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Electric power transmission

Miszaina Osman

Izham Zainal Abidin

Tuan Ab Rashid Tuan Abdullah

Marayati Marsadek College of Engineering, Universiti Tenaga Nasional, Jalan IKRAM-UNITEN, Selangor Darul Ehsan, Malaysia

Abstract

This chapter covers the basic components used for modeling a transmission network. The models start with the basic resistance (R), inductance (L), and capacitance (C) derivation, which proceed toward the effect of bundling of conductors. Combining the RLC elements, the short, medium, and long line models were derived, which were later simplified to a two-port network equivalent. In addition, since the long transmission network suffers from inductance and capacitance effects, DC transmission alternatives (high voltage DC, HVDC) are introduced, which include the required converter and filter configuration.

Keywords

transmission line models

short line model

medium line model

long line model

HVDC

converter stations

harmonic filters

17.1. Introduction

The purpose of the transmission system in a power grid is to transmit electrical energy from the generating stations to the distribution networks. The transmission network also plays a role as an interconnection system with neighboring countries’ power grids, which allows economic dispatch of power within regions during normal and emergency conditions.

Because of AC flowing in the conductors, transmission lines exhibit the electrical properties of resistance, inductance, capacitance, and conductance. Inductance and capacitance is a result of magnetic and electric fields around the current carrying conductors. Transmission line models can be developed with these two parameters. As for leakage currents flowing across the insulators and ionized pathways in the air, these are represented as shunt conductance in the circuit.

17.2. Overhead transmission lines

A transmission line consists of conductors, support structures, and other equipment such as insulators, spacers, jumpers, etc. A few examples of typical structures of a transmission line are poles, lattice structures, and H-frame structures. The conductors are hung from these towers or structures, which are usually made of steel, wood, or reinforced concrete. Most transmission lines are also installed with shield wires and lightning protection devices such as the surge arrestors. Voltages above 110 kV are usually considered to be the transmission level voltages. Voltages less than 33 kV are usually used for distribution, voltages above 230 kV are considered extra-high voltage, and voltages above 765 kV are usually referred to as ultra-high voltage.

The most commonly used conductor materials for high voltage transmission lines are ACSR (aluminum conductor steel-reinforced), AAC (all-aluminum conductor), AAAC (all-aluminum alloy conductor), and ACAR (aluminum conductor alloy reinforced). These types of conductors are widely used in transmission and some distribution systems due to their relatively low cost and high strength to weight ratio. Detail characteristics of these conductors can be found in manufacturers’ data sheets. The conductors are stranded to increase their stability and flexibility and consist of a center core of steel strands surrounded by layers of aluminum strands. For purposes of heat dissipation, overhead power line conductors are bare (no insulating cover). ACSR conductors have long been widely used as overhead high tension power lines and have an established reputation for economy and dependability. Typical ACSR conductors and their standard sizes are shown in Figure 17.1.

Figure 17.1 Cross-sectional view of typical standard sizes and stranding pattern for ACSR conductors.

17.3. Transmission line parameters

For power system analysis, a particular transmission line can be represented by its resistance, inductance or inductive reactance, capacitance or capacitive reactance, and leakage resistance.

17.3.1. Line resistance

Resistance of the line will determine the performance of the line in terms of its efficiency and costing. The DC resistance of a solid round conductor at a specified temperature is given by:

$R_{DC} = \frac{ρ l}{A} Ω$ $R_{DC} = \frac{ρ l}{A} Ω$

(17.1)

where ρ is the conductor resistivity, l is the conductor length, and A is the conductor cross sectional area.

The conductor resistance is affected by three factors, which are frequency, spiraling, and temperature. When AC flows in a conductor, the phenomenon of skin effect happens. This is where the current distribution is not uniformly distributed over the conductor cross-sectional area, and the current density is greatest at the surface of the conductor. Thus, AC resistance is somewhat higher than DC resistance. Spiraling a stranded conductor will make the strand longer than the finished conductor, thus creating a higher AC resistance.

The conductor resistance increases as temperature increases. This change can be considered linear over the range of temperatures normally encountered and is calculated as follows:

$R_{2} = R_{1} \frac{T + t_{2}}{T + t_{1}}$ $R_{2} = R_{1} \frac{T + t_{2}}{T + t_{1}}$

(17.2)

where R₂ and R₁ are conductor resistances at t₂ and t₁, respectively. T is a temperature constant and depends on the conductor material. For example, for hard-drawn copper T is 241, and for annealed copper T is 234.5.

17.3.2. Line inductance

17.3.2.1. Single-phase overhead lines

Figure 17.2 shows a single-phase overhead line, consisting of two solid round conductors with radius r and separated by a distance D.

Figure 17.2 Single-phase with two wire line.

Assume that the current flows out from conductor X and returns in conductor Y. The currents will result in magnetic field lines that link between the conductors. Thus, the inductance of the conductor is expressed as:

$L = 0.2 \ln \frac{D}{D_{s}} mH / km$ $L = 0.2 \ln \frac{D}{D_{s}} mH / km$

(17.3)

where D is the distance between the two conductors and D_s is the geometric mean radius (GMR).

The GMR for stranded conductors is usually provided by the manufacturers, as calculations for conductors with a large number of strands can be very tedious. For a solid cylindrical conductor, GMR is calculated as re^−1/4 or 0.7788r.

