Cheers guys for you help. It was the last part of this example that has sent me of track slightly. The text at the end in green.
When a number of correction factors applies
In some cases all the correction factors will need to be applied because there are parts of the cable which are subject to all of them. For example, if a mineral insulated cable with p.v.c. sheath protected by a circuit breaker and with a tabulated rated current of 34 A is run within the insulated ceiling of a boiler house with an ambient temperature of 45°C and forms part of a group of four circuits, derating will be applied as follows:
Actual current rating (lz)
= tabulated current (It) x ambient temperature factor(Ca) x group factor (Cg) x thermal
insulation factor (Ci)
= 34 x 0.77 x 0.65 x 0.5A =
8.5A
In this case, the current rating is only one quarter of its tabulated value due to the application of correction factors. A reduction of this sort will only occur when all the correction factors apply at the same time. There are many cases where this is not so. If, for example, the cable above were clipped to the ceiling of the boiler house and not buried in thermal insulation, the thermal insulation factor would not apply.
Then, Iz = It x Ca x Cg = 34 x 0.77 x 0.65 A =
17.0 A
The method is to calculate the overall factor for each set of cable conditions and then to use the lowest only. For example, if on the way to the boiler house the cable is buried in thermal insulation in the wall of a space where the temperature is only 20°C and runs on its own, not grouped with other circuits, only the correction factor for thermal insulation would apply. However, since the cable is then grouped with others, and is subject to a high ambient temperature, the factors are:
Ci = 0.5
Ca x Cg = 0.77 x 0.65 = 0.5
The two factors are the same, so either fbut not both) can be applied. Had they been different, the smaller would have been used.
