Methodology for selecting the cross-section of current-carrying conductors of high-voltage flexible cables
Abstract Generated abstract
This paper proposes a procedure for choosing the conductor cross section of high-voltage flexible cables used with mobile equipment, where insulation life is affected not only by heating but also by mechanical stresses during bending. The method compares the nominally selected section with smaller standard sections by constructing annual temperature duration curves, calculating equivalent insulation temperature, mechanical stress, and electric field strength under overvoltage statistics, and then estimating insulation service life. An example for a 6 kV KShVGM excavator cable shows that, under the stated operating assumptions, a smaller conductor section can yield longer insulation life in a bent cable because higher temperature reduces mechanical stress.
Full Text
Methodology for Selecting the Cross-Section of Current-Carrying Conductors of High-Voltage Flexible Cables
V. S. Dmitrevsky
(Presented by the scientific seminar of the Department of EIKT)
The cross-section of the current-carrying conductors of cables is selected on the basis of the load current and the permissible current density [1]. This principle for selecting the cross-section of current-carrying conductors is based on the dependence of the service life of insulation on temperature. Usually, to estimate the service life of insulation at different temperatures, the Arrhenius equation [2] is used, which shows that as the temperature rises the lifetime of the insulation decreases exponentially. It follows from this rule that lowering the temperature of cable insulation increases its durability. Such a regularity is observed very clearly for cables of stationary installation.
In cables used to supply power to mobile consumers—for example, excavators—internal mechanical stresses in the insulation become of substantial importance. When the cable is bent, mechanical stresses arise in the insulation, the magnitude of which is greater the lower the temperature. With an increase in mechanical stresses, the durability of the insulation decreases.
Thus, an increase in temperature, on the one hand, causes accelerated failure of the insulation if there is no deformation of it, and, on the other hand, by reducing mechanical stresses, increases its durability.
Below a methodology is described for selecting the cross-section of current-carrying conductors of high-voltage flexible cables of the KShVGM type. From the tables of permissible currents, the cross-section of the current-carrying conductors of the cable is selected under the condition that the rated current of the installation is equal to the rated current of the cable. At the same time, the two nearest smaller standard cross-sections of the current-carrying conductors are selected. For example, if, according to the condition that the rated current of the installation is equal to the rated current of the cable, a cross-section of \(50\ \mathrm{mm}^2\) is adopted, then additionally, from the tables, two smaller cross-sections are selected—\(35\) and \(25\ \mathrm{mm}^2\).
Using meteorological data, an annual graph of ambient temperatures over the year is constructed. On the same graph, the curve of the current load of the installation over the year is plotted, using the methods set forth in [1, 3]. For each current and ambient temperature, the insulation temperature is determined [4]:
\[ T_{\mathrm{н}} = T_{0} + I^{2} n R_{\mathrm{ж}} S_{\mathrm{т}}, \tag{1} \]
where
\(T_o\) — ambient temperature;
\(I\) — load current;
\(n\) — number of current-carrying conductors;
\(R_{\mathrm{zh}}\) — resistance of 1 cm of length of a current-carrying conductor;
\(S_{\mathrm{t}}\) — thermal resistance of the cable elements from the conductor to the surrounding air, thermal ohm cm.
The thermal resistance is calculated as
\[ S_{\mathrm{t}}=\frac{0.159\,S_{\mathrm{tr}}G_1}{n}+\frac{3.2}{D_k B}. \tag{2} \]
In this expression
\(G_1\) — geometric factor;
\(S_{\mathrm{tr}}\) — specific thermal resistance of the rubber and braid, taken equal to 650 thermal ohm·mm;
\(D_k\) — cable diameter, mm;
\(B\) — coefficient of heat transfer from the cable to air; for \(D_k\) greater than 44.5 mm it is taken as
\[ \frac{1}{1200}\ \frac{1}{\text{thermal ohm}\cdot \text{cm}^2}, \]
and for \(D_k\) less than 44.5 mm it is calculated by the formula
\[ \frac{1}{12.6D_k+650}\cdot \frac{1}{\text{thermal ohm}\cdot \text{mm}^2}. \]
The geometric factor is selected from Fig. 1.
