NEVUnderground

= Underground Distribution Lines =

Underground lines will be modeled in a flexible way such that not only neutral conductors can be included but also any number of phases and/or circuits.

Series Impedance
Initially neglecting shunt admittances, the voltage/current relationship between two 'nodes' 1 and 2 (corresponding to physical terminal locations in the network) with $$m$$ phases can be expressed in matrix form:
 * $$\displaystyle{}\begin{bmatrix}V_{1_{ag}}\\\downarrow \\V_{1_{mg}}\end{bmatrix}-\begin{bmatrix}V_{2_{ag}}\\\downarrow \\V_{2_{mg}}\end{bmatrix}=\begin{bmatrix}\hat z_{1-2_{aa}}&\rightarrow &\hat z_{1-2_{am}}\\\downarrow &\searrow &\downarrow \\\hat z_{1-2_{ma}}&\rightarrow &\hat z_{1-2_{mm}}\end{bmatrix}\begin{bmatrix}I_{1-2_{ag}}\\\downarrow \\I_{1-2_{mg}}\end{bmatrix}$$

For example, consider a distribution line with two electrically isolated feeders sharing one neutral phase. One of the feeders has all three phases $$(a, b, c)$$ present, while the other has only two phases $$(a, c)$$ present. For clarity, the two phases on the second feeder are renamed $$(d, e)$$. Then, the voltage drop on the line can be expressed by:
 * $$\displaystyle{}\begin{bmatrix}V_{1_{ag}}\\V_{1_{bg}}\\V_{1_{cg}}\\V_{1_{dg}}\\V_{1_{eg}}\\V_{1_{ng}}\end{bmatrix}

- \begin{bmatrix}V_{2_{ag}}\\V_{2_{bg}}\\V_{2_{cg}}\\V_{2_{dg}}\\V_{2_{eg}}\\V_{2_{ng}}\end{bmatrix} = \begin{bmatrix} \hat z_{1-2_{aa}}&\hat z_{1-2_{ab}}&\hat z_{1-2_{ac}}&\hat z_{1-2_{ad}}&\hat z_{1-2_{ae}}&\hat z_{1-2_{an}}\\ \hat z_{1-2_{ba}}&\hat z_{1-2_{bb}}&\hat z_{1-2_{bc}}&\hat z_{1-2_{bd}}&\hat z_{1-2_{be}}&\hat z_{1-2_{bn}}\\ \hat z_{1-2_{ca}}&\hat z_{1-2_{cb}}&\hat z_{1-2_{cc}}&\hat z_{1-2_{cd}}&\hat z_{1-2_{ce}}&\hat z_{1-2_{cn}}\\ \hat z_{1-2_{da}}&\hat z_{1-2_{db}}&\hat z_{1-2_{dc}}&\hat z_{1-2_{dd}}&\hat z_{1-2_{de}}&\hat z_{1-2_{dn}}\\ \hat z_{1-2_{ea}}&\hat z_{1-2_{eb}}&\hat z_{1-2_{ec}}&\hat z_{1-2_{ed}}&\hat z_{1-2_{ee}}&\hat z_{1-2_{en}}\\ \hat z_{1-2_{na}}&\hat z_{1-2_{nb}}&\hat z_{1-2_{nc}}&\hat z_{1-2_{nd}}&\hat z_{1-2_{ne}}&\hat z_{1-2_{nn}} \end{bmatrix} \begin{bmatrix}I_{1-2_{ag}}\\I_{1-2_{bg}}\\I_{1-2_{cg}}\\I_{1-2_{dg}}\\I_{1-2_{eg}}\\I_{1-2_{ng}}\end{bmatrix}$$

