**2.3. High voltage**

In the Scandinavian countries, the development of overhead lines that use covered conductors began in the mid-eighties. The first country to introduce covered conductors was Finland. The finish distribution company FingridOjv with the cooperation of Eltel Networks began with the use of covered conductor in high voltage networks. They build a 110 kV DV power line Mätäkivi-Sula with a length of 6 km, half of which was equipped with covered conductors. In the construction of the transmission lines, composite insulators were used.

In General Public Utilities/ Pennsylvania Electric Company they researched the electric and mechanic effects of covered conductors insulated with polyethylene. Those cables had high specific weight and were installed on several insulators and clamps. They conducted two experiments with the voltage of 70 kV. The first was to determine the feasibility of the existent spacer (a device for spacing bundled conductors). The second was done on a distribution cable with higher voltage, to determine the criteria for building new compact high-voltage transmission lines in small corridors, especially in densely populated areas.

Since 1980, the use of covered conductors increased worldwide. The reason for this is that the covered conductors are more compact and environmentally friendlier than traditional non-insulated conductors. Also the number of failures is much lower. This development has also an impact on the characteristics of voltage drops and is an important aspect of working with clients, sensitive to such decrease in voltage.

### **3. Using covered conductors at the highest voltage levels**

The space around an electrically charged body is in a special state. This special state acts only on particles that have an electric charge. If we introduce a small electric charge that does not significantly alter the state of that space, we find that there is force acting on that small charge. This force is proportional to the electric charge *q* and the vector quantity that defines the state of the space. It is denoted as *E*. The vector *E* of the electric field intensity has the same direction as the force.

#### **3.1. The electric field: Cylinder of charge**

For the understanding of the distribution of voltage and electric field intensity in a covered conductor, we first look at the electric field in a cylinder of charge (Voršič J., Pihler J., 2005). The electric field intensity *E*(r) of an isolated cylindrical Gaussian surface is shown in figure 1.

In an infinitely long charged cylinder two things are true:


From the equation,

$$\mathbf{Q} = \mathbf{q} \cdot \mathbf{l} = \mathbf{D} \cdot \mathbf{2} \cdot \boldsymbol{\pi} \cdot \mathbf{r} \cdot \mathbf{l},\tag{1}$$

where:

382 Polyurethane

electrically.

**2.3. High voltage** 

**2.1. Covered conductors** 

**2.2. The use of covered conductors** 

covered conductors are better safety, ecology (fewer disturbances in the nature, especially

The most widespread structure of covered conductors consists of the core made from hard, compact aluminium alloy and a watertight/waterproof mantel made from a cross-linked polyethylene (XLPE). The use has shown the reliability of this type of conductor in very difficult conditions. It will withstand the weight of a fallen tree for days, mechanically and

Because of the outer mantel, the covered conductors are not that vulnerable when touching each other, or in contact with tree branches. This enables the spacing between phases/cables to shrink to one third of the space between normal over ground power lines. The platform

Slovenia began to introduce covered conductors in 1992. In the same year Elektro Gorenjska performed reconstruction of the transmission line Savica – Komna, where they replaced the bare conductors with covered ones. All other parts of the power lines stayed the same. In the year 1939, Elektro Ljubljana build the first power line based on finish technology. They decided for the finish 20kV system because of positive experience from a more than 10000 km of build power lines with covered conductors. These experiences show us that the use of such system reliability has increased by 5 times. The number of failures per year on a 100 km long section is 4.5 for bare conductors and 0.9 for covered conductors in the SN network (Tičar I., Zorič T., 2003). Because the two mentioned power lines successfully passed the adverse weather conditions (sleet, an additional burden of winter), the importer and representative (C&G d.o.o. Ljubljana) for the company (Ensto), obtain an expert opinion on the imported equipment. It was found that the covered conductors, the corresponding hanging and insulation materials and the overvoltage protection used in the 20kV distribution network conform to the current JUS standard. The material and equipment that is not included in the

In the Scandinavian countries, the development of overhead lines that use covered conductors began in the mid-eighties. The first country to introduce covered conductors was Finland. The finish distribution company FingridOjv with the cooperation of Eltel Networks began with the use of covered conductor in high voltage networks. They build a 110 kV DV power line Mätäkivi-Sula with a length of 6 km, half of which was equipped with covered conductors. In the construction of the transmission lines, composite insulators were used.

