I have an existing 8kw shower with 6mm cable. However I am now fitting a shower rated at 9.8kw at 240v or 9kw at 240v.
The 6mm cable runs from consumer box, clipped to the surface of the wall, then into empy ceiling joists, then inside metal covering chased into a solid brick wall, then into the roof resting on the surface of the roof insulation.
What installation method from the 17th regs is this ??
Then ultimately can I still use the 6mm or do I need 10mm
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None of the Appendix 4 methods exactly matches this. The anally minded would say use ref. method 100 or 101, depending on the insulation thickness, giving ratings of 34 or 27 A respectively [table 4D5A]. For the new shower Ib is 39 A so method 100 would need 10 mm^2 cable and method 101 would need 16 mm^2!
That's a bit OTT though and I'd suggest that Method A and Table 4D2A would be OK - see the 6th note in section 7.2 of appendix 4 - so the ratings become 32 A for 6 mm^2 and 43 A for 10 mm^2 and the latter size is still required.
The best option though is to re-route the loft section so it's not in contact with thermal insulation - see reg. 523.7, first paragraph. Then revert to method C, table 4D5A and 6 mm^2, voltage drop and Zs permitting.
It is one that came up at work. The NICEIC were quite happy with method C as the majority of the cable was in free air. ie only one side was in contact with the insulation. Obviously there were no clips!
Yes, ideally the cable would not be in contact with the insulation, or in a position where it may be totally covered with insulation later eg more loft isulation added.
I never really thought about this in depth before...
All refs below to 17th Edition:
Appendix 12 refers to 5% of the "nominal voltage of the installation".
Nominal voltage is defined (in Part II definitions) as a range with the actual value differing from the notional nominal value by a quantity "within normal tolerances".
This suggests that *any given installation* has a nominal voltage which might be defined as "the long term-average voltage at the meter tails of a specific installation (i.e. taking into account daily, weekly and seasonal cyclic and peak supply voltage variations and cyclic and peak load variations), providing that such a long term average is within normal tolerances".
i.e. under this definition, nominal voltage is specific to the installation.
However, Appendix II, states that UK nominal voltage is 230V with "permitted tolerance" of +10/-6%
Finally, we all "know" that the nominal voltage in the UK is 240V not 230V.
So, which figure do you take as the nominal voltage from which you calculate permissible voltage drop?
230V, 240V or the actual nominal of the specific installation?
I'm amazed by that. Prior to the 17th ed. 'one side in contact with thermal insulation' meant a 25% derating. I suppose though that if it snakes unclipped across the loft much of it we be in clear air above the insulation, just touching in places. Now suppose someone goes in the loft and deposits a box or old duvets (etc.) on top of the cable...
Have you got a reference you the 25% derating. All I can find is sheathed cables clipped direct or lying on a non-metallic surface is method 1.
There is always the danger of stuff being piled on top of the cable. Not only could this act as thermal insulation it could also bury the cable into the insulation by pushing down on it so that it sinks into the insulation.
And there seems to be little regulation in the insulation business with regards to covering cables. With all the insulation that is thrown into lofts these days via the grant schemes I suspect that I will see many overheated cables in the future.
The fact that the appliance is rated at 9.8kW at 240V or 9kW at 230V tells us (R = U^2/P) that its resistance is 5.878 Ohm.
If the cable drops 5% of the voltage, then the appliance will "see" only
95% of 240V or 230V as the case may be, and so the current through the appliance will also be only 95% of what it would be if it were connected via a zero resistance cable. So the current won't be 40.83A or 39.13A, but only 38.79A or 37.17A, respectively for nominal 240V or 230V.
For 240V I get L = 12V / ((7.3mV/A/m) * 38.79A) = 42.4m, and for 230V I get L = 11.5V / ((7.3mV/A/m) * 37.17A) = 42.4m, and it's no coincidence that the lengths are the same for both nominal voltages. This is because the "mV/A/m" figure is basically the cable's resistance in milli-Ohm per metre. To drop 5% of the voltage in the cable, the cable's resistance must be 5% of the combined resistance of cable plus appliance, so the cable resistance must be (5/95 or) 5.263% of that of the appliance. Since the latter is 5.878 Ohm, the former must be 309.4 milli-Ohm.
Divide that by 7.3 milli-Ohm/m and what so you get? 42.4m!
Obviously if you do use the full maximum cable length, and thus get the full 5% voltage (and hence current) drop, the appliance's power will drop by nearly 10% to 8845W at 240V or 8123W at 230V. Note that you would be dissipating (wasting) 466W or 428W in the cable, or about 10-11 W/m. Those sound like pretty good reasons, if the cable can't be any shorter, to upgrade to a thicker one.
is so that you can work out if the diameter and length of cable is suitable for the intended appliance. It is not there to work out the actual running current or power of the appliance. For a domestic appliance you do not need not know the actual running current.
I (in this equation) is the design current of the appliance not the actual current. The actual current will be lower than the figures you have worked out when you add on the external resistance of the supply.
I didn't use the formula to work out the actual current. I worked out the actual current by simply deducting 5% from the design current, based on the assumption that the maximum permitted length of cable was going to be used, thus giving the maximum permitted voltage drop of 5%.
Using the formula in reverse then tells us what length of cable would give the 5% voltage drop at that current.
But it would make for a more accurate result if you did use the actual current. On the other hand, it would give the wrong answer anyway, because the 7.3mV/A/m figure is actually the wrong value for working out the actual voltage drop, since it's really more of a rating, which incorporates various fiddle factors for safety.
I would expect the external resistance to be negligible, giving a voltage drop between substation and the cable in question very much lower than that in the cable itself. In fact, I suspect that the actual current will be
*higher* than the figures I've worked out, because the voltage drop in the cable will almost certainly be *less* than that calculated on the basis of
7.3mV/A/m.
The resistivity of copper is about 17 nano-Ohm-metres and so a conductor of
6 mm^2 CSA should give about 2.83 milli-Ohm per m. Doubling this for flow and return means that a twin 6mm^2 cable should drop only about 5.7 mV/A/m at 20 degrees C.
For temperature coefficient to bump that up to 7.3, the conductor would need to get bloody hot first. Copper's TC is about .004 per K, so we're talking in excess of 90 degrees C.
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