507 6. Determining conductor crosssectional areas Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 508 6. DETERMINING CONDUCTOR CROSS-SECTIONAL AREAS Owing to the respective characteristics of LV and MV conductors, they have been dealt with in separate paragraphs. 6.1. Determining conductor cross-sectional areas and choosing protective devices in low voltage n definition of terms relating to low voltage wiring systems (Insulated) cable Assembly comprising: - one or more insulated conductors - their eventual individual screening - any eventual assembly protection - any eventual protective shielding It may also comprise one or several bare conductors. Multi-core cable Cable comprising more than one conductor, which may eventually include bare conductors. Note: the term three-core cable is used to designate the cable making up the phases of a three-phase system. Single-core cable Cable comprising a single insulated conductor. Note: the term single-core cable is especially used to designate a cable making up one of the phases of a three-phase system. Wiring system Assembly made up of one or more electric conductors and the devices ensuring their fixation and, if necessary, their mechanical protection. Cable channel Ventilated or enclosed duct located above or in the ground, having dimensions preventing persons from moving around inside it but allowing access to the cables over their entire length during and after installation. Note: a cable channel may or may not form part of the building construction. Cable tray Holder made up of a base and sides but no cover. Note: A cable tray may be perforated or unperforated. Electrical circuit (of an installation) All the electrical equipment of the installation fed from the same source and protected against overcurrents by the same protective device(s). Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 509 (Insulated) conductor Assembly comprising the conductor, its insulating envelope and eventual screens. (Circular) conduit Enclosed envelope, having a circular cross-section, designed for the installation or the replacement of insulated conductors or cables by capstan, in electrical installations. Ducting Assembly of closed envelopes having a non circular cross-sectional area, designed for the installation or the replacement of insulated conductors or cables by capstan, in electrical installations. Brackets Horizontal cable supports fixed at one of their ends, arranged from point to point and on which the cables rest. Design current of a circuit Current to be carried in a circuit in normal service (Continuous) current carrying capacity of a conductor Maximum value of the current that, in given conditions, can continuously flow in a conductor without its steady-state operating temperature being higher than the specified value. Cable ladder Cable support made up of a series of non-touching elements firmly fixed to main vertical rods. Sleeve (or tube) Element surrounding wiring and providing it with extra protection in building passages (walls, partitions, floor, ceiling) or in buried passages. Sheath Enclosure located above ground level having dimensions preventing persons from moving around inside it but allowing access to the cables over their entire length. A sheath may or may not be built into the masonry. Trough Assembly of envelopes closed by a cover and ensuring mechanical protection of insulated conductors or cables not installed or removed by a capstan and which allow other electrical equipment to be added . Building void Space in a structure or building parts which is only accessible at certain places. Note: - spaces in walls, supported floors, ceilings and certain types of window or door frames and jamb linings are examples of building voids. - specially built building voids are also called "ducts". Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 510 6.1.1. Method principle In compliance with the recommendations of IEC 364-4-43, the cross-sectional area of wiring systems and the protective device must be chosen to meet several conditions necessary for the security of the installation. The wiring system must: - carry the maximum design current and its normal transient peaks - not generate voltage drops above the allowed values. The protective device must: - protect the wiring system against any overcurrents up to the short-circuit current - ensure the protection of persons against indirect contact. The logigram in figure 6-1 sums up the principle of the method which may be described by the following stages: 1st stage: - using the load power, the maximum design current I B is calculated and the rated current I n of the protective device is deduced from this - the maximum short-circuit current Isc at the origin of the circuit is calculated and the breaking capacity of the protective device is deduced from this. 2nd stage: - depending on the installation conditions (installation method, ambient temperature, etc.), the overall correction factor f is determined - the suitable conductor cross-sectional area is chosen in relation to I n and f . Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 511 3rd stage: - the maximum voltage drop is checked - the thermal withstand of the conductors in the event of a short circuit is checked - for TN and IT systems, the maximum length relating to the protection of persons against indirect contact is checked. The conductor cross-sectional area meeting all these conditions is then chosen. Note: an economic cross-sectional area larger than the cross-sectional area determined above may be chosen if necessary (see § 6.3). apparent power to be carried upstream or downstream network short-circuit power at the origin of the circuit short-circuit current design current I sc IB rated current of protective device protective device breaking capacity In choice of protective device choice of protective device installation conditions wiring system conductor cross-sectional area check of thermal withstand in case of short-circuit IT or TN earthing system maximum voltage drop check maximum wiring system length chek TT earthing system conductor crosssectional area determination confirmation of the choice of wiring system cross-sectional area and its electrical protection economic cross-sectional area possibly chosen Figure 6-1: wiring system cross-sectional area and protective device choice logigram Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 512 6.1.2. Determining the maximum design current The maximum design current ( I B ) is defined according to the type of installation fed by the wiring system. In the case of individual power supply to a device, the current I B will be equal to the rated current of the device being fed. On the other hand, if the wiring system feeds several devices, the current I B will be equal to the sum of currents absorbed, taking into account the installation utilisation and coincidence factors. In the case of motor starting or cyclical operating conditions of loads (spot welding station, see § 3.4.2), current inrushes must be taken into account when their thermal effects are cumulated. Some installations are subject to future extensions. The current corresponding to this extension will be added to the existing value. In direct current: I= In alternating current: I = S U P power consumed (in W ) U duty voltage (in V ) S S in three-phase. in single-phase and I = U U 3 : apparent power consumed (VA) : . voltage between the two conductors for a single-phase power supply . phase-to-phase voltage for a three-phase power supply When high harmonic currents circulate in the conductor, they must be taken into account. In order to choose the cross-sectional area, the following must therefore be taken: ∞ I r .m.s. = I p2 p =1 1 ∑ (see § 8) I1 : current value at 50 Hz (or 60 Hz) I p : value of harmonic current of order p For example, for a speed variator Ir .m.s. ≅ 1.7 I1 When there are compensation capacitors downstream of the wiring system, the design current is determined as follows: - assuming that compensation is in operation: in case of failure of the capacitors, the wiring system is placed out of service - assuming that compensation is out of service; in case of failure of the capacitors, the conductor cross-sectional area is sufficient and availability is thus improved. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 513 n factor taking into account the power factor and efficiency: a The apparent power of a load is: S= P η × Fp in kVA P : active power in kW η : efficiency Fp : power factor We define the coefficient: a = 1 η × Fp When a current stripped of harmonics flows through the conductor, Fp = cos ϕ . n load utilisation factor: b In an industrial installation, it is assumed that loads will never be used at their full power level. A utilisation factor ( b ) is therefore introduced which generally varies from 0.3 to 1. Without knowing the accurate values, we may take: - b = 0.75 for motors - b =1 for lighting and heating Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 514 c n coincidence factor: In an industrial installation, the loads (of a workshop, for example) fed by the same wiring system do not operate simultaneously in all cases. To take this phenomenon, which is linked to the operating conditions of the installation, into account, the coincidence factor is applied to the sum of the load powers in conductor sizing. In the absence of precise indications resulting from experience of standard installations, the values of tables 6-1 et 6-2 may be applied: Use Coincidence factor c Lighting 1 Lighting and air conditioning 1 Power outlets 0.1 to 0.2 (for a number > 20) Table 6-1: coincidence factor for an administrative building Number of circuits having similar nominal currents Coincidence factor 2 and 3 0.9 4 and 5 0.8 5 to 9 0.7 10 and more 0.6 Table 6-2: coincidence factor for industrial distribution switchboards n factor taking into account possible future extensions: d The value of factor installation. d must be estimated according to the foreseeable extensions of the In the absence of precise indications, the value 1.2 is often used. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 515 n power conversion factor in current: e The power conversion factor in current is: - e = 8 in single-phase 127 V e = 2.5 in three-phase 230 V - e = 4.35 in single-phase 230 V e = 1.4 in three-phase 400 V n maximum design current The maximum design current is thus: IB = P × a × b × c × d × e P : active power in kW 6.1.3. Choosing the protective device Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 516 n general rule In compliance with IEC 364, a protective device (circuit-breaker or fuse) correctly fulfils its function if the conditions outlined below are met. o nominal or setting current This must be between the design current and the current carrying capacity I a of the wiring system: I B ≤ I n ≤ I a , which corresponds to zone a in figure 6.2. o conventional tripping current This must meet the following relation: I2 ≤ 1.45 I a , which corresponds to zone b in figure 6.2. case of circuit-breakers - For domestic circuit-breakers, standard IEC 898 specifies: I2 = 1.45 In - For industrial circuit-breakers, standard IEC 947-2 specifies: I2 = 1.30 I set we thus have I2 ≤ 1.45 I n (or I set ) while I n ≤ I a (above condition) The condition I2 ≤ 1.45 I a (zone b ) is thus automatically met. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 517 case of fuses Standard IEC 269-1 specifies that I 2 is the current which ensures that the fuse fuses in the conventional time (1 h or 2 h); I 2 is referred to as the conventional fusing current (see § 6.3.1 of the Protection guide). I 2 = k2 × I n where k2 = 1.6 to 1.9 depending on the fuses Let us define the coefficient k3 such that: k3 = k2 1.45 Thus, the condition I2 ≤ 1.45 I a is met if: I In ≤ a k3 For gG fuses: - I n ≤ 10 A 10 A < I n ≤ 25 A I n > 25 A à à à k3 = 1.31 k3 = 1.21 k3 = 1.10 o breaking capacity This must be higher than the three-phase maximum short-circuit current ( Isc 3 ) at its installation point: Breaking capacity ≥ Isc 3 , which corresponds to zone c in figure 6.2. