2391 sum practical help

[brucelee]

HI mate
Buy IEE guidance note 3 learn it back to front
Cristopher kitchers guide to inspection and testing
GS38
memorandum of guidance on electricity at work regs 1989
learn all tests and how to explain in bullet points with or without diagrams
practice drawing earth fault loop diagrams for all systems TN-C-S, TN-S,TT.IT and fully label using correct terminology
C&G are sticklers for correct terminology
use the search bar for 2391 there are loads of threads on the subject
good luck mate

read all amberleafs posts especially the thread section 301 it has loads of stuff on the 2391 and some past papers and questions

 

[jackos]

The IET guidance note 3 has everything you need to know for the 2391 it is the bible for this course. Buy it, read it and read it again until you can recite it in your sleep. Get all the correct terminology into your mind. Also 2391 exam success book will tell you how to construct your answers, how to interpret what they are asking and the best way to answer it.

Cheers...........Howard

thanks a meg jackos well useful.

 

[SPARTYKUS]

yeah you got me there but hey if you aint got the questions you cant find the answers ......... by the way totally happy i am a senior wow happy days BTW that does not mean i am old honest my kids will tell you that oooops ha ha yey happy days plod on also on a personal note i used most of the information here on bullit points both as a reminder and as a good practical aid to help me pass the 2391 and it did.
stud study, then get as many 17th edition 2391 past papers as poss study please note when buying from ebay please contact buyer to find out if it its 17th cos i got duped many times and yet the chancers are still selling past papers for 16 th edition which dont get me wrong can help but when messing with new terms and terminoligies (not sure i spelt the last word right but f it iam slightly p 1 sse d yep) good luck 2 u all happy days ....PS PASS MARK FOR 2391 IS AROUND 80 % SO I HAVE BEEN LED TO BELIVE MAJOR POINTERS EFLI DIAGRAMS AND INS CALCS ARE GOOD 2 KNOW ALSO GET A COPY OF JUNE 2391 PAST PAPER THAT WOS THE BOMB LEARN VOLT DROP AND PHASE ROTATION, DO THE SEARCH FOR THE 2391 ALL THE INFO IS THERE IF YOU SEEK THE ANSWER THE QUESTION IS NOT THE PROBLEM......................ONLY FOOLS TEST WITH THEIR TOUNGE OR LUCKY IDIOTS .......
REMEMBER IAM TRAINED IDIOT OR AN IDIOT THAT WOS TRAINED ARF ARF....OOHH HAAPY DAYS I WISH YOU ALL WELL GOOD LUCK. ALLS WELL THAT BLOWS SO I KEEP TELLING THE WIFE . ARF ARF .

That was the one I took .... yeah the volt drop question threw me (I thought) but passed so mustve got something right.

 

[i=p/u]

good post

 

[brucelee]

Hi
the june exam i took and the volt drop question threw me also but i also passed, this was the only question that had me for the june exam

 

[234567891]

What was the question about V.D.? was it transposed or something funky ? I just started my course this week and need all the help i can get by the sounds of it !! Cheers guys

 

[brucelee]

Hi mate
the volt drop had to be worked out for a 10mm conductor being connected in parallel from given data which was a table for resistance of csa of different size cables the 10mm had a resistance of 1.83m/ohm /per/metre at 20C the load was 45A, the length was 80Metres work out the volt drop show all calculations
then after working out the volt drop state wether the volt drop is acceptable or not
if not give two actions that could be taken if above is not acceptable thats basically it but not worded like C&G

 

[234567891]

Nice one Ninja,The dirty *******s eh!, so they don't give you the mV/A/M for the V.D. calc? would I be right in thinking you can work out the V.D. for whole length of the cable by 1.83 x 80/1000 = 0.1464 ohm. Then 0.1464 x 45A =6.588V ? that complies with 3% and 5% of 230V( guessing its not a big --- lighting circuit though!) remedial if not could be decrease length of run or increase csa.

 

[brucelee]

but dont forget as its in paralell ie the Line and the neutral multiply by 2 right here is the full question mate itsall about ohms law and transforming the equation using the data and info given

Q26
During the course of a electrical inspection of the single phase radial circuits supplying an item of fixed equipment is to be checked for voltage drop compliance the circuit has the following charicteristics
. Load current = 45 A
.circuit length = 80 m
. Live conductors = 10mm
a). using the info in figure 1 determine wether the circuits meets the requirements for voltage dro.
Show all calculations

Fig 1 has conductor CSA size for 10mm as 1.83 m/ohm/m at 20C

Answer as i did i forgot to use a correction factor for increase of conductor temp from 20C to 70C
I answered as following as ihad only ever done volt drop from mv/A/m X Ib X L /1000

