Hi I've a few questions regarding volt drop. Apologies if its a little longwinded but I'd like to get my head around everything.
An example scenario is a section board with a TP sub main to lighting and power board (DB2)
Calculations for circuit :
Supply Voltage = 400 Volts
Ib - Design current = 63 Amps
Protective Device Type = 63A MCB type B (BS EN 60898)
Cable Type : Thermosetting ARMOURED 90°C - Multicore
Length of run of cable = 100 metres
Lets assume there are no correction factors.
I've decided to go with a 16mm SWA so from TABLE 4E4B mV/A/m = = 2.5
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 2.5 x 100 x 63 ) / 1000
Voltdrop = 15.75 Volts
Now my first set of questions are:
- Do I use 63amp for the volt drop (as this is what the cable COULD carry at some point)? or use actual figures from DB2 which would be less?
- If I use actual values how do I work this out when its unbalanced loads?
- I've read that as a rule of thumb only 1.5% VD should be lost for sub mains. So 400/100 * 1.5 = 6v
So if this 1.5% rule is correct I deceide to put a bigger cable in, let's say a 70mm
From TABLE 4E4B mV/A/m = = 0.87
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 0.87 x 100 x 63 ) / 1000
Voltdrop = 5.48 Volts
So far does this look correct?
Now at DB2 I have a few circuits
1/L1 - (20 x 33w LED panels. So 660/230 = 2.9A
1/L2 - (30 x 33w LED panels. So 990/230 = 4.3A
1/L3 - (25 x 33w LED panels. So 825/230 = 3.6A
I've been told that when working VD from a 400v to 230v supply I use the following correction factors.
(x 0.866) Assumes there is some current in the neutral.
(÷ 2 Assumes a balanced circuit with no current in the neutral.
So its an unbalanced load (x 0.866)
Volt drop for the lighting circuit would be 230/100 * 3 = 6.9v
Circuit 1 is 20m long
Again assuming there are no correction factors:
TABLE 4D5 For 1 mm²: mV/A/m = 44
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 44 x 20 x 2.9 ) / 1000
Voltdrop = 2.55 Volts
Total volt drop from orogin to end light
Sub main = 5.48 x 0.866 = 4.7v
Lighting circuit = 2.55
Total VD = 4.7 + 2.55 = 7.25v (too high)
Next size up
TABLE 4D5 For 1.5 mm²: mV/A/m = 29
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 29 x 20 x 2.9 ) / 1000
Voltdrop = 1.68 Volts
Total volt drop from orogin to end light
Sub main = 5.48 x 0.866 = 4.7v
Lighting circuit = 1.68
Total VD = 4.7 + 1.68 = 6.38v
Does this look correct?
A few other general questions:
- I've read if its designed at only 30%. so 3A for a circuit thats 10A mcb no correcion factors apply. Is this correct?
- Also how is a socket circuit designed? Becuase its very unlikely that 32Amps will be used,
An example scenario is a section board with a TP sub main to lighting and power board (DB2)
Calculations for circuit :
Supply Voltage = 400 Volts
Ib - Design current = 63 Amps
Protective Device Type = 63A MCB type B (BS EN 60898)
Cable Type : Thermosetting ARMOURED 90°C - Multicore
Length of run of cable = 100 metres
Lets assume there are no correction factors.
I've decided to go with a 16mm SWA so from TABLE 4E4B mV/A/m = = 2.5
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 2.5 x 100 x 63 ) / 1000
Voltdrop = 15.75 Volts
Now my first set of questions are:
- Do I use 63amp for the volt drop (as this is what the cable COULD carry at some point)? or use actual figures from DB2 which would be less?
- If I use actual values how do I work this out when its unbalanced loads?
- I've read that as a rule of thumb only 1.5% VD should be lost for sub mains. So 400/100 * 1.5 = 6v
So if this 1.5% rule is correct I deceide to put a bigger cable in, let's say a 70mm
From TABLE 4E4B mV/A/m = = 0.87
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 0.87 x 100 x 63 ) / 1000
Voltdrop = 5.48 Volts
So far does this look correct?
Now at DB2 I have a few circuits
1/L1 - (20 x 33w LED panels. So 660/230 = 2.9A
1/L2 - (30 x 33w LED panels. So 990/230 = 4.3A
1/L3 - (25 x 33w LED panels. So 825/230 = 3.6A
I've been told that when working VD from a 400v to 230v supply I use the following correction factors.
(x 0.866) Assumes there is some current in the neutral.
(÷ 2 Assumes a balanced circuit with no current in the neutral.
So its an unbalanced load (x 0.866)
Volt drop for the lighting circuit would be 230/100 * 3 = 6.9v
Circuit 1 is 20m long
Again assuming there are no correction factors:
TABLE 4D5 For 1 mm²: mV/A/m = 44
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 44 x 20 x 2.9 ) / 1000
Voltdrop = 2.55 Volts
Total volt drop from orogin to end light
Sub main = 5.48 x 0.866 = 4.7v
Lighting circuit = 2.55
Total VD = 4.7 + 2.55 = 7.25v (too high)
Next size up
TABLE 4D5 For 1.5 mm²: mV/A/m = 29
Voltdrop = (mV/A/m x Length x Design current) / 1000
Voltdrop = ( 29 x 20 x 2.9 ) / 1000
Voltdrop = 1.68 Volts
Total volt drop from orogin to end light
Sub main = 5.48 x 0.866 = 4.7v
Lighting circuit = 1.68
Total VD = 4.7 + 1.68 = 6.38v
Does this look correct?
A few other general questions:
- I've read if its designed at only 30%. so 3A for a circuit thats 10A mcb no correcion factors apply. Is this correct?
- Also how is a socket circuit designed? Becuase its very unlikely that 32Amps will be used,