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 518 o associating protective devices The use of a protective device having a breaking capacity below the short-circuit current at the point where it is installed is permitted by standard IEC 364 under the following conditions: - there is another device upstream having at least the necessary breaking capacity - the energy that the device placed upstream lets through is lower than the energy that the downstream device and wiring systems protected by these devices can withstand without being damaged. This possibility is implemented: . in circuit-breaker/fuse associations . in the cascading technique which uses the high current limitation capacity of certain circuit-breakers (e.g. the Compact). The possible associations resulting from actual tests performed in a laboratory are given in manufacturer catalogues. 6.1.4. Current-carrying capacity of wiring systems This is the maximum current that the wiring system can continuously carry without this being prejudicial to its lifetime. To determine this current, it is necessary to carry out the following: - using tables 6-3 to 6-5, define the installation method, its associated selection number and letter, and correction factors to be applied - using the installation conditions, the correction factor values which must be applied are determined (see tables 6-6 to 6-15) - calculate the overall correction factor f equal to the product of the correction factors - using table 6-16 for selection letters B, C, E, F and table 6-17 for selection letter D, the maximum current I 0 that the wiring system can carry under standard conditions ( f0 to f10 = 1 ) is determined - calculate the maximum current that the wiring system can carry in relation to its installation conditions: I a = f I 0 . Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 519 n installation methods Tables 6-3 to 6-5 give the main installation methods used in industrial networks. For each installation method, the following is given: - its associated selection number and letter - the correction factors to be applied. Factor f 0 corresponds to the installation method; factors f1 to f10 are explained below (see tables 6-6 to 6-15). Example Description N° Selection Correction factors letter f0 to be applied 11 C 1 f1 f4 f5 11A C 0.95 f1 f4 f5 12 C 1 f1 f4 f5 Single or multi-core cables with or without armour - fixed on a wall - fixed to a ceiling - on unperforated trays cables multi-core singlecore - on perforated trays run horizontally or vertically 13 E F 1 f1 f4 f5 - on brackets 14 E F 1 f1 f4 f5 - on ladders 16 E F 1 f1 f4 f5 Table 6-3: installation methods for selection letters C, E and F Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 520 Example Description Single or multi-core cables in building voids Single or multi-core cables in conduits in building voids N° Selection Correction factors letter f0 21 B 0.95 f1 f4 f5 -- 22A B 0.865 f1 f4 f5 f6 23A B 0.865 f1 f4 f5 f6 24A B 0.865 f1 f4 f5 f7 25 B 0.95 f1 f4 f5 -- 31A B 0.9 f1 f4 f5 -- 32A B to be applied Single or multi-core cables in ducting in building voids Single or multi-core cables in ducting built into the masonry Single or multi-core conductors : - in false ceilings - in suspended ceilings Single or multi-core cables in troughs fixed to walls: - run horizontally - run vertically 0.9 f1 f4 f5 -- Table 6-4: installation methods for selection letter B Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 521 Example Description N° Selection Correction factors letter f0 to be applied Single or multi-core cables in troughs built into floors 33A B 0.9 f1 f4 f5 -- Single or multi-core cables in suspended troughs 34A B 0.9 f1 f4 f5 -- Multi-core cables in enclosed channels run horizontally or vertically 41 B 0.95 f1 f4 f5 -- Single or multi-core cables in open or ventilated channels 43 B 1 f1 f4 f5 -- Table 6-4 (cont.): installation methods for selection letter B Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 522 Example Description N° Selection Correction factors letter f0 to be applied Single or multi-core cables in conduits or in buried ducting 61 D 0.8 f2 f3 f8 f9 Single or multi-core cables buried without any extra mechanical protection 62 D 1 f2 f3 f10 -- Single or multi-core cables buried with extra mechanical protection 63 D 1 f2 f3 f10 -- Table 6-5: installation methods for selection letter D Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 523 n correction factors for ambient temperatures other than 30 °C (wiring systems above ground): f1 When electrical wiring systems are built into walls having heating elements, it is generally necessary to reduce current-carrying capacities by applying the reduction factors in table 6-6. This supposes that the distribution of temperatures inside the heated walls in contact with the electrical wiring system is known. When the air temperature is other than 30 °C, the correction coefficient to be applied is given in the formula: f1 = θ p −θ0 θ p − 30o θ p : maximum temperature permitted by the insulating material under steady-state conditions, °C θ 0 : air temperature, °C The value of f1 is given in table 6-6 for different values of θ p and θ 0 . Insulation Ambient temperatures (°C) PVC XLPE and EPR θ0 Elastomers (rubber) θ p = 60 °C θ p = 70 °C θ p = 90 °C 10 15 20 25 35 40 45 50 55 60 65 70 75 80 85 90 95 1.29 1.22 1.15 1.07 0.93 0.82 0.71 0.58 - 1.22 1.17 1.12 1.06 0.94 0.87 0.79 0.71 0.61 0.50 - 1.15 1.12 1.08 1.04 0.96 0.91 0.87 0.82 0.76 0.71 0.65 0.58 0.50 0.41 - Table 6-6: correction factors for ambient temperatures other than 30 °C (above ground wiring systems) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 524 n correction factors for ground temperatures other than 20 °C (buried wiring systems): f 2 When the ground temperature is other than 20°C, the correction coefficient to be applied is given in the formula: f2 = θ p −θ0 θ p − 20 θ p : maximum temperature permitted by the insulating material under steady-state conditions, °C θ 0 : ground temperature, °C The value of f 2 is given in table 6-7 for different values of θ p and θ 0 . Ground temperature θ 0 (°C) Insulation θ p = 70 °C PVC XLPE and EPR θ p = 90 °C 10 1.10 1.07 15 1.05 1.04 25 0.95 0.96 30 0.89 0.93 35 0.84 0.89 40 0.77 0.85 45 0.71 0.80 50 0.63 0.76 55 0.55 0.71 60 0.45 0.65 65 - 0.60 70 - 0.53 75 - 0.46 80 - 0.38 Table 6-7: correction factor for ground temperatures other than 20 °C (buried wiring systems) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 525 n correction factors for buried wiring systems, in relation to the soil thermal resistivity: f 3 The soil thermal resistivity depends on the type and humidity of the ground. The correction factor to be applied according to the soil resistivity is given in table 6-8. Soil thermal resistivity K.m/W Correction factor 0.40 1.25 underwater installation marshes 0.50 1.21 very moist soil sand 0.70 1.13 moist soil clay 0.85 1.05 normal soil and 1.00 1.00 dry soil chalk 1.20 1.50 0.94 0.86 very dry soil 2.00 0.76 and 2.50 0.70 clinker 3.00 0.65 Humidity Observations Type of soil ash Table 6-8: correction factors for buried wiring systems in relation to the soil thermal resistivity n correction factors for a group of several multi-core cables or groups of single-core cables The circuits or cables may be: - touching; the correction factor f 4 must be applied - arranged in several layers; the correction factor f 5 must be applied - both touching and arranged in several layers (see fig. 6-3); correction factors must then be applied. f 4 and f5 Figure 6-3: 6 multi-core cables - 2 layers of 3 touching cables Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 526 o touching multi-core or groups of single-core cables: f 4 The factors in table 6-9 are to be applied to homogenous groups of cables, equally loaded, for the given installation methods. When the horizontal distance between neighbouring cables is greater than twice their external diameter, no reduction factor is necessary. The same correction factors are applicable: - to groups of two or three single-core cables - to multi-core cables. N° of installation methods Number of touching multi-core cables or groups of single-core cables 1 2 3 4 5 6 7 8 9 12 16 20 21, 22A, 23A, 24A, 25, 31, 31A, 32, 32A, 33A, 34A, 41, 43 1.00 0.80 0.70 0.65 0.60 0.55 0.55 0.50 0.50 0.45 0.40 0.40 11, 12 1.00 0.85 0.79 0.75 0.73 0.72 0.72 0.71 0.70 No extra 11A 1.00 0.85 0.76 0.72 0.69 0.67 0.66 0.65 0.64 reduction 13 1.00 0.88 0.82 0.77 0.75 0.73 0.73 0.72 0.72 factor for 14, 16 1.00 0.88 0.82 0.80 0.80 0.79 0.79 0.78 0.78 more than 9 cables Table 6-9: correction factors for touching multi-core cables or groups of single-core cables o multi-core cables or groups of single-core cables arranged in several layers: f 5 When cables are arranged in several layers, the correction factors in table 6-10 must be applied. Number of layers Correction factors f 5 2 3 4 or 5 6 to 8 9 plus 0.80 0.73 0.70 0.68 0.66 table 6-10: correction factors for a group of multi-core cables or groups of single-core cables arranged in several layers Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 527 n correction factors in relation to the number of conduits in air and their arrangement (see table 6-11): f 6 Number of Number of conduits arranged horizontally conduits arranged vertically 1 2 3 4 5 6 1 1 0.94 0.91 0.88 0.87 0.86 2 0.92 0.87 0.84 0.81 0.80 0.79 3 0.85 0.81 0.78 0.76 0.75 0.74 4 0.82 0.78 0.74 0.73 0.72 0.72 5 0.80 0.76 0.72 0.71 0.70 0.70 6 0.79 0.75 0.71 0.70 0.69 0.68 Table 6-11: correction factors in relation to the number of conduits in the air and their arrangement n correction factors in relation to the number of conduits buried or built into concrete and their arrangement (see table 6-12): f 7 Number of conduits arranged vertically Number of conduits arranged horizontally 1 2 3 4 5 6 1 1 0.87 0.77 0.72 0.68 0.65 2 0.87 0.71 0.62 0.57 0.53 0.50 3 0.77 0.62 0.53 0.48 0.45 0.42 4 0.72 0.57 0.48 0.44 0.40 0.38 5 0.68 0.53 0.45 0.40 0.37 0.35 6 0.65 0.50 0.42 0.38 0.35 0.32 Table 6-12: correction factors in relation to the number of conduits buried or built into concrete and their arrangement Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 528 n correction factors for non-touching buried conduits run horizontally or vertically on the basis of one cable or group of 3 single-core cables per conduit (see table 6-13) : f 8 Distance between conduits (a) Number of conduits 0.25 m 0.5 m 1.0 m 2 0.93 0.95 0.97 3 0.87 0.1 0.95 4 0.84 0.9 0.94 5 0.81 0.7 0.93 6 0.79 0.6 0.93 Table 6-13: correction factors for non-touching buried conduits run horizontally or vertically on the basis of one cable or group of 3 single-core cables per conduit The distances between conduits are measured as shown in figure 6-4. a multi-core cables a single-core cables Figure 6-4: distance between conduits (a) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 529 n correction factors in the case of several circuits or cables in the same buried conduit (see table 6-14): f 9 This is applicable to groups of cables with varying cross-sectional areas but having the same allowable maximum temperature. Correction factors Arrangement of touching circuits or cables Number of circuits or multi-core cables Installed in a buried conduit 1 2 3 4 5 6 7 8 9 12 16 20 1 0.71 0.58 0.5 0.45 0.41 0.38 0.35 0.33 0.29 0.25 0.22 Table 6-14: correction factors in the case of several circuits or cables in the same buried conduit n correction factors for a group of several cables installed directly in the ground - single or multi-core cables arranged horizontally or vertically (see table 6-15): f10 Distance between cables or groups of 3 single-core cables (a) Number of cables or circuits Zero (touching cables) One cable diameter 0.25 m 0.5 m 1.0 m 2 0.76 0.79 0.84 0.88 0.92 3 0.64 0.67 0.74 0.79 0.85 4 0.57 0.61 0.69 0.75 0.82 5 0.52 0.56 0.65 0.71 0.80 6 0.49 0.53 0.60 0.69 0.78 Table 6-15: correction factors for a group of several cables installed directly in the ground single or multi-core cables arranged horizontally or vertically The distances between cables are measured as shown in figure 6-5. a a a multi-core cables single-core cables Figure 6-5: distance between cables (a) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 530 n current-carrying capacities (in amps) of wiring systems in standard installation conditions for selection letters B, C, E, F The current carrying capacities given in table 6-16 are valid for simple circuits made up of the following number of conductors: Selection letter B: - two insulated conductors or two single-core cables or one two-core cable - three insulated conductors or three single-core cables or one three-core cable Selection letter C: - two single-core cables or one two-core cable - three single-core cables or one three-core cable Selection letters E and F (see fig. 6-6): - one two-core or three-core cable for letter E - two or three single-core cables for letter F . E E F F Figure 6-6: illustration of installation methods for selection letters E and F The number of conductors to be considered in a circuit is that of the conductors through which the current actually flows. When, in a three-phase circuit, the currents are assumed to be balanced, it is not necessary to take into account the corresponding neutral conductor. When the current value of the neutral conductor is close to that of the phases, a reduction factor of 0.84 is to be applied. Such currents may, for example, be due to the presence of third harmonic currents in the phase conductors (see § 6.2). Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 531 Selection letter B Insulating material and number of loaded conductors PVC 3 C PVC 2 XLPE 3 PVC 3 PVC 2 E PVC 3 F XLPE 2 XLPE 3 PVC 2 PVC 3 XLPE 2 XLPE 3 XLPE 2 PVC 2 XLPE3 XLPE2 Copper crosssection (mm²) 1.5 2.5 4 6 10 16 25 35 50 70 95 120 150 185 240 300 400 500 630 15.5 21 28 36 50 68 89 110 134 171 207 239 17.5 24 32 41 57 76 96 119 144 184 223 259 299 341 403 464 18.5 25 34 43 60 80 101 126 153 196 238 276 319 364 430 497 19.5 27 36 48 63 85 112 138 168 213 258 299 344 392 461 530 22 30 40 51 70 94 119 147 179 229 278 322 371 424 500 576 656 749 855 23 31 42 54 75 100 127 158 192 246 298 346 395 450 538 621 754 868 1005 24 33 45 58 80 107 138 169 207 268 328 382 441 506 599 693 825 946 1088 26 36 49 63 86 115 149 185 225 289 352 410 473 542 641 741 16.5 22 28 39 53 70 86 104 133 161 186 18.5 25 32 44 59 73 90 110 140 170 197 227 259 305 351 19.5 26 33 46 61 78 96 117 150 183 212 245 280 330 381 21 28 36 49 66 83 103 125 160 195 226 261 298 352 406 23 31 39 54 73 90 112 136 174 211 245 283 323 382 440 526 610 711 24 32 42 58 77 97 120 146 187 227 263 304 347 409 471 600 694 808 26 35 45 62 84 101 126 154 198 241 280 324 371 439 508 663 770 899 28 38 49 67 91 108 135 164 211 257 300 346 397 470 543 161 200 242 310 377 437 504 575 679 783 940 1083 1254 Aluminium crosssection (mm²) 2.5 4 6 10 16 25 35 50 70 95 120 150 185 240 300 400 500 630 121 150 184 237 289 337 389 447 530 613 740 856 996 Table 6-16: current carrying capacities (in amps) of wiring systems in standard installation conditions ( f0 to f10 = 1) for selection letters B, C, E, F Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 532 n current-carrying capacities (in amps) of wiring systems in standard installation conditions for selection letter D (buried wiring systems) (see table 6-17) The number of conductors to be considered in a circuit is that of the conductors through which the current actually flows. When, in a three-phase circuit, the currents are assumed to be balanced, it is not necessary to take into account the corresponding neutral conductor. When the current value of the neutral conductor is close to that of the phases, a reduction factor of 0.84 is to be applied. Such currents may, for example, be due to the presence of third harmonic currents in the phase conductors (see § 6.2). Selection letter D Insulating material and number of loaded conductors PVC 3 PVC 2 XLPE 3 XLPE 2 26 34 44 56 74 96 123 147 174 216 256 290 328 367 424 480 32 42 54 67 90 116 148 178 211 261 308 351 397 445 514 581 31 41 53 66 87 113 144 174 206 254 301 343 387 434 501 565 37 48 63 80 104 136 173 208 247 304 360 410 463 518 598 677 57 74 94 114 134 167 197 224 254 285 328 371 68 88 114 137 161 200 237 270 304 343 396 447 67 87 111 134 160 197 234 266 300 337 388 440 80 104 133 160 188 233 275 314 359 398 458 520 Copper cross-sectional area (mm²) 1.5 2.5 4 6 10 16 25 35 50 70 95 120 150 185 240 300 Aluminium cross-sectional area (mm²) 10 16 25 35 50 70 95 120 150 185 240 300 Table 6-17: current carrying capacities (in amps) of wiring systems in standard installation conditions ( f0 to f10 = 1) for selection letter D (buried wiring systems) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 533 6.1.5. Practical method for determining the minimum cross-sectional area of an LV wiring system conductor installation conditions design current IB determination of the protective device rated current I n or setting current I set taken to be just higher than the design current: I n or I set I B I n or I set determination of the selection letter and overall correction factor f (see tab. 8-3 to 8-5) determination of current I z of the wiring system to be protected by the protective device fuse circuit-breaker Iz 1.31 I n if I n 10 A Iz 1.21 I n if I n 10 A and I n 25 A 1.10 I n if I n 25 A Iz Iz I n or I set I z2 I z1 determination of the cross-sectional area S of the wiring system conductors able to carry I z1 or I z2 : I z1 I or z 2 (1) f f ' - determine the cross-sectional area able to carry I z in standard installation conditions, depending on the insulating material, the number of loaded conductors and the type of conductor (copper or aluminium) (see tab. 8-16 and 8-17) ' - calculate the equivalent current I z S check of other required conditions: - maximum voltage drop - maximum length for protection against indirect contact (IT and TN earthing systems) - check of thermal withstand in case of short circuit (1) I z' is an equivalent current which, in standard installation conditions, causes the same thermal effect as I z1 or I z2 in actual installation conditions Figure 6-7: logigram for determining the cross-sectional area of a LV wiring system Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 534 6.1.6. Cross-sectional area of protective conductors (PE), equipotential bonding conductors and neutral conductors (IEC 364) In a low voltage installation, the protective conductors ensure that the exposed conductive parts of loads are interconnected and insulation fault currents are evacuated to the ground. The equipotential bonding conductors allow the exposed conductive parts and extraneous conductive parts to be set at the same potential, or similar potentials. In this chapter, we will limit ourselves to conductor sizing rules. Refer to paragraph 2 for the protection and connection rules. n cross-sectional area of protective conductors between MV/LV transformer and main LV switchboard (see fig. 6-8) main LV switchboard PE Figure 6-8: PE conductors between transformer and main switchboard Table 6-18 gives the protective conductor cross-sectional areas (in mm²) in relation: - to the nominal power of the MV/LV transformer - to the operating time t (in seconds) of the MV protection. When protection is ensured by a fuse, the cross-sectional area to be taken into account corresponds to t = 0.2 s - to the insulating material and type of conductor metal. In an IT earthing system, if an overvoltage limiter is inserted between the neutral and earth, the same sizing is applied to its connecting conductors. In the case where several transformers operate in parallel, the sum of their nominal powers will be used to determine the cross-sectional area. 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Industrial electrical network design guide T&D 6 883 427/AE 535 Transformer power (kVA) LV voltage 127/220 V 230/400 V Type of conductors Copper t (s) Aluminium Bare conductors 0.2 s 0.5 s - PVC-insulated conductors - 0.2 s 0.5 s 0.2 s 0.5 s - - 0.2 s 0.5 s XLPE-insulated conductors 0.2 s 0.5 s - - 0.2 s 0.5 s ≤ 63 ≤ 100 25 25 25 25 25 25 25 25 25 100 160 25 25 35 25 25 50 25 25 35 125 200 25 35 50 25 35 50 25 25 50 160 250 25 35 70 35 50 70 25 35 50 200 315 Potective conductor 35 50 70 35 50 95 35 50 70 250 400 cross-sectional area 50 70 95 50 70 95 35 50 95 315 500 S PE (mm²) 50 70 120 70 95 120 50 70 95 400 630 70 95 150 70 95 150 70 95 120 500 800 70 120 150 95 120 185 70 95 150 630 1 000 95 120 185 95 120 185 95 120 150 800 1 250 95 150 185 120 150 240 95 120 185 Table 6-18: cross-sectional area of protective conductors between MV/LV transformer and main LV switchboard n cross-sectional areas of low voltage exposed conductive part protective conductors: (PE) The cross-sectional area of the PE conductor is defined in relation to the cross-sectional area of the phases (for the same metal conductor) as follows: - for S phase ≤ 16 mm² , S PE = S phase - for 16 mm² < S phase ≤ 35 mm² , S PE = 16 mm² - for S phase > 35 mm² , S PE = (1) S phase 2 (1) when the protective conductor is not part of the wiring system, it must have a cross-sectional area of at least: - 2.5 mm² if it comprises a mechanical protection - 4 mm² if it does not comprise a mechanical protection Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 536 In the TT earthing system, the protective conductor cross-sectional area may be limited to: - 25 mm² for copper - 35 mm² for aluminium on condition that the neutral and exposed conductive part earth electrodes are separate, otherwise the conditions of the TN earthing system are applicable (in a TT earthing system, there may be an involuntary connection via the metal structure or other part between the two earth electrodes; the earth fault current is then high). n cross-sectional area of equipotential bonding conductors o main equipotential bonding conductor Its cross-sectional area must be at least equal to half the cross-sectional area of the installation's largest protective conductor, with a minimum of 6 mm². However, it may be limited to 25 mm² for copper or 35 mm² for aluminium. o supplementary equipotential bonding conductor If it connects two exposed conductive parts, its cross-sectional area must not be smaller than the smallest of the protective conductors connected to these parts (see fig. 6-9-a). If it connects an exposed conductive part to an extraneous conductive part, its cross-sectional area must not be smaller than half the cross-sectional area of the protective conductor connected to this exposed conductive part (see fig. 6-9-b). S PE1 ≤ S PE 2 If S S LS = PE 2 S LS = S PE1 S PE1 S PE2 S PE S LS P1 (*) S LS P2 a) between two exposed conductive parts P b) between an exposed conductive part and a structure Figure 6-9: cross-sectional area of supplementary equipotential bonding conductors (*) with a minimum of: - 2.5 mm² if the conductors are mechanically protected - 4 mm² if the conductors are not mechanically protected Conductors which are not incorporated in a cable are mechanically protected when they are installed in conduits, troughs or casing or protected in a similar way. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 537 n cross-sectional area of PEN conductors In the case of a TNC earthing system, the protective conductor also plays the role of the neutral conductor. In this case, the cross-sectional area of the PEN must be at least equal to the greatest value resulting from the following requirements: - S PEN ≥ − 10 mm 2 for copper − 16 mm 2 for aluminium - meet the conditions relating to the PE conductor - meet the conditions required for the neutral conductor cross-sectional area. n cross-sectional area of the neutral conductor - The neutral conductor must have the same cross-sectional area as the phase conductors in the following cases: . single-phase circuit . three-phase circuit having phase cross-sectional areas smaller than or equal to 16 mm² for copper or 25 mm² for aluminium. - For three-phase circuits having a phase cross-sectional area greater than 16 mm² for copper or 25 mm² for aluminium, the neutral cross-sectional area may be smaller than that of the phases as long as the following conditions are met: . the maximum current likely to continuously circulate in the neutral is lower than the current-carrying capacity of the chosen cross-sectional area. The unbalance of singlephase loads and third and multiples of third harmonics which may require the use of a cross-sectional area greater than the phases must be taken into account (see § 8.2 neutral conductor heating). . the neutral conductor is protected against overcurrent by a fuse or a circuit-breaker trip setting suitable to its cross-sectional area. . the cross-sectional area of the neutral conductor is at least equal to 16 mm² for copper or 25 mm² for aluminium. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 538 6.1.7. Checking voltage drops The voltage drop over a wiring system is calculated using the following formula: L ∆V = b ρ1 cos ϕ + λ L sin ϕ × I B S ∆V : voltage drop, in volts = 1 for three - phase circuit = 2 for single - phase circuit b : coefficient ρ1 : conductor resistivity during normal service, i.e. 1.25 times that at 20 °C ρ1 = 0.0225 Ω mm²/m for copper; ρ1 = 0.036 Ω mm²/m for aluminium L S cos ϕ IB λ : length of wiring system, in metres : cross-sectional area of conductors, in mm² : power factor, in the absence of specific indications we can take cos ϕ = 0.8 ( sin ϕ = 0.6) : maximum design current, in amps : reactance per unit length of the conductors, in Ω/m The values of λ in LV are: - 0.08 × 10 −3 Ω / m for three-core cables - 0.09 × 10 −3 Ω / m for single-core cables in a flat formation - 0.15 × 10 −3 Ω / m for single-core cables spaced by d = 8 r d r or triangular formation : mean distance between conductor : radius of conductor cores The relative voltage drop is defined as: ∆V Vn for phase-to-neutral fed three-phase or single-phase circuits ∆V Un for phase-to-phase fed single-phase circuits (in this case, ∆V represents a phase-to-phase voltage drop) Vn : nominal single-phase voltage U n : nominal phase-to-phase voltage Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 539 In accordance with IEC 364-5-52, in the absence of other considerations, it is recommended that in practice the voltage between the origin of consumer's installation and the equipment should not be greater than 4% of the nominal voltage of the installation. n circuits feeding motors The voltage drop is calculated by replacing the design current I B by the motor starting current. Taking into account all the motors able to start simultaneously, the voltage drop must be lower than 10% to ensure correct motor starting and not disturb the rest of the installation too much. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 540 6.1.8. Maximum lengths of wiring systems for protection against indirect contact ( TN and IT earthing system) Standard IEC 364 specifies that the fault current for TN and IT earthing systems must be cleared in a time compatible with the protection of persons. This time is determined by a curve in relation to the prospective touch voltage; it is based on the physiological effects of the electrical current on the human body. To simplify matters, using this curve, it is possible to determine a maximum disconnecting time in relation to the nominal voltage of the installation (see table 6-20 and 6-21). Nominal a.c. voltage Vn / U n (Volts) Disconnecting time (seconds) (*) non-distributed neutral distributed neutral 120/240 0.8 5 230/400 0.4 0.8 400/690 0.2 0.4 580/1000 0.1 0.2 Table 6-20: maximum disconnecting times in the IT earthing system (second fault) Nominal a.c. voltage Vn (Volts) (**) Disconnecting time (seconds) (*) 120 0.8 230 0.4 277 0.4 400 0.2 > 400 0.1 Table 6-21: maximum disconnecting times in the TN earthing system (*) these values are not valid in premises containing a bath or shower. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 541 Note 1: if the disconnecting time is more than the time t0 , but less than 5 seconds, protection is allowed by IEC 364 (§ 413.1.3.5) in the following cases: - in distribution circuits when the protective conductor at the downstream end of the circuit is directly connected to the main equipotential bonding. - in terminal circuits supplying stationary equipment only and having a protective conductor that is connected to the main equipotential bonding and which is located in the area that is influenced by the main equipotential bonding. Note 2 : in the TT earthing system, protection is in general ensured by residual current devices which are set to meet the following condition (see IEC 364, § 413.1.4.2): RA IA ≤ 50 V RA : resistance of the earth electrode of the exposed conductive parts IA : rated residual current of the circuit-breaker If selectivity is seen to be necessary, an operating time at the most equal to 1 second is allowed in the distribution circuits without taking into account the touch voltage Note 3 : in an IT earthing system, when the exposed conductive parts are earthed individually or in groups, the conditions of the TT earthing system given in Note 2 must be met (see IEC 364, § 413.1.5.3). n circuit-breaker protection IEC 364 specifies that the magnetic tripping threshold of the circuit-breaker in TN and IT earthing systems must be lower than the minimum short-circuit current. Furthermore, any eventual circuit-breaker time delay must be shorter than the maximum disconnecting time defined in tables 6-20 and 6-21. For a given circuit-breaker and cross-sectional area, there is thus a maximum circuit length not to be exceeded in order to comply with the requirements concerning the protection of persons against indirect contact. In the following part of the chapter, we will apply the conventional method for determining maximum circuit lengths. This is more restrictive than the impedance method, but can be applied by carrying out the calculations by hand. In the conventional method, we neglect the influence of the reactance of the conductors for cross-sectional areas smaller than 150 mm². For large cross-sectional areas, we will take into account the influence of the reactance by dividing Lmax by: - 1.15 for a cross-sectional area of 150 mm² 1.20 for a cross-sectional area of 185 mm² 1.25 for a cross-sectional area of 240 mm² 1.30 for a cross-sectional area of 300 mm². Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 542 Note: for minimum short-circuit current calculations, refer to the "Industrial network protection guide" § 4.4.1. o TN earthing system The maximum length of a circuit in a TN earthing system is: Lmax = 0.8 × Vn × S ph ρ × (1 + m) × Im Lmax : maximum length in m Vn : single-phase voltage in volts S ph : cross-sectional area of the phases in mm² ρ : resistivity of the conductors taken to be equal to 1.5 times that at 20 °C ( ρ copper; = 0.027 Ω mm 2 / m for ρ = 0.043 Ω mm 2 / m for aluminium) m S ph : cross - sectional area of phases = SPE : cross - sectional area of protective conductor Im : circuit-breaker magnetic trip operating current o IT earthing system The maximum length of a circuit in an IT earthing system is: - if the neutral conductor is not distributed: Lmax = 0.8 × 3 × Vn × S ph 2 ρ × (1 + m) × I m - if the neutral conductor is distributed: Lmax = = Sph S1 : 0.8 × Vn × S1 2 ρ × (1 + m) × Im if the outgoing feeder considered does not have a neutral = Sneutral if the outgoing feeder considered has a neutral o TT earthing system No condition on the wiring system length is specified since the protection of persons is ensured by the residual current device. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 543 n fuse protection Using the fuse fusing curve, we can determine the current I a ensuring fusion of the fuse in the time t 0 specified in tables 6-20 and 6-21 (see fig. 6-10). We can then calculate the maximum length of the wiring system in the same way as for the circuit-breaker replacing I m by I a . t t0 I Ia Figure 6-10: fuse fusing curve n application In practice, checking the cross-sectional area of the wiring system in relation to the protection of persons against indirect contact consists in making sure that the length of the wiring system is less than Lmax for a given arrangement. If the wiring system length is greater than Lmax , we can take the following measures: - choose a circuit-breaker (or trip relay) with a lower magnetic threshold if the selectivity requirements permit this - install a residual current circuit-breaker for TNS earthing system it is not possible to use a RCD) and IT earthing system (in a TNC - take larger phase and protective conductor cross-sectional areas meeting the maximum length condition. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 544 6.1.9. Checking the thermal withstand of conductors When a short-circuit current flows through the conductors of a wiring system for a very short time (up to five seconds), the heating of the conductors is considered to be adiabatic; this means that the energy stored remains in the metal of the core and is not transmitted to the insulating material. It is therefore necessary to check that the short-circuit thermal stress is lower than the conductor thermal withstand: 2 tdis I sc ≤ k2 S2 tdis : protective device disconnecting time in seconds S Isc : cross-sectional area of conductors in mm² : short-circuit current in A The value of k (see table 6-22). depends on the core metal and the type of insulating material Insulating material PVC XLPE Copper 115 135 Aluminium 74 87 Core Table 6-22: value of factor k in accordance with IEC 364-4-43 If the disconnecting time is given, the cross-sectional area must comply with: I S ≥ sc × tdis k Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 545 n circuit-breaker protection The check must be carried out for the maximum short-circuit current at the circuit-breaker location. The curves in manufacturers' catalogues give the maximum disconnecting time of the circuitbreaker. When circuit-breaker tripping is time delayed, the disconnecting time is taken to be equal to the time delay. To check the thermal withstand, the short-circuit current value must be calculated with a resistivity ρ of the conductors taken to be equal to 1.5 times that at 20 °C : - ρ = 0.027 Ω mm 2 / m for copper - ρ = 0.043 Ω mm 2 / m for aluminium o case of current-limiting circuit-breakers On occurrence of a short circuit, current-limiting circuit-breakers only let a current below the prospective fault current through (see fig. 6-11). Isc prospective peak I sc prospective Isc limited peak I sc t Figure 6-11: current limiting curve The wiring system protected by this type of device is not therefore subjected to the (prospective) calculated Isc thermal stress, but a much smaller stress defined by manufacturers' limiting curves for each type of circuit-breaker. 2 The limiting curves give the thermal stress tdis I sc expressed in A 2 × second . Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 546 o example We want to check the thermal withstand of a PVC-insulated 6 mm² copper conductor protected by a Compact NS 80H-MA 380/415 V circuit-breaker fitted with an LR2-D33 63 thermal relay. The themal withstand of the cable is: k 2 S 2 = (115) 2 × 6 2 = 4.76 × 10 5 A2 × s . The limiting curves in figure 6-12 give the maximum thermal stress of the circuit-breaker: 2 × 105 A2 × s . The cable is thus protected up to the circuit-breaker breaking capacity. The curves are in the table order Figure 6-12: thermal stress limiting curves for Compact NS 80H-MA-380/415V circuit-breakers n fuse protection The current causing the most stress is the minimum short-circuit current at the end of the wiring system. The fusing time t f of the fuse corresponding to I sc min must comply with the relation: 2 2 2 t f Isc min ≤ k S The method for calculating I sc min is given in paragraph 4.4.1 of the Protection guide. