back too the books instead of trying to keep up with yous.

[pushrod]

I=P/U have alook at table 7.1 in the OSG . when you are designing circuits the length is not usually something that you can change.

 

[i=p/u]

so what your saying is you need to know the length and then design to that.. its like when the current goes up the zs value comes away down.

is that where distribution circuits come in then???? ps i ask myself more questions than i need to probably thanks.

 

[Silly Sausage]

Re: back too the books instead of tryoing to keep up with yous.

You haven’t downloaded it then Archy? I had a look at the site and got as far as them wanting an e-mail, postal address, phone No……… I always get cold feet at that point, I just don’t like giving it out.

Tony, Downloaded it late last night.
Free version limited to SP Radials & Rings, flat 2 & 3 core PVC.

On first playing with it for about 30mins it seemed quite good, but, trying it again it doesn't seem consistent, it might, stress might, have bugs.
I've neither the time nor inclination to investigate further...

2nd time I've d/led stuff from Castline, never had email etc hassle from them.

 

[pushrod]

so what your saying is you need to know the length and then design to that.. .

You are putting a circuit in- You have worked out the likely max load and circuit breaker and cable size. Then you check the length of cable you are putting in against table 7.1 and if it it is less than the max length shown for the conditions you are using you then you should be ok for volt drop. If it wasn't ok you would think about changing the design, eg increasing the cable size or maybe dividing the circuit up etc

is that where distribution circuits come in then???? ps i ask myself more questions than i need to probably thanks.
If you are putting your new circuit in from a sub main you would have to calculate what size Vd you were getting from the main CU to there and take that figure from your 5% or 3% figure. You could not use table 7.1 for a circuit from a sub main. As the distribution circuit to the sub main will use some of your allowable volt drop.

Hope that helps a bit.

 

[i=p/u]

it does. ty . i think what i left out and confused my swlf with my questions was, the bigger the cable = less resistance..

 

[thekingiam]

Thanks for answering this not that I understand the question or answer as I just do it . Plus this time of night I've had a few cobras , but it's nice that he got an answer hic... Cheers

 

[thekingiam]

woke up and thought more on this i think the question was calculate max length for a perticular circuit?

page 38 of my amicus 16th ed to good electrical practises states

cable length x design current x mv / A /m
---------------------------------------- = volt drop
1000
so please correct my first attempt at this,but i think using transposition of formulae thingy u could say
L = Vd
---------------- x 1000
Ib x mv/A/m
so if you know the max volt is 4 % of 230v = 9.2 volts
input your design current and cable voltage drop rating you can calculate max permitted length of cable ( this does not take into account grouping factors wirring methods etc.)

please correct any mistakes.... as its been a while and had a few cobra beers last night.

 

[i=p/u]

When calculating the minimum value of the cross-sectional area of a protective conductor the following information is available Fault current = 650A Operating time of protective device = 0.3s Constant k, for protective conductor material = 115 The minimum cross-sectional area of the protective conductor must be
a. 2.5mm2
b. 1.5mm2
c. 6.0mm2
d. 4.0mm2


cant figure this 1, if you want to help

 

[i=p/u]

I x R = V

woke up and thought more on this i think the question was calculate max length for a perticular circuit?

page 38 of my amicus 16th ed to good electrical practises states

cable length x design current x mv / A /m
---------------------------------------- = volt drop
1000
so please correct my first attempt at this,but i think using transposition of formulae thingy u could say
L = Vd
---------------- x 1000
Ib x mv/A/m
so if you know the max volt is 4 % of 230v = 9.2 volts
input your design current and cable voltage drop rating you can calculate max permitted length of cable ( this does not take into account grouping factors wirring methods etc.)

please correct any mistakes.... as its been a while and had a few cobra beers last night.

 

[Richard Burns]

When calculating the minimum value of the cross-sectional area of a protective conductor the following information is available Fault current = 650A Operating time of protective device = 0.3s Constant k, for protective conductor material = 115 The minimum cross-sectional area of the protective conductor must be
a. 2.5mm2
b. 1.5mm2
c. 6.0mm2
d. 4.0mm2


cant figure this 1, if you want to help

basic adiabatic equation s= root(Isquaredt)/k
=root(650*650*0.3)/115=3
therefore min csa 4mm2

 

[i=p/u]

so you need the calculator todo the number crunching, no tricks up your sleeve, but ty aswell

 

[Richard Burns]

Always have a calculator with a square root button!
Could work it roughly in your head
0.3*650 is about 215
I would then half this to about 100 and multiply by 2*650 >=125000
Square root in your head is a bit UHH for me but 300*300=90000 400*400=160000 so about half way say 350 divide by just over 100 (115) gives less than 3.5 hey min CSA must be 4!