In General Public Utilities/ Pennsylvania Electric Company they researched the electric and mechanic effects of covered conductors insulated with polyethylene. Those cables had high specific weight and were installed on several insulators and clamps. They conducted two

for an over ground power line in a wooden area can therefore be smaller/narrower.

JUS standard was covered and conformed in the international IEC standards.

less clearing of trees), better operational reliability and lower operating costs.

*q –*is the electric charge, gathered on the length of the cylinder,

*r –* is the radius of the equipotential surface,

*l –* is the length of the cylinder,

*D–*is the electric displacement field and

*Q–* is the electric charge;

we obtain the absolute value for the electric displacement field,

$$D = \frac{Q}{A} = \frac{q \cdot l}{2 \cdot \pi \cdot r \cdot l} = \frac{q}{2 \cdot \pi \cdot r} \tag{2}$$

And it's vector quantity

$$
\bar{D} = D \cdot \bar{\mathbf{1}}\_{\mathbf{r}} = \frac{q}{2 \cdot \pi \cdot r} \cdot \bar{\mathbf{1}}\_{\mathbf{r}}.\tag{3}
$$

Polyurethane as an Isolation for Covered Conductors 385

(8)

(9)

2 1

*r*

*r*

ln

( )

*<sup>U</sup> E r*

 

We get the highest value of the electric field intensity on the surface of the inner cylinder.

1

*r*

( )

**Figure 2.** The electric field of two concentric cylinders

**3.2. Double-layer single-wire cable** 

layers of the dielectric.

*r*

1 max 2

ln *<sup>U</sup> E r <sup>E</sup> r*

 

1

*r*2

2

3

2 3

*<sup>r</sup>* (10)

*<sup>r</sup>* (11)

*UUU* 1 2 (12)

(13)

0 r1 1 ln

0 r2 2 ln

0 r1 1 r2 2 1 1 ln ln

 *r r*

*r*

*r*1

The double-layer single-wire cable (figure 3) is a typical example of the use of double-layer dielectrics. The voltage between the core of the cable and the mantel is distributed over the

> *<sup>q</sup> <sup>r</sup> <sup>U</sup>*

*<sup>q</sup> <sup>r</sup> <sup>U</sup>* 

*<sup>q</sup> <sup>r</sup> <sup>r</sup> <sup>U</sup>*

1

2

2

 

2

2

The corresponding electric field intensity is

$$
\bar{E} = \frac{\bar{D}}{\varepsilon} = \frac{q}{2 \cdot \pi \cdot \varepsilon \cdot r} \cdot \bar{1}\_{\text{r}} \tag{4}
$$

The Electric potential *V*(*r*) on an equipotential surface (concentric cylinder) with a radius *r* is

$$V(r) = \int\_{r}^{r\_0} E \cdot \mathbf{dr} = \frac{q}{2 \cdot \pi \cdot \varepsilon} \cdot \int\_{r}^{r\_0} \frac{\mathbf{dr}}{r} = \frac{q}{2 \cdot \pi \cdot \varepsilon} \cdot \ln \frac{r\_0}{r} \tag{5}$$

where *r*0 is the radius of the equipotential surface, on which we chose the potential s starting point. Because we assume that the cylinder is infinitely long, we cannot select the starting point of the potential to be in infinity.

**Figure 1.** Electric field intensity *E*(r) of an isolated cylindrical Gaussian surface

Between two equipotential surfaces with the radius *r*1 and *r*2 there is a potential difference

$$\mathbb{U}I\_{12} = \mathbb{U}I = V\_1 - V\_2 = \frac{q}{2 \cdot \pi \cdot \varepsilon} \cdot \ln{\frac{r\_2}{r\_1}}\tag{6}$$

Two concentric cylinders with the length *l,* radius *r*1 and *r*2 and a dielectric between them they form a cylindrical capacitor with the capacitance

$$C = \frac{Q}{U} = \frac{q \cdot l}{U} = \frac{2 \cdot \pi \cdot \varepsilon \cdot l}{\ln \frac{r\_2}{r\_1}}\tag{7}$$

Because we don't know the exact electric charge, but only the charge between the electrodes, we denote the expression for the electric field intensity at any random point between the two cylindrical electrodes (figure 2) in the form of

Polyurethane as an Isolation for Covered Conductors 385

$$E(r) = \frac{\mathcal{U}}{r \cdot \ln{\frac{r\_2}{r\_1}}} \tag{8}$$

We get the highest value of the electric field intensity on the surface of the inner cylinder.