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 547 6.1.10. Application example n hypotheses Let us consider the diagram in figure 6-13 the data of which is given below. Since the installation feeds loads requiring good continuity of service the IT earthing system without distributed neutral is chosen. o W2 wiring system This is made up of a PVC insulated copper three-core cable which is installed touching 3 other multi-core cables on perforated trays in an ambient temperature of 40°C. It is protected by fuses. It feeds a load having the following characteristics: - active power P = 15 kW - efficiency η = 0.89 - cos ϕ = 0.85 - utilisation factor b = 0.9 . o W1 wiring system This is made up of 3 XLPE-insulated copper single-core cables in a triangular formation. The cables are buried alone, without any extra mechanical protection, in soil which has a thermal resistivity of 0.85 K.m/W and a temperature of 35 °C. They are protected by a circuit-breaker. The wiring system feeds load L1 and 3 other outgoing feeders the I B current values of which are given in figure 6-13. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 548 250 kVA U sc = 4 % unearthed neutral 400 V W1 L1 = 100 m cos = 0.8 400 V 25 A 50 A 40 A IB W2 L2 = 15 m R1 Figure 6-13: diagram of the installation n determining the maximum design current o W2 wiring system - P = 15 kW 1 = 1.32 η cos ϕ - the utilisation factor b = 0.9 - the factor a = - for a single load the coincidence factor is c = 1 - no extension is planned, thus d = 1 - for a 400 V three-phase network, the power conversion factor in current is e = 1.4 . We then have: I B = P × a × b × c × d × e = 15 × 1.32 × 0.9 × 1 × 1.4 = 24.9 A Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 549 o W1 wiring system The maximum design current of the W1 wiring system is obtained by calculating the sum of currents ( I B ) of all the outgoing feeders fed by W1 and by applying a coincidence factor estimated at 0.8 (see table 6-2): I B = (25 + 50 + 40 + 24.9) × 0.8 = 115.9 A n correction factors o W2 wiring system Table 6-3 gives the installation method N° 13 and the selection letter E . The correction factors to be applied are: - ambient temperature cable group (see table 6-6) (see tables 6-9 et 6-10) : f1 = 0.87 : f 4 = 0.77 and f 5 = 1 The overall correction factor is: f = 0.87 × 0.77 × 1 = 0.67 o W1 wiring system Table 6-3 gives the installation method N° 62 and the selection letter D . The correction factors to be applied are: - ground temperature soil thermal resistivity cable group (see table 6-7) (see table 6-8) (see table 6-15) : f 2 = 0.89 : f 3 = 1.05 : f10 = 1 The overall correction factor is: f = 0.89 × 1.05 × 1 = 0.935 Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 550 n determining the cross-sectional area and choosing the protective device o W2 wiring system I B = 24.9 A f = 0.67 The fuse nominal current must comply with the condition I n ≥ I B . The fuse with a rating of I n = 25 A is chosen. For 10 A < I n ≤ 25 A , the current I z of the wiring system protected by this fuse is: I z = k3 In = 1.21 In = 30.3 A The equivalent current that the wiring system must be able to carry in standard installation I . A conditions is: I 'z = z = 451 f Table 6-16 (selection letter E , PVC3, copper) gives a minimum cross-sectional area of S = 10 mm2 which has a current-carrying capacity of I 0 = 60 A . o W1 wiring system I B = 115.9 A f = 0.935 For an adjustable circuit-breaker, the setting current must comply with the condition Iset ≥ I B ; I set = 120 A is chosen. The current I z of the wiring system protected by this setting is: I z = I n = 120 A The equivalent current that the wiring system must be able to carry in standard installation I conditions is: I 'z = z = 128.3 A f Table 6-17 (selection letter D , XLPE3, copper) gives a minimum cross-sectional area of S = 25 mm2 which has a current-carrying capacity of I 0 = 144 A . Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 551 n maximum length of the wiring system o W2 wiring system For S ph = 10 mm² , we have S PE = S ph = 10 mm² whence m= S ph S PE =1 Table 6-20 gives a maximum disconnecting time of t = 0.4 s for a network with non-distributed neutral. The time-current characteristic for a 25 A rated fuse gives us a current of I a = 200 A for a disconnecting time of 0.4 s. The neutral is not distributed and we thus have: Lmax = 0.8 × 3 × Vn × S ph 0.8 × 3 × 230 × 10 = 147.5 m = 2 ρ (1 + m ) Ia 2 × 0.027 × 2 × 200 The length of the W2 wiring system (15 m) is far smaller than Lmax and the protection of persons against indirect contact is thus ensured. o W1 wiring system For 16 mm² < S ≤ 35 mm² , we have S PE = 16 mm² whence m= S ph SPE = 25 = 1.56 16 The circuit-breaker chosen is a Compact NS 125E with an STR 22SE trip relay having a magnetic tripping threshold set at I m = 1 250 A because of the selectivity. The neutral is not distributed and we thus have: Lmax = 0.8 × 3 × Vn × S ph 2 ρ (1 + m) I m = 0.8 × 3 × 230 × 25 = 46.1 m 2 × 0.027 × 2.56 × 1 250 The length of the W1 wiring system (100 m) is greater than Lmax . By taking cross-sectional areas greater, i.e. S ph = 35 mm² and S PE = 35 mm² ( m = 1) , we find Lmax = 82.6 m < 100 m ; which is not sufficient. So as not to oversize the conductors, it is decided that the outgoing feeder should be fitted with a residual current device which ensures the protection of persons against indirect contact. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 552 n checking the voltage drop o W2 wiring system S = 10 mm² , L = 15 m , I B = 24.9 A The cable is three-core and we thus have λ = 0.08 × 10 −3 Ω / m . The power factor is cos ϕ = 0.85 , whence sin ϕ = 0.53 . For a three-phase circuit b = 1 . For copper ρ1 = 0.0225 Ω mm 2 / m . We deduce from this that 15 ∆ V = 0.0225 × × 0.85 + 0.08 × 10 −3 × 15 × 0.53 × 24.9 10 ∆ V = 0.73 V whence ∆ V 0.73 = = 0.3 % Vn 230 The total voltage drop is 4.2 % (the voltage drop in the W1 wiring system is 3.9 %, see below). o W1 wiring system S = 25 mm² , L = 100 m , I B = 115.9 A The 3 single-core cables are in a flat formation and we thus have: λ = 0.09 × 10 −3 Ω / m The overall power factor of the installation is cos ϕ = 0.8 , whence sin ϕ = 0.6 . 100 We deduce from this that ∆ V = 0.0225 × × 0.8 + 0.09 × 10 −3 × 100 × 0.6 × 115.9 25 ∆ V = 8.97 V whence ∆ V 8.97 = = 3.9 % Vn 230 Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 553 n checking the thermal stress o W2 wiring system For fuse protection, the current to be taken into account is the minimum short-circuit current at the end of the wiring system. For the IT earthing system, this is the short-circuit current for a double phase-earth fault. By applying the conventional method (see § 4.4.1.2 of the Protection guide), we can calculate: Isc min = 3 × Vn × 0.8 3 × 230 × 0.8 = = 1.97 kA 1 1 2 × 15 × 0.027 1 + 1 + 2 L 2 ρ 10 10 S ph SPE The time-current characteristic of the 25 A rated fuse gives us a fusing time of t f = 5 ms for a current of 1.97 kA. The maximum thermal stress is thus: ( 2 3 Isc min × t = 1.97 × 10 ) 2 × 5 × 10 −3 = 19.4 × 10 3 A2 × s 2 The permitted cable thermal withstand is: k 2 S 2 = (115) × 102 = 1322 × 103 A2 × s . The cross-sectional area of S = 10 mm2 is thus largely able to withstand to the fuse thermal stress. o W1 wiring system The maximum short-circuit current of the circuit-breaker (neglecting the connection linking the circuit-breaker to the transformer) is: Isc = Sn 1 250 × 103 100 × = × = 9.02 kA 4 3 Un Usc 3 × 400 We assume that the circuit-breaker trip relay is delayed by 0.1 second, the maximum shortcircuit thermal stress is then: ( 2 Isc t = 9.02 × 10 3 ) 2 × 0.1 = 8.14 × 10 6 A2 × s The permitted cable thermal withstand is: k 2 × S 2 = 1432 × 252 = 12.78 × 10 6 A2 × s The cross-sectional area of S = 25 mm2 is thus largely able to withstand the circuit-breaker thermal stress. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 554 n conclusion The cross-sectional areas to be chosen are: - W1 wiring system: 3 × 35 mm 2 + 1 × 16 mm 2 copper - W2 wiring system: 3 × 10 mm 2 + 1 × 10 mm 2 copper Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/A 555 6.2. Determining conductor cross-sectional areas in medium voltage 6.2.1. Method principle The method for determining the cross-sectional area of conductors in medium voltage consists in: - determining the maximum design current I B of the loads to be supplied - determining the cross-sectional area S1 complying with the heating of the cable core under normal operating conditions, which may be continuous or discontinuous. To do this, it is necessary to know: . the actual installation conditions of the wiring system and consequently the overall correction factor f . the current-carrying capacities of the different types of cable in standard installation conditions. - determining the cross-sectional area S 2 required for the thermal withstand of the cable in the event of a three-phase short circuit - determining the cross-sectional area screen in the event of an earth fault S3 required for the thermal withstand of the cable - possibly checking the voltage drop in the wiring system for the chosen cross-sectional area S. The technical cross-sectional area S to be selected is the maximum value among cross-sectional areas S1 , S 2 and S3 . - possibly calculating and choosing the economical cross-sectional area. 6.2.2. Determining the maximum design current The maximum design current I B is determined on the basis of the sum of powers of the loads fed, applying if necessary utilisation and coincidence coefficients (see § 6.1.2.). In medium voltage, a wiring system most often feeds a single load (transformer, motor, furnace, steam generator), in this case I B is taken to be equal to the rated current of the device. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 556 6.2.3. Current-carrying capacities in wiring systems n general rules The current-carrying capacity is the maximum current that a wiring system can continuously carry without this affecting its life span. The current-carrying capacities of cables are given in standards or by manufacturers for standard installation conditions. To determine the current-carrying capacity of a wiring system in actual installation conditions, the following must be carried out: - using table 6-23, define the installation method, its associated table column number and correction factors to be applied - using the installation conditions, determine the correction factor values which must be applied (see tables 6-24 to 6-28) - calculate the overall correction factor f equal to the product of the correction factors - using table 6-29 for impregnated paper-insulated cables and tables 6-30 to 6-34 for synthetically-insulated cables, determine the maximum current that the wiring system can carry in standard conditions ( f0 to f6 = 1) - calculate the maximum current-carrying capacity of the wiring system in relation to its installation conditions: I a = f I 0 . Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 557 n installation methods Table 6-23 gives, for each installation method, the current-carrying capacity table column to be used for choosing the cross-sectional area of the conductors (see tables 6-29 to 6-34). Factor f 0 corresponds to the installation method; factors f1 to f 6 are explained below Table Correction factors to be applied f0 (see tables 6-24 to 6-28). Installation methods Example column A Conduits on wall (3) 0.90 f1 f5 B Flush mounted conduits (3) 0.90 f1 f5 F Installed on cable trays (3) 1 f1 f5 G Installed on brackets or cable ladders (3) 1 f1 f6 H Troughs (enclosed) (3) 0.90 f1 f5 (3) 1 f1 f6 (3) 0.80 f1 f5 J Ducts (open troughs) L1 Conduits in open or ventilated channels Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 558 Installation method Example L3 Directly installed in open or ventilated channels L4 Directly installed in enclosed channels L5 Directly installed in channels filled with sand Table Correction factors to be applied column f0 (3) 0.90 f1 -- f5 (3) 0.80 f1 -- f5 (3) 0.80 f1 -- f5 (3) 0.90 f1 -- f5 (3) 0.90 f1 -- f5 1 f2 f3 f4 1 f2 f3 f4 N Troughs (in masonry) P Manufactured blocks S1 Directly buried (armoured cables) S2 Buried with mechanical protection P : steady-state operating conditions D : discontinuous operating conditions P D ____ ___ (1) (2) (1) (2) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 559 Installation method Example Table column f0 Correction factors to be applied P D ____ ___ (1) (2) S3 Buried in sleeves 0.8 f2 f3 f4 S4 Cables installed in trefoil formation in a prefabricated channel, buried directly in the ground, possibly with extra backfill (1) (2) 0.8 f2 f3 f4 S5 Single-core cables installed in individual channels, buried directly in the ground, possibly with extra backfill (1) (2) 0.8 f2 f3 f4 (1) (2) 0.8 f2 f3 f4 (3) 1,1 f1 -- -- Single-core cables in a flat formation spaced out in a prefabricated channel, buried directly in the ground, possibly with extra backfill V Overhead lines P : steady-state operating conditions D : discontinuous operating conditions Table 6-23: installation methods Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 560 n correction factors for ambient temperatures other than 30 °C (cables installed in air): f1 Temperature Type of insulating material °C PVC PE XLPE EPR 10 1.22 1.15 15 1.17 1.12 20 1.12 1.08 25 1.06 1.04 30 1.00 1.00 35 0.94 0.96 40 0.87 0.91 45 0.79 0.87 50 0.71 0.82 55 0.61 0.76 Table 6-24: correction factors for ambient temperatures other than 30 °C (cables installed in air) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 561 n correction factors for ground temperatures other than 20 °C (buried cables): f 2 Temperature Type of insulating material °C PVC PE XLPE EPR 0 1.18 1.13 5 1.14 1.10 10 1.10 1.07 15 1.05 1.04 20 1.00 1.00 25 0.95 0.96 30 0.89 0.93 35 0.84 0.89 40 0.77 0.85 45 0.71 0.80 50 0.63 0.76 60 0.45 0.65 65 - 0.60 70 - 0.53 75 - 0.46 80 - 0.38 Table 6-25: correction factors for ground temperatures other than 20 °C (buried cables) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 562 n correction factors for soil thermal resistivities other than 1 K.m/W (buried cables): f 3 Soil thermal resistivity (K.m/W) Humidity 0.5 Assembly of three singlecore cables Three-core cables Very moist soil 1.25 1.20 0.7 Moist soil 1.14 1.10 0.85 Normal soil Clay 1.06 1.05 Dry soil and 1.00 1.00 Chalk 0.93 0.95 Ash 0.85 0.88 and 0.75 0.79 Clinker 0.68 0.72 0.62 0.68 1 Type of soil 1.2 Sand 1.5 Very dry soil 2 2.5 3 Tableau 6-26: correction factors for soil thermal resistivities other than 1 K.m/W (buried cables) n correction factors for a group of several wiring systems (buried cables): f 4 Number of circuits Distance between cables "a" Zero (cables touching) One cable diameter 0.125 m 0.25 m 0.5 m 2 0.75 0.80 0.85 0.90 0.90 3 0.65 0.70 0.75 0.80 0.85 4 0.60 0.60 0.70 0.75 0.80 5 0.55 0.55 0.65 0.70 0.80 6 0.50 0.55 0.60 0.70 0.80 Determination of the distance "a" in the case of single-core cables installed in a flat or trefloid formation and three-core cables. single-core cables three-core cables a a a Table 6-27: correction factors for a group of several wiring systems (buried cables ) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 563 n correction factors for a group of several circuits or several cables (cables installed in air and away from direct sunlight): f 5 , f 6 Installation method Arrangement f5 On unperforated horizontal trays......................... f6 On perforated horizontal trays or on brackets .................................................... Number of circuits or multi-core cables 2 3 4 6 >9 0.85 0.80 0.75 0.70 0.70 0.90 0.80 0.80 0.75 0.75 Table 6-28: correction factors for a group of several circuits or several cables (cables installed in air and away from direct sunlight) n current-carrying capacities of cables in standard installation conditions ( f0 to f6 = 1) References (1), (2) and (3) of tables 6-29 to 6-34 correspond to the column number given in table 6-23. o impregnated paper-insulated cables Impregnated paper-insulated cables have stopped being manufactured for several years. However, for calculation purposes for existing installations, the current-carrying capacities may be calculated to an approximate value of ± 5% using the following formula: I = 10 B × S A I : current-carrying capacity, in A S : nominal cross-sectional area of the cable, in mm² A and B : are coefficients given for each type of cable (see table 6-29) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 564 Wiring systems Columns Copper Aluminium A B A B Three-core (1) 0.540 1.446 0.549 1.321 collectively (2) 0.543 1.492 0.544 1.386 screened cable (3) 0.588 1.371 0.598 1.293 3 single-core (1) 0.556 1.269 0.571 1.130 cables (2) 0.567 1.286 0.573 1.179 (3) 0.587 1.196 0.605 1.064 Three-core individually (1) 0.581 1.215 0.594 1.089 screened cables (2) 0.573 1.264 0.578 1.155 (3) 0.600 1.117 0.608 1.004 Table 6-29: values of coefficients A and B for impregnated paper-insulated cables o synthetically-insulated cables The detailed calculation method for current-carrying capacities of cables under steady-state operating conditions is given in IEC publication 287. The current-carrying capacities are given in tables 6-30 to 6-34, according to the type of conductor, the type of insulating material and the rated voltage. The rated voltage for which a cable is designed is expressed by a set of three values, in kV, as U 0 / U (Um) , where: - U0 : U : Um : voltage between the conductor core and a reference potential (screen or earth) voltage between the cores of two phase conductors maximum voltage which may occur between the network phases in normal operating conditions The expression of the rated voltage differs depending on whether the cable is an individually screened type or not (see fig. 2.2.a and 2.2.b). For an individually screened cable, U 0 is different from U , both values being generally in the ratio of 3 . However, due to the way it is made, a collectively screened cable has an equivalent insulation level between two phases and between one phase and the screen. This results in U 0 and U having identical values. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. 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Industrial electrical network design guide T&D 6 883 427/AE 565 PVC-insulated (*) Nominal crosssectional area (mm²)* EPR or XLPE-insulated (1) (2) (3) Copper (1) (2) (3) 72 78 62 10 86 94 78 94 100 81 16 110 120 100 120 130 105 25 145 155 130 145 160 130 35 170 190 165 185 205 165 50 215 240 205 225 250 205 70 260 295 255 270 300 250 95 315 355 310 310 345 290 120 360 405 360 345 390 330 150 405 455 410 385 430 370 185 450 505 460 445 500 440 240 525 590 550 (1) (2) (3) Aluminium (1) (2) (3) 56 61 48 10 67 73 60 72 79 62 16 86 94 79 94 100 82 25 110 120 105 115 125 100 35 135 145 125 145 160 130 50 165 185 160 175 195 160 70 205 230 195 210 235 195 95 245 275 240 240 270 225 120 280 315 280 270 300 255 150 315 355 320 300 335 285 185 350 395 360 350 390 345 240 410 460 430 Above 50 mm², the values are calculated for sector conductors Table 6-30: current-carrying capacities in three-core collectively screened cables having a rated voltage lower than or equal to 6/6 (7.2) kV Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. 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Industrial electrical network design guide T&D 6 883 427/AE 566 Copper 10 16 25 35 50 70 95 120 150 185 240 300 400 500 630 800 1 000 1 200 1 400 1 600 Aluminium 10 16 25 35 50 70 95 120 150 185 240 300 400 500 630 800 1 000 1 200 1 400 1 600 (*) PE-insulated* PVC-insulated Nominal crosssectional area (mm²) (1) 80 105 135 160 190 235 285 320 360 410 475 540 610 680 770 850 930 980 1 030 1 080 (1) 62 80 105 125 150 180 220 250 280 320 370 420 480 540 620 700 780 840 890 940 (2) 89 115 150 180 215 265 320 365 410 470 540 610 700 780 880 980 1 070 1 130 1 190 1 250 (2) 69 89 115 140 170 205 250 285 320 365 425 485 550 630 720 810 900 970 1 030 1 080 (3) 71 95 125 150 180 230 280 320 370 425 500 580 670 760 870 990 1 110 1 210 1 290 1 360 (3) 55 73 96 115 140 175 215 250 285 330 390 455 530 610 710 820 940 1 030 1 110 1 180 (1) 86 110 140 170 200 245 295 335 375 425 490 550 600 700 790 870 950 1 000 1 050 1 100 (1) 67 86 110 130 160 190 230 260 290 330 385 435 495 560 640 720 800 860 910 950 (2) 97 125 160 195 230 285 340 385 435 490 570 640 690 810 920 1 010 1 100 1 160 1 220 1 280 (2) 76 97 125 150 180 220 265 300 335 380 445 500 580 650 750 840 930 1 000 1 060 1 110 EPR or XLPE-insulated (3) 76 100 130 160 190 240 295 340 385 445 530 600 700 790 920 1 040 1 160 1 260 1 350 1 420 (1) 99 125 165 195 230 285 340 385 430 485 560 630 720 800 910 1 000 1 100 1 160 1 220 1 280 (3) 59 78 100 125 150 185 230 265 300 345 410 470 550 640 750 860 980 1 080 1 160 1 230 (1) 77 98 125 150 180 220 260 300 335 380 440 500 570 640 740 830 920 990 1 050 1 100 (2) (3) 110 145 185 225 265 325 390 445 500 560 650 730 840 940 1 060 1 170 1 270 1 350 1 420 1 480 (2) 93 120 160 200 235 295 360 420 475 550 650 740 860 990 1 140 1 300 1 450 1 570 1 680 1 770 (3) 87 110 145 175 205 250 300 345 385 440 510 580 660 750 860 970 1 070 1 150 1 230 1 290 72 95 125 150 185 230 280 325 370 425 510 580 680 790 920 1 070 1 220 1 340 1 450 1 530 For cables having high density polythene insulation, the values are to be multiplied by: 1.05 for columns (1) and (2) 1.06 for column (3) Table 6-31: current-carrying capacities in cables made up of three single-core cables having a rated voltage lower than or equal to 6/10 (12) kV Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. 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Industrial electrical network design guide T&D 6 883 427/AE 567 PE-insulated* Nominal crosssectional area (mm²) (1) (2) (3)* 110 140 170 200 250 295 335 375 425 490 550 630 700 790 870 960 1 010 1 070 1 110 125 160 195 230 280 335 385 430 490 560 640 720 810 920 1 010 1 100 1 170 1 240 1 290 105 135 165 200 250 300 350 395 455 530 610 710 810 930 1 050 1 180 1 270 1 360 1 430 (1) (2) (3) 86 110 130 155 190 230 260 290 330 385 435 495 560 640 720 800 860 920 960 96 125 150 180 220 260 300 335 380 445 500 570 650 740 830 930 1 000 1 060 1 110 81 105 130 155 190 235 270 305 355 420 480 560 650 750 860 990 1 090 1 170 1 240 (*) Copper 16 25 35 50 70 95 120 150 185 240 300 400 500 630 800 1 000 1 200 1 400 1 600 Aluminium 16 25 35 50 70 95 120 150 185 240 300 400 500 630 800 1 000 1 200 1 400 1 600 EPR or XLPE-insulated (1) (2) (3) 125 165 195 230 280 335 385 430 490 560 640 720 810 910 1 010 1 110 1 180 1 240 1 290 140 185 220 260 320 385 440 495 560 650 730 830 940 1 060 1 170 1 280 1 360 1 440 1 500 130 170 200 245 305 375 425 485 560 660 750 870 1 000 1 150 1 300 1 470 1 590 1 700 1 790 (1) (2) (3) 98 125 150 180 220 260 300 335 380 440 500 570 640 740 830 930 1 000 1 060 1 110 110 140 170 205 250 300 340 385 435 510 570 660 740 850 960 1 070 1 160 1 230 1 290 99 130 160 190 235 290 330 375 430 510 590 680 790 930 1 060 1 230 1 350 1 450 1 540 For cables having high density polythene insulation, the values are to be multiplied by: 1.05 for columns (1) and (2) 1.06 for column (3) Table 6-32: current-carrying capacities in cables made up of three single-core cables having a rated voltage greater than 6/6 (7.2) kV and lower than or equal to 18/30 (36) kV Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. 