Discuss 3 phase voltage drop tables in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

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If you have a 50A 3 phase MCB, that's 50 per phase - when calculating voltage drop using the tables do you calculate using 50A or 150A?

ie 10mm cable carrying 50A 3 phase over 30m

VD=3.8x50x30 / 1000 = 5.7V

or

VD=3.8x150x30 / 1000 = 17.1V

I think it has to be the first but am a bit mixed up, need to brush up on some 3 phase theory I think.
 
Voltage drop(in volts)
Run Length (ft) X Conductor Current (amps) X Factor
----------------------------------------------------------------
1000

Voltage drop (in percent)
Voltage Drop (volts) X 100
----------------------------------------
Circuit Voltage (volts)


IIRCC
When calculating the voltage drop in 3 phase circuits, do the same as in the single phase but the voltage drop must be multiplied by 0.866





Below was taken from >>>Clicky<<< and has some useful tables and formulas - worth having a look at
Voltage Drop Calculations

The voltage drop of any insulated cable is dependent upon the route length under consideration (in meters), the required current rating (in amperes) and the relevant total impedance per unit length of the cable. The maximum impedance and voltage drop applicable to each cable at maximum conductor temperature and under a.c. conditions is given in the tables. For cables operating under dc conditions, the appropriate voltage drops may be calculated using the formula:
2 x route length x current x resistance x 10¯³.
The values detailed in the tables are given in m/V/Am, (volts/100 per ampere per metre), and the nominal
maximum acceptable volt drop specified by the IEE Regulations is 2.5% of the system voltage, i.e. 0.025 x 415
= 10.5 volts for 3 phase working or 0.025 x 240 = 6.0 volts for single phase working.
Consider a 3 phase system
The requirement may be for a load of 100A to be transmitted over a route length of 150m, the cable to be
clipped to the wall and close protection provided. The rating tables in the IEE Regulations indicate that a
35mm copper conductor PVC SWA PVC cable would be suitable for the loading required, but the voltage drop
must be checked.
Volt drop = Y x current x length
= 1.1 x 100 x 150 millivolts
= 1.1 x 100 x 150 volts/1000
= 16.5 volts
where Y = value from tables in mV/A/m Unless a particular value of voltage drop, acceptable to the user, is
specified, the IEE Regulations figure of 10.5 volts must be adhered to.
Thus: total volt drop = 10.5 volts
10.5 = Y x 100 x 150
Therefore Y = 10.5/100 x 150
= 0.7/1000 volts/ampere/meters
Reference to the voltage drop tables indicates that the cable size with a voltage drop of 0.7/1000 V/A/m
(0.7mV/A/m) OR LESS is a 70mm copper conductor.
Therefore, in order to transmit a 3 phase current of 100A per phase over a route length of 150m, with a total
voltage drop equal to or less than the statutory maximum 10.5 volts, the use would require a
70mm (cu.) multicore PVC.
Conversely
The user may have 150m of 35mm (Cu.) multicore PVC cable and require to know what maximum current
rating can be applied without exceeding the allowable voltage drop. The method is exactly the same as above,
viz:total drop = 16.6
= YxAxM
= 1.1 x A x 150/1000
from the tables Y = 1.1mV/A/m
=1.1/1000V/A/m
therefore A = 10.5 x 1000/1.1.x 150
=64 amperes
From the foregoing, it is apparent that knowing any two values of Y, A or m, the remaining, unknown value can
readily be calculated.
The advice is always available to check, clarify or suggest the most suitable size and type of cable for any particular, specified requirements.
 
Last edited by a moderator:
Thanks, I'd seen that on the net but got lost before the "100A per phase" bit! so sticking with my original example you're saying

10mm cable carrying 50A 3 phase over 30m

VD= (3.8 x 50A x 3phases x 30m / 1000) x 0.866 = 14.8V drop over all 3 phases so 14.8V / 400V = 3.7%

When calculated per phase

VD=3.8 x 50A x 30m / 1000 = 5.7V so 5.7V/230V = 2.5%

What am I missing? Something (or everything!)'s got to be wrong here
 
my interpretation, right or wrong is to calc. as for single phase e.g.your example VD=3.8 x 50 x 30/1000 =5.7 then x 0.866 = 4.94v
 
All the above can only be used with balanced 3ph supply with no 3rd harmonics if you have current in the neutral or harmonics caused by electronics or discharge lighting etc then thing get very complicated but must not be ignored or your calculations will be way off mark.
 
good comment. must admit I've not had much to do with unbalanced loads or harmonics lately
 
my interpretation, right or wrong is to calc. as for single phase e.g.your example VD=3.8 x 50 x 30/1000 =5.7 then x 0.866 = 4.94v

So that would mean the total VD over a given distance would be less for a 3 phase than for a single phase? I suppose that might make sense as there are 3 conductors and the voltage over all phases obviously isn't 3 times as much. Where has this 0.866 figure come from? Can't find it in the book. Will have a look at GN1 later see if there's anything in there.
 
I would recommend you purchase the IEE Electrical Installation Design Guide as this explains all the above and much more and with a bit of background is easy to interpret.
 
Just bumping some threads in the Electrical Wiring Theories and Regulations forum category here on our Electricians Forums. If this specific topic isn't current, just ignore it, it'll drop off the list in no time. However, if it's something you'd like to contribute to, feel free to reply and bring it back into current discussion.
 

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