Proper cable sizing is critical for ensuring an electrical installation's safety, efficiency, and dependability. It can aid in the prevention of voltage drops, overheating, and fire dangers. A properly designed cable can help reduce energy loss and extend the life of your electrical gadgets.

Considerations for Cable Sizing

Current Carrying Capacity; the highest current a cable can carry without exceeding its thermal constraints is referred to as its current-carrying capability. Consider the load current as well as any safety margins you may require.

Voltage drop: as current passes via a cable, the voltage steadily decreases due to conductor resistance. Voltage drop is critical in ensuring that the correct voltage is maintained at the receiving end of the cable. Cable length, conductor material, and load current are all factors that influence voltage drop.

Fault Withstand: the ability of a cable to tolerate electrical stress induced by defects such as short circuits or ground faults without experiencing major damage is referred to as cable fault withstand.

Key Variables

Conductor size and material: the size and material of the conductor have a direct impact on the cable's current carrying capacity, voltage drop, and cost. Copper and aluminium are two common conductivity materials.

Ambient temperatures: high ambient temperatures can have an effect on a cable's current-carrying capacity. When selecting a cable, keep in mind the surrounding environment, as greater temperatures can reduce the cable's capacity.

Installation conditions: how the cable is installed influences its heat dissipation, affecting its current-carrying capacity. The thermal performance of cables can be influenced by whether they are installed in conduits, directly buried, or on a tray.

The Effects of Improper Cable Sizing

Using incorrectly sized cables can result in a variety of issues, including:

• Voltage drop: excessive voltage drop can result in poor performance, decreased efficiency, and even damage to sensitive electrical equipment.
• Overheating: due to high current, undersized cables can overheat, providing a fire danger and lowering the lifespan of the cable and connected equipment.
• Energy loss: inadequately sized cables can result in lost energy, increased operational expenses, and a larger carbon footprint.
• Equipment malfunction: Incorrectly sized cables can cause unstable power delivery, causing devices to malfunction or fail prematurely.

Understanding cable size concepts is critical for assuring your electrical installations' safety, efficiency, and lifespan. When picking the suitable cable for your project, you may make informed judgements by carefully examining parameters such as current-carrying capability, voltage drop, and fault withstand.

Many standards, manufacturers and power authorities give guidance on determining how much current flow in a cable by using the idea of derating factors.  While each standard has variations and particularities, the general principals are the same. 

Other knowledge base notes discussed particular standards in more detail.  This note outlines the general principals relevant to the method.

Derating By Factors Method

Depending on whether overload protection is required or not, the required current rating of a cable is given by:

$Ir=\text{ the greater of }{I}_b\text{ or }{I}_n$

where:
I_r        - required current rating, A
I_b        - cable design current (that intended to be carried in normal service, load current), A
I_n        - rated current or current setting of the protective device (if required), A

Each standard will maintain a table of base cables ratings, I_t,for a cable under a defined set of installation and operating environmental conditions.  Depending on the standard, several derating factors are then applied to the base rating.  The following are the most commonly applied derating factors:

C_a        - for ambient temperature (either air or ground)
C_c        - for circuits buried in the ground (BS 7671 factor only for buried circuits)
C_d        - for depth of burial (circuits directly buried or buried in ducts)
C_f        - for semi-enclosed fuse to BS 3036 (BS 7671 factor for this type of protective device)
C_g        - for grouping (number of cable circuits run together)
C_i         - for thermal insulation (if a circuit is encased inside any thermal insulating material)
C_s        - for thermal resistivity of soil (for direct buried or cables buried in ducts)

The required current rating and base rating of the cable are related by:

$I_t\ge \frac{I_r}{C_a{C}_c{C}_d{C}_f{C}_g{C}_i{C}_s}$

Typical input data required for cable sizing:

General Information

General reference information to facilitate identification of the cable, not used for calculations:

•  cable designation
•  cable description
•  cable source/starting point

System, Load and Protection Details

Details of the type of system (a.c., d.c. etc.), what the cable is supplying (delivering) and details on protective devices:

•  system voltage
•  a.c. or d.c. and frequency if an a.c. supply
•  phase if an a.c. system (1-ph, 3-ph, 3-ph+neutral
•  earthing arrangement for low voltage a.c. systems (TN-S, TN-C-S, TT, IT)
•  
•  load type and details
  •  load designation
  •  distribution circuit or final equipment supply 
  •  distribution/load current and power factor if a.c. system
•  
•  protective device details (if any)
  •  circuit breaker or relay
  •  frame size, trip rating (o/c, e/f) if a circuit breaker
  •  primary trip current (o/c, e/f) if a relay
•  RCD protected if low voltage cable