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Industrial electrical network design guide T&D 6 883 427/AE 568 Copper PE-insulated* PVC-insulated Nominal crosssectional area (mm²) (1) (2) (3) (1) (2) EPR or XLPE-insulated (3) (1) (2) (3) 10 80 87 71 85 94 75 97 110 92 16 100 115 90 110 120 98 125 140 120 25 130 145 120 140 155 125 160 180 155 35 160 175 145 165 190 155 190 215 190 50 185 205 175 195 220 185 225 250 225 70 230 255 215 240 270 230 275 310 280 95 275 305 260 285 320 275 330 370 340 120 310 345 300 325 365 315 370 420 385 150 345 385 340 365 415 365 420 475 445 185 390 435 385 410 465 410 470 535 510 240 450 500 450 475 530 485 540 610 590 300 500 560 520 530 605 560 610 690 680 Aluminium (1) (2) (3) (1) (2) (3) (1) (2) (3) (*) 10 62 68 55 66 73 58 75 84 71 16 79 87 71 84 94 76 96 110 92 25 100 115 93 110 120 99 125 140 120 35 120 135 115 130 145 120 150 165 145 50 145 160 135 150 170 140 175 195 175 70 180 195 165 185 210 175 215 240 215 95 210 235 205 220 250 215 255 285 260 120 240 270 235 250 285 245 290 325 300 150 270 300 265 285 325 280 325 370 345 185 305 340 300 320 360 320 365 415 395 240 350 390 355 370 420 380 425 480 465 300 395 440 405 420 475 435 480 540 530 For cables having high density polythene insulation, the values are to be multiplied by: 1.05 for columns (1) and (2) 1.06 for column (3) Table 6-33: current-carrying capacities in three-core individually screened cables having a rated voltage lower than or equal to 6/10 (12) kV Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 569 Nominal crosssectional area (mm²) Copper EPR or XLPE-insulated (1) (2) (3) 16 125 140 125 25 160 175 160 35 190 210 195 50 225 250 230 70 270 305 280 95 330 370 345 120 370 420 395 150 415 465 450 185 465 525 510 240 540 610 600 Aluminium (1) (2) (3) 16 96 105 95 25 125 135 125 35 145 165 150 50 175 195 175 70 210 235 220 95 255 285 265 120 290 325 305 150 320 360 345 185 360 410 395 240 420 475 470 Table 6-34: current-carrying capacities in three-core individually screened cables having a rated voltage greater than 6/6 (7,2) kV and lower than or equal to 18/30 (36) kV Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 570 6.2.4. Thermal withstand of conductors in the event of a short circuit and determination of the cross-sectional area S 2 The thermal withstand of live conductors must be checked for the maximum short-circuit current at the origin of the cable. It is calculated using the impedance method taking into account the participation of all the network elements (motors, generators, etc., see Protection guide § 4.2). In the case of an installation with an internal generator set, the thermal withstand is established on the basis of the short-circuit current during the transient period, this approximately corresponding to the short-circuit clearance time (see Protection guide § 4.1.2). For a short-circuit time less than 5 seconds, cable heating is considered to be adiabatic; this means that the energy stored stays in the core and is not transmitted to the insulating material. The thermal calculations are then simplified. They are given below. Note: to check the thermal withstand of protective and equipotential bonding conductors, the earth fault current must be taken into account (see § 4.2.2 of the Protection guide) n general method The heating calculation results are shown by the curves in figure 6-14. They give the current density withstands δ 0 in different types of cable for a short-circuit time of one second, in relation to the cable temperature before the short circuit. The minimum conductor cross-sectional area complying with heating in the case of a short circuit is determined by the fomula: I S = sc δ Isc : maximum short-circuit current, in A δ : current density withstand, in A / mm² for a short-circuit time other than one second, we have: δ= t δ0 t : short-circuit time Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 571 Figure 6-14: short circuit in the core Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 572 n simplified method This assumes that the cable temperature before the short circuit is equal to the temperature allowed in steady-state operating conditions. In this case, the conductor cross-sectional area must meet the following condition: I S ≥ sc k t Isc : maximum short-circuit current t : short-circuit time k : coefficient the value of which is given in table 6-35 For protective conductors, the current to be taken into account is the earth fault current I f . Insulating material PVC PE XLPE EPR 115 143 74 94 Live conductors - in copper - in aluminium Protective conductors a b a b 143 115 176 143 - in aluminium 95 75 116 94 - in steel 52 _ 64 _ - in copper a b protective conductors not incorporated in cables protective conductors incorporated in cables Table 6-35: coefficient k values Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 573 6.2.5. Short-time withstand currents in cable screens with extruded synthetic insulation (determination of S3 ) In the event of a phase-to-screen short circuit, the thermal withstand resulting from the passage of the fault current I f for a time t , must not exceed the thermal withstand of the cable screen. If is the earth fault current and the method for determining its value is described in the Protection guide, paragraph 4-2. The calculation of the overcurrent permitted in the cable screens depends on what the screen is made of and the type of cable. In the absence of precise indications, the values of tables 6-37, 6-38 and 6-39 can be used. These values correspond to a screen made up of a copper band 0.1 mm thick wrapped around the insulating material with an overlap of 15 %. Table 6-36 gives, for each type of insulating material, the temperatures during normal service and at the end of overcurrents used for calculating cable screen heating. Type of insulating material Temperature on the screen during service (°C) Final temperature following overcurrent (°C) XLPE 70 250 EPR 70 250 PE 60 150 PVC 60 160 Table 6-36: temperature conditions used for the calculation Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 574 o overcurrent values permitted in cable screens with extruded synthetic insulation See tables 6-37, 6-38 and 6-39. Rated voltage Short-circuit time 6/10 (12) kV 0.5 s 1s 8.7/15 (17.5) kV 2s 0.5 s 1s 2s 12/20 (24) kV 0.5 s 1s 2s 18/30 (36) kV 0.5 s 1s 2s Conductor cross-sectional area in mm² 16 1 100 900 650 1 350 1 000 800 1 800 1 400 1 100 25 1 200 950 700 1 400 1 050 800 1 800 1 400 1 100 35 1 400 1 000 50 1 600 1 150 1 000 1 750 1 350 1 050 1 950 1 450 1 150 2 500 1 950 1 550 70 1 750 1 250 1 050 1 900 1 450 1 150 2 100 1 600 1 250 2 700 2 050 1 650 95 1 850 1 350 1 100 2 050 1 550 1 200 2 200 1 700 1 300 2 800 2 150 1 700 120 1 900 1 400 1 150 2 150 1 650 1 300 2 500 1 950 1 550 3 100 2 400 1 900 150 2 150 1 650 1 300 2 400 1 850 1 500 2 600 2 000 1 600 3 150 2 450 1 950 185 2 400 1 850 1 450 2 600 2 000 1 600 2 750 2 150 1 700 3 350 2 600 2 100 240 2 700 2 050 1 650 2 800 2 150 1 700 3 100 2 400 1 950 3 600 2 750 2 200 300 2 800 2 150 1 750 3 150 2 450 1 950 3 300 2 550 2 050 3 800 2 950 2 350 400 3 050 2 350 1 800 3 450 2 650 2 150 3 650 2 800 2 250 4 200 3 300 2 650 500 3 400 2 550 1 950 3 800 2 950 2 350 4 100 3 200 2 550 4 550 3 550 2 850 630 3 750 3 000 2 300 4 250 3 300 2 650 4 450 3 450 2 800 4 950 3 850 3 100 800 4 400 3 400 2 600 4 650 3 600 2 900 4 850 3 750 3 000 5 300 4 150 3 300 1 000 5 100 3 900 3 050 5 200 4 050 3 250 5 350 4 200 3 350 5 850 4 550 3 650 1 200 5 350 4 100 3 300 5 450 4 250 3 400 5 650 4 400 3 550 6 150 4 800 3 850 1 400 5 600 4 400 3 550 5 900 4 550 3 650 6 050 4 700 3 800 6 550 5 100 4 100 1 600 6 000 4 700 3 800 6 200 4 850 3 900 6 400 5 000 4 000 6 900 5 350 4 300 900 1 650 1 250 1 000 1 850 1 400 1 100 Table 6-37: single-core or three-core individually screened cables with XLPE or EPR insulation short-circuit current permitted in the screen (A) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. 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Industrial electrical network design guide T&D 6 883 427/AE 575 Rated voltage Short-circuit time 6/10 (12) kV 0.5 s 1s 8.7/15 (17.5) kV 2s 0.5 s 1s 2s 12/20 (24) kV 0.5 s 1s 2s 18/30 (36) kV 0.5 s 1s 2s Conductor cross-sectional area in mm² 16 800 650 490 1 000 740 560 1 200 870 660 25 900 700 510 1 000 750 570 1 200 870 660 35 1 000 750 540 1 100 800 600 1 200 880 660 50 1 100 800 580 1 150 840 640 1 250 1 000 770 1 750 1 300 70 1 300 920 700 1 350 990 760 1 450 1 100 820 1 750 1 300 1 000 95 1 350 1 000 750 1 450 1 050 820 1 550 1 150 880 2 050 1 550 1 200 120 1 450 1 050 800 1 500 1 150 860 1 650 1 200 930 2 150 1 650 1 230 150 1 550 1 100 840 1 600 1 200 910 1 700 1 300 1 000 2 250 1 700 1 300 185 1 650 1 150 900 1 700 1 250 970 2 000 1 500 1 200 2 350 1 800 1 400 240 1 800 1 450 1 100 2 000 1 550 1 200 2 150 1 650 1 250 2 650 2 050 1 600 300 2 000 1 550 1 200 2 150 1 650 1 300 2 300 1 750 1 350 2 800 2 150 1 700 400 2 300 1 750 1 400 2 600 2 000 1 550 2 650 2 050 1 600 3 000 2 300 1 800 500 2 550 1 900 1 500 2 900 2 200 1 750 3 050 2 350 1 850 3 400 2 600 2 050 630 2 750 2 050 1 550 3 000 2 300 1 800 3 150 2 400 1 900 3 500 2 650 2 050 800 3 000 2 250 1 700 3 300 2 500 2 000 3 450 2 600 2 100 3 700 2 800 2 200 1 000 3 300 2 400 1 800 3 500 2 700 2 100 3 650 2 800 2 200 3 950 3 000 2 400 1 200 3 550 2 550 1 900 3 700 2 850 2 200 3 850 2 950 2 300 4 200 3 200 2 550 1 400 3 650 2 750 2 000 3 900 3 000 2 350 4 050 3 100 2 450 4 350 3 350 2 650 1 600 3 750 2 850 2 100 4 000 3 100 2 400 4 150 3 200 2 500 4 500 3 400 2 700 990 Table 6-38: single-core or three-core individually screened cables with PE insulation short-circuit current permitted in the screen (A) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 576 Conductor crosssectional area Short-cicuit time mm² 0.5 s 1s 2s 10 1 550 1 200 980 16 1 700 1 300 1 050 25 1 950 1 450 1 200 35 2 050 1 550 1 250 50 2 150 1 600 1 300 70 2 300 1 700 1 400 95 2 550 1 900 1 550 120 2 750 2 100 1 650 150 2 900 2 200 1 750 185 3 350 2 450 2 050 240 3 500 2 650 2 200 Table 6-39: PVC-insulated three-core collectively screened cables with a rated voltage of 6/6 (7.2 kV) short-circuit current permitted in the screen (A) o example Let us consider a PE-insulated single-core cable in a 10 kV network having an earth fault current I f limited to 1 000 A. According to table 6-38, the minimum cross-sectional area of the conductor depends on the short-circuit time: - for - for - for t = 0.5 s t= 1s t= 2s , , , S min = 35 mm² S min = 95 mm² S min = 240 mm² . The cross-sectional area S3 is selected in relation to I f and the short-circuit time, which is taken to be equal to the longest time needed to clear the fault (e.g., the back-up protection time delay). Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 577 6.2.6. Checking voltage drops Voltage drops in medium voltage cables in industrial networks are in general negligible. However, it seems useful to give the calculation method able to be applied notably for very long wiring systems. For a three-phase circuit, the voltage drop (single-phase voltage) is calculated by the fomula: L ∆V = ρ1 cos ϕ + λ L sin ϕ I B S ρ1 : conductor resistivity during normal service, i.e. 1.25 times that at 20 °C ρ1 = 0.0225 Ω mm 2 / m for copper; ρ1 = 0.036 Ω mm 2 / m for aluminium L S cos ϕ : length of wiring system, in metres : conductor cross-sectional areas, in mm² IB λ : maximum design current in : power factor; in the absence of precise indications, we may take cos ϕ = 0.8 (sin ϕ = 0.6) A : reactance per unit length of the wiring system, in Ω/m . The values of λ in MV are: - 0.08 × 10 −3 Ω / m for three-core cables - 0.15 × 10 −3 Ω / m for single-core cables We define the relative voltage drop as: ∆V Vn Vn : nominal single-phase voltage Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 578 6.2.7. Practical determination of the minimum cross-sectional area of a medium voltage cable (see fig. 6-15) cable installation conditions determination of maximum design current I B determination of the cable column and overall correction factor (see tab. 8-23) equivalent current (1) Iz determination of the cross-sectional area S1 of the cable able to carry I z in standard installation conditions in relation to the type of cable, its insulation and rated voltage (see tab. 8-29 to 8-34) IB f thermal withstand S2 I sc max S3 function I f , t (see tab. 8-37 to 8-39) screen thermal withstand: If voltage drop check Isc max t k S max ( S1 , S 2 , S3 ) economic cross-sectional area possibly chosen (1) I z is an equivalent current which, in standard installation conditions, causes the same thermal effect as I B in actual installation conditions Figure 6-15: logigram for determining the minimum cross-sectional area of a medium voltage cable Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 579 6.2.8. Cable screen earthing conditions n single-core cables The passage of a current in the cable core produces an induced voltage in the screen. This voltage depends on the geometrical arrangement of the cables, the length and the current carried: 2a E0 = 0.145 × log10 × l × I d a d l I : : : : distance between cable axes (mm) mean diameter of the screen (mm) connection length (km) current carried in the core (A). For very long cables, E0 may reach dangerous values for persons. The standard recommends screen earthing at both ends when E0 is likely to exceed the limit of 50 V under steady-state operating conditions. However, screen earthing at both ends produces currents continuously circulating in the screen. For screen earthing at one end only, on occurrence of a short circuit, the potential induced on the second end may be high and cause a breakdown of the screen insulation where it is connected. The necessary precautions must therefore be taken. o calculation of the current circulating in screens earthed at both ends In balanced steady-state operating conditions (or during a three-phase short circuit), the induced voltage in screens earthed at both ends causes a three-phase current to circulate. This current is given by the formula: I0 = where Rs Xs l E0 Rs2 + Xs2 2a Xs = 0.145 × log10 × l d : screen resistance (Ω) : screen reactance (Ω) : length of cable or line Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 580 o example Let us consider a 20 kV aluminium single-core cable with a cross-sectional area of 300 mm² , with PE insulation and a length of l = 3 km , buried in soil having a resistivity of ρ = 100 Ω ⋅ m , the characteristics of which are as follows: - Icapacity = 500 A - d = 33.5 mm a = 38.5 mm Rs = 0.45 Ω / km It is installed in a network such that: - I B = 400 A Isc = 8 kA The induced voltage under steady-state operating conditions is: 2a E0 = 0.145 × log10 × I B × l = 63 V d The 50 V limit is exceeded and the screen must therefore be earthed at both ends. The circulation current in the screen is in this case: I0 = E0 Rs2 + Xs2 Rs = 1.35 Ω 2a X s = 0.145 × log10 × l = 0.157 Ω d whence Note: I0 = 46.4 A the circulation current in the screen is independent of the cable length. