Sim's Electrical Companion
Basic Cable Calculations - Ib<In<Iz
All electrical installations must satisfy:
Ib - Design Current - The maximum current the circuit is designed to carry
In - Protective device - Must be equal to or greater than the design current
Iz - The current carrying capacity of the cable - must be equal to or greater than the protective device.
Ib ≤ In ≤ Iz
* Smaller diameter cables may be used if the installation is in mineral insulated copper cable. Conductors may need to be larger than stated depending upon installation reference methods and environmental factors.
Single Device Radial Circuits
​
Examples include heating devices, showers, instantaneous water heaters, car charging points and cookers etc. These devices will be fed by a single cable that serves no other loads. To determine the design current the power of the device must be divided by the supply voltage to determine the current rating.
Power law I = P / V.
Multiple Load Circuits
​
The total intended load should be added up to determine the design current. Historically standard lighting circuits would be counted as 100W per light fitting. In modern lighting if a type is being installed with no user serviceable parts then the manufacturers power consumption may be stated per fitting instead of 100W per fitting.
​
In - Protective Device
In must exceed Ib to prevent nuisance tripping or fuse operation. The common values are listed below:
Iz - Current Carrying Capacity of Cables
Current Carrying Capacity: Iz
Mixed Reference Methods
​
Single and three phase installations in different reference methods must use the lowest rating to ensure compliance.
If the phases are mixed some single phase and some three phase use the lowest figure to allow for the magnetic effects of three phase cables in the same containment.
SWA - Steel Wire Armour
​
Steel wire armour is commonly used for low voltage electrical distribution. Notably the table left shows it is most effective in free air single phase.
For more tables including (MICC)mineral insulated copper cables, aluminium cables, flexible cables and a greater range of cross sectional areas reference Appendix 4 of BS 7671.
Iz - Correction Factors
Heat is the greatest inhibitor of a cables current carrying capacity. When surrounded by thermal insulation, bunched together or in high ambient air temperature a cable cannot shed its heat so its resistance to current flow is increased.
If you consider heat to be movement at an atomic level it makes sense that a lot of atomic movement will hinder the passing of electrons.
​
The correction factors are:
Ca = Ambient Air temperature
Ci = Cables in thermal insulation
Cg = Cables grouped or bunched together
Cf = Fusing factor allows for a slower acting protective device a BS3036 semi-enclosed fuse. The rating factor is x 0.725
As can be seen by the table that 30 degrees Centigrade is considered to be par for a cable and no derating is applied. When it is at least 5 degrees lower the rating factor actually increases the cables ability to carry current.
Cables Grouped or Bunched Together - Cg
Application of Correction Factors
​
Once a cable has been selected that satisfies Ib ≤ In ≤ Iz the selected cable must then have its current carrying capacity multiplied by all the appropriate correction factors.
The selected tabulated current symbol is: It
​
To satisfy the requirements the tabulated current carrying capacity must exceed the selected protective device rating divided by the products of all applicable correction factors:
It ≥ In / (Ca x Ci x Cf x Cg)
​
Worked Example 1
​
A single phase low voltage motor is rated at 11.5 kW. It is to be installed in a boiler house that has an ambient air temperature of 35°. The cable is a 90° thermosetting three core steel wire armour cable which will be secured to a perforated cable tray & grouped with one other circuit. The cable does not encounter any thermal insulation throughout its forty five meter run.
Design current can be found by applying power law:
Ib = P / V 11500W / 230V = Ib = 50 Amperes
​
In Protective device can be equal to or higher than Ib so I select a 50A circuit breaker Type C for motors.
​
It Tabulated current carrying capacity must be higher than 50 Amperes. The table for SWA thermosetting reference method E, cable tray supported, single phase cable:
It = 53 A for a 6mm² conductor
​
This satisfies:
Ib ≤ In ≤ Iz 50 ≤ 50 ≤ 53
​
Apply the correction factors to ensure that the cable can carry more than the rating of the protective device.
Ca The boiler house ambient air temperature of 35° will incur a factor of 0.96 for PVC thermosetting SWA.
Ci There is no thermal insulation in direct contact with the cable so the factor = 1
Cf The protective device is a 50 Ampere B type circuit breaker not a BS 3036 fuse so the factor = 1
Cg A single layer multicore cable grouped with one other cable on a perforated cable tray has a factor of 0.88
It = 53 50 / ( 0.96 x 1 x 1 x 0.88) ≈ 59.2A
​
It is not greater or equal and will not safely operate in these conditions.
​
Going back to the table and selecting a cable that can carry more than 59.2 Amperes.
A 10mm² Steel armoured cable on a perforated tray can carry 67 Amperes which satisfies:
​
It ≥ In / (Ca x Ci x Cf x Cg)
67 > 50 / (0.96 x 1 x 1 x 0.88)
so: 67 > 59.2
​
This satisfies the requirements of:
​
Ib ≤ In ≤ Iz
​
​
Worked Example 2
​
A domestic single phase 8kW shower circuit is grouped with two other circuits. For ease of installation the circuit is run through the non insulated walls and into the loft space. In the loft space it is under 200mm of thermal insulation and is touching the surface of the ceiling board. The circuit is to be protected by an RCBO with arc fault protection. The ambient air temperature on a hot summer day in the loft is often as high as 40°. The proposed cable is 35 meters of PVC/PVC Multicore (twin and earth) with thermoplastic insulation and a sheath.
