1. Problem formulation

Power flow analysis is a fundamental study discussed in any power system analysis textbook such as [1–6]. The objective of a power flow study is to calculate the voltages (magnitude and angle) for a given load, generation, and network condition. Once voltages are known for all buses, line flows and losses can be calculated. The starting point of solving power flow problems is to identify the known and unknown variables in the system. Based on these variables, buses are classified into three types: slack, generation, and load buses as shown in Table 1.

The slack bus is required to provide the mismatch between scheduled generation the total system load including losses and total generation. The slack bus is commonly considered as the reference bus because both voltage magnitude and angles are specified; therefore, it is called the swing bus. The rest of generator buses are called regulated or PV buses because the net real power is specified and voltage magnitude is regulated. Most of the buses in practical power systems are load buses. Load buses are called PQ buses because both net real and reactive power loads are specified.

For PQ buses, both voltage magnitudes and angles are unknown, whereas for PV buses, only the voltage angle is unknown. As both voltage magnitudes and angles are specified for the Slack bus, there are no variables that must be solved for. In a system with n buses and g generators, there are 2(n-1)-(g-1) unknowns. To solve these unknowns, real and reactive power balance equations are used. To write these equations, the transmission network is modeled using the admittance matrix (Y-bus).


Table 1.

Type of buses in the power flow problem.

#### 2. Admittance matrix and power flow equation

The admittance matrix of a power system is an abstract mathematical model of the system. It consists of admittance values of both lines and buses. The Y-bus is a square matrix with dimensions equal to the number of buses. This matrix is symmetrical along the diagonal.

$$Y = \begin{bmatrix} Y\_{11} & \cdots & Y\_{1n} \\ \vdots & \ddots & \vdots \\ Y\_{n1} & \cdots & Y\_{nn} \end{bmatrix} \tag{1}$$

The values of diagonal elements (Yii) are equal to the sum of the admittances connected to bus i. The off-diagonal elements (Yij) are equal to the negative of the admittance connecting the two buses i and j. It is worth noting that with large systems, Y-bus is a sparse matrix.

$$Y\_{ii} = \sum\_{\substack{j=0\\j\neq i}}^n y\_{ij} \tag{2}$$

$$Y\_{\vec{\eta}} = Y\_{\vec{\mu}} = -\mathcal{y}\_{\vec{\eta}} \tag{3}$$

The net injected power at any bus can be calculated using the bus voltage (Vi), neighboring bus voltages (Vj), and admittances between the bus and its neighboring buses (yij) as shown in Figure 1.

$$I\_i = V\_i y\_{i0} + (V\_i - V\_1)y\_{i1} + (V\_i - V\_2)y\_{i2} + \dots + (V\_i - V\_j)y\_{ij}$$

Rearranging the elements as a function of voltages, the current equation becomes as follows:

$$I\_i = V\_i \left(\mathbf{y}\_{i0} + \mathbf{y}\_{i1} + \mathbf{y}\_{i2} + \dots + \mathbf{y}\_{ij}\right) - V\_1 \mathbf{y}\_{i1} - V\_2 \mathbf{y}\_{i2} - \dots - V\_j \mathbf{y}\_{ij}$$

$$I\_i = V\_i \sum\_{\substack{j=0 \\ j \neq i}} \mathbf{y}\_{ij} - \sum\_{\substack{j=1 \\ j \neq i}} \mathbf{y}\_{ij} \mathbf{V}\_j = V\_i Y\_{ii} + \sum\_{\substack{j=1 \\ j \neq i}} \mathbf{Y}\_{ij} \mathbf{V}\_j$$

The power equation at any bus can be written as follows:

$$\mathcal{S}\_i = P\_i + jQ\_i = V\_i I\_i^\*$$

Power Flow Analysis DOI: http://dx.doi.org/10.5772/intechopen.83374

Figure 1. Net injected power.

Or

$$\mathbf{S}\_i{}^\* = P\_i - j\mathbf{Q}\_i = \mathbf{V}\_i{}^\* I\_i$$

Substituting the expression on the current in Si <sup>∗</sup> equation results in the following formula:

$$\mathbf{S}\_i^\* = \mathbf{V}\_i^\* \left( \mathbf{V}\_i \sum\_{\substack{j-0\\j \neq i}} \mathbf{y}\_{ij} - \sum\_{\substack{j-1\\j \neq i}} \mathbf{y}\_{ij} \mathbf{V}\_j \right) = \mathbf{V}\_i^\* \left( \mathbf{V}\_i \mathbf{Y}\_{ii} + \sum\_{\substack{j-1\\j \neq i}} \mathbf{Y}\_{ij} \mathbf{V}\_j \right)$$

Real and reactive power can be calculated from the following equations:

$$\begin{aligned} P\_i &= \operatorname{Re} \left\{ V\_i^\* \left( V\_i \sum\_{\substack{j=0 \\ j \neq i}} y\_{ij} - \sum\_{\substack{j=1 \\ j \neq i}} y\_{ij} V\_j \right) \right\} = \operatorname{Re} \left\{ V\_i^\* \left( V\_i Y\_{ii} + \sum\_{\substack{j=1 \\ j \neq i}} Y\_{ij} V\_j \right) \right\} \\ Q\_i &= -\operatorname{Im} \left\{ V\_i^\* \left( V\_i \sum\_{\substack{j=0 \\ j \neq i}} y\_{ij} - \sum\_{\substack{j=1 \\ j \neq i}} y\_{ij} V\_j \right) \right\} = -\operatorname{Im} \left\{ V\_i^\* \left( V\_i Y\_{ii} + \sum\_{\substack{j=1 \\ j \neq i}} Y\_{ij} V\_j \right) \right\} \end{aligned}$$

Or

$$P\_i = \sum\_{j=1}^{n} |V\_i| \left| V\_j \right| \left| Y\_{ij} \right| \cos \left( \theta\_{ij} - \delta\_i + \delta\_j \right) \tag{4}$$

$$Q\_i = -\sum\_{j=1}^{n} |V\_i| \left| V\_j \right| \left| Y\_{ij} \right| \sin \left( \theta\_{\vec{y}} - \delta\_i + \delta\_j \right) \tag{5}$$

And the current (Ii) can be written as a function of the power as follows:

$$\frac{P\_i - j\mathbb{Q}\_i}{V\_i^\*} = V\_i \sum\_{\substack{j=0 \\ j \neq i}} \mathbb{y}\_{ij} - \sum\_{\substack{j=1 \\ j \neq i}} \mathbb{y}\_{ij} V\_j = V\_i Y\_{ii} + \sum\_{\substack{j=1 \\ j \neq i}} Y\_{ij} V\_j \tag{6}$$

Figure 2. Impedance diagram.

Figure 3. Admittance diagram.

Example 1: Admittance matrix formation.

For the below 4-bus system in Figure 2, the admittance matrix is constructed by converting all impedances in the system into admittances as shown in Figure 3. Then, diagonal and off-diagonal elements are calculated using Eqs. (2) and (3).

$$Y = \begin{bmatrix} -j7.5 & j4 & j2.5 & 0\\ j4 & -j7.75 & 0 & j2.5\\ j2.5 & 0 & -j4.5 & j2\\ 0 & j2.5 & j2 & -j4.5 \end{bmatrix} = \begin{bmatrix} 7.5\angle{-90^{\circ}} & 4\angle{90^{\circ}} & 2.5\angle{90^{\circ}} & 0\\ 4\angle{90^{\circ}} & 7.75\angle{-90^{\circ}} & 0 & 2.5\angle{90^{\circ}}\\ 2.5\angle{90^{\circ}} & 0 & 4.5\angle{-90^{\circ}} & 2\angle{90^{\circ}}\\ 0 & 2.5\angle{90^{\circ}} & 2\angle{90^{\circ}} & 4.5\angle{-90^{\circ}} \end{bmatrix}$$

#### 3. Gauss-Seidel technique

The Gauss-Seidel (GS) method, also known as the method of successive displacement, is the simplest iterative technique used to solve power flow problems. In general, GS method follows the following iterative steps to reach the solution for the function f xð Þ¼ 0:

• Rearrange the function into the form x ¼ g xð Þ to calculate the unknown variable.


