Current conductors cross-section

Type of protection EEx e II Certificate KEMA Ex-95.D.4411U Rated voltage 726 V Rated current 350 A Conductor cross-section (flexible) 70-185 mm2 Insulating material CTI 600. The lines in dark grey indicate the creepage distances. 

Figure 6.40(a)-(d) Overtemperatures at branching points and cable entries versus thermal losses. Low voltage switchgear EEx de I. The thermal losses indicate the heat dissipation of fuses, magnetic coils and transformers in the d main part. Two cables (I, II) with 3 AC, 50cps, conductor cross-section 6 mm2, one cable with 3 AC, 50 cps, conductor cross-section 70 mm2, at different current densities a. 

Type of protection EEx e II Certificate PTB 98 ATEX 3132 U Rated voltage 550 V Rated current 68 A Conductor cross-section 0.5... 16 mm2 Insulating material ... 

Figure 6.52 Flameproof conductor bushings (connection between e and d ). Marking Ex II 2G, EEx de II, IM2 EEx de I Type of protection EEx de l/EEx de II C Certificate PTB 98 ATEX 1067 U Rated voltage 6.6 kV and 11 kV Rated current 400 A Conductor cross-sections up to 300 mm2 Conductor diameter 16 mm Thread M 80 x 1.5 and M 110 x 1.5 Insulating material cast resin.

Type of protection EEx dl Certificate INERIS 95.D.7027X Rated voltage 1100 V 3 AC Rated current 450 A Cable conductor cross-section up to 185 mm2 Auxiliary contacts 4 mm2 Cable conductor cross-section 6 mm2 (Top to bottom) two plugs for different cable diameters socket with mounting flange longitudinal section of a plug. 

Type of protection EEx dl Certificates - Plug MECS 93 C 5502 U Socket MECS 93 C 5503 U Plug coupler MECS 93 C 5504 Rated voltage 1100V 3 AC Rated current 300 A Cable conductor cross-section up to 120 mm2. 

For currents exceeding approx. 500 mA it shall be taken into account that inadmissible high temperatures may occur for inadequate small conductor cross-sections (cable near to breakoff ) or for poor contact making (terminal tightened insufficiently). This effect is well known as ignition by incandescence. 

Figure 3.2 shows the net flow of electrons in a conductor cross-sectional area A in the presence of an applied field E. Notice that the direction of electron motion is opposite to that of the electric field and of conventional current, because the electrons experience a coulombic force qE, in the X direction, due to their negative charge. We know that the conduction electrons are actually moving around randomly in the metal, but we will assume that, as a result of the application of the electric field E, they all acquire a net velocity in the x direction. Otherwise, there would be no net flow of charge through area A. 

The conductor can thus be represented as an anisotropic medium with interrelated nonlinear magnetic and electrical properties. The current pattern that results in the conductor from the application of time-dependent external magnetic fields and excitation (transport) currents can be determined in principle, from Maxwell s equations. An integration of pJ and the /-dependent magnetic hysteresis over the entire conductor cross section can then be used to calculate the loss. 

When a circuit is operating at high frequencies, the skin effect causes the current to be redistributed over the conductor cross-section in such a way as to make most of the current flow where it is encircled by the smallest number of flux lines. This general principle controls the distribution of current, regardless of the shape of the conductor involved. With a flat-strip conductor, the current flows primarily along the edges, where it is surrounded by the smallest amount of flux. 

Current vs. Cross Section for 10°C. Temp. Rise of Etched Copper Conductors... 

Electric current ampere A Magnitude of the current that, when flowing through each of two straight parallel conductors of infinite length, of negligible cross-section, separated by 1 meter in a vacuum, results in a force between the two wires of 2 X 10 newton per meter of length. 

Ampere. The ampere is that constant current which, if maintained in two straight, parallel conductors of infinite length, of negligible circular cross section, and placed one meter apart in a vacuum, would produce between these conductors a force equal to 2 x 10 newton per meter of length. 

In smaller cross-sectional area,s of the current-carrying conductors of the distribution network, i.e. for low-capacity networks where R/Xl is high, series compensation may be redundant. 

In a d.c. system the current distribution through the cross-section of a current-canying conductor is uniform as it consists of only the resistance. In an a.c. system the inductive effect caused by the induced-electric field causes skin and proximity effects. These effects play a complex role in determining the current distribution through the cross-section of a conductor. In an a.c. system, the inductance of a conductor varies with the depth of the conductor due to the skin effect. This inductance is further affected by the presence of another current-carrying conductor in the vicinity (the proximity effect). Thus, the impedance and the current distribution (density) through the cross-section of the conductor vaiy. Both these factors on an a.c. system tend to increase the effective... 

We can derive the same inference from Tables 30.2, 30.4 and 30.5, specifying current ratings for different cross-sections. The current-carrying capacity varies with the cross-section not in a linear but in an inconsistent way depending upon the cross-section and the number of conductors used in parallel. It is not possible to define accurately the current rating of a conductor through a mathematical expression. This can be established only by laboratory tests. 

As discussed later, the enclosure of an IPB may carry induced currents up to 95% of the current through the main conductors. Accordingly, the enclosure is designed to carry longitudinal parasitic currents up to 90-95% of the rated current of the main busbars. The cross-sectional area of the enclosure is therefore maintained almost equal to and even more than the main conductors to account for the dissipation of heat of the main conductors through the enclosure only, unless an additional forced cooling system is also adopted. The outdoors part of the enclosure exposed to atmospheric conditions is also subjected to solar radiation. Provision must be made to dissipate this additional heat, from the enclosure. 

The series resistance of a transmission line is closely related to the losses that will be dissipated when current passes through the line (proportional to the square of the current magnitude). The resistance is proportional to the length of the line but inversely proportional to the cross-sectional area of the conductor. 

Conductivity is a very important parameter for any conductor. It is intimately related to other physical properties of the conductor, such as thermal conductivity (in the case of metals) and viscosity (in the case of liquid solutions). The strength of the electric current I in conductors is measured in amperes, and depends on the conductor, on the electrostatic field strengtfi E in tfie conductor, and on the conductor s cross section S perpendicular to the direction of current flow. As a convenient parameter that is independent of conductor dimensions, the current density is used, which is the fraction of current associated with the unit area of the conductor s cross section i = I/S (units A/cnF). 

In the above relationship p is an intrinsic property called the specific resistance (or resistivity) of the conductor. The definition of the specific resistance of any given conductor follows from this relationship. It is the resistance in ohms of a specimen of the material, 1 cm long and 1 cm2 in cross-sectional area (units ohm cm-1), the length being in the direction of the current and the cross-section normal to it. In other words, the specific resistance p of a conductor is the resistance of a cube of 1 centimeter edge. If the conductance is denoted by C = 1 /R, then the specific conductance (or conductivity) K, is given by JC= 1/a (units ohm-1 cm-1, mho cm-1, reciprocal ohm cm-1). Therefore, the relationship R = aL/A may be written as R = L/KA (units ohms) and the conductance can be expressed as C = 1/R = KA/l (units reciprocal ohms). 

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