I have two questions.

[IQ Electrical]

What I was asking (sorry to jump on to someone elses thread), was when calculating volts drop, is do you need to double the length to take into account the return via the neutral in your (mV/A/M)/1000 calculation ?

I cannot find where the calcs are for this at the minute, I cannot see it in the BRB or GN3, it may be in unites red book

If you're asking for 2391-10 then you need to look at what values you are given:

1 The resistance per Metre of the cable OR

2 The mV/A/M figure from BS7671:2008

With option 1, you need to double the figure to allow for the neutral conductor, with 2 the figure is already included in the mV/A/M figure.

 

[spark 68]

Thanks IQ,

I have a big thick book of calcs somewhere, but it is fairly heavyweight mathematics.

I wanted to Know this for both the 2391-10 and to understand it for if I have to design for this.

As always there is more than one method, and I don't want to mix them up

 

[JUD]

What I was asking (sorry to jump on to someone elses thread), was when calculating volts drop, is do you need to double the length to take into account the return via the neutral in your (mV/A/M)/1000 calculation ?

I cannot find where the calcs are for this at the minute, I cannot see it in the BRB or GN3, it may be in unites red book

No. The mV/A/m take into account the neutral.

They're worked out using the resistance of both line and neutral.

(R"1 + R"n) x Ct

R"1 = resistance per meter of line conductor
R"n = resistance per meter of neutral conductor
Ct = rating factor for operating temperature of conductor

Example

2.5mm² cable

R"1 = R"n = 7.41mΩ/meter at 20°C (Table 9A OSG)
Ct = 1.2 (corrects resistance for 70°C) (Table 9C OSG)

(7.41 + 7.41) x 1.2 = 17.78 (rounded up to 18)

mV/A/m for 2.5mm² is 18 (Table 6D2 OSG)

 

[spark 68]

Thanks Jud,

That clarifies things, I was just working it out another way, (possibly wrong) as I was not sure if the neutral return was taken into account.

The OSG (6D2) is the easier method.

 

[IQ Electrical]

Thanks IQ,

I have a big thick book of calcs somewhere, but it is fairly heavyweight mathematics.

I wanted to Know this for both the 2391-10 and to understand it for if I have to design for this.

As always there is more than one method, and I don't want to mix them up

It caught a few people out on one of last years 2391-10 papers, just giving conductor resistance values.

 

[JUD]

Thanks Jud,

That clarifies things, I was just working it out another way, (possibly wrong) as I was not sure if the neutral return was taken into account.

The OSG (6D2) is the easier method.

Yes. Still use 6D2.

My post above is just to let you know how the values in 6D2 are worked out.

 

[MarkieSparkie]

It's worth pointing out to apprentice and new sparks that GN3 does not recognise direct measurement of voltage drop with a voltmeter. It assumes voltage drop will be determined by calculation, either from the design data if available or from the measured circuit impedance. The calculated value is then compared with the limits in BS7671, Appendix 12. The values in Appendix 12 may be exceed in periods of motor starting or high inrush current, this may acceptable if it is within the relevant product standards or the manufacturers data.

 

[malcolmsanford]

Thanks, folks. I have not designed this circuit but it was already in place and I have done a CU change as part of assessment; I needed to find this out for the record. The R1+R2 = 0.57 ohms. Thanks

I think the Mathematics have side tracked us somewhat here, so back to you Goody ................why.

What record do you need to find this out for, where are you recording this for your CU change.