The rate of convergence can be improved using acceleration factors by modifying the step size α.

$$\boldsymbol{\mathfrak{x}}^{[k+1]} = \boldsymbol{\mathfrak{x}}^{[k]} + \alpha \left[ \mathbf{g} \left( \boldsymbol{\mathfrak{x}}^{[k+1]} \right) - \boldsymbol{\mathfrak{x}}^{[k]} \right] \tag{7}$$

In the context of a power flow problem, the unknown variables are voltages at all buses, but the slack. Both voltage magnitudes and angles are unknown for load buses, whereas voltage angles are unknown for regulated/generation buses.

The voltage Vi at bus i can be calculated using either equations:

$$\begin{aligned} V\_i &= \frac{1}{\sum\_{j} \mathbf{1}\_{j\bar{j}}} \left( \frac{\mathbf{P}\_i^{sch} - j\mathbf{Q}\_i^{sch}}{V\_i^\*} + \sum\_{j} \mathbf{y}\_{i\bar{j}} \mathbf{V}\_j \right) \\ &\quad j \neq i \end{aligned} \tag{8}$$

$$\mathbf{V}\_i = \frac{\mathbf{1}}{\mathbf{Y}\_{ii}} \left( \frac{\mathbf{P}\_i^{sch} - j\mathbf{Q}\_i^{sch}}{\mathbf{V}\_i^\*} - \sum\_{\substack{j=1 \\ j \neq i}} \mathbf{Y}\_{i\bar{j}} \mathbf{V}\_j \right) \tag{9}$$

where yij is the admittance between buses i and j, Yij is the Y-bus element, Pi sch the net scheduled injected real power, Q <sup>i</sup> sch is the net scheduled injected reactive power, and Vi <sup>∗</sup> is the conjugate of Vi. The net injected quantities are the sum of the generation minus load. Typically, the initial estimates of Vi <sup>¼</sup> <sup>1</sup>∠0° .

The iterative voltage equation is as follows:

$$\boldsymbol{V}\_{i}\boldsymbol{V}\_{i}^{[k+1]} = \frac{\mathbf{1}}{\mathbf{Y}\_{ii}} \left( \frac{\mathbf{P}\_{i}^{\text{sch}} - \boldsymbol{j}\mathbf{Q}\_{i}^{\text{sch}}}{\mathbf{V}\_{i}^{\ast [k]}} - \sum\_{j \neq i} \mathbf{Y}\_{ij} \mathbf{V}\_{j}^{[k \text{ or } k+1]} \right) \tag{10}$$

Or

$$\mathbf{V}\_{i}^{[k+1]} = \frac{\mathbf{1}}{\sum\_{j=0} \mathbf{y}\_{ij}} \left( \frac{P\_{i}^{sch} - jQ\_{i}^{sch}}{\mathbf{V}\_{i}^{\*}} + \sum\_{j \neq i} \mathbf{y}\_{ij} V\_{j}^{[k \text{ or } k+1]} \right) \tag{11}$$

Both real and reactive powers are scheduled for the load buses, and Eq. 6 is used to determine both voltage magnitudes and angles ( Vi j j∠δ) for every iteration (Vi ½ � <sup>k</sup>þ<sup>1</sup> ).

For regulated buses, only real power is scheduled. Therefore, net injected reactive power is calculated based on the iterative voltages (Vi ½ � <sup>k</sup>þ<sup>1</sup> ) using either equations:

$$\mathcal{Q}\_i^{\left[k+1\right]} = -Im\left\{ V\_i^{\*\left[k\right]} \left( V\_i^{\left[k\right]} \sum\_{j=0}^n \mathcal{Y}\_{ij} - \sum\_{j=1 \atop j \neq i}^n \mathcal{Y}\_j^{\left[k\right] \text{ or } k+1\right]} \right) \right\} \tag{12}$$

$$\left|Q\_{i}\right|^{[k+1]} = -\sum\_{j=1}^{n} \left|V\_{i}\right|^{[k]} \left|V\_{j}\right|^{[k\text{ or }k+1]} \left|Y\_{ij}\right| \sin\left(\theta\_{ij} - \delta\_{i}^{[k]} + \delta\_{j}^{[k]}\right) \tag{13}$$

where Vi j j and Vj � � � � are the magnitudes of the voltage at buses i and j, respectively. δ<sup>i</sup> and δ<sup>j</sup> are the associated angles. yij is the admittance between buses i and j. Yij � � � � is the magnitude of the Y-bus element between the two buses; and θij is the corresponding angle.

Since the voltage magnitude ( Vi j j) is specified at regulated/PV buses, Eqs. (8) or (9) will be used to determine the voltage angles only. To achieve this, two options can be used:


$$Re\_i^{new} = \sqrt{\left|V\_i\right|^2 - \left|m\_i\right|^2} \tag{14}$$

The iterative process stops when the voltage improvement reaches acceptable limits: Vi ½ � <sup>k</sup>þ<sup>1</sup> � Vi ½ � <sup>k</sup> � � � �≤ ϵ.

Example 2: Gauss-Seidel power flow solution.

Figure 4 below shows a 3-bus system. Perform 2 iterations to obtain the voltage magnitude and angles at buses 2 and 3. Impedances are given on 100 MVA base. Solution:

The admittance values of the transmission network and the injected power in per unit at buses 2 and 3 are calculated as shown in Figure 5. Note that net injected power at the load bus is negative while that of the PV bus is positive. Per units values are obtained by diving actual values (MW and MVAR) by the base (100 MVA).

Iteration #1: assume V<sup>2</sup> ½ � <sup>0</sup> <sup>¼</sup> <sup>1</sup>:00∠0° and <sup>V</sup><sup>3</sup> ½ � <sup>0</sup> <sup>¼</sup> <sup>1</sup>:<sup>03</sup> <sup>∠</sup>0° .

$$V\_{2}|{\!|{\!}} = \frac{1}{\mathcal{Y}\_{21} + \mathcal{Y}\_{23}} \left( \frac{P\_{2}{\!{\!}}^{\ \!{\!}} - jQ\_{2}{\!\!{\!}}^{\ \!{\!}}}{V\_{2}{\!\!{\!\!{\!}}}^{\ \!{\!\!{\!}}}} + \mathcal{Y}\_{21}V\_{1} + \mathcal{Y}\_{23}V\_{3}{\!\!{\!{\!}}}^{\ \!{\!\!{\!}}} \right)$$

$$= \frac{1}{(5 - j15) + (15 - j50)} \left( \frac{-2 + j0.5}{1.00 \angle 0} + (5 - j15)1.02 \angle 0^{\ \!{\!{\!}}} + (15 - j50)1.00 \angle 0^{\ \!{\!{\!{\!}}}} \right)$$

$$V\_{2}{\!\!{\!{\!{\!{\!{\!}}}}}} = 1.0120 - j0.0260 = 1.0123 \angle -1.4717^{\prime}$$

As Q3 is not given, it is calculated based on the latest available information using Eqs. (12) or (13).

$$\begin{aligned} Q\_3^{[1]} &= -Im \left\{ V\_3^\* {}^{[0]} \left( V\_3 {}^{[0]} \sum\_{j=0}^n \mathcal{Y}\_{3j} - \sum\_{j=1 \atop j \neq i}^n \mathcal{Y}\_{3j} V\_j {}^{[k \text{ or } k+1]} \right) \right\} \\ Q\_3^{[1]} &= -Im \left\{ V\_3^\* {}^{[0]} \left( V\_3 {}^{[0]} \left( \mathcal{Y}\_{31} + \mathcal{Y}\_{32} \right) - \left[ \mathcal{Y}\_{31} V\_1 + \mathcal{Y}\_{32} V\_2 {}^{[1]} \right] \right) \right\} \end{aligned}$$

Power Flow Analysis DOI: http://dx.doi.org/10.5772/intechopen.83374

Figure 4. 3-Bus power system.

