help needed with 17th edition please

[1983]

Bit of help needed please

A single phase load of 13a is supplied via a 15a bs 1361 fuse using single core 70c thermoplastic copper cables installed to method 4 (ref method b) the rating fator for grouping is 0.7 and for the ambient tem 0.87, and overload protection is to be provuded. the minimum acceptale size cale would be?

Can someone please explain how this is caculated.

Q2 which of the following statments regarding exhibitions,shows and stands is incorrect?

A- all socket outlets not excceeding 32a shall be protcted by an rcd

B- the norminal supply v shall not exceed 230/400v a.c

C- all electric motors shall have means of isolation.

D- signs shall be construceted of materials having an adequate heat resistance.

I thought the answer was d as i could not find it in the regs but i was told the answer is c but 711.55.41 states where an electric motor gives rise to a hazard the motor shall be provided with an effective means of isolation on all ploes and such means shall be adjacent to the motor which it controls. so which one is right?

Q3 the space under a bath, unless accessible only with a tool is considered to be in zone?

A-0
B-1
C-2
D-3

I understand that the space under a bath tub or shower basin is zone 1 and if its only accessible with a tool it is considered to be outside of the zone. is this question asking me which zone the space under a bath is with or without the use of a tool?

Thank you for those who take take the time to read my questions and answer them much appreciated

 

[Sparky Ninja]

Bit of help needed please

A single phase load of 13a is supplied via a 15a bs 1361 fuse using single core 70c thermoplastic copper cables installed to method 4 (ref method b) the rating fator for grouping is 0.7 and for the ambient tem 0.87, and overload protection is to be provuded. the minimum acceptale size cale would be?

Can someone please explain how this is caculated.

Ib < In < Iz

So far,

Ib = 13A, and In = 15A

It < Iz

It = In/Ca x Cc x Ci x Cg

Ca = 0.87
Cg = 0.7
Cc & Ci = 1 (as they are not applicable)

Therefore:

It = 15/0.87 x 1 x 1 x 0.7 = 24.6A

Refer to Table 4D1A of BS 7671 for the required cable size. Reference method B, single phase shows a 2.5mm cable will carry 24 A, however we need to select the next size up from It, so we will have to use the 4mm with a CCC (Iz) of 32 A.

Without the length, voltage drop cannot be calculated, so this is as far as it can go.

 

[Sparky Ninja]

Bit of help needed please
Q2 which of the following statments regarding exhibitions,shows and stands is incorrect?

A- all socket outlets not excceeding 32a shall be protcted by an rcd

B- the norminal supply v shall not exceed 230/400v a.c

C- all electric motors shall have means of isolation.

D- signs shall be construceted of materials having an adequate heat resistance.

I thought the answer was d as i could not find it in the regs but i was told the answer is c but 711.55.41 states where an electric motor gives rise to a hazard the motor shall be provided with an effective means of isolation on all ploes and such means shall be adjacent to the motor which it controls. so which one is right?

711.422.4.2 States:

Showcases and signs shall be constructed of material having an adequate heat-resistance.

711.55.41 States:

Where an electric motor might give rise to a hazard...

Option C suggests ALL motors, whilst BS 7671 clearly suggest only if a hazard is to be considered, so it does not apply to all motors.

 

[Deleted member 26818]

The easiest method, is to divide the rating of the protective device, by the correction factors, then look in the tables in Appendix 4 for the appropriate cable and refference method, and choose a cable CSA that has a rating greater than the corrected rating of the protective device.
15A / 0.7 = 21.428571... / 0.87 = 24.63054... . So you need to find a value in the table, which is greater than 24.6A.

