Calculations:Cable Sizing Calculations

Cable Sizing Calculation

Introduction

Cables are vital components in any electrical installation. The cost impact of cables on the total project cost is significant. The selection of the correct cables for the electrical installation is crucial to the project safety, operational efficiency, and economy. Under-sized cables pose danger to people and properties, oversize cables will increase project costs, and improperly selected cable sizes will impact the operational efficiency of the electrical system.

Cable sizing methodologies differ across international standards (e.g. IEC, NEC, BS, etc) however the general principles underlying any cable sizing calculation remain the same.

Importance of cable sizing calculation

The correct size of electrical cables will ensure that the cable can:

• Under normal operating conditions, the cable can operate continuously on full load without damaged
• Withstand the short circuits currents
• Voltage drop on the cable is within acceptable limits
• Cable impedance as affected by cable length can provide sufficiently the tripping current for protective devices in the event of an earth fault.

When to do the calculation?

This calculation can be done individually for each power cable that needs to be sized, or alternatively, it can be used to produce cable sizing waterfall charts for groups of cables with similar characteristics (e.g. cables installed on ladder feeding induction motors).

General Methodology

All cable sizing methods more or less follow the same basic six step process:

1) Gathering data about the cable, its installation conditions, the load that it will carry, etc

2) Determine the minimum cable size based on continuous current carrying capacity

3) Determine the minimum cable size based on voltage drop considerations

4) Determine the minimum cable size based on short circuit temperature rise

5) Determine the minimum cable size based on earth fault loop impedance

6) Select the cable based on the highest of the sizes calculated in step 2, 3, 4 and 5

Step 1: Data Gathering

The first step is to collate the relevant information that is required to perform the sizing calculation. Typically, you will need to obtain the following data:

Load Details

The characteristics of the load that the cable will supply, which includes:

• Load type: motor or feeder
• Three phase, single phase or DC
• System / source voltage
• Full load current (A) - or calculate this if the load is defined in terms of power (kW)
• Full load power factor (pu)
• Locked rotor or load starting current (A)
• Starting power factor (pu)
• Distance / length of cable run from source to load - this length should be as close as possible to the actual route of the cable and include enough contingency for vertical drops / rises and termination of the cable tails

Cable Construction

The basic characteristics of the cable's physical construction, which includes:

• Conductor material - normally copper or aluminium
• Conductor shape - e.g. circular or shaped
• Conductor type - e.g. stranded or solid
• Conductor surface coating - e.g. plain (no coating), tinned, silver or nickel
• Insulation type - e.g. PVC, XLPE, EPR
• Number of cores - single core or multicore (e.g. 2C, 3C or 4C)

Installation Conditions

How the cable will be installed, which includes:

• Above ground or underground
• Installation / arrangement - e.g. for underground cables, is it directly buried or buried in conduit? for above ground cables, is it installed on cable tray / ladder, against a wall, in air, etc.
• Ambient or soil temperature of the installation site
• Cable bunching, i.e. the number of cables that are bunched together
• Cable spacing, i.e. whether cables are installed touching or spaced
• Soil thermal resistivity (for underground cables)
• Depth of laying (for underground cables)
• For single core three-phase cables, are the cables installed in trefoil or laid flat?

Step 2: Cable Selection Based on Current Rating

Current flowing through a cable generates heat through the resistive losses in the conductors, dielectric losses through the insulation and resistive losses from current flowing through any cable screens / shields and armouring.

The component parts that make up the cable (e.g. conductors, insulation, bedding, sheath, armour, etc) must be capable of withstanding the temperature rise and heat emanating from the cable. The current carrying capacity of a cable is the maximum current that can flow continuously through a cable without damaging the cable's insulation and other components (e.g. bedding, sheath, etc). It is sometimes also referred to as the continuous current rating or ampacity of a cable.

Cables with larger conductor cross-sectional areas (i.e. more copper or aluminium) have lower resistive losses and are able to dissipate the heat better than smaller cables. Therefore a 16 $mm^{2}$ cable will have a higher current carrying capacity than a 4 $mm^{2}$ cable.

