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Worked example of cable calculation

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General rules of electrical installation design
Connection to the MV utility distribution network
Connection to the LV utility distribution network
MV and LV architecture selection guide for buildings
LV Distribution
Protection against electric shocks and electrical fires
Sizing and protection of conductors
LV switchgear: functions and selection
Overvoltage protection
Energy Efficiency in electrical distribution
Power Factor Correction
Power harmonics management
Characteristics of particular sources and loads
PhotoVoltaic (PV) installation
Residential premises and other special locations
ElectroMagnetic Compatibility (EMC)
Measurement

Contents


Worked example of cable calculation

(see Fig. G69)

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 G69 below. The results of a computer study for the circuit from transformer T1 down to the cable C7 is reproduced on Figure G70. 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.

Fig. G69Example of single-line diagram

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

Fig. G70Partial results of calculation carried out with Ecodial software (Schneider Electric). The calculation is performed according to Cenelec TR50480 and IEC 60909

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:

I_b=\frac{630 \times 10^3}{\sqrt 3 \times 400}=909\,A 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:

R=\frac{18.51 \times 5}{240\times 2}=0.19\,m\Omega (cable resistance: 18.51 mΩ.mm2/m at 20 °C)

X = 0.08 / 2 \times 5 = 0.2\,m\Omega (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:

I_b=\frac{310 \times 10^3}{\sqrt 3 \times 400 \times 0.85}=526\,A

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:

R=\frac{18.51\times 20}{120\times 2}=1.54\,m\Omega (cable resistance: 18.51 mΩ.mm2/m at 20 °C)

X = 0.08/ 2 \times 20 = 1.6\,m\Omega (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:

I_b=\frac{140 \times 10^3}{\sqrt 3 \times 400 \times 0.85}=238\,A

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:

R=\frac{18.51\times 5}{95}=0.97\,m\Omega (cable resistance: 18.51 mΩ.mm2/m)

X = 0.08 \times 5 = 0.4\,m\Omega (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

Fig. G71Example of short-circuit current evaluation

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):

S_{PE}=\frac{\sqrt {I^2 . t} }{k}

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:

S_{PE}=\frac{\sqrt {I^2 . t} }{k}=\frac {21000 \times \sqrt {0.5} }{143}=104\,mm^2

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:

L_{max}=\frac{0.8 \times U_0 \times S_{ph} }{\rho \times \left ( 1 + m \right )\times I_a }

where m = Sph / SPE

For circuit C3, this gives:

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

(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

Fig. G72Voltage drop introduced by the different cables