2391 exam question. pls give an answer

[morph]

A single phase power circuit supplying an item of equipment,having a full load current of 25 A,is to be checked to confirm voltage drop compliance.The circuit is installed using 6.0 mm2 cooper conductors,with a circuit length of 35m.
a) Calculate the maximum permissible voltage drop for this circuit.

230 x 5% = 11.5v

b) The line and neutral combined resistance for the circuit is 0.26 Ohms . Determine whether the circuit meets the voltage drop requirements if the voltage drop to the local distribution board 4.5 V.

25 x .26 x 1.2 = 7.8 v + 4.5 v distribution = 12.3 v, i would assume this was a measured value at 20 C so temp factor needs applying.


c)State one other method which could be used to confirm voltage drop.

It may be evaluated using site design information for example

correct

 

[morph]

Yes, for exam purposes, you need to use the 1.2 factor but I'm talking about my design calculations,-If I have a cable with a CCP of say 87 Amps, my Ib will be well below this level and as a result, my conductor operating temperature will be nowhere near maximum.

It's the difference between a rigid factor and some real world commercial acumen/experience to supplement the regulations.

"Where it is known that the actual operating temperature under normal load is less than the maximum permissible value for the type of cable insulation concerned (as given in the tables of current carrying capacity) the multipliers given in Table 9C may be reduced accordingly"

OSG Page 167

i know what you are saying , but it was an exam question and the 1.2 factor does need to be used

 

[ringer]

Sorry, I still don't get it. I see no reference to a sub-board in the original question, and even if there was, surely the VD will only be along the L that supplies the sub-board. I still don't see how VD along the neutral comes in to play.

Does current flow in the neutral in a single phase circuit?

if it didn't then we could do away with it and just use single core cable.

Sorry guys, I still don't see how any volt drop that may happen in the neutral is of concern. I accept that there is volt drop in the neutral, but that is the return path, and by the time the current reaches the neutral cable it has already passed through the load, and it is the voltage at the load that we are trying to maintain.

VD = (mV/A/m) x Ib x L / 1000 where L is the length of the circuit from CU to the appliance. If we were to calculate volt drop over both the live and the neutral, we would have to use 2L instead of L in this calculation.

I am not professing to be an expert in these matters - am just trying to better understand why you are taking the neutral into account when calculating volt drop.

 

[telectrix]

ringer, iv'e done the calc as in the OP. using 0.26 ( L+E ) and got 6.5 as others have done. then did it using mV/A/m as in tables ( 7.3mV for 6mm O) and got 6.387. that tells me that no matter how weird it seems, you have to add the L and N resistances to get the right answer.

 

[ringer]

okay - so it sounds then like the OSG table 6F values for mV/A/m take into account the return path as well. I think I am starting to understand now, though I still struggle, logically, to see why VD over the neutral is of consequence. Am doing the 17th at college atm - will ask the tutor for his take on this.

 

[trebor]

A single phase power circuit supplying an item of equipment,having a full load current of 25 A,is to be checked to confirm voltage drop compliance.The circuit is installed using 6.0 mm2 cooper conductors,with a circuit length of 35m.
a) Calculate the maximum permissible voltage drop for this circuit.
230 x 5% = 11.5V
b) The line and neutral combined resistance for the circuit is 0.26 Ohms . Determine whether the circuit meets the voltage drop requirements if the voltage drop to the local distribution board 4.5 V.
25 x 0.26 = 6.5V + 4.5V = 11V therefore complies
c)State one other method which could be used to confirm voltage drop.

direct measurement under load conditions

 

[gamblor6]

sorry but i think ringers right. vd is only calculated on the line conductor. mvam x ib x l / 1000. not lx2.

 

[IQ Electrical]

sorry but i think ringers right. vd is only calculated on the line conductor. mvam x ib x l / 1000. not lx2.

You're missing the point, the mV/A/M values include the return path but if you take conductor resistance values as in the original question then you MUST double the value.

 

[gamblor6]

You're missing the point, the mV/A/M values include the return path but if you take conductor resistance values as in the original question then you MUST double the value.

i stand corrected, do they. i never knew ty iq

 

[pushrod]

okay - so it sounds then like the OSG table 6F values for mV/A/m take into account the return path as well. I think I am starting to understand now, though I still struggle, logically, to see why VD over the neutral is of consequence. Am doing the 17th at college atm - will ask the tutor for his take on this.

Try this - all the voltage is dropped over the total length of the circuit.
The current will only flow if the circuit is complete ie line , load and neutral.
All 3 have a resistance - so voltage must be dropped over all 3.
You can not discount voltage being dropped on the neutral because like the line, it is there and part of the circuit.

Any good?

 

[brucelee]

IQ spot on as usual and a few others i did the june exam and this question was the only one that had me
i had only ever been shown Vd = mv/a/m x ib x l
so to see resistance there had me all the time ohms law was ther in my mind

i did the exam, the question we were given the resistance of 10mm cable of 1.83m/Ohm which we had to x 2 because 10mm was being connected in parallel also the length which i believe was 80metres and the table for the resistance was at 20 C the and then given Ib of 45a i think
i had completed rest of exam in 1 hr 30 this one had me for about 30 mins as i kept going back at first i just kept looking at it in disbelief lol i even drew the triangle for ohms law but in the end had a go still dont know if i got it completley correct but must have got some marks for it or the examiner felt sorry lol i read through all my questions and answers about 3 times this and one other i wasnt sure on
i have the paper somewhere with the question

still gives me nightmares that exam lol
no only joking would do it again tomorrow but they wont let me as i passed

 

[morph]

vd calcs are very simple and you must have done them a million times during the 2360

 

[brucelee]

didnt mate as i dont have 2360

 

[morph]

did you do the 2330 ? that should be basically the same course,
i think some colleges seem to teach more than others ,as the vd question is a common question on the 2391 all colleges should teach this .
more grey areas i guess

 

[IQ Electrical]

did you do the 2330 ? that should be basically the same course,
i think some colleges seem to teach more than others ,as the vd question is a common question on the 2391 all colleges should teach this .
more grey areas i guess

From a 2391-10 point of view, the first volt drop calculation question was June 2010=-you will not find one on past papers before this.

I guess it takes some training providers a while to catch on!