Figure 5. Power flow input data.

$$\begin{split} Q\_3^{[1]} &= -Im\{\mathbf{1}.03 \angle 0^{\circ} (\mathbf{1}.03 \angle 0^{\circ} (10 - j40 + 15 - j50) - [(10 - j40)\mathbf{1}.02 \angle 0^{\circ}]) \\ &+ (\mathbf{15} - j50)(\mathbf{1}.0120 - j0.0260)] \} \\ &= -Im\{\mathbf{1}.7201 - j0.9373\} = \mathbf{0.9373} \,\mathrm{pu} \end{split}$$

Now that Q<sup>3</sup> ½ � <sup>1</sup> is calculated, the voltage V<sup>3</sup> ½ � <sup>1</sup> can be calculated:

$$\begin{split} V\_{3}|^{[1]} &= \frac{1}{\mathcal{Y}\_{31} + \mathcal{Y}\_{32}} \left( \frac{P\_{3}^{\text{self}} - jQ\_{3}^{[1]}}{V\_{3}^{[0]\uparrow}} + \mathcal{Y}\_{31}V\_{1} + \mathcal{Y}\_{32}V\_{2}^{[1]} \right) \\ &= \frac{1}{10 - j40 + 15 - j50} \left( \frac{1.5 - j0.9373}{1.03 \angle 0^{\circ}} + (10 - j40) \text{1.02 } \angle 0^{\circ} + (15 - j50)(1.0120 - j0.0260) \right) \\ &= 1.0294 - j0.0022 \, pu \end{split} $$

Since the magnitude of V<sup>3</sup> is specified, we retain the imaginary part of V<sup>3</sup> ½ � <sup>1</sup> and calculate the real part using Eq. (14).

$$Re\_i^{new} = \sqrt{\mathbf{1.03^2} - \mathbf{0.0022^2}} = \mathbf{1.03} \,\mu\nu$$

Therefore,

$$V\_3^{[1]} = V\_3^{[1]} = \mathbf{1.03} - j0.0022 = \mathbf{1.03} \angle -0.1226^\circ pu$$

Iteration #2: considering V<sup>2</sup> ½ � <sup>1</sup> <sup>¼</sup> <sup>1</sup>:<sup>0120</sup> � <sup>j</sup>0:0260 and <sup>V</sup><sup>3</sup> ½ � <sup>1</sup> <sup>¼</sup> <sup>1</sup>:<sup>03</sup> � <sup>j</sup>0:0022.

$$\begin{split} V\_{2}|^{[2]} &= \frac{1}{y\_{21} + y\_{23}} \left( \frac{P\_{2}^{\text{self}} - jQ\_{2}^{\text{self}}}{V\_{2}|^{[1]}} + y\_{21}V\_{1} + y\_{23}V\_{3}|^{[1]} \right) \\ &= \frac{1}{(5 - j15) + (15 - j50)} \left( \frac{-2 + j0.5}{1.0120 - j0.0260} + (5 - j15) 1.02 \angle 0^{\circ} + (15 - j50)(1.03 - j0.0022) \right) V\_{2}|^{[2]} \\ &= 1.0115 - j0.0270 = 1.0119 \angle -1.5273 \end{split}$$

Q<sup>3</sup> ½ � <sup>2</sup> calculation is given below:

$$\begin{aligned} Q\_3^{[2]} &= -Im\left\{ V\_3^{\*[1]} \left( V\_3^{[1]} (\mathcal{y}\_{31} + \mathcal{y}\_{32}) - \left[ \mathcal{y}\_{31} V\_1 + \mathcal{y}\_{32} V\_2^{[2]} \right] \right) \right\} \\ Q\_3^{[2]} &= -Im\{ (1.03 - j0.0022)((1.03 - j0.0022)(10 - j40 + 15 - j50)) \} \\ &- \left[ (10 - j40) \mathbf{1}.02 \angle \mathbf{0}^\circ + (15 - j50)(\mathbf{1}.0115 - j\ \mathbf{0}.0270) \right] \right) \\ &= \mathbf{1}.0000 \ p u \end{aligned}$$

The voltage V<sup>3</sup> ½ � <sup>2</sup> is calculated as follows:

$$\begin{split} |V\_{3}|^{[2]} &= \frac{1}{\mathcal{Y}\_{31} + \mathcal{Y}\_{32}} \left( \frac{P\_{3}^{\text{sch}} - jQ\_{3}^{[2]}}{V\_{3}^{[1]} ^ \*} + \mathcal{Y}\_{31}V\_{1} + \mathcal{Y}\_{32}V\_{2} ^ [2]} \right) \\ &= \frac{1}{10 - j40 + 15 - j50} \left( \frac{1.5 - j1}{1.03 \angle 0^{\circ}} + (10 - j40) \text{1.02 } \angle 0^{\circ} + (15 - j50)(1.0115 - j0.0270) \right) \\ &= 1.0298 - j0.0030 \text{ } pu \end{split} $$

Only imaginary value of the calculated V<sup>3</sup> ½ � <sup>2</sup> is retained and the real part is calculated based on the retained imaginary values and the scheduled j j V<sup>3</sup> :

$$Re\_i^{new} = \sqrt{\mathbf{1.03^2 - 0.0030^2}} = \mathbf{1.03 pu}$$

Therefore,

$$V\_3^{[2]} = \mathbf{1.03} - j0.003 = \mathbf{1.03} \angle -0.1644^\circ \,\mu\mathrm{u}$$

The iterative solution is presented in Table 2.


Table 2. Gauss-Seidel iterative solution.

## 4. Newton-Raphson technique

The Newton-Raphson (N-R) technique, also known as the method of successive approximation, is based on Taylor's expansion approximation. The unknown x in the function f xð Þ¼ c can be determined using Taylor's expansion approximation. Starting with an initial estimate x½ � <sup>0</sup> , the deviation from the correct solution is Δx½ � <sup>0</sup> . Applying Taylor's expansion, the function can be written as follows:

$$f\left(\mathbf{x}^{[0]} + \Delta \mathbf{x}^{[0]}\right) = \mathbf{c} \tag{15}$$

$$f\left(\mathbf{x}^{[0]} + \Delta \mathbf{x}^{[0]}\right) = f\left(\mathbf{x}^{[0]}\right) + f'\left(\mathbf{x}^{[0]}\right)\Delta \mathbf{x}^{[0]} + \frac{1}{2!}f''\left(\mathbf{x}^{[0]}\right)\Delta \mathbf{x}^{[0]^2} + \frac{1}{3!}f'''\left(\mathbf{x}^{[0]}\right)\Delta \mathbf{x}^{[0]^3} + \dots = c \tag{16}$$

