Calculating A Cable Size
WORK IN PROGRESS
Choosing the right sized cable is not always as easy as it looks.
Selecting the correct cable for the application is imperative to ensure a satisfactory life of conductors and insulation subjected to the thermal effects of carrying current for prolonged periods of time in normal service.
Choosing the mimimum size cross sectional area of the conductors is essential to meet the requirements for
- Protection against electric shock
- Protection against thermal effects
- Overcurrent protection
- voltage drop
- Limiting temperatures for terminals of equipment to which the conductors are connected
How to size a cable
Differences between overcurrent and overload protection
All circuits are designed to protect the cables against overcurrent (eg a short circuit, aka "fault current"). However full overload protection is not always required or possible on some circuits.
Overload protection requires the cable be able to carry an overload that is 1.45 times the MCBs rated current. A typical example of this would be a lighting circuit with a 6 amp MCB. The circuit is designed to carry a maximum of 6 amps but the MCB will not trip instantly at 6 amps. In fact the MCB will take around 1 hour to trip when the current is (6 x 1.45)= 8.7 amps. Designing the circuit to carry 8.7A gives the overload protection needed should a householder change the light fittings for higher powered ones and start to exceed the 6A maximum design current.
Overcurrent protection is not provided when we know the fixed current of the appliance and we install a cable that is capable of carrying this current. A typical example of this would be an electric shower. We will show in the worked examples later on that it is possible to safely install an MCB that has a higher rating than the cable's maximum current rating.
Sizing conductors for your circuit
How to calculate Iz
- Iz is defined as the rated current carrying capacity of the chosen cable, for continuous service, under the particular installation conditions.
There are some other standard terms we can define, which we will need shortly:
- Ib - The design current of the circuit. This is the starting point for all the calculations. Ib is calculated by dividing the power of the appliance (W) by 230 (the nominal voltage).
- In - The rated current of the protective device. This is usually the MCB with the closest rating to Ib where In > Ib
- It - The tabulated value of the current carrying capacity of the cable. For T&E cable It is taken from this table.
- I2 - The actual operating current of the protective device. For a MCB, I2 is 1.45 x In. Note I2 is only needed if overload protection is required
Having calculated the easy bits: the design current (Ib) and then protective device rating (In) we now need to calculate Iz.
Iz may be found either by reference to the tables in BS7671 (the wiring regs) or the IEE On Site Guide (a subset of which are reproduced here for common domestic cables sizes), or, by calculation based on the various factors that affect the installtion.
If full overload protection is required then all we need to ensure is that Iz must be >= I2
Where overload protection is not required then Iz should ideally be greater than or equal to the MCBs nominal rating, (i.e. Iz >= In). However if this can't easily be achieved, then it is also acceptable to opt for Iz > Ib even if I is actually less than the nominal rating of the MCB (i.e. Ib <= Iz <= In). Warning: If the latter design option is used, then it should be remembered that the cable size will have been verified as adequate only for the selected appliance, and it may not be adequate for a more powerful appliance even if the MCB could in theory support it.
Method 1 With direct reference to the table. If the cable is installed ungrouped, in an ambient temperature of 30C, and is protected by a B type MCB, then Iz = It and the cable size is chosen by looking down your reference method column to find Iz.
Method 2 Iz is calculated by using the formula Iz = It x Ca x Cg x Ci x Cr Where It is column C of the table. Ca is a correction factor due to the ambient temperature (values from table 4B1) Cg is a correction value for cables grouped with other circuits (values from table 4C1) Ci is a correction value for cables in insulation (Table 52.2) Cr is a correction factor of 0.725 for BS3036 fuses
Table 4B1
| Ambient temperature | Dereating factor Ca |
|---|---|
| 25 | 1.03 |
| 30 | 1.00 |
| 35 | 0.94 |
| 40 | 0.87 |
| 45 | 0.79 |
| 50 | 0.71 |
| 55 | 0.61 |
| 60 | 0.50 |
Table 52.2
| Length in insulation (mm) | Dereating factor Ci |
|---|---|
| 50 | 0.88 |
| 100 | 0.78 |
| 200 | 0.63 |
| 400 | 0.51 |
| >500 | 0.50 |
Table 4C1
| Arrangement (cables touchimg) | Number of circuits | Applicable reference method | |||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 12 | 16 | 20 | ||
| Bunched in air, on a surface, embedded or enclosed | 1.0 | 0.80 | 0.70 | 0.65 | 0.60 | 0.57 | 0.54 | 0.52 | 0.50 | 0.45 | 0.41 | 0.38 | A to F |
| Single layer on a wall | 1.0 | 0.85 | 0.79 | 0.75 | 0.73 | 0.72 | 0.72 | 0.71 | 0.70 | 0.70 | 0.70 | 0.70 | C |
Checking Voltage Drop
For lighting circuits the maximum volts drop is 3% (6.9V) and for all other circuits the maximum voltage drop is 5% (11.5V).