**Figure 2.** The electric field of two concentric cylinders

#### **3.2. Double-layer single-wire cable**

384 Polyurethane

And it's vector quantity

The corresponding electric field intensity is

point of the potential to be in infinity.

r r 1 1. 2

*<sup>r</sup>*

r 1

 *r r* (5)

2

*<sup>r</sup>* (6)

(7)

1 ln

(3)

(4)

*<sup>q</sup> D D*

2 *<sup>D</sup> <sup>q</sup> <sup>E</sup>* 

*r r*

*r r*

**Figure 1.** Electric field intensity *E*(r) of an isolated cylindrical Gaussian surface

they form a cylindrical capacitor with the capacitance

two cylindrical electrodes (figure 2) in the form of

Between two equipotential surfaces with the radius *r*1 and *r*2 there is a potential difference

*<sup>q</sup> <sup>r</sup> U UV V*

Two concentric cylinders with the length *l,* radius *r*1 and *r*2 and a dielectric between them

*Q l q l <sup>C</sup> U U r*

Because we don't know the exact electric charge, but only the charge between the electrodes, we denote the expression for the electric field intensity at any random point between the

2

2

 

 

> 2 1

*r*

ln

12 1 2

The Electric potential *V*(*r*) on an equipotential surface (concentric cylinder) with a radius *r* is

0 0 <sup>d</sup> <sup>0</sup> () d ln , 2 2

where *r*0 is the radius of the equipotential surface, on which we chose the potential s starting point. Because we assume that the cylinder is infinitely long, we cannot select the starting

*q q <sup>r</sup> <sup>r</sup> Vr E r* 

 *<sup>r</sup>*

> The double-layer single-wire cable (figure 3) is a typical example of the use of double-layer dielectrics. The voltage between the core of the cable and the mantel is distributed over the layers of the dielectric.

$$
\Delta U\_1 = \frac{q}{2 \cdot \pi \cdot \varepsilon\_0 \cdot \varepsilon\_{r1}} \cdot \ln \frac{r\_2}{r\_1} \tag{10}
$$

$$\mathcal{U}\_2 = \frac{q}{2 \cdot \pi \cdot \varepsilon\_0 \cdot \varepsilon\_{r2}} \cdot \ln \frac{r\_3}{r\_2} \tag{11}$$

$$
\mathcal{U}\mathcal{U}\_1 = \mathcal{U}\_1 + \mathcal{U}\_2 \tag{12}
$$

$$\Delta U = \frac{q}{2 \cdot \pi \cdot \varepsilon\_0} \left( \frac{1}{\varepsilon\_{\rm r1}} \cdot \ln \frac{r\_2}{r\_1} + \frac{1}{\varepsilon\_{\rm r2}} \cdot \ln \frac{r\_3}{r\_2} \right) \tag{13}$$

**Figure 3.** A double-layer single-wire cable

We calculate the electric charge and then both voltage levels.

$$\mathcal{U}\_1 = \frac{\mathcal{U} \cdot \ln \frac{r\_2}{r\_1}}{\varepsilon\_{r1} \cdot \left(\frac{1}{\varepsilon\_{r1}} \cdot \ln \frac{r\_2}{r\_1} + \frac{1}{\varepsilon\_{r2}} \cdot \ln \frac{r\_3}{r\_2}\right)}\tag{14}$$

Polyurethane as an Isolation for Covered Conductors 387

(18)

**3.3. Polarized conductors and the electric field intensity** 

field intensity in the air is as small as possible: *E*air = *E*2(*r*2) .

**Figure 4.** Coaxial cylindrical arrangement of partial isolation

variable *r*2 as *x* we derive the function *y*(*x*) from *x*.

Taking into account that *ε*r1 =*ε*r and *ε*r2 = 1 in the equation (20), we get:

2

2

*r*

*<sup>U</sup> <sup>E</sup>*

Symbols in figure represent:

*E*2*–*the electric field intensity of air.

