Electrical Networks: An Introduction

Abstract

A world without electricity is beyond our imagination. Starting from the prehistoric times, man has made much progress in every walk of life. We have become accustomed to getting everything at the flick of a switch, touch of a button, or turn of a knob. While we have become so used to enjoying the benefits of electricity, it is not easy to imagine how electricity travels from its source to our homes and offices. It sometimes has to cover large distances through a complex network of transmission lines and power substations to provide us the facilities and entertainment that we take for granted. This network which transports electricity from the source to the consumers is called the electrical network. The electrical network is a collective term for different components such as transformers, transmission lines, substations, and different stages and sub-networks devoted to generation, transmission, and distribution. Sometimes, there may be sub-transmission and secondary distribution networks too. A simple schematic of an electric network is shown in Fig. 8.1. In the past decade, analysis of the electrical power system as a complex network has been an evolving and challenging topic of research.

Appendix: Electrical Network Terminology and Models for Analysis

1.1 Phasor Representation

The voltage and current in electrical power systems are sinusoidal quantities that vary with time at the same frequency. A sinusoidal voltage $v(t)$ and a sinusoidal current $i(t)$ are expressed as follows:

$$\begin{aligned} v(t)&= V_{m}cos(\omega t + \delta )\end{aligned}$$

(8.1)

$$\begin{aligned} i(t)&= I_{m} cos(\omega t + \beta ) \end{aligned}$$

(8.2)

where $V_{m}$ and $I_{m}$ are the maximum voltage and current, $\omega$ is the angular speed, and $\delta$ and $\beta$ are the phase shift of the voltage and current, respectively.

The voltage is expressed in Volt. However, power systems operate on voltages that range from several 1,000s to 100,000s of Volt. Consequently, it is more convenient to express the voltage in KiloVolt (KV). Electrical current is measured in Ampere (A), the angular speed in radian per second (rad/sec), and the phase shift in radian (rad). The angular speed is proportional to the electrical frequency $f$, which is the number of cycles per second, expressed in Hertz (Hz). The relationship between the angular speed and the frequency is

$$\begin{aligned} \omega = 2 \pi f. \end{aligned}$$

(8.3)

As the voltage and current have the sinusoidal form at steady state, it is convenient to express the magnitude and phase angle of the voltage in a complex number form called a phasor. A phasor is developed using the Euler’s identity as follows:

$$\begin{aligned} e^{\pm j\phi }=cos\phi \pm j sin\phi . \end{aligned}$$

(8.4)

The terms $cos\phi$ and $sin\phi$ are the real and imaginary parts and they are denoted by $Re\{e^{j\phi }\}$ and $Im\{e^{j\phi }\}$, respectively. Therefore, the voltage and current can be written in the phasor form as follows

$$\begin{aligned} v(t)&= Re\{V_{m}e^{j(\omega t + \delta )}\} = Re\{V_{m}e^{j\omega t} e^{j\delta }\} \end{aligned}$$

(8.5)

$$\begin{aligned} i(t)&= Re\{I_{m}e^{j(\omega t + \beta )}\} = Re\{I_{m}e^{j\omega t} e^{j\beta }\}. \end{aligned}$$

(8.6)

Since both the voltage and current have the same frequency, the component $e^{j\omega t}$ becomes less important, and for convenience it is enough to express the voltage and current in terms of their magnitude and phase shift using the following form

$$\begin{aligned} V&= V_{m}e^{j\delta } = V_{m} \angle \delta \end{aligned}$$

(8.7)

$$\begin{aligned} I&= I_{m}e^{j\beta } = I_{m} \angle \beta . \end{aligned}$$

(8.8)

The voltage and current are usually represented through their effective values, called the root-mean-square (rms) values. The effective phasor representations of the voltage and current are

$$\begin{aligned} V&= \frac{V_{m}}{\sqrt{2}} e^{j\beta } = |V|e^{j\beta } \end{aligned}$$

(8.9)

$$\begin{aligned} I&= \frac{I_{m}}{\sqrt{2}} e^{j\delta } = |I|e^{j\delta } \end{aligned}$$

