calculation question

[stv292]

hi

the osg say's i can use 56m of cable of 1.0/1.0 installation method 103,101 e.tc. for lighting, on a radial at 6A. the mV/A/m is 44.
so 44x6x56/1000=14.7v that is 6% of lighting not 3%.
what am i missing

 

[telectrix]

you are using the rated current of the MCB rather than the actual load current, which may be quite a bit lower.

 

[stv292]

yes but you can have 13 100w bulbs on a 6A, that would be 5.65A

 

[Richard Burns]

The resistance of a single 1mm2 copper conductor at 20oC is 19.5 mohm/m (possibly 18.1 in OSG) this gives a volt drop of 6.55 V over 56m at 6A. You may be looking at R1 + R2 values instead which are two conductors in series so your length would then be 112m.

 

[telectrix]

true. i've not got my osg to hand, but in the older version ( 16th,2004, it gives 38m for 1.0/1.0 on a 6A.

 

[stv292]

how do you derive the volt drop from the resistance value's, i was using mV/A/m

 

[Richard Burns]

Yes sorry not thinking what you were using. I did V=IR current is 6A, resistance is (19.5mohm/m*56m) =1.092 ohms so volt drop = 6*1.092 =6.552V

 

[stv292]

that helped a lot, thanks

 

[i=p/u]

and I x R = v

 

[Chr!s]

Yes sorry not thinking what you were using. I did V=IR current is 6A, resistance is (19.5mohm/m*56m) =1.092 ohms so volt drop = 6*1.092 =6.552V

And what about the volt drop in the neutral ?

Regards Chris

 

[Chr!s]

hi

the osg say's i can use 56m of cable of 1.0/1.0 installation method 103,101 e.tc. for lighting, on a radial at 6A. the mV/A/m is 44.
so 44x6x56/1000=14.7v that is 6% of lighting not 3%.
what am i missing

Your missing the fact the load is distributed along its length.

Regards Chris

 

[telectrix]

well spotted, that man.

 

[i=p/u]

isit not just in line conductor you work out vd???

 

[Chr!s]

isit not just in line conductor you work out vd???

Does current flow in the neutral ?

Regards Chris

 

[i=p/u]

it does but when you doing the sums , is tha mv/ counting in neutral aswell. u very clever chris or have you been sparking since you 5

 

[telectrix]

the volt drop in the tables allows for the resistance of the line and neutral added together. what chris is saying is that the load is not all at 1 end of the circuit. say you've got 6 lights, each drawing 1A, evenly spaced. the first section of the cable carrys 6A, the 2nd carrys 5A , the 3rd carrys 4A etc., so the VD calculation is a bit more complicated than if it were a single load at the end of the cable. as a rough guide, you could calculate the VD for a single load, then divide by 2 to get an approximate value.

 

[i=p/u]

so when you installing a load of lights spread around cheshire, you actualy have more allowance for vd than allowed at the extremity of the circuit.... ive never done this sum, as you say ive only been taught the single load one

 

[i=p/u]

so i get 194v with higher resistance neutral conductor...

 

[i=p/u]

yes yes i see, so less voltage with more resistance, which i understand