An Awful Feeling This

[IQ Electrical]

another one for you

state the purpose of an initial verification of a newly installed circuit

To conform that the circuit complies with the designers intentions
The circuit has been designed, installed and tested in accordance with the requirements of BS7671
It is not damaged so as to impair safety or otherwise defective

ok but guidance note 3 also says

initial verification shall be carried out to verify:

all installed electrical equipment and material is of the correct type and complies with applicable British Standards or acceptable equivalents
all parts of the fixed installation are correctly selected and erected
no part of the fixed installation is visibly damaged or otherwise defective

so which one is right?

They're all. 'right' so now you need to check how many marks the question is worth and tailor your answer to that.

 

[gap30]

The fault path diagrams were in the exam success book, have they been removed?

there is one drawing for a tt installation

 

[gap30]

They're all. 'right' so now you need to check how many marks the question is worth and tailor your answer to that.

but the exam success book is written by the examiners so surely i would go with their way of thinking?

 

[brucelee]

Hi mate

MED

materials of the corrct type and Bs
erected and selected properly
damage and defects are not present

this will be a 3 mark question so 1 % lol if you get them all right
the question will say according to BS7671 or Gn3

I asked the same of my tutor lol he gave me some useful answers to some of my many questions he was a good tutor and would recommend to anyone near bury college business solutions couldnt help enough

 

[gap30]

Hi mate

MED

materials of the corrct type and Bs
erected and selected properly
damage and defects are not present

tis will be a 3 mark question so 1 % lol if you get them all right
the question will say according to BS7671 or Gn3

I asked the same of my tutor lol he gave me some useful answers to some of my many questions he was a good tutor and would recommend to anyone near bury college businees solutions couldnt help enough

MED thats a good one, thanks for the drawings pal

 

[gap30]

a) Determine the expected value for (Zs ) for circuit 3 , assume the ambient temperature is 20ºC ( **** )
b) State ↔TWO↔ likely conditions that may have an effect on the value of earth fault loop impedance ( **** )
c) State ↔ ONE ↔ method that can be used to compensate for the conditions in ↔ b ) ( *** )
d) Determine whether the value in ↔ a ) is acceptable if the max ( Zs ) for a 10A BS-EN 60898 circuit breaker is 4.6Ω ( **** )

dont shout at me here, especially bruce lee lol


3) ………. 10A ….. B ……… Lights ……………… ....1.5 …………… 1.0 …………………. 10

C.S.A ………. mΩ per metre at 20ºC
1.0 ……………………….. 18.10
1.5 ……………………….. 12.10
2.5 ……………………….. 7.41

a) mohmspermetre x Ib x L/1000 i think

12.10 x 10/1000 - 121/1000 = 0.121
18.10 x 10/1000 - 181/1000 = 0.181

i am struggling with this question as a whole, if someone could talk me through it please - i do not mean answer it for me just help me with it please, btw far too much revision

 

[brucelee]

Hi mate to calculate Zsyou need to know ZE and when calculating Zs as a designer its Zs = Ze + m/ohm/m X MF X length divide 1000
as installer its Zs =Ze+(R1 +R2)
the MF is the Multiplying figure which takes in to account increase of resistance due to operating temp due to load current
which is 1.2 MF as max is 70C for and for 90C cables its 1.28
and when you tested you tested the circuit at 20C ambient temp unloaded

 

[gap30]

The supply together with the installation forms a 230v 50hZ TN-S system , the measured ( Ze ) is 0.55Ω and the PFC is 0.6kA

 

[gap30]

Zs = Ze + m/ohm/m x mf x length/1000

Zs = 0.55 + 12.10 x 10 x 1.2/1000 for 1.5mm

0.55 + 12.10 x 10 x 1.2 = 145.75

145.75/1000 = 0.14575

or is it the complete sum =0.6952

Zs = 0.55 + 18.10 x 10 x 1.2/1000 for 1.0mm

0.55 + 18.10 x10 x 1.2 =217.75

or is it the complete sum = 2.722

i do have to do the 2 sums right? as the line conductor is 1.5 and the cpc is 1.0mm

 

[malcolmsanford]

Zs = Ze + ( R1 + R2) so you know your Ze is 0.55

so Zs = 0.55 + ( R1 + R2 ) therefore 0.55 + m/ohm/m x mf x 10mt /1000 your m/ohm/m is in table 9A of the OSG page 166 and also the same table is in your GN 3 so for 1.5/1.0 the m/ohm/m factor is 30.20

so the formula is 30.20 x 1.2 x 10/1000 which is 0. 36

Zs = 0.55 + 0.36 = 0.91

 

[gap30]

you have lost me there completley mate

C.S.A ………. mΩ per metre at 20ºC
1.0 ……………………….. 18.10
1.5 ……………………….. 12.10
2.5 ……………………….. 7.41

this is the table on the test sheet the 2391 exam in question refers to

 

[gap30]

3) ………. 10A ….. B ……… Lights ……………… ....1.5 …………… 1.0 …………………. 10

and this is the circuit in question 1.5 line 1.0 cpc

 

[malcolmsanford]

yes 18.10 + 12.10 = 30.20

 

[malcolmsanford]

Zs = Ze + m/ohm/m x mf x length/1000

Zs = 0.55 + 12.10 x 10 x 1.2/1000 for 1.5mm

0.55 + 12.10 x 10 x 1.2 = 145.75

145.75/1000 = 0.14575

or is it the complete sum =0.6952

Zs = 0.55 + 18.10 x 10 x 1.2/1000 for 1.0mm

0.55 + 18.10 x10 x 1.2 =217.75

or is it the complete sum = 2.722

i do have to do the 2 sums right? as the line conductor is 1.5 and the cpc is 1.0mm

That is right for the 1.5mm = 0.145

Where as that figure come from

 

[telectrix]

you have lost me there completley mate

C.S.A ………. mΩ per metre at 20ºC
1.0 ……………………….. 18.10
1.5 ……………………….. 12.10
2.5 ……………………….. 7.41

this is the table on the test sheet the 2391 exam in question refers to

your res is 18.1+12.1 = 30.2 as malcolm says.