Next question:
You have to choose keword/s when taking this exam, otherwise you could be wasting time reading the whole book looking for the answer.
Obviously you will be looking for the chapter relating to exhibitions, shows and stands.
You must then choose an appropriate Keyword/s for each statement, and look up the section that refers.
Statement 'A' mentions RCDs, which are considered addition protection, so look up additional protectin in the Chapter.
Statement 'B' mentions nominal supply, so look up supply.
Statement 'C' mentions isolation and electric motors, so look up both isolation and electric motors.
Statement 'D' mentions heat, and in the Regulations heat is referd to as thermal effect, so look up thermal effect.
The answer is in fact 'C', as the statement is incorrect, all motors do not have to have a means of isolation, only those that may give rise to a hazard.

With the last question, yes it is asking what zone the space under a bath is considered to be, if the space is accessible without the use of a tool.

 

[Sparky Ninja]

Q3 the space under a bath, unless accessible only with a tool is considered to be in zone?

A-0
B-1
C-2
D-3

I understand that the space under a bath tub or shower basin is zone 1 and if its only accessible with a tool it is considered to be outside of the zone. is this question asking me which zone the space under a bath is with or without the use of a tool?

I'll rephrase the question for you.

If I can get to the space under my bath without the use of a key or tool, then it is zone?


(A good clue with interpreting this question is reviewing the answers. You clearly know what it is trying to ask, but am not sure which way round it is.
If you look at the optional answers you will see that "outside zones" is not here, so by default.. the question can only be asking one thing)

 

[1983]

thanks for the answers but i have to say widdler thank you very much for the tip that was so usefull one more im having a bit of a problem with is finding the external influnce code ACM ip rating

 

[Deleted member 26818]

External influence AC is altitude, AM is Electromagnetic, electrostatic or ionizing.
There are no specific IP ratings for these influences.
I am not aware there is an ACM code?

 

[i=p/u]

with out the the legth, vd can not be calulated', but im sure a man with your skills can give you a max length???? and a we calculation will be lovely

Ib < In < Iz

So far,

Ib = 13A, and In = 15A

It < Iz

It = In/Ca x Cc x Ci x Cg

Ca = 0.87
Cg = 0.7
Cc & Ci = 1 (as they are not applicable)

Therefore:

It = 15/0.87 x 1 x 1 x 0.7 = 24.6A

Refer to Table 4D1A of BS 7671 for the required cable size. Reference method B, single phase shows a 2.5mm cable will carry 24 A, however we need to select the next size up from It, so we will have to use the 4mm with a CCC (Iz) of 32 A.

Without the length, voltage drop cannot be calculated, so this is as far as it can go.

 

[1983]

my q is the external influence code ACM requires rated equipment to

A-ipx4
B-ipx1
C-ip0
D-ipx2

I have a past exam paper and the answer is a but i cant seem to find where in the regs it says that

 

[Sparky Ninja]

with out the the legth, vd can not be calulated', but im sure a man with your skills can give you a max length???? and a we calculation will be lovely

well yes of course, but it's beer o'clock?

I tell you what, if nobody else pops it up I'll do it later.

 

[i=p/u]

i have few 17th exam papers in front of me.. well as it says external, i would be drawn straight to A as is outside.. external influences=ip44... or there abouts

 

[1983]

lol no need for vd as that was the only question but if i wanted to cal vd it b mv/a/m x l x ib/1000. say the L was 10m would vd be 1.43v?

 

[Marnik]

you don't need to know thength to calculate the voltage drop of a circuit!
The simple formula ; r1 + rn x 1.2 x Ib = vd

 

[i=p/u]

wheres 1.2 come from if you dont mind me asking, isit operating tempeture,

 

[Marnik]

correct i=p/u . it is indeed the operating temperature correcting factor.

 

[i=p/u]

so when would you do this adding of 1.2 in the real working world, if you have we example

you don't need to know thength to calculate the voltage drop of a circuit!
The simple formula ; r1 + rn x 1.2 x Ib = vd

 

[i=p/u]

i just know IxR=V

 

[telectrix]

well that's what he means . V ( volt drop) = I x R, where R is R1+R2, then correct for temp. with the 1.2 factor.

 

[i=p/u]

what reg is this telectrix? so when would you use this, would you just not measure it, give you awnser ... or is this when designing