5. Shunt Admittance of Overhead and Underground Lines

Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

The shunt admittance of a line consists of the conductance and the capacitive susceptance. The conductance is usually ignored because it is very small compared to the capacitive susceptance. The capacitance of a line is the result of the potential difference between conductors. A charged conductor creates an electric field that emanates outward from the center of the conductor. Lines of equipotential are created that are concentric to the charged conductor. This is illustrated in Figure 5.1.

In Figure 5.1, a difference in potential between two points (P1 and P2) is a result of the electric field of the charged conductor. When the potential difference between the two points is known, the capacitance between the two points can be computed. If there are other charged conductors nearby, the potential difference between the two points will be a function of the distance to the other conductors and the charge on each conductor. The principle of superposition is used to compute the total voltage drop between two points and then the resulting capacitance between the points. Understand that the points can be points in space or the surface of two conductors or the surface of a conductor and the ground.

5.1 General Voltage Drop Equation

Figure 5.2 shows an array of N positively charged solid round conductors. Each conductor has a unique uniform charge density of q cb/m.

The voltage drop between conductor i and conductor j as a result of all of the charged conductors is given by:

V_{i j} = \frac{1}{2 π ε} (q_{1} \ln \frac{D_{1 j}}{D_{1 i}} + \dots + q_{i} \ln \frac{D_{i j}}{R D_{i}} + \dots + q_{j} \ln \frac{R D_{j}}{D i j} + \dots + q_{N} \ln \frac{D_{N j}}{D_{N i}})

$V_{i j} = \frac{1}{2 π ε} (q_{1} \ln \frac{D_{1 j}}{D_{1 i}} + \dots + q_{i} \ln \frac{D_{i j}}{R D_{i}} + \dots + q_{j} \ln \frac{R D_{j}}{D i j} + \dots + q_{N} \ln \frac{D_{N j}}{D_{N i}})$ (5.1)

images

FIGURE 5.1
Electric field of a charged round conductor.

images

FIGURE 5.2
Array of round conductors.

Equation 5.1 can be written in general form as:

V_{i j} = \frac{1}{2 π ε} \sum_{n = 1}^{N} q_{n} \ln \frac{D_{n j}}{D_{n i}}

$V_{i j} = \frac{1}{2 π ε} \sum_{n = 1}^{N} q_{n} \ln \frac{D_{n j}}{D_{n i}}$ (5.2)

where ε = ε₀ε_r = permittivity of the medium

ε₀ = permittivity of free space = 8.85·10^–12 μF/m

ε_r = relative permittivity of the medium

q_n = charge density on conductor n cb/m

$D_{n i}$ $D_{n i}$ = distance between conductor n and conductor i (ft)

$D_{n j}$ $D_{n j}$ = distance between conductor n and conductor j (ft)

$D_{n n}$ $D_{n n}$ = radius (RD_n) of conductor n (ft)

5.2 Overhead Lines

The method of conductors and their images is employed in the calculation of the shunt capacitance of overhead lines. This is the same concept that was used in Chapter 4 in the general application of Carson’s equations. Figure 5.3 illustrates the conductors and their images, and it will be used to develop a general voltage drop equation for overhead lines.

In Figure 5.3, it is assumed that:

\begin{array}{l} q_{i}^{'} = - q_{i} \\ q_{j}^{'} = - q_{j} \end{array}

$\begin{array}{l} q_{i}^{'} = - q_{i} \\ q_{j}^{'} = - q_{j} \end{array}$ (5.3)

Appling Equation 5.2 to Figure 5.3:

V_{i i} = \frac{1}{2 π ε} (q_{i} \ln \frac{S_{i i}}{R D_{i}} + q_{i}^{'} \ln \frac{R D_{i}}{S_{i i}} + q_{j} \ln \frac{S_{i j}}{D_{i j}} + q_{j}^{'} \ln \frac{D_{i j}}{S_{i j}})

$V_{i i} = \frac{1}{2 π ε} (q_{i} \ln \frac{S_{i i}}{R D_{i}} + q_{i}^{'} \ln \frac{R D_{i}}{S_{i i}} + q_{j} \ln \frac{S_{i j}}{D_{i j}} + q_{j}^{'} \ln \frac{D_{i j}}{S_{i j}})$ (5.4)

Because of the assumptions of Equation 5.3, Equation 5.4 can be simplified to:

\begin{array}{l} V_{i i} = \frac{1}{2 π ε} (q_{i} \ln \frac{S_{i i}}{R D_{i}} - q_{i} \ln \frac{R D_{i}}{S_{i i}} + q_{j} \ln \frac{S_{i j}}{D_{i j}} - q_{j} \ln \frac{D_{i j}}{S_{i j}}) \\ V_{i i} = \frac{1}{2 π ε} (q_{i} \ln \frac{S_{i i}}{R D_{i}} + q_{i} \ln \frac{S_{i i}}{R D_{i}} + q_{j} \ln \frac{S_{i j}}{D_{i j}} + q_{j} \ln \frac{S_{i j}}{D_{i j}}) \\ V_{i i} = \frac{1}{2 π ε} (2 \cdot q_{i} \ln \frac{S_{i i}}{R D_{i}} + 2 \cdot q_{j} \ln \frac{S_{i j}}{D_{i j}}) \end{array}

$\begin{array}{l} V_{i i} = \frac{1}{2 π ε} (q_{i} \ln \frac{S_{i i}}{R D_{i}} - q_{i} \ln \frac{R D_{i}}{S_{i i}} + q_{j} \ln \frac{S_{i j}}{D_{i j}} - q_{j} \ln \frac{D_{i j}}{S_{i j}}) \\ V_{i i} = \frac{1}{2 π ε} (q_{i} \ln \frac{S_{i i}}{R D_{i}} + q_{i} \ln \frac{S_{i i}}{R D_{i}} + q_{j} \ln \frac{S_{i j}}{D_{i j}} + q_{j} \ln \frac{S_{i j}}{D_{i j}}) \\ V_{i i} = \frac{1}{2 π ε} (2 \cdot q_{i} \ln \frac{S_{i i}}{R D_{i}} + 2 \cdot q_{j} \ln \frac{S_{i j}}{D_{i j}}) \end{array}$ (5.5)

where S _ii = distance from conductor i to its image i′ (ft)

S _ij = distance from conductor i to the image of conductor j (ft)

D _ij = distance from conductor i to conductor j (ft)

RD _i = radius of conductor i in ft.

Equation 5.5 gives the total voltage drop between conductor i and its image. The voltage drop between conductor i and the ground will be one-half of that given in Equation 5.5.

V_{i g} = \frac{1}{2 π ε} (q_{i} \ln \frac{S_{i i}}{R D_{i}} + q_{j} \ln \frac{S_{i j}}{D_{i j}})

$V_{i g} = \frac{1}{2 π ε} (q_{i} \ln \frac{S_{i i}}{R D_{i}} + q_{j} \ln \frac{S_{i j}}{D_{i j}})$ (5.6)

images

Figure 5.3
Conductors and images.

Equation 5.6 can be written in a general form as:

V_{i g} = P_{i i} \cdot q_{i} + P_{i j} \cdot q_{j}

$V_{i g} = P_{i i} \cdot q_{i} + P_{i j} \cdot q_{j}$ (5.7)

where $P_{i i}$ $P_{i i}$ and $P_{i j}$ $P_{i j}$ are the self- and mutual “potential coefficients.”

