.7):

$\begin{array}{l}F\left(x\right)=F\left(x_0\right)+\frac{\left(x-x_0\right)}{1!}{F}^{\prime }\left(x_0\right)+\frac{{\left(x-x_0\right)}^2}{2!}{F}^{″}\left(x_0\right)+\dots \\ +\frac{{\left(x-x_0\right)}^n}{n!}{F}^{\left(n\right)}\left(x_0\right)+\dots \text{,}\end{array}$

  (7-47)

or

$\begin{array}{l}F\left(x_0+h\right)=F\left(x_0\right)\\ +\frac{h}{1!}{F}^{\prime }\left(x_0\right)+\frac{h^2}{2!}{F}^{″}\left(x_0\right)+\dots +\frac{h^n}{n!}{F}^{\left(n\right)}\left(x_0\right)+\dots \text{,}\end{array}$

  (7-48)

$F\left(x_0+h\right)\approx F\left(x_0\right)+\frac{h}{1!}{F}^{\prime }\left(x_0\right)$

  (7-49a)

$F\left(x_1\right)\approx F\left(x_0\right)+\left(x_1-x_0\right)F^{\prime }\left(x_0\right)$

  (7-49b)

$\mathrm{f}\mathrm{o}\mathrm{r}F\left(x_1\right)=0x_1=x_0-\frac{F\left(x_0\right)}{F^{\prime }\left(x_0\right)}\text{.}$

  (7-49c)

iii) Taylor series for two variables (replace “d” by “$\partial$”, Fig. 7.6):

$\begin{array}{l}F\left(x_0+h,y_0+k\right)=F\left(x_0,y_0\right)+\left\{\frac{\partial F\left(x_0,y_0\right)}{\partial x}\right.h\left.+\frac{\partial F\left(x_0,y_0\right)}{\partial y}k\right\}\\ +\frac{1}{2}\left\{\frac{{\partial }^{2}F\left(x_0,y_0\right)}{\partial x^2}\right.h^2+\left.2\frac{{\partial }^{2}F\left(x_0,y_0\right)}{\partial x\partial y}hk+\frac{{\partial }^{2}F\left(x_0,y_0\right)}{\partial y^2}{k}^2\right\}\\ +\frac{1}{6}\left\{\dots \right\}+\dots +\frac{1}{h!}\left\{\dots \right\}+R_n\text{,}\end{array}$

  (7-50)

where R_n is the remainder.

Equation 7-50 can be written in matrix form neglecting higher order terms:

$\begin{array}{l}F\left(x_0+h,y_0+k\right)=F\left(x_0,y_0\right)\\ +\left[\frac{\partial F\left(x_0,y_0\right)}{\partial x}\frac{\partial F\left(x_0,y_0\right)}{\partial y}\right]\left[\begin{array}{c}\hfill k\hfill \\ \hfill k\hfill \end{array}\right]\text{,}\end{array}$

  (7-51)

where

$h=\left(x_1-x_0\right)\mathrm{and}k=\left(y_1-y_0\right)$

or, for F(x₁, y₁) = 0,

$\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill y_1\hfill \end{array}\right]=\left[\begin{array}{c}\hfill x_0\hfill \\ \hfill y_0\hfill \end{array}\right]-\frac{F\left(x_0,y_0\right)}{\left[\frac{\partial F\left(x_0,y_0\right)}{\partial x}\frac{\partial F\left(x_0,y_0\right)}{\partial y}\right]}\text{,}$

  (7-52)

or

$\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill y_1\hfill \end{array}\right]=\left[\begin{array}{c}\hfill x_0\hfill \\ \hfill y_0\hfill \end{array}\right]-{\left[\frac{\partial F\left(x_0,y_0\right)}{\partial x}\frac{\partial F\left(x_0,y_0\right)}{\partial y}\right]}^{-1}F\left(x_0,y_0\right)\text{.}$

  (7-53)

Therefore,

$\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill y_1\hfill \end{array}\right]=\left[\begin{array}{c}\hfill x_0\hfill \\ \hfill y_0\hfill \end{array}\right]-{\left(\overline{J}\right)}^{-1}F\left(x_0,y_0\right)\text{,}$

  (7-54)

where the two-dimensional Jacobian $\overline{J}$ is

$\overline{J}=\left[\frac{\partial F\left(x_0,y_0\right)}{\partial x}\frac{\partial F\left(x_0,y_0\right)}{\partial y}\right]\text{.}$

  (7-55)

Residuals of an Equation System

Given the equation system:

$\begin{array}{c}{x}_1-2x_2-4x_3=0\\ 3x_1+x_2+7x_3=0.\\ 7x_1-x_2+x_3=0\end{array}$

The approximate (guessed) solutions are x₁₀, x₂₀, x₃₀. Introducing approximate solutions in the equation system:

$x_{10}–2x_{20}–4x_{30}=\mathrm{residual}1$

$3x_{10}+x_{20}+7x_{30}=\mathrm{residual}2$

$7x_{10}–x_{20}+x_{30}=\mathrm{residual}3$

If ${\displaystyle \sum _{i=1}^3\left|\mathrm{residual}i\right|=0}$, then solution of the equation system has been obtained.

