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

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 in Conventional_method 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

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

	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