17.3.2.2. Three-phase overhead lines

Practically, transmission lines are not able to maintain a symmetrical spacing between their conductors due to construction constraints. A three-phase line conductor with asymmetrical spacing is shown in Figure 17.3.

Figure 17.3 Three-phase line with asymmetrical spacing.

For a given conductor configuration, the average values of the inductance and capacitance can be found by representing them with equivalent equilateral spacing. Geometric mean distance (GMD) is the equivalent conductor spacing and it is calculated as:

$GMD = \sqrt[3]{D_{ab} D_{bc} D_{ca}}$ $GMD = \sqrt[3]{D_{ab} D_{bc} D_{ca}}$

(17.4)

In practice, the conductors of a transmission line are transposed. The transposition is an operation of exchanging the conductor positions, and is usually carried out at the switching stations. Therefore, the average inductance per phase is:

$L = 0.2 \ln \frac{GMD}{D_{s}} mH / km$ $L = 0.2 \ln \frac{GMD}{D_{s}} mH / km$

(17.5)

Typically, high voltage transmission lines are constructed with bundled conductors and consist of two, three, or four subconductors. The subconductors within the bundle are separated using spacer-dampers. The GMR calculations for the bundled conductors are as follows:

${GMR}_{L} = D_{s}^{b} = \sqrt{D_{s} \times d} (for two-subconductor bundle)$ ${GMR}_{L} = D_{s}^{b} = \sqrt{D_{s} \times d} (for two-subconductor bundle)$

(17.6)

${GMR}_{L} = D_{s}^{b} = \sqrt[3]{D_{s} \times d^{2}} (for three-subconductor bundle)$ ${GMR}_{L} = D_{s}^{b} = \sqrt[3]{D_{s} \times d^{2}} (for three-subconductor bundle)$

(17.7)

${GMR}_{L} = D_{s}^{b} = 1.09 \sqrt[4]{D_{s} \times d^{3}} (for four-subconductor bundle)$ ${GMR}_{L} = D_{s}^{b} = 1.09 \sqrt[4]{D_{s} \times d^{3}} (for four-subconductor bundle)$

(17.8)

17.3.3. Line capacitance

17.3.3.1. Single-phase overhead lines

In Figure 17.2, the conductors of the single-phase line have a radius r and are separated by a distance D. If conductors X and Y are carrying charges, the presence of the second conductor and ground will disturb the field of the first conductor. However, the charge is assumed to be uniformly distributed since the separation distance D is larger with respect to the radius r and the height of the conductors from ground is larger compared with D. The line-to-neutral capacitance is given by:

$C = \frac{0.0556}{\ln \frac{D}{r}} μ F / km$ $C = \frac{0.0556}{\ln \frac{D}{r}} μ F / km$

(17.9)

where D is the distance between the two conductors and r is the radius of the conductor.

17.3.3.2. Three-phase overhead lines

Similar to the inductance calculation, for capacitance of a three-phase line, the value of the capacitance will take into account the GMD of the three phases as follows:

$C = \frac{0.0556}{\ln \frac{GMD}{r}} μ F / km$ $C = \frac{0.0556}{\ln \frac{GMD}{r}} μ F / km$

(17.10)

In capacitance calculation, the effect of bundling will introduce an equivalent radius r^b. The r^b calculations for the bundled conductors are as follows:

$r^{b} = \sqrt{r \times d} (for two-subconductor bundle)$ $r^{b} = \sqrt{r \times d} (for two-subconductor bundle)$

(17.11)

$r^{b} = \sqrt[3]{r \times d^{2}} (for three-subconductor bundle)$ $r^{b} = \sqrt[3]{r \times d^{2}} (for three-subconductor bundle)$

(17.12)

$r^{b} = 1.09 \sqrt[4]{r \times d^{3}} (for four-subconductor bundle)$ $r^{b} = 1.09 \sqrt[4]{r \times d^{3}} (for four-subconductor bundle)$

(17.13)

Example 17.1

A 500-kV three-phase transposed transmission line is 100 km long and consists of one ACSR 1,272,000-cmil, 45/7 Bittern conductor per phase in a flat horizontal configuration, with a spacing of 10 m. The conductors have a diameter of 3.4160 cm and a GMR of 1.3560 cm. Calculate the inductance per phase in mH/km and the capacitance per phase in μF/km for the configuration.

Solution

From (17.4),

$\begin{array}{l} GMD & = \sqrt[3]{D_{12} D_{23} D_{13}} = \sqrt[3]{(10) (10) (20)} = 12.5992 m \\ r & = \frac{3.4160}{2} = 1.708 cm = 0.01708 m \end{array}$ $\begin{array}{l} GMD & = \sqrt[3]{D_{12} D_{23} D_{13}} = \sqrt[3]{(10) (10) (20)} = 12.5992 m \\ r & = \frac{3.4160}{2} = 1.708 cm = 0.01708 m \end{array}$

Using (17.5):

$L = 0.2 \ln \frac{GMD}{D_{S}} mH / km = 0.2 \ln \frac{12.5992}{0.013560} = 1.3669 mH / km$ $L = 0.2 \ln \frac{GMD}{D_{S}} mH / km = 0.2 \ln \frac{12.5992}{0.013560} = 1.3669 mH / km$