The calculations carried out make it possible to construct an annual duration curve of insulation temperatures for three selected cable cross sections. On the basis of the temperature duration curve, the equivalent temperature is calculated [4]:
\[ T_{\mathrm{e}}= \frac{D}{ 2k\ln\displaystyle\sum_{i=1}^{i=n}\frac{\tau_i}{\tau} \exp\left(-\frac{D}{2kT_i}\right) }. \tag{3} \]
Our experiments showed that, in a bent cable, the mechanical stress in the insulation depends on temperature according to the law
\[ \gamma\sigma=\gamma_0\sigma_0 e^{-b_1T}. \tag{4} \]
The equivalent value \(\gamma\sigma\) is found [4]:
\[ (\gamma\sigma)_{\mathrm{e}}\ln\frac{2De}{(\gamma\sigma)_{\mathrm{e}}} = D+2kT_{\mathrm{e}}\ln \sum_{i=1}^{i=n} q\,\frac{\tau_1}{\tau} \exp\left[ -\frac{ D-(\gamma\sigma)_i\ln\frac{2De}{(\gamma\sigma)_i} }{ 2kT_i } \right]. \tag{5} \]
The equivalent field strength is determined from the expression
\[ Ae^{-bT_{\mathrm{e}}}\beta fE_{\mathrm{e}} \ln\frac{2De}{Ae^{-bT_{\mathrm{e}}}\beta fE_{\mathrm{e}}} = D+ \]
\[ +2kT_{\mathrm{e}}\ln \sum_{i=1}^{i=n} \frac{\tau_i}{\tau} e^{-\frac{D}{2kT_i}} \sum_{j=1}^{j=m} \frac{\tau_j}{\tau_i} \exp\left[ \frac{ Ae^{-bT_i}\beta fE_j \ln\frac{2De}{Ae^{-bT_i}\beta fE_j} }{ 2kT_i } \right]. \tag{6} \]
To calculate the equivalent field strength it is necessary to find the field strength under overvoltages, which is computed by the known formula:
\[ E_j=\frac{U_j}{R-r}. \tag{7} \]
The integral distribution curve of overvoltage-amplitude probabilities is divided into intervals. \(U_j\) corresponds to the mean value
Fig. 1. Geometric factor \(G_1\) for cables: \(\delta_{\mathrm{iz}}\) — insulation thickness, \(\delta_{\mathrm{ob}}\) — sheath thickness, \(d_{\mathrm{zh}}\) — diameter of the current-carrying conductor.
Visible labels in the graph: \(G_1\) for a four-core cable, \(\delta_{\mathrm{ob}}/\delta_{\mathrm{iz}}=1.0\), \(0.5\), \(0.0\); \(G_1\) for a three-core cable, \(1.0\), \(0.8\), \(0.6\), \(0.4\), \(0.2\), \(0.0\); \(G_1\) for a two-core cable, \(1.0\), \(0.5\), \(0.0\). Bottom-axis label: ratio \((\delta_{\mathrm{iz}}+\delta_{\mathrm{ob}})/d_{\mathrm{zh}}\). Right-side axis labels: geometric factor of a four-core cable; geometric factor of a three-core cable; geometric factor of a two-core cable.
of the \(j\)-th interval. \(U_1\) is equal to the operating voltage. The duration of the voltage \(U_j\) is found as
\[ \tau_j=\tau_n\frac{N}{\tau}\tau_1\left[\Phi(U_{j n})-\Phi(U_{j l})\right]. \tag{8} \]
In these expressions the following notation is used:
$T_{\mathrm{e}}$ — equivalent insulation temperature;
$k$ — Boltzmann constant;
$D$ — dissociation energy of the C—C bond in rubber;
$T_i$ — temperature of the $i$-th interval of the graph of temperatures versus duration;
$\tau_i$ — duration of action of $T_i$;
$\tau$ — period of variation of the insulation temperatures, taken as $3.16 \cdot 10^7$ sec;
$\sigma$ — internal mechanical stress in the insulation of a bent cable;
$\gamma$ — structure-sensitive coefficient;
$\gamma_0$, $\sigma_0$ — taken at the temperature of absolute zero;
$q$ — probability of operation of the cable in the bent state;
$b_1$ — constant coefficient;
$(\gamma\sigma)_{\mathrm{e}}$ — equivalent value in the insulation of a bent cable;
$(\gamma\sigma)_i$ — value of $\gamma\sigma$ at temperature $T_i$;
$n$ — number of temperature intervals;
$A$, $b$ — constants dependent on the structure of the polymer or rubber;
$\beta$ — coefficient of concentration of the local field intensity in the dielectric, depending on the characteristics of nonuniformity and the dimensions of the insulation;
$f$ — coefficient of nonuniformity for cable insulation, equal to
\[ \frac{R-r}{r \ln \dfrac{R}{r}}; \]
$R$ — outer radius of the insulation;
$r$ — inner radius of the insulation;
$m$ — number of intervals into which the integral curve of the distribution of overvoltage amplitudes is divided;
$N$ — number of overvoltages per year;
$\Phi(U_{j\mathrm{r}})$, $\Phi(U_{j\mathrm{l}})$ — probabilities of occurrence of overvoltage amplitudes corresponding to the right and left boundaries of the $j$-th interval;
$\tau_{\mathrm{p}}$ — average wave length of one overvoltage. From the equivalent values $T_{\mathrm{e}}$, $(\gamma\sigma)_{\mathrm{e}}$, and $E_{\mathrm{e}}$, the average service life of the insulation is determined for each variant of the current-carrying conductor cross-section by the formula
\[ \tau = \frac{1}{\tau_0} \exp \left\{ \frac{ D - \sqrt{(Ae^{-b/T_{\mathrm{e}}}\beta f E_{\mathrm{e}})^2 + (\gamma\sigma)_{\mathrm{e}}^2} \ln \frac{ 2De }{ \sqrt{(Ae^{-b/T_{\mathrm{e}}}\beta f E_{\mathrm{e}})^2 + (\gamma\sigma)_{\mathrm{e}}^2} } }{ 2kT_{\mathrm{e}} } \right\}, \tag{9} \]
where $\tau_0$ — bond relaxation time, taken as $10^{-13}$ sec.