For overhead lines, the relative permittivity of air is assumed to be 1.0 so that:

ε_{air} = 1.0 \times 8.85 \times 10^{- 12} F/m ε_{air} = 1.4240 \times 10^{- 2} mF/mile

$ε_{air} = 1.0 \times 8.85 \times 10^{- 12} F/m ε_{air} = 1.4240 \times 10^{- 2} mF/mile$ (5.8)

Using the value of permittivity in μF/mile, the self- and mutual potential coefficients are defined as:

{\hat{P}}_{i i} = 11.17689 \cdot \ln \frac{S_{i i}}{R D_{i}} mile/ μ F

${\hat{P}}_{i i} = 11.17689 \cdot \ln \frac{S_{i i}}{R D_{i}} mile/ μ F$ (5.9)

{\hat{P}}_{i j} = 11.17689 \cdot \ln \frac{S_{i j}}{D_{i j}} mile/ μ F

${\hat{P}}_{i j} = 11.17689 \cdot \ln \frac{S_{i j}}{D_{i j}} mile/ μ F$ (5.10)

Note In applying Equations 5.9 and 5.10, the values of RD _i, S _ii, S _ij, and D _ij must be in the same units. For overhead lines, the distances between conductors are typically specified in feet, whereas the value of the conductor diameter from a table will typically be in inches. Care must be taken to ensure that the radius in feet is used in applying the two equations.

For an overhead line of n-cond conductors, the “primitive potential coefficient matrix” $[{\hat{P}}_{primitive}]$ $[{\hat{P}}_{primitive}]$ can be constructed. The primitive potential coefficient matrix will be an n-cond × n-cond matrix. For a four-wire grounded wye line, the primitive coefficient matrix will be of the form:

[{\hat{P}}_{primitive}] = [\begin{matrix} {\hat{P}}_{a a} & {\hat{P}}_{a b} & {\hat{P}}_{a c} & • & {\hat{P}}_{a n} \\ {\hat{P}}_{b a} & {\hat{P}}_{b b} & {\hat{P}}_{b c} & • & {\hat{P}}_{b n} \\ {\hat{P}}_{c a} & {\hat{P}}_{c b} & {\hat{P}}_{c c} & • & {\hat{P}}_{c n} \\ • & • & • & • & • \\ {\hat{P}}_{n a} & {\hat{P}}_{n b} & {\hat{P}}_{n c} & • & {\hat{P}}_{n n} \end{matrix}]

$[{\hat{P}}_{primitive}] = [\begin{matrix} {\hat{P}}_{a a} & {\hat{P}}_{a b} & {\hat{P}}_{a c} & • & {\hat{P}}_{a n} \\ {\hat{P}}_{b a} & {\hat{P}}_{b b} & {\hat{P}}_{b c} & • & {\hat{P}}_{b n} \\ {\hat{P}}_{c a} & {\hat{P}}_{c b} & {\hat{P}}_{c c} & • & {\hat{P}}_{c n} \\ • & • & • & • & • \\ {\hat{P}}_{n a} & {\hat{P}}_{n b} & {\hat{P}}_{n c} & • & {\hat{P}}_{n n} \end{matrix}]$ (5.11)

The dots ( $•$ $•$ ) in Equation 5.11 are partitioning the matrix between the third and fourth rows and columns. In partitioned form, Equation 5.11 becomes:

[{\hat{P}}_{primitive}] = [\begin{matrix} [{\hat{P}}_{i j}] & [{\hat{P}}_{i n}] \\ [{\hat{P}}_{n j}] & [{\hat{P}}_{n n}] \end{matrix}]

$[{\hat{P}}_{primitive}] = [\begin{matrix} [{\hat{P}}_{i j}] & [{\hat{P}}_{i n}] \\ [{\hat{P}}_{n j}] & [{\hat{P}}_{n n}] \end{matrix}]$ (5.12)

Because the neutral conductor is grounded, the matrix can be reduced using the “Kron reduction” method to an n-phase × n-phase phase potential coefficient matrix [P_abc].

[P_{a b c}] = [{\hat{P}}_{i j}] - [{\hat{P}}_{i n}] \cdot {[{\hat{P}}_{n n}]}^{- 1} \cdot [{\hat{P}}_{j n}]

$[P_{a b c}] = [{\hat{P}}_{i j}] - [{\hat{P}}_{i n}] \cdot {[{\hat{P}}_{n n}]}^{- 1} \cdot [{\hat{P}}_{j n}]$ (5.13)

The inverse of the potential coefficient matrix will give the n-phase × n-phase capacitance matrix [C_abc].

[C_{a b c}] = {[P_{a b c}]}^{- 1} μ F

$[C_{a b c}] = {[P_{a b c}]}^{- 1} μ F$ (5.14)

For a two-phase line, the capacitance matrix of Equation 5.14 will be 2 × 2. A row and column of zeros must be inserted for the missing phase. For a single-phase line, Equation 5.14 will result in a single element. Again, rows and columns of zero must be inserted for the missing phase. In the case of the single-phase line, the only nonzero term will be that of the phase in use.

Neglecting the shunt conductance, the phase shunt admittance matrix is given by:

[y_{a b c}] = 0 + j \cdot ω \cdot [C_{a b c}] μ S/mile

$[y_{a b c}] = 0 + j \cdot ω \cdot [C_{a b c}] μ S/mile$ (5.15)

where $ω = 2 \cdot π \cdot f = 376.9911 .$ $ω = 2 \cdot π \cdot f = 376.9911 .$

Example 5.1

Determine the shunt admittance matrix for the overhead line in Example 4.1. Assume that the neutral conductor is 25 ft above the ground.

The diameters of the phase and neutral conductors from the conductor table (Appendix A) are:

\begin{array}{l} Conductor : 336, 400 26 / 7 A C S R d_{c} = 0.721 in., R D_{c} = 0.03004 ft \\ 4 / 0 6 / 1 A C S R : d_{s} = 0.563 in., R D_{s} = 0.02346 ft \end{array}

$\begin{array}{l} Conductor : 336, 400 26 / 7 A C S R d_{c} = 0.721 in., R D_{c} = 0.03004 ft \\ 4 / 0 6 / 1 A C S R : d_{s} = 0.563 in., R D_{s} = 0.02346 ft \end{array}$

Using the Cartesian coordinates in Example 4.1, the image distance matrix is given by:

\begin{array}{l} S_{i j} = | d_{i} - d_{j}^{*} | \end{array}

$\begin{array}{l} S_{i j} = | d_{i} - d_{j}^{*} | \end{array}$

where d*_j = the conjugate of d_j.

