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

[Bear]

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

 

[Lister1987]

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.
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Try with Vd = (R1+Rn)x1.2xIb

 

[R-fur]

V = IxR
V = 29.6 x 0.4
V = 11.84

 

[Julie.]

11.84V

But, is this one of the sample tests from the C&G? If so they are renown for incorrect answers!

 

[Bear]

exactly what i though, 11.84v

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

 

[R-fur]

Multiply the resistance by 1.2 for temperature correction under load ?

 

[Julie.]

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!)
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Multiply the resistance by 1.2 for temperature correction under load ?

that looks the case to me.

 

[R-fur]

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?

 

[AJshep]

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.

 

[Bear]

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.


AJshep,
I think you cracked it, thank you.

This was driving me nuts,