We now need to test that the chosen cable will be big enough to supply the circuit without dropping the maximum allowed voltage drop. The table below shows the voltage drop in mili Volts (mV) per amp, per meter. So to find the total drop simply multiply this value by the design current Ib, and the length of the cable L:
Total voltages drop = (Voltage Drop x Ib x L) This will give the voltage drop in mV. Divide by 1000 to convert to Volts
| Conductor CSA (mm²) | PVC (max 70° C)
Voltage drop mV/A/m |
|---|---|
| 1.0 | 44 |
| 1.5 | 29 |
| 2.5 | 18 |
| 4.0 | 11 |
| 6 | 7.3 |
| 10 | 4.4 |
| 16 | 2.8 |
Worked Example Say we have an electric shower with a design current Ib of 41A, and a cable run of 19m, installed using Method C. Initial checks would suggest that 6mm² T&E will be adequate. From the table above we know that 6mm² will drop 7.3 mV/A/m. So: Total Drop = 7.3 x 41 x 19 / 1000 = 5.67V which is acceptable
Checking the Maximum Earth Loop Impedance
Table 41.3
| Type B circuit breakers | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Rating (amps) | 6 | 10 | 16 | 20 | 25 | 32 | 40 | 45 | 50 | In |
| Zs (ohms) | 7.67 | 4.60 | 2.87 | 2.30 | 1.84 | 1.44 | 1.15 | 1.02 | 0.92 | 46/In |
Having now chosen a cable that is suitable to carry our design current and meet the required voltage drop requirements we now need to check that the cable will allow the MCB , fuse or RCBO to trip the circuit quickly enough in the event of a fault.
A fault may be caused by either a line to neutral fault or a line to earth fault. A prospective fault current (PFC) is a line to earth fault limited by the Earth Loop Impedance (ELI) and a prospective short circuit (PSC) is a line to neutral fault limited by the Line Impedance.
Non time delayed RCD protected circuits automatically protect the circuit against line earth faults however it may be advisable to design the circuit so that the MCB or overcurrent characteristics of an RCBO still clears the fault in the case of a faulty RCD.
For TN supplies disconnection times are 0.4 seconds for circuits up to 32A and 5 seconds for other circuits. Fortunately for a B type MCB or RCBO the trip current for both 0.4 and 5 seconds disconnection are the same. Table 41.3 above gives the maximum impedance to meet the disconnection times.
To check the disconnection times, we need to know the ELI or LI at the far end of the supply cable. To calculate this we in turn need to know the Earth Loop Impedance or Line Impedance at the consumer unit, this can be found by measurement if you have the appropriate test gear. Alternatively use the default figures of 0.8 for a TN-S earthing system and 0.35 for a TN-C-S system (for TT systems see notes below). Add to this the cable resistance. You can find this by multiplying the length of the cable by the appropriate value from the table below.
| Wire CSA/CPC (mm²) | L + N (mOhms/metre) |
L + CPC Round trip (mOhms/metre) |
|---|---|---|
| 1.0 / 1 | 36.20 | 36.20 |
| 1.5 / 1 | 24.20 | 30.20 |
| 2.5 / 1.5 | 14.82 | 19.51 |
| 4.0 / 1.5 | 9.22 | 16.71 |
| 6.0 / 2.5 | 7.39 | 12.59 |
| 10.0 / 4 | 3.66 | 6.44 |
| 16.0 / 6 | 2.30 | 4.23 |
Table Notes
- The Wire Cross Section Area (CSA) column also indicates the typical CSA of the CPC wire used in a modern cable.
- The values in this table are for copper cables at 70degC
If the calculated impedance is less than in table 41.3 then the disconnection times are met.
Worked Example Say we have an electric shower with a design current Ib of 41A, and a cable run of 19m, installed using Method C. Initial checks would suggest that 6mm² T&E will be adequate. We have already confirmed the voltage drop is acceptable so now we need to check the disconnection times. From the above table ELI = 19 x 12.59 / 1000 = 0.24 ohms and LI = 19 x 7.39 / 1000 = 0.13 ohms For a TN-S supply with a Ze of 0.8 ohms then the calculated maximum resistances are ELI is 0.8 + 0.24 = 1.04 ohms LI is 0.8 + 0.14 = 0.94 ohms As the ELI of 1.04 ohms is greater than the maximum 1.02 allowed for a 45A MCB then this able installation is not suitable for use on a non- RCD protected TN-S supply and a bigger cable will need to be used, the cable route shortened or an RCD used.
Worked Examples
More to follow...