(8.10)

where $|V|=\frac{V_{m}}{\sqrt{2}}$ and $|I|=\frac{I_{m}}{\sqrt{2}}$ are the rms values for the sinusoidal form of the voltage and current. The rms voltage phasor and rms current phasor can be written in the rectangular form as follows

$$\begin{aligned} V&= |V|(cos \delta + j sin \delta ) \end{aligned}$$

(8.11)

$$\begin{aligned} I&= |I|(cos \beta + j sin \beta ). \end{aligned}$$

(8.12)

1.2 Instantaneous Power

The electrical power is the work done by the electrical system in unit time. It is a function of both the voltage and current. The unit of electrical power is Watt, however it is convenient to use MegaWatt (MW) when dealing with large amounts of power generation and loads. In a closed circuit with a voltage source and a load, the instantaneous power $p(t)$ that is absorbed by the load is the product of the instantaneous voltage difference across the load and the instantaneous current passing through the load. Mathematically, the instantaneous power is evaluated as follows

$$\begin{aligned} p(t)&= v(t) i(t)\\&= V_{m}cos(\omega t + \delta )I_{m}cos(\omega t+\beta )\nonumber \\&= \frac{V_{m}I_{m}}{2}[cos(\delta -\beta )+cos(2(\omega t+\delta )-(\delta -\beta ))] \nonumber \\&= \frac{V_{m}I_{m}}{2}cos(\delta -\beta )[1+cos(2(\omega t+\delta ))]\nonumber \\&\quad +\frac{V_{m}I_{m}}{2}sin(\delta -\beta )[sin(2(\omega t+\delta ))]\nonumber \end{aligned}$$

(8.13)

The instantaneous power is composed of two components as shown in Eq. (8.13). Assume that the phase angle difference $\delta -\beta$ is constant. The first component is sinusoidal function with a frequency that is twice the frequency of the voltage and current. The maximum value equals $V_{m}I_{m}cos(\delta -\beta )$ and the minimum value equals zero. The constant term $\frac{V_{m}I_{m}}{2}cos(\delta -\beta )$ represents the average power, while the time-varying sinusoidal function has zero average. We refer to the first component as the instantaneous active power. The second component is time-varying sinusoidal function with zero mean value, twice the frequency of the voltage and current, and maximum value of $\frac{V_{m}I_{m}}{2}sin(\delta -\beta )$. This component is called the reactive power, which represents the power that oscillates with twice the frequency of the voltage and current between the reactive components in the power systems that stores the electrical energy and the power generation. In other words, the component of complex power, that averaged over a complete cycle of the AC waveform, causes a net transfer of energy in one direction is known as real power. The component of complex power due to stored energy, which returns to the source in each cycle, is known as reactive power.

Using the rms values for the voltage and current, the active power $P$ and the reactive power $Q$ are as follows

$$\begin{aligned} P&= |V||I|cos(\delta -\beta ) \end{aligned}$$

(8.14)

$$\begin{aligned} Q&= |V||I|sin(\delta -\beta ). \end{aligned}$$

(8.15)

The cosine of the phase angle difference $cos(\delta -\beta )$ is called the power factor. The unit of the active power is Watt, while the unit of the reactive power is Volt-Ampere Reactive (VAR). Let us assume that there are three cases for loads:

• Resistive load: There is no phase angle difference between the voltage and current. Therefore, the power factor is 1, and the active power is $|V||I|$, while the reactive power equals zero because there is no reactive load elements that can store the electrical energy.

• Inductive load: The voltage phase angle leads the current phase angle by $90\,^{\circ }$ i.e. $\delta -\beta = \frac{\pi }{2}$. The reactive power is $|V||I|$, while the active power is zero because there is not resistive load elements that can consume the active power.

• Capacitive load: The voltage phase angle lags the current phase angle by $90\,^{\circ }$ i.e. $\delta -\beta = -\frac{\pi }{2}$. The reactive power $-|V||I|$, while the active power is zero.

