Voltage Drop Calculation

[Debfam]

I'm connecting 24 pieces of 240V, 250W and 3A halogen lamp as a street lights. The cable to be used is 450m of 4x10mm2 meaning 8 halogens will be connected to each phase. therefore what will be the voltage drop. I need the calculation.

 

[Dennis Alwon]

Voltage Drop is a function of distance so without that info we cannot help. Here is a simple online calculator http://www.elec-toolbox.com/calculators/voltdrop.htm

 

[jumper]

Google says your nominal voltage is 230/400V. Is this correct.

230V, 250W, and 3A for a halogen fixture does not compute mathmatically.

 

[mivey]

Voltage Drop is a function of distance so without that info we cannot help.

450m

 

[Debfam]

Google says your nominal voltage is 230/400V. Is this correct.

230V, 250W, and 3A for a halogen fixture does not compute mathmatically.

230/400 is correct. Is there any help

 

[Ingenieur]

Are the fixtures evenly spaced?
450/24 = 18 m apart

a 3 ph 240/400 4 wire circuit
each ph 8 fixtures to neutral
each fixture 250/240 or 1.1A

 

[topgone]

I'm connecting 24 pieces of 240V, 250W and 3A halogen lamp as a street lights. The cable to be used is 450m of 4x10mm2 meaning 8 halogens will be connected to each phase. therefore what will be the voltage drop. I need the calculation.

If the load is at the end, the voltage drop would be 6.5%. but since you mentioned they're strret lights, it should be lower than that.

 

[Ingenieur]

Rough estimate
d between fixtures on a given phase 200'
Z = 0.6 ohm /1000'
i = 1.2 / fixture
branch....fixtures
1....8
2....7
...
8....1

V drop per branch = d/1000 x Z x fixt qty x i
V drop total = branch 1 + branch 2 + .....branch 8
= 200/1000 x 0.6 x (8 x 1.2 + 7 x 1.2 + 6 x 1.2 ..... + 1 x 1.2)
= 0.12 x (9.6 + 8.4 + 7.2 .... + 1.2)
= 0.12 x (43.2) = 5.2 V
x 2 for run out/return
11V
4.5%

I estimated the 10 sq mm Z from 7 and 8 awg

 

[Dennis Alwon]

450m

I thought that was cable size--doh

 

[Phil Corso]

Debfam,

Try the solution presented in the thread ID 13648, dealing with "Evenly-Spaced" Loads!

Regards, Phil Corso

 

[mivey]

I thought that was cable size--doh

You don't want to try pulling that!

 

[Julius Right]

Since 240 V phase-to-neutral it is actually 415/sqrt(3) I think you are in a British Standard rule. So you have to consult BS7671 [IET Wiring Regulations].
According to BS7671 Table 4D2B Voltage drop per A per m for 4*10 mm^2 copper conductors, PVC insulated cable the maximum voltage drop is 3.8 mV/A/m.
I agree with Ingenieur : if the fixtures would be even spaced you could consider only a half of voltage drop. So DV=3.8*8.333*450/1000/2=7.125 V [DV%=7.125/415*100=1.72%].

 

[Phil Corso]

DebFam...

If you (or anyone else) would like a solution, contact me via the "Notification" function!

Phil

 

[topgone]

Since 240 V phase-to-neutral it is actually 415/sqrt(3) I think you are in a British Standard rule. So you have to consult BS7671 [IET Wiring Regulations].
According to BS7671 Table 4D2B Voltage drop per A per m for 4*10 mm^2 copper conductors, PVC insulated cable the maximum voltage drop is 3.8 mV/A/m.
I agree with Ingenieur : if the fixtures would be even spaced you could consider only a half of voltage drop. So DV=3.8*8.333*450/1000/2=7.125 V [DV%=7.125/415*100=1.72%].

1.72% is too low a voltage drop, IMO.

Following your idea of using symmetry in loading, why not analyze just three lamp positions (complete three-phase, balanced loading) and solve for that group voltage drop and then multiply by 8 (24 lamps divided by 3 = 8)?