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electricitycity
Protection of feeders alghoritm
| Question | Answer | 3 |
|---|---|---|
| choose the type of cable (conductive material, insulation material, the type of conductor) | 1 | |
| choose the type of the installation according to the environment | 2 | |
| choose the cross section according to the load current Ib | 3 | |
| verify the voltage drop dU [V] | 4 | dU=k*Z*Ib |
| choose the method of the installation in accordance with the load current | 5 | |
| Ambient air temperature, not 30 degrees? choose the correction factor k1 | 6 | choose the correction factor k1 |
| similar cross sections of cables? choose the correction factor k2 | 7 | choose the correction factor k2 |
| single layer? multi-core cable? choose the correction factor k2 | 8 | choose the correction factor k2 |
| calculate ktot=k1*k2 | 9 | ktot=k1*k2 |
| calculate the value of the current I’b=Ib/ktot | 10 | I’b=Ib/ktot |
| determine the cross section of the cable with I0 ≥ I’b, according to the method of installation, the insulation and conductive material and the number of live conductors | 11 | Io ≥ I’b ? |
| calculate the actual cable carrying capacity | 12 | Iz=Io*ktot Io-cable carrying capacity |
| Current load less then actual cable carrying capacity | 13 | I'b<In<Iz |
| If third harmonic is contained reduction factor must be used | 14 | K3 (koefficient) |
| dU. use coefficient k equal ---for single-phase and two-phase systems; ---- for three-phase systems; | 4.1 | - 2 for single-phase and two-phase systems; -1,732 for three-phase systems; |
| dU. calculate Ib [A] is the load current; if no information are available, the cable carrying capacity Iz shall be considered; | 4.2 | Ib=P/k*Ur*cosf |
| dU. calculate Z (full wire resistanse) | 4.3 | Z=L*(r*cosf+x*sinf)/n r -phase active resistance per unit of length of the cable at 80 °C [Ω/km]; x- -phase REactive resistance per unit of length of the cable at 80 °C [Ω/km]; n-quantity of parallel wires on one phase L - cable length [km]. |
| dU. calculate sinf | 4.4 | sinf=√(1-cos²f) |
| calculate dU in % and determine with allowed value | 4.5 finish | dU%=dU*100/Ur Ur-rated voltage |
| Calculation of short-circuit current at end of the conductor | 15 | Ikmin |
| Ikmin with non-distributed neutral conductor (единая нейтраль) | 15.1а | Ikmin=(0,8*Ur*ksec*kpar)/(1,5*ρ*2L/S) |
| with distributed neutral conductor (распределённая нейтраль) | 15.1b | Ikmin=(0,8*Uo*ksec*kpar)/[1,5*ρ*(1+m)*L/S] |
| Ur-supply voltage [V]; Uo-phase to earth supply voltage [V]; | ||
| ρ is the resistivity at 20 °C of the material of the conductors in Ωmm2/m | - 0.018 for copper; - 0.027 for aluminium; | |
| L - length of the protected conductor [m]; S - cross ection of the conductor [mm2]; ksec-correction factor which takes into account the reactance of the cables with cross section larger than 95 mm2 |
Created by:
Marina Shvedova
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