The complex power is defined as a complex number with a real part representing the active power and an imaginary part representing the reactive power as follows

$$\begin{aligned} \mathbf {S}&= P+jQ \end{aligned}$$

(8.16)

$$\begin{aligned} \mathbf {S}&= \mathbf {VI^{*}} \end{aligned}$$

(8.17)

where $I^{*}$ is the complex conjugate of the current $\mathbf {I}$. The apparent power $S$ is the magnitude of the complex power

$$\begin{aligned} S&= \sqrt{P^{2}+Q^{2}} \end{aligned}$$

(8.18)

$$\begin{aligned}&= |V||I|(cos^{2}(\delta -\beta ) + sin^{2}(\delta -\beta )) \end{aligned}$$

(8.19)

$$\begin{aligned}&= |V||I|. \end{aligned}$$

(8.20)

The complex power and the apparent power are related through the following equation

$$\begin{aligned} \mathbf {S}=S(cos({\delta -\beta })+sin(\delta -\beta )) \end{aligned}$$

(8.21)

The unit of both the complex power and the apparent power is Volt-Ampere (VA).

1.3 Per Unit System

Any power grid is composed of 100s of electrical elements such as transmission lines, transformers, circuit breakers and shunt impedances. Every element can be represented using the ideal form in which it is lossless; however, the ideal form hides many details that influence the performance of a power system. On the other hand, a detailed representation of each element will account for the amount of electrical power loss. Computationally, analysis of detailed representation of power grids is not trivial. Therefore, voltage, current, and power are normalized with respect to their base values, and they become “per unit values”. The “per unit” method is a very powerful method for analyzing the power grid because (1) it can be applied to a detailed representation of a power grid, thus reducing the error, and (2) it can be systematically applied to different circuits throughout the power grid, and each circuit has its voltage value close to the normal value. The per unit value is defined as follows

$$\begin{aligned} \text{ Per } \text{ unit } \text{ value }&= \frac{\text {Actual value}}{\text {Base value}}. \end{aligned}$$

(8.22)

Both the actual value and the base value have the same dimension, while the per unit value is dimensionless. Traditionally, the base value of the complex power is arbitrarily chosen, and the per unit value becomes as follows

$$\begin{aligned} \frac{\mathbf {S}}{S_{base}}&= \frac{\mathbf {VI^{*}}}{S_{base}}\end{aligned}$$

(8.23)

$$\begin{aligned} \frac{S\angle \theta }{S_{base}}&= \frac{V\angle \delta I\angle -\beta }{S_{base}}. \end{aligned}$$

(8.24)

The base complex power is defined as

$$\begin{aligned} S_{base}&= V_{base} I_{base}. \end{aligned}$$

(8.25)

In addition to the base complex power, either the base voltage or the base current is arbitrarily chosen. Because a power grid is composed of multiple circuits, each has a voltage level, the base voltage is usually proposed, and the based current is evaluated using Eq. (8.25). Using the base values for the complex power, voltage, and current, Eq. (8.24) becomes as follows

$$\begin{aligned} \mathbf {S_{pu}}&= \mathbf {V_{pu}} \mathbf {I_{pu}^{*}}. \end{aligned}$$

(8.26)

We notice that the phase angles do not change using the per unit system, showing that the per unit system is only applied to the magnitude values. The base impedance becomes as follows

$$\begin{aligned} Z_{base}&= \frac{V_{base}}{I_{base}}\end{aligned}$$

(8.27)

$$\begin{aligned}&= \frac{V_{base}^{2}}{S_{base}}. \end{aligned}$$

(8.28)

We further obtain the per unit impedance as follows

$$\begin{aligned} \mathbf {Z}&= \frac{\mathbf {V}}{\mathbf {I}} \end{aligned}$$

(8.29)

$$\begin{aligned} \frac{\mathbf {Z}}{Z_{base}}&= \frac{\mathbf {V}/V_{base}}{\mathbf {I}/I_{base}}\end{aligned}$$

(8.30)

$$\begin{aligned} \mathbf {Z_{pu}}&= \frac{\mathbf {V_{pu}}}{\mathbf {I_{pu}}} \end{aligned}$$

(8.31)

$$\begin{aligned} \mathbf {Z_{pu}}&= \frac{R+jX}{Z_{base}} \end{aligned}$$

(8.32)

$$\begin{aligned} \mathbf {Z_{pu}}&= R_{pu} + jX_{pu}. \end{aligned}$$

(8.33)

We notice that the resistance and the reactance have the same base value, which is base impedance

$$\begin{aligned} Z_{base} = R_{base} = X_{base}. \end{aligned}$$

(8.34)

Similarly, the active power and the reactive power have the same base value as follows

$$\begin{aligned} P_{base} = Q_{base} = S_{base}. \end{aligned}$$

(8.35)

The base complex power is usually expressed in MVA, while the base voltage is expressed in KV. Therefore, it is worth noticing that the base current is in KA, and the base impedance is in Ohm.

