07_sizing Busbar Trunkings V1 [Compatibility Mode]

Setting of the rated current (Ib)

Set the nominal current (In):

Of each machines in cable:– In Lathe = 3 kW -> In = 6 A

– In Milling machine = 11 kW -> In = 20 A

Of a several machines in Canalis:– In = 3 x 11 + 3 x 3 = 42 kW -> In = 80 A

Take into account the Simultaneity factor (Ks)

Canalis: In = 80 x 0.7 = 56 A

N b o f K s

Schneider Electric / E&S – Bernard Genoulaz – 14/12/2005 - FR 12

Rq: for a machine which works continuously / for lighting K=1.

Take into account the expansion factor: generally, in the case of an industrial installations, it needs at least a factor of 1.2 (= leave a margin of 20%)

Canalis: In = 56 x 1.2 = 68 APower sum

N b o fm a c h in e s

K s

2 e t 3 0 ,94 e t 5 0 ,86 à 9 0 ,71 0 à 3 9 0 ,6

4 0 a n d m o r e 0 ,5

Choice of the electrical distribution architecture

Choice of materials according to external influences

Power sum

Sizing of wires

Chronology of a design


OverloadsPermissive current

Conductors are always sized from the value of the permissive current

It corresponds to the maximal current which can go through the conductor

Beyond this value, the insulation qualities are jeopardised

premature ageing


A cable which is submitted to a 10%overload current will have a reduced lifetime up to 50%

Sizing

OverloadsPermissive current

The permissive current of a cable is:

Iz = (K x Ib) / (f1 x f2 x f3 x f4 x f5 x n x fs)

The nominal rated current is:

Inc = (K x Ib) / (f1 x f3)

– With:


– In: Nominal current of the circuit

– K: Type of protection (fuse / circuit breaker)

– f1: Ambient temperature

– f2: Number of conductors loaded in a circuit

– f3: Reference method and fixing mode

– n: Number of parallel cables

– fs: Symmetrical factor

– f4: Number of layers

– f5: Several circuit grouping (joined cables)Sizing

Constant

parameters

whenever

Factors

depending on the evolutions

Iz = (K x In) / (f1 x f2 x f3 x f4 x f5 x fs x n)

Protection by circuit breaker K = 1

Ambient temperature = 30°C f1 = 1

Neutral not loaded f2 = 1

T

R R

Example 1: 5 machines - P = 9 kW/ machine(Cos ϕ = 0.87 / Ib = 15 A)


Cable with 3 conductors (Méthode E) + cable fixed on cable tray on a horizontal

way (Mode 13) f3 = 1

1 layer f4 = 1

5 non joined circuits f5 = 0,1

No parallel cables n et fs = 1

Iz = (1 x 15) / (1x1x1x1x1x1x1) = 15 A

R R R

S theoric = 1.08 mm² S chosen = 1.5 mm²

Example 2: 9 machines - P = 9 kW / machine(Cos ϕ = 0.87 then Ib = 15 A)



Ambient temperature de 30°C f1 = 1


Cable with 3 conductors (Méthode E) + cable fixed on cable tray on a horizontalway (Mode 13) f3 = 1R R R

TD

R R R RR

R


way (Mode 13) f3 = 1

1 layer f4 = 1

9 non joined circuits f5 = 0.72


Iz = (1 x 15) / (1x1x1x0,8x0,72x1x1) = 21 A

S theoric = 1.83 mm² S chosen = 2.5 mm²

R R R R

4 machines more:

9 joined circuits f5 = 0.72

Sz1 = 2.5 mm²

Setting up 5 machines in t0:

5 non joined circuits f5 = 1

Sz0 = 1.5 mm²

Evolution from 5 to 9 machines


Results:

the 5 circuits installed in t0 are overloaded :

no conformity to installation standards

Consequences :

the circuits installed in t0 have to be replaced by 2.5 mm²

material and labour extra-costs and a long production stopping, ….

if the installation remains the same, the cable lifetime will considerably

be reduced Fire risks ….

Overloads Application: Canalis 3 lathes + 3 milling machines

Inc = (K x Ib) / (f1 x f3)

Choice of the range of Canalis:

Ib = 68 A

IP < 55

KN is chosen


KN is chosen

Choice of the rating of the Canalis:



Edgewise mounting = Standard f3 = 1

Inc = (1 x 68) / (1 x 1) = 68 A

KN 100 A is chosenSizing

Overloads Application: cable supplying Canalis





Cable with 3 conductors (Method E) + cable fixed on cable tray


Cable with 3 conductors (Method E) + cable fixed on cable tray

on a horizontal way (Mode 13) f3 = 1

1 layer f4 = 1

2 joined circuits f5 = 1


Iz = (1 x 100) / (1 x 1 x 1 x 1 x 1 x 1 x 1) = 100 A

S chosen = 16 mm²

Sizing

Ligthing Otheruse

Installation power directly supplied by alow voltage public distribution network

3% 5%

Voltage drops

The voltage drop between the origin and any point of use mustn’t

exceed the values below:


Sizing

Setting of the voltage drop

In cable (it needs a calculation):

u = b (ρL/S x Ib x cos ϕ)

– u: voltage drop in V

– b: 1 in single phase and 2 in 3 phases

– ρ: resistivity of phase conductor in ?.mm²/m

– L: length of the cable

– S: section of the phase conductor in mm²

– Ib: rated current in A

– cos ϕ: power factor (use 0.8 when there is no indication)


Sizing

In Canalis (catalogue data):

u = Val catalogue * Ib * L / 100

Example : Page "CharactéristicsKN from 40 until 100 A"

KN 40 A KN 63 A KN 100 A

With Cos ϕ = 0,8 V/100m/A 0.389 0.163 0.075

With Cos ϕ = 0,9 V/100m/A 0.433 0.180 0.080

With Cos ϕ = 1 V/100m/A 0.473 0.191 0.080

Checking of the voltage dropApplication: cable of the lathe

u = b (ρL/S x Ib x cos ϕ)

3 phases machines b = 2

Copper cable ρcopper = 22.5 x 10-3 Ω.mm²/m

– Rq: ρaluminium = 36 x 10-3 Ω.mm²/m

LLathe (The worst case) = 40 m

Section of a phase conductor 1.5 mm²


Sizing

Rated current 8 A

cos ϕ = 0.8 (No specification)

u = 2 (22.5 x 10-3 x 40 / 1.5 x 8 x 0.8) = 7.68 V

∆U = u x 100 / U

u = 7.68 V

U = 400 V

∆∆∆∆U = 0.768 x 100 / 400 = 1.92 %

Checking of the voltage dropApplication: Canalis

Supplying cable of the busway:

u = 2 (22.5 x 10-3 x 18 / 16 x 100 x 0.8) = 4.05 V

Canalis:

Canalis KN 100 A, Cos ϕ = 0.8 Val cat 0.075 V / 100 m/A

Ib = 100 A (the worst case - it’s not the actual rated current)

L = 20 m


L = 20 m

u = 0.075 x 100 x 20 /100 = 1.5 V

Supplying cable of the lathe:

u = 2 (22.5 x 10-3 x 6 /1.5 x 8 x 0.8) = 1.15 V

∆Ucable + canalis = u x 100 / U

u = 4.05 + 1.5 + 1.15 = 6.7 V

U = 400 V

∆∆∆∆U = 6.7 x 100 / 400 = 1.68 %

Sizing

Short circuits currents (Isc)

Generally, Canalis is over sized and can support the short circuits.

Particular cases (building > 5000 m²):

Transformers in parallel

Canalis of low power near a transformer of high power


Sizing

Practice …


Power distribution

60 m

45 m

L: Lathe 4 kW

M: Milling machine 4 kW

B: Buffing wheel 0,75 kW

C: Circular saw 4 kW

DT: Drilling/Tapper machine 2.2 kW


45 m

Exercise

D: Drilling machine 1.5 kW

DT: Drilling/Tapper machine 2.2 kW

DM: Drilling/milling machine 3 kW

RD: Radial drilling machine 7.5 kW

V: Vice filer 2.2 kW

Fluo Tubes 2 x 58 W

45 m

60 m

Ligthing distribution


Structure: 1 beam every 2.50 m

Exercise

Practice

Power


Power - Choice of the materials

Calculation of the current needed by each machine:

I = P / √3 x U x cosϕ– L: Lathe = 4 kW I = 7.2 A

– M: Milling machine = 4 kW I = 7.2 A

– B: Buffing wheel = 0.75 kW 1.4

– C: Circular saw= 4 kW I = 7.2 A

– DT: Drilling / Tapper machine = 2,2 kW I = 4 A


– DT: Drilling / Tapper machine = 2,2 kW I = 4 A

– DM: Drilling / Milling machine = 3 kW I = 5.4 A

– RD: Radial drilling machine = 7.5 kW I = 13.5 A

– V: Vice filer = 2.2 kW I = 4 A

– D: Drilling machine = 1.5 kW I = 2.7 A

Exercise

L1 L2 L3 L4

L0

20 m

Power - Tracking of the runsSolution 1


L5 L6 L7 L8

L0

25 m

Exercise

Power - Power sumSolution 1

IL1 = (3 IEL + 3 IF) x Ks x 1.2 = (3x4 + 3x7.2)x0.7x1.2 = 28 A

IL2 = IL1 = 28 A

IL3 = (3 IT + 3 IF) x Ks x 1.2 = (3x7.2 + 3x7.2)x0.7x1.2 = 36 A

IL4 = IL3 = 36 A

IL5 = (6 IF + 2 IM) x Ks x 1.2 = (6x7.2 + 2x1.4)x0.7x1.2 = 39 A

IL6 = IL5 = 39 A


L6 L5

IL7 = (3 IP + 3 IF) x Ks x 1.2 = (3x2.7 + 3x7.2)x0.7x1.2 = 25 A

IL8 = (2 Itr + IPT + IPF + IPR) x Ks x 1.2 = (2x7.2 + 4 + 5.4 + 13.5)x0.8x1.2 = 36 A