Using (17.10):

$C = \frac{0.0556}{\ln \frac{GMD}{r}} μ F / km = \frac{0.0556}{\ln \frac{12.5992}{0.01708}} μ F / km = 0.00842 μ F / km$ $C = \frac{0.0556}{\ln \frac{GMD}{r}} μ F / km = \frac{0.0556}{\ln \frac{12.5992}{0.01708}} μ F / km = 0.00842 μ F / km$

Example 17.2

The line in Example 17.1 is now replaced with a two-conductor bundle of ACSR 636,000-cmil, 24/7 Rook conductors per phase in a flat horizontal configuration, with a spacing of 10 m measured from the center of the bundles. The spacing between the conductors in the bundle is 45 cm. The conductors have a diameter of 2.4816 cm and a GMR of 1.0028 cm. Calculate the inductance per phase in mH/km and the capacitance per phase in μF/km for the configuration.

Solution

$r = \frac{2.4816}{2} = 1.2408 cm = 0.012408 m$ $r = \frac{2.4816}{2} = 1.2408 cm = 0.012408 m$

For inductance calculation from (17.6),

{GMR}_{L} = \sqrt{D_{s} \times d} = \sqrt{0.010028 \times (0.45)} = 0.0672 m

${GMR}_{L} = \sqrt{D_{s} \times d} = \sqrt{0.010028 \times (0.45)} = 0.0672 m$

$L = 0.2 \ln \frac{GMD}{{GMR}_{L}} mH / km = 0.2 \ln \frac{12.5992}{0.0672} = 1.0467 mH / km$ $L = 0.2 \ln \frac{GMD}{{GMR}_{L}} mH / km = 0.2 \ln \frac{12.5992}{0.0672} = 1.0467 mH / km$

For inductance calculation from (17.11),

$\begin{array}{l} {GMR}_{C} = r^{b} = \sqrt{r \times d} = \sqrt{0.012408 \times 0.45} = 0.0747 m \\ C = \frac{0.0556}{\ln \frac{GMD}{{GMR}_{c}}} μ F / km = \frac{0.0556}{\ln \frac{12.5992}{0.0747}} μ F / km = 0.0108 μ F / km \end{array}$ $\begin{array}{l} {GMR}_{C} = r^{b} = \sqrt{r \times d} = \sqrt{0.012408 \times 0.45} = 0.0747 m \\ C = \frac{0.0556}{\ln \frac{GMD}{{GMR}_{c}}} μ F / km = \frac{0.0556}{\ln \frac{12.5992}{0.0747}} μ F / km = 0.0108 μ F / km \end{array}$

17.4. Transmission line representation

This section presents a transmission line representation using a π model. An alternative representation of the line using a T model is briefly discussed. The model is based on formulae to calculate the voltage, current, and power along a transmission line. In a transmission line, a set of values such as voltage and current are usually known at one end of the line. Those values are required to calculate other unknown electrical quantities at another point in the line. The model may be used to evaluate the design and operational performance of a power system such as line efficiency, losses, and limits of power flow over a line both under steady-state and transient conditions. It can also be used to study the effect of the parameters of the line on bus voltages and the flow of the power.

For discussion purposes, consider the representation of an electric system consisting of a generator supplying a balanced three-phase load Z_L through a power transmission line with a lump parameter Z = R + jωL. The three-phase electrical connection of this system is Y-connected. A schematic of the generator, transmission line, and load is shown in Figure 17.4. The line parameters are represented by lump parameters and the line capacitor is omitted.

Figure 17.4 Generator supplying a balance three-phase load.

Normally, the transmission lines are operated in a balanced three-phase load. A single-phase equivalent circuit of the three-phase schematic can be used to simplify the discussion and calculation presented in this section. A single-phase equivalent circuit with shunt capacitors added at the sending end and receiving end of the transmission line is shown in Figure 17.5. The total capacitance of the line, C, is divided equally at the sending and receiving end and labeled as C/2. To differentiate the total series impedances of a line and the series impedance per unit length, the following symbols are used:

z = series impedance per unit length per phase,

y = shunt admittance per unit length per phase to neutral,

l = length of line,

Z = zl = total series admittance per phase,

Y = yl = total shunt admittance per phase to neutral,

γ = α + jβ = a propagation constant. Its real part is called attenuation constant α measured in nepers per unit length. Its quadrature part is called phase constant β measured in radians per unit length,

Z_c = characteristic impedance of a line.

Figure 17.5 A single-phase equivalent of the circuit with capacitance to neutral added.

An equivalent circuit of a short-length transmission line is depicted in Figure 17.6, the sending and receiving end currents I_S and I_R are the same since there is no shunt arm. The V_S and V_R are the sending and receiving end line-to-neutral voltages. The line impedance is modeled as a lumped parameter impedance Z = R + jωL.

Figure 17.6 Equivalent circuit of a short transmission line.

The formulae to calculate the currents and voltages for the short power transmission line are:

$I_{S} = I_{R}$ $I_{S} = I_{R}$

(17.14)

$V_{S} = V_{R} + I_{R} Z$ $V_{S} = V_{R} + I_{R} Z$

(17.15)

Given a set of currents and voltages at one end of the line and the line parameters, the value of the currents and voltages at the other end can be calculated from these two equations, which modeled the short-length power transmission line.