​
Design current can be found by applying power law:
Ib = P / V 7000W / 230V = Ib = 30.4 Amperes
​
In Protective device can be equal to or higher than Ib so I select a 32A B Type RCBO.
It Tabulated current carrying capacity must be higher than 32A Amperes.
The table for PVC/PVC thermoplastic reference method 101 single phase cable:
It = 36A for a 10mm² conductor.
​
This satisfies: Ib ≤ In ≤ Iz 30.4 ≤ 32 ≤ 36
​
Apply the correction factors to ensure that the cable can carry more than the rating of the protective device.
​
Ca The roof space ambient air temperature of 40° will incur a factor of 0.87 for PVC/PVC thermoplastic.
Ci The thermal insulation is 200mm so the factor is 0.63
Cf The protective device is not a BS3036 fuse a 32A B type RCBO was selected so the factor = 1
Cg The cable is embedded or enclosed so the grouping factor for three cables = 0.65
It ≥ In / (Ca x Ci x Cf x Cg)
32 / ( 0.87 x 0.63 x 1 x 0.65) ≈ 89.9A
​
At 36A It is not greater or equal to In & will not safely operate in these conditions.
​
Going back to the table it is clearly impossible to fit a cable that can handle nearly 90Amps. When most entire domestic installations are rated at 100A or less this also points to the installation being impossible in the prescribed manner.
​
This is where the designer needs to consider what factors can be changed. If you run the cable through the fabric of the building and not into the loft you lose both Ca and Ci factors.
Buried in the building fabric will be counted as clipped direct - Reference method C one of the most agreeable methods for ensuring a cable can dissipate its heat.
​
So this would now look like: 32 / ( 1 x 0.63 x 1 x 1) = 50.8A
Selecting a 10mm² cable which has a tabulated current carrying capacity (It) of 64A which exceeds the protective device value (In) after applying the factors.
It ≥ In / (Ca x Ci x Cf x Cg)
64 > 32 / (1 x 0.63 x 1 x 1)
so: 64 > 50.8A
This satisfies the requirements of:
​
Ib ≤ In ≤ Iz
As can be seen from the scenarios the designer needs to be acutely aware of the intended method of installation to correctly decide upon he most suitable cable choice. Even a simplest domestic circuit like in the second worked example can prove to be problematic. The tables give the designer a clear direction for selecting the smallest diameter cable to safely carry the desired current.
Simply put:
Ca Install in the coolest areas possible Factor can be >1
Ci Don't install in thermal insulation Factor = 1
Cf Don't install BS3036 fuses Factor = 1
Cg Don't group your cables! Factor = 1
​
If this were the case for both the scenarios the cable sizes would be reduced from both being 10mm2 down to 6mm² & 4mm² respectively. In both cases the cost and ease of installation would be greatly improved.
If installing in a loft consider that if it is not insulated but it may be so later.
Volt Drop - mV/A/m
Every meter of cable can be considered to be a very low value resistor. For every Ampere of current being drawn additional resistance is added due to the atoms being more active. This is in effect volt drop, the higher the resistance the higher the volt drop.
With regard to low voltage installations the recommended maximum amount of volt drop allowed is:
​
3% for lighting 230V x 3% = 6.9V
5% Power / other 230V x 5% = 11.5V
​
If these values are exceeded the results may be poor lighting, motors running slowly, equipment malfunction or an increase in maintenance factor of installed equipment.
​
The equation for volt drop takes into account the table left (mv) multiplied by the current being drawn (A) and the length of the circuit in metres (m). That is all divided by 1000 to get it out of milli volts and into Volts.
Extended Example 1
​
On the correction factor section there are two examples of basic cable calculations. The first was an 11.5kW single phase motor that drew 50 Amperes of current. It was 45 meters long and determined after correction factors that it would need to be installed with a 10mm² cable.
With these figures we can determine if it satisfies the minimum requirements for volt drop which is 5% or 11.5 V
​
vd = (mv A m) / 1000
so vd = (4.4 x 50 x 45) / 1000 = 9.9V
which complies with the requirements.
The maximum length of the circuit can be determined by:
​
Max' length = (Max' volt drop / (mv x A)) x 1000
so: ((11.5 / (4.4 x 50) * 1000 = 52.27m
Extended Example 2
​
In the second example there is a 32A Circuit Breaker protecting an 8kW shower. The circuit was 35 meters of single phase AC PVC/PVC multicore and was determined to be 10mm2 with the applied correction values.
​
vd = (mv A m) / 1000 so
vd = (4.4 x 32 x 35) / 1000 = 4.9V which complies.
​
In the conclusion I stated that the circuits could be 6mm² and 4mm² respectively for scenarios 1 & 2 if there was no correction values to be added.
If this were the case example 1 would be:
​
vd = (7.5 x 50 x 45) / 1000 = 16.87V
which does not comply so the cable size would need to be increased to 10mm²
​
vd = (4.4 x 50 x 45) / 1000 = 9.9V
Example 2 would be:
vd = (11 x 32 x 35) / 1000 = 12.3 V which does not comply
so the cable would need to be increased to 6mm²
vd = (7.5 x 32 x 35) / 1000 = 8.4V
​
In the case of non compliance the options are to increase the conductor size, reduce the load or reduce the length of the circuit. The designer will hopefully make these decisions before installing and finding out the hard way.