Worked example of cable calculation

Worked example of cable calculation

(see Fig. G65)
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 G65 below. The results of a computer study for the circuit from transformer T1 down to the cable C7 is reproduced on Figure G66. 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. G65: Example of single-line diagram

Calculation using software Ecodial

General network characteristics				Number of poles and protected poles	4P4d
Earthing system		TN-S		Tripping unit	Micrologic 2.3
Neutral distributed		No		Overload trip Ir (A)	510
Voltage (V)		400		Short-delay trip Im / Isd (A)	5100
Frequency (Hz)		50		Cable C3
Upstream fault level (MVA)		500		Length	20
Resistance of MV network (mΩ)		0.0351		Maximum load current (A)	509
Reactance of MV network (mΩ)		0.351		Type of insulation	PVC
Transformer T1				Ambient temperature (°C)	30
Rating (kVA)		630		Conductor material	Copper
Short-circuit impedance voltage (%)		4		Single-core or multi-core cable	Single
Transformer resistance RT (mΩ)		3.472		Installation method	F
Transformer reactance XT (mΩ)		10.64		Phase conductor selected csa (mm2)	2 x 95
3-phase short-circuit current Ik3 (kA)		21.54		Neutral conductor selected csa (mm2)	2 x 95
Cable C1				PE conductor selected csa (mm2)	1 x 95
Length (m)		5		Cable voltage drop ΔU (%)	0.53
Maximum load current (A)		860		Total voltage drop ΔU (%)	0.65
Type of insulation		PVC		3-phase short-circuit current Ik3 (kA)	19.1
Ambient temperature (°C)		30		1-phase-to-earth fault current Id (kA)	11.5
Conductor material		Copper		Switchboard B6
Single-core or multi-core cable		Single		Reference	Linergy 800
Installation method		F		Rated current (A)	750
Number of layers		1		Circuit-breaker Q7
Phase conductor selected csa (mm2)		2 x 240		Load current (A)	255
Neutral conductor selected csa (mm2)		2 x 240		Type	Compact
PE conductor selected csa (mm2)		1 x 120		Reference	NSX400F
Voltage drop ΔU (%)		0.122		Rated current (A)	400
3-phase short-circuit current Ik3 (kA)		21.5		Number of poles and protected poles	3P3d
Courant de défaut phase-terre Id (kA)		15.9		Tripping unit	Micrologic 2.3
Circuit-breaker Q1				Overload trip Ir (A)	258
Load current (A)		860		Short-delay trip Im / Isd (A)	2576
Type		Compact		Cable C7
Reference		NS1000N		Length	5
Rated current (A)		1000		Maximum load current (A)	255
Number of poles and protected poles		4P4d		Type of insulation	PVC
Tripping unit		Micrologic 5.0		Ambient temperature (°C)	30
Overload trip Ir (A)		900		Conductor material	Copper
Short-delay trip Im / Isd (A)		9000		Single-core or multi-core cable	Single
Tripping time tm (ms)		50		Installation method	F
Switchboard B2				Phase conductor selected csa (mm2)	1 x 95
Reference		Linergy 1250		Neutral conductor selected csa (mm2)	-
Rated current (A)		1050		PE conductor selected csa (mm2)	1 x 50
Circuit breaker Q3				Cable voltage drop ΔU (%)	0.14
Load current (A)		509		Total voltage drop ΔU (%)	0.79
Type		Compact		3-phase short-circuit current Ik3 (kA)	18.0
Reference		NSX630F		1-phase-to-earth fault current Id (kA)	10.0
Rated current (A)		630

Fig. G66: Partial results of calculation carried out with Ecodial software (Schneider Electric). The calculation is performed according to Cenelec TR50480

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: [math]\displaystyle{ I_B=\frac{630 \times 10^3}{\sqrt 3 \times 420}=866 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 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:

[math]\displaystyle{ R=\frac{23.7 \times 5}{240\times 2}=0.25 m\Omega }[/math]  (cable resistance: 23.7 mΩ.mm²/m)

[math]\displaystyle{ X = 0,08 \times 5 = 0.4 m\Omega }[/math] (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:  [math]\displaystyle{ I_B=\frac{300 \times 10^3}{\sqrt 3 \times 400 \times 0.85}=509 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 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:

[math]\displaystyle{ R=\frac{23.7\times 20}{95\times 2}=2.5 m\Omega }[/math] (cable resistance: 23.7 mΩ.mm²/m)

[math]\displaystyle{ X = 0.08 \times 20 = 1.6m \Omega }[/math] (cable reactance: 0.08 mΩ/m)

Dimensioning circuit C7

Circuit C7 supplies one 150kW load with cos φ = 0.85, so the total load current is:  [math]\displaystyle{ I_B=\frac{150 \times 10^3}{\sqrt 3 \times 400 \times 0.85}=255 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 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:

[math]\displaystyle{ R=\frac{23.7\times 5}{95}=1.25m\Omega }[/math]  (cable resistance: 23.7 mΩ.mm²/m)

[math]\displaystyle{ X = 0.8 \times 5 = 0.4m\Omega }[/math] (cable reactance: 0.08 mΩ/m)

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

Circuit components	R (mΩ)	X (mΩ)	Z (mΩ)	Ikmax (kA)
Upstream MV network, 500MVA fault level (see Fig. G34)	0,035	0,351
Transformer 630kVA, 4% (see Fig. G35)	2.9	10.8
Cable C1	0.23	0.4
Sub-total	3.16	11.55	11.97	20.2
Cable C3	2.37	1.6
Sub-total	5.53	13.15	14.26	17
Cable C7	1.18	0.4
Sub-total	6.71	13.55	15.12	16

Fig. G67: Example of short-circuit current evaluation

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:

[math]\displaystyle{ S_{PE}=\frac{\sqrt {I^2 . t} }{k} }[/math]

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:

[math]\displaystyle{ S_{PE}=\frac{\sqrt {I^2 . t} }{k}=\frac {20200 \times \sqrt {0.5} }{143}=100 mm^2 }[/math]

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 and F40, 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:

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

[math]\displaystyle{ L_{max}=\frac{0.8 \times 230 \times 2 \times 95}{23.7 \times 10^{-3}\times \left ( 1 + 2 \right )\times 630\times 11 } = 71 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 G28, for balanced three-phase circuits, motor power normal service (cos φ = 0.8).
The results are summarized on figure G68:

	C1	C3	C7
c.s.a.	2 x 240mm2	2 x 95mm2	1 x 95mm2
∆U per conductor (V/A/km) see Fig. G28	0.22	0.43	0.43
Load current (A)	866	509	255
Length (m)	5	20	5
Voltage drop (V)	0.48	2.19	0.55
Voltage drop (%)	0.12	0.55	0.14

Fig. G68: Voltage drop introduced by the different cables

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