For the configuration, the distances between conductors and images in matrix form are:

[S] = [\begin{matrix} 58 & 58.0539 & 58.4209 & 54.1479 \\ 58.0539 & 58 & 58.1743 & 54.0208 \\ 58.4209 & 58.1743 & 58 & 54.0833 \\ 54.1479 & 54.0208 & 54.0833 & 50 \end{matrix}] ft

$[S] = [\begin{matrix} 58 & 58.0539 & 58.4209 & 54.1479 \\ 58.0539 & 58 & 58.1743 & 54.0208 \\ 58.4209 & 58.1743 & 58 & 54.0833 \\ 54.1479 & 54.0208 & 54.0833 & 50 \end{matrix}] ft$

The self-primitive potential coefficient for phase a and the mutual primitive potential coefficient between phases a and b are:

{\hat{P}}_{a a} = 11.17689 \ln \frac{58}{0.03004} = 84.5600 mile/ μ F

${\hat{P}}_{a a} = 11.17689 \ln \frac{58}{0.03004} = 84.5600 mile/ μ F$

{\hat{P}}_{a b} = 11.17689 \ln \frac{58.0539}{2.5} = 35.1522 mile/ μ F

${\hat{P}}_{a b} = 11.17689 \ln \frac{58.0539}{2.5} = 35.1522 mile/ μ F$

Using Equations 5.9 and 5.10, the total primitive potential coefficient matrix is computed to be:

[{\hat{P}}_{primitive}] = [\begin{matrix} 84.5600 & 35.1522 & 23.7147 & 25.2469 \\ 35.1522 & 84.5600 & 28.6058 & 28.3590 \\ 23.7147 & 28.6058 & 84.5600 & 26.6131 \\ 25.2469 & 28.3590 & 26.6131 & 85.6659 \end{matrix}] mile/ μ F

$[{\hat{P}}_{primitive}] = [\begin{matrix} 84.5600 & 35.1522 & 23.7147 & 25.2469 \\ 35.1522 & 84.5600 & 28.6058 & 28.3590 \\ 23.7147 & 28.6058 & 84.5600 & 26.6131 \\ 25.2469 & 28.3590 & 26.6131 & 85.6659 \end{matrix}] mile/ μ F$

Because the fourth conductor (neutral) is grounded, the Kron reduction method is used to compute the “phase potential coefficient matrix.” Because only one row and column need to be eliminated, the $[{\hat{P}}_{n n}]$ $[{\hat{P}}_{n n}]$ term is a single element, so that the Kron reduction equation for this case can be modified to:

P_{i j} = {\hat{P}}_{i j} - \frac{{\hat{P}}_{i n} \cdot {\hat{P}}_{j n}}{{\hat{P}}_{44}}

$P_{i j} = {\hat{P}}_{i j} - \frac{{\hat{P}}_{i n} \cdot {\hat{P}}_{j n}}{{\hat{P}}_{44}}$

where i =1, 2, 3 and j =1, 2, 3.

For example, the value of P_cb is computed to be:

P_{c b} = {\hat{P}}_{32} - \frac{{\hat{P}}_{34} \cdot {\hat{P}}_{42}}{{\hat{P}}_{44}} = 28.6058 - \frac{26.6131 \cdot 28.359}{85.6659} = 19.7957

$P_{c b} = {\hat{P}}_{32} - \frac{{\hat{P}}_{34} \cdot {\hat{P}}_{42}}{{\hat{P}}_{44}} = 28.6058 - \frac{26.6131 \cdot 28.359}{85.6659} = 19.7957$

Following the Kron reduction, the phase potential coefficient matrix is:

[P_{a b c}] = [\begin{matrix} 77.1194 & 26.7944 & 15.8714 \\ 26.7944 & 75.1720 & 19.7957 \\ 15.8714 & 19.7957 & 76.2923 \end{matrix}] mile/ μ F

$[P_{a b c}] = [\begin{matrix} 77.1194 & 26.7944 & 15.8714 \\ 26.7944 & 75.1720 & 19.7957 \\ 15.8714 & 19.7957 & 76.2923 \end{matrix}] mile/ μ F$

Invert [P_abc] to determine the shunt capacitance matrix:

[C_{a b c}] = {[P]}^{- 1} = [\begin{matrix} 0.015 & - 0.0049 & - 0.0019 \\ - 0.0049 & 0.0159 & - 0.0031 \\ - 0.0019 & - 0.0031 & 0.0143 \end{matrix}] μ F/mile

$[C_{a b c}] = {[P]}^{- 1} = [\begin{matrix} 0.015 & - 0.0049 & - 0.0019 \\ - 0.0049 & 0.0159 & - 0.0031 \\ - 0.0019 & - 0.0031 & 0.0143 \end{matrix}] μ F/mile$

Multiply [C_abc] by the radian frequency to determine the final three-phase shunt admittance matrix.

[y_{a b c}] = j \cdot 376.9911 \cdot [C_{a b c}] = [\begin{matrix} j 5.6711 & - j 1.8362 & - j 0.7033 \\ - j 1.8362 & j 5.9774 & - j 1.169 \\ - j 0.7033 & - j 1.169 & j 5.3911 \end{matrix}] μ S/mile

$[y_{a b c}] = j \cdot 376.9911 \cdot [C_{a b c}] = [\begin{matrix} j 5.6711 & - j 1.8362 & - j 0.7033 \\ - j 1.8362 & j 5.9774 & - j 1.169 \\ - j 0.7033 & - j 1.169 & j 5.3911 \end{matrix}] μ S/mile$

5.2.1 The Shunt Admittance of Overhead Parallel Lines

The development of the shunt admittance matrix for parallel overhead lines is similar to the steps taken to create the phase impedance matrix. The numbering of the conductors must be the same as that which was used in developing the phase impedance matrix. To develop the shunt admittance matrix for overhead lines, it is necessary to know the distance from each conductor to the ground, and it will be necessary to know the radius in feet for each conductor.

The first step is to create the primitive potential coefficient matrix. This will be an n-cond × n-cond matrix, where n-cond is the total number of phase and ground conductors. For the lines in Figure 4.8, n-cond will be 7; for two lines each with its own grounded neutral, n-cond will be 8.

The elements of the primitive potential coefficient matrix are given by:

\begin{matrix} {\hat{P}}_{i i} = 11.17689 \cdot \ln \frac{S_{i i}}{R D_{i}} \\ {\hat{P}}_{i j} = 11.17689 \cdot \ln \frac{S_{i j}}{D_{i j}} \end{matrix} mile/ μ F

$\begin{matrix} {\hat{P}}_{i i} = 11.17689 \cdot \ln \frac{S_{i i}}{R D_{i}} \\ {\hat{P}}_{i j} = 11.17689 \cdot \ln \frac{S_{i j}}{D_{i j}} \end{matrix} mile/ μ F$ (5.16)

where S_ii = distance in ft from a conductor to its image below ground

S_ij = distance in ft from a conductor to the image of an adjacent conductor

D_ij = distance in ft between two overhead conductors

RD_i = radius in ft of conductor i

The last one or two rows and columns of the primitive potential coefficient matrix are eliminated by using Kron reduction. The resulting voltage equation is:

[\begin{matrix} V 1_{_{a g}} \\ V 1_{_{b g}} \\ V 1_{_{c g}} \\ V 2_{_{a g}} \\ V 2_{_{b g}} \\ V 2_{_{c g}} \end{matrix}] = [\begin{matrix} P 11_{_{a a}} & P 11_{_{a b}} & P 11_{_{a c}} & P 12_{_{a a}} & P 12_{_{a b}} & P 12_{_{a c}} \\ P 11_{_{b a}} & P 11_{_{b b}} & P 11_{_{b c}} & P 12_{_{b a}} & P 12_{_{b b}} & P 12_{_{b c}} \\ P 11_{_{c a}} & P 11_{_{c b}} & P 11_{_{c c}} & P 12_{_{c a}} & P 12_{_{c b}} & P 12_{_{c c}} \\ P 21_{_{a a}} & P 21_{_{a b}} & P 21_{_{a c}} & P 22_{_{a a}} & P 22_{_{a b}} & P 22_{_{a c}} \\ P 21_{_{b a}} & P 21_{_{b b}} & P 21_{_{b c}} & P 22_{_{b a}} & P 22_{_{b b}} & P 22_{_{b c}} \\ P 21_{_{c a}} & P 21_{_{c b}} & P 21_{_{c c}} & P 22_{_{c a}} & P 22_{_{c b}} & P 22_{_{c c}} \end{matrix}] \cdot [\begin{matrix} q 1_{_{a}} \\ q 1_{_{b}} \\ q 1_{_{c}} \\ q 2_{_{a}} \\ q 2_{_{b}} \\ q 2_{_{c}} \end{matrix}] V