Base Current Ratings

Table 1. Example of base current rating table (Excerpt from IEC 60364-5-52)

International standards and manufacturers of cables will quote base current ratings of different types of cables in tables such as the one shown on the right. Each of these tables pertain to a specific type of cable construction (e.g. copper conductor, PVC insulated, 0.6/1kV voltage grade, etc) and a base set of installation conditions (e.g. ambient temperature, installation method, etc). It is important to note that the current ratings are only valid for the quoted types of cables and base installation conditions.

In the absence of any guidance, the following reference based current ratings may be used.

Installed Current Ratings

When the proposed installation conditions differ from the base conditions, derating (or correction) factors can be applied to the base current ratings to obtain the actual installed current ratings.

International standards and cable manufacturers will provide derating factors for a range of installation conditions, for example ambient / soil temperature, grouping or bunching of cables, soil thermal resistivity, etc. The installed current rating is calculated by multiplying the base current rating with each of the derating factors, i.e.

$I_{c}=I_{b}.k_{d}\,$

where $I_{c}\,$ is the installed current rating (A)

$I_{b}\,$ is the base current rating (A)

$k_{d}\,$ are the product of all the derating factors

For example, suppose a cable had an ambient temperature derating factor of $k_{amb}=0.94$ and a grouping derating factor of $k_{g}=0.85$, then the overall derating factor $k_{d}=0.94x0.85=0.799$. For a cable with a base current rating of 42A, the installed current rating would be $I_{c}=0.799x42=33.6A$.

In the absence of any guidance, the following reference derating factors may be used.

Cable Selection and Coordination with Protective Devices

Feeders

When sizing cables for non-motor loads, the upstream protective device (fuse or circuit breaker) is typically selected to also protect the cable against damage from thermal overload. The protective device must therefore be selected to exceed the full load current, but not exceed the cable's installed current rating, i.e. this inequality must be met:

$I_{l}\leq I_{p}\leq I_{c}\,$

Where $I_{l}\,$ is the full load current (A)

$I_{p}\,$ is the protective device rating (A)

$I_{c}\,$ is the installed cable current rating (A)

Motors

Motors are normally protected by a separate thermal overload (TOL) relay and therefore the upstream protective device (e.g. fuse or circuit breaker) is not required to protect the cable against overloads. As a result, cables need only to be sized to cater for the full load current of the motor, i.e.

$I_{l}\leq I_{c}\,$

Where $I_{l}\,$ is the full load current (A)

$I_{c}\,$ is the installed cable current rating (A)

Of course, if there is no separate thermal overload protection on the motor, then the protective device needs to be taken into account as per the case for feeders above.

Step 3: Voltage Drop

A cable's conductor can be seen as an impedance and therefore whenever current flows through a cable, there will be a voltage drop across it, which can be derived by Ohm’s Law (i.e. V = IZ). The voltage drop will depend on two things:

• Current flow through the cable – the higher the current flow, the higher the voltage drop
• Impedance of the conductor – the larger the impedance, the higher the voltage drop

Cable Impedances

The impedance of the cable is a function of the cable size (cross-sectional area) and the length of the cable. Most cable manufacturers will quote a cable’s resistance and reactance in $\Omega$/km. The following typical cable impedances for low voltage AC and DC single core and multicore cables can be used in the absence of any other data.

Calculating Voltage Drop

For AC systems, the method of calculating voltage drops based on load power factor is commonly used. Full load currents are normally used, but if the load has high startup currents (e.g. motors), then voltage drops based on starting current (and power factor if applicable) should also be calculated.

For a three phase system:

$V_{3\phi }={\frac {{\sqrt {3}}I(R_{c}\cos \phi +X_{c}\sin \phi )L}{1000}}\,$

Where $V_{3\phi }\,$ is the three phase voltage drop (V)

$I\,$ is the nominal full load or starting current as applicable (A)

$R_{c}\,$ is the ac resistance of the cable ($\Omega$/km)

$X_{c}\,$ is the ac reactance of the cable ($\Omega$/km)

$\cos \phi \,$ is the load power factor (pu)

$L\,$ is the length of the cable (m)

For a single phase system:

$V_{1\phi }={\frac {2I(R_{c}\cos \phi +X_{c}\sin \phi )L}{1000}}\,$

Where $V_{1\phi }\,$ is the single phase voltage drop (V)

