Help With Circuit Design

[App_Rentice]

Guys, I have an assignment which I am struggling with but which should be easy and have read loads of books I keep going around in circles. Can you please put me back on the tracks.

The task is for the design of a RFC in a garage with the routing clipped a few centimetres below the ceiling and taken down single 20mm conduit from this height to each of the 4 double outlets. The conduit is to provide additional mechanical protection.

The Ref method is method B because of the conduit limiting the capacity for the 2,5mm cable to 23A (From 30A) [Table7.1(ii) or Table F5(i) OSG], as this is a ring to be protected by a 30 breaker this is still ok. However, as both cables go down the same conduit to the outlet there should be a grouping factor to consider. Looking at Appendix F (OSG) there is a general equation It=In/(CaxCgxCixCf), but I have three issues.

Firstly there appears to be no Cg factor for a conduit delivery in Table F3 OSG. Where can I find this info.

Secondly the numbers in table F3 get smaller the more cables, and as they go on the bottom line of the equation they will increase the It value where I would have expected it to decrease.

Thirdly once I get the grouping factor how do I apply it to the Ref Method B data to get a revised capacity.

Thanks

 

[telectrix]

i don't think that grouping factor comes into it as it's only 1 circuit in the conduit.

 

[Spoon]

As what Tel says.

 

[Richard Burns]

The Ref method is method B because of the conduit limiting the capacity for the 2,5mm cable to 23A (From 30A) [Table7.1(ii) or Table F5(i) OSG], as this is a ring to be protected by a 30 breaker this is still ok. However, as both cables go down the same conduit to the outlet there should be a grouping factor to consider. Looking at Appendix F (OSG) there is a general equation It=In/(CaxCgxCixCf), but I have three issues.

Firstly there appears to be no Cg factor for a conduit delivery in Table F3 OSG. Where can I find this info.

Secondly the numbers in table F3 get smaller the more cables, and as they go on the bottom line of the equation they will increase the It value where I would have expected it to decrease.

Thirdly once I get the grouping factor how do I apply it to the Ref Method B data to get a revised capacity.

Thanks

I would agree with the above: single circuit, but just to explain.

The grouping factors in the case are the first row of the table "bunched in air, on a surface, embedded or enclosed" (the cable is enclosed).

As you have more cables close together any specific cable can take less current before overheating. Therefore when designing the circuit the current carrying capacity of the cable needs to be higher so that the current it is carrying does not overheat the cable.
When you have a grouping factor as a smaller number with more cables, then the calculation gives an increase in the minimum permitted current carrying capacity of the cable.
e.g. if you have a design current of 19A and a cable that can take 23A and a breaker of 20A this is fine as a single radial circuit. If you then have two circuits in the conduit, so grouping of 0.8, the current carrying capacity of the cable needs to be 20A/0.8 = 25A. So your cable is now too small (as it can take only 23A) so you need to increase the size of the cable to compensate.