*2.2.2.1. Normal jacketed shield power cable HF loss calculations*

The conductor and ground shield layers of shielded distribution cables help provide a smooth interface between conductors and insulation. Since the losses caused by the conductor shield are pretty small compared to the losses in other components, these losses are sometimes ignored in analyses. Figure 7 shows a typical jacketed shielded cable geometry and its simplified computation model [11 and 17]. The current flows from the conductor through the insulation (dielectric) passing through the ground semiconducting shield before it reaches the neutral, resulting in big losses in the shield at high frequency. The current passing through the ground shield is determined by the conductor voltage level and dielectric impedance, and can be considered as a "current source."

**Figure 7.** A typical jacketed shielded cable geometry and its simplified computation model [11 and 17]. The current flows from the conductor through the insulation (dielectric) passing the ground semiconducting shield before it reach‐ es the neutral resulting in big losses in the shield at high frequency.

The loss caused by the current in the ground shield, including the component caused by the propagation of the current in a circumferential direction to reach a neutral wire, can be calculated with known cable parameters.

With this simplified model, the ground shield current and voltage distributions of such a system can be derived [11 and 17]:

power cable can have different geometries, configurations, and number of neutral wires. The chapter won't cover all cable configurations. Two typical shielded cables will be studied.

The conductor and ground shield layers of shielded distribution cables help provide a smooth interface between conductors and insulation. Since the losses caused by the conductor shield are pretty small compared to the losses in other components, these losses are sometimes ignored in analyses. Figure 7 shows a typical jacketed shielded cable geometry and its simplified computation model [11 and 17]. The current flows from the conductor through the insulation (dielectric) passing through the ground semiconducting shield before it reaches the neutral, resulting in big losses in the shield at high frequency. The current passing through the ground shield is determined by the conductor voltage level and dielectric impedance, and can

**Figure 7.** A typical jacketed shielded cable geometry and its simplified computation model [11 and 17]. The current flows from the conductor through the insulation (dielectric) passing the ground semiconducting shield before it reach‐

The loss caused by the current in the ground shield, including the component caused by the propagation of the current in a circumferential direction to reach a neutral wire, can be

es the neutral resulting in big losses in the shield at high frequency.

calculated with known cable parameters.

*2.2.2.1. Normal jacketed shield power cable HF loss calculations*

be considered as a "current source."

100 Advanced Electromagnetic Waves

$$\begin{aligned} V(\mathbf{x}) &= -\frac{V\_o \left\{ \exp\left(x\sqrt{\frac{K\_1}{K\_2}}\right) + \exp\left[(d-x)\sqrt{\frac{K\_1}{K\_2}}\right] \right\}}{\left[1 + \exp(d)\sqrt{\frac{K\_1}{K\_2}}\sqrt{K\_2}\right]} \\\ I(\mathbf{x}) &= \frac{V\_o \left\{ \exp\left[(d-x)\sqrt{\frac{K\_1}{K\_2}}\right] - \exp(x\sqrt{\frac{K\_1}{K\_2}}) \right\}}{\left[1 + \exp(d)\sqrt{\frac{K\_1}{K\_2}}\right] \sqrt{K\_1 K\_2}} \\\ K\_1 &= \frac{1}{\left(\sigma\_3 + j\alpha \kappa\_0 \kappa\_3\right)\Gamma} - \frac{2j\pi R\_s}{\alpha \kappa \Gamma} \\\ K\_2 &= \frac{T\_2}{\sigma\_4 + j\alpha \kappa\_0 \kappa\_4} - \frac{2j\pi R\_s}{\alpha \kappa \Gamma} \end{aligned} (6)$$

where *d* is the circumferential distance between two neutral wires; *V*<sup>0</sup> is the applied external voltage; *x* is the position along the circumference of the ground shield; *σ*<sup>3</sup> is the conductivity of the ground shield; *ε*<sup>3</sup> is the dielectric constant of the ground shield; *σ*<sup>4</sup> is the conductivity of the conductor shield; *ε*4 is the dielectric constant of the conductor shield; *ω* is the angular frequency of the applied voltage; *C* is the capacitance of the insulation per meter; *R*s is the radius of the ground shield; *T* is the thickness of the ground shield; and *T*2 is the thickness of the conductor shield. From Figure 7, it is the current that passes through the resistive compo‐ nent of the grounding shield that causes losses. To find the loss, the current through the resistive component needs to be derived, which is given by [11 and 17]