$[\begin{matrix} V 1_{_{a g}} \\ V 1_{_{b g}} \\ V 1_{_{c g}} \\ V 2_{_{a g}} \\ V 2_{_{b g}} \\ V 2_{_{c g}} \end{matrix}] = [\begin{matrix} P 11_{_{a a}} & P 11_{_{a b}} & P 11_{_{a c}} & P 12_{_{a a}} & P 12_{_{a b}} & P 12_{_{a c}} \\ P 11_{_{b a}} & P 11_{_{b b}} & P 11_{_{b c}} & P 12_{_{b a}} & P 12_{_{b b}} & P 12_{_{b c}} \\ P 11_{_{c a}} & P 11_{_{c b}} & P 11_{_{c c}} & P 12_{_{c a}} & P 12_{_{c b}} & P 12_{_{c c}} \\ P 21_{_{a a}} & P 21_{_{a b}} & P 21_{_{a c}} & P 22_{_{a a}} & P 22_{_{a b}} & P 22_{_{a c}} \\ P 21_{_{b a}} & P 21_{_{b b}} & P 21_{_{b c}} & P 22_{_{b a}} & P 22_{_{b b}} & P 22_{_{b c}} \\ P 21_{_{c a}} & P 21_{_{c b}} & P 21_{_{c c}} & P 22_{_{c a}} & P 22_{_{c b}} & P 22_{_{c c}} \end{matrix}] \cdot [\begin{matrix} q 1_{_{a}} \\ q 1_{_{b}} \\ q 1_{_{c}} \\ q 2_{_{a}} \\ q 2_{_{b}} \\ q 2_{_{c}} \end{matrix}] V$ (5.17)

In short hand form Equation 5.17 is:

[V_{L G}] = [P] \cdot [q]

$[V_{L G}] = [P] \cdot [q]$ (5.18)

The shunt capacitance matrix is determined by:

[q] = {[P]}^{- 1} \cdot [V_{L G}] = [C] \cdot [V_{L G}]

$[q] = {[P]}^{- 1} \cdot [V_{L G}] = [C] \cdot [V_{L G}]$ (5.19)

The resulting capacitance matrix is partitioned between the third and fourth rows and columns.

[C] = {[P]}^{- 1} = [\begin{matrix} [C 11] & [C 12] \\ [C 21] & [C 22] \end{matrix}]

$[C] = {[P]}^{- 1} = [\begin{matrix} [C 11] & [C 12] \\ [C 21] & [C 22] \end{matrix}]$ (5.20)

The shunt admittance matrix is given by:

[y] = j ω \cdot [C] \cdot 10^{- 6} = [\begin{matrix} [y 11] & [y 12] \\ [y 21] & [y 22] \end{matrix}] S

$[y] = j ω \cdot [C] \cdot 10^{- 6} = [\begin{matrix} [y 11] & [y 12] \\ [y 21] & [y 22] \end{matrix}] S$ (5.21)

where $ω = 2 \cdot π \cdot frequency .$ $ω = 2 \cdot π \cdot frequency .$

Example 5.2

Determine the shunt admittance matrix for the parallel overhead lines in Example 4.2.

The position coordinates for the seven conductors and the distance matrix are defined in Example 4.2. The diagonal terms of the distance matrix (Example 4.2) must be the radius in feet of the individual conductors. For this example:

D_{1, 1} = D_{2, 2} = D_{3, 3} = \frac{d_{1}}{24} = \frac{0.721}{24} = 0.030 0^{'}

$D_{1, 1} = D_{2, 2} = D_{3, 3} = \frac{d_{1}}{24} = \frac{0.721}{24} = 0.030 0^{'}$

D_{4, 4} = D_{5, 5} = D_{6, 6} = \frac{d_{2}}{24} = \frac{0.567}{24} = 0.023 6^{'}

$D_{4, 4} = D_{5, 5} = D_{6, 6} = \frac{d_{2}}{24} = \frac{0.567}{24} = 0.023 6^{'}$

D_{7, 7} = \frac{d_{n}}{24} = 0.023 5^{'}

$D_{7, 7} = \frac{d_{n}}{24} = 0.023 5^{'}$

The resulting distance matrix is:

[D] = [\begin{matrix} 0.0300 & 2.5000 & 7.0000 & 3.2016 & 7.2801 & 2.0000 & 7.2111 \\ 2.5000 & 0.0300 & 4.5000 & 2.0000 & 4.9244 & 3.2016 & 6.1847 \\ 7.0000 & 4.5000 & 0.0300 & 4.9244 & 2.0000 & 7.2801 & 6.7082 \\ 3.2016 & 2.0000 & 4.9244 & 0.0236 & 4.5000 & 2.5000 & 4.2720 \\ 7.2801 & 4.9244 & 2.0000 & 4.5000 & 0.0236 & 7.0000 & 5.0000 \\ 2.0000 & 3.2016 & 7.2801 & 2.5000 & 7.0000 & 0.0236 & 5.6569 \\ 7.2111 & 6.1847 & 6.7082 & 4.272 & 5.0000 & 5.6569 & 0.0235 \end{matrix}] ft.

$[D] = [\begin{matrix} 0.0300 & 2.5000 & 7.0000 & 3.2016 & 7.2801 & 2.0000 & 7.2111 \\ 2.5000 & 0.0300 & 4.5000 & 2.0000 & 4.9244 & 3.2016 & 6.1847 \\ 7.0000 & 4.5000 & 0.0300 & 4.9244 & 2.0000 & 7.2801 & 6.7082 \\ 3.2016 & 2.0000 & 4.9244 & 0.0236 & 4.5000 & 2.5000 & 4.2720 \\ 7.2801 & 4.9244 & 2.0000 & 4.5000 & 0.0236 & 7.0000 & 5.0000 \\ 2.0000 & 3.2016 & 7.2801 & 2.5000 & 7.0000 & 0.0236 & 5.6569 \\ 7.2111 & 6.1847 & 6.7082 & 4.272 & 5.0000 & 5.6569 & 0.0235 \end{matrix}] ft.$

The distances between conductors and conductor images (image matrix) can be determined by:

S_{i, j} = | d_{i} - d_{j}^{*} |

$S_{i, j} = | d_{i} - d_{j}^{*} |$

For this example, the image matrix is:

[S] = [\begin{matrix} 70.000 & 70.045 & 70.349 & 68.046 & 68.359 & 68.000 & 64.125 \\ 70.045 & 70.000 & 70.145 & 68.000 & 68.149 & 68.046 & 64.018 \\ 70.349 & 70.145 & 70.000 & 68.149 & 68.000 & 68.359 & 64.070 \\ 68.046 & 68.000 & 68.149 & 66.000 & 66.153 & 66.047 & 62.018 \\ 68.359 & 68.149 & 68.000 & 66.153 & 66.000 & 66.370 & 62.073 \\ 68.000 & 68.046 & 68.359 & 66.047 & 66.370 & 66.000 & 62.129 \\ 64.125 & 64.018 & 64.070 & 62.018 & 62.073 & 62.129 & 60.000 \end{matrix}] ft.

