Current carrying capacity of ring circuit

[specks]

I am happy with calculating current carrying capacity for straight runs of cable. I am not to sure how the imact of a ring effects the cable calcs? Its hard to see what the possible loads might be. As of this do you take the center point and use maximum demand?

So for example on a 35 meter ring with say 8 sockets on a load of 5kW (21.7A) would you assume socket 4 (assuming half way) is the maximum demand point? or am I way off here?

Cheers

 

[joe-sims]

best way to think of a ring is if it was one cable, the basic point of it is that the farthest point is half the distance and you are using twice as much cable to get there is its up to 4X better than just the one cable going the hole way... so yes the mid point would be the highest resistance

 

[dim_bulb]

I thought theoretically it should be equal all the way round I.e same at each point

 

[joe-sims]

if you are doing a r1 + r2 test then yes, because it is a big figure of eight from of the way you should join the ends up but. doing a loop test under normal load conditions a socket that might be right under the consumer unit for example is going to have a better reading than the one that is on the opposite side of the house. so guess it just comes down to witch ever way you do it. (Zs = Ze+ r1+r2)

 

[HappyHippyDad]

So what is the current carrying capacity of a 2.5mm[SUP]2[/SUP] ring?

For all other circuits we know the current carrying capacity of the cable so we use the appropriate MCB to protect the cable. Why do we use a 32A MCB for a ring, this means the conductors must be able to handle more than 32A, but how do we know this? The regs have a nice set of tables for all other size conductors but not for a ring.

 

[Deleted member 26818]

Whenever we design a ring, the conductors should have a minimum CCC of 20A.
This would technically give us a CCC of 40A at the point in the middle of the ring.
The 40A is split between the two conductors 20A on one leg and 20A on the other.
However, at a point a quarter of the way around the ring, ¾ of the current will be on the short leg and ¼ of the current on the long leg.
Which would mean 10A on the short leg and 30A on the long.
BS1363 double socket-outlets are rated at 20A (although some manufacturers' rate theirs higher), so we should only have a maximum of 20A at each socket-outlet.
That's 10A on each leg in the middle of the ring, and 5A and 15A at a quarter way round the ring.
Obviously the closer the first socket-outlet is to the start of the ring, the greater the current that will be on the short leg.
30A (now 32A) was chosen, because it's halfway between 20A the CCC of each cable and 40A the combined CCC of the two cables

 

[BigAndyB]

So, Spinlondon, ever heard of Kerchoff's current law, it's works a bit different to your theory,

 

[telectrix]

no, spin is correct. assume for easy maths, that a socket close to the CU has a resistance on the short leg of 1 ohm and on the long leg of 4 ohms. assume the load on that socket is 20A. the current in the short leg will be 16A and the long leg 4A.

 

[Deleted member 26818]

So, Spinlondon, ever heard of Kerchoff's current law, it's works a bit different to your theory,

Kirchoff's current law:
At any node in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.

 

[brman]

Kirchoff's current law:
At any node in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.

or his voltage law....
"The sum of the potential differences around any closed network is zero"

still not sure what that has to do with your earlier statement though :lol:

 

[BigAndyB]

Ok, hand up!
I miss read something, been talking crap.