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This is the first time I have used this stack site so am not 100% sure this question is on topic (based on the tags I couldn't find) but thought someone may still be able to help me.

For those not familiar, in the UK we have the IET Wiring Regulations which set out regulations for electrical installations. This document also includes disconnection times for different fuse types so that the correct fuse can be selected for a system based on the PFC of the system and the required disconnection time.

Here is an example of one of the graphs for British Standard BS88 fuses; enter image description here

Now I'm no mathematician but looking at the graph I'm sure there must be a way to deduce formula for the different fuse ratings (no formula are provided in the regulations). I'm currently updating an electrical design process where currently, the designer needs to manually refer to the graph to see which fuse should be used. I want to automate this process, hence wanting to find the formula for the graph.

Now, actually getting to the question, is it likely that the graph has been created using a formula (assuming it even has one), or could it just be data recorded during testing of the fuses with a "line of best fit" chucked over the top? I find it strange that I can't seem to find any formula for the graph anywhere online or even any clues.

I tried plugging in the values in the provided table into Excel and used the "show formula" option (I know, very crude) but using the formula provided didn't give me correct results when I manually checked them on the graph.

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  • \$\begingroup\$ It's a LOG-LOG plot, so if they were straight lines it would be a power law. You could approximate the curves with straight lines and fit with a power law... but I don't think that is what you want. \$\endgroup\$ – George Herold Sep 17 '14 at 18:06
  • \$\begingroup\$ It's a potential option but I don't really want to be doing any approximating if possible. Thanks for the input though. \$\endgroup\$ – Jonny Wright Sep 17 '14 at 18:09
  • \$\begingroup\$ Please let me be "Devil's advocate". 1. How much time does the designer spend looking this up? 2. How long will it take you to automate it and you+designer to test it to ensure it is correct for all plausible values? 3. If there is a mistake in your code, would you be liable for any damage or loss of life cause by fitting the wrong fuse? 4. How will you know when the program might be updated, i.e. the specification has changed? I asked 4 because [IET](electrical.theiet.org/wiring-matters/38/fuse-standards.cfm?type=pdf) says your graphs in "Bs 88-2.2 and Bs 88-6 were withdrawn on 1 March 2010" \$\endgroup\$ – gbulmer Sep 17 '14 at 18:12
  • \$\begingroup\$ Very valid points. Admittedly, it is not a massively time consuming task, most of the time for this aspect would be spent finding the graph in the book! However with the amount of designs we do, any time saved would be beneficial. The designs are also always checked by A.N Other, who would hopefully pick up any errors. Another option is having the graph as part of the document to check once the value has been calculated. Admittedly makes the formula seem redundant but would still be a time saver. RE the graph, it was just one I grabbed off Google. Being a member of the IET we are always... \$\endgroup\$ – Jonny Wright Sep 17 '14 at 18:21
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    \$\begingroup\$ If you or your company are members of the IET, then ask them what the formula is. At least that way, you are using definitive information, reducing the number of ways errors can be introduced, and hence reduce the number of ways you might become liable. Of course, their might be an IET member who exactly knows this stuff on here. So I am not suggesting 'instead of'; I am only suggesting "as well as". \$\endgroup\$ – gbulmer Sep 17 '14 at 18:25
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The adiabatic equation gives earth sizes for disconnection times of 5 seconds

\$S = \frac{\sqrt{I^2}\times{T}}{k}\$

where \$T\$ is the disconnection time and \$k\$ is the factor chosen from tables in BS7671

But why square something then square root it? So just \$I\$.

So

\$\frac{S\times k}{ I} = T\$

where,

  • \$S\$ is size of conductor
  • \$I\$ is fault current
  • \$T\$ is disconnection time
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