$[S] = [\begin{matrix} 70.000 & 70.045 & 70.349 & 68.046 & 68.359 & 68.000 & 64.125 \\ 70.045 & 70.000 & 70.145 & 68.000 & 68.149 & 68.046 & 64.018 \\ 70.349 & 70.145 & 70.000 & 68.149 & 68.000 & 68.359 & 64.070 \\ 68.046 & 68.000 & 68.149 & 66.000 & 66.153 & 66.047 & 62.018 \\ 68.359 & 68.149 & 68.000 & 66.153 & 66.000 & 66.370 & 62.073 \\ 68.000 & 68.046 & 68.359 & 66.047 & 66.370 & 66.000 & 62.129 \\ 64.125 & 64.018 & 64.070 & 62.018 & 62.073 & 62.129 & 60.000 \end{matrix}] ft.$

The distance and image matrices are used to compute the 7 ×7 potential coefficient matrix by:

P p_{i, j} = 11.17689 \cdot \ln (\frac{S_{i, j}}{D_{i, j}})

$P p_{i, j} = 11.17689 \cdot \ln (\frac{S_{i, j}}{D_{i, j}})$

The primitive potential coefficient matrix is partitioned between the sixth and seventh rows and columns, and the Kron reduction method produces the 6 ×6 potential matrix. This matrix is then inverted and multiplied by $ω = 376.9911$ $ω = 376.9911$ to give the shunt admittance matrix. The final shunt admittance matrix in partitioned form is:

[y 11] = [\begin{matrix} j 6.2992 & - j 1.3413 & - j 0.4135 \\ - j 1.3413 & j 6.5009 & - j 0.8038 \\ - j 0.4135 & - j 0.8038 & j 6.0257 \end{matrix}] μ S/mile

$[y 11] = [\begin{matrix} j 6.2992 & - j 1.3413 & - j 0.4135 \\ - j 1.3413 & j 6.5009 & - j 0.8038 \\ - j 0.4135 & - j 0.8038 & j 6.0257 \end{matrix}] μ S/mile$

[y 12] = [\begin{matrix} - j 0.7889 & - j 0.2992 & - j 1.6438 \\ - j 1.4440 & - j 0.5698 & - j 0.7988 \\ - j 0.5553 & - j 1.8629 & - j 0.2985 \end{matrix}] μ S/mile

$[y 12] = [\begin{matrix} - j 0.7889 & - j 0.2992 & - j 1.6438 \\ - j 1.4440 & - j 0.5698 & - j 0.7988 \\ - j 0.5553 & - j 1.8629 & - j 0.2985 \end{matrix}] μ S/mile$

[y 21] = [\begin{matrix} - j 0.7889 & - j 1.4440 & - j 0.5553 \\ - j 0.2992 & - j 0.5698 & - j 1.8629 \\ - j 1.6438 & - j 0.7988 & - j 0.2985 \end{matrix}] μ S/mile

$[y 21] = [\begin{matrix} - j 0.7889 & - j 1.4440 & - j 0.5553 \\ - j 0.2992 & - j 0.5698 & - j 1.8629 \\ - j 1.6438 & - j 0.7988 & - j 0.2985 \end{matrix}] μ S/mile$

[y 22] = [\begin{matrix} j 6.3278 & - j 0.6197 & - j 1.1276 \\ - j 0.6197 & j 5.9016 & - j 0.2950 \\ - j 1.1276 & - j 0.2950 & j 6.1051 \end{matrix}] μ S/mile

$[y 22] = [\begin{matrix} j 6.3278 & - j 0.6197 & - j 1.1276 \\ - j 0.6197 & j 5.9016 & - j 0.2950 \\ - j 1.1276 & - j 0.2950 & j 6.1051 \end{matrix}] μ S/mile$

5.3 Concentric Neutral Cable Underground Lines

Most underground distribution lines consist of one or more concentric neutral cables. Figure 5.4 illustrates a basic concentric neutral cable with the center conductor being the phase conductor, and the concentric neutral strands displaced equally around a circle of radius R _b.

Referring to Figure 5.4, the following definitions apply:

R _b = Radius of a circle passing through the centers of the neutral strands

d _c = Diameter of the phase conductor

d _s = Diameter of a neutral strand

k = Total number of neutral strands

images

Figure 5.4
Basic concentric neutral cable.

The concentric neutral strands are grounded so that they are all at the same potential. Because of the stranding, it is assumed that the electric field created by the charge on the phase conductor will be confined to the boundary of the concentric neutral strands. In order to compute the capacitance between the phase conductor and ground, the general voltage drop of Equation 5.2 will be applied. Because all of the neutral strands are at the same potential, it is only necessary to determine the potential difference between the phase conductor p and strand 1.

V_{p 1} = \frac{1}{2 π ε} (q_{p} \ln \frac{R_{b}}{R D_{c}} + q_{1} \ln \frac{R D_{s}}{R_{b}} + q_{2} \ln \frac{D_{12}}{R_{b}} + \dots + q_{i} \ln \frac{D_{1 i}}{R_{b}} + \dots + q_{k} \ln \frac{D_{k 1}}{R_{b}})

$V_{p 1} = \frac{1}{2 π ε} (q_{p} \ln \frac{R_{b}}{R D_{c}} + q_{1} \ln \frac{R D_{s}}{R_{b}} + q_{2} \ln \frac{D_{12}}{R_{b}} + \dots + q_{i} \ln \frac{D_{1 i}}{R_{b}} + \dots + q_{k} \ln \frac{D_{k 1}}{R_{b}})$ (5.22)

where $R D_{c} = \frac{d_{c}}{2}$ $R D_{c} = \frac{d_{c}}{2}$ $R D_{s} = \frac{d_{s}}{2}$ $R D_{s} = \frac{d_{s}}{2}$

It is assumed that each of the neutral strands carries the same charge such that:

q_{1} = q_{2} = q_{i} = q_{k} = - \frac{q_{p}}{k}

$q_{1} = q_{2} = q_{i} = q_{k} = - \frac{q_{p}}{k}$ (5.23)

Equation 5.22 can be simplified:

V_{p 1} = \frac{1}{2 π ε} [q_{p} \ln \frac{R_{b}}{R D_{c}} - \frac{q_{p}}{k} (\ln \frac{R D_{s}}{R_{b}} + \ln \frac{D_{12}}{R_{b}} + \dots + \ln \frac{D_{1 i}}{R_{b}} + \dots + \ln \frac{D_{1 k}}{R_{b}})]

$V_{p 1} = \frac{1}{2 π ε} [q_{p} \ln \frac{R_{b}}{R D_{c}} - \frac{q_{p}}{k} (\ln \frac{R D_{s}}{R_{b}} + \ln \frac{D_{12}}{R_{b}} + \dots + \ln \frac{D_{1 i}}{R_{b}} + \dots + \ln \frac{D_{1 k}}{R_{b}})]$

