Appendix 4, section 6.1 equation 6

That is a formula to conjure with, though I do not think I ever have done so in that form …

We can check it by asking what do we expect, well we know that the resistance of the copper rises with temperature, & the voltage  drop in millivolts per amp is just milliohms by another name.  

So,  if we changed  the Cg, Cs Cd etc are in the direction that they are less than one when the conditions are worse - hotter, then the formulation also needs to be the way up that reflects a higher resistance when for example ambient is hotter.

The 230 and 30 figures are just rather clunky ways of fitting a line to the not very fast changing plot of copper resistance versus temperature. (you will often see it published as +0.39 % to + 0.4%   per degree C, for temperatures around 20C-50C so a 30 degree rise is about a 11% -12% increase in resistance.)

If you really are running a mere 2A through a 20A cable, you can probably drop most of the formula into the sea and just consider the effect of the higher ambient temperature, as the self heating part will be very small.

It should come out dashed close to the fully-calculated answer.

regards mike.

Thanks Mike that is very helpful, I should have mentioned that the cable is part of a temporary installation and is 300m long fed from a 2A MCB, in a high ambient temperature that's why I was running through the Calculations.

You are using In as Ib. Is that the case? Do you not have issues with Ud?

2.5 mm 

https://www.tlc-direct.co.uk/Technical/Charts/VoltageDrop.html

The text says multiply, but that would mean the resistance decreases with the increase in temperature (or have I got the sum wrong?)

Isn't the overall factor increasing from 0.82 (for just the high ambient temperature) to 0.95 (now accounting for the reduced conductor temperature, due to the reduced load, as well) - which seems to be going in the right direction to me.

   - Andy.

Hi Lyedunn, ud is what I am ultimately calculating, actual draw is less than 2, but the breaker is tha limit at 2, so I was looking at what ud would be if there was a fault and in was in a hot location

Thanks Andy, my questioning was over the wording of the application of factor to  the mA/V/M figure  - multiplying it by 0.95 make it smaller, dividing by 0.95 makes it larger

Andrew James Skinner: 
 

 

Hi I am working through equation 6 for an ambient temperature of 50 degrees for 1mm 90degC thermoplastic cable (Table 4E2A)

I get a correction factor of 0.95 for a 2A load. 

Do i divide the mV/A/m by this factor ? 

The text says multiply, but that would mean the resistance decreases with the increase in temperature (or have I got the sum wrong?)

I'd just round up the 0.952619 to 1.0 and call it a day. We ain't that precise in the real world of chucking in cables and “power” electrics.

Please read the last paragraph on page 381.

Also, please read the NOTE in 6.1 regarding the APPROXIMATE  resistance temperature coefficient. The calculation is very approximate.

Z.

Andrew James Skinner: 
 

Thanks Andy, my questioning was over the wording of the application of factor to  the mA/V/M figure  - multiplying it by 0.95 make it smaller, dividing by 0.95 makes it larger

The mv/A/m tabulated figure is multiplied by 0.95 making it smaller and thus gaining advantage in terms of overall Ud. However, as Zoom indicates, there is no real advantage to be had so just use the tabulated value. 
Ud is calculated using Ib, so depending on what limits you are working to use that value rather than In as it may result in a smaller csa. 

Thanks for point out that note Z, I had not taken notice of that.