$$I\_{r1} = \frac{j\sigma\_3 I(\mathbf{x})}{j\sigma\_3 - \alpha \mathfrak{e}\_0 \mathfrak{e}\_3} \tag{7}$$

and similarly, the current passing through the resistive component of the conductor shield can be found as

$$I\_{r2} = \frac{j\sigma\_4 I\_c(\mathbf{x})}{j\sigma\_4 - \alpha \varepsilon\_0 \varepsilon\_4} \tag{8}$$

With these two currents known, the losses in the conductor and ground shields can be found by calculating the power dissipation, i.e. the product of *I*r1 and *I*r2 and the resistance of the ground and conductor shields. With some math software such as Maple program and integrating the product of current and the resistive components, the amount of power dissipated can be found. To make it easy for the reader to perform such calculations, the Maple program is attached in the Appendix.

The losses in the ground shield with some typical parameters (neutral wire conductor radius of 4 mm, dielectric thickness of 7 mm, conductor shield thickness of 0.2 mm, and ground shield thickness of 0.5 mm, the relative dielectric constant of 200, and conductivity of 0.1 S/m) can be seen in Figure 8 for a different number of neutral wires. Finite analysis element calculations are also conducted and the results of which are compared with the analytic results. They agree with each other well.

**Figure 8.** Losses in the ground shield for different numbers of neutral wires [11 and 17]. Finite analysis element calcu‐ lations are also conducted and the results of which are compared with the analytic results. They agree with each other well.

To give the reader a better idea on the total HF losses in the shielded power cable, one more plot is given in Figure 9 for a cable with six neutral wires [11 and 17]. From Figure 9, it can be seen that for high frequencies (from 5 to 20 MHz), the ground shield loss, including the losses caused by the interaction of the ground shield with the neutral wires, dominates the HF loss for this six neutral-wire cable. The interaction of the neutral wires with the ground shield plays a critical role for the total loss, as can be seen by comparison with the line showing the ground shield loss with and without the effect of the circumferential current.

### *2.2.2.2. Unjacketed shield power cable HF loss calculations*

To save costs, sometimes, unjacketed cables are installed. A cable jacket ensures intimate contact between ground shield and neutral wires. For unjacketed cables, this contact is not ensured resulting in the loss caused by the ground shield as a function of the separation between the neutral wires and the ground shield and the distance between points of contact of the neutral wires with the ground shield [13]. This section will address the loss as a function of neutral wire separation from the ground shield, contact interval between the neutral wires, Electromagnetic Waves Propagation and Detection in Shielded Dielectric Power Cables http://dx.doi.org/10.5772/61055 103

The losses in the ground shield with some typical parameters (neutral wire conductor radius of 4 mm, dielectric thickness of 7 mm, conductor shield thickness of 0.2 mm, and ground shield thickness of 0.5 mm, the relative dielectric constant of 200, and conductivity of 0.1 S/m) can be seen in Figure 8 for a different number of neutral wires. Finite analysis element calculations are also conducted and the results of which are compared with the analytic results. They agree

**Figure 8.** Losses in the ground shield for different numbers of neutral wires [11 and 17]. Finite analysis element calcu‐ lations are also conducted and the results of which are compared with the analytic results. They agree with each other

To give the reader a better idea on the total HF losses in the shielded power cable, one more plot is given in Figure 9 for a cable with six neutral wires [11 and 17]. From Figure 9, it can be seen that for high frequencies (from 5 to 20 MHz), the ground shield loss, including the losses caused by the interaction of the ground shield with the neutral wires, dominates the HF loss for this six neutral-wire cable. The interaction of the neutral wires with the ground shield plays a critical role for the total loss, as can be seen by comparison with the line showing the ground

To save costs, sometimes, unjacketed cables are installed. A cable jacket ensures intimate contact between ground shield and neutral wires. For unjacketed cables, this contact is not ensured resulting in the loss caused by the ground shield as a function of the separation between the neutral wires and the ground shield and the distance between points of contact of the neutral wires with the ground shield [13]. This section will address the loss as a function of neutral wire separation from the ground shield, contact interval between the neutral wires,

shield loss with and without the effect of the circumferential current.