V_{p 1} = \frac{q_{p}}{2 π ε} [\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} (\ln \frac{R D_{s} \cdot D_{12} \cdot D_{1 i} \cdot \dots \cdot D_{1 k}}{R_{b}^{k}})]

$V_{p 1} = \frac{q_{p}}{2 π ε} [\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} (\ln \frac{R D_{s} \cdot D_{12} \cdot D_{1 i} \cdot \dots \cdot D_{1 k}}{R_{b}^{k}})]$ (5.24)

The numerator of the second ln term in Equation 5.24 needs to be expanded. The numerator represents the product of the radius and the distances between strand i and all of the other strands. Referring to Figure 5.4, the following relations apply:

\begin{array}{l} θ_{12} = \frac{2 \cdot π}{k} \\ θ_{13} = 2 \cdot θ_{12} = \frac{4 \cdot π}{k} \end{array}

$\begin{array}{l} θ_{12} = \frac{2 \cdot π}{k} \\ θ_{13} = 2 \cdot θ_{12} = \frac{4 \cdot π}{k} \end{array}$

In general, the angle between strand #1 and any other strand #i is given by:

θ_{1 i} = (i - 1) \cdot θ_{12} = \frac{(i - 1) \cdot 2 π}{k}

$θ_{1 i} = (i - 1) \cdot θ_{12} = \frac{(i - 1) \cdot 2 π}{k}$ (5.25)

The distances between the various strands are given by:

\begin{array}{l} D_{12} = 2 \cdot R_{b} \cdot \sin (\frac{θ_{12}}{2}) = 2 \cdot R_{b} \cdot \sin (\frac{π}{k}) \\ D_{13} = 2 \cdot R_{b} \cdot \sin (\frac{θ_{13}}{2}) = 2 \cdot R_{b} \cdot \sin (\frac{2 π}{k}) \end{array}

$\begin{array}{l} D_{12} = 2 \cdot R_{b} \cdot \sin (\frac{θ_{12}}{2}) = 2 \cdot R_{b} \cdot \sin (\frac{π}{k}) \\ D_{13} = 2 \cdot R_{b} \cdot \sin (\frac{θ_{13}}{2}) = 2 \cdot R_{b} \cdot \sin (\frac{2 π}{k}) \end{array}$ (5.26)

The distance between strand 1 and any other strand i is given by:

D_{1 i} = 2 \cdot R_{b} \cdot \sin (\frac{θ_{1 i}}{2}) = 2 \cdot R_{b} \cdot \sin [\frac{(i - 1) \cdot π}{k}]

$D_{1 i} = 2 \cdot R_{b} \cdot \sin (\frac{θ_{1 i}}{2}) = 2 \cdot R_{b} \cdot \sin [\frac{(i - 1) \cdot π}{k}]$ (5.27)

Equation 5.27 can be used to expand the numerator of the second log term of Equation 5.24.

\begin{array}{l} R D_{s} \cdot D_{12}, \dots, D_{1 i}, \dots, D_{1 k} = \\ R D_{s} \cdot R_{b}^{k - 1} [2 \sin (\frac{π}{k}) \cdot 2 \sin (\frac{2 π}{k}), \dots, 2 \sin \times {(\frac{(i - 1) π}{k})}, \dots, 2 \sin {(\frac{(k - 1)}{k})}] \end{array}

$\begin{array}{l} R D_{s} \cdot D_{12}, \dots, D_{1 i}, \dots, D_{1 k} = \\ R D_{s} \cdot R_{b}^{k - 1} [2 \sin (\frac{π}{k}) \cdot 2 \sin (\frac{2 π}{k}), \dots, 2 \sin \times {(\frac{(i - 1) π}{k})}, \dots, 2 \sin {(\frac{(k - 1)}{k})}] \end{array}$ (5.28)

The term inside the bracket in Equation 5.28 is a trigonometric identity that is merely equal to the number of strands k [1]. Using that identity, Equation 5.18 becomes:

V_{p 1} = \frac{q_{p}}{2 π ε} [\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} (\ln \frac{k \cdot R D_{s} \cdot R_{b}^{k - 1}}{R_{b}^{k}})]

$V_{p 1} = \frac{q_{p}}{2 π ε} [\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} (\ln \frac{k \cdot R D_{s} \cdot R_{b}^{k - 1}}{R_{b}^{k}})]$

V_{p 1} = \frac{q_{p}}{2 π ε} [\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} (\ln \frac{k \cdot R D_{s}}{R_{b}})]

$V_{p 1} = \frac{q_{p}}{2 π ε} [\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} (\ln \frac{k \cdot R D_{s}}{R_{b}})]$ (5.29)

Equation 5.29 gives the voltage drop from the phase conductor to neutral strand #1. Care must be taken such that the units for the various radii are the same. Typically, underground spacings are given in inches; so the radii of the phase conductor (RD _c) and the strand conductor (RD_s) should be specified in inches.

Because the neutral strands are all grounded, Equation 5.29 gives the voltage drop between the phase conductor and ground. Therefore, the capacitance from the phase to ground for a concentric neutral cable is given by:

C_{p g} = \frac{q_{p}}{V_{p 1}} = \frac{2 π ε}{\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} \ln \frac{k \cdot R D_{s}}{R_{b}}} μ F/mile

$C_{p g} = \frac{q_{p}}{V_{p 1}} = \frac{2 π ε}{\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} \ln \frac{k \cdot R D_{s}}{R_{b}}} μ F/mile$ (5.30)

where ε = ε₀ε_r = permittivity of the medium

ε₀ = permittivity of free space = 0.01420 μF/mile

ε_r = relative permittivity of the medium

The electric field of a cable is confined to the insulation material. Various types of insulation material are used, and each will have a range of values for the relative permittivity. Table 5.1 gives the range of values of relative permittivity for four common insulation materials [2].

TABLE 5.1 Typical Values of Relative Permittivity (ε_r)

Cross-linked polyethlyene is a very popular insulation material. If the minimum value of relative permittivity is assumed (2.3), the equation for the shunt admittance of the concentric neutral cable is given by:

y_{a g} = 0 + j \frac{77.3619}{\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} \ln \frac{k \cdot R D_{s}}{R_{b}}} μ S/mile

$y_{a g} = 0 + j \frac{77.3619}{\ln \frac{R_{b}}{R D_{c}} - \frac{1}{k} \ln \frac{k \cdot R D_{s}}{R_{b}}} μ S/mile$ (5.31)

Example 5.3

Determine the three-phase shunt admittance matrix for the concentric neutral line in Example 4.3 in Chapter 4.

From Example 4.3:

R_{b} = R = 0.0511 ft = 0.631 in.

$R_{b} = R = 0.0511 ft = 0.631 in.$

Diameter of the 250,000 AA phase conductor =0.567 in.

R D_{c} = \frac{0.567}{2} = 0.2835 in.

$R D_{c} = \frac{0.567}{2} = 0.2835 in.$

Diameter of the #14 CU concentric neutral strand =0.0641 in.

R D_{s} = \frac{0.0641}{2} = 0.03205 in.

