Worked example of cable calculation
Worked example of cable calculation
(see Fig. G68)
The installation is supplied through a 630 kVA transformer. The process requires a high degree of supply continuity and part of the installation can be supplied by a 250 kVA standby generator. The global earthing system is TN-S, except for the most critical loads supplied by an isolation transformer with a downstream IT configuration.
The single-line diagram is shown in Figure G68 below. The results of a computer study for the circuit from transformer T1 down to the cable C7 is reproduced on Figure G69. This study was carried out with Ecodial (a Schneider Electric software).
This is followed by the same calculations carried out by the simplified method described in this guide.
Calculation using software Ecodial
| General network characteristics | Cable C3 | |||
|---|---|---|---|---|
| Earthing system | TN-S | Length | 20 | |
| Neutral distributed | No | Maximum load current (A) | 518 | |
| Voltage (V) | 400 | Type of insulation | PVC | |
| Frequency (Hz) | 50 | Ambient temperature (°C) | 30 | |
| Upstream fault level (MVA) | 500 | Conductor material | Copper | |
| Resistance of MV network (mΩ) | 0.035 | Single-core or multi-core cable | Single | |
| Reactance of MV network (mΩ) | 0.351 | Installation method | F31 | |
| Transformer T1 | Phase conductor selected csa (mm2) | 2 x 120 | ||
| Rating (kVA) | 630 | Neutral conductor selected csa (mm2) | 2 x 120 | |
| Short-circuit impedance voltage (%) | 4 | PE conductor selected csa (mm2) | 1 x 120 | |
| Load-losses (PkrT) (W) | 7100 | Cable voltage drop ΔU (%) | 0.459 | |
| No-load voltage (V) | 420 | Total voltage drop ΔU (%) | 0.583 | |
| Rated voltage (V) | 400 | 3-phase short-circuit current Ik3 (kA) | 21.5 | |
| Cable C1 | 1-phase-to-earth fault current Ief (kA) | 18 | ||
| Length (m) | 5 | Distribution Board B6 | ||
| Maximum load current (A) | 909 | Reference | Prisma Plus G | |
| Type of insulation | PVC | Rated current (A) | 630 | |
| Ambient temperature (°C) | 30 | Circuit breaker Q7 | ||
| Conductor material | Copper | Load current (A) | 238 | |
| Single-core or multi-core cable | Single | Type | Compact | |
| Installation method | 31F | Reference | NSX250B | |
| Number of layers | 1 | Rated current (A) | 250 | |
| Phase conductor selected csa (mm²) | 2 x 240 | Number of poles and protected poles | 3P3d | |
| Neutral conductor selected csa (mm²) | 2 x 240 | Tripping unit | Micrologic 5.2 E | |
| PE conductor selected csa (mm²) | 1 x 240 | Overload trip Ir (A) | 238 | |
| Voltage drop ΔU (%) | 0.124 | Short-delay trip Im / Isd (A) | 2380 | |
| 3-phase short-circuit current Ik3 (kA) | 21.5 | Cable C7 | ||
| Earth fault current Ief (kA) | 18 | Length | 5 | |
| Circuit breaker Q1 | Maximum load current (A) | 238 | ||
| Load current (A) | 909 | Type of insulation | PVC | |
| Type | Masterpact | Ambient temperature (°C) | 30 | |
| Reference | MTZ2 10N1 | Conductor material | Copper | |
| Rated current (A) | 1000 | Single-core or multi-core cable | Single | |
| Number of poles and protected poles | 4P4d | Installation method | F31 | |
| Tripping unit | Micrologic 5.0X | Phase conductor selected csa (mm²) | 1 x 95 | |
| Overload trip Ir (A) | 920 | Neutral conductor selected csa (mm²) | 1 x 95 | |
| Short-delay trip Im / Isd (A) | 9200 | PE conductor selected csa (mm²) | 1 x 95 | |
| Tripping time tm (ms) | 50 | Cable voltage drop ΔU (%) | 0.131 | |
| Switchboard B1 | Total voltage drop ΔU (%) | 0.714 | ||
| Reference | Prisma Plus P | 3-phase short-circuit current Ik3 (kA) | 18.0 | |
| Rated current (A) | 1000 | 1-phase-to-earth fault current Ief (kA) | 14.2 | |
| Circuit breaker Q3 | ||||
| Load current (A) | 518 | |||
| Type | Compact | |||
| Reference | NSX630F | |||
| Rated current (A) | 630 | |||
| Number of poles and protected poles | 4P4d | |||
| Tripping unit | Micrologic 5.3 E | |||
| Overload trip Ir (A) | 518 | |||
| Short-delay trip Im / Isd (A) | 1036 | |||
The same calculation using the simplified method recommended in this guide
Dimensioning circuit C1
The MV/LV 630 kVA transformer has a rated voltage of 400 V. Circuit C1 must be suitable for a current of:
[math]\displaystyle{ I_b=\frac{630 \times 10^3}{\sqrt 3 \times 400}=909\,A }[/math] per phase
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 31F.
Each conductor will therefore carry 455 A. Figure G21 indicates that for 3 loaded conductors with PVC insulation, the required c.s.a. is 240 mm².
