BS7671 : Help Needed on the Example Question

[clanky]

Hello, I took the opportunity over the lockdown to do 18th Ed course but theres a few bits im struggling with. Because its distance learning, the college i signed up with havent exactly been that helpful. So hopefully someone on here knows the answers!

Its to do with the example working out for the 4kw heating load. On the second part of the that question (determining the size of the CPC) theres a table showing various Conductor CSA's with appropriate Conductor resistance @20 degC. I cant find that table in BS7671! the course narrative says table 52.3 but thats nothing like that one. Am i missing something?

Secondly, using the adiabatic formula, the factor of 1.2 is used in the calculation for the temp difference in conductors. Where is that info as it just seems to have appeared!

Thirdly, and i know this is probably a bit stupid, but for the final calc of earth fault loop impedance, the Ze figure is 0.2. Fair enough, but again, where has this come from?

Thanks in advance for the help!

 

[telectrix]

the 1.2 factor is to correct resistance from 20deg. to70 deg.

the 0.2.ohm Ze is given so you can work out the Zs. in real lfe thie Ze would have been ofbiained by measurement.

 

[clanky]

the 1.2 factor is to correct resistance from 20deg. to70 deg.

the 0.2.ohm Ze is given so you can work out the Zs. in real lfe thie Ze would have been ofbiained by measurement.

Hi, thanks for that. But, i know what they are, i just dont know where in 7671 this info is. Im not happy about the way these figures/factors are just thrown in on the course without ref to the book. Or am i overthinking this?

 

[timhoward]

heres a table showing various Conductor CSA's with appropriate Conductor resistance @20 degC. I cant find that table in BS7671! the course narrative says table 52.3 but thats nothing like that one. Am i missing something?

Secondly, using the adiabatic formula, the factor of 1.2 is used in the calculation for the temp difference in conductors. Where is that info as it just seems to have appeared!

Well both are technically in there, but they are both communicated in an obscure fashion that is far from obvious. In fact I hesitated before replying because you won't really need any of this stuff in the exam and I don't want to add any confusion.
But to satisfy your curiosity....

BS7671 gives values for volt drop in tables in appendix 4, e.g. table 4D5 on page 409 for our famous twin and earth.
Volt drop is given for each csa size in milli-volt/ampare / m
Ohms law V=IR , so I=V/R
So instead of milli-volt / ampare / m you could write milli-ohm/ m (which is also resistance / m in milli-ohms.)

In real life - you'd probably use table 11 in Appendix I of On Site Guide, or the ProCert app on your phone, but I don't suggest you look at OSG yet until you have passed 18th as the layout is different.
The volt drop values are given at the max operating temp of the cable btw.
It sounds as if the question just wanted you to trust that information and do something else with it.

Second part - If you look at section 6.1 on page 382 there is a horrendous formula. Ignore it and look at the note underneath saying there is approx resistance temperature co-efficient of 0.004 per degree at 20 degrees. It doesn't actually complete the logic for you here, but the conversion formula is 1+(0.004 x temp difference)
So to correct from 20 degrees to 70 degrees is 50 degrees difference = 1+(0.004 * 50) = 1.2
I'm assuming this makes some sense in the context of the question.

Final part - the Ze
This changes with every installation out there. They have simply given you a value for it so you can do other things as Tel said.

The exam isn't as hard as those questions are making out and I'm sure you'll be fine.

 

[clanky]

Well both are technically in there, but they are both communicated in an obscure fashion that is far from obvious. In fact I hesitated before replying because you won't really need any of this stuff in the exam and I don't want to add any confusion.
But to satisfy your curiosity....



The exam isn't as hard as those questions are making out and I'm sure you'll be fine.

Ah-ha! massive thumbs up for that, ill work through your info later. I know this is probably part of my inherent engineering OCD but it was annoying me that these figures just 'appeared' I was concerned that i would have to do a similar example in the exam and without knowing where those factors come from (or whether they are variable depending on the conditions) then i just couldnt see how i could answer them properly. I should take solace in the fact that you think the exam isnt as hard as im imagining it!

Thankyou!

 

[telectrix]

Ah-ha! massive thumbs up for that, ill work through your info later. I know this is probably part of my inherent engineering OCD but it was annoying me that these figures just 'appeared' I was concerned that i would have to do a similar example in the exam and without knowing where those factors come from (or whether they are variable depending on the conditions) then i just couldnt see how i could answer them properly. I should take solace in the fact that you think the exam isnt as hard as im imagining it!

Thankyou!

the Ze value of 0.2 is given as a typical value for a TN-S or a TNC-S earthing system. it's given in the question so you can work out the Zs by adding (R1+R2) to Ze.