View source for Worked example of cable calculation

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== Worked example of cable calculation  ==

(see {{FigRef|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 {{FigureRef|G69}} below. The results of a computer study for the circuit from transformer T1 down to the cable C7 is reproduced on {{FigureRef|G70}}. This study was carried out with [http://www.schneider-electric.com/products/ww/en/5100-software/5110-electrical-design-software/61013-ecodial-advanced-calculation Ecodial] (a Schneider Electric software).

This is followed by the same calculations carried out by the simplified method described in this guide.

{{FigImage|DB431011_EN|svg|G69|Example of single-line diagram}}

== Calculation using software Ecodial ==

{{tb-start|id=Tab1243|num=G70|title=Partial results of calculation carried out with Ecodial software (Schneider Electric). The calculation is performed according to Cenelec TR50480 and IEC 60909|cols=5}}
{| class="wikitable"
|-
! colspan="2" | General network characteristics 
| rowspan="51" |
! colspan="2" |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
|-
! colspan="2" | 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
|-
! colspan="2" | Cable C1 
|1-phase-to-earth fault current Ief (kA)
|18
|-
| Length (m) 
| 5 
! colspan="2" |Distribution Board B6
|-
| Maximum load current (A) 
| 909 
|Reference
|Prisma Plus G
|-
| Type of insulation 
| PVC 
|Rated current (A)
|630
|-
| Ambient temperature (°C) 
| 30 
! colspan="2" |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 
! colspan="2" |Cable C7
|-
| Earth fault current Ief (kA) 
| 18 
|Length
|5
|-
! colspan="2" | 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
|-
! colspan="2" | 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
|-
! colspan="2" | Circuit breaker Q3 
| rowspan="9" colspan="2" |
|-
| 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>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. [[Recommended simplified approach for cable sizing#Tab1193|{{FigureRef|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>R=\frac{18.51 \times 5}{240\times 2}=0.19\,m\Omega</math> (cable resistance: 18.51 mΩ.mm<sup>2</sup>/m at 20 °C)

<math>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>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. [[Recommended simplified approach for cable sizing#Tab1193|{{FigureRef|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>R=\frac{18.51\times 20}{120\times 2}=1.54\,m\Omega</math> (cable resistance: 18.51 mΩ.mm<sup>2</sup>/m at 20 °C) 

<math>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>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. [[Recommended simplified approach for cable sizing#Tab1193|{{FigureRef|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>R=\frac{18.51\times 5}{95}=0.97\,m\Omega</math> (cable resistance: 18.51 mΩ.mm<sup>2</sup>/m) 

<math>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 {{FigRef|G71}})

{{tb-start|id=Tab1244|num=G71|title=Example of short-circuit current evaluation|cols=5}}
{| class="wikitable"
|-
! Circuit components 
! R (mΩ) 
! X (mΩ) 
! Z (mΩ) 
! Ikmax (kA)
|-
| Upstream MV network, 500MVA fault level (see {{FigRef|G36}}) 
| 0,035 
| 0,351 
| 
|
|-
| Transformer 630kVA, 4% (see {{FigRef|G37}}) 
| 2.90
| 10.8 
| 
| 
|-
| Cable C1 
| 0.19
| 0.20
|
| 
|-
| {{tb-HC2}} | '''Sub-total'''
| {{tb-HC2}} | '''3.13''' 
| {{tb-HC2}} | '''11.4'''
| {{tb-HC2}} | '''11.8''' 
| {{tb-HC2}} | '''21'''
|-
| Cable C3 
| 1.54 
| 0.80 
| 
| 
|-
| {{tb-HC2}} | '''Sub-total''' 
| {{tb-HC2}} | '''4.67''' 
| {{tb-HC2}} | '''12.15''' 
| {{tb-HC2}} | '''13.0''' 
| {{tb-HC2}} | '''19'''
|-
| Cable C7 
| 0.97 
| 0.40 
| 
| 
|-
| {{tb-HC2}} | '''Sub-total'''
| {{tb-HC2}} | '''5.64 '''
| {{tb-HC2}} | '''12.55''' 
| {{tb-HC2}} | '''13.8''' 
| {{tb-HC2}} | '''18'''
|}

=== The protective conductor ===

Generally, for circuits with phase conductor c.s.a. Sph &ge; 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 [[Sizing of protective earthing conductor#Tab1219|{{FigRef|G59}}]]):

<math>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>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).  [[TN system - Protection against indirect contact#Conventional_method|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>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>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 [[Calculation of voltage drop in steady load conditions#Tab1200|{{FigureRef|G30}}]], for balanced three-phase circuits, motor power normal service (cos φ = 0.8).

The results are summarized on {{FigRef|G72}}:

The total voltage drop at the end of cable C7 is then: '''0.73 %'''.

{{tb-start|id=Tab1245|num=G72|title=Voltage drop introduced by the different cables|cols=3}}
{| class="wikitable"
|-
!
! C1 
! C3 
! C7
|-
| c.s.a.
| 2 x 240mm<sup>2</sup> 
| 2 x 120mm<sup>2</sup> 
| 1 x 95mm<sup>2</sup>
|-
| ∆U per conductor&nbsp;(V/A/km)<br>see {{FigRef|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
|}