Circuit Length

[Deleted member 26818]

I take the measured value of resistance (0.7Ω), divide it by the value given in Table A.1 in GN3 or Table 9A in the OSG (36.2mΩ), and then multiply it by 1000, to change mΩ to Ω.
0.7 ÷ 36.2 = 0.01933 x 1000 = 19.33.

 

[telectrix]

or you could just use a tape measure. only thing is i don't know how to correct the tape for expansion when warm.

 

[Richard Burns]

I have been trying to work this out and, as in the case of others, have not seen this equation before. However the calculation does work out but the number 29.4 should be 27.6 by my calculations.

The proportional constant (in this case given as 29.4)

For Length L, resistance R and resistance per meter Rm
L = R / Rm
Or L = R * 1/Rm

When measuring at the db with L and N connected at the load you are measuring the resistance of twice the length of the circuit; once for the line conductor and then back for the neutral conductor.

So L = (R/2) * 1/Rm
This can be rearranged to give
L= R* 1/(2*Rm)


Given the resistance per meter of 1mm2 copper at 18.1mohm/m= 0.0181ohm/m
(From OSG table 9A appx 9 for a single 1mm2 cable (not the R1+R2 value)

1/(2*Rm) = 1/ 2*0.0181 = 27.6

So for a 1mm2 cable the length = R * 27.6

The length of the cable for any given resistance will be proportional to the csa.
I.e if the resistance R for a 1mm2 copper conductor gives a cable length of 1000m
Given the same resistance R but for a 2.5mm2 copper conductor the length must be increased by 2.5 i.e. 2500m

Therefore for any size cable the length L in relation to the resistance measured R and the csa S is as follows.

L = R * S * 27.6

Unfortunately this does not answer the original question where did 29.4 come from!
But at least it is closer

 

[Deleted member 26818]

Yes, I got 27.6 as well.
It would be nice if there were a table of values, that you could then using a simple calculation X multiplied by the measured resistance = the length.
I generally end up doing the calculation two, three or how ever many times, just to confirm my results.
It doesn't help if you hit the + key on the calculator when you should have hit the x key, or you miss off one of the zeros when dividing by 1000.
Then there's those bloody scientific calculators!

 

[topquark]

or you could just use a tape measure. only thing is i don't know how to correct the tape for expansion when warm.

Using the coefficient of linear expansion of steel (assuming it's a steel tape)

 

[Richard Burns]

Unless it is assuming it is a hot summer day and you are measuring at 35 degrees!

 

[telectrix]

Using the coefficient of linear expansion of steel (assuming it's a steel tape)

yes but you've also got to allow for the expansion of the cable. copper and steel have different coefficients. where's me slide rule!

 

[Deleted member 26818]

yes but you've also got to allow for the expansion of the cable. copper and steel have different coefficients. where's me slide rule!

In the fridge, so it doesn't expand?

 

[Mark.W]

If you've measured the l/n resistance, why do you need to work out the circuit length?

V=IR so you have R you just need the design current.

So with a L/N loop of .7 as per your example, if the design current was say 6 amps the volt drop would be .7X6 =4.2 volts. Job done.

Hi

This is the entire question, examples and answers as per my notes.

Loop Resistance (L/N) of a Lighting circuit is 0.7 ohms using 1mm2 cable. Work out the cable length?
R=0.7 Ohms S=1mm2
Answer in my notes is Circuit Length = 29.4*R*S
= 29.4*0.7*1
=20.6 metres

The second part to the question is to calculate voltage drop assuming that the circuit is carrying a current when fully loaded of 5A.

The answer it gives is Vd = Ib*L*(mV/A/m) / 1000
= 5*20.6*44 / 1000
= 4.53V

I just want to know what the formula is for Circuit Length I know the formula for Voltage drop. ie what is the 29.4 ?? I thought is was resistivity but then I am not so sure!

Cheers for everyones replies,

Mark

 

[topquark]

Just rearrange the formula you have Vd = Ib * L * (mV/A/m) / 1000

So Vd * 1000 = Ib * L * (mV/A/m)

and (Vd * 1000) / (Ib * mV/A/m) = L

 

[Richard Burns]

OK
The value given is a correction factor based on the resistivity of copper the csa and the fact that you are measuring twice the length.

R is resistance in ohms, rho is resistivity of copper in ohm.m, L = length in m, A = csa in m[SUP]2 [/SUP],S = csa in mm[SUP]2[/SUP]
You have R in ohms, S in mm2
To Convert A to S multiply by 1,000,000 (or S to A divide by 1,000,000)
rho= 1.7*10[SUP]-8[/SUP]=0.000000017 ohm.m

Calculation is R = rho*L /A
Therefore L = R*A /rho
L= R * S / 1,000,000*rho
L= R *S * (1/(1,000,000*rho))
Therefore L= R * S * (1/1,000,000 * 0.000000017)
L= R * S * (1/0.017)
L= R * S * 58.8

Because you are measuring twice the length of the circuit the value becomes half of 58.8 =29.4

Hope this answers the question, I think the values I used before, (from the OSG) are modified for various factors to simplify calculations.

 

[Mark.W]

If you've measured the l/n resistance, why do you need to work out the circuit length?

V=IR so you have R you just need the design current.

So with a L/N loop of .7 as per your example, if the design current was say 6 amps the volt drop would be .7X6 =4.2 volts. Job done.

Hi Andyb

The correct answer would be .7 * 6 * 1.29 = 5.41V :yes:

Cheers for the advice though.

Mark