E
ecservices
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
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