An equivalent circuit of a medium-length transmission line is depicted in Figure 17.7. The sending and receiving end currents I_S and I_R are not necessarily the same since there are two shunt arms. The shunts arms represent the total admittance of the line divided into two equal parts and placed at the sending and receiving ends of the line. The circuit is called a nominal π. An expression for V_S can be represented by noting the capacitance at the receiving end V_RY/2 and the current in the series arm is I_R + V_RY/2. The formulae to compute the current and voltage are:

$V_{S} = (V_{R} \frac{Y}{2} + I_{R}) Z + V_{R}$ $V_{S} = (V_{R} \frac{Y}{2} + I_{R}) Z + V_{R}$

(17.16)

$V_{S} = (\frac{Z Y}{2} + 1) V_{R} + Z I_{R}$ $V_{S} = (\frac{Z Y}{2} + 1) V_{R} + Z I_{R}$

(17.17)

Figure 17.7 Nominal circuit of a medium-length transmission line.

Noting that the current in the shunt capacitance at the sending end is

V_{S} (Y / 2)

$V_{S} (Y / 2)$

added to the current in the series arm gives I_S, thus:

$I_{S} = V_{S} \frac{Y}{2} + V_{R} \frac{Y}{2} + I_{R}$ $I_{S} = V_{S} \frac{Y}{2} + V_{R} \frac{Y}{2} + I_{R}$

(17.18)

Substituting V_S given by equation (17.17) into equation (17.18) gives:

$I_{S} = V_{R} Y (1 + \frac{Z Y}{4}) + (\frac{Z Y}{2} + 1) I_{R}$ $I_{S} = V_{R} Y (1 + \frac{Z Y}{4}) + (\frac{Z Y}{2} + 1) I_{R}$

(17.19)

Corresponding equations can also be derived for the nominal T. In the latter case, all the shunt admittances of the line are lumped in the shunt arm of the T and the series impedance is divided equally between the two series arms.

In the previous two cases, lump parameters were used to represent the line constants in the short-length and medium-length power transmission lines. However, the nominal π model may not represent a long-length power transmission line exactly since the nominal model does not take into account the uniformly distributed parameters of the long-length line. Despite that, an equivalent circuit of a long line can be established by modifying the model for the medium-length transmission line.

For discussion purposes, denote the series equivalent π circuit Z′ and the shunt arms Y′/2 to distinguish them from the nominal π parameters. Then, replace the respective Z and Y/2 in equation (17.17), thus obtaining the sending voltage as:

$V_{S} = (\frac{Z^{'} Y^{'}}{2} + 1) V_{R} + Z^{'} I_{R}$ $V_{S} = (\frac{Z^{'} Y^{'}}{2} + 1) V_{R} + Z^{'} I_{R}$

(17.20)

We note that the generalized circuit constant for a long-length power transmission line is:

$A = \cosh γ l$ $A = \cosh γ l$

(17.21)

$B = Z_{c} \sinh γ l$ $B = Z_{c} \sinh γ l$

(17.22)

$C = \frac{\sinh γ l}{Z_{c}}$ $C = \frac{\sinh γ l}{Z_{c}}$

(17.23)

$D = \cosh γ l$ $D = \cosh γ l$

(17.24)

And the line equations are:

$V_{S} = A V_{R} + B I_{R}$ $V_{S} = A V_{R} + B I_{R}$

(17.25)

$I_{S} = C V_{R} + D I_{R}$ $I_{S} = C V_{R} + D I_{R}$

(17.26)

Inspecting and equating the coefficients of Equations (17.20) and (17.25), we can deduce that:

$Z^{'} = Z_{c} \sinh γ l$ $Z^{'} = Z_{c} \sinh γ l$

(17.27)

$Z^{'} = \sqrt{\frac{z}{y}} \sinh γ l = z l \frac{\sinh γ l}{\sqrt{z l} l}$ $Z^{'} = \sqrt{\frac{z}{y}} \sinh γ l = z l \frac{\sinh γ l}{\sqrt{z l} l}$

(17.28)

$Z^{'} = Z \frac{\sinh γ l}{γ l}$ $Z^{'} = Z \frac{\sinh γ l}{γ l}$

(17.29)

Recall that Z is equal to the zl, the total series impedance of the line. For the shunt arm of the equivalent π circuit, equate the coefficient of V_R in Equation (17.20) with circuit constant A in Equation (17.21), thus:

$\frac{Z^{'} Y^{'}}{2} + 1 = \cosh γ l$ $\frac{Z^{'} Y^{'}}{2} + 1 = \cosh γ l$

(17.30)

Use Equation (17.27) to eliminate Z′ in Equation (17.30) and obtain:

$\frac{Y^{'} Z_{c} \sinh γ l}{2} + 1 = \cosh γ l$ $\frac{Y^{'} Z_{c} \sinh γ l}{2} + 1 = \cosh γ l$

(17.31)

Rearrange Equation (17.31) and obtain:

$\frac{Y^{'}}{2} = \frac{1}{Z_{c}} \frac{\cosh γ l - 1}{\sinh γ l}$ $\frac{Y^{'}}{2} = \frac{1}{Z_{c}} \frac{\cosh γ l - 1}{\sinh γ l}$

(17.32)

From series impedance Equation (17.29) and the shunts arms Equation (17.32), we can deduce the equivalent π circuit for the long-length power transmission line as shown in Figure 17.8.

