3 phase voltage drop calculation?

[bbudding]

Hi!

Once you've calculated the voltage drop in Volts - you divide by 400V for three phase right?

For example: three phase 30kW appliance, 10mm 90 degree SWA, 45 meter run.

30,000/sqrt3/400 = 43.3A = Ib
from the table 4E4B, a 10mm cable gives 4.7
So then 43.3 x 4.7 x the 45 meter run gives = 9157.95 mV
gives 9.157 V
9.157/400 = 0.02289 = 2.289% voltage drop

Is that right?

Thanks!

Ben

 

[bbudding]

Hi!

Once you've calculated the voltage drop in Volts - you divide by 400V for three phase right?

For example: three phase 30kW appliance, 10mm 90 degree SWA, 45 meter run.

30,000/sqrt3/400 = 43.3A = Ib
from the table 4E4B, a 10mm cable gives 4.7
So then 43.3 x 4.7 x the 45 meter run gives = 9157.95 mV
gives 9.157 V
9.157/400 = 0.02289 = 2.289% voltage drop

Is that right?

Thanks!

Ben

Sorry - 4.0 from the table! But - still - how's the working?!

 

[Speeka]

Sorry - 4.0 from the table! But - still - how's the working?!

You do not divide the voltage by the the drop. The drop as you have calculated it, is accurate & you must now compare that it's within the limits permissible, which is 5% for power circuits ( 3% for lighting).
5% of 400v is 20v. Your circuit is within the limits with enough to spare for any additional Vd if your circuit is supplied from a secondary distribution board.

 

[pc1966]

Hi!

Once you've calculated the voltage drop in Volts - you divide by 400V for three phase right?

Yes

For example: three phase 30kW appliance, 10mm 90 degree SWA, 45 meter run.

30,000/sqrt3/400 = 43.3A = Ib

Correct

from the table 4E4B, a 10mm cable gives 4.7

No, it should be 4.0 from column 4 as you are looking for the three-phase drop.

So then 43.3 x 4.7 x the 45 meter run gives = 9157.95 mV
gives 9.157 V
9.157/400 = 0.02289 = 2.289% voltage drop

Is that right?

Almost, you would be looking at:
43.3 * 4.0 * 45 = 7.79V
Then percent drop is 7.79V / 400V = 1.95%

 

[Speeka]

Correct

No It isn't correct! You do not divide the Line to Line voltage by the voltage drop!

Why would you? The voltage drop is per individual cable & since when do you run 3 phases through 1 cable? You have already factored in the additional phases when working out Ib. (design current)

 

[pc1966]

No It isn't correct! You do not divide the Line to Line voltage by the voltage drop!

Why would you? The voltage drop is per individual cable & since when do you run 3 phases through 1 cable? You have already factored in the additional phases when working out Ib. (design current)

Because the VD values given in the related column are for that case.

The alternative is to consider 3 pairs of single phase cables, for which you would have:

VD = 43.3 x 4.7 x 45 = 9.157 V = 9.157 / 230 = 3.98%

Now as it is actually a balanced 3-phase load the neutral currents cancel, so you only get drop in the line conductors and thus half the percentage drop in total = 3.98 / 2 = 1.99%

Given that column 3 and column 4 in Table 4E4B are only quoted to 1 significant digit (from 0.1/4.0 = 2.5% etc) you see the approaches both give the same answer of around 1.9-something (as 1.95 * 1.025 = 1.99875 etc)