7. Slack bus power and losses calculations

The main objective of power flow calculations is to determine the voltages (magnitude and angle) for a given load and generation conditions. Once voltages are known for all buses, slack bus power, as well as line flows and losses, can be calculated. The slack bus real and reactive power are calculated using Eqs. (4) and (5), respectively. Overall system losses are the difference between generation and load.

$$\mathbf{S}\_{L} = \mathbf{S}\_{Gen} - \mathbf{S}\_{Load} \tag{34}$$

Specific branch losses are calculated using branch power flow. For example, the losses in the line i – j are the algebraic sum of the power flow.

$$\mathcal{S}\_{L\ \vec{\eta}} = \mathcal{S}\_{\vec{\eta}} + \mathcal{S}\_{\vec{\mu}} \tag{35}$$

Sij and Sji are defined as follows:

$$\mathbf{S}\_{\vec{\eta}} = \mathbf{V}\_i \, I\_{\vec{\eta}}\, ^\* \tag{36}$$

$$\mathbf{S}\_{\circ i} = \mathbf{V}\_{j} \, I\_{\circ i}^\* \tag{37}$$

The current between buses Iij is a function of the voltages and the admittance between Vi and Vj. It is worth noting that Iji ¼ �Iij

$$I\_{\vec{\eta}} = (\mathbf{V}\_i - \mathbf{V}\_j)\mathbf{y}\_{\vec{\eta}} \tag{38}$$

Example 6: Slack bus power and losses.

For the 3-bus system shown in Figure 3, the voltages at buses 2 and 3 were iteratively calculated: <sup>V</sup><sup>1</sup> <sup>¼</sup> <sup>1</sup>:02∠0° , <sup>V</sup><sup>2</sup> <sup>¼</sup> <sup>1</sup>:0118<sup>∠</sup> � <sup>1</sup>:5871° and <sup>V</sup><sup>3</sup> <sup>¼</sup> <sup>1</sup>:03<sup>∠</sup> � <sup>0</sup>:2005°


Solution:

The polar form of the Y-bus is used.

$$Y = \begin{bmatrix} 57.01 \angle -74.74^{\circ} & 15.81 \angle 108.43^{\circ} & 41.23 \angle 104.04^{\circ} \\ 15.81 \angle 108.43^{\circ} & 68.01 \angle -72.90^{\circ} & 52.20 \angle 106.70^{\circ} \\ 41.23 \angle 104.04^{\circ} & 52.20 \angle 106.70^{\circ} & 93.41 \angle -74.48^{\circ} \end{bmatrix}$$

a. The slack bus power

The slack bus power can be calculated using real and reactive power equations:

$$\begin{split} P\_1 &= \sum\_{j=1}^{n} |V\_1||V\_j||Y\_{\bar{y}}|\cos\left(\theta\_{\bar{y}} - \delta\_1 + \delta\_{\bar{\delta}}\right) \\ &= |V\_1|^2 |Y\_{11}|\cos\left(\theta\_{11}\right) + |V\_1||V\_2||Y\_{12}|\cos\left(\theta\_{12} - \delta\_1 + \delta\_2\right) + |V\_1||V\_3||Y\_{13}|\cos\left(\theta\_{13} - \delta\_1 + \delta\_3\right) \\ &= 0.5195 \, p u \\ Q\_1 &= -\sum\_{j=1}^{n} |V\_1||V\_j| |Y\_{\bar{y}}| \sin\left(\theta\_{\bar{y}} - \delta\_1 + \delta\_{\bar{\delta}}\right) \\ &= -|V\_1|^2 |Y\_{11}| \sin\left(\theta\_{11}\right) - |V\_1||V\_2||Y\_{12}| \cos\left(\theta\_{12} - \delta\_1 + \delta\_2\right) - |V\_1||V\_3||Y\_{13}| \sin\left(\theta\_{13} - \delta\_1 + \delta\_3\right) \\ &= -0.4572 \, p u \end{split}$$

### b.The total system losses

The total real power losses can be calculated using the total net injected real a power at all buses:

$$P\_L = P\_1 + P\_2 + P\_3$$

$$P\_L = P\_1 + P\_2 + P\_3 = 0.5195 - 2 + 1.5 = 0.0195 \, pu$$

Similarly, the reactive power losses can be calculated.

$$Q\_L = Q\_1 + Q\_2 + Q\_3$$

However, Q<sup>3</sup> is unknown and can be calculated based on the given bus voltages:

$$\begin{split} Q\_3 &= -\sum\_{j=1}^{n} |V\_3|| \left| V\_j \right| \left| Y\_{3j} \right| \sin \left( \theta\_{\tilde{y}} - \delta\_3 + \delta\_{\tilde{l}} \right) \\ &= -|V\_3|| |V\_1|| |Y\_{31}| \sin \left( \theta\_{31} - \delta\_3 + \delta\_1 \right) - |V\_3|| |V\_2|| |Y\_{32}| \sin \left( \theta\_{32} - \delta\_3 + \delta\_2 \right) - |V\_3|^2 |Y\_{33}| \sin \left( \theta\_{33} \right) \\ &= 1.0220 \, pu \end{split}$$