*2.2.2.2. Unjacketed shield power cable HF loss calculations*

with each other well.

102 Advanced Electromagnetic Waves

well.

**Figure 9.** The total losses in a cable with six neutral wires. The losses from the current flowing through the ground losses contribute significantly to the total loss.

ground shield dielectric properties, frequency, etc. Figure 10 [13] gives a simplified model for the unjacketed cable.

**Figure 10.** A simplified model for the unjacketed cable. Due to the nature of unjacketed cables, the neutral wires only have contacts with certain points, A, B and C.

Due to the nature of unjacketed cables, the neutral wires only have contacts with certain points, A, B, and C. Similar to a normal jacketed cable, the current flowing through resistive compo‐ nent of the ground shield leads to loss that dominates the loss for high frequencies. The voltage and current distributions as well as the current passing through the resistive component of such a system can be found as [13]

$$V(Y) = -\frac{V\_0 \left(Z\_c + Z\_c \exp(-d\sqrt{K\mathbf{1}}) + Z\_{c1} \left[\exp\left(-x\sqrt{K\mathbf{1}}\right) + \exp\left[(-d\sqrt{K\mathbf{1}})\right]\right]\right)}{\left(Z\_{c1} + Z\_c\right)\left[1 + \exp\left(-d\sqrt{K\mathbf{1}}\right)\right]},$$

$$I(\mathbf{x}) = \frac{V\_0 \sqrt{Z\_{c1}(Z\_c + Z\_{c1})} \left[\exp\left(-x\sqrt{K\mathbf{1}}\right) - \exp\left[(-d\sqrt{K\mathbf{1}})\right]\right]}{\left[\left(Z\_{c1} + Z\_c\right)\left(1 + \exp\left(-d\sqrt{K\mathbf{1}}\right)\right)\right]\sqrt{Z\_c Z\_c}},\tag{9}$$

$$\begin{aligned} K1 &= \frac{Z\_c(Z\_{c1} + Z\_c)}{Z\_{c1}Z\_c} \\ Z\_c &= \frac{-j\ln\left(\frac{R}{R\_c}\right)}{2\alpha\pi x\_0} \\ Z\_c &= \frac{-j\alpha}{\alpha\mathcal{C}\_{\text{eff}}} \\ Z\_c &= \frac{-j\alpha\kappa\_{\text{eff}}}{8\pi\mathcal{R}\_d T \left[\sigma^2 + \left(\alpha\kappa\_{\text{eff}}\right)^2\right]} \end{aligned}\tag{9}$$

$$\begin{aligned} &\sigma^2 - j\alpha\kappa\_{\text{eff}} \le M(\sigma) \end{aligned}$$

$$I\_{r1} = \frac{\left(\sigma^2 - j\sigma\alpha\varepsilon\_0\varepsilon\_1\right)I(\mathbf{x})}{\sigma^2 + \left(\alpha\varepsilon\_0\varepsilon\_1\right)^2} \tag{10}$$

Similarly to a normal jacketed cable, the losses can be found by integrating the product of the resistive current and the resistance. The Maple program is similar to the normal jacketed shield cable and can be found in paper [13], and thus is not given in this chapter. The ground shield losses in the unjacketed cable, which dominates the HF loss, can vary a lot for different shield dielectric constants, conductivity, and thickness of the ground shield. The thickness of the ground shield plays an important role because the displacement current from the conductor to the ground shield flows longitudinally through the resistance of the ground shield before it reaches a neutral wire-ground shield contact. This loss is maximized under the condition that the resistive impedance of this path is comparable to the capacitive impedance of the dielectric, which is significantly affected by the thickness of the ground shield. For a typical ground shield thickness of 1 mm, and conductor radius, 3.62 mm; insulation thickness, 4.83 mm; capacitance between ground shield and neutral wires, 150 pF/m; insulation dielectric constant, 2.2; distance between two neutral wire-ground shield contacts, 3 cm; frequency, 20 MHz, the loss of such a jacketed cable vs. dielectric constant and conductivity is shown in Figure 11 [13]. From Figure 11, it can be seen that the losses can be pretty large due to the long distance needed for the displacement current to flow before it reaches the neutral wires, which means that the HF electromagnetic PD wave can be attenuated significantly in the unjacketed cable.