1.4 Transformers and Transmission Lines

Electrical power is generated at low voltage level leading to increase in the power loss which is proportional to $I^{2}$ in the transmission systems. On the other hand, loads do not require high voltages for operation. Transformers are used to step up the voltage from the generation side to the transmission side. Similarly, transformers step down the voltage from the transmission side to the distribution side. Below, we discuss the operation and the representation of the transformers and the transmission lines in more details.

1.4.1 Transformers

A transformer is composed of a primary side and a secondary side. Each side is connected with a winding coil that generates magnetic field, which in turn creates electric current and voltage across the secondary coil. The equivalent circuit of a practical transformer is composed of winding resistance and leakage reactance on each side in which the reactance is added in series with the resistance. In addition, there is power loss in the magnetizing equivalent circuit due to hysteresis current losses. In an ideal transformer, the internal resistances, reactances, and the magnetization circuits are neglected, and the transformer becomes lossless. A practical representation of the transformer is to neglect the magnetization circuit because the magnetizing current is very small compared to the rated current, and to consider the resistances and the reactances of the primary and secondary sides. For transformers that handle large power, the internal resistances become very small compared to the reactance. Thus the internal resistance can be neglected.

Denote the voltages across the primary and secondary coils as $\mathbf {E_{1}}$ and $\mathbf {E_{2}}$, respectively. In addition, denote the currents in the primary and secondary sides as $\mathbf {I_{1}}$ and $\mathbf {I_{2}}$, respectively. Let the ratio between number of turns in the primary side to number of turns in the secondary side be $n$. The fact that the complex power at each side of the transformer is preserved, the voltages and currents at both sides are related as follows

$$\begin{aligned} \frac{\mathbf {E_{1}}}{\mathbf {E_{2}}} = \frac{\mathbf {I_{2}}}{\mathbf {I_{1}}} = n. \end{aligned}$$

(8.36)

The reactance of the secondary side $x_{2}$ seen from the primary side is $n^{2} x_{2}$. Therefore, the equivalent reactance of the transformer seen at the primary side is the sum of the reactance at the primary side and $n^{2} x_{2}$. The transformer can be represented in terms of per unit as follows

$$\begin{aligned} \frac{V_{base1}}{V_{base2}}&= n \end{aligned}$$

(8.37)

$$\begin{aligned} \mathbf {E_{1pu}}&= \frac{\mathbf {E_{1}}}{V_{base1}} \end{aligned}$$

(8.38)

$$\begin{aligned} \mathbf {E_{2pu}}&= \frac{\mathbf {E_{2}}}{V_{base2}} \end{aligned}$$

(8.39)

$$\begin{aligned} \mathbf {E_{2pu}}&= \frac{\mathbf {E_{1}}/n}{V_{base1}/n} \end{aligned}$$

(8.40)

$$\begin{aligned} \mathbf {E_{2pu}}&= \mathbf {E_{1pu}}. \end{aligned}$$

(8.41)

Similarly, the per unit currents at each side of the transformers are equal. To study the per unit representation of the reactance in the primary side, we have

$$\begin{aligned} x_{2}&= \frac{x_{1}}{n^{2}} \end{aligned}$$

(8.42)

$$\begin{aligned} x_{1pu}&= \frac{x_{1}}{Z_{base1}} \end{aligned}$$

(8.43)

$$\begin{aligned} x_{1pu}&= \frac{x_{1}}{V_{base1}^{2}/S_{base}} \end{aligned}$$

(8.44)

$$\begin{aligned} x_{1pu}&= \frac{x_{2} n^{2}}{V_{base1}^{2}/S_{base}} \end{aligned}$$

(8.45)

$$\begin{aligned} x_{1pu}&= \frac{x_{2}}{Z_{base2}} = x_{2pu}. \end{aligned}$$

(8.46)

Therefore, the per unit value of the reactance on one side of the transformer is used when studying the integration of the transformer in the single phase diagram.