*r*1*–*the radius of the core, *r*2*–*the radius of the mantel, *r*3*–*the radius of insulation,

A covered conductor can be regarded as a two layered insulated conductor. In this case the inner electrode is insulated and the space to the outer electrode is air. The insulation has a much higher dielectric strength than air. Because of that, it is irrelevant that the electric field intensity is small on the inner electrode (figure 4). The important thing is that the electric

> 2 3 1 2

*r r r r*

1

ln ln

 

r

We get the minimum field intensity *E*2 as a function of *r*2, when the denominator in equation (18) is at its highest value (Beyer M., 1986). When we denote the denominator with y and the

$$IL\_2 = \frac{U \cdot \ln{\frac{r\_3}{r\_2}}}{\varepsilon\_{r2} \cdot \left(\frac{1}{\varepsilon\_{r1}} \cdot \ln{\frac{r\_2}{r\_1}} + \frac{1}{\varepsilon\_{r2}} \cdot \ln{\frac{r\_3}{r\_2}}\right)}\tag{15}$$

We get the highest electric field intensity in material 1 at the radius *r*1,

$$E\_{1\text{max}} = \frac{q}{2 \cdot \pi \cdot \varepsilon\_0 \cdot \varepsilon\_{\text{r}1}} \cdot \frac{1}{r\_1} = \frac{U \cdot \frac{1}{r\_1}}{\varepsilon\_{\text{r}1} \cdot \left(\frac{1}{\varepsilon\_{\text{r}1}} \cdot \ln\frac{r\_2}{r\_1} + \frac{1}{\varepsilon\_{\text{r}2}} \cdot \ln\frac{r\_3}{r\_2}\right)}.\tag{16}$$

and the highest electric field intensity in material 2 at the radius *r*<sup>2</sup>

$$E\_{2\max} = \frac{q}{2 \cdot \pi \cdot \varepsilon\_0 \cdot \varepsilon\_{t2}} \cdot \frac{1}{r\_2} = \frac{U \cdot \frac{1}{r\_2}}{\varepsilon\_{t2} \cdot \left(\frac{1}{\varepsilon\_{t1}} \cdot \ln\frac{r\_2}{r\_1} + \frac{1}{\varepsilon\_{t2}} \cdot \ln\frac{r\_3}{r\_2}\right)}.\tag{17}$$

## **3.3. Polarized conductors and the electric field intensity**

A covered conductor can be regarded as a two layered insulated conductor. In this case the inner electrode is insulated and the space to the outer electrode is air. The insulation has a much higher dielectric strength than air. Because of that, it is irrelevant that the electric field intensity is small on the inner electrode (figure 4). The important thing is that the electric field intensity in the air is as small as possible: *E*air = *E*2(*r*2) .

Symbols in figure represent:

386 Polyurethane

**Figure 3.** A double-layer single-wire cable

1max

2 max

We calculate the electric charge and then both voltage levels.

1

2

r1

r2

We get the highest electric field intensity in material 1 at the radius *r*1,

 

and the highest electric field intensity in material 2 at the radius *r*<sup>2</sup>

 

*<sup>r</sup> <sup>U</sup>*

*<sup>r</sup> <sup>U</sup>*

*<sup>q</sup> <sup>r</sup> <sup>E</sup>*

*<sup>q</sup> <sup>r</sup> <sup>E</sup>*

*r1* -*q*

+*q*

*r*3

*U*<sup>1</sup> *U*<sup>2</sup> *U*

*r*2

2 1

ln

*<sup>r</sup> <sup>U</sup>*

 

r1 1 r2 2

 

3 2

ln

*<sup>r</sup> <sup>U</sup>*

 

r1 1 r2 2

 

0 r1 1 2 3

0 r2 2 2 3

<sup>1</sup> . <sup>2</sup> 1 1 ln ln

<sup>1</sup> . <sup>2</sup> 1 1 ln ln

r1

r2

1 1 ln ln

1 1 ln ln

2 3

2 3

1

 

1

*U*

*r r r*

r1 1 r2 2

2

 

1

*U*

*r r r*

r1 1 r2 2

 

*r r*

 

*r r*

*r r r r* (14)

(15)

(16)

(17)

*r r r r* *r*1*–*the radius of the core, *r*2*–*the radius of the mantel, *r*3*–*the radius of insulation, *E*2*–*the electric field intensity of air.