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 581 The induced voltage in the event of a short circuit is: 2a E0 sc = 0.145 × log10 × I sc × l = 1 260 V d The circulation current in the screen is then: I0 sc = 927 A This current must be withstood by the screen for the maximum short-circuit time. This is the case since it can withstand 1 350 A for 2 s (see table 6-38). Note: if the cable length was 2 km, the screen would be earthed at one end only. The voltage induced in the screen on occurrence of the short circuit will then be equal to 840 V. In this case it is necessary to check that the screen insulation at the point where the terminal box is located is sufficient. Evaluation of Ws losses in the screen Ws = Rs I02 for Rs = 0.45 Ω / km , l = 3 km and I0 = 46.4 A Ws = 0.45 × 3 × (46.4 )2 = 2.9 kW The losses in the core are: Wc = Rc × I B2 Rc : core resistance For an aluminium conductor with a cross-sectional area of S = 300 mm² , Rc = 0.1 Ω / km whence Wc = 0.1 × 3 × ( 400)2 = 48 kW We determine the ratio Ws =6% Wc The screen losses represent 6 % of the core losses. They must therefore be taken into account when determining the maximum current-carrying capacity of the cable. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 582 o thermal effect in the cable screens As we saw in the previous example, when the screen is earthed at both ends, the continuous circulation of induced current in the screen causes extra heating in the cable and consequently reduces its current-carrying capacity. Generally, this phenomenon is only to be taken into account for cables with a cross-sectional area greater than 240 mm². We can apply the following rule: - thin screen without armour, for S > 1 000 mm² the current-carrying capacity is reduced by 5 % - non-thin screen without armour, the current-carrying capacity is to be reduced by: . 5 % for . 10 % for 240 mm² ≤ S ≤ 800 mm² S > 800 mm² - cables with screen and armour, the current-carrying capacity is to be reduced by: . 5 % for . 10 % for . 15 % for 240 mm² ≤ S ≤ 400 mm² 500 mm² ≤ S ≤ 800 mm² S > 800 mm² n three-core cables For three-core collectively screened cables, the electromagnetic field is zero in balanced operating conditions. In normal operating conditions, there is no circulation current in the screen. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 583 6.2.9. Application example Let us determine the conductor cross-sectional area of the W1 wiring system inserted into the network illustrated in figure 6-17. The W1 wiring system is made up of three single-core three-phase 6/10 (12) kV aluminium cables with XLPE insulation, directly installed in a enclosed channel in a temperature of 35 °C. The time delay of the protection against phase-to-phase short circuits is: t = 0.2 s . 20 kV T1 1000 A S n = 10 MVA U sc = 8 % U n = 5.5 kV W1 wiring system l = 1200 m T2 S n = 630 kVA U sc = 4 % 400 V Figure 6-17: diagram of the installation Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 584 n determining the maximum design current I B The W1 wiring system only feeds the 630 kVA power transformer T2 . The current I B is thus taken to be equal to the nominal transformer current: I B = In = Sn 630 × 10 3 = 66 A = 3 Un 3 × 5.5 × 10 3 n correction factors and choice of S1 The direct installation in an enclosed channel corresponds to installation type L4 table 6-23). Column (3) in the current-carrying capacity tables must be used. (see The correction factors to be applied are: - installation method: f0 = 0.8 - ambient temperature (see table 6-24): f1 = 0.96 - group of several cables (see table 6-28): f 5 = 1 The overall correction factor is: f = 0.8 × 0.96 = 0.77 The equivalent current that the cable must be able to carry in standard installation conditions is: I I z = B = 86 A f Table 6-31 (column (3), XLPE, aluminium) gives a minimum cross-sectional area of S1 = 16 mm2 which has a current-carrying capacity of I 0 = 95 A . Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 585 (S 2 ) n checking thermal withstand Neglecting the impedance upstream of the transformer and the impedance of the transformerbusbar connection, the maximum short-circuit current at the origin of the cable is equal to the short-circuit current of the transformer. The impedance of the transformer T1 is: ( ) 5.5 × 10 3 Un2 U sc × = ZT 1 = Sn 100 10 × 10 6 2 × 8 = 0.242 Ω 100 The maximum short-circuit current is thus: Isc = 11 . Un 5.5 × 10 3 = 11 = 14.4 kA . × 3 ZT 1 3 × 0.242 (see Protection guide § 4.2.1) The cross-sectional area complying with the short-circuit requirement is: I S2 ≥ sc k t k = 94 : value of the coefficient corresponding to a XLPE-insulated aluminium conductor (see table 6-35) t = 0.2 s : short-circuit time equal to the protection time delay whence S 2 ≥ 69 mm2 The minimum cross-sectional area to be chosen is thus S 2 = 70 mm2 Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 586 n checking the cable screen thermal withstand ( S3 ) The 5.5 kV distribution network has an earthing system with a 1 000 A current limiting resistor. The fault current is then: If = Vn + IC RN (see Protection guide § 4.3.2) Vn : single-phase network voltage RN : limiting resistance I C : 5.5 kV network capacitive current ( I C = 3 j C ω Vn ) The capacitive current of an industrial network is of the order of several amps to several dozen amps and it can thus be neglected in relation to the 1 000 A limiting current. We thus have I f = 1 000 A We assume that the screen must be able to withstand the fault current for 2 seconds, in order to take into account the maximum time delay of the protection against phase-earth faults and eventual reclosing. The cross-sectional area of the conductor complying with the thermal withstand of the cable screen is then: S3 = 50 mm2 (see table 6-37) Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. 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Industrial electrical network design guide T&D 6 883 427/AE 587 n checking voltage drops The voltage drop is given by the formula: l ∆V = ρ1 cos ϕ + λl sin ϕ I B S l = 1 200 m ; S = 70 mm² ; λ = 0.15 × 10 −3 Ω / m ; I B = 66 A ; ρ1 = 0.036 Ω ⋅ mm 2 / m We assume that the cable load has a cos ϕ = 0.6 (sin ϕ = 0.8) whence 1 200 ∆V = 0.036 × 0.8 + 0.15 × 10 −3 × 1 200 × 0.6 × 66 50 ∆V = 53 V The relative voltage drop is: 53 ∆V = = 1.7 % Vn 5500 3 In spite of a very long cable length for an industrial network, the voltage drop is acceptable. n choosing the technical cross-sectional area The calculations carried out give the following cross-sectional areas: S1 = 16 mm² S 2 = 70 mm² S3 = 50 mm² The technical cross-sectional area to be chosen is thus: S = 70 mm² Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 588 6.3. Calculating the economic cross-sectional area The methods described in chapters 6-1 and 6-2 lead to the choice of technical cross-sectional areas of wiring systems, complying with the different thermal withstands, voltage drops and protection of persons. But it may be useful to take into account the economic criterion, based on the cost of investment and the operating costs, when looking for the optimum cross-sectional area. The investment cost is essentially composed of: - the cable cost, linear function of the cross-sectional area S and length L , i.e. K1 L + K2 L S - the cost of civil engineering and installation, depending on the length and regardless of the cross-sectional area in a limited interval, i.e. K3 L . The operating costs comprise: - the Joule losses in the cable - the maintenance costs. To calculate the economic cross-sectional area, only the cost of the Joule losses w relative to the wiring system is taken into account: w=n×ρ n ρ L HC × I2 S 1 000 Euros. : number of live conductors : resistivity of the live conductor during normal service, i.e. 1.25 times that at 20 °C. ρ = 0.0225 Ω mm ²/ m for copper; ρ = 0,036 × Ω mm² / m for aluminium L S I H C : : : : : cable length cross-sectional area of conductors current carried, assumed to be constant, in A number of cable operating hours (for a year H = 8 760) cost of kWh, Euro/kWh. The cost of investment and the cost of losses w do not have the same term of payment. It is necessary to change in order to carry out the sum of their values. This can be done by converting the operating costs paid at the end of consecutive years to current value, i.e. by converting them to the period in which the cable is purchased. 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Industrial electrical network design guide T&D 6 883 427/AE 589 If N (years) is the amortizement time forecast for the cable, and if the price of energy and the cable load are assumed to be constant for the entire period, the sum of converted values of Joule losses is: 1 1 1 W =w + + ....... + 2 1 + t (1 + t ) (1 + t ) N (1 + t ) N − 1 = w× N t (1 + t ) t being the forecast conversion to current value rate. t (1 + t ) w We can write W = , where A = A (1 + t ) N − 1 N The total cost is therefore: P ( S) = K3 L + K1 L + K2 L S + n ρ L 2 HC I S 1 000 × A ∂ P The function P ( S ) goes via a minimum = 0 ∂ S for a cross-sectional area of S0 = I nρHC K2 A × 1 000 For an approximate calculation we can use the following formula: S0 = KI 100 HC , mm² A where K = 2.56 for copper and 4.61 for aluminium. The value of the economic cross-sectional area to be chosen is the closest standard value to S 0 . Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE 590 n example Taking the elements from the application example of § 6.2.9: - design current I B = 66 A - energy cost: C = 0.061 Euro/kWh - aluminium conductor, K = 4.61 - conversion to current value rate of 8 % - amortizement time N = 20 years - number of operating hours H = 3 800 hours. A= 0.08 × (1 + 0.08) S0 = 20 (1 + 0.08)20 − 1 = 0.102 66 × 4.61 3 800 × 0.061 = 145 mm ² 100 0.102 The economic cross-sectional area is the closest standard value to S 0 , i.e. S = 150 mm2 . In practice, the economic cross-sectional area is often greater than the technical crosssectional area. n advantages of cable oversizing - Improved voltage quality under normal operating conditions and reduced amplitude of voltage surges during motor or other machine starting. - Presence of reserve power offering the possibility of future extensions. Publication, traduction et reproduction totales ou partielles de ce document sont rigoureusement interdites sauf autorisation écrite de nos services. The publication, translation and reproduction, either wholly or partly, of this document are not allowed without our written consent. Industrial electrical network design guide T&D 6 883 427/AE