1.4.2 Transmission lines

Transmission lines are responsible for transferring the generated power from the generation side to the loads. Depending on the length of the transmission line, the operating voltage is set to reduce the amount of power loss in the lines. Transmission lines with short length require lower voltages than long transmission lines. Transmission lines are classified to short-length, medium-length and long transmission line. A transmission line has an equivalent resistance, inductance, and capacitance. The equivalent $\pi$-model is used to represent the transmission lines in the grid. In $\pi$-model, the resistance and the inductance are connected in series, and the equivalent capacitance is connected in parallel at the sending and receiving ends of the lines. In short-length transmission lines, the capacitance is neglected, and the transmission line is represented using the series connection of the resistance and reactance. In medium-length transmission lines, half of the total equivalent capacitance is represented at each end of the line, while the series resistance and reactance connection exists between the two ends of the line.

The analysis of the electrical power grid has to be done through a power flow model which can solve the optimal load flow problem considering all the elements described above. In the next two subsections, we discuss two such models for solving the power flow problem.

1.5 AC Power Flow Model

To study the power flow in the power grid, assume that the generators, transmission lines and loads locations are given. First we would like to classify the buses into three groups:

• Slack bus: A slack bus produces enough active and reactive power to match the system needs. The voltage and angle at the slack bus are 1 p.u. and zero, respectively, while the generated power $P$ and $Q$ are unknown.

• Load bus: Load bus connects load(s) with the grid. There is no generator connected with the load bus. The amount of active and reactive power needed at the loads are given.

• Voltage controlled bus: Bus that connects a generator with the power grid. Load can be connected on the same bus. The bus voltage and generated active power are known, while the voltage phase shift angle and the reactive power are unknown.

To find the power flow in each transmission line, we first apply Kirchhoff’s current law (KCL) at each bus by assuming that the algebraic sum of the currents at any bus is equal to zero. We obtain a group of equations representing the relationship between the voltages and currents, which can be written in a matrix form as follows

$$\begin{aligned} \left[ \begin{array}{cccc} \mathbf {Y_{11}} &{} \mathbf {Y_{12}} &{} \dots &{} \mathbf {Y_{1N}} \\ \mathbf {Y_{21}} &{} \mathbf {Y_{22}} &{} \dots &{} \mathbf {Y_{2N}} \\ \dots &{} \dots &{} \dots &{} \dots \\ \dots &{} \dots &{} \dots &{} \dots \\ \mathbf {Y_{N1}} &{} \dots &{} \mathbf {Y_{N(N-1)}} &{} \mathbf {Y_{NN}} \end{array}\right] \left[ \begin{array}{c} \mathbf {V_{1}} \\ \mathbf {V_{2}} \\ . \\ .\\ \mathbf {V_{N}} \end{array}\right]&= \left[ \begin{array}{c} \mathbf {I_{1}} \\ \mathbf {I_{2}} \\ . \\ . \\ \mathbf {I_{N}} \end{array}\right] \end{aligned}$$

(8.47)

Where $I_{k}$ is the current that enters the bus from the generator/load side. The first matrix is called the admittance matrix or the $\mathbf {Y_{bus}}$ matrix. Each diagonal element $\mathbf {Y_{kk}}$ equals the sum of the admittances of all branches connected to bus $k$. Every off-diagonal element $\mathbf {Y_{jk}}$ where $j \ne k$ is the sum of admittances of all branches between bus $j$ and bus $k$ multiplied by $-1$. Using Eq. (8.47), we obtain the following equation at bus $k$

$$\begin{aligned} \mathbf {V_{1}Y_{k1}} + \mathbf {V_{2}Y_{k2}} + \dots + \mathbf {V_{k}Y_{kk}} + \dots + \mathbf {V_{N}Y_{kN}} = \mathbf {I_{k}} = \frac{P_{k}-jQ_{k}}{\mathbf {V_{k}^{*}}}. \end{aligned}$$

