just started the 2391

[i=p/u]

this thread will go on for quite a while (14 WEEKS), to exams... ive nearly read gn3 for the first time and there is a few things i dont fully understand but im sure i will for when the time comes from help from you guys/dudes/punks who know just a little to much.

 

[i=p/u]

is that it

 

[i=p/u]

this one...the insulation resistance reading take between neutral and earth is 50M ohms(10m of 2.5mm twin cable) what would the resistance of 20m of the same cable? id say if 10m = 50Mohms then 20m = 25Mohms

 

[i=p/u]

calculate the maximum resistance(R1+R2) of a lighting circuit in a domestic installation with the last ceiling rose placed 32m from cu.switch drops are 3m.

1.5/1.0mm = 30.20/1000 x 35m= 1.057 ohms

 

[i=p/u]

calculate the maximum resistance (R1+R2) of an oven circuit in a domestic installation 7m from ccu.(3Kw)

would you just use the lowes cable you can get away with for 3Kw , 13.04A so 1.5mm

 

[i=p/u]

be back later... 2391 night , 40 miles away, im on course to success

 

[Richard Burns]

Q. periodic inspection involves?
Possibly:
Inspection and testing of an existing installation to provide for:
The safety of persons and livestock against electric schok and burns, regulation 131.1
the protection of property against damage by fir and heat from defects in the installation
Confirmation that the installation is not damaged or deteriorated so as to impair safety
Identification of defects/ non compliances which may cause danger.

what about this one... the resistance of 100m of 4.5mm cable is 0.36 ohms, what is the resistance of 600m of this cable, isit just a matter of 0.36x6=2.16 ohms directly proportional Yes

..this one...the insulation resistance reading take between neutral and earth is 50M ohms(10m of 2.5mm twin cable) what would the resistance of 20m of the same cable?id say if 10m = 50Mohms then 20m = 25Mohms Yes

calculate the maximum resistance(R1+R2) of a lighting circuit in a domestic installation with the last ceiling rose placed 32m from cu.switch drops are 3m.

1.5/1.0mm = 30.20/1000 x 35m= 1.057 ohms

calculate the maximum resistance (R1+R2) of an oven circuit in a domestic installation 7m from ccu.(3Kw)

would you just use the lowes cable you can get away with for 3Kw , 13.04A so 1.5mm
Not certain what they want on these two, there is not enough information, for C&G to ask you to guess cable size is not likely. Maximum resistance is a bit odd too, since you are calculating expected not maximum resistance.
A domestic cooker over 2kW would be wired in 6mm on a B32 breaker.

 

[i=p/u]

spoke to tudor about these questions and he said very likely wouldnt come up but he trying to show us the relationship between cables i.e

if the resisttance per 100m of 2,5mm csa cable is 0.06 ohms what is the resistance of 100m of a cable of the same conductor material with a csa of 6mm

R2 = R1 x A1 / A2
R2 = 0.06 x 2.5 / 6
R2 = 0.06 x 0.416 = 0.025 ohms



if the resistance of 100m of 4mm cable is0.08 ohms, what will be the resistance of 700 of 10mm cable of the same material?


i used the book values and got10mm=1.83m/ohms per meter /1000 x 700 = 1.28 ohms


R2 = R1 x L2 / L1 x A1 / A2

= 0.08 x 700 / 100 x 4 / 10

= 0.08 x 7 x 0.4 = 0.224 ohms

 

[i=p/u]

where supplementry bonding is provided as a dditional protection, it must include all simutaneously accessible exposed conductive parts, extrabeous conductive parts and the protective conductors of all equipment in the location being applied.
the effectivness of supplemenrty bonding can be verified by satisfying the following condition R<50/Ia

where R is the resistance between simultaneously accessible exposed conductive parts,
50 is the maximum safe touch voltage,
Ia is the current causing automatic operation of the protective devices within 5 secs or IAn raiting for rcds

now i understand if 30mA rcd is used you get 1666 ohms , now what happens if was no rcd what value for Ia would you use.., say was a bathroom with 40A shower R= 50/200=0.25 ohms, 6A lighting circuit R=50/30=1.7 , a radiator and towel rail
and a hotpress with copper heating tank

so does all the circuits need to be bonded together..

i know if main protective bonding is in place you dont need all this... just put me straight

 

[Richard Burns]

now i understand if 30mA rcd is used you get 1666 ohms , now what happens if was no rcd what value for Ia would you use.., say was a bathroom with 40A shower R= 50/200=0.25 ohms, 6A lighting circuit R=50/30=1.7 , a radiator and towel rail
and a hotpress with copper heating tank

so does all the circuits need to be bonded together..
so R must be less than 0.25 ohms and all extraneous / exposed conductive parts must be connected to the CPCs of the circuits involved.
i know if main protective bonding is in place you dont need all this... just put me straight

supplementary bonding can only be omitted if the following are in place:
Suitable main bonding connection for all extraneous parts
disconnection times are met
30mA RCD additional protection is in place

The Ia value used if an RCD is not in place is the current causing disconnection time of 5s for the protective device used (5x for B type MCB) or look them up in time current characteristics in BGB Appx 3.

 

[i=p/u]

in my kitchen, the water pipe has main protective bonding, and supplementry bonding..... in my bathroom the raidator has one pipe earthed, which i think is tied into the light and reason i think this is, is because yes the shower is on a rcd but the lights aint.... and the lights is within 2.5m of the raidator

 

[i=p/u]

and when you look at reg 701.415.2 it more less says if main bonding is in place water , heating etc there is no need for supplementry bonding

 

[lennox]

hi there

i have just started the 2391 also and my tutor has stated that one of the conditions to not install supplementry bonding is that there must be a continuity reading of within 0.05 ohms between all conductive parts and extraneous conductive parts.page 35 of guidance note 3,paragraph 4.aswell as it having a rcd protection on all circuits in the bathroom and also main protective bonding

 

[i=p/u]

so would you say if main bonding is in place , i.e water,central heating/gas, and rcd installed no need for supplementry bonding... what is the test you do with insulation resistance, is that to see if something needs bonded/in contact with earth potential

 

[Richard Burns]

Replied a bit fast there and forgot to say all accessible extraneous / exposed conductive parts.

There is some discussion that simultaneously accessible could be as low as 1.25m, but using 2.5m (within arms reach) is a safe approach.

It is important to remember that there are 3 requirements not just an RCD and main bonding, but also that the circuit meets the ADS requirements (even though it is unlikely that an RCD protected circuit would not meet these requirements).

The measurement of continuity <0.05 ohm between parts is a test to ensure that inaccessible supplementary bonding is in place. If the continuity is <0.05 ohm then it may be assumed to be in place.

To test to see if extraneous conductive parts are extraneous you do an IR test between the part and earth (CPC or earthed metalwork) and if the result is >22kohm (some say 16kohm for 30mA not 10mA protection) then the part is not extraneous as it will not be able to introduce an earth potential.

 

[i=p/u]

what is the significance of the value obtained and recorded from step 3 of continuity of ring final circuit conductor test

id say is part of you earth fault loop impedance value and has to be below a set value.. and can calculate voltdrop from this reading for the circuit..

maybe im saying this all wrong so if i am share you cool , il go cheeers