Taking into account that *ε*r1 =*ε*r and *ε*r2 = 1 in the equation (20), we get:

**Figure 4.** Coaxial cylindrical arrangement of partial isolation

We get the minimum field intensity *E*2 as a function of *r*2, when the denominator in equation (18) is at its highest value (Beyer M., 1986). When we denote the denominator with y and the variable *r*2 as *x* we derive the function *y*(*x*) from *x*.

Denominator:

$$y(\mathbf{x}) = \left(\frac{\ln\frac{\mathbf{x}}{r\_1}}{\varepsilon\_r} + \frac{\ln\frac{r\_3}{\mathbf{x}}}{\mathbf{1}}\right) = \mathbf{x} \cdot \left(\frac{\mathbf{1}}{\varepsilon\_r} \cdot \ln\frac{\mathbf{x}}{r\_1} - \ln\frac{\mathbf{x}}{r\_3}\right) = \frac{\mathbf{x}}{\varepsilon\_r} \cdot \left(\ln\frac{\mathbf{x}}{r\_1} - \varepsilon\_r \cdot \ln\frac{\mathbf{x}}{r\_3}\right) \tag{19}$$

We find the maximum value of the denominator and get the optimal value of the radius,

$$r\_{2\text{opt}} = r\_1 \cdot \frac{1}{\mathbf{e}} \cdot \left(\frac{r\_3}{r\_1}\right)^{\frac{\mathcal{E}\_r}{\mathcal{E}\_r - 1}} \text{ .} \tag{20}$$

Polyurethane as an Isolation for Covered Conductors 389

**Figure 5.** The sketch of a typical 220 kV transmission tower for an overhead power line

**Figure 6.** The optimal radius of the insulation depending on the relative permittivity

where the electric field intensity in the air is at its smallest value.

$$E\_2 = E\_{2\text{ min}} = \frac{U \cdot \mathbf{e}}{r\_1 \cdot \left(\frac{r\_3}{r\_1}\right)^{\frac{\mathcal{E}\_r}{\mathcal{E}\_r - 1}} \cdot \left(1 - \frac{1}{\mathcal{E}\_t}\right)}\tag{21}$$

With the transmission tower, the chain of insulators (the vertical string of discs, l=2.25 m) and the cable (490/65 Al/Fe, r = 15.3 mm) of the existing 220kV over ground power line (Figure 5) the geometry is already set. The only remaining variable is the relative permittivity. If we attempt to calculate the relative permittivity of polyurethane (*ε*r = 3.4), we get the optimal radius where the electric field intensity in the surrounding air is at its smallest value:

$$r\_{2\text{opt}} = r\_1 \cdot \frac{1}{\mathbf{e}} \cdot \left(\frac{r\_3}{r\_1}\right)^{\frac{\mathbf{c}\_r}{\mathbf{c}\_r - 1}} = 0,\\ 0153 \cdot \frac{1}{2,71828} \cdot \left(\frac{2,25}{0,0153}\right)^{\frac{3,4}{3,4 - 1}} = 6,62\text{ m}$$

But this is unrealistic. This is why we try to use other materials with higher values of relative permittivity. Figure 6 shows the dependence of the optimal radius (equation 20) from the relative permittivity. Despite the clear advantages of materials with larger values of relative permittivity we try to use polyurethane.

If we use the same radius as the radius in the current 220kV overhead conductor (r = 15.3mm), the relative permittivity εr = 3.4 and an insulation that is 15 mm thick, together with the voltage of 400kV, we get:

$$E\_{\rm air} = \frac{\mathcal{U}}{r\_2 \cdot \varepsilon\_{r2} \cdot \left(\frac{1}{\varepsilon\_{r1}} \cdot \ln\frac{r\_2}{r\_1} + \frac{1}{\varepsilon\_{r2}} \cdot \ln\frac{r\_3}{r\_2}\right)} = 2,39 \text{ MV/m} \tag{22}$$

The value is smaller from the Dielectric Strength of air in normal conditions (3 MV/m).

Denominator:

smallest value:

2 opt 1

*r r*

with the voltage of 400kV, we get:

1

*r*

of relative permittivity we try to use polyurethane.

air

2 2

*r*

*<sup>U</sup> <sup>E</sup>*

 

*r r <sup>r</sup>*

3

where the electric field intensity in the air is at its smallest value.

