Calculation help!

[chrisgc]

Hi,

With reference to BS 7671 on Page 128:

Using the mathmatic formula can anyone tell me how to punch in a calculator the workings out of the following:

When calculating the minimum cross-sectional area of a protective conductor, the following is available:

Fault current = 650A
Op time of protective device = 0.35s
Constant K for protective conductor material = 115

The selected size of the protective conductor should be????

How do i type in on calculator the formula as dont understand the little maths diagram on page 128 of BS7671.

Thanks

 

[ian.settle1]

Hi,

With reference to BS 7671 on Page 128:

Using the mathmatic formula can anyone tell me how to punch in a calculator the workings out of the following:

When calculating the minimum cross-sectional area of a protective conductor, the following is available:

Fault current = 650A
Op time of protective device = 0.35s
Constant K for protective conductor material = 115

The selected size of the protective conductor should be????

How do i type in on calculator the formula as dont understand the little maths diagram on page 128 of BS7671.

Thanks

Are you doing this for college work?

If you are you need to show all your working out as the marks are not just for the answer, if the question has 5 marks then you will only get either a half or one mark for the answer.

All you need to do is Square the fault current multiply this answer by the time and then divide the resultant answer by k.

Then what ever the answer is you find the Square Root of this which will give you the minimum CSA.

 

[chrisgc]

With the method you have given I come up with:

0.077

I am obviously doing it wrong as the answer on a sample sheet for 17the edition is saying the protective conductor should be one of the following:

A) 2.5mm
B) 4.0mm
C) 6.0mm
D)10mm

The answer sheet says B) 4.0mm

How is the math worked out on that if I am coming up with 0.077

Chris

 

[ian.settle1]

With the method you have given I come up with:

0.077

I am obviously doing it wrong as the answer on a sample sheet for 17the edition is saying the protective conductor should be one of the following:

A) 2.5mm
B) 4.0mm
C) 6.0mm
D)10mm

The answer sheet says B) 4.0mm

How is the math worked out on that if I am coming up with 0.077

Chris

I apologise it is fault current squared multiplied by time then this answer is square rooted with this divided by k

 

[V]

s=650x650x0.35 all squared rooted / 115=3.34 therefore 4mm

i think that is it mate

 

[Doomed]

giving you 3.343871181752, so the next size up will be 4mm

edit: Vince beat me to it!

 

[The Truth]

Hi Chris,

i just went through the equation to find the answer its very easy to mess this up ( it took me a couple of attemtps ) . you only square root the top part of the sum and then divide that answer by K.. try this way..


square root of ( fault current squared x disconnection time )

so

( 650 squared = 422500 x 0.35 = 147875 )


square root of 147875 = 384.54

Now Divide your answer by K

384.54 divided by 115 = 3.34 mm2

3.34mm2 is the minimum size of conductor allowed... so the closest size you can use is???

yep 4 mm2


hope that helps

 

[ian.settle1]

By the way what are you doing to get 0.077

 

[chrisgc]

Re Ian.Settle

Dont ask me to do it again lol.

Seriously thanks a lot.

I did manage to work it out myself but its great that you guys are on the ball with the replies and obviously confirmed that I was doing the right method in the end at least.

Thanks again all of you.

Chris

 

[V]

i did one now you do one
chris or anyone else
zs test on an oven cct gives 0.68 ohms the previous cpc continuity test gave (R1+R2) as 0.35 ohmns ambient temp at time of test was 25c (factor 0.98) and the cable is 70cpvc(factor1.2)
if max tabulated value of zs for the cct is 0.86
show by calculation wheather measured zs vaule is acceptable