$I\,$ is the nominal full load or starting current as applicable (A)

$R_{c}\,$ is the ac resistance of the cable ($\Omega$/km)

$X_{c}\,$ is the ac reactance of the cable ($\Omega$/km)

$\cos \phi \,$ is the load power factor (pu)

$L\,$ is the length of the cable (m)

For a DC system:

$V_{dc}={\frac {2IR_{c}L}{1000}}\,$

Where $V_{dc}\,$ is the dc voltage drop (V)

$I\,$ is the nominal full load or starting current as applicable (A)

$R_{c}\,$ is the dc resistance of the cable ($\Omega$/km)

$L\,$ is the length of the cable (m)

Maximum Permissible Voltage Drop

It is customary for standards (or clients) to specify maximum permissible voltage drops, which is the highest voltage drop that is allowed across a cable. Should your cable exceed this voltage drop, then a larger cable size should be selected.

Maximum voltage drops across a cable are specified because load consumers (e.g. appliances) will have an input voltage tolerance range. This means that if the voltage at the appliance is lower than its rated minimum voltage, then the appliance may not operate correctly.

In general, most electrical equipment will operate normally at a voltage as low as 80% nominal voltage. For example, if the nominal voltage is 230VAC, then most appliances will run at >184VAC. Cables are typically sized for a more conservative maximum voltage drop, in the range of 5 – 10% at full load.

Calculating Maximum Cable Length due to Voltage Drop

It may be more convenient to calculate the maximum length of a cable for a particular conductor size given a maximum permissible voltage drop (e.g. 5% of nominal voltage at full load) rather than the voltage drop itself. For example, by doing this it is possible to construct tables showing the maximum lengths corresponding to different cable sizes in order to speed up the selection of similar type cables.

The maximum cable length that will achieve this can be calculated by re-arranging the voltage drop equations and substituting the maximum permissible voltage drop (e.g. 5% of 415V nominal voltage = 20.75V). For a three phase system:

$L_{max}={\frac {1000V_{3\phi }}{{\sqrt {3}}I(R_{c}\cos \phi +X_{c}\sin \phi )}}\,$

Where $L_{max}\,$ is the maximum length of the cable (m)

$V_{3\phi }\,$ is the maximum permissible three phase voltage drop (V)

$I\,$ is the nominal full load or starting current as applicable (A)

$R_{c}\,$ is the ac resistance of the cable ($\Omega$/km)

$X_{c}\,$ is the ac reactance of the cable ($\Omega$/km)

$\cos \phi \,$ is the load power factor (pu)

For a single phase system:

$L_{max}={\frac {1000V_{1\phi }}{2I(R_{c}\cos \phi +X_{c}\sin \phi )}}\,$

Where $L_{max}\,$ is the maximum length of the cable (m)

$V_{1\phi }\,$ is the maximum permissible single phase voltage drop (V)

$I\,$ is the nominal full load or starting current as applicable (A)

$R_{c}\,$ is the ac resistance of the cable ($\Omega$/km)

$X_{c}\,$ is the ac reactance of the cable ($\Omega$/km)

$\cos \phi \,$ is the load power factor (pu)

For a DC system:

$L_{max}={\frac {1000V_{dc}}{2IR_{c}}}\,$

Where $L_{max}\,$ is the maximum length of the cable (m)

$V_{dc}\,$ is the maximum permissible dc voltage drop (V)

$I\,$ is the nominal full load or starting current as applicable (A)

$R_{c}\,$ is the dc resistance of the cable ($\Omega$/km)

$L\,$ is the length of the cable (m)

Step 4: Short Circuit Temperature Rise

During a short circuit, a high amount of current can flow through a cable for a short time. This surge in current flow causes a temperature rise within the cable. High temperatures can trigger unwanted reactions in the cable insulation, sheath materials and other components, which can prematurely degrade the condition of the cable. As the cross-sectional area of the cable increases, it can dissipate higher fault currents for a given temperature rise. Therefore, cables should be sized to withstand the largest short circuit that it is expected to see.