Figure 17.8 Equivalent π circuit of a long-length transmission line.

17.5. Transmission line as a two-port network and power flow

A transmission line is very important in ensuring energy is transported from energy sources to loads. From a power system analysis point of view, transmission line models are very important to ensure better representation of transmission lines. In the previous section, detailed line models were derived. Since a detailed model is assumed to be the best, however, from a power system simulation point of view, simplification of models is recommended. In general, there are three different types of transmission line models based upon line distance. They are as follows:

1. Short line model (below 80 km)

2. Medium line model (80–250 km)

3. Long line model (above 250 km)

17.5.1. The two-port model

The two-port model is a very simplified approach to determine the correct model parameters to be used for different types of transmission lines. The two-port model is shown in Figure 17.9.

The two-port model assumes that the transmission line model is a black box with sending end current (I_S) and voltage (V_S) at the input and the receiving end current (I_R) and voltage (V_R) at the output. Based upon Figure 17.9, the derived equations are as follows:

$V_{S} = A V_{R} + B I_{R}$ $V_{S} = A V_{R} + B I_{R}$

(17.33)

$I_{S} = C V_{R} + D I_{R}$ $I_{S} = C V_{R} + D I_{R}$

(17.34)

In matrix form:

$[\begin{array}{c} V_{S} \\ I_{S} \end{array}] = [\begin{array}{c} A & B \\ C & D \end{array}] [\begin{array}{c} V_{R} \\ I_{R} \end{array}]$ $[\begin{array}{c} V_{S} \\ I_{S} \end{array}] = [\begin{array}{c} A & B \\ C & D \end{array}] [\begin{array}{c} V_{R} \\ I_{R} \end{array}]$

(17.35)

17.5.2. Short line model

The short line model is reserved for lines that are less than 80 km in distance. This is because the impact of line capacitance is considered to be insignificant. Therefore the short line model is simply modeled with a unit series impedance (per km) multiplied by the total line distance. This is shown in Equation (17.36), and the circuit diagram representing the short line model is shown in Figure 17.6.

$Z_{line} = (R + j ω L) l_{line} = R + j X Ω$ $Z_{line} = (R + j ω L) l_{line} = R + j X Ω$

(17.36)

where R is the per phase resistance per unit distance, L is the per phase inductance per unit distance, and l_line is the total line distance (usually in km).

The two-port model representation of a short line is as follows:

$A = 1, B = Z, C = 0, D = 1$ $A = 1, B = Z, C = 0, D = 1$

(17.37)

17.5.3. Medium line model

For lines that are between 80 km and 250 km, the medium line model is used. This is because at this distance, shunt capacitance will have some impact on the overall impedance of the line. The circuit representation is shown in Figure 17.7. The two-port ABCD model for medium lines is as follows:

$A = (1 + \frac{Z Y}{2}), B = Z, C = Y (1 + \frac{Z Y}{4}), D = (1 + \frac{Z Y}{2})$ $A = (1 + \frac{Z Y}{2}), B = Z, C = Y (1 + \frac{Z Y}{4}), D = (1 + \frac{Z Y}{2})$

(17.38)

17.5.4. Long line model

If the lines are longer than 250 km, distributed line parameters must be considered in order to determine the overall line impedance. The circuit representation is shown in Figure 17.10.

The two-port ABCD model for long lines is as follows:

$A = \cosh (γ l), B = Z_{c} \sinh (γ l), C = \frac{1}{Z_{c}} \sinh (γ l), D = \cosh (γ l)$ $A = \cosh (γ l), B = Z_{c} \sinh (γ l), C = \frac{1}{Z_{c}} \sinh (γ l), D = \cosh (γ l)$

(17.39)

where

γ = \sqrt{z y}

$γ = \sqrt{z y}$

(propogation constant),

Z_{c} = \sqrt{\frac{z}{y}}

$Z_{c} = \sqrt{\frac{z}{y}}$

(characteristic impedance), and l = line length.

The previously mentioned model can be further simplified by assuming a lossless model where the two-port ABCD model for this is as follows:

$A = \cos (β l), B = j Z_{c} \sin (β l), C = j \frac{1}{Z_{c}} \sin (β l), D = \cos (β l)$ $A = \cos (β l), B = j Z_{c} \sin (β l), C = j \frac{1}{Z_{c}} \sin (β l), D = \cos (β l)$

(17.40)

where l = line length,

β = ω \sqrt{L C}

$β = ω \sqrt{L C}$

(phase constant), and

Z_{c} = \sqrt{\frac{z}{y}}

$Z_{c} = \sqrt{\frac{z}{y}}$

(characteristic impedance).

Example 17.3

A 500-kV, 60-Hz, 3-Ø line has a length of 180 km and delivers 1600 MW at 475 kV and at 0.95 power factor leading to the receiving end at full load. The series impedance (z) is given as 0.0201 + j 0.335 Ω/Ø/km and the shunt admittance (y) is j 4.807 × 10⁻⁶ S/Ø/km. Using the nominal π- circuit for the problem stated, determine the ABCD parameters for this model, sending-end voltage and current, sending-end power and power factor, full-load line losses and efficiency, and voltage regulation.