From the specified variants of the current-carrying conductor cross-section, the one is selected for which the service life of the insulation is maximal.
The proposed methodology for selecting the cross-section of current-carrying conductors makes it possible to take sufficiently fully into account the operating conditions of the cable. Correct selection of the cross-section of current-carrying conductors will make it possible to substantially increase the service life of the cable (see Appendix).
APPENDIX
Example of selecting the cross-section of the current-carrying conductor of a KShVGM cable—
6 kV
Initial data: 1) nominal current of the excavator, 102 A; 2) variation of the current over the working cycle is shown in Fig. 2; 3) the excavator operates
according to a three-shift schedule during the year, 4) the annual temperature graph is shown in Fig. 3, 5) the probability of occurrence of overvoltage amplitudes obeys a normal law with the parameters: mathematical expectation
Fig. 2. Load-current graph of the excavator over the working cycle
Fig. 3. Annual temperature graph: ————— ambient medium; – – – – – – cable insulation for a current-carrying conductor cross-section of 25 mm²
expectation \(U = 4.25\,U_{\phi}\) and standard deviation \(s = 1.08\,U_{\phi}\), 6) the number of overvoltages per year \(N = 2000\), 7) the average length of the overvoltage wave is \(0.004\) sec, 8) the cable is constantly operated in a bent state \((q = 1)\), 9) the insulation characteristics determined experimentally are given in Table P-1.
Table P-1
| \(D\), \(\dfrac{\text{J}}{\text{s}\cdot\text{V}}\) | \(A\), \(\dfrac{\text{J}\cdot\text{V}}{\text{s}\cdot\text{V}}\) | \(\varepsilon\), \(\dfrac{1}{^\circ\mathrm{K}}\) | \(\gamma_0\sigma_0\), \(\dfrac{\text{J}}{\text{s}\cdot\text{V}}\) | \(b_1\), \(\dfrac{1}{^\circ\mathrm{K}}\) | \(\beta\) |
|---|---|---|---|---|---|
| \(5.51\cdot10^{-19}\) | \(0.4\cdot10^{-26}\) | \(0.00351\) | \(5.28\cdot10^{-18}\) | \(0.0172\) | \(3.5\) |
According to the nominal current of the excavator, we select a current-carrying conductor cross-section of \(25\ \mathrm{mm}^2\), and additionally \(16\) and \(10\ \mathrm{mm}^2\). From the graph of the current variation over the operating cycle of the excavator, we find the equivalent current causing heating of the cable,
\[ I_{\mathrm{e}}=\sqrt{\frac{1}{\tau_{\mathrm{p}}}\sum I_k^2\Delta\tau_k}, \tag{P-1} \]
where
\[ \begin{aligned} \tau_{\mathrm{p}}&\text{ is the duration of the operating cycle;}\\ I_k&\text{ is the current in the }k\text{th interval of the operating cycle;}\\ \Delta\tau_k&\text{ is the time of its action.} \end{aligned} \]
Formula (P-1) is sufficiently accurate when the heating time constant of the cable is much greater than \(\tau_{\mathrm{p}}\).
From the graph in Fig. 1 and formula (P-1), \(T_{\mathrm{e}}=0.57T_{\mathrm{nom}}\) was obtained. Since the excavator operates continuously throughout the entire year, it may be approximately assumed that \(T_{\mathrm{e}}\) is constant.
Substituting \(I_{\mathrm{e}}\) into formula (1), we calculate the insulation temperature of the cable at different moments of time. In Fig. 3 the insulation-temperature curve is plotted for a cable with a cross-section of \(25\ \mathrm{mm}^2\). In the same way, we calculate the insulation temperature for current-carrying conductor cross-sections of \(16\) and \(10\ \mathrm{mm}^2\). For the calculation, we divide the insulation-temperature graph into intervals such that in each interval the deviation of the temperature from the mean value does not exceed \(\pm 5^\circ\). Table P-2 gives the values of the mean temperatures of the intervals, with the relative duration of the interval indicated.