Assuming Δx½ � <sup>0</sup> is small, the higher order terms ( <sup>1</sup> 2! f 00 x½ � <sup>0</sup> � �Δx½ � <sup>0</sup> <sup>2</sup> þ 1 3! f 000 x½ � <sup>0</sup> � �Δx½ � <sup>0</sup> <sup>3</sup> þ …) are neglected and the function can be approximated by the first two terms.

$$f\left(\mathbf{x}^{[0]} + \Delta \mathbf{x}^{[0]}\right) \approx f\left(\mathbf{x}^{[0]}\right) + f'\left(\mathbf{x}^{[0]}\right) \Delta \mathbf{x}^{[0]} = \mathbf{c} \tag{17}$$

Based on x½ � <sup>0</sup> , the deviation from the correct solution can be iteratively calculated.

$$
\Delta \mathbf{x}^{[0]} = \frac{\mathbf{c} - f\left(\mathbf{x}^{[0]}\right)}{f'\left(\mathbf{x}^{[0]}\right)} = \frac{\Delta f\left(\mathbf{x}^{[0]}\right)}{f'\left(\mathbf{x}^{[0]}\right)}\tag{18}
$$

The improved solution can be calculated iteratively.

$$\mathfrak{x}^{[1]} = \mathfrak{x}^{[0]} + \Delta \mathfrak{x}^{[0]} \tag{19}$$

The iterative process is stopped when the mismatch between scheduled and calculated value (Δf ½ � <sup>k</sup> <sup>¼</sup> <sup>c</sup> � f x½ � <sup>k</sup> � �) is within acceptable limits <sup>Δ</sup><sup>f</sup> ½ � <sup>k</sup> � � � � � �<sup>≤</sup> <sup>ϵ</sup>.

In the context of power flow problems, unknown variables x are both voltage magnitude and angles ( Vi j j∠ δi) at load buses, as well as voltage angles (δi) at regulated buses. The scheduled (specified) quantities (c) are both net real (Pi sch) and reactive power (jQ <sup>i</sup> sch) values at load buses and real power at generation buses as shown in Table 1. The iterative values of reactive power are calculated using Eqs. (12) or (13). Similarly, the iterative values of real power are calculated using Eqs. (20) or (21):

$$P\_i^{\left[k+1\right]} = \operatorname{Re}\left\{ \left. \boldsymbol{V}\_i^{\*\left[k\right]} \left( \boldsymbol{V}\_i^{\left[k\right]} \sum\_{j=0}^n \boldsymbol{\mathcal{y}}\_{ij} - \sum\_{j=1 \atop j \neq i}^n \boldsymbol{\mathcal{y}}\_{ij} \boldsymbol{V}\_j^{\left[k\right] \text{ or } k+1\right]} \right) \right\} \tag{20}$$

$$P\_i^{\left[k+1\right]} = \sum\_{j=1}^n \left| V\_i \right|^{\left[k\right]} \left| V\_j \right|^{\left[k \text{ or } k+1\right]} \left| Y\_{ij} \right| \cos \left( \theta\_{\vec{\eta}} - \delta\_i^{\left[k\right]} + \delta\_j^{\left[k\right]} \right) \tag{21}$$

The Newton-Raphson power flow formulation can be written as follows:

$$\boldsymbol{\alpha}^{[k]} = \begin{bmatrix} \boldsymbol{\delta}\_{i}^{[k]} \\ \boldsymbol{|V\_{i}|}^{[k]} \end{bmatrix}, \boldsymbol{\varepsilon} = \begin{bmatrix} \boldsymbol{P}\_{i}^{\operatorname{sch}} \\ \boldsymbol{Q}\_{i}^{\operatorname{sch}} \end{bmatrix}, \boldsymbol{\mathcal{f}} = \begin{bmatrix} \boldsymbol{P}\_{i}^{[k]} \\ \boldsymbol{Q}\_{i}^{[k]} \end{bmatrix}, \text{and} \begin{bmatrix} \boldsymbol{P}\_{i}^{\operatorname{sch}} \\ \boldsymbol{Q}\_{i}^{\operatorname{sch}} \end{bmatrix} - \begin{bmatrix} \boldsymbol{P}\_{i}^{[k]} \\ \boldsymbol{Q}\_{i}^{[k]} \end{bmatrix} = \begin{bmatrix} \boldsymbol{\Delta} \boldsymbol{P}\_{i}^{[k]} \\ \boldsymbol{\Delta} \boldsymbol{Q}\_{i}^{[k]} \end{bmatrix}.$$

When more variables are used, the derivative f <sup>0</sup> is replaced by partial derivatives with respect to different variables. The partial derivative matrix is called the Jacobian matrix.

$$f' = \begin{bmatrix} \frac{\partial P\_i}{\partial \delta\_i} & \frac{\partial P\_i}{\partial |V\_i|}\\ \frac{\partial Q\_i}{\partial \delta\_i} & \frac{\partial Q\_i}{\partial |V\_i|} \end{bmatrix} = \begin{bmatrix} J\_{P\delta} & J\_{P|V|}\\ J\_{Q\delta} & J\_{Q|V|} \end{bmatrix} \tag{22}$$

Therefore, the Newton-Raphson power flow formulation can be solved using the below equation:

$$
\begin{bmatrix} P\_i^{sch} \\ Q\_i^{sch} \end{bmatrix} - \begin{bmatrix} P\_i^{[k]} \\ Q\_i^{[k]} \end{bmatrix} = \begin{bmatrix} \Delta P\_i^{[k]} \\ \Delta Q\_i^{[k]} \end{bmatrix} = \begin{bmatrix} \frac{\partial P\_i^{[k]}}{\partial \delta\_i} & \frac{\partial P\_i^{[k]}}{\partial |V\_i|} \\\\ \frac{\partial Q\_i^{[k]}}{\partial \delta\_i} & \frac{\partial Q\_i^{[k]}}{\partial |V\_i|} \end{bmatrix} \begin{bmatrix} \Delta \delta\_i^{[k]} \\ \Delta |V\_i|^{[k]} \end{bmatrix} \tag{23}
$$

To solve for the deviation, the inverse of the Jacobian matrix is required for every iteration.

$$
\begin{bmatrix}
\Delta \delta^{[k]} \\
\Delta |V\_i|^{[k]}
\end{bmatrix} = \begin{bmatrix}
J\_{P\delta}^{[k]} & J\_{P[V]}^{-[k]} \\
J\_{Q\delta}^{[k]} & J\_{Q[V]}^{-[k]}
\end{bmatrix}^{-1} \begin{bmatrix}
\Delta P\_i^{[k]} \\
\Delta Q\_i^{[k]}
\end{bmatrix} \tag{24}
$$

Then, new values are calculated:

$$
\begin{bmatrix} \delta\_i^{[k]} \\ |V\_i|^{[k]} \end{bmatrix} = \begin{bmatrix} \delta\_i^{[k-1]} \\ |V\_i|^{[k-1]} \end{bmatrix} + \begin{bmatrix} \Delta \delta\_i^{[k]} \\ \Delta |V\_i|^{[k]} \end{bmatrix} \tag{25}
$$

The iterative process stops when the mismatch between calculated and sched-" #

$$\text{uled quantities is within } \begin{bmatrix} \Delta P\_i^{[k]} \\ \Delta Q\_i^{[k]} \end{bmatrix} \le \epsilon.$$

Example 3: Newton-Raphson power flow solution.

Solve the power flow problem shown in Figure 3 using the Newton-Raphson technique. Perform two iterations.