Cable Selection

•  type of cable (copper, aluminium, pvc, xlpe, armoured, unarmoured, etc.)
•  single-core or multi-core cable
•  prefered cable cross-sectional-area if applicable (including maximum to be considered)
•  length of cable

Sustained Current Capacity

•  cable standard to size to (agreed with client/owner)
•  installation method (air, buried, detailed arrangements, spacing, etc.)
•  buried cables: depth, free laid or in ducts, soil thermal resistivity
•  number of circuits using the same containment/installation path (grouping)
•  ambient temperature (air or ground as appropriate)
•  cable source/starting point

Voltage Drop

•  required maximum allowable voltage drop (per cable, complete circuit or both)

Fault Withstand

•  cable source short circuit and earth fault levels
  •  fault level (kA or Ze, power factor)
  •  maximum duration of anticipated faults
•  any external CPC consideration if low voltage cable

Many countries and organisations produce cable sizing standards. Standards, allow engineers to size cables with a  consistent approach. Using a standard also gives the engineer a fullback with which to validate any design should the need arise.

myCableEngineering

myCableEngineering incorporates the following sustained current sizing standards:

BS 7671 is the UK requirements for electrical installations standard, also known as the Wiring Regulations. Appendix 4 (BS 7671:2018) gives guidance determining the current capacity and voltage drop for cables.

Current Capacity

Cable current capacity calculations in BS 761 are based on the derating factor method, see .  The derating tables have been derived from IEC 60287 and aligned with IEC 60364 (although there is some fundamental difference).

The parameters considered in the standard are, ambient temperature, soil thermal resistivity, the installation methods and grouping of cables.

Voltage Drop

ERA 69-30 sustained current ratings is a nine-volume set of recommendations, published by Edif ERA.  The calculation of cable rating follow the derating factor method, see . 

ERA is a UK based organisation, and these standards reflect this.  The user of ERA is particularly popular amongst DNO (district network operators) and supply authorities. 

Each volume covers a particular cable type, for example, part V is for 600/100 V and 1900/33000 V cables with thermosetting insulation. Typically the recommendation cover both three-phase a.c. 50 Hz and d.c. are for in air and buried cable installations.

In addition to current capacity, the recommendations provide estimates of power dissipation and a guide to average soil temperatures within the UK.

IEC 60502 is an International Electrotechnical Commission standard, which gives current ratings for medium voltage cables.  The calculation of cable rating follow the derating factor method, see . 

The IEC 60502 standard is in two parts; part 1 for voltages up to 1 kV and part 2 for voltages from 1 kV to 30 kV.  Part 1 does not contain any current capacity sizing, as this is intended to be covered by IEC 60364.  Part 2 Annex B, has a calculation method for determining current capacities for voltages covered by this part.

The current rating tables include three phase cables both installed in the air and buried.  Derating factors considered are the ambient temperature, depth of laying and soil thermal resistivity for buried cables, and the grouping of cables.

Traditionally the thickness of cable coverings has been related to nominal cable diameters by step tables.  As there can be variations in calculated nominal diameter, this can cause variations in thickness of layers for cables of the same design.  IEC 60502-2 'Cables for rated voltages from 6 kV up to 30 kV', introduces the concept of fictitious calculation to overcome these issues.

Fictitious Method

Conductors - them fictitious diameter, d_L (irrespective of shape or compactness) is given by:

Cores - for cable cores without semi-conducting layers, the ficititious core diameter D_c is given by:

$D_c=d_L+2t_i$

and for cables with semi-conducting layers:

$D_c=d_L+2t_i+3.0$

where:
D_c - core diamter in mm
t_i - is the nomiman insulation thickness, in mm

Diameter over laid-up Cores - is given by:

$D_f=k{D}_c$

where: k is the assembly coefficient

k = 1 for single core cables
k = 2 for two core cables
k = 2.16 for three core cables
k = 2.42 for four core cables
k = 2.70 for five core cables

Inner Covering/ Bedding - the fictitious diamter D_B is given by:

$D_B=D_f+2t_B$

where:
D_B - diamter or inner covering/bedding in mm
t_B - bedding thickness in mm
= 0.4 mm for D_f <= 40 mm
= 0.6 mm for D_f > 40 mm

Concentric condcutors and metallic screens - increase in diamter due to screens is given by:

Note: if the cross-sectional area lies between two values, that the largest value as the increase in diameter.

