B

bobby101

did the exam last nite i thought the exam was not to bad still quite tough and the dreaded VD question came up by VD i mean voltage drop i think i managed to work it all out but i made the stupid stupid stupid mistake of saying it was within six percent instead of five dont know how heavily i will get penilised for that but was quite impressed with myself for being one of the very few to work it out dont get me wrong i did make a few more schoolboy errors on the exam too but fingers crossed got my practical on mon at 8am aarrgghh had a few beers last nite though how did the rest of you find it and what questions did you think were nuts cause i think there were a few
 
cheers mate i thought it was but in Gn3 its 1.83 but it doesnt really make a difference because of rounding up when working resistance out you still get same answer
should have not panacked as now thinking rationally its pretty easy to do just never been shown that way
all the time iwas thinking its something to do with V=IXR
i would like to see the paper again to check where i messed up
 
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i really didnt under stand the first question.... lol... think i put accsesoys comply to bs or equilavent standard... installation complys to bs7671 and not defected as to cause danger,,, or something along those line... any one else remember the question... or what they put???

I spent about 5 mins trying to interpret that first question and i still couldn't work out what they were asking me. I started writing about check for damage ie overload and crossed it out wondering if it was supply characteristics so skipped it and left a mark on question book thinking i'll come back to it later on. Left another one or two out in the same way thinking i'll come back to them after answering section 2. I still managed to spend over an hour on section A even though i had time to spare in any mock papers i've done.

I ran out of time just finished anwering qu 25 so i'm at least 20+ marks down already, hopefully the ones i did answer i got top marks otherwise a resit awaits.

I actually spent more time reading the questions than answering them, this exam is crazily worded, for those who pass this i have every respect, but i think soemtimes they want rocket scientists only to pass this course. I've revised like mad but nothing could prepare me for the purposefully silly way they ask questions.

I remember some questions that were asked but remember the wording in most of the questions was not wording that ought to be used for an exam in any conditions, unless there's an exam out there for 'interprating really awkwardly worded questions'!
What makes it even sillier is that City and Guilds have the cheek to constantly say in each of there examiners reports that students fail to grasp and understand what is asked of them in the question!

Well bugger me change the wording into practical easy to understand ENGLISH!!!!
 
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could not have put it better my mate when we get results we should all send them a email telling them exactly what we thik of there wording:eek: this was how i felt after?
 
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could not have put it better my mate when we get results we should all send them a email telling them exactly what we thik of there wording:eek: this was how i felt after?

I'm of the same feeling. In fact i want to send them a letter right now, i wonder if any folks on here have before? It is rediculous. However i daren't send any views now incase it has any negative bearing on my result. Can you remember many of the questions to reproduce them on here anyone? I will post a few on here that i remember another time after i've drowned me sorrows over that GREEN moment!:(
 
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----------------------R1 ( 5 Ω )--------- Rload ( 10 Ω )--------- RN ( 5 Ω )

-----------------|=======|---------|========|-------|=======|------------

R1=Line Conductor
Rload = Load
RN = Neutral conductor

If we put 100V through this circuit, then 5A would flow through it 100/(5+10+5).... So we know therefore that V=IXR.

VDrop at R1 = 25V, Rload = 50V....

Leaving us with a PD from the point of connection with the load-neutral connection, all the way back to the centre point of the transformer Of 25V .......




Volt drop in any circuit is 100% therefore, the Vdrop must be 25V across neutral in this circuit



This in my opinion, is underpinning knowledge....That's why they tested it.


the volt drop if for the end of line of the circuit. wher you would read a ZS not thfor the neutral as no load is drawn off the neutral side.

you also said there is always a voltage in the nutral interesting comment what about ballanced three phase supplyes where no neutral is used?
 
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I think it should be doubled, the reason it's not normally in this calc is that the volt drop tables in BS7671 already allow for 2-conductors whereas the resistance values given in the exam were for single conductors.


what volt drop tables? never seen them can you point them out. or are you thinking of the tables in appendix 4 which give a resistive value in mA?
 
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you also said there is always a voltage in the nutral interesting comment what about ballanced three phase supplyes where no neutral is used?

When a circuit is energised, there is a potential in neutral..... Without it, no current can flow.

You know neutral has a resistance (remember little Rn) So lets say neutral = 1 ohm but V=0 do the calcs yourself mate.........

NO CURRENT FLOWS.... But we know equal current flows in all parts of circuit.... think of the water analogy !!!

