Worked example of cable calculation

The same calculation using the simplified method recommended in this guide

• Dimensioning circuit C1

The MV/LV 630 kVA transformer has a rated no-load voltage of 420 V. Circuit C1 must be suitable for a current of:

 

 

Two single-core PVC-insulated copper cables in parallel will be used for each phase.These cables will be laid on cable trays according to method F.
Each conductor will therefore carry 433A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 240mm².
The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 5 metres, are:

 

                                         (cable resistance: 22.5 mΩ.mm2/m)

X = 0,08 x 5 = 0,4 mΩ (cable reactance: 0.08 mΩ/m)

• Dimensioning circuit C3

Circuit C3 supplies two 150kW loads with cos φ = 0.85, so the total load current is:

 

 

 

Two single-core PVC-insulated copper cables in parallel will be used for each phase. These cables will be laid on cable trays according to method F.
Each conductor will therefore carry 255A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 95mm².
The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 20 metres, are:

 

 

• Dimensioning circuit C7

Circuit C7 supplies one 150kW load with cos φ = 0.85, so the total load current is:

 

 

 

One single-core PVC-insulated copper cable will be used for each phase. The cables will be laid on cable trays according to method F.
Each conductor will therefore carry 255A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 95mm².
The resistance and the inductive reactance for a length of 20 metres is:

 

                                           (cable resistance: 22.5 mΩ.mm2/m)

                                           (cable reactance: 0.08 mΩ/m)

 

 

• Calculation of short-circuit currents for the selection of circuit-breakers Q1, Q3, Q7 (seeFig. G67)

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Fig. G67:Example of short-circuit current evaluation

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• The protective conductor

When using the adiabatic method, the minimum c.s.a. for the protective earth conductor (PE) can be calculated by the formula given in Figure G58:

 

 

For circuit C1, I = 20.2kA and k = 143.
t is the maximum operating time of the MV protection, e.g. 0.5s
This gives:

 

 

A single 120 mm² conductor is therefore largely sufficient, provided that it also satisfies the requirements for indirect contact protection (i.e. that its impedance is sufficiently low).
Generally, for circuits with phase conductor c.s.a. Sph ≥ 50 mm2, the PE conductor minimum c.s.a. will be Sph / 2. Then, for circuit C3, the PE conductor will be 95mm², and for circuit C7, the PE conductor will be 50mm2.

• Protection against indirect-contact hazards

 

 

For circuit C3 of Figure G65, Figures F41 andF40, or the formula given page F25 may be used for a 3-phase 4-wire circuit.
The maximum permitted length of the circuit is given by:

(The value in the denominator 630 x 11 is the maximum current level at which the instantaneous short-circuit magnetic trip of the 630 A circuit-breaker operates).
The length of 20 metres is therefore fully protected by “instantaneous” over-current devices.

• Voltage drop

The voltage drop is calculated using the data given inFigure G28, for balanced three-phase circuits, motor power normal service (cos φ = 0.8).
The results are summarized on figure G68:

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Fig. G68:Voltage drop introduced by the different cables

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The total voltage drop at the end of cable C7 is then: 0.77%.

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