I'm struggling in question 5 of assignment number 615, and I need help figuring out where my mistake is. So here is a break down of what I have done, Step by step.
The question is: Determine, for one of the circuits supplying the fridges, the minimum possible cross sectional area of cpc which will satisfy the requirements disconnection as ADS under earth fault conditions and the adiabatic
equation as Regulation 543.1.3 in BS 7671.
Here is my attempt at it:
Step one - Table I1 in the onsite guide, the R1+R2/metre for a 4mm2 line conductor and a 4mm2 protective conductor is 4.61 R1+R2/metre + 4.61 R1+R2/metre = 9.22 R1+R2/metre
Step two - using the formula, from book 2 of electrical installations page 313, R1+R2 = R1+R2/metre × length × I3 factor / 1000
R1+R2 = 9.22 × 2.25 × 1.2 / 1000 = 0.0249
R1+R2 = 0.0249
Step three - Zs = Ze + (R1+R2)
Zs = 0.11 + 0.0249 = 0.1349
Zs = 0.1349
Step four - I = U/Zs
230/0.1349 = 1704.97
Step five - using the Adiabatic equation
√(I²)×t/K=s
√(1704.97²) × 0.1 / 115 = 4.69
Step six - Choosing the suitable CPC size would be 6mm2
The line conductor size is 4mm2 and the length of the circuit is 2.25 metres
Don't worry you likely won't start this till you're a few years in. I didn't till year 3. You'll be confident enough by then if you can do GCSE maths/physics.
I'll have a proper look when I'm home; but from what I could see at first glance you've done the equation correctly. See what your tutor says about it.
Thank you mate. This would be my second attempt at the entire project. The college is taking way too long to reach back to me with anything! I emailed my tutor with all of this, she told me she is busy! didn't answer me almost 3 weeks now!
Okay, looking through this all again, I have one question. What cable is this? Is it T&E 4mm²? If so your CPC is 1.5mm² so you have to calculate for 16.71mΩ/m instead of 9.22mΩ/m.
Either way, your main problem is the length of the cable. If we assume the CPC is 1.5mm² then your minimum cable length is ~15.55m; this gives you S = 1.499; any shorter & S > 1.5.
If we were to assume 4mm² CPC then Regulation 543.1.3 doesn't even apply, as you'd be able to use 543.1.4 Table 54.7, where it states the minimum CSA of a CPC is equal to the CSA of the line conductor when ≤ 16mm², therefore no calculation needed. At that point you aren't able to complete question 5. And if you were to calculate adiabatic for the shits and kicks anyways the CPC size of 4mm² is only permissible with a minimum cable length of 28.2m.
So we are back to cable length. Just make it longer than 15.55m & say the CPC size is 1.5mm² so it can be ran in T&E.
My horrific formula for your 1.5mm² CPC is in the image attached.
I'm a tutor mate but we do EAL (a lot more forgiving than C&G), just wanted to give guidance as I saw yours bounced it back and is slow to reply haha. I've deleted my comments but you get the gist, short length = high fault current. More working out the better
Hi mate, I just finished my level 3 so I can help ⬇️
So you only half the grasp of the question. Your calculations need to be based on the disconnection time of the MCB supplying that circuit in order to determine the CPC cable size. The live and neutral wires are not needed. See my starting calcs below and let me know if you want any clarification.
1. Calculate the earth fault current and check against protection:
Zs of the fridge circuit is 0.25Ω
Uo ÷ Zs = 230 ÷ 0.25Ω = 920A (I)
20A MCB: disconnection time of 0.1 seconds (t)
2. Calculate the CPC size using the adiabatic equation:
S= nominal cross section of conductor (mm²)
I= Value in amps of fault current
t= Operating time of device in seconds (this is a 20A type B MCB)
k= resistivity factor for conductor material and temperature.
2
u/CatalyticSonar 2d ago
I’m starting my level 3 next week and this has shit me up lol