Minimum Cable Size Due to Short Circuit Temperature Rise

The minimum cable size due to short circuit temperature rise is typically calculated with an equation of the form:

$A={\frac {\sqrt {i^{2}t}}{k}}\,$

Where $A\,$ is the minimum cross-sectional area of the cable ($mm^{2}$)

$i\,$ is the prospective short circuit current (A)

$t\,$ is the duration of the short circuit (s)

$k\,$ is a short circuit temperature rise constant

The temperature rise constant is calculated based on the material properties of the conductor and the initial and final conductor temperatures (see the derivation here). Different international standards have different treatments of the temperature rise constant, but by way of example, IEC 60364-5-54 calculates it as follows:

$k=226{\sqrt {\ln \left(1+{\frac {\theta _{f}-\theta {i}}{234.5+\theta _{i}}}\right)}}\,$ (for copper conductors)

$k=148{\sqrt {\ln \left(1+{\frac {\theta _{f}-\theta {i}}{228+\theta _{i}}}\right)}}\,$ (for aluminium conductors)

Where $\theta _{i}\,$ is the initial conductor temperature (deg C)

$\theta _{f}\,$ is the final conductor temperature (deg C)

Initial and Final Conductor Temperatures

The initial conductor temperature is typically chosen to be the maximum operating temperature of the cable. The final conductor temperature is typically chosen to be the limiting temperature of the insulation. In general, the cable's insulation will determine the maximum operating temperature and limiting temperatures.

As a rough guide, the following temperatures are common for the different insulation materials:

Material	Max Operating Temperature oC	Limiting Temperature oC
PVC	75	160
EPR	90	250
XLPE	90	250

Short Circuit Energy

The short circuit energy $i^{2}t\,$ is normally chosen as the maximum short circuit that the cable could potentially experience. However for circuits with current limiting devices (such as HRC fuses), then the short circuit energy chosen should be the maximum prospective let-through energy of the protective device, which can be found from manufacturer data.

Step 5: Earth Fault Loop Impedance

Sometimes it is desirable (or necessary) to consider the earth fault loop impedance of a circuit in the sizing of a cable. Suppose a bolted earth fault occurs between an active conductor and earth. During such an earth fault, it is desirable that the upstream protective device acts to interrupt the fault within a maximum disconnection time so as to protect against any inadvertent contact to exposed live parts.

Ideally the circuit will have earth fault protection, in which case the protection will be fast acting and well within the maximum disconnection time. The maximum disconnection time is chosen so that a dangerous touch voltage does not persist for long enough to cause injury or death. For most circuits, a maximum disconnection time of 5s is sufficient, though for portable equipment and socket outlets, a faster disconnection time is desirable (i.e. <1s and will definitely require earth fault protection).

However for circuits that do not have earth fault protection, the upstream protective device (i.e. fuse or circuit breaker) must trip within the maximum disconnection time. In order for the protective device to trip, the fault current due to a bolted short circuit must exceed the value that will cause the protective device to act within the maximum disconnection time. For example, suppose a circuit is protected by a fuse and the maximum disconnection time is 5s, then the fault current must exceed the fuse melting current at 5s (which can be found by cross-referencing the fuse time-current curves).

By simple application of Ohm's law:

$I_{A}={\frac {V_{0}}{Z_{s}}}\,$

Where $I_{A}\,$ is the earth fault current required to trip the protective device within the minimum disconnection time (A)

$V_{0}\,$ is the phase to earth voltage at the protective device (V)

$Z_{s}\,$ is the impedance of the earth fault loop ($\Omega$)

It can be seen from the equation above that the impedance of the earth fault loop must be sufficiently low to ensure that the earth fault current can trip the upstream protection.

The Earth Fault Loop

The earth fault loop can consist of various return paths other than the earth conductor, including the cable armour and the static earthing connection of the facility. However for practical reasons, the earth fault loop in this calculation consists only of the active conductor and the earth conductor.

