18-1 | Current Carrying Capacity of Conductor
About current carrying capacity
When current flowing through a conductor, it will generate heat. The higher the current, more heat is produced and when insulator material reaches hottest peak temperature, insulation material will fail.
Current carrying capacity is defined as the amperage a conductor can carry.
This is very important to calculate correct sizing of the conductor before installation of the circuit is been done. If the value of current is higher than design current it may cause the following condition:
the cable will melt the conductor or the insulation can then cause fault.
may risk to creating short circuit fault.
may cause fire.
According wiring regulation, the cable current carrying capacity must be equal to or greater than the circuit design current .These requirements are to protect the cable, in case of overloaded circuit. These other influences are:
high ambient temperature
cables grouped together closely, uncleared overcurrents
contact with thermal insulation.
Definitions or terms:
Iz - is the current carrying capacity of the cable in the situation where it is installed.
Ib - is the design current, the actual current to be carried by the cable In is the rating of the protecting fuse or circuit breaker.
In - is the circuit protection rating of overcurrent device, fuse and miniature circuit breaker (mcb)
I2 - is the operating current for the fuse or circuit breaker (the current at which the fuse blows or the circuit breaker opens)
It - is the tabulated current for a single circuit at an ambient temperature of 30°C
Formula calculation
Current carrying capacity ;
Formula 1: Iz = In ÷ (Ca x Cg x Ci)
Formula 2: Iz = In ÷ (Ca x Cg x Ci x 0.725) (If In is using rewirable fuse type)
Refer to formula 2, the correction factor for protection by a semi-enclosed (rewirable) fuse is not given a symbol but has a fixed value of 0.725.
Tabulated Current ; It = Ca x Cg x Ci
Operating current to blow or trip cb ; I2 = Iz x 1.45
Design current ; Ib = P ÷ 240V or Ib = P ÷ ( √3 x 415V )
Important Rule for Conductor , Cable
Iz > Ib - current carrying capacity of conductor must be larger than current design.
In > Ib - current rating protection must be larger than current design.
Correctional Factor on Ambient Temperature (Ca)
The transfer of heat, whether by conduction, convection or radiation, depends on temperature difference. Ambient temperature factor is taken depending on site, example a hot room environment, consider take temperature reading to determine the factor.
Example a boiler room has ambient temperature of 46 deg.Cel. The type of cable used for the circuit wiring is rubber cable and the protection rating used is miniature circuit breaker. Therefore the value of Ca taken referring table on the right is 0.80 (50 deg.Cel, rubber selected).
Correctional Factor on Cable Grouping (Cg)
When cables is installed together inside a conduits or trunking metal these conductors is carrying current, they will heat up due to current flowing in the circuit.
Refer to figure (i) when cable is install inside a metal trunking is less cable and have free space then heat dissipate easily.
However figure (ii) showing many cables is put inside the metal trunking and no much free space is available then heat dissipate is slower. When more cable is group together the overall cables will be hotter.
(i) widely spaced cables dissipate heat easily.
(ii) A closely packed cable cannot easily dissipate heat and so its temperature rises.
Example a circuit is laid with two(2) other circuits in the premise, all are installed in PVC conduits (enclosed), referring to the table Ci above the value of Cg is 0.7 (refer to enclose column 2+1 total 3).
Correctional Factor on Thermal Insulation (Ci)
The use of thermal insulation in buildings, in the forms of cavity wall filling, roof space blanketing is to limit the transfer of heat from rooftop to ceiling space. This affects the ability of a cable to dissipate the heat buildup quickly if cable is installed nearby the roofing material.
Figure (i) Figure (ii)
Figure (i) is showing when daytime sunlight during in the morning or afternoon will cause heat to the concrete walls of a house or building. The wall if it is hollow brick then there is air flow and this will reduce the heating effect toward the cables.
Figure (ii) is where cables is installed near to roof top, if the roof top is metal then the heat effecting the cable is greater compare to cement tiles type or other heat insulated roofing materials. If the cable installed have a gap distance between the metal roof and the cable then the heat effects is lesser.
Figure (iii) on the left showing a ceiling space either is concrete or wooden ceiling in an shoplot or office environment. The spacing of ceiling toward the false ceiling need to take measurements and refer it with the table Ci factor below. The Ci value is 0.55 because the spacing is 345mm and we take the value from 400mm.
Example Calculation
Example:
1) A lighting circuit with total power of 520 watts/240v, the cable is using PVC cable 1.5mm.sq. The circuit is protected by a miniature circuit breaker (mcb). The correctional factor for Ca is given 0.8 and, Cg is 0.75 and Ci is 0.55.
a) Calculate the current design (Ib) and find the protection rating, In ?
Solution: Ib = P ÷ V
Ib = 520w ÷ 240v = 2.166 ampere
MCB rating for lighting circuit is 6 A.
b) Calculate the current carrying capacity of conductor for this circuit, Iz.
Solution: Iz = In ÷ (Ca x Cg x Ci)
Iz = 6A ÷ (0.8 x 0.75 x 0.55) = 18.18 ampere
c) Calculate the operating current to trip mcb, I2?
Solution: I2 = Iz x 1.45
I2 = 18.18 x 1.45 = 26.36 ampere
2) A machine with total power of 6200 watts/415v, the cable is using PVC cable 6.0mm². The circuit is protected by a rewirable fuse. The correctional factor for Ca is given 0.75 and, Cg value is 0.8.
a) Calculate the current design (Ib) and find the protection rating, In ?
Solution: Ib = P ÷ V
Ib = 6200w ÷ 415 v = 14.93 ampere
The rewirable fuse rating for the machine is 20 A.
b) Calculate the current carrying capacity of conductor for this circuit, Iz.
Solution: Iz = In ÷ ( 0.725 x Ca x Cg x Ci )
Iz = 20 A ÷ ( 0.725 x 0.75 x 0.8 ) = 45.97 ampere
c) Calculate the operating current to blow the fuse, I2?
Solution: I2 = Iz x 1.45
I2 = 45.97 A x 1.45 = 66.65 ampere