- tape screen: 

$\text{cross-sectional area}=n_t\times t_t\times w_t$

where:
n_t - number of tapes
t_t - nominal thickness of individual tape in mm
w_t - nominal width of indvidual tape in mm

for lapped tape with overlap, the total thickness is twice that of one tape

for longitudinally applied type:
 - overlap < 30%, total tickness = thickness of tape
- overlap >= 30%, total thichness - 2 x thickness of tape

- wire screen:

$\text{cross-sectional area}=\frac{n{}_w\times d_w^2\times \pi }{4}+n_h\times t_h\times W_h$

where:
n_w - number of wires
d_w - diameter of wire in mm
n_h - number of counter helix
t_h - thickness of coutner helix in mm (if greater than 3 mm)
W_h - width of counter helix in mm

Lead sheath - fictitious diamter of lad shearth , D_pb is given by :

$D_{pb}=D_g+2t_{pb}$

where:

D_g - fictitious diameter under lead sheath in mm
t_pb - thickness of lead sheath in mm

Seperation sheath - fictitious diameter D_s given by:

$D_s=D_u+2t_s$

where:
D_u - fictitious diameter under separation sheath in mm
t_s - thickness of separation sheath in mm

Lapped bedding - fictitious diameter D_lb given by:

$D_{lb}=D_{ulb}+2t_{lb}$

where:
D_ulb - fictitious diameter under lapped beding in mm
t_lb  - thickness of lapped bedding in mm

Addition bedding for tape-armoured cables - provide dover inner covering:

 - fictitious diameter under addition bedding <= 29mm,  increase 1.0 mm
- fictitious diameter under additional beddign > 29 mm, increase 1.6 mm

Armour - fictitious diameter over the armour, D_x is given by:

- flat or round wire armour:

$D_x=D_A+2t_A+2t_w$

where:
D_A - diameter under armour in mm
t_A - thickness or diameter of armour wire, in mm
t_w - thickness of counter helix (if any) in mm (if > 3 mm)

 - dobuble tape armour:

$D_x=D_A+4t_A$

where:
D_a - diameter under armour in mm
t_A - thickness of armour tape in mm

IEC 60287 "Calculation of the continuous current rating of cables (100% load factor)" is the International Standard which defines the procedures and equations to be used in determining the current carry capacity of cable.  The standard is applicable to all alternating current voltages and direct current cables up to 5kV.

This note will introduce the concepts adopted by the standard, provide some guidance on using the standard and direct the reader to further resources.

Thermal Problem

Principle- simple wire in
homogeneous material The methodology taken to the sizing of cables is that of treating the issue as a thermal problem. 

Losses within a cable will create heat.  Depending on the installation conditions this heat will be dissipated to the surrounding environment at a given rate.  As the cable heats up rate of heat dissipation will increase. 

At some temperature the rate at which heat is being dissipated to the environment will be the same as the rate at which it is generated (due to loses).  The cable is then in thermal equilibrium.

The losses (and heat generated) are dependent on the amount of current flowing within the cable.  As the current increases, the losses increase and the thermal equilibrium temperature of the cable will increase. 

At some given current level, the cable temperature at thermal equilibrium will equal the maximum allowable temperature for the cable insulation.  This is the maximum current carrying capacity of the cable for the installation conditions depicted by the calculation. 

To illustrate the principle, we can consider a simplistic scenario of a d.c. cable (as shown in the illustration), surrounded by an insulating material and placed in a homogeneous thermal conducting material. 

Given:
I - conductor current, A
R' - d.c. resistance of the conductor per unit length, Ω/m
θ - maximum conductor operating temperature, °C
θ_a - ambient temperature, °C
Δθ - temperature difference (θ-θ_a), K
T - thermal resistance per unit length between conductor and surrounding, K.m/W

The losses (watts per unit length) generated by the conductor is given by:

$I^2{R}^{\text{'}}$

The heat flow (watts per unit length) from the conductor is given by:

$\Delta \theta /T$

At thermal equilibrium these will be equal and can be rearranged to give the cable current carrying capacity (in Ampere):

$I=\sqrt{\