(8.48)

To find all unknown active power, reactive power, voltages, voltage angles, a famous method called Gauss-Seidel iterative approach is used by assuming flat initial solutions for all voltages and voltage angles equal 1 p.u. and zero, respectively. For bus $k$ at iteration $i+1$, the following iterative equation is used to find the solution of the unknown variables

$$\begin{aligned} \mathbf {V_{k}^{i+1}}&= \frac{1}{\mathbf {Y_{kk}}}\Bigg [\mathbf {I_{k}^{i}} - \sum _{n=1}^{k-1} \mathbf {V_{n}^{i+1}Y_{kn}} - \sum _{n=k+1}^{N}\mathbf {V_{n}^{i}Y_{kn}}\Bigg ] \end{aligned}$$

(8.49)

1.6 DC power flow model

A power grid can be considered as a complex network with $N$ nodes and $L$ links. Nodes represent the generation and transmission substations, and links represents the transmission lines. To simplify the power flow analysis in the power grids, the DC Power Flow model has been originally introduced as DC Power Flow in the DC network analyzer [84] as suggested by [85]. In the original work, the network branch is represented by a resistance and the resistance value is proportional to the reactance that is connected in series with the resistance and each DC current is proportional to the power flow. The DC power flow model represents a linearization of the full AC model. In the AC model, let $V_{i}$ and $V_{j}$ represent the voltage at the buses $i$ and $j$, respectively. In addition, let $Y_{ij}$ represent the admittance of the transmission line between buses $i$ and $j$. The relation between real power, complex voltages and line impedance is expressed through the following equation which describes the amount of real power flowing through a transmission line

$$\begin{aligned} P_{ij}&= |V_{i}| |V_{j}| |Y_{ij}| cos(\delta _{i} - \delta _{j} + \theta _{ij}) \end{aligned}$$

(8.50)

where $\theta _{ij}$ is the phasor angle of the admittance $Y_{ij}$. To obtain the DC power flow model, the following assumptions are applied to Eq. (8.50) as follows

• Voltage angle differences are small, i.e. $sin(\delta _{ij}) \approx \delta _{ij}$.

• Flat Voltage profile: All voltages are considered 1 $p.u$.

• Line resistance is negligible i.e. $R << X$.

Applying Taylor expansion on Eq. (8.50) around the operating voltage, and neglect the coupling between the power flow and the voltage, we obtain

$$\begin{aligned} P_{ij} = \frac{\delta _{ij}}{x_{ij}} \end{aligned}$$

(8.51)

where $\delta _{ij}$ is the difference in phase shift angle between the voltages at the sending and receiving buses, and $x_{ij}$ is the reactance of the transmission line. The DC power flow Eq. (8.51) can be written in matrix form where $P$ is the $N\times N$ matrix of power flows between each node $i$ and $j$ in the network, $\delta$ is the $N\times 1$ vector of phase angles and $X$ is the $N\times N$ weighted adjacency matrix, each element of which represents the reactance of a transmission line. It is a real number if a line is present between two nodes, and zero otherwise. In matrix form,

$$\begin{aligned}{}[P]=[b][\delta ] \end{aligned}$$

(8.52)

The matrix $[b]$ represents the imaginary part of the $Y_{bus}$ matrix of the power grid, where $b_{ij} = -\frac{1}{x_{ij}}$ and $b_{ii} = \sum _{i\in N}-b_{ij}$ for $i \ne j$. We usually assume that there is a reference node with voltage angle equals 0. The power handled by each node is the net sum of all the ingoing and outgoing power flows at that node as follows:

$$\begin{aligned} P_{i}=\sum _{j=1}^{N} P_{ij} = \sum _{j=1}^{N} (-b_{ij}\delta _{ij}) \end{aligned}$$

(8.53)

The total load at each node is given, while the phase angles are computed using the following equation:

$$\begin{aligned}{}[\delta ] = [b]^{-1}[P]. \end{aligned}$$

(8.54)