Help with equations and college work needed

[Luke123]

[FONT=&quot] [/FONT] [FONT=&quot]Hey guys, iam looking for help on the question below!

Q) A 12mm[/FONT][FONT=&quot]²[/FONT][FONT=&quot] [/FONT][FONT=&quot]cable with 90C thermosetting rubbers is to be used in conjunction with lighting, small power, data and fire alarm installations. It is protected by a 40A fuse to BS 88 part 2 in an installation where the loop impedance between lines of phase where the fuse is installed is 0.14Ω. If the supply is 415V three phase, what will the protective short circuit (PSC) be? Also determine the time for the current to raise the conductor temperature to the permissible level?

so far i have this

PSC - I = UL / Z - 415 / 0.14 = 2.96KA

so i believe the psc to be 2.96KA but iam not sure if i have to times this by 3 for the three phase supply or something along thos lines?

For [/FONT][FONT=&quot] time for the current to raise the conductor temperature to the permissible level [/FONT]i am abit stuck on, i know it has something to do with this eqaution T = Ksq Ssq / Isq, but still finding it hard to work out :banghead:

A breakdown of this equation or any help would be would be great

many thanks

Luke

 

[Knobhead]

Why would you multiply by 3?

 

[Luke123]

at Tony not sure its just something my boss said, i looked in my books but could not find anything, so thats why i posted on this forum to check/confirm.

thank you for your reply

 

[Guest123]

Fair play Luke, at least you've had a go at it yourself first before asking questions. Most expect to be spoonfed an answer without attempting anything themselves.

I'll shed a bit of light on it for you..........as you have the line-line loop impedance, there is no need to multiply the result of the sum by (2). This is only needed where you only have the line-neutral loop measurement for use in the sum.

 

[Richard Burns]

For time taken to raise the temperature this equation is described in 434.5.2 and also gives you the k values you need.

t= (k² x S²) / I²

t is the time taken that you want to determine
k is the factor for the type of cable (you have this in the question 90°C thermosetting (presumably copper)) the value is 143 from table 43.1.
S it the cross sectional area that you also have in the questions (12mm², I would check this as it is non standard)
I is the fault current that you worked out in the first part.

 

[Luke123]

[FONT=&quot]alright guys, cheers for your help!!!

I think ive cracked it, does this look right to you guys!!

T = K² S² / I² 143² x 12² / 2960² = 0.34s[/FONT]
[FONT=&quot] [/FONT] [FONT=&quot]The working out for the above equation is as follows[/FONT]
[FONT=&quot] [/FONT] [FONT=&quot]143 x 143 = 20449[/FONT] [FONT=&quot] [/FONT] [FONT=&quot]12 x 12 = 144[/FONT] [FONT=&quot] [/FONT] [FONT=&quot]20449 x 144 = 2944656[/FONT] [FONT=&quot] [/FONT] [FONT=&quot]2960 x 2960 = 8761600[/FONT] [FONT=&quot] [/FONT] [FONT=&quot]2944656 / 8761600 = 0.34 s [/FONT]

 

[Richard Burns]

I haven't checked the actual values but the way you have calculated is correct, well done, always good to work it out yourself as you will learn much better and it may stick in mind.

 

[Luke123]

at richard, lenny and tony, cheers for your help guys! its very much appreciated

 

[Geordie Spark]

at richard, lenny and tony, cheers for your help guys! its very much appreciated

Good lad

Always remember: "Please and thank you costs nothing but can open a lot of doors for you".

 

[Spudnik]

As all the lads have said, if you try and work it for yourself but are still confused or unsure then ask here!!

It is a bit annoying when new comers are expecting their 'homework' done for them.

Good luck!