Solution:

The first step in Newton-Raphson Power Flow technique is building Y-bus using Eqs. (2) and (3).

$$\begin{aligned} \boldsymbol{Y} &= \begin{bmatrix} 15 - j55 & -5 + j15 & -10 + j40 \\ -5 + j15 & 20 - j65 & -15 + j50 \\ -10 + j40 & -15 + j50 & 25 - j90 \end{bmatrix} \\ &= \begin{bmatrix} 57.01 \angle -74.74^{\circ} & 15.81 \angle 108.43^{\circ} \\ 15.81 \angle 108.43^{\circ} & 68.01 \angle -72.90^{\circ} & 52.20 \angle 106.70^{\circ} \\\\ 41.23 \angle 104.04^{\circ} & 52.20 \angle 106.70^{\circ} & 93.41 \angle -74.48^{\circ} \end{bmatrix} \end{aligned}$$

Since the unknown variables are δ2, δ3, and j j V<sup>2</sup> ; the scheduled quantities are P2 sch, P<sup>3</sup> sch, and Q<sup>2</sup> sch, the following problem formulation can be written.

Power Flow Analysis DOI: http://dx.doi.org/10.5772/intechopen.83374

$$
\begin{bmatrix} P\_{2}^{\text{sch}} \\ P\_{3}^{\text{sch}} \\ Q\_{2}^{\text{sc}} \end{bmatrix} - \begin{bmatrix} P\_{2}^{[k]} \\ P\_{3}^{[k]} \\ Q\_{2}^{[k]} \end{bmatrix} = \begin{bmatrix} \frac{\partial P\_{2}[k]}{\partial \delta\_{2}} & \frac{\partial P\_{2}[k]}{\partial \delta\_{3}} & \frac{\partial P\_{2}}{\partial [V\_{2}]} \\\\ \frac{\partial P\_{3}[k]}{\partial \delta\_{2}} & \frac{\partial P\_{3}[k]}{\partial \delta\_{3}} & \frac{\partial P\_{3}}{\partial [V\_{2}]} \\\\ \frac{\partial Q\_{2}[k]}{\partial \delta\_{2}} & \frac{\partial Q\_{2}}{\partial \delta\_{3}} & \frac{\partial Q\_{2}}{\partial [V\_{2}]} \end{bmatrix} \begin{bmatrix} \Delta \delta\_{2}^{[k]} \\ \Delta \delta\_{3}^{[k]} \\ \Delta \left| V\_{2} \right|^{[k]} \end{bmatrix}
$$

To calculate the Jacobian matrix elements, P2, P3, and Q<sup>2</sup> equations are obtained using (4) and (5).

P<sup>2</sup> ¼ ∑ n j¼1 j j V<sup>2</sup> Vj � � � � Y2<sup>j</sup> � � � �cos θ2<sup>j</sup> � δ<sup>2</sup> þ δ<sup>j</sup> � � <sup>¼</sup> j j <sup>V</sup><sup>2</sup> j j <sup>V</sup><sup>1</sup> j j <sup>Y</sup><sup>21</sup> cosðθ<sup>21</sup> � <sup>δ</sup><sup>2</sup> <sup>þ</sup> <sup>δ</sup>1Þ þ j j <sup>V</sup><sup>2</sup> <sup>2</sup> j j Y<sup>22</sup> cosð Þþ θ<sup>22</sup> j j V<sup>2</sup> j j V<sup>3</sup> j j Y<sup>23</sup> cosð Þ θ<sup>23</sup> � δ<sup>2</sup> þ δ<sup>3</sup> ∂P<sup>2</sup> ∂δ2 ¼ j j V<sup>2</sup> j j V<sup>1</sup> j j Y<sup>21</sup> sin ðθ<sup>21</sup> � δ<sup>2</sup> þ δ1Þ þ j j V<sup>2</sup> j j V<sup>3</sup> j j Y<sup>23</sup> sin ð Þ θ<sup>23</sup> � δ<sup>2</sup> þ δ<sup>3</sup> ∂P<sup>2</sup> ∂δ3 ¼ �j j V<sup>2</sup> j j V<sup>3</sup> j j Y<sup>23</sup> sin ð Þ θ<sup>23</sup> � δ<sup>2</sup> þ δ<sup>3</sup> ∂P<sup>2</sup> <sup>∂</sup>j j <sup>V</sup><sup>2</sup> ¼ j j V<sup>1</sup> j j Y<sup>21</sup> cosðθ<sup>21</sup> � δ<sup>2</sup> þ δ1Þ þ 2j j V<sup>2</sup> j j Y<sup>22</sup> cosð Þþ θ<sup>22</sup> j j V<sup>3</sup> j j Y<sup>23</sup> cosð Þ θ<sup>23</sup> � δ<sup>2</sup> þ δ<sup>3</sup> P<sup>3</sup> ¼ ∑ n j¼1 j j V<sup>3</sup> Vj � � � � Y3<sup>j</sup> � � � �cos θ3<sup>j</sup> � δ<sup>3</sup> þ δ<sup>j</sup> � � <sup>¼</sup> j j <sup>V</sup><sup>3</sup> j j <sup>V</sup><sup>1</sup> j j <sup>Y</sup><sup>31</sup> cosðθ<sup>31</sup> � <sup>δ</sup><sup>3</sup> <sup>þ</sup> <sup>δ</sup>1Þ þ j j <sup>V</sup><sup>3</sup> j j <sup>V</sup><sup>2</sup> j j <sup>Y</sup><sup>32</sup> cosðθ<sup>32</sup> � <sup>δ</sup><sup>3</sup> <sup>þ</sup> <sup>δ</sup>2Þ þ j j <sup>V</sup><sup>3</sup> <sup>2</sup> j j Y<sup>33</sup> cosð Þ θ<sup>33</sup> ∂P<sup>3</sup> ∂δ2 ¼ �j j V<sup>2</sup> j j V<sup>3</sup> j j Y<sup>32</sup> sin ð Þ θ<sup>32</sup> � δ<sup>3</sup> þ δ<sup>2</sup> ∂P<sup>3</sup> ∂δ3 ¼ j j V<sup>3</sup> j j V<sup>1</sup> j j Y<sup>31</sup> sin ðθ<sup>31</sup> � δ<sup>3</sup> þ δ1Þ þ j j V<sup>3</sup> j j V<sup>2</sup> j j Y<sup>32</sup> sin ð Þ θ<sup>32</sup> � δ<sup>3</sup> þ δ<sup>2</sup> ∂P<sup>3</sup> <sup>∂</sup>j j <sup>V</sup><sup>2</sup> ¼ j j V<sup>3</sup> j j Y<sup>32</sup> cosð Þ θ<sup>32</sup> � δ<sup>3</sup> þ δ<sup>2</sup> Q<sup>2</sup> ¼ �∑ n j¼1 j j V<sup>2</sup> Vj � � � � Y2<sup>j</sup> � � � � sin θ2<sup>j</sup> � δ<sup>2</sup> þ δ<sup>j</sup> � � ¼ �j j <sup>V</sup><sup>2</sup> j j <sup>V</sup><sup>1</sup> j j <sup>Y</sup><sup>21</sup> sin <sup>ð</sup>θ<sup>21</sup> � <sup>δ</sup><sup>2</sup> <sup>þ</sup> <sup>δ</sup>1Þ � j j <sup>V</sup><sup>2</sup> <sup>2</sup> j j Y<sup>22</sup> sin ð Þ� θ<sup>22</sup> j j V<sup>2</sup> j j V<sup>3</sup> j j Y<sup>23</sup> sin ð Þ θ<sup>23</sup> � δ<sup>2</sup> þ δ<sup>3</sup> ∂Q<sup>3</sup> ∂δ2 ¼ j j V<sup>2</sup> j j V<sup>1</sup> j j Y<sup>21</sup> cosðθ<sup>21</sup> � δ<sup>2</sup> þ δ1Þ þ j j V<sup>2</sup> j j V<sup>3</sup> j j Y<sup>23</sup> cosð Þ θ<sup>23</sup> � δ<sup>2</sup> þ δ<sup>3</sup> ∂Q<sup>3</sup> ∂δ3 ¼ �j j V<sup>2</sup> j j V<sup>3</sup> j j Y<sup>23</sup> cosð Þ θ<sup>23</sup> � δ<sup>2</sup> þ δ<sup>3</sup> ∂Q<sup>3</sup> <sup>∂</sup>j j <sup>V</sup><sup>2</sup> ¼ �j j V<sup>1</sup> j j Y<sup>21</sup> sin ðθ<sup>21</sup> � δ<sup>2</sup> þ δ1Þ � 2j j V<sup>2</sup> j j Y<sup>22</sup> sin ð Þ� θ<sup>22</sup> j j V<sup>3</sup> j j Y<sup>23</sup> sin ð Þ θ<sup>23</sup> � δ<sup>2</sup> þ δ<sup>3</sup> Iteration #1: assume V<sup>2</sup> ½ � <sup>0</sup> <sup>¼</sup> <sup>1</sup>:00∠0° and <sup>V</sup><sup>3</sup> ½ � <sup>0</sup> <sup>¼</sup> <sup>1</sup>:<sup>03</sup> <sup>∠</sup>0° . The calculated quantities: P2 ½ � 1 P3 ½ � 1 Q2 ½ � 1 2 6 4 3 7 5 ¼ �0:5500 0:5665 �1:8000: 2 6 4 3 7 5 sch 2 3 2 3

