CSA on a Radial, what length would you reach with a 4mm?

So I have been down a rabbit hole for some time now and need a bit of clarification.

Possibly someone might be able to explain Fig 15B - Radial final circuit arrangement

it's suggested that 4.0/1.5mm cable could serve a 75 m2 on a 30A or 32A CB

what would the length be of that cable?

when I perform a voltage drop calculation the length of the cable is 32m which is no where near 75m.

The OSG mentions that the max length for a radial on a 32A CB is 43m with an assumed load of 26A.

Anyone able explain how 75 min fig 15B has been worked out, surely there should be a not stating what length cable its equivalent to.

on a separate note what if the radial circuit was supplying 1 x 13A socket, is there an equation to work out the overload allowance of a circuit or is it simply Ib < In < Iz?

if so then what would the maximum load be expected on a 13A socket? there's nothing stopping someone plugging in an extension lead or adaptor to increase the load.

Parents
  • The OSG mentions that the max length for a radial on a 32A CB is 43m with an assumed load of 26A.

    I can confirm that the length in Table 7.1(ii) in the OSG is based on the shortest length given by calculating for:

    • EFLI (Tables 41.2 to 41.5 as appropriate)
    • Voltage drop for the assumed load
    • Where fuses are used, adiabatic (where overcurrent protection is provided by mcb or RCBO, then let-through energy for a cross-section of typical devices on the market is addressed - hence the 'NPad' under Types C and D)
    Anyone able explain how 75 min fig 15B has been worked out, surely there should be a not stating what length cable its equivalent to.

    There is no longer an actual floor area served "limit". The floor area recommendation is based on average usage of power per floor area in typical situations. Note that older data is likely not relevant to a modern dwelling (or perhaps even office block).

    Regarding the length of run for the floor area, this typically depends on a number of factors. If you think about it, to "serve" an area the shape of a rectangle, you would typically need to run along 2 walls (perhaps 3 if the dimensions are above 4 m on the shortest side). To calculate a rough length of cable for the 75 sq m, the table below shows cable length required for 2 sides of a rectangle with area 75 sq m with sides a and b (in m):

    a (m) b (m) Total area (sq m) Total cable length required for 2 sides (m)
    1.00 75.00 75 76.00 (NOTE this is too long for 4 sq mm)
    2.00 37.50 75 39.50 (NOTE this is getting near the max length of 43 m for 4 sq mm)
    3.00 25.00 75 28.00
    4.00 18.75 75 22.75
    5.00 15.00 75 20.00
    6.00 12.50 75 18.50
    7.00 10.71 75 17.71
    8.00 9.38 75 17.38
    9.00 8.33 75 17.33
    10.00 7.50 75

    17.50

    Of course, there are no 'vertical runs' in the above, but it does show that provided the rectangle has one side of at least 2 to 3 m, there is enough cable length with 43 m for some vertical runs as well as horizontal runs. If the floor is raised, permitting you to wire under-floor, you might have more possibilities.

    So the conclusions are:

    (a) Floor area served has nothing directly to do with cable length, because the length of run required to serve the area depends on the shape of the perimeter (unless you can wire under-floor to floor boxes).

    (b) It's possible to serve 75 sq mm with 43 m of 4.0/1.5 cable, BUT it depends on the shape of the perimeter of the area served (and whether you can use floor boxes in a raised floor).

  • Agree with your comment regarding floor area, it's a shame they haven't made a note to clarify it in that figure.

    Is it fair to say that VD determines the length of a circuit and changing the load on that circuit could have a significant impact on its length?

    I recalculated using a fixed 10A load and manged to get the VD below 5% with length of 104 meters on a 4 sq mm.

    (vd x 1000) / ((mv/m/a) x lb) = L

    (11.5 × 1000) / (11 x 10) = 104m

    so if the CB was changed to a type B 10A the length of the circuit could be increased as long as Zs was achived.

    What would the assumed load be for a single socket outlet?

    is there a formula to allow for overload current?

  • so if the CB was changed to a type B 10A the length of the circuit could be increased as long as Zs was achived.

    To achieve Zs, you could use an RCD (or RCBO), then the Zs target is Table 41.5 (not Tables 41.2 to 41.4) ... in which case volt-drop is usually the limiting factor.

    What would the assumed load be for a single socket outlet?

    13 A, or, if you assume an unfused adaptor might be used, 20 A.

    This illustrates a good (and related) point, though ...

    Assume I have a house in which I decide to have 20 circuits for individual socket-outlets, rather than two circuits with 10 socket-outlets each. I would need to apply the same type of rule for each circuit (I don't know which, if any, circuit might be overloaded). So, the demand for each individual circuit will be 13 A or 20 A (per above), BUT would the maximum demand of the property be any different ... probably not, which is how the industry got to using Wm-2 as a benchmark for overall demand.

    It also shows that you can't determine maximum demand by adding up the ratings of individual circuits.

  • Regarding the length of run for the floor area, this typically depends on a number of factors. If you think about it, to "serve" an area the shape of a rectangle, you would typically need to run along 2 walls (perhaps 3 if the dimensions are above 4 m on the shortest side). To calculate a rough length of cable for the 75 sq m, the table below shows cable length required for 2 sides of a rectangle with area 75 sq m with sides a and b (in m):

    I take the point, but does this not assume that the DB is adjacent to the area being served?

    A couple of examples might be a DB by the front door serving a conservatory at the back of the house, or quite commonly, a garage. The area served might be well under 75m², but much of the length is used up getting there.

Reply
  • Regarding the length of run for the floor area, this typically depends on a number of factors. If you think about it, to "serve" an area the shape of a rectangle, you would typically need to run along 2 walls (perhaps 3 if the dimensions are above 4 m on the shortest side). To calculate a rough length of cable for the 75 sq m, the table below shows cable length required for 2 sides of a rectangle with area 75 sq m with sides a and b (in m):

    I take the point, but does this not assume that the DB is adjacent to the area being served?

    A couple of examples might be a DB by the front door serving a conservatory at the back of the house, or quite commonly, a garage. The area served might be well under 75m², but much of the length is used up getting there.

Children
  • I take the point, but does this not assume that the DB is adjacent to the area being served?

    Almost certainly not very far away... OSG and Appendix 15 are predominantly aimed at 100 A single-phase domestic and similar installations, where this is true.

    If you look at the total length permitted for a 32 A radial run in 4.0/1.5, which is 43 m, it doesn't allow much of a run before the intended room - but that's just physics.

    Given the constraint in terminal capacity (at least in the BS 1363-2 standard ... there's nothing stopping a manufacturer exceeding the requirements of course), a larger CSA cable starts to become quite a challenge also!

    Such is physics.