The hat notation, taken from [Kersting], indicates that the return path, i.e. the ground impedance, has been folded into the other impedances. According to Carson's equations, the elements of the primitive impedance matrix can be calculated by:
 * $$\displaystyle{}\hat z_{1-2_{ii}}=r_i+4\omega P_{ii}G + j\left(X_i + 2\omega G\ln{\frac{S_{ii}}{RD_i}} + 4\omega Q_{ii}G\right)\Omega /mi$$
 * $$\displaystyle{}\hat z_{1-2_{ij}}=4\omega P_{ii}G + j\left(2\omega G\ln{\frac{S_{ij}}{D_{ij}}} + 4\omega Q_{ij}G\right)\Omega /mi$$

Wherein:
 * $$X_i = 2\omega{}G\ln{\frac{RD_i}{GMR_{i}}}\Omega /mi$$
 * $$P_{ij} = \frac{\pi}{8}-\frac{1}{3\sqrt{2}}k_{ij}\cos{\theta_{ij}}+\frac{k_{ij}^2}{16}\cos{2\theta_{ij}}\left(0.6728+\ln{2}{k_{ij}}\right)+\frac{k_{ij}^2}{16}\theta_{ij}\sin{2\theta_{ij}}$$
 * $$Q_{ij} = -0.0386 + \frac{1}{2}\ln{\frac{2}{k_{ij}}}+\frac{1}{3\sqrt{2}}k_{ij}\cos{\theta_{ij}}$$
 * $$k_{ij} = 8.565\times 10^{-4}S_{ij}\sqrt{\frac{f}{\rho}}$$

The primitive series admittance matrix can be inverted to yield the primitive series admittance matrix for the line. For a more complete model of the line, the primitive shunt admittance matrix can then be added.

The effective resistance and GMR of the tape-shield and concentric neutral ring need to be calculated in order to use the previously defined equations.

Concentric Neutral
The effective geomentric mean of the concentric neutral ring can be found using the following equation.
 * $$\displaystyle{}GMR_{i,cn}=(GMR_{i,s}k_{i}R_{i}^{k_{i}-1})^{\frac{1}{k_{i}}} ft$$

$$R$$ is the radius of the circle passing through the center of the concentric neutral strands in ft and can be found using the equation below.
 * $$\displaystyle{}R_{i}=\frac{d_{i,od}-d_{i,s}}{24} ft$$

The effective resistance of the concentric neutral ring is calculated using the following equation.
 * $$\displaystyle{}r_{i,cn}=\frac{r_{i,s}}{k_{i}} \Omega/mi$$

Because the distance between cables is much greater than $$R_{i}$$ it is a good approximation to treat the concentric neutral strands as a single conductor located a distance $$R$$ above the center of the cable when determining distances between adjacent conductor cables.

Tape-Shielded
The GMR of the tape shield is given the equation below.
 * $$\displaystyle{}GMR_{i,sh}=\frac{\frac{d_{i,sh}}{2}-\frac{T_{i}}{2000}}{12} ft$$

The resistance of the tape sheild is the given in the equation below.
 * $$\displaystyle{}r_{i,sh}=\frac{C_{3}\rho_{i,sh}}{(C_{1}C_{2}d_{i,sh}T{i}+(C_{2}T_{i})^2)} \Omega/mi$$

The distance between the tape shield and it's own phase conductor is $$GMR_{sh}$$.

Shunt Admittance
The equations presented below assume that the electric field created by the charge on the phase conductor is is confined to the boundary of the insulation. Because of this assumption there is no cross coupling of the admittances between conductors.

Concentric Neutral
The shunt admittance between a conductor and the concentric neutral ring for a single cable is defined by the equation below.
 * $$\displaystyle{}y_{in}=0+j\frac{2\pi\omega\epsilon_{0}\epsilon_{i,r}}{10^{6}(ln(\frac{2R_{i}}{d_{i,c}})-\frac{1}{k_{i}}ln(k_{i}\frac{d_{i,s}}{2R_{i}}))} S/mi$$

Tape-Shield
The shunt admittance between a conductor and the tape-shield for a single cable is defined by the equation below.
 * $$\displaystyle{}y_{in}=0+j\frac{2\pi\omega\epsilon_{0}\epsilon_{i,r}}{10^{6}ln(\frac{2R_{i}}{d_{i,c}})} S/mi$$