$$\text{The scheduled quantities:} \begin{bmatrix} P\_2^{\text{sch}} \\ P\_3^{\text{sch}} \\ Q\_2^{\text{sch}} \end{bmatrix} = \begin{bmatrix} -2 \\ \mathbf{1.5} \\ -\mathbf{0.5} \end{bmatrix}$$

$$\begin{aligned} \text{The mismatch power matrix:} \begin{bmatrix} \Delta P\_2^{[1]} \\ \Delta P\_3^{[1]} \\ \Delta Q\_2^{[1]} \end{bmatrix} = \begin{bmatrix} -2 \\ 1.5 \\ -0.5 \end{bmatrix} - \begin{bmatrix} -0.5500 \\ 0.5665 \\ -1.8000 \end{bmatrix} = \begin{bmatrix} -1.4500 \\ 0.9335 \\ 1.3000 \end{bmatrix} \\ \text{The Jacobian matrix (J) elements for iteration } \neq \mathbf{1} \text{ are as follows:} \end{aligned}$$

$$J = \begin{bmatrix} \ 66.8 & -51.5 & 19.45 \\ -51.5 & 93.52 & -15.45 \\ -20.55 & 15.45 & 63.2 \end{bmatrix}$$

Newton-Raphson formulation is as follows:

�1:4500 :9335 :3000 ¼ :8 �51:5 19:45 �51:5 93:52 �15:45 �20:55 15:45 63:2 Δδ<sup>2</sup> ½ � 1 Δδ<sup>3</sup> ½ � 1 <sup>Δ</sup>j j <sup>V</sup><sup>2</sup> ½ � <sup>1</sup> Δδ<sup>2</sup> ½ � 1 Δδ<sup>3</sup> ½ � 1 <sup>Δ</sup>j j <sup>V</sup><sup>2</sup> ½ � <sup>1</sup> ¼ :8 �51:5 19:45 �51:5 93:52 �15:45 �20:55 15:45 63:2 �<sup>1</sup> �1:<sup>4500</sup> :9335 :3000 ¼ �0:0279 rad �0:0033 rad :0123 pu δ2 ½ � 1 δ3 ½ � 1 j j <sup>V</sup><sup>2</sup> ½ � <sup>1</sup> ¼ δ2 ½ � 0 δ3 ½ � 0 j j <sup>V</sup><sup>2</sup> ½ � <sup>0</sup> þ Δδ<sup>2</sup> ½ � 1 Δδ<sup>3</sup> ½ � 1 <sup>Δ</sup>j j <sup>V</sup><sup>2</sup> ½ � <sup>1</sup> δ2 ½ � 1 δ3 ½ � 1 j j <sup>V</sup><sup>2</sup> ½ � <sup>1</sup> ¼ :00 þ �0:0279 rad �0:0033 rad :0123 pu ¼ �0:0279 rad �0:0033 rad :0123 pu ¼ �1:5986° �0:1891° :0123 pu 

Iteration #2: Consider V<sup>2</sup> ½ � <sup>1</sup> <sup>¼</sup> <sup>1</sup>:<sup>0123</sup> <sup>∠</sup> � <sup>1</sup>:5986° and <sup>V</sup><sup>3</sup> ½ � <sup>1</sup> <sup>¼</sup> <sup>1</sup>:03<sup>∠</sup> � <sup>0</sup>:1891° .

$$\begin{aligned} \text{The calculated quantities:} \begin{bmatrix} P\_2 |^2 \\ P\_3 |^2 \\ Q\_2 |^2 \end{bmatrix} &= \begin{bmatrix} -2.0109 \\ 1.5202 \\ -0.4621 \end{bmatrix} \\ \text{The mismatch power matrix:} \begin{bmatrix} \Delta P\_2 |^2 \\ \Delta P\_3 |^2 \\ \Delta Q\_2 |^2 \end{bmatrix} &= \begin{bmatrix} -2 \\ 1.5 \\ -0.5 \end{bmatrix} - \begin{bmatrix} -2.0109 \\ 1.5202 \\ -0.4621 \end{bmatrix} = \begin{bmatrix} 0.0109 \\ -0.0202 \\ -0.0379 \end{bmatrix} \\ \text{The Jacobian matrix elements are calculated as follows:} \end{aligned}$$

The Jacobian matrix elements are calculated as follows:

$$J = \begin{bmatrix} 67.07 & -51.74 & 18.26 \\ -52.50 & 94.49 & -14.18 \\ -22.51 & 16.91 & 65.34 \end{bmatrix}$$

$$\begin{bmatrix} \Delta \delta\_2^{[2]} \\ \Delta \delta\_3^{[2]} \\ \Delta |V\_2|^{[2]} \end{bmatrix} = \begin{bmatrix} 67.07 & -51.74 & 18.26 \\ -52.50 & 94.49 & -14.18 \\ -22.51 & 16.91 & 65.34 \end{bmatrix}^{-1} \begin{bmatrix} 0.0109 \\ -0.0202 \\ -0.0379 \end{bmatrix}$$

$$\begin{bmatrix} \Delta \delta\_2^{[2]} \\ \Delta \delta\_3^{[2]} \\ \Delta |V\_2|^{[2]} \end{bmatrix} = \begin{bmatrix} 0.1272 \times 10^{-3} \, rad \\ -0.2154 \times 10^{-3} \, rad \\ -0.4810 \times 10^{-3} \, pu \end{bmatrix}$$

Power Flow Analysis DOI: http://dx.doi.org/10.5772/intechopen.83374

$$
\begin{bmatrix}
\delta\_2^{[2]} \\
\delta\_3^{[2]} \\
\end{bmatrix} = \begin{bmatrix}
\delta\_2^{[1]} \\
\delta\_3^{[1]} \\
\end{bmatrix} + \begin{bmatrix}
\Delta\delta\_2^{[2]} \\
\Delta\delta\_3^{[2]} \\
\Delta|V\_2|^{[2]}
\end{bmatrix}
$$

$$
\begin{bmatrix}
\delta\_2^{[2]} \\
\delta\_3^{[2]} \\
\end{bmatrix} = \begin{bmatrix}
1.0123 \, pu
\end{bmatrix} + \begin{bmatrix}
0.1272 \times 10^{-3} \, rad \\
\end{bmatrix}
$$

$$
\begin{bmatrix}
\delta\_2^{[2]} \\
\delta\_3^{[2]} \\
\end{bmatrix} = \begin{bmatrix}
1.0118 \mu
\end{bmatrix} = \begin{bmatrix}
1.0118 \, pu
\end{bmatrix}
$$

It is worth noting that the mismatch between calculated and scheduled quantities diminishes very quickly.