And 3 phase dosent need a neutral, you're right. Because it uses the potential difference between line conductors. No offence meant here, but this is basic stuff... Guess what though, single phase dosent need a neutral either..... it'll quite happily use your butt as a path to earth.
 
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i have done the calcs using the information. i passed 2391-10 in 2008 first time and 2391-20 first time and an playing the devils advocate a bit.

I mentioned the 3 phase to try and prove a point. you use the tables in appendix 4 and the lenghty of the cable. you have no need to double the lenght of run for voltage drop.
 
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i have done the calcs using the information. i passed 2391-10 in 2008 first time and 2391-20 first time and an playing the devils advocate a bit.

I mentioned the 3 phase to try and prove a point. you use the tables in appendix 4 and the lenghty of the cable. you have no need to double the lenght of run for voltage drop.

I will not argue that... My argument is the neutral puts up a resistance... So it must therefore be built into the calculations in the tables. There is no need to seperate them, after all L and N conductors must be equal size.

BUT, we were not given tabulated values of Vdrop.
 
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What pi55es me off most, is that it is a requirement that we carry BS7671 and OSG, so why not give us the info out of em ???
 
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I doubled the length of the conductors, but now I'm confused:

Table 4D1B in Appendix 4 states that 10mm sq cable in conduit drops 4.4mV/A/m at 70 [FONT=Tahoma, sans-serif]°[/FONT]C
The title at the top states:[FONT=Tahoma, sans-serif] 'Single-core ... cables', [/FONT]so I'd assume the values are for one conductor only.

The table in the question paper is different. 'Figure 1 below shows information on the resistance of conductors in mΩ/A/m.'[FONT=Tahoma, sans-serif]
The bottom row says nothing more than: 1m of a 10mm² conductor has a resistance of 1.83m[/FONT]Ω at 20[FONT=Tahoma, sans-serif]°C.
At 1amp, this gives a voltage drop of 1.83mV.
The scenario also mentions single cables in trunking and conduit.

Sadly, when I compare the figures with BS7671, logic and common-sense cease to apply.
The values on the exam paper should be the same as the values listed in BS7671, table 4D1B, but reduced a certain amount because they're at [/FONT]20[FONT=Tahoma, sans-serif]°C[/FONT][FONT=Tahoma, sans-serif] instead of 70.

BS7671 says a 1m length of 10mm [/FONT][FONT=Tahoma, sans-serif]²[/FONT][FONT=Tahoma, sans-serif] cable drops 4.4V per amp,
C&G say that the same cable drops 1.83mV.

Temperature correction is 4% increase in resistance per 10[/FONT][FONT=Tahoma, sans-serif]°C[/FONT][FONT=Tahoma, sans-serif]increase in temperature, i.e. 20% for a 50[/FONT][FONT=Tahoma, sans-serif]°C difference.


The difference in these two figures is 58%.

What's going on? [/FONT][FONT=Tahoma, sans-serif] [/FONT]
icon8.gif

[FONT=Tahoma, sans-serif]
[/FONT]
 
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back pages of OSG about page 120ish gives you resistive values of cables as well as values for CPCs the vold drop calcs mainly use the values in appendix4 so the info is there int both books they want to know if you can adjust calcs to suit the info. 75% of the exam is interpritation of the questions. once yu can do that the rest is easy i completed mine in less than 2 hours and still passed
 
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I doubled the length of the conductors, but now I'm confused:

Table 4D1B in Appendix 4 states that 10mm sq cable in conduit drops 4.4mV/A/m at 70 [FONT=Tahoma, sans-serif]°[/FONT]C
The title at the top states:[FONT=Tahoma, sans-serif] 'Single-core ... cables', [/FONT]so I'd assume the values are for one conductor only.

The table in the question paper is different. 'Figure 1 below shows information on the resistance of conductors in mΩ/A/m.'
[FONT=Tahoma, sans-serif]The bottom row says nothing more than: 1m of a 10mm² conductor has a resistance of 1.83m[/FONT]Ω at 20[FONT=Tahoma, sans-serif]°C.[/FONT]
[FONT=Tahoma, sans-serif]At 1amp, this gives a voltage drop of 1.83mV.[/FONT]
[FONT=Tahoma, sans-serif]The scenario also mentions single cables in trunking and conduit.[/FONT]

[FONT=Tahoma, sans-serif]Sadly, when I compare the figures with BS7671, logic and common-sense cease to apply.[/FONT]
[FONT=Tahoma, sans-serif]The values on the exam paper should be the same as the values listed in BS7671, table 4D1B, but reduced a certain amount because they're at [/FONT]20[FONT=Tahoma, sans-serif]°C[/FONT][FONT=Tahoma, sans-serif] instead of 70.[/FONT]