Solution

Using the medium line model:

$\begin{array}{l} A & = D = 1 + \frac{Y Z}{Z} = 1 + \frac{1}{2} (0.336 \times 180 ∠ 86.6 °) (4.807 \times 10^{- 6} 180 ∠ 90 °) \\ = 0.9739 ∠ 0.0912 ° pu \\ B & = Z = z ℓ = 0.336 (180) ∠ 86.6 ° = 60.48 ∠ 86.6 ° Ω \\ C & = Y (1 + \frac{YZ}{4}) = (4.307 \times 10^{- 6} \times 180 ∠ 90 °) (1 + 0.0131 ∠ 176.6 °) = 8.54 \times 10^{- 4} ∠ 90.05 ° S \end{array}$ $\begin{array}{l} A & = D = 1 + \frac{Y Z}{Z} = 1 + \frac{1}{2} (0.336 \times 180 ∠ 86.6 °) (4.807 \times 10^{- 6} 180 ∠ 90 °) \\ = 0.9739 ∠ 0.0912 ° pu \\ B & = Z = z ℓ = 0.336 (180) ∠ 86.6 ° = 60.48 ∠ 86.6 ° Ω \\ C & = Y (1 + \frac{YZ}{4}) = (4.307 \times 10^{- 6} \times 180 ∠ 90 °) (1 + 0.0131 ∠ 176.6 °) = 8.54 \times 10^{- 4} ∠ 90.05 ° S \end{array}$

To calculate the sending-end voltage and current:

$\begin{array}{l} V_{R} & = \frac{475}{\sqrt{3}} ∠ 0 ° = 274.24 ∠ 0 ° kV, I_{R} = \frac{P_{R} ∠ \cos^{- 1} (p f)}{\sqrt{3} \times V_{R - LL (pf)}} = \frac{1600 MVA ∠ \cos^{- 1} (0.95)}{\sqrt{3} \times 475 kV (0.95)} \\ = 2.047 ∠ 18.19 ° kA \\ V_{S} & = A V_{R} + B I_{R} = (0.9739 ∠ 0.0912 °) (274.24) + (60.48 ∠ 86.6 °) (2.047 ∠ 18.19 °) \\ = 264.4 ∠ 27.02 ° {kV}_{3 \emptyset} \\ So, V_{S} & = \sqrt{3} \times 264.4 = 457.9 {kV}_{3 \emptyset} \\ I_{S} & = C V_{R} + D I_{R} = (8.54 \times 10^{- 4} ∠ 90.05 °) (274.24) + (0.9739 ∠ 0.0912 °) (2.047 ∠ 18.19 ° kA) \\ = 2.079 ∠ 24.42 ° kA \end{array}$ $\begin{array}{l} V_{R} & = \frac{475}{\sqrt{3}} ∠ 0 ° = 274.24 ∠ 0 ° kV, I_{R} = \frac{P_{R} ∠ \cos^{- 1} (p f)}{\sqrt{3} \times V_{R - LL (pf)}} = \frac{1600 MVA ∠ \cos^{- 1} (0.95)}{\sqrt{3} \times 475 kV (0.95)} \\ = 2.047 ∠ 18.19 ° kA \\ V_{S} & = A V_{R} + B I_{R} = (0.9739 ∠ 0.0912 °) (274.24) + (60.48 ∠ 86.6 °) (2.047 ∠ 18.19 °) \\ = 264.4 ∠ 27.02 ° {kV}_{3 \emptyset} \\ So, V_{S} & = \sqrt{3} \times 264.4 = 457.9 {kV}_{3 \emptyset} \\ I_{S} & = C V_{R} + D I_{R} = (8.54 \times 10^{- 4} ∠ 90.05 °) (274.24) + (0.9739 ∠ 0.0912 °) (2.047 ∠ 18.19 ° kA) \\ = 2.079 ∠ 24.42 ° kA \end{array}$

To calculate sending-end power and power factor:

$\begin{array}{l} P_{S} = \sqrt{3} \times V_{S - 3 \emptyset} I_{S} (p f) = \sqrt{3} \times 457.9 (2.079) \cos (27.02 ° - 24.42 °) = 1647 MW \\ p f = \cos (27.02 ° - 24.42 °) = 0.999 lagging \end{array}$ $\begin{array}{l} P_{S} = \sqrt{3} \times V_{S - 3 \emptyset} I_{S} (p f) = \sqrt{3} \times 457.9 (2.079) \cos (27.02 ° - 24.42 °) = 1647 MW \\ p f = \cos (27.02 ° - 24.42 °) = 0.999 lagging \end{array}$

Calculating full-load line losses and efficiency:

Full load line losses = P_S – P_R = 1647 – 1600 = 47 MW

Efficiency =

(\frac{P_{R}}{P_{S}}) 100 = (\frac{1600}{1647}) 100 = 97.1 %

$(\frac{P_{R}}{P_{S}}) 100 = (\frac{1600}{1647}) 100 = 97.1 %$

Voltage regulation is as follows:

$\begin{array}{l} V_{R - NL} = \frac{V_{S}}{A} = \frac{457.9}{0.9739} = 470.2 {kV}_{3 \emptyset} \\ V_{R} = \frac{V_{R - NL} - V_{R - FL}}{V_{R - FL}} = \frac{470.2 - 475}{475} \times 100 % = - 1 % \end{array}$ $\begin{array}{l} V_{R - NL} = \frac{V_{S}}{A} = \frac{457.9}{0.9739} = 470.2 {kV}_{3 \emptyset} \\ V_{R} = \frac{V_{R - NL} - V_{R - FL}}{V_{R - FL}} = \frac{470.2 - 475}{475} \times 100 % = - 1 % \end{array}$

17.6. High voltage DC transmission

The improvement and enhancement of power transfer capability is made possible with the advent of power electronics technologies. High voltage DC (HVDC) transmission connects two AC systems via a DC link and utilizes a power electronics-based converter. Figure 17.11 shows the typical arrangement of an HVDC system. Power transfer using HVDC offers the following advantages:

1. Fewer line losses

2. Less expensive for long distance transmission

3. Reduced system disturbance

4. Enables power transmission between two asynchronous AC systems

5. Power flow can be controlled

17.6.1. High voltage DC transmission converter

A converter is one of the main components in an HVDC system. The converter used in an HVDC station can either be a 6-pulse or 12-pulse converter. The typical circuit arrangement for a six-pulse converter is shown in Figure 17.12. In a six-pulse converter, the thyristor valve is fired every 60°. The average output voltage of the converter (i.e., V_d) can be varied by varying the firing delay angle α of the thyristor. The mode of operation of the converter is determined by α, which is summarized in Table 17.1. Figure 17.13 shows the output voltage of the converter when α is 0°, 30°, and 120°, where curves ab, ba, ac, ca, bc, and cb represent the line-to-line voltage. The average output voltage of the converter can be calculated as follows:

$V_{d} = V_{do} cos (α)$ $V_{d} = V_{do} cos (α)$

(17.41)

Table 17.1

Mode of operation of a converter

Range of α	Mode of operation	Remarks
0° < α < 90°	Rectification	V_d has positive value
α = 90°	–	V_d is zero
90° < α < 180°	Inversion	V_d has negative value

Figure 17.13 Output voltage of converter.

where:

$V_{do} = \frac{3 \sqrt{3}}{π} V_{m}$ $V_{do} = \frac{3 \sqrt{3}}{π} V_{m}$

In a 12-pulse converter, the thyristor valve is fired every 30° and it is obtained by arranging two 6-pulse converters in series. Figure 17.14 shows a simplified diagram of a 12-pulse converter.

17.6.2. AC and DC filter

The converters in an HVDC system introduce harmonics to the system. The harmonics generated by the converter are fed into the AC system and DC link. Harmonics must be sufficiently filtered otherwise they may cause problems like excessive power losses, overvoltage, and heating in the machine. From an AC system point of view, the converter acts as a source of current harmonics while in the DC link the converter acts as a source of voltage harmonics. AC and DC filter arms are installed on the AC system and DC link, respectively, in order to reduce the harmonics order that penetrates into the system. The functions and order of harmonics that exist in the HVDC system are summarized in Table 17.2.

Figure 17.14 Simplified diagram of a 12-pulse converter.

Table 17.2

Filter in HVDC system

Type of filter	Function	Order of harmonics, n
AC filter	• To suppress the harmonics generated by the converter • To supply reactive power to the converter	n = 6k ± 1 (for six-pulse converter) n = 12k ± 1 (for 12-pulse converter) where k = 1, 2, …
DC filter	• To suppress the harmonics generated by the converter	n = 6k (for six-pulse converter) n = 12k (for 12-pulse converter) where k = 1, 2, …

Type of filter

Function

Order of harmonics, n

AC filter

• To suppress the harmonics generated by the converter

• To supply reactive power to the converter

n = 6k ± 1 (for six-pulse converter)

n = 12k ± 1 (for 12-pulse converter)

where k = 1, 2, …

DC filter

• To suppress the harmonics generated by the converter

n = 6k (for six-pulse converter)

n = 12k (for 12-pulse converter)

where k = 1, 2, …

17.6.3. Control of HVDC

The power flow in an HVDC system can be inherently controlled. The control strategies in an HVDC system should fulfill the following criteria:

1. The maximum DC current should be limited to 1.2 pu.

2. The maximum DC voltage should be maintained in order to reduce line losses.

3. The converter should minimize reactive power consumption.

4. The minimum extinction angle should be maintained so as to avoid commutation failure.

The general control mode of a converter is summarized in Table 17.3. During normal voltage, the converter in the rectification process will operate in CC (constant current) control mode while the converter in the inversion process operates in CEA (constant extinction angle) control mode. In the event of disturbance where voltage at the rectifier side drops to a level at which the minimum firing delay angle has been reached and no further voltage increase is possible, the control mode of the converter in the rectification process switches from CC to CIA (constant ignition angle) control mode. When operating at this lower voltage, the DC current will reduce to a value within the current margin I_m and the converter in the inversion process operates in CC control mode. Figure 17.15 shows the V_d – I_d characteristic of converters in an HVDC system. The intersection of control mode between rectifier and inverter signifies the operating point. The analysis of HVDC control mode can be carried out by considering the simplified converter circuit shown in Figure 17.16.