The iterative solution is presented in Table 3.


Table 3.

Newton-Raphson iterative solution.

#### 5. Fast-decoupled technique

In high voltage transmission systems, the voltage angles between adjacent buses are relatively small. In addition, X=R ratio is high. These two properties result in a strong coupling between real power and voltage angle and between reactive power and voltage magnitude. In contrary, the coupling between real power and voltage magnitude, as well as reactive power and voltage angle, is weak. Considering adjacent buses, real power flows from the bus with a higher voltage angle to the bus with a lower voltage angle. Similarly, reactive power flows from the bus with a higher voltage magnitude to the bus with a lower voltage magnitude.

Fast-decoupled power flow technique includes two steps: (1) decoupling real and reactive power calculations; (2) obtaining of the Jacobian matrix elements directly from the Y-bus.

$$
\begin{bmatrix}
\Delta P\\\Delta Q
\end{bmatrix} = \begin{bmatrix}
J\_{P\delta} & \mathbf{0} \\\mathbf{0} & J\_{Q|V|}
\end{bmatrix} \begin{bmatrix}
\Delta \delta \\\Delta |V|
\end{bmatrix} \tag{26}
$$

or

$$[\Delta P] = [J\_{P\delta}][\Delta \delta] \tag{27}$$

$$\begin{bmatrix} \Delta Q \end{bmatrix} = \begin{bmatrix} J\_{Q|V|} \end{bmatrix} \begin{bmatrix} \Delta |V| \end{bmatrix} \tag{28}$$

Next step is to obtain JP<sup>δ</sup> and JQ Vj j from the Y-bus as flowing:

$$\left[\Delta\delta\right] = -\left[B'\right]^{-1}\left[\frac{\Delta P}{|V|}\right] \tag{29}$$

$$\left[\Delta|V|\right] = -\left[\boldsymbol{\mathcal{B}}^{\prime}\right]^{-1}\left[\frac{\Delta Q}{|V|}\right] \tag{30}$$

Where the B<sup>0</sup> ½ � and B � �<sup>00</sup> are relevant imaginary part of the Y-bus matrix elements. B<sup>0</sup> ½ � is related to the buses at which real power is scheduled (δ is unknown) and B � �<sup>00</sup> is related to the buses at which reactive power is scheduled (j j V is unknown).

$$B\_{\vec{\eta}} = \left| Y\_{\vec{\eta}} \right| \sin \left( \theta\_{\vec{\eta}} \right) \tag{31}$$

The fast-decoupled technique requires more iterations to converge compared to the Newton-Raphson power flow formulation, especially if X=R ratio is not high. Another advantage of this method is that the Jacobian matrix has constant term elements which are obtained and inverted once at the beginning of the iterative process.

Example 4: Fast-decoupled power flow solution.

Solve the power flow problem shown in Figure 3 using fast-decoupled power flow technique. Perform two iterations.

Solution:

The first step in fast-decoupled power flow technique is obtaining B<sup>0</sup> ½ � and B � �<sup>00</sup> from the Y-bus.

$$Y = \begin{bmatrix} \mathbf{15} - j\mathbf{55} & -\mathbf{5} + j\mathbf{15} & -\mathbf{10} + j\mathbf{40} \\ -\mathbf{5} + j\mathbf{15} & \mathbf{20} - j\mathbf{65} & -\mathbf{15} + j\mathbf{50} \\ -\mathbf{10} + j\mathbf{40} & -\mathbf{15} + j\mathbf{50} & \mathbf{25} - j\mathbf{90} \end{bmatrix}$$

Since the unknown voltage angles are δ<sup>2</sup> and δ3, elements for B<sup>0</sup> ½ � are obtained from the Y-bus (intersection of columns numbers 2 & 3 and rows numbers 2 & 3) as follows:

$$\begin{bmatrix} \mathbf{B'} \end{bmatrix} = \begin{bmatrix} -\mathbf{65} & \mathbf{50} \\ \mathbf{50} & -\mathbf{90} \end{bmatrix}$$

Since the unknown voltage magnitude is j j V<sup>2</sup> at bus 2, B � �<sup>00</sup> contains one element only (intersection of the column and the row number 2):

$$\begin{bmatrix} \mathbf{B'} \end{bmatrix} = \begin{bmatrix} -\mathbf{6S} \end{bmatrix}$$

The fast-decoupled power flow formulation becomes as follows:

$$
\begin{bmatrix}
\Delta \delta\_2 \\
\Delta \delta\_3
\end{bmatrix} = -\left[\boldsymbol{\mathcal{B}}\prime\right]^{-1} \begin{bmatrix}
\frac{\Delta P\_2}{|V\_2|} \\
\frac{\Delta P\_3}{|V\_3|} \\
\hline
\end{bmatrix} \text{and} \begin{bmatrix}
\Delta |V\_2|
\end{bmatrix} = -\left[\boldsymbol{\mathcal{B}}\prime\prime\right]^{-1} \begin{bmatrix}
\frac{\Delta Q\_2}{|V\_2|} \\
\hline
\end{bmatrix} .
$$

Or

Power Flow Analysis DOI: http://dx.doi.org/10.5772/intechopen.83374

$$
\begin{bmatrix}
\Delta\delta\_2^{[k]} \\
\Delta\delta\_3^{[k]}
\end{bmatrix} = -[B']^{-1} \begin{bmatrix}
\frac{\Delta P\_2^{[k]}}{|V\_2|^{[k-1]}} \\
\frac{\Delta P\_3^{[k]}}{|V\_3|^{[k-1]}}
\end{bmatrix} \text{and} \begin{bmatrix}
\Delta|V\_2|^{[k]}
\end{bmatrix} = -\begin{bmatrix}
B' \end{bmatrix}^{-1} \begin{bmatrix}
\frac{\Delta Q\_2^{[k]}}{|V\_2|^{[k-1]}}
\end{bmatrix} .
$$

Iteration #1: assume V<sup>2</sup> ½ � <sup>0</sup> <sup>¼</sup> <sup>1</sup>:00∠0° and <sup>V</sup><sup>3</sup> ½ � <sup>0</sup> <sup>¼</sup> <sup>1</sup>:<sup>03</sup> <sup>∠</sup>0° . The calculated quantities:

$$
\begin{bmatrix} P\_2^{[1]} \\ P\_3^{[1]} \\ Q\_2^{[1]} \end{bmatrix} = \begin{bmatrix} -0.5500 \\ 0.5665 \\ -1.8000 \end{bmatrix}
$$

The mismatch power matrix:

$$
\begin{bmatrix}
\Delta P\_2^{[1]} \\
\Delta P\_3^{[1]} \\
\Delta Q\_2^{[1]}
\end{bmatrix} = \begin{bmatrix}
1.5 \\
\end{bmatrix} - \begin{bmatrix}
0.5665 \\
\end{bmatrix} = \begin{bmatrix}
0.9335 \\
1.3000
\end{bmatrix}
$$

