Warehouse design, lights, cable calculations

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Warehouse Front view
Warehouse Top view

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Front entrance- shutter doors
Metal containment

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Metal containment
With Lights

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With connections:

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Creating and inserting the blocks:

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Paper Space Mode:

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Design sequence
1. Determine load characteristics, load centres, maximum current demands of load centres.
2. Determine supply charcateristics and agree tariffs with electricity company.
3. Position main and sub-distribution boards.
4. Prepare installation outline of a distribution system.
5. Design distribution system:
a) Estimate maximum current(Ib) demands of each distribution board.
b) Select protective devices and rating(In).
c) Select cable type and size (s).
d) Calculate the voltage drop.
e) Calculate prospective fault currents.
f) Calculate earth fault loop impedances.
g) Select earthing and protective conductor size (s).
6. Size final circuits:
a) Determine circuit design current (Ib).
b) Select protective devices and rating(In).
c) Select cable type and size (s).
d) Calculate the voltage drop.
e) Calculate prospective fault currents.
f) Calculate earth fault loop impedances.
g) Select earthing and protective conductor size (s).
7. Check isolation and switching requirements.
8. Carry out final assessment.
Cable selection:
Nr. Circuit Lenght Protective device Cable
1. Outside lights
1L1
28m 6A Type B MCB 1.0mm2
line;
1.0mm2
cpc
2. 1L2, 400W
Lights
16A 2.5mm2
line;
1.5mm2
cpc
3. 1L3, 400W
Lights
16A 2.5mm2
line;
1.5mm2
cpc
4. 2L1, Fluorescent
Lights 38W
2.5mm2
line;
1.5mm2
cpc
5. 2L2, Fluorescent
Lights 38W
2.5mm2
line;
1.5mm2
cpc
6. 2L3, Fluorescent
Lights 38W
2.5mm2
line;
1.5mm2
cpc

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Calculation example:
Outside lights 1L1
1. Calculate design current from KW:
Ib= P/ U0=1000W/230V= 4.347A
Select protective device rating: In> Ib => In=6A
2. Protection against overload and short circuit:
It≥In/(CgCaCiCf)= 6/ (1.03x 0.6x 1x 1)= 9.708 A
Using the information contained in Appendix 4 of BS 7671, the method of installation is C
from Table 4A2.
Ca= 1.03 from Table 4B1 for 25°C and thermoplastic cable
Cg= 0.6 from Table 4C1
Ci=1 as no thermal insulation exists
Cf= 1 as the device is not to BS 3036
Using Table 4D1A BS7671, select cable size 1mm2
.
3. Cable selection for voltage drop:
Vd=(mV/A/m) x Ib x L / 1000= 44 x 4.35 x 28/ 1000= 5.36 V < 6.9 V (3% Vd for light )
4. Determining values of R1+ R2:
Using Table I1, OSG (r1+r2) = 36.2 for 1mm2
Using Table I3, OSG Cr(correction factor)= 1.20
R1+R2= (r1+r2) x L x Cr /1000= 36.2 x 28 x 1.20 / 1000= 1.216 Ω
Zs= Ze + (R1+R2) = 0.35 + 1.216= 1.566 Ω
Calculated Zs < Max Zs
1.56Ω < 7.67 Ω => disconnection times are satisfied.
From Table 41.3, BS 7671, Zs= 7.67 Ω (for 6A type B MCB)