Table 17.3

Typical HVDC control mode

Converter operation mode	Control mode	Description
Rectification	CC	• Rectifier maintains constant current by changing α
Rectification	CIA	• Rectifier operates with minimum firing delay angle (i.e., α_min) and no further voltage increase is possible
Inversion	CEA	• Inverter maintains constant extinction angle (i.e., γ)
Inversion	CC	• The CC control mode in the inverter is provided when there is a reduction in voltage that may cause the current to drop to zero

Figure 17.15 V_d – I_d characteristic of converters.

Figure 17.16 Simplified converter diagram.

17.7. Summary

An electric transmission line has important parameters, namely resistance, inductance, and capacitance. These parameters will affect the electrical design and performance of the line. The line parameters are uniformly distributed along the whole line and are functions of line geometry, construction material, and operational frequency. These parameters and the load current and power factor will determine the electrical performance of the line.

Problems

1. A three-phase, 500-kV, 50-Hz, transposed line is composed of four ACSR 1,033,525, 54/7 Curlew conductors per phase with flat horizontal spacing of 14 m. The Curlew conductors have a diameter of 3.162 cm and a GMR of 0.5715 cm. The bundle spacing is 45 cm and the line is 260 km long. For the purpose of this problem, a lossless line is assumed.

a. Based on the parameters given, calculate the ABCD constant.

b. If the line delivers a load of 1500 MVA at 0.7 lagging power factor at 500 kV, determine the sending end quantities (V_S, I_S, and S_S (three-phase)) and voltage regulation.

c. If the line now delivers a purely resistive load of 238 Ω, discuss what will happen to the voltage regulation and the receiving end current (I_R) as compared to load and condition in (b). Give reasons to your answer.

2. A three-phase transposed transmission line with voltage rating of 765 kV, 60 Hz is composed of four ACSR 1,431,000 cimil, 45/7 Bobolink-type conductors per phase. This configuration has a flat horizontal spacing of 14 m. The conductors have a diameter of 3.625 cm and a GMR of 1.439 cm. The bundle has a spacing of 45 cm. The length of the line is 400 km.

a. From the structure of the transmission line conductors described earlier, calculate the inductance (L) in mH/km and the capacitance (C) in μF/km of the line.

b. Using the long line lossless model, determine the transmission line ABCD constants.

c. Determine the sending-end quantities, sending-end current (I_S), sending-end line voltage (V_S(LL)), sending-end power (S_S(3Ø)), and voltage regulation if this line delivered 2000 MVA at 0.8 lagging power factor at 735 kV.

3. What is the fraction in error on the calculation of the series impedance of the long-length power transmission line using the equivalent π model for the medium-length power transmission line as compared to the equivalent π model for the long-length power transmission line?

4. Show that the shunt arm of admittance for the equivalent π model for the long-length power transmission line is:

\frac{Y^{'}}{2} = \frac{Y}{2} \frac{\tanh (\frac{γ l}{2})}{(\frac{γ l}{2})}

$\frac{Y^{'}}{2} = \frac{Y}{2} \frac{\tanh (\frac{γ l}{2})}{(\frac{γ l}{2})}$

Hint:

tanh (\frac{γ l}{2}) = \frac{\cosh γ l - 1}{\sinh γ l}

$tanh (\frac{γ l}{2}) = \frac{\cosh γ l - 1}{\sinh γ l}$

5. Derive an equivalent T circuit model for the long-length power transmission line. Note: An equivalent T circuit model for the long-length power transmission line has all the shunt admittance of the line lumped in the shunt arm of the T and the series impedance is divided equally between the two series arms.

6. Figures 17.17 and 17.18 show the V_d – I_d characteristics and equivalent circuit of converters in an HVDC station, respectively. The rectifier and inverter are initially operated with ignition delay angle and extinction angle of 20° and 15°, respectively. Point “P” is the converter’s operating point at normal voltage. Answer the following questions:

a. If the converters operate at a normal voltage, determine the value of the rectifier no-load voltage V_dor and rectifier end voltage V_dr required to transmit a load at 1000 A. The rectifier ignition angle and inverter extinction angle is given as 15° and 18°, respectively.

b. What is the maximum reduction of voltage (in %) on the rectifier side in order for the converters to remain in the existing control mode if the minimum firing delay angle is given as 5°?

c. Describe what will happen to the control mode of each converter if further voltage reduction on the rectifier side occurred due to some disturbances.

Figure 17.17 V_d – I_d characteristic of converters used in Problem 6.

Figure 17.18 Simplified converter diagram.

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Table of Contents for 17: Electric power transmission

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Abstract

Keywords

17.1. Introduction

17.2. Overhead transmission lines

17.3. Transmission line parameters

17.3.1. Line resistance

17.3.2. Line inductance

17.3.2.1. Single-phase overhead lines

17.3.2.2. Three-phase overhead lines

17.3.3. Line capacitance

17.3.3.1. Single-phase overhead lines

17.3.3.2. Three-phase overhead lines

17.4. Transmission line representation

17.5. Transmission line as a two-port network and power flow

17.5.1. The two-port model

17.5.2. Short line model

17.5.3. Medium line model

17.5.4. Long line model

17.6. High voltage DC transmission

17.6.1. High voltage DC transmission converter

17.6.2. AC and DC filter

17.6.3. Control of HVDC

17.7. Summary

Table of Contents for
17: Electric power transmission