26.a
R = m/ohm/m x L / 1000
= 1.83 x 2 x 80 / 1000 the 2 because of the two live conductors in paralell i.e L and N
= 292.8/1000
=0.2928 rounded up
=0.3 Ohms
I should have included a correction factor for increase in temp due to operating temp
Vd = 0.3 X 45
= 13.5v
b).
State a suitable observation to be recorded on the periodic inspection report if the voltage drop fails to comply

b) answer
volt drop not acceptible as does not comply with the max 5% value of 11.5v

c) one action which could be taken to rectify the the situation of b) above
answer Increase csa of conductors

this was only one of two questions that had me i did exam in 1 hr 40 but stalled on this question and spent about half hour going over it lol as my head was telling me mv/A/m x Ibx l / 1000
and ohms law kept coming into my head someone of the forum told meishould have included the correction factor from 20C to 70C but i passed first time only 1 of 4 in our class and 1 of the 39% thatpassed in all the country so it was happy days after the 3 month i waited for results and the 5 months it took for certificate as our practical results didnt get dsent of until october so the practical bit i did on the 15th june now has a new date of october lol
hope that some help mate good luck

 

[234567891]

Sorry for the delayed reply, i'd like to say i was studying hard with my head in a book!? (no been on me jollys) Thanks for the breakdown it all makes sense now but just wait till the exam and it will feel like i've had my brain formatted! This might seem like a stupid question but- the question says conductors in parallel (l +N) but isn't the V.D. only relevant to the line conductor supplying the outlets? The correction factor for the temperature is to take into account the difference between the table values of resistance (@20c)and the actual resistance during operation(70c) and this is 1.2? Does it seriously take so long to find out if you pass??, good old 7671 took 10 minutes and i had the printout in my hand ready to spill beer on!

 

[brucelee]

C&G say it takes 42 days to get results but can take 12 week as mine did lol

but some get it in 8 week or less

hi mate when working vd then the neutral is the return of the circuit and also a live conductor so has to be included into calc as its connected in parallel to the line
EG
R =1.83m/ohm/m at 20C for 10mm 45A load 80 metres length of circuit
R=m/ohm/m x Length / 1000
R=1.83 x 2 x 1.2 x 80 / 1000
R= 292.8/1000
R=0.2928
R = rounded up to
0.3 ohms
vd = R X I
vd = 0.3 x 45
vd = 13.5v
max of 5% 11.5 v so 13.5v not acceptible as greater than max

increase CSA of conductors

somebody like IQ will be able to explain better
this is how i worked it out but i forgot to include MF for increase in conductor temp due to operating temp

 

[SirKit Breaker]

Dont know if this thread is still running, but with the calcs for the 2391, you will get marks for showing all calcs even if you get the wrong answer, it is the way you come to the answer that is as important as the answer itself. so even if you are unsure like i was have a go. this is because in the real world you have access to all the books, and they are looking to see if you know what and how you should be working things out.

Cheers.......Howard

 

[234567891]

Good luck tonight 2391 peeps, 1 hour and counting! better sober up and get going........

 

[La Poste]

This is how I saw the question:

During the course of the electrical inspection one of the single phase radial circuits supplying an item of fixed equipment is to be checked for voltage drop compliance. The circuit has the following characteristics:
. Load current = 45 A
.circuit length = 80 m
. Live conductors = 10mm
a). using the info in figure 1 determine whether the circuits meets the requirements for voltage drop.
Show all calculations

Fig 1 has conductor resistance per metre for a cable with CSA size 10mm as 1.83 m/ohm/m at 20C

resistance of line conductor to piece of equipment:
= L*R/1000 = 80 * 1.83/1000 = .146 Ohms.

Correct for operational temperature of cable using rule of thumb multiply by 1.2.
= .146 * 1.2 = 0.175 Ohms.
This is the resistance of the cable from the DB to the item of equipment.
Since we are only interested in the Volt drop at the piece of equipment then this is the only figure we will need.
Volt drop is I * R = 45 * 0.175 = 7.875 Volts.

Now I am fairly certain this is wrong but I can not see why.
Line and neutral cables are not in parallel they are in series, the sum of their resistances will be added together if you were to include the two cables in any calculation for instance if you wanted to find the potential current flowing in the line and neutral cable of a circuit under short circuit conditions.
In this particular question we are asked for the volt drop at the point where the piece of equipment it attached to the circuit and hence we are only interested in the resistance of the line conductor to that point.
With a volt drop of 7.875 Volts, this circuit will comply as it is a lower figure than 11.5 Volts which is the maximum volt drop allowed in BS7671 for this type of circuit.

Someone please correct me if this is wrong as this type of question may come up again in my exam in July.
Thanks alot.

 

[La Poste]

Bump.
I have read a few threads about this and it seems that to calculate volt drop at the piece of installed equipment you need to take the resistance of the line and neutral together in series, I do not really understand why but it seems to be set in stone so I will accept that it is correct.
If BS7671 says it is the correct way to do it then who am I to argue.