voltage drop calculations

Note that these calculation represent under the following considerations .
1. Maximum voltage drop 2.5% from Authority’s Supply
2. Working Voltage 380 volt / 3 PH / 50 Hz.
3. Ambient Air Temperature 50 º C
4. Cable to be installed fully in Air
5. Cable type DUCAB / XLPE / SWA / PVC
6. Route length 70 Mtr.
7. Two cables will connected parallel in order to carry out one chiller load .
8. Cable Should Cary out Maximum 370.5 Amp. / phase / at 380 volt as per electrical data sheet for our chiller attached.

A - Cable selection for above condition.

- From table 13 . Rating factor for XLPE at 50 º C = 0.82, which mean that we have to consider that our current capacity should be 370.5 / 0.82 = 452Amp. at 30º C .
- Appropriate volt drop ( mV ) equation as follows

Vd x 1000
mV = in mV / Amp. / Meter
I x L

- Where Vd : Maximum acceptable volt drop ( in Volt )
- I : Current per phase ( in Amp. )
- L : Route Length ( in Meter )


2.5% x 380 x 1000
mV = = 0.3002 mV / Amp. / Meter
452 x 70

- From table 30. for voltage drop less than 0.3002 and in same time representing the 452Amp. Load required, it will be seen that at voltage drop of 0.255 mV/amp./meter and current of 463Amp., Cable selection could be 4c x 185mm2 .

B – voltage drop calculation in case of using cable size 4c x 300mm2.

- In case of using cable with size 4c x 300mm2 and if we consider a current capacity in range of 452Amp. From table 30. , mV value will be 0.176 mv / amp./ meter ,
- Using same equation above


mV x L x I
Vd = =
1000

0.176 x 70 x 465
Vd = = 5.7288 volt
1000
Vd = (5.7288 volt ) it is 1.50% from total supply , and it is compliance with regulation 525 of the 16th edition of the IEE wiring regulation and with local Authority’ rules