2 2 min

r 13 1 3

3

 

1 <sup>1</sup> , <sup>e</sup>

*r*

 

1 *<sup>r</sup> r r*

*rr r r*

 

1

*r r <sup>r</sup>*

1

 

*r r*

1 r

<sup>1</sup> <sup>1</sup>

3,4

2,39 MV/m

(22)

e

3 1

*r r r*

With the transmission tower, the chain of insulators (the vertical string of discs, l=2.25 m) and the cable (490/65 Al/Fe, r = 15.3 mm) of the existing 220kV over ground power line (Figure 5) the geometry is already set. The only remaining variable is the relative permittivity. If we attempt to calculate the relative permittivity of polyurethane (*ε*r = 3.4), we get the optimal radius where the electric field intensity in the surrounding air is at its

<sup>1</sup> 3,4 1 <sup>3</sup>

e 2,71828 0,0153

 

But this is unrealistic. This is why we try to use other materials with higher values of relative permittivity. Figure 6 shows the dependence of the optimal radius (equation 20) from the relative permittivity. Despite the clear advantages of materials with larger values

If we use the same radius as the radius in the current 220kV overhead conductor (r = 15.3mm), the relative permittivity εr = 3.4 and an insulation that is 15 mm thick, together

2 3

*r r*

*r r*

1122

 

1 1 ln ln *<sup>r</sup> r r*

The value is smaller from the Dielectric Strength of air in normal conditions (3 MV/m).

 

1 1 2,25 0,0153 6,62 m

(20)

(21)

(19)

*r x x xx x x*

 

ln ln <sup>1</sup> ( ) ln ln ln ln

We find the maximum value of the denominator and get the optimal value of the radius,

2 opt 1

*<sup>U</sup> E E*

*r r*

1

*y x x*

*x r*

**Figure 5.** The sketch of a typical 220 kV transmission tower for an overhead power line

**Figure 6.** The optimal radius of the insulation depending on the relative permittivity

For other insulation thicknesses, the electric field intensity (according to equation 21) is shown in figure 7. Figure 8 shows the highest electric field intensities on the edge of the insulation for different values of relative permittivity. The difference becomes apparent only at greater thicknesses of the mantel. You can also see that better dielectric displace more electric field in the worse dielectric - the air.

Polyurethane as an Isolation for Covered Conductors 391

**Figure 8.** The highest field intensity on the edge of the insulation for different values of relative

**Figure 9.** The highest electric field intensity on the edge of the conductor as the function of insulation

permittivity

thickness

Figure 9 shows the electric field intensity at the edge of the conductor as the function of the insulation thickness. As expected it is substantially lower than the dielectric strength, the reason being that the majority of the electric field is displaced into the worse dielectric – the surrounding which is also more abundant (the length of the string of insulating discs).

According to survey results, we find that the use of covered conductors with insulation made from polyurethane enables the preservation of the existing 220 kV power line platforms and the transition to 400kV power lines. Furthermore we decided to reduce the weight by using the thinnest possible insulation that still meets all requirements – thickness of 15 mm.

**Figure 7.** The highest electric field intensity in air

discs).

of 15 mm.

electric field in the worse dielectric - the air.

**Figure 7.** The highest electric field intensity in air

For other insulation thicknesses, the electric field intensity (according to equation 21) is shown in figure 7. Figure 8 shows the highest electric field intensities on the edge of the insulation for different values of relative permittivity. The difference becomes apparent only at greater thicknesses of the mantel. You can also see that better dielectric displace more

Figure 9 shows the electric field intensity at the edge of the conductor as the function of the insulation thickness. As expected it is substantially lower than the dielectric strength, the reason being that the majority of the electric field is displaced into the worse dielectric – the surrounding which is also more abundant (the length of the string of insulating

According to survey results, we find that the use of covered conductors with insulation made from polyurethane enables the preservation of the existing 220 kV power line platforms and the transition to 400kV power lines. Furthermore we decided to reduce the weight by using the thinnest possible insulation that still meets all requirements – thickness

> **Figure 8.** The highest field intensity on the edge of the insulation for different values of relative permittivity

**Figure 9.** The highest electric field intensity on the edge of the conductor as the function of insulation thickness