Table P-2
| Conductor cross-section, \(\mathrm{mm}^2\) | \(\tau_i/\tau\): 0.1667 | \(\tau_i/\tau\): 0.25 | \(\tau_i/\tau\): 0.1667 | \(\tau_i/\tau\): 1.1667 | \(\tau_i/\tau\): 0.25 |
|---|---|---|---|---|---|
| 25 | 268 | 274 | 285 | 298 | 305 |
| 16 | 278 | 284 | 295 | 308 | 315 |
| 10 | 296 | 302 | 313 | 326 | 333 |
Using the temperature values from Table P-2, the insulation characteristics from Table P-1, and formulas [3, 4, 5], we find the equivalent quantities \(T_{\mathrm{e}}\) and \((\gamma\sigma)_{\mathrm{e}}\), which are given in Table P-4.
We divide the integral probability distribution function into equal intervals, taking their width equal to \(2\ \mathrm{kV}\). By formula (7), for the mean value \(U_i\) we calculate \(E_i\), taking into account that \(R-r=4\ \mathrm{mm}\) for all cables. By formula (8), we find the ratio \(\dfrac{\tau_j}{\tau_i}\), which remains unchanged for any temperature. The probability values
Table P-3
| 1 | 9,5·10^-10 | 2,47·10^-9 | 5,7·10^-9 | 1,15·10^-8 | 2,01·10^-8 | 3,04·10^-8 | |
|---|---|---|---|---|---|---|---|
| $\dfrac{\tau_j}{\tau_i}$ | 1 | 9,5·10^-10 | 2,47·10^-9 | 5,7·10^-9 | 1,15·10^-8 | 2,01·10^-8 | 3,04·10^-8 |
| $E_j \cdot 10^6\ \mathrm{V/m}$ | 1,225 | 1,5 | 2 | 2,5 | 3 | 3,5 | 4 |
| $\dfrac{\tau_j}{\tau_i}$ | 3,98·10^-8 | 4,52·10^-8 | 4,43·10^-8 | 3,84·10^-8 | 2,76·10^-8 | 1,83·10^-8 | 1,03·10^-8 |
| $E_j \cdot 10^6\ \mathrm{V/m}$ | 4,5 | 5 | 5,5 | 6 | 6,5 | 7 | 7,5 |
| $\dfrac{\tau_j}{\tau_i}$ | 4,9·10^-9 | 2,18·10^-9 | 6,96·10^-10 | 2,74·10^-10 | |||
| $E_j \cdot 10^6\ \mathrm{V/m}$ | 8 | 8,5 | 9 | 9,5 |
are taken from tables of the Laplace integral [5]. Table P-3 gives \(-\dfrac{\tau_j}{t_i}\) and \(E_j\). The data of Tables P-2 and P-3 make it possible to calculate \(E_{\mathrm{e}}\) from (6) for different cross sections of the cable’s current-carrying conductors. Table P-4 gives the values of \(E\).
Table P-4
| Cross section of the current-carrying conductor, mm² | \(T_{\mathrm{e}}\), °K | \((\gamma^\circ)_{\mathrm{e}}\), \(\dfrac{\Omega}{\text{cm}}\) | \(E_{\mathrm{e}}\), \(\dfrac{\text{V}}{\text{m}}\) | Insulation service life, years: bent cable | Insulation service life, years: straight cable |
|---|---|---|---|---|---|
| 25 | 300 | \(3.49 \cdot 10^{-20}\) | \(2.30 \cdot 10^{6}\) | 0.4 | 8900 |
| 16 | 309 | \(2.92 \cdot 10^{-20}\) | \(2.00 \cdot 10^{6}\) | 1 | 4000 |
| 10 | 326 | \(2.05 \cdot 10^{-20}\) | \(1.60 \cdot 10^{6}\) | 4 | 1000 |
We choose the most advantageous variant, giving the greatest longevity; that is, the cross section of the current-carrying conductor should be \(10\ \text{mm}^2\).
REFERENCES
-
L. N. Baptidanov, V. I. Tarasov. Electrical Equipment of Electric Power Stations and Substations. Vol. I, Gosenergoizdat, 1960.
-
P. M. Glupushkin, A. E. Saakyan, D. P. Shcherbakov. Cable Rubbers. Energiya, 1966.
-
V. S. Livshits. Fundamentals of the theory of summation of electrical loads of heterogeneous industrial electrical receivers. Izvestiya Vuzov, “Elektromekhanika,” No. 8, 1970, pp. 900–904.
-
V. S. Dmitrevskii. Calculation of the insulation of high-voltage cables. Kabel’naya tekhnika, in press.
-
E. S. Venttsel. Probability Theory. Nauka, 1964.