Voltage drop calculation format

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VOLTAGE DROP CALCULATION FORMAT

hi my friends,
is anybody have voltage drop calculation format?

AND THERE BEEN SOME CHANGING OF OPINIONS AND ADVICES AS FOLLOW

1ST REPLY
Shabu there's a few ways of calculating volt drop, the vast majority provide you with an "average" assuming the load is at the furthest point of the cable.

Much depends on the load and whether it is evenly spread over the circuit.

If you look around or follow the FIA guides you can get some much more accurate formulas than just the "length x mV/A/mt x current".

To be honest its quite simple to knock up a spreadsheet to allow you to work out all the volt drops in each leg of cable to take account variable lengths of run and load at each point and be very accurate


2ND REPLY

Hi

INTRESTING QUESTION

I work it out this way

1 mm cable will carry max load of 5amp over for 1ooo meter

at start of voltage it will be 24o volt at end will be 234 volt

for every 1oo meter u will loose 0.6 v

but ather factors has effect ,local weather temp,Frequency,thickness of outer insullation ....etc
and how much the copper pure

3RD REPLY
Sorry ECG how do you work that out ?

1mm cable is approx 18.1 ohms per 1000m and Vdrop = IR......

or 44mv/A/m

so (44 x 5A x 1000)/1000......

......doesn't seem to add up ?

4TH REPLY
Hi ESP

SORRY ITS MY MISTAKE ITS FOR EVERY 100METER

its mentioned in some cables manufacturers data sheets

also its basic rule when u design wiring for any electrical circuits

u need to determine max load

POWER=v x I

Pardon me Iam getting old to do my math.

regards
Sorry ECG how do you work that out ?

1mm cable is approx 18.1 ohms per 1000m and Vdrop = IR......

or 44mv/A/m

so (44 x 5A x 1000)/1000......

......doesn't seem to add up ?
[/quote]



Can any one help old timer like me for more specifiec details

 

[sparks1234]

Use the guide in the regs if you have a fixed load, amps x distance x volt drop (mVolts)

 

[trebor]

VOLTAGE DROP CALCULATION FORMAT

hi my friends,
is anybody have voltage drop calculation format?

AND THERE BEEN SOME CHANGING OF OPINIONS AND ADVICES AS FOLLOW

1ST REPLY
Shabu there's a few ways of calculating volt drop, the vast majority provide you with an "average" assuming the load is at the furthest point of the cable.

Much depends on the load and whether it is evenly spread over the circuit.

If you look around or follow the FIA guides you can get some much more accurate formulas than just the "length x mV/A/mt x current".

To be honest its quite simple to knock up a spreadsheet to allow you to work out all the volt drops in each leg of cable to take account variable lengths of run and load at each point and be very accurate


2ND REPLY

Hi

INTRESTING QUESTION

I work it out this way

1 mm cable will carry max load of 5amp over for 1ooo meter

at start of voltage it will be 24o volt at end will be 234 volt

for every 1oo meter u will loose 0.6 v

but ather factors has effect ,local weather temp,Frequency,thickness of outer insullation ....etc
and how much the copper pure

3RD REPLY
Sorry ECG how do you work that out ?

1mm cable is approx 18.1 ohms per 1000m and Vdrop = IR......

or 44mv/A/m

so (44 x 5A x 1000)/1000......

......doesn't seem to add up ?

4TH REPLY
Hi ESP

SORRY ITS MY MISTAKE ITS FOR EVERY 100METER

its mentioned in some cables manufacturers data sheets

also its basic rule when u design wiring for any electrical circuits

u need to determine max load

POWER=v x I

Pardon me Iam getting old to do my math.

regards
Sorry ECG how do you work that out ?

1mm cable is approx 18.1 ohms per 1000m and Vdrop = IR......

or 44mv/A/m

so (44 x 5A x 1000)/1000......

......doesn't seem to add up ?

Can any one help old timer like me for more specifiec details[/QUOTE]

Volt Drop Calcs

Method 1 Refer to BS7671appendix4 and get the mV/A/m figure for the relevant cable

Vd = design current x mV/A/m x length/1000

10mm2 = 4.4Mv/A/m 40 amp load 50mtrs = 40 x 4.4 x 50/1000 = 8.8V
percentage = 8.8 x 100/230 = 3.82%

Method 2 miiliohms per metre corrected for temperature if required x length/ 1000 = ohms

then multiply the load by the resistance = volts

1.83 milliohms x 2 (line and neutral) x ct say 1.20 x 50/1000 = 0.21 ohms

40amps x 0.21 = 8.4V

NB 1.20 is the cf for 20 to 70 degrees conductor temperature for t/plastic pvc eg t&e

 

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Thanks trober

very good explain