[FONT=Tahoma, sans-serif]BS7671 says a 1m length of 10mm [/FONT][FONT=Tahoma, sans-serif]²[/FONT][FONT=Tahoma, sans-serif] cable drops 4.4V per amp,[/FONT]
[FONT=Tahoma, sans-serif]C&G say that the same cable drops 1.83mV.[/FONT]

[FONT=Tahoma, sans-serif]Temperature correction is 4% increase in resistance per 10[/FONT][FONT=Tahoma, sans-serif]°C[/FONT][FONT=Tahoma, sans-serif]increase in temperature, i.e. 20% for a 50[/FONT][FONT=Tahoma, sans-serif]°C difference.[/FONT]


[FONT=Tahoma, sans-serif]The difference in these two figures is 58%.[/FONT]

[FONT=Tahoma, sans-serif]What's going on? [/FONT]
icon8.gif


the single core reffert to the type of wiring if you look through all cable types have different values for different cables.

when using the tables use the corrrect refrence e.g on tray,clipped direct,burried etc.

as for the differnces in the resistances its related to the temp. i dont have books nearby most info is from memory but i thin the No in OSG are for 20degc and the BRB are at 70c
 
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Thanks Scotsparky, OSG seems to be clear in saying that values are for a 'length of run in metres'. It does make more sense to give figures for voltage dropped on the cable length, rather than individual conductors. I assume it's the same for the values in BS7671, which would explain the 58% difference in the values given in the exam.
The values in the exam are for one conductor. The remaining 8% is the temperature difference (I thought it would be 20% for temperature, but percentages do weird things depending on which way you divide/multiply them).
 
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I will not argue that... My argument is the neutral puts up a resistance... So it must therefore be built into the calculations in the tables. There is no need to seperate them, after all L and N conductors must be equal size.

BUT, we were not given tabulated values of Vdrop.

I agree. Values of voltage drop make sense if you include both conductors, but the question gave resistance of conductors, and needed you to recognise that there are two cables presenting this resistance. You needed to double the length.
 
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Hi Guys

Also did this paper and would like to get a copy, to help me when I resit it in October!!

Many Thank

PS I also did not understand first question and most of the others

Steve
 
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Hi Guys

Also did this paper and would like to get a copy, to help me when I resit it in October!!

Many Thank

PS I also did not understand first question and most of the others

Steve


Join the club!!!! If you want a copy pm me your email and I will send it out to you
 
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There seems to be a unanimous conclusion that these 2391 exams aren't worded very well. How many folks took this exam on here? I'm thinkin we should start a revoloution or something, at the least get a petition sent out to ****** and no thrills telling them that whoever writes out this exam wants sacking!!!
 
Last edited by a moderator:
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I doubled the length of the conductors, but now I'm confused:

Table 4D1B in Appendix 4 states that 10mm sq cable in conduit drops 4.4mV/A/m at 70 °C
The title at the top states: 'Single-core ... cables', so I'd assume the values are for one conductor only.

The table in the question paper is different. 'Figure 1 below shows information on the resistance of conductors in mΩ/A/m.'
The bottom row says nothing more than: 1m of a 10mm² conductor has a resistance of 1.83mΩ at 20°C.
At 1amp, this gives a voltage drop of 1.83mV.
The scenario also mentions single cables in trunking and conduit.

Sadly, when I compare the figures with BS7671, logic and common-sense cease to apply.
The values on the exam paper should be the same as the values listed in BS7671, table 4D1B, but reduced a certain amount because they're at 20°C instead of 70.

BS7671 says a 1m length of 10mm ² cable drops 4.4V per amp,
C&G say that the same cable drops 1.83mV.

Temperature correction is 4% increase in resistance per 10°Cincrease in temperature, i.e. 20% for a 50°C difference.


The difference in these two figures is 58%.

What's going on?

The 4.4 mV/A/m is for 2 cables (L+N) single phase 10mm copper conductor at 70 degrees.
if you divide that by 2 you get 2.2 mV/A/m for one 10mm conductor at 70 degrees

The mV/A/m value is doubled to add the neutral conductor so you dont need to double the length. Just use the length of the run.

The 1.83mΩ/m is at 20 degrees.
Temperature correction is 4% increase in resistance per 10°Cincrease in temperature, i.e. 20% for a 50°C difference.

so 1.83 x 1.2 (20%) = 2.2mΩ/m (at 70 degrees)

mV/A/m is the same as mΩ/m
 
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