Due to the nature of unjacketed cables, the neutral wires only have contacts with certain points, A, B, and C. Similar to a normal jacketed cable, the current flowing through resistive compo‐ nent of the ground shield leads to loss that dominates the loss for high frequencies. The voltage and current distributions as well as the current passing through the resistive component of

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Similarly to a normal jacketed cable, the losses can be found by integrating the product of the resistive current and the resistance. The Maple program is similar to the normal jacketed shield cable and can be found in paper [13], and thus is not given in this chapter. The ground shield losses in the unjacketed cable, which dominates the HF loss, can vary a lot for different shield dielectric constants, conductivity, and thickness of the ground shield. The thickness of the ground shield plays an important role because the displacement current from the conductor to the ground shield flows longitudinally through the resistance of the ground shield before it reaches a neutral wire-ground shield contact. This loss is maximized under the condition that the resistive impedance of this path is comparable to the capacitive impedance of the dielectric, which is significantly affected by the thickness of the ground shield. For a typical ground shield thickness of 1 mm, and conductor radius, 3.62 mm; insulation thickness, 4.83 mm; capacitance between ground shield and neutral wires, 150 pF/m; insulation dielectric constant, 2.2; distance between two neutral wire-ground shield contacts, 3 cm; frequency, 20 MHz, the loss of such a jacketed cable vs. dielectric constant and conductivity is shown in

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such a system can be found as [13]

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104 Advanced Electromagnetic Waves

*I x*

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1

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As stated at the beginning of this section, shield cables have different geometries and config‐ urations. The authors won't address too many configurations and only two typical shielded cables, i.e. jacketed normal shielded cable and unjacketed shielded cables, are studied in this section. The reader should keep in mind that when designing and analyzing the HF electro‐ magnetic PD wave signals, it's very important to know the shield dielectric properties and cable actual geometry and configurations to get a rough idea of the cable's HF attenuation properties. The readers should also know that it's not easy to do the HF dielectric property measurements and use caution when conducting such measurements. Furthermore, the reader needs to know that the shields in the cables are under pressure whereas the shields under test have no pressure. Some compensation might be necessary to offset the pressure difference.

**Figure 11.** Losses vs. dielectric constant and conductivity for typical parameters.

### **2.3. HF electromagnetic PD wave propagation**

The HF electromagnetic PD wave propagates in the shield cable and the wave is attenuated while it travels. How significant the wave is is mainly determined by the HF loss properties of the cable, which has been discussed extensively in Section 2.2. Besides the losses of the cable, the cable splices and termination can also affect the wave propagation and attenuation. Figure 12 shows one typical PD wave propagation. The PD wave propagates in both directions passing through the cable splices before it reaches the cable termination and reflects back if the waves are not attenuated completely. When it travels along the cable, HF components are most attenuated due to the facts discussed in Section 2.2, because the displacement current increased with frequency resulting in higher losses for higher frequencies. This provides the electromagnetic wave pulse a big pulse width as shown in Figure 12. If the pulse is not attenuated completely, it will reach the cable termination and reflect backward toward the PD source as shown in Figure 12. To analyze the effect of the cable losses, the electromagnetic wave pulse can be treated as a Gaussian pulse. If a Gaussian pulse in time domain is applied at the left end of the cable to simulate the PD source signal, it will propagate through the splices before it can reach the termination. For simplicity, splices are thought in parallel with loads having characteristic impedance. Since we apply a Gaussian pulse in time domain, we need to transform it to the frequency domain, multiply it with attenuation factor caused by splices in frequency domain, then transform it back to time domain. The output voltage is attenuated by a factor of Vatt which is

