Volt drop for ring circuit

[Albert]

When calculating the volt drop for a radial circuit we take in consideration the design current Ib, (imagine that we do not have correction factors),
according to OSG table 8A page 151, for a radial circuit we can use 2.5 mm2 with protective device of 20A, as we do not know the load (socket outlets), I will assume Ib=20A.
Using the same conditions with a ring circuit I can use 2.5mm2 cable with 30-32A protective device, (I know that for socket outlets it is recommended to use ring circuit). If I want to use ring circuit what will be the value of Ib for the volt drop calculation?
Thanks, Albert

 

[plugwash]

32A

but the resistance is halved becuase of the paralel paths

 

[Albert]

32A

but the resistance is halved becuase of the paralel paths

Thanks
just to clarify, OSG page 122 table 6D2 gives the values of volt drop according to the method and the size, if for example i will take the 2.5mm2 method 3 the voltage drop is 18 mv/A/m, if I understood correctly the value of the voltage drop for a ring will be 9 mv/A/m (obviously I need to take in consideration the difference in the length of the cable, or not?)
Albert

 

[plugwash]

let Rp be the resistance to your point and Rt be the total resistance of the cable from the CU back to the CU and X be a variable beteen 0 and RT representing how far along the ring you are

[code:1]
Rp=(X*(Rt-X))/(X+(rt-X));
=(X*(Rt-X))/rt

DRp/DX=-X+rt-X
=rt-2X

at the point of greatest resistance DRp/DX=0
Rp=2X
X=1/2Rp
[/code:1]
so the point of highest resistance is halfway found the ring

therefore you should analyse voltage drop as the drop at the midpoint of the ring
as you have two paralell paths half the current flows each way so the voltage drop is half what it would be with a single conductor to that of the same size and carying the same current

 

[Albert]

Thanks
Albert

 

[Albert]

let Rp be the resistance to your point and Rt be the total resistance of the cable from the CU back to the CU and X be a variable beteen 0 and RT representing how far along the ring you are

[code:1]
Rp=(X*(Rt-X))/(X+(rt-X));
=(X*(Rt-X))/rt

DRp/DX=-X+rt-X
=rt-2X

at the point of greatest resistance DRp/DX=0
Rp=2X
X=1/2Rp
[/code:1]
so the point of highest resistance is halfway found the ring ...etc.

Thanks
So the cable length to consider is half way meaning the same length as if it would be radial circuit, the Ib will be the 32A and the Vd in mv/A/m will be half of what in the table.
(Sorry to drive you made, but i asked this question in college and could not get a clear answer)
Albert

 

[Big_Spark]

Albert, everything said above is true, however the actual volt drop of a given size cable remains the same, this is dictated by physics and making the circuit into a Ring will not mean the volt drop will reduce as a result..be blooming nice if it did.

The volt drop is mV/A/M/1000, whether the circuit is a ring is irrelevent.

The difference is that when calculating it for a Ring circuit, you must half the distance. ie: If the TOTAL circuit length is 66 meters from the board, round the circuit and back to the board, then the voltage drop can be calculated using half that distance, as in 33 meters.

However this is notoriously difficult as rings can be unevenly loaded very easilly. The usual method is to take the maximum circuit length, do the volt-drop calculation, then halve it to ensure compliance.

 

[Albert]

Albert, everything said above is true, however the actual volt drop of a given size cable remains the same, this is dictated by physics and making the circuit into a Ring will not mean the volt drop will reduce as a result..be blooming nice if it did.

The volt drop is mV/A/M/1000, whether the circuit is a ring is irrelevent.
.

Hi FWL_Engineer, long time no hear,

I understand what the Vd values in the table is based on (cable material, csa etc...) so I understand as well that it does not make a difference whether it is a radial or Ring, what I tied to do it to make an analogy, If we use a radial circuit the circuit will end at the last socket, right?, if we use ring the circuit will start to return from the last socket, assuming that the last socket is the first half of the ring, this is the length I will put in for the calculation, where this will be the case with the Radial.
Up to this stage am I wrong, please enlighten me with your wisdom...
Albert

 

[Big_Spark]

The usual method is to take the maximum circuit length, do the volt-drop calculation, then halve it to ensure compliance.

You can do it that way, or the way you stated.

Check out my sig

 

[Albert]

I wonder if this argument makes sense,
A friend electrician for many years, claims that he never uses ring circuits, because for 2.5mm2 you can use 32A CB, the problem (according to his theory) if the circuit breaks somewhere, you will still have the circuit under power, but you will have now 2 radial circuits of 2.5mm2 wire, and 32A CB, which sounds wrong. What do you think?
Albert

 

[Big_Spark]

Hi thinking is flawed.

Whilst he is correct that if your split the ring you end up with two seperate Radial circuits, this can be easily rectified.

Regarding 2.5mm T&E, Table 6D2A shows that this has a MAXIMUM rating of 30A if clipped on tray.

And bear in mind that the entire circuit would be required to be on the tray or the 30A would be subject to be derated to the rating of the lowest installation method of the circuit.

ie: If any part of the circuit were to run through insulation in a wall, be buried in a wall or something similar, you MUST use the rating for this installation method to do your calculations, and you must still derate using correction factors. For Domestic Installation, and many commercial ones too, the Reference Installation Method used is 4, thus giving 2.5mm a rating of 18.5A, BEFORE the correction factors are accounted for.

 

[Albert]

FWL_Engineer

I looked at the AE3 site, the idea sounds great, (based on my experience as an engineer in a different field). As I will be qualified as an electrician in coming June it seems to be at the right time. I thing that you must make sure, that this will not go down as other sites, as all start nice and neat.
Hope You have the stamina, to keep it so, this is not easy, good luck
Albert

 

[Big_Spark]

FWL_Engineer

I looked at the AE3 site, the idea sounds great, (based on my experience as an engineer in a different field). As I will be qualified as an electrician in coming June it seems to be at the right time. I thing that you must make sure, that this will not go down as other sites, as all start nice and neat.
Hope You have the stamina, to keep it so, this is not easy, good luck
Albert

I have the stamina and the drive. This has been brewing for a number of years, never had the time before, now I am MAKING the time

 

[pegasus]

old topic I know....... but could someone please re-clarify how you can calculate voltage drop in a ring circuit.

FWL Engineer mentions that the ...."The usual method is to take the maximum circuit length, do the volt-drop calculation, then halve it to ensure compliance."

However, in another thread

//www.diynot.com/forums/viewtopic.php?t=15482&highlight=ring+circuit

it states that you work out the voltage drop for quarter of the length.

The two methods seem to contradict each other.

Many thanks

 

[Damocles]

It is a quarter. If you had 60m of cable all together, the worst place is half way round. Then distance from CU is 30m (half), but there are two paths back. So half again.

The discussion is right about the theoretical risks. If the circuit breaks near one end, then everything is on a single 2.5mm with 32A breaker. And worst of all, all the sockets would be working.