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I

irishsailor

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Can anyone shed a little light on this please

How is the cable factors worked out on a ring??

32A Circuit breaker 2.5 T&E clipped direct 27A

How is the calc worked out for the above as I wa on the understanding that the cable capacity has to be greater than the Fuse rating??

Can you please explain

Irish
 
W

WayneL

  • Thread Starter Thread Starter
  • #2
Hi mate,

If it was a 2.5 t&e radial the capacity would be 27A and you would use
a 20A breaker.

Being a ring, the current carrying capacity is higher (not sure how much-haven't got
my regs on me) - but its more than 32 A.

Hope that helps,

Cheers,

Wayne
 

ian.settle1

-
Mentor
Arms
Hi mate,

If it was a 2.5 t&e radial the capacity would be 27A and you would use
a 20A breaker.

Being a ring, the current carrying capacity is higher (not sure how much-haven't got
my regs on me) - but its more than 32 A.

Hope that helps,

Cheers,

Wayne

It is because you have 2 x 2.5mm T&E in parallel that is why you can use a 32A breaker on a ring circuit.
 
I

irishsailor

  • Thread Starter Thread Starter
  • #4
So does that effectively mean that u have 54 amp in the 2.5
 
A

amberleaf

  • Thread Starter Thread Starter
  • #6
general domestic premises there should be a separate ring

Final circuit for every 100mm2 of floor area. must remember
That the maximum load that can be connected to a ring protected
By a 30A / 32A fuse is just over 7 kW, and the number of circuits Selected accordingly

rule of thumb


( 433.1.5 p : 74

Ring final circuit volt drop
Ring final circuit : ( cable doubled )

<<< Cable length to farthest point >>>
is ½ ( cable length )
2.5mm2 Line conductor :
1.5mm2 Protective conductor :

to the farthest point in effect two cables are run , so ( mV / A / m ) is halved and the length
of doubled cable to the farthest point is half the length of cable in the ring , given a multiplier of ( 4 )

the design current ( Ib ) in each leg of the cable is not to exceed 20A ,
for ring circuits the equation becomes :


Multiplier
4 x 11.5 x 1000
Lvd = -------------------------------------- for ring circuits
Ib x ( mV / A / m ) x Ct

Ring final circuit wired with 2.5mm2 regs : 4D5 , p: 282 ( mV / A / m ) = 18

mV / A / mand assumingIbis 32A and Ct is 1 then
*
( The maximum circuit length is then the smallest of 2 ) to 4 )
Multiplier5% amp mV meters
Lvd = 4 x 11.5 x 1000 / 32 / 18 = 79.9 m
*
Multiplier
Answer 4 x 11.5 x 1000 div 32 div 18 = 79.9 m
Answer 2 x 11.5 x 1000 div 32 div 18 = 39.9 m

However , in a ring circuit the current in a ring will not be the same all round the ring .
If it is assumed to be 20A at the far end and additional 12A is evenly distributed ,
The average current is ( 32A + 20 ) / 2 = 26A , and
Answer ( 32 + 20 = 52 div 2 = 26amp


Multiplier5% amp mV meters
Lvd = 4 x 11.5 x 1000 / 26 / 18 = 98.3 m
*

If Ct is to be calculated assuming Cg is 1 , cable Ib is 26 / 2 , conductors It is 20 and tp is 70oC ,
Then Ct = 0.923 and Lvd = 106.5m
 
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