The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 5 metres, are:
[math]\displaystyle{ R=\frac{18.51 \times 5}{240\times 2}=0.19\,m\Omega }[/math] (cable resistance: 18.51 mΩ.mm2/m at 20 °C)
[math]\displaystyle{ X = 0.08 / 2 \times 5 = 0.2\,m\Omega }[/math] (cable reactance: 0.08 mΩ/m, 2 cables in parallel)
Dimensioning circuit C3
Circuit C3 supplies two loads, in total 310 kW with cos φ = 0.85, so the total load current is:
[math]\displaystyle{ I_b=\frac{310 \times 10^3}{\sqrt 3 \times 400 \times 0.85}=526\,A }[/math]
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 263 A. Figure G21 indicates that for 3 loaded conductors with PVC insulation, the required c.s.a. is 120 mm².
The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 20 metres, are:
[math]\displaystyle{ R=\frac{18.51\times 20}{120\times 2}=1.54\,m\Omega }[/math] (cable resistance: 18.51 mΩ.mm2/m at 20 °C)
[math]\displaystyle{ X = 0.08/ 2 \times 20 = 1.6\,m\Omega }[/math] (cable reactance: 0.08 mΩ/m, 2 cables in parallel)
Dimensioning circuit C7
Circuit C7 supplies one 140kW load with cos φ = 0.85, so the total load current is:
[math]\displaystyle{ I_b=\frac{140 \times 10^3}{\sqrt 3 \times 400 \times 0.85}=238\,A }[/math]
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 238 A. Figure G21 indicates that for 3 loaded conductors with PVC insulation, the required c.s.a. is 95 mm².
The resistance and the inductive reactance for a length of 5 metres is:
[math]\displaystyle{ R=\frac{18.51\times 5}{95}=0.97\,m\Omega }[/math] (cable resistance: 18.51 mΩ.mm2/m)
[math]\displaystyle{ X = 0.08 \times 5 = 0.4\,m\Omega }[/math] (cable reactance: 0.08 mΩ/m)
Calculation of short-circuit currents for the selection of circuit-breakers Q1, Q3, Q7
(see Fig. G71)
| Circuit components | R (mΩ) | X (mΩ) | Z (mΩ) | Ikmax (kA) |
|---|---|---|---|---|
| Upstream MV network, 500MVA fault level (see Fig. G36) | 0,035 | 0,351 | ||
| Transformer 630kVA, 4% (see Fig. G37) | 2.90 | 10.8 | ||
| Cable C1 | 0.19 | 0.20 | ||
| Sub-total | 3.13 | 11.4 | 11.8 | 21 |
| Cable C3 | 1.54 | 0.80 | ||
| Sub-total | 4.67 | 12.15 | 13.0 | 19 |
| Cable C7 | 0.97 | 0.40 | ||
| Sub-total | 5.64 | 12.55 | 13.8 | 18 |
The protective conductor
Generally, for circuits with phase conductor c.s.a. Sph ≥ 50 mm², the PE conductor minimum c.s.a. will be Sph / 2.
The proposed c.s.a. of the PE conductor will thus be 1x240 mm² for circuit C1, 1x120 mm² for C3, and 1x50 mm² for C7.
The minimum c.s.a. for the protective earth conductor (PE) can be calculated using the adiabatic method (formula given in Fig. G59):
[math]\displaystyle{ S_{PE}=\frac{\sqrt {I^2 . t} }{k} }[/math]
For circuit C1, I = 21 kA and k = 143.
t is the maximum operating time of the MV protection, e.g. 0.5 s
This gives:
[math]\displaystyle{ S_{PE}=\frac{\sqrt {I^2 . t} }{k}=\frac {21000 \times \sqrt {0.5} }{143}=104\,mm^2 }[/math]
A single 120 mm² conductor is therefore largely sufficient, provided that it also satisfies the requirements for fault protection (indirect contact), i.e. that its impedance is sufficiently low.
Fault protection (protection against indirect contact)
For TN earthing system, the minimum value of Lmax is given by phase to earth fault (highest impedance). Conventional_method details the calculation of typical phase-to-earth fault and maximum circuit length calculation.
In this example (3-phase 4-wire circuit), the maximum permitted length of the circuit is given by the formula:
[math]\displaystyle{ L_{max}=\frac{0.8 \times U_0 \times S_{ph} }{\rho \times \left ( 1 + m \right )\times I_a } }[/math]
where m = Sph / SPE
For circuit C3, this gives:
[math]\displaystyle{ L_{max}=\frac{0.8 \times 230 \times 2 \times 120}{23.7 \times 10^{-3}\times \left ( 1 + 2 \right )\times 630\times 11 } = 90\,m }[/math]
(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 in Figure G30, for balanced three-phase circuits, motor power normal service (cos φ = 0.8).
The results are summarized on Fig. G72:
The total voltage drop at the end of cable C7 is then: 0.73 %.
| C1 | C3 | C7 | |
|---|---|---|---|
| c.s.a. | 2 x 240mm2 | 2 x 120mm2 | 1 x 95mm2 |
| ∆U per conductor(V/A/km)
see Fig. G30 |
0.22 | 0.36 | 0.43 |
| Load current (A) | 909 | 526 | 238 |
| Length (m) | 5 | 20 | 5 |
| Voltage drop (V) | 0.50 | 1.89 | 0.51 |
| Voltage drop (%) | 0.12 | 0.47 | 0.13 |
