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Discuss Hello guys. I have a question for 302 written test. in the Electrical Forum area at ElectriciansForums.net

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Martin.h.lee

Active EF Member
I am preparing 302 written test.
One thing is confusing.

8kw 230 v cooker using T&E pvc cable directly clipped. the circuit is prtected by BSEN 60898 type B circuit breaker.
the run covers 18m. the ambient tempurature is 35 C.

With this given condition, I need to cable size, design current, fuse rating etc.
I did get the answers.
the design current is 34.8A(Ib)
the rating of circuit breaker is 40A(In)
minimum size of cable is 6 mm2

However, one of questions is that "the eath fault loop impedance Zs is 1.12 ohm; does the circuit comply with Bs7671?

I got Zs(max) value 1.15 ohm (from table 41.3 of bs7671), and the measure Zs value is 1.12
Does it complies? The given answer is 'YES'
But I think I need to apply the rule of thumb : Zs> 0.8 X Zs(measured).
If I apply the rule of thumb, then it doesn't comply.

What is your opinion?

I hope I made myself clear for the question.
Thanks
 

telectrix

Scouser and Proud of It
Respected Member
you have not applied diversity. use a 32A MCB. Zs will then comply.
 
P

Plonker 3

In BS7671 does it say anything about the "Rule of thumb" in it?
 

malcolmsanford

Electrician's Arms
Measured value to me would mean that the circuit is energized and therefore working at it operational temperature and so negates the rule of thumb co-efficient of 80%
 
D

Dave 85

you have not applied diversity. use a 32A MCB. Zs will then comply.
Interesting one this, I am about to price a kitchen and need to run circuits to a hob rated 27amps and an oven rated 23amps (thats all the info i've been given.) So im gonna put the hob on a 30amp MCB sure, but I was gonna do the same with the oven, I do understand diversity but have never been sure it if applies to the max current of a single peice of fixed equipment (double oven). I do appreciate that you're unlikely to run both ovens at 250 degrees so, my question is, can I put the 23amp oven on a 20 amp MCB?
 

malcolmsanford

Electrician's Arms
If you use the OSG guide, which IMO is spot on with diversity for cookers, other areas IMO is not so good, then with diversity your looking at 19amps and that 20amp for me would be near the bone
 
G

Geoffsd

If it is a private house.

27 (-10 * 0.3 +10) = 15.1A
23 (-10 * 0.3 +10) = 13.9A
50 (-10 * 0.3 +10) = 22.0A
 

malcolmsanford

Electrician's Arms
If it is a private house.

27 (-10 * 0.3 +10) = 15.1A
23 (-10 * 0.3 +10) = 13.9A
50 (-10 * 0.3 +10) = 22.0A
And 13.9amps is right rather than the 19 amp drivel I wrote ......................teach me to be flash and try and do it in my head
Thanks geoff
 

telectrix

Scouser and Proud of It
Respected Member
And 13.9amps is right rather than the 19 amp drivel I wrote ......................teach me to be flash and try and do it in my head
Thanks geoff
at least you were erring on the cautious side. :santa_cheesy:
 

Martin.h.lee

Active EF Member
If it is a private house.

27 (-10 * 0.3 +10) = 15.1A
23 (-10 * 0.3 +10) = 13.9A
50 (-10 * 0.3 +10) = 22.0A
You are talking about the Ib(design current)?
I thought I need to apply diversity (according to table 1A and 1B on OSG) for design current. and that is what learn from teacher.
But in every questions about cable sellection, they don't apply diversity on OSG, they just do simple calculation
design current(Ib)= kw/v
that is why I got Ib=34.8A and type MCB of 40A.

I wonder I should apply diversity or not.
which is right?
 
G

Geoffsd

You are talking about the Ib(design current)?
Yes. see definitions in BGB (current carried in normal service).
I thought I need to apply diversity (according to table 1A and 1B on OSG) for design current. and that is what learn from teacher.
Not sure what you mean. I have used 1B.
But in every questions about cable sellection, they don't apply diversity on OSG, they just do simple calculation
design current(Ib)= kw/v
that is why I got Ib=34.8A and type MCB of 40A.
Diversity only applies (and varies) with certain appliances.
I wonder I should apply diversity or not.
If relevant you can.

Show all calculations.

In your example you could run both appliances on 2.5mm² with 25A mcb but normally 6mm² with 32A would be installed.

It depends what they want. Is it just calculations to show you know how or what would be done in real life?



Also, with my comment about Ze.

Do they just want the calculations or do they want you to realise that with a Zs of 1.12Ω on the specified circuit that it is far too high and indicates a problem?

I suppose it depends how far along you are.
 
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G

Geoffsd

deleted, confused about different threads.
 
Last edited by a moderator:

Martin.h.lee

Active EF Member
This question is what I am struggling with.
I got this question from this forum.
Actually, this question doesn't provide any clue of type of premises(to refer OSG table 1B).
If I assume this installation is for individual houses and there is an incorporated socket-outlet, then diversity applied design current would be
10A+[(8000/230)-10]X0.3+5A= 10+7.43+5=22.43A
therefore fuse rating will be 25A, and Zs(max)=1.84ohm, tabulated current carrying capacity (It) = 27A, cable size 2.5mm2..
The first step of cable sellection is Design current(Ib), from this everything goes differently.
Very confusing.
19 -

An 8kW 230V cooker is supplied using T&E pvc cable clipped direct. The run covers a distance of 18m. The circuit is protected by a BS EN 60898 Type B circuit breaker. The ambient temperature is 35º C.
Determine:
a) The installation method
b) The design current
c) The rating of the circuit breaker
d) The correction factors to be used (if any)
e) The tabulated current carrying capacity It of the cable
f) The minimum cable size
g) The voltage drop
h) The maximum voltage drop allowed
i) The earth fault loop impedance Zs is 1.12Ω; does the circuit comply with BS 7671?
j) The minimum size of cpc required


[h=2]Answer[/h]

The Tables referred to in the answers are from BS 7671:2008
a) Installation Method C (Table 4A2)
b) 34.8A
c) 40A Type B (Table 41.3a)
d) Ca = 0.94 (Table 4B1)
e) It = In / Ca = (40 / 0.94) = 42.6A
f) 6mm[SUP]2[/SUP] (Table 4D5 column 6) can carry 47A
g) 7.3mV x 34.8 x 18 (Table 4D5) = 4573 / 1000 = 4.57V
h) 5% of 230V = 11.5V (4.57V acceptable)
i) Max allowable Zs = 1.15 (Table 41.3) (complies)
j) The minimum size of the c.p.c. is determined by using the Adiabatic Equation S = √(I[SUP]2[/SUP] x t)/K
Where I = the earth fault current, calculated from Uoc/Zs = 230/1.12 = 205A
t = the disconnection time in seconds = 0.1 (Fig. 3.4)
K = the factor determined from Table 54.3 = 115
S = √(I[SUP]2[/SUP] x t)/ K = √(2052 x 0.1)/115 = 64.8/115 = 0.56mm[SUP]2

[/SUP]
 
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