$R D_{s} = \frac{0.0641}{2} = 0.03205 in.$

Substitute into Equation 5.24:

y_{a g} = j \frac{77.3619}{\ln (\frac{R_{b}}{R D_{c}}) - \frac{1}{k} \cdot \ln (\frac{k \cdot R D_{s}}{R_{b}})}

$y_{a g} = j \frac{77.3619}{\ln (\frac{R_{b}}{R D_{c}}) - \frac{1}{k} \cdot \ln (\frac{k \cdot R D_{s}}{R_{b}})}$

y_{a b} = j \frac{77.3619}{\ln (\frac{0.6132}{0.2835}) - \frac{1}{13} \cdot \ln (\frac{13 \cdot 0.03205}{0.6132})} = j 96.6098 μ S/mile

$y_{a b} = j \frac{77.3619}{\ln (\frac{0.6132}{0.2835}) - \frac{1}{13} \cdot \ln (\frac{13 \cdot 0.03205}{0.6132})} = j 96.6098 μ S/mile$

The phase admittance for this three-phase underground line is:

[y_{a b c}] = [\begin{matrix} j 96.6098 & 0 & 0 \\ 0 & j 96.6098 & 0 \\ 0 & 0 & j 96.6098 \end{matrix}] μ S/mile

$[y_{a b c}] = [\begin{matrix} j 96.6098 & 0 & 0 \\ 0 & j 96.6098 & 0 \\ 0 & 0 & j 96.6098 \end{matrix}] μ S/mile$

5.4 Tape-Shielded Cable Underground Lines

A tape-shielded cable is shown in Figure 5.5.

Referring to Figure 5.5, R_b is the radius of a circle passing through the center of the tape shield. As with the concentric neutral cable, the electric field is confined to the insulation so that the relative permittivity of Table 5.1 will apply.

images

Figure 5.5
Tape-shielded conductor.

The tape-shielded conductor can be visualized as a concentric neutral cable where the number of strands k has become infinite. When k in Equation 5.24 approaches infinity, the second term in the denominator approaches zero. Therefore, the equation for the shunt admittance of a tape-shielded conductor becomes:

y_{a g} = 0 + j \frac{77.3619}{\ln \frac{R_{b}}{R D_{c}}} μ S/mile

$y_{a g} = 0 + j \frac{77.3619}{\ln \frac{R_{b}}{R D_{c}}} μ S/mile$ (5.32)

Example 5.4

Determine the shunt admittance of the single-phase tape-shielded cable in Example 4.4 in Chapter 4. From Example 4.4, the outside diameter of the tape shield is 0.88 in. The thickness of the tape shield (T) is 5 mils. The radius of a circle passing through the center of the tape shield is:

T = \frac{5}{1000} = 0.005

$T = \frac{5}{1000} = 0.005$

R_{b} = \frac{d_{s} - T}{2} = \frac{0.88 - .005}{2} = 0.4375 in.

$R_{b} = \frac{d_{s} - T}{2} = \frac{0.88 - .005}{2} = 0.4375 in.$

The diameter of the 1/0 AA phase conductor =0.368 in.

R D_{c} = \frac{d_{p}}{2} = \frac{0.368}{2} = 0.1840 in.

$R D_{c} = \frac{d_{p}}{2} = \frac{0.368}{2} = 0.1840 in.$

Substitute into Equation 5.25:

y_{b g} = j \frac{77.3619}{\ln (\frac{R_{b}}{R D_{c}})} = j \frac{77.3619}{\ln (\frac{0.4375}{0.184})} = j 89.3179

$y_{b g} = j \frac{77.3619}{\ln (\frac{R_{b}}{R D_{c}})} = j \frac{77.3619}{\ln (\frac{0.4375}{0.184})} = j 89.3179$

The line is on phase b so that the phase admittance matrix becomes:

[y_{a b c}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & j 89.3179 & 0 \\ 0 & 0 & 0 \end{matrix}] μ S/mile

$[y_{a b c}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & j 89.3179 & 0 \\ 0 & 0 & 0 \end{matrix}] μ S/mile$

5.5 Sequence Admittance

The sequence admittances of a three-phase line can be determined in much the same manner as the sequence impedances were determined in Chapter 4. Assume that the 3 × 3 admittance matrix is given in S/mile. Then the three-phase capacitance currents as a function of the line-to-ground voltages are given by:

[\begin{matrix} I cap_{_{a}} \\ I cap_{_{b}} \\ I cap_{_{c}} \end{matrix}] = [\begin{matrix} y_{a a} & y_{a b} & y_{a c} \\ y_{b a} & y_{b b} & y_{b c} \\ y_{c a} & y_{c b} & y_{c c} \end{matrix}] \cdot [\begin{matrix} V_{a g} \\ V_{b g} \\ V_{c g} \end{matrix}]

$[\begin{matrix} I cap_{_{a}} \\ I cap_{_{b}} \\ I cap_{_{c}} \end{matrix}] = [\begin{matrix} y_{a a} & y_{a b} & y_{a c} \\ y_{b a} & y_{b b} & y_{b c} \\ y_{c a} & y_{c b} & y_{c c} \end{matrix}] \cdot [\begin{matrix} V_{a g} \\ V_{b g} \\ V_{c g} \end{matrix}]$ (5.33)

[I cap_{_{a b c}}] = [y_{a b c}] \cdot [V L G_{_{a b c}}]

$[I cap_{_{a b c}}] = [y_{a b c}] \cdot [V L G_{_{a b c}}]$ (5.34)

Applying the symmetrical component transformations:

[I cap_{_{012}}] = {[A_{s}]}^{- 1} \cdot [I cap_{_{a b c}}] = {[A_{s}]}^{- 1} \cdot [y_{a b c}] \cdot [A_{s}] \cdot [V L G_{_{012}}]

$[I cap_{_{012}}] = {[A_{s}]}^{- 1} \cdot [I cap_{_{a b c}}] = {[A_{s}]}^{- 1} \cdot [y_{a b c}] \cdot [A_{s}] \cdot [V L G_{_{012}}]$ (5.35)

From Equation 5.35, the sequence admittance matrix is given by:

[y_{012}] = {[A_{s}]}^{- 1} \cdot [y_{a b c}] \cdot [A_{s}] = [\begin{matrix} y_{00} & y_{01} & y_{02} \\ y_{10} & y_{11} & y_{12} \\ y_{20} & y_{21} & y_{22} \end{matrix}]

$[y_{012}] = {[A_{s}]}^{- 1} \cdot [y_{a b c}] \cdot [A_{s}] = [\begin{matrix} y_{00} & y_{01} & y_{02} \\ y_{10} & y_{11} & y_{12} \\ y_{20} & y_{21} & y_{22} \end{matrix}]$ (5.36)

For a three-phase overhead line with unsymmetrical spacing, the sequence admittance matrix will be full. That is, the off-diagonal terms will be non-zero. However, a three-phase underground line with three identical cables will only have the diagonal terms since there is no “mutual capacitance” between phases. In fact, the sequence admittances will be exactly the same as the phase admittances.

5.6 The Shunt Admittance of Parallel Underground Lines

For underground cable lines using either concentric neutral cables or tape-shielded cables, the computation of the shunt admittance matrix is quite simple. The electric field created by the charged phase conductor does not link to adjacent conductors because of the presence of the concentric neutrals or the tape shield. As a result, the shunt admittance matrix for parallel underground lines will consist of diagonal terms only.