Once Q<sup>2</sup> is calculated, the total reactive power losses are calculated:

$$Q\_L = Q\_1 + Q\_2 + Q\_3 = -0.4572 - 0.5 + 1.0220 = 0.0648 \, pu$$

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

Figure 6. Power flow results.

c. Branch losses

To calculate the losses in line 1–2, SL <sup>12</sup> is calculated by summing S<sup>12</sup> and S21. These flows are calculated as follows: <sup>S</sup><sup>12</sup> <sup>¼</sup> <sup>V</sup><sup>1</sup> <sup>I</sup>12<sup>∗</sup> and <sup>S</sup><sup>21</sup> <sup>¼</sup> <sup>V</sup><sup>2</sup> <sup>I</sup>21<sup>∗</sup>

$$I\_{12} = (V\_1 - V\_2)y\_{12} = \left(1.02\angle 0^\circ - 1.0118\angle -1.5871^\circ \right)(-5 + j15) = 0.4635 + j0.0121 \, pu$$

$$I\_{21} = (V\_2 - V\_1)y\_{21} = -I\_{12} = -0.4635 - j0.0121 \, pu$$

$$S\_{12} = V\_1 \, I\_{12}{^\circ} = 1.02\angle 0^\circ \times (0.4635 - j0.0121) = 0.4728 - j0.0123 \, pu$$

$$S\_{21} = V\_2 \, I\_{21}{^\circ} = 1.0118\angle -1.5871^\circ \times (-0.4635 + j0.0121) = -0.4685 + j0.0252 \, pu$$

$$S\_{12} = S\_{12} + S\_{12} = (0.4728 - j0.0123) + (-0.4685 + j0.0252) = 0.0043 + j0.0129 \, pu$$

Similarly, power flow and losses in other branches are calculated. Line 1–3:

$$\mathbf{S}\_{31} = -\mathbf{0}.\mathbf{0}45\mathbf{6} + j\mathbf{0}.\mathbf{4}4\mathbf{9}\mathbf{4}$$

$$\mathbf{S}\_{13} = \mathbf{0}.\mathbf{0}46\mathbf{7} - j\,\mathbf{0}.\mathbf{4}44\mathbf{9}$$

$$\mathbf{S}\_{L\,13} = \mathbf{0}.\mathbf{0}\mathbf{0}\mathbf{1}\mathbf{1} + j\,\mathbf{0}.\mathbf{0}\mathbf{0}\mathbf{4}\mathbf{5}$$

Line 2–3:

$$\begin{aligned} \mathbf{S\_{23}} &= -\mathbf{1.5315} - j \mathbf{0.5252} \\ \mathbf{S\_{32}} &= \mathbf{1.5456} + j \mathbf{0.5722} \\ \mathbf{S\_{L\\_23}} &= \mathbf{0.0141} + j \mathbf{0.0470} \end{aligned}$$

Total losses:

$$\mathbf{S}\_{L} = \mathbf{S}\_{L\ 13} + \mathbf{S}\_{L\ 23} + \mathbf{S}\_{L\ 12} = \mathbf{0}.019\mathbf{S} + j0.0648\begin{bmatrix} pu\\v \end{bmatrix}$$

The generation and load at different buses is shown in Figure 6.

#### 8. Conclusions

Power flow analysis, or load flow analysis, has a wide range of applications in power systems operation and planning. This chapter presents an overview of the power flow problem, its formulation as well as different solution methods. The

power flow model of a power system can be built using the relevant network, load, and generation data. Outputs of the power flow model include voltages (magnitude and angles) at different buses. Once nodal voltages are calculated, real and reactive power flows in different network branches can be calculated. The calculation of branch power flows enables technical loss calculation in different network branches, as well as the total system technical losses.

Power flow analysis is performed by solving nodal power balance equations. Since these equations are nonlinear, iterative techniques such as the Gauss-Seidel, the Newton-Raphson, and the fast-decoupled power flow methods are commonly used to solve this problem. In general, the Gauss-Seidel method is simple but converges slower than the Newton-Raphson method. However, the latter method required the Jacobian matrix formation of at every iteration. The fast-decoupled power flow method is a simplified version of the Newton-Raphson method. This simplification is achieved in two steps: 1) decoupling real and reactive power calculations; 2) obtaining of the Jacobian matrix elements directly from the Y-bus matrix. The DC power method is an extension to the fast-decoupled power flow formulation. In DC power flow method, the voltage is assumed constant at all buses and the problem becomes linear.

#### Acknowledgements

The author would like to acknowledge the financial support and thank the Rural Area Electricity Company and Sultan Qaboos University for sponsoring the publication of this chapter under project number CR/ENG/ECED/16/02.

## Author details

Mohammed Albadi Sultan Qaboos University, Muscat, Oman

\*Address all correspondence to: mbadi@squ.edu.om

© 2019 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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