Calculation of δ<sup>2</sup> and δ3:

$$
\begin{split}
\begin{bmatrix}
\Delta\boldsymbol{\delta\_{2}}^{[1]} \\
\Delta\boldsymbol{\delta\_{3}}^{[1]}
\end{bmatrix} &= -\left[\boldsymbol{B}^{\prime}\right]^{-1} \begin{bmatrix}
\frac{\Delta\boldsymbol{P}\_{2}^{[1]}}{|\boldsymbol{V}\_{2}|^{[0]}} \\
\frac{\Delta\boldsymbol{P}\_{3}^{[1]}}{|\boldsymbol{V}\_{3}|^{[0]}}
\end{bmatrix} = -\begin{bmatrix}
50 & -90
\end{bmatrix}^{-1} \begin{bmatrix}
\frac{-1.4500}{1.00} \\
\frac{0.9335}{1.03} \\
\frac{0.9335}{1.03}
\end{bmatrix} \\
&= \begin{bmatrix}
\end{bmatrix} = \begin{bmatrix}
\end{bmatrix} \\
\begin{bmatrix}
\delta\_{2}^{[1]} \\
\delta\_{3}^{[1]}
\end{bmatrix} &= \begin{bmatrix}
\delta\_{2}^{[0]} \\
\delta\_{3}^{[0]}
\end{bmatrix} + \begin{bmatrix}
\Delta\delta\_{2}^{[1]} \\
\Delta\delta\_{3}^{[1]}
\end{bmatrix} = \begin{bmatrix}
0 \\
0
\end{bmatrix} + \begin{bmatrix}
\end{bmatrix} = \begin{bmatrix}
\end{bmatrix}
\end{split}
$$

Calculation of j j V<sup>2</sup> :

$$
\left[\Delta|V\_2|^{[1]}\right] = -[-\mathsf{65}]^{-1}\left[\frac{\mathsf{1.3}}{\mathsf{1.00}}\right] = \mathsf{0.0200}
$$

$$
\left[\Delta|V\_2|^{[1]}\right] = |V\_2|^{[0]} + \Delta|V\_2|^{[1]} = \mathsf{1.00} + \mathsf{0.0200} = \mathsf{1.020}\,pu
$$

Iteration #2: considering V<sup>2</sup> ½ � <sup>1</sup> <sup>¼</sup> <sup>1</sup>:020<sup>∠</sup> � <sup>1</sup>:4569° and <sup>V</sup><sup>3</sup> ½ � <sup>0</sup> <sup>¼</sup> <sup>1</sup>:03<sup>∠</sup> � <sup>0</sup>:2324° . The calculated quantities:

$$
\begin{bmatrix} P\_2^{[2]} \\ P\_3^{[2]} \\ Q\_2^{[2]} \end{bmatrix} = \begin{bmatrix} -1.6671 \\ 1.2133 \\ -0.0239 \end{bmatrix}
$$

The mismatch power matrix:

$$
\begin{bmatrix}
\Delta P\_2^{[2]} \\
\Delta P\_3^{[2]} \\
\Delta Q\_2^{[2]}
\end{bmatrix} = \begin{bmatrix}
1.5 \\
\end{bmatrix} - \begin{bmatrix}
1.2133 \\
\end{bmatrix} = \begin{bmatrix}
0.2867 \\
\end{bmatrix}
$$

Calculation of δ<sup>2</sup> and δ3:

$$
\begin{bmatrix}
\Delta\delta\_2^{[2]} \\
\Delta\delta\_3^{[2]}
\end{bmatrix} = -[B^\prime]^{-1} \begin{bmatrix}
\frac{\Delta P\_2^{[2]} \\
\Delta V\_3^{[1]} \\
\end{bmatrix} = -\begin{bmatrix}
50 & -90
\end{bmatrix}^{-1} \begin{bmatrix}
\frac{-0.3329}{1.02} \\
\frac{0.2867}{1.03} \\
\frac{0.2867}{1.03}
\end{bmatrix}
$$

$$
\begin{bmatrix}
\Delta\delta\_2^{[2]} \\
\Delta\delta\_3^{[2]}
\end{bmatrix} = \begin{bmatrix}
0.0005 \, rad
\end{bmatrix} = \begin{bmatrix}
0.0286^\circ \\
0.0286^\circ
\end{bmatrix}
$$

$$
\begin{bmatrix}
\delta\_2^{[2]} \\
\delta\_3^{[3]}
\end{bmatrix} = \begin{bmatrix}
\delta\_2^{[1]} \\
\delta\_3^{[1]}
\end{bmatrix} + \begin{bmatrix}
\Delta\delta\_2^{[2]} \\
\Delta\delta\_3^{[2]}
\end{bmatrix} = \begin{bmatrix}
\end{bmatrix} + \begin{bmatrix}
0.0286^\circ
\end{bmatrix} = \begin{bmatrix}
\end{bmatrix}
$$

Calculation of j j V<sup>2</sup> :

$$
\left[\Delta|V\_2|^{[2]}\right] = -[-65]^{-1} \left[\frac{-0.4761}{1.02}\right] = -0.0072
$$

$$
\left[|V\_2|^{[2]}\right] = |V\_2|^{[1]} + \Delta|V\_2|^{[2]} = 1.02 - 0.0072 = 1.0128 \,\mu\text{m}
$$

The remaining five iterations are shown in Table 4. It shows that this method converges slower than Newton-Raphson method.


Table 4. Fast-decoupled iterative solution.

## 6. DC power flow technique

DC power is an extension to the Fast-decoupled power flow formulation. In DC power flow method, the voltage is assumed constant at all buses; therefore, theðΔj j V ,ΔQ) equation is neglected. The (Δδ,ΔP) equation can be further simplified to a linear problem that does not require iterative solution:

$$-[B][\delta] = [P] \tag{32}$$

Or

$$\begin{bmatrix} \delta \end{bmatrix} = -[\mathcal{B}]^{-1}[P] \tag{33}$$

Example 5: DC power flow.

Solve the power flow problem shown in Figure 3 using the DC power flow technique.

Solution:

The first step in DC power flow technique is obtaining ½ � B from the Y-bus.

$$Y = \begin{bmatrix} 15 - j55 & -5 + j15 & -10 + j40 \\ -5 + j15 & 20 - j65 & -15 + j50 \\ -10 + j40 & -15 + j50 & 25 - j90 \end{bmatrix}$$

For this problem, Since the unknown voltage angles are δ<sup>2</sup> and δ3, elements for ½ � B are obtained from the Y-bus as follows:

$$[B] = \begin{bmatrix} -65 & 50 \\ 50 & -90 \end{bmatrix}$$

$$\begin{aligned} \begin{bmatrix} \delta\_2 \\ \delta\_3 \end{bmatrix} &= -\begin{bmatrix} -65 & 50 \\ 50 & -90 \end{bmatrix}^{-1} \begin{bmatrix} P\_2 \\ P\_3 \end{bmatrix} = -\begin{bmatrix} -65 & 50 \\ 50 & -90 \end{bmatrix}^{-1} \begin{bmatrix} -2 \\ 1.5 \end{bmatrix} \\ &= \begin{bmatrix} -0.0313 \, rad \\ -0.0007 \, rad \end{bmatrix} = \begin{bmatrix} -1.7958^\circ \\ -0.0428^\circ \end{bmatrix} \end{aligned}$$