The diagonal terms for concentric neutral cables are given by:

y_{i i} = 0 + j \frac{77.3619}{\ln \frac{R_{b}}{R D_{i}} - \frac{1}{k} \cdot \ln \frac{k \cdot R D_{s}}{R_{b}}} \cdot 10^{- 6} S/mile

$y_{i i} = 0 + j \frac{77.3619}{\ln \frac{R_{b}}{R D_{i}} - \frac{1}{k} \cdot \ln \frac{k \cdot R D_{s}}{R_{b}}} \cdot 10^{- 6} S/mile$ (5.37)

where R_b = radius in ft of circle going through the center of the neutral strands

RD_i = radius in ft of the center phase conductor

RD_s = radius in ft of the neutral strands

k = number of neutral strands

The diagonal terms for tape-shielded cables are given by:

y_{i i} = 0 + j \frac{77.3619}{\ln \frac{R_{b}}{R D_{i}}} \cdot 10^{- 6} S/mile

$y_{i i} = 0 + j \frac{77.3619}{\ln \frac{R_{b}}{R D_{i}}} \cdot 10^{- 6} S/mile$ (5.38)

where R_b = radius in ft of circle passing through the center of the tape shield

RD_i = radius in ft of the center phase conductor

Example 5.5

Compute the shunt admittance matrix (6 ×6) for the concentric neutral underground configuration in Example 4.5.

From Example 4.5:

Diameter of the central conductor: $d_{c} = 0.56 7^{″}$ $d_{c} = 0.56 7^{″}$

Diameter of the strands: $d_{s} = 0.64 1^{″}$ $d_{s} = 0.64 1^{″}$

Outside diameters of concentric neutral strands: $d_{od} = 1.2 9^{″}$ $d_{od} = 1.2 9^{″}$

Radius of circle passing through the strands:

R_{b} = \frac{d_{od} - d_{s}}{24} = 0.051 1^{'}

$R_{b} = \frac{d_{od} - d_{s}}{24} = 0.051 1^{'}$

Radius of central conductor: $R D_{c} = \frac{d_{c}}{24} = \frac{0.567}{24} = 0.23 6^{'}$ $R D_{c} = \frac{d_{c}}{24} = \frac{0.567}{24} = 0.23 6^{'}$

Radius of the strands: $R D_{s} = \frac{d_{s}}{24} = \frac{0.0641}{24} = 0.002 7^{'}$ $R D_{s} = \frac{d_{s}}{24} = \frac{0.0641}{24} = 0.002 7^{'}$

Because all cables are identical, the shunt admittance of a cable is:

\begin{array}{l} y_{c} = 0 + j \cdot \frac{77.3619}{\ln (\frac{R_{b}}{R D_{c}}) - \frac{1}{k} \cdot \ln (\frac{k \cdot R D_{s}}{R_{b}})} = 0 + j \cdot \frac{77.3619}{\ln (\frac{0.0511}{0.0236}) - \frac{1}{13} \cdot \ln (\frac{13 \cdot 0.0027}{0.0511})} \\ y_{c} = 0 + j \cdot 96.6098 μ S/mile \end{array}

$\begin{array}{l} y_{c} = 0 + j \cdot \frac{77.3619}{\ln (\frac{R_{b}}{R D_{c}}) - \frac{1}{k} \cdot \ln (\frac{k \cdot R D_{s}}{R_{b}})} = 0 + j \cdot \frac{77.3619}{\ln (\frac{0.0511}{0.0236}) - \frac{1}{13} \cdot \ln (\frac{13 \cdot 0.0027}{0.0511})} \\ y_{c} = 0 + j \cdot 96.6098 μ S/mile \end{array}$

The phase admittance matrix is:

{[y]}_{a b c} = [\begin{matrix} j 96.6098 & 0 & 0 & 0 & 0 & 0 \\ 0 & j 96.6098 & 0 & 0 & 0 & 0 \\ 0 & 0 & j 96.6098 & 0 & 0 & 0 \\ 0 & 0 & 0 & j 96.6098 & 0 & 0 \\ 0 & 0 & 0 & 0 & j 96.6098 & 0 \\ 0 & 0 & 0 & 0 & 0 & j 96.6098 \end{matrix}] μ S/mile

${[y]}_{a b c} = [\begin{matrix} j 96.6098 & 0 & 0 & 0 & 0 & 0 \\ 0 & j 96.6098 & 0 & 0 & 0 & 0 \\ 0 & 0 & j 96.6098 & 0 & 0 & 0 \\ 0 & 0 & 0 & j 96.6098 & 0 & 0 \\ 0 & 0 & 0 & 0 & j 96.6098 & 0 \\ 0 & 0 & 0 & 0 & 0 & j 96.6098 \end{matrix}] μ S/mile$

5.7 Summary

Methods for computing the shunt capacitive admittance for overhead and underground lines have been presented in this chapter. The development of computing the shunt admittance matrix for parallel overhead and underground lines is included.

Distribution lines are typically so short that the shunt admittance can be ignored. However, there are cases of long, lightly loaded overhead lines where the shunt admittance should be included. Underground cables have a much higher shunt admittance per mile than overhead lines. Again, there will be cases where the shunt admittance of an underground cable should be included in the analysis process. When the analysis is being done using a computer, the approach to take is to model the shunt admittance for both overhead and underground lines, rather than making a simplifying assumption when it is not necessary.

Problems

5.1 Determine the phase admittance matrix $[y_{a b c}]$ $[y_{a b c}]$ and sequence admittance matrix $[y_{012}]$ $[y_{012}]$ in μS/mile for the three-phase overhead line of Problem 4.1.

5.2 Determine the phase admittance matrix in μS/mile for the two-phase line of Problem 4.2.

5.3 Determine the phase admittance matrix in μS/mile for the single-phase line of Problem 4.3.

5.4 Verify the results of Problems 5.1, 5.2, and 5.3 using WindMil.

5.5 Determine the phase admittance matrix and sequence admittance matrix in μS/mile for the three-phase line of Problem 4.5.

5.6 Determine the phase admittance matrix in μS/mile for the single-phase concentric neutral cable of Problem 4.9.

5.7 Determine the phase admittance matrix and sequence admittance matrix for the three-phase concentric neutral line of Problem 4.10.

5.8 Verify the results of Problems 5.6 and 5.7 using WindMil.

5.9 Determine the phase admittance matrix in μS/mile for the single-phase tape-shielded cable line of Problem 4.12.

5.10 Determine the phase admittance for the three-phase tape-shielded cable line of Problem 4.13.

5.11 Verify the results of Problem 5.9 and 5.10 using WindMil.

5.12 Determine the shunt admittance matrix for the parallel overhead lines of Problem 4.15.

5.13 Determine the shunt admittance matrix for the underground concentric neutral parallel lines of Problem 4.16.

WindMil Assignment

Add to the WindMil System 1 a single-phase line connected to Node 2. Call this “System 2.” The single-phase line is on phase b and is defined in Problem 4.3. Call this line OH-2. At the end of the line, connect a node and call it Node 3. The load at Node 3 is 200 kVA at a 90% lagging power factor. The load is modeled as a constant impedance load.

Determine the voltages at the nodes on a 120-V base and the currents flowing on the two lines.

References

1. Glover, J. D. and Sarma, M., Power System Analysis and Design, 2nd Edition, PWS-Kent Publishing, Boston, MA, 1995.

2. T. P. Arnold and Mercier, C. D.(eds), Power Cable Manual,2nd Edition, Southwire Company, Carrollton, GA, 1997.

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for
5. Shunt Admittance of Overhead and Underground Lines