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Select (Scottish) Certification Scheme Question on volt drop from 2391

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Bear

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Hi all,
I have worked out the volt drop on other questions using the formula mVxAxM.
had no issues, but this question below has just fried whats left of my brain, I have the answer because i got it wrong,
but how do you work this one out??

Being Dyslexic i do sometimes read the questions all wrong and am sure i am missing something simple here.

Voltage drop of a single-phase circuit supplying a bread oven is to be verified as part of periodic inspection and testing within a bakery.

The circuit has a measured R1+Rn value of 0.40 Ω and an Ib of 29.6 A. The circuit protective device has an In of 32 A.
The installation forms part of a public 400/230 V TN-S system. What is the calculated value of voltage drop?

a) 11.8 V
b) 13.5 V
c) 14.2 V
d) 14.8 V
 
If I had to guess I'd say it's to do with cable resistance times Length (You're given R1 RN) however there is no distance listed, nor cable type used so I'm a bit stumped.

Design current 29.6A, Voltage 230/400 gives 7.7Ω @230v and 13.51Ω @400v (R=V/I)

Using the resistances above, check against (purely a guess here) resistence per metre to get the length and possibly cable type, then you should have the required info to calc volt drop.
[automerge]1596551079[/automerge]
Try with Vd = (R1+Rn)x1.2xIb
 
11.84V

But, is this one of the sample tests from the C&G? If so they are renown for incorrect answers!
 
Last edited:
exactly what i though, 11.84v

but the answer sheet says i was wrong, it is 14.2v

Is it the right answer sheet to the paper? - which paper is it, I have just looked at C&G web site, question 27 on paper 1.3 appears to be the same question.

The 1.3 answer sheet gives it as answer b (which is also wrong!)
[automerge]1596553244[/automerge]
Multiply the resistance by 1.2 for temperature correction under load ?

that looks the case to me.
 
When I said multiply by 1.2 for temperature, that was a shot in the dark, having looked at the books I can find nothing to substantiate a factor of 1.2. What does your Tutor say?
 
some of the questions are strange, how can you have a earth fault loop impedance test of a Radial to a local isolator? what it stops there?

sorry I was looking 2391-051
 
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Since when do we apply the 1.2 value to calculate volt drop.
The volt drop will alter based on the load changing.
It’s giving you the load in amps so why do you need to adjust the temperature?
It’s not the same as calculating R1+R2 where we need to adjust to 70 degrees.
There is a complex looking formula in the appendix of bs7671 for temperature adjustment of VD where it is known that the cable will not be operating at its maximum temperature but this is not it.
 
Last edited:
R1+Rn x design current = volt drop at nominal temp

(R1xRn x design current) x1.2 = volt drop at 70°c

So yes correct answer is 14.2 volts.
But the question has no information to state the cables will be running at 70C, nor the information (if any) to calculate what temperature it might be operating at.

What if it were XLPE SWA at 90C instead?

What if a long run of 16mm that would barely show any rise?

The examiners deserves a boll0cking for that.
 
Absolutely, this city and guilds course is the bane of my life right now, There is little evidence in my history at school or collage to say i have dyslexia so there not giving me more time to understand the questions, The primary school has one teacher left that said she remembers me going to support, but i got that in an email and city and guilds want letter proving it, Am 40 years old id need to get retested and that hard to do right now.

74.23, failed the first one i did due to time, and some of the questions were ridicules and tried to catch me out due to the way they where worded...
 
That is doggy the 1.2 value is only really used for design unless the temperature is stated in the question.

again on 2391-051
and question 23 ignores sequence of tests I know you can have limitations on Periodic inspection but it says to use Earth fault loop impedance test to prove continuity of CPC.
[automerge]1596703388[/automerge]
sorry OP thats not very helpful of me and paper 2391-52 isn't that bad, it's pretty good.
 
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