$$\text{Vatt} = \frac{\text{ZZ}}{\text{Z} + \text{Z}\_c}, \text{Z} = \frac{\text{Z}\_s \text{Z}\_c}{\text{Z}\_s + \text{Z}\_c} \tag{11}$$

where *Z*s is splice effective impedance, *Z*c is characteristic impedance. By doing so, we get loss vs. number of splices as shown in Figure 13. Parameters used in computation are given in the caption of Figure 13. We also can treat the problem by calculating the energy dissipated by the cable. Current flows through semicon will result in loss. By calculating the loss, we can get a voltage attenuation ratio in frequency domain, Vatt1, which is:

$$Vatt1 = \sqrt{1 - \frac{P\_L}{P\_I}}\tag{12}$$

where *P*L is the power dissipated by splice whereas *P*I is the input power. A Gaussian pulse multiplied by this ratio results in an attenuated Gaussian pulse as seen in Figure 14.

**Figure 12.** One typical PD wave propagation pattern. Note that the wave propagates in both directions although only the right section is shown here.

the cable splices and termination can also affect the wave propagation and attenuation. Figure 12 shows one typical PD wave propagation. The PD wave propagates in both directions passing through the cable splices before it reaches the cable termination and reflects back if the waves are not attenuated completely. When it travels along the cable, HF components are most attenuated due to the facts discussed in Section 2.2, because the displacement current increased with frequency resulting in higher losses for higher frequencies. This provides the electromagnetic wave pulse a big pulse width as shown in Figure 12. If the pulse is not attenuated completely, it will reach the cable termination and reflect backward toward the PD source as shown in Figure 12. To analyze the effect of the cable losses, the electromagnetic wave pulse can be treated as a Gaussian pulse. If a Gaussian pulse in time domain is applied at the left end of the cable to simulate the PD source signal, it will propagate through the splices before it can reach the termination. For simplicity, splices are thought in parallel with loads having characteristic impedance. Since we apply a Gaussian pulse in time domain, we need to transform it to the frequency domain, multiply it with attenuation factor caused by splices in frequency domain, then transform it back to time domain. The output voltage is attenuated

> <sup>2</sup> , *s c c s c*

*Vatt Z*

*Vatt*

voltage attenuation ratio in frequency domain, Vatt1, which is:

*Z Z Z*

where *Z*s is splice effective impedance, *Z*c is characteristic impedance. By doing so, we get loss vs. number of splices as shown in Figure 13. Parameters used in computation are given in the caption of Figure 13. We also can treat the problem by calculating the energy dissipated by the cable. Current flows through semicon will result in loss. By calculating the loss, we can get a

1 1 *<sup>L</sup>*

multiplied by this ratio results in an attenuated Gaussian pulse as seen in Figure 14.

where *P*L is the power dissipated by splice whereas *P*I is the input power. A Gaussian pulse

**Figure 12.** One typical PD wave propagation pattern. Note that the wave propagates in both directions although only

*I P*

*Z Z Z Z* = = <sup>+</sup> <sup>+</sup> (11)

*<sup>P</sup>* = - (12)

by a factor of Vatt which is

106 Advanced Electromagnetic Waves

the right section is shown here.

**Figure 13.** Amplitude as a function of number of splice number. Splices have the following parameters: length, 60 cm; semicon conductivity, 5 s/m; conductor radius, 4 mm; thickness of the insulation, 4 mm; thickness of semicon, 6 mm; dielectric constant of semicon, 300; dielectric constant of insulation, 2.2.

**Figure 14.** Attenuated Gaussian pulse. The splice has the following parameters: radius of conductor, 1 cm; thickness of insulation, 0.5 cm; thickness of semicon, 0.3 cm; conductivity of semicon, 10 S/m; dielectric constant of semicon, 300; dielectric constant of insulation, 3.3; distance of splice, 40 cm.

The reader should keep in mind that the studied case is a simulation result based on some assumed shield dielectric properties. To assess the losses caused by the splices, some field measurements as well as further computations are necessary. Normally, the splices have smaller losses compared with the long cable.