IL0 = (IL1 + IL2 + IL3 + IL4 + IL5 + IL6 + IL7 + IL8 )x Ks x 1.2 =

(2x28 + 2x36 + 2x39 + 25 + 36 )x0.7x1.2 = 225 A

Exercise

Power - Choice of the buswaysSolution 1

Workshop: IPbusway > 20

Inc = (K x Ib) / (f1 x f3)

Protection of runs by circuit breaker K = 1


Edgewise mounting = standard f3 = 1


Exercise


Inc L1/L2 = 28 A KNA 40 A



Inc L7 = 25 A KNA 40 A


Inc L0 = 225 A KSA 250 A

Power - Choice of Canalis elements (1/2) Solution 1

Runs L1 to L8:

Canalis KNA 40 A, 3L+N+PE - length 20 m:– 1 End feed box 63 A

– 7 straight elements 40 A - 3m - 1 tap-off outlet / m

– 8 universal brackets

Connectors KNA 3L+N+PE,for fuses:


Exercise

Connectors KNA 3L+N+PE,for fuses:– Runs L1/L2: 6 connectors 25 A

– Runs L3/L4: 6 connectors 25 A

– Runs L5/L6: 8 connectors 25 A

– Run L7: 6 connectors 25 A


Power - Choice of Canalis Elements (2/2) Solution 1

Run L0 :

Canalis KSA 250 A, 3L+N+PE - length 55 m:– 1 end feed box 250 A + 1 end cover

– 11 straight elements 250 A - 5m - 2 tap-off outlets / m


Connectors KSA 3L+N+PE, for circuit-breakers:


Exercise


– 8 connectors 50 A

L0

Power - Tracking of the runs Solution 2


L1 L2 L3 L4

45 m

Exercise

Power - Power sumSolution 2

IL1 = (3 IEL + 9 IF + 2 IM)x Ks x 1.2 = (3x4 + 9x7.2 + 2x1.4)x0.6x1,2 = 57 A

IL2 = IL1 = 57 A

IL3 = (3 IT + 6 IF + 3 IP) x Ks x 1.2 = (3x7.2 + 6x7.2 + 3x2.7)x0.6x1.2 = 52 A


Exercise

IL4 = (3 IF + 3 IT + 2 Itr + IPT + IPF + IPR)x Ks x 1.2 =

(3x7.2 + 6x7.2 + 2x7.2 + 4 + 5.4 + 13.5)x0.6x1.2 = 74 A

IL0 = (IL1 + IL2 + IL3 + IL4 )x Ks x 1.2 = (57x2 + 52 + 74)x0.8 x1,2 = 230 A

Power - Choice of the buswaysSolution 2

Work-shop: IPbusway > 20

Inc = (K x Ib) / (f1 x f3)

Protection of runs by circuit-breaker K = 1








Inc L4 = 74 A KNA 100 A

Inc L0 = 230 A KSA 250 A

Exercise


Runs L1, L2 and L3:

Canalis KNA 63 A, 3L+N+PE - Length 41 m:– 1 end feed box 63 A

– 13 straight elements 63 A - 3 m - 1 tap-off outlet / m


Connectors KNA 3L+N+PE, for fuses:– Run L1: 14 connectors 25 A





Run L4:

Canalis KNA 100 A, 3L+N+PE length 41 m:– 1 end feed box 100 A

– 13 straight elements 100 A de 3 m - 1 tap-off outlet / m

– 14 universal outlets

Connectors KNA 3L+N+PE, for fuses:– Run L1: 11 connectors 25 A

Exercise


Run L0:

Canalis KSA 250 A, 3L+N+PE length 55 m:– 1 end feed box 250 A

– 11 straight elements 250 A - 5m - 2 tap-off outlets / m




Connectors KSA 3L+N+PE, for circuit-breakers:– 4 connectors 100 A

Exercise

Practice

Power

Lighting


Lighting - Tracking of the runsSolution 1


Exercise

Lighting - Power sumSolution 1

Iluminaire = (Pluminaire+Pballast) / (U0xcos ϕ) = (2*58+5)/(230*0.86)

Iluminaire = 0.65

Iéclairage = 12 ILuminaire x Ks x 1.2 = (12x0.65)x1x1.2 = 9,36 A

I = 14 I x Ks x 1.2= (14x9,36)x1x1.2 = 157 A


Ifeeder = 14 Iéclairage x Ks x 1.2= (14x9,36)x1x1.2 = 157 A

Exercise

Lighting - Choice of the buswaysSolution 1

Work-shop: IPbusways > 20

Inc = (K x Ib) / (f1 x f3)

Protection of runs by circuit-breakers K = 1





Inc Llighting = 9,36 A KBA 25 A - 1L+N+PE

Inc Lfeeder = 157 A KSA 160 A

Exercise

Lighting - Choice of the trackingSolution 2


Exercise

Documents

07_sizing Busbar Trunkings V1 [Compatibility Mode]