Fundamental Mismatch Power

The power flow algorithm is based on the Newton–Raphson approach forcing the mismatch powers (at system buses) to zero. The mismatch powers (real and reactive powers) correspond to the residuals of the equation system as discussed above. Specifically the term “mismatch power” at bus i refers to the summation of the complex powers leaving via lines connected to bus i and the specified complex load (demand) power at this bus. Note that the bus itself cannot produce or dissipate power and therefore the mismatch power for a given bus must be zero. Therefore, the mismatch power vector is

$\mathrm{\Delta }\overline{W}={\left[P_2+F_{r,2},Q_2+F_{i,2},\dots P_n+F_{r,n},Q_n+F_{i,n}\right]}^t\text{,}$

  (7-56)

where P_j and Q_j are the real and reactive load powers at bus j, respectively, and F_r,j and F_i,j are the real and reactive line powers computed later in this section. Figure 7.8 illustrates the definition of power at a given bus. Note the following:

• P₂ > 0 demand (load) of real power,

• P₂ < 0 generation of real power,

• Q₂ > 0 demand (inductive load) of reactive power, and

• Q₂ < 0 generation (capacitive load) of reactive power.

For an n-bus system consisting of the swing bus and PQ buses (inclusion of PV buses is considered later), the mismatch power vector has 2(n – 1) rows corresponding to buses 2 through n (bus 1 is assumed to be the swing bus). Clearly $\mathrm{\Delta }\overline{W}$ is a function of $\overline{x}$, the bus voltage vector (Eq. 7-42).

The Fundamental Newton–Raphson Power Flow Algorithm

The power flow problem simplifies to finding $\overline{x}$ such that

$\mathrm{\Delta }\overline{W}\left(\overline{x}\right)=0\text{.}$

  (7-57)

Let ${\overline{x}}_s$ be the bus solution vector and expand Eq. 7-57 in a Taylor series about $\overline{x}$ = ${\overline{x}}_s$ where ${\overline{x}}_s$ is the (guessed) solution (see Eqs. 7-46 and 7-52):

$\mathrm{\Delta }\overline{W}\left(\overline{x}\right)=\mathrm{\Delta }\overline{W}\left({\overline{x}}_s\right)+\overline{J}\left(\overline{x}-{\overline{x}}_s\right)\text{,}$

  (7-58)

or

${\overline{J}}^{-1}\mathrm{\Delta }\overline{W}\left(\overline{x}\right)={\overline{J}}^{-1}\mathrm{\Delta }\overline{W}\left({\overline{x}}_s\right)+\left(\overline{x}-{\overline{x}}_s\right)\text{,}$

  (7-59)

where $\overline{J}$ is the Jacobian.

With $\mathrm{\Delta }\overline{W}\left(\overline{x}\right)$ = 0 because $\overline{x}$ is the (converged) solution vector,

$\overline{x}={\overline{x}}_s-{\overline{J}}^{-1}\mathrm{\Delta }\overline{W}\left({\overline{x}}_s\right)\text{.}$

  (7-60)

Due to the linearization (neglect of higher order terms), the solution ${\overline{x}}_s$ of Eq. 7-60 is an approximation which only can be improved by iterations. Therefore,

${\overline{x}}^{\xi +1}={\overline{x}}^{\xi }-{\overline{J}}^{-1}\mathrm{\Delta }\overline{W}\left({\overline{x}}^{\xi }\right)$

  (7-61)

or with

$\mathrm{\Delta }{\overline{x}}^{\xi }={\overline{J}}^{-1}\mathrm{\Delta }\overline{W}\left({\overline{x}}^{\xi }\right)$

  (7-62)

${\overline{x}}^{\xi +1}={\overline{x}}^{\xi }-\mathrm{\Delta }{\overline{x}}^{\xi }$

  (7-63)

where superscript ξ indicates the iteration number and $\overline{J}$ is a 2(n – 1) × 2(n – 1) matrix consisting of partial derivatives of mismatch powers with respect to the bus voltages. Note that all quantities are complex (thus the 2 in front of (n – 1)) and the swing bus voltage (magnitude and phase angle) is known (therefore n – 1). If $\mathrm{\Delta }\overline{W}\left({\overline{x}}_s\right)$ = 0 then

$\overline{x}={\overline{x}}_s-{\overline{J}}^{-1}\mathrm{\Delta }\overline{W}\left({\overline{x}}_s\right)={\overline{x}}_s\text{.}$

  (7-64)

The evaluation of Eq. 7-61 (the solution) is done by computer using triangular factorization and forward/backward substitution. This works for a few thousands of buses quite well. In case of hand calculations, where there are just a few buses, we use matrix inversion.