The earth fault loop impedance can be found by:

$Z_{s}=Z_{c}+Z_{e}\,$

Where $Z_{s}\,$ is the earth fault loop impedance ($\Omega$)

$Z_{c}\,$ is the impedance of the active conductor ($\Omega$)

$Z_{e}\,$ is the impedance of the earth conductor ($\Omega$)

Assuming that the active and earth conductors have identical lengths, the earth fault loop impedance can be calculated as follows:

$Z_{s}={\frac {L}{1000}}{\sqrt {(R_{c}+R_{e})^{2}+(X_{c}+X_{e})^{2}}}\,$

Where $L\,$ is the length of the cable (m)

$R_{c}\,$ and $R_{e}\,$ are the ac resistances of the active and earth conductors respectively ($\Omega$/km)

$X_{c}\,$ and $X_{e}\,$ are the reactances of the active and earth conductors respectively ($\Omega$/km)

Maximum Cable Length

The maximum earth fault loop impedance can be found by re-arranging the equation above:

$Z_{s,max}={\frac {V_{0}}{I_{A}}}\,$

Where $Z_{s}\,$ is the maximum earth fault loop impedance ($\Omega$)

$V_{0}\,$ is the phase to earth voltage at the protective device (V)

$I_{A}\,$ is the earth fault current required to trip the protective device within the minimum disconnection time (A)

The maximum cable length can therefore be calculated by the following:

$L_{max}={\frac {1000V_{0}}{I_{A}{\sqrt {(R_{c}+R_{e})^{2}+(X_{c}+X_{e})^{2}}}}}\,$

Where $L_{max}\,$ is the maximum cable length (m)

$V_{0}\,$ is the phase to earth voltage at the protective device (V)

$I_{A}\,$ is the earth fault current required to trip the protective device within the minimum disconnection time (A)

$R_{c}\,$ and $R_{e}\,$ are the ac resistances of the active and earth conductors respectively ($\Omega$/km)

$X_{c}\,$ and $X_{e}\,$ are the reactances of the active and earth conductors respectively ($\Omega$/km)

Note that the voltage $V_{0}$ at the protective device is not necessarily the nominal phase to earth voltage, but usually a lower value as it can be downstream of the main busbars. This voltage is commonly represented by applying some factor $c\,$ to the nominal voltage. A conservative value of $c\,$ = 0.8 can be used so that:

$V_{0}=cV_{n}=0.8V_{n}\,$

Where $V_{n}$ is the nominal phase to earth voltage (V)

Worked Example

In this example, we will size a cable for a 415V, 37kW three-phase motor from the MCC to the field.

Step 1: Data Gathering

The following data was collected for the cable to be sized:

• Cable type: Cu/PVC/GSWB/PVC, 3C+E, 0.6/1kV
• Operating temperature: 75C
• Cable installation: above ground on cable ladder bunched together with 3 other cables on a single layer and at 30C ambient temperature
• Cable run: 90m (including tails)
• Motor load: 37kW, 415V three phase, full load current = 61A, power factor = 0.85
• Protection: aM fuse of rating = 80A, max prospective fault $I^{2}t$ = 90 $A^{2}s$ , 5s melt time = 550A

Step 2: Cable Selection Based on Current Rating

Suppose the ambient temperature derating is 0.89 and the grouping derating for 3 bunched cables on a single layer is 0.82. The overall derating factor is 0.89 $\times$ 0.82 = 0.7298. Given that a 16 $mm^{2}$ and 25 $mm^{2}$ have base current ratings of 80A and 101A respectively (based on Reference Method E), which cable should be selected based on current rating considerations?

The installed current ratings for 16 $mm^{2}$ and 25 $mm^{2}$ is 0.7298 $\times$ 80A = 58.38A and 0.7298 $\times$ 101A = 73.71A respectively. Given that the full load current of the motor is 61A, then the installed current rating of the 16 $mm^{2}$ cable is lower than the full load current and is not suitable for continuous use with the motor. The 25 $mm^{2}$ cable on the other hand has an installed current rating that exceeds the motor full load current, and is therefore the cable that should be selected.

Step 3: Voltage Drop

Suppose a 25 $mm^{2}$ cable is selected. If the maximum permissible voltage drop is 5%, is the cable suitable for a run length of 90m?

