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2330 302 Question!

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My exams been moved forward to tomorrow and this is the only question im not 100% on!

1: A single phase load of 30A is supplied by 6mm2 single core 70oC thermoplastic pcv cables in conduit and trunking. If the cable is 24 metres long calculate the circuit voltage drop

2: if the above cable is subjected to a fault current of 125A which causes disconnection in 5 seconds, what is the minimum size cpc which could be used

1: Using the tables from the 17th ed ap 4, i worked out VD to be 5.26....

2: Adiabatic equation? Or is there just a table, my answer im getting in the region of 2590mm which is a chunky cpc!

Any help would be great, exam is in 20 hours !
 
My exams been moved forward to tomorrow and this is the only question im not 100% on!

1: A single phase load of 30A is supplied by 6mm2 single core 70oC thermoplastic pcv cables in conduit and trunking. If the cable is 24 metres long calculate the circuit voltage drop

2: if the above cable is subjected to a fault current of 125A which causes disconnection in 5 seconds, what is the minimum size cpc which could be used

1: Using the tables from the 17th ed ap 4, i worked out VD to be 5.26....

2: Adiabatic equation? Or is there just a table, my answer im getting in the region of 2590mm which is a chunky cpc!

Any help would be great, exam is in 20 hours !

1.i got 5.688 VD ,,,,, (7.9 x 30 x 24)/1000. 2. adiabatic gave me 2.43sq.mm.


you missed the last step. look at tel's example and you will see
 
I dont get the second part, im looking at table 54.5? My question doesnt give the conductor material, I was putting in the value S= (sq root) 125a2 x 5 / 47

Am I right in I=125A, t=5 and K=47? (from T54.5)
Really not getting through my head this...
 
Conductor will be a 6mm single in conduit and trunking (it is not likely that the cpc would be the conduit and trunking as the CPC size calculation would be irrelevant).

I = 125 A
t = 5 s
k = 115 (copper cable bunched with other cables)

s = √(125² * 5) / 115 = √(15625 * 5) / 115 = √78125 /115 = 279.5 / 115 = 2.43 mm²
 
Conductor will be a 6mm single in conduit and trunking (it is not likely that the cpc would be the conduit and trunking as the CPC size calculation would be irrelevant).

I = 125 A
t = 5 s
k = 115 (copper cable bunched with other cables)

s = √(125² * 5) / 115 = √(15625 * 5) / 115 = √78125 /115 = 279.5 / 115 = 2.43 mm²

Thats where I went wrong! Using the 54.2 table instead of 54.3... thanks guys its clear now
 
Wow the exact same question with same values turned up... confident I got a good mark the only question I couldnt answer was to do with working out conduit sizes, never has our tutor mentioned this or had I come across the question type.
We had the OSG with the sizes in but I had no idea how to use it, asking to work our conduit size for 6x6mm cables in a 10m run with four bends....
 

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