Worked example of cable calculation

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. G69 – Example of single-line diagram

Calculation using software Ecodial

Fig. G70 – Partial results of calculation carried out with Ecodial software (Schneider Electric). The calculation is performed according to Cenelec TR50480 and IEC 60909

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)

Fig. G71 – Example of short-circuit current evaluation

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 %.

Fig. G72 – Voltage drop introduced by the different cables

	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