A 25 $mm^{2}$ cable has an ac resistance of 0.884 $\Omega$/km and an ac reactance of 0.0895 $\Omega$/km. The voltage drop across the cable is:

$V_{d}={\frac {90}{1000}}\times {\sqrt {3}}\times 61\times \left[0.884\times 0.85+0.0895\times \sin(\cos ^{-1}(0.85)\right]=7.593V\,$

A voltage drop of 7.593V is equivalent to ${\frac {7.593}{415}}=1.83\%$, which is lower than the maximum permissible voltage dorp of 5%. Therefore the cable is suitable for the motor based on voltage drop considerations.

Step 4: Short Circuit Temperature Rise

The cable is operating normally at 75C and has a prospective fault capacity ($I^{2}t$) of 90,000 $A^{2}s$. What is the minimum size of the cable based on short circuit temperature rise?

PVC has a limiting temperature of 160C. Using the IEC formula, the short circuit temperature rise constant is 111.329. The minimum cable size due to short circuit temperature rise is therefore:

$A={\frac {\sqrt {90,000}}{111.329}}=2.695mm^{2}\,$

In this example, we also use the fuse for earth fault protection and it needs to trip within 5s, which is at the upper end of the adiabatic period where the short circuit temperature rise equation is still valid. Therefore, it's a good idea to also check that the cable can withstand the short circuit temperature rise for for a 5s fault. The 80A motor fuse has a 5s melting current of 550A. The short circuit temperature rise is thus:

$A={\frac {\sqrt {550^{2}\times 5}}{111.329}}=11.047mm^{2}\,$

Therefore, our 25 $mm^{2}$ cable is still suitable for this application.

Step 5: Earth Fault Loop Impedance

Suppose there is no special earth fault protection for the motor and a bolted single phase to earth fault occurs at the motor terminals. Suppose that the earth conductor for our 25 $mm^{2}$ cable is 10 $mm^{2}$. If the maximum disconnection time is 5s, is our 90m long cable suitable based on earth fault loop impedance?

The 80A motor fuse has a 5s melting current of 550A. The ac resistances of the active and earth conductors are 0.884 $\Omega$/km and 2.33 $\Omega$/km) respectively. The reactances of the active and earth conductors are 0.0895 $\Omega$/km and 0.0967 $\Omega$/km) respectively.

The maximum length of the cable allowed is calculated as:

$L_{max}={\frac {(1000)(0.8)(240)}{550{\sqrt {(0.884+2.33)^{2}+(0.0895+0.0967)^{2}}}}}=108.43m\,$

The cable run is 90m and the maximum length allowed is 108m, therefore our cable is suitable based on earth fault loop impedance. In fact, our 25 $mm^{2}$ cable has passed all the tests and is the size that should be selected.

Waterfall Charts

Table 2. Example of a cable waterfall chart

Sometimes it is convenient to group together similar types of cables (for example, 415V PVC motor cables installed on cable ladder) so that instead of having to go through the laborious exercise of sizing each cable separately, one can select a cable from a pre-calculated chart.

These charts are often called "waterfall charts" and typically show a list of load ratings and the maximum of length of cable permissible for each cable size. Where a particular cable size fails to meet the requirements for current carrying capacity or short circuit temperature rise, it is blacked out on the chart (i.e. meaning that you can't choose it).

Preparing a waterfall chart is common practice when having to size many like cables and substantially cuts down the time required for cable selection.

International Standards

IEC

IEC 60364-5-52 (2009) "Electrical installations in buildings - Part 5-52: Selection and erection of electrical equipment - Wiring systems" is the IEC standard governing cable sizing.

NEC

NFPA 70 (2020) "National Electricity Code" is the equivalent standard for IEC 60364 in North America and includes a section covering cable sizing in Article 300.

BS

BS 7671 (18th edition) "Requirements for Electrical Installations - IEE Wiring Regulations" is the equivalent standard for IEC 60364 in the United Kingdom.

AS/NZS

AS/NZS 3008.1 (2017) "Electrical installations - Selection of cables - Cables for alternating voltages up to and including 0.6/1 kV" is the standard governing low voltage cable sizing in Australia and New Zealand. AS/NZS 3008.1.1 is for Australian conditions and AS/NZS 3008.1.2 is for New Zealand conditions.

What next?

Having sized the power / load-bearing cables, the cable schedule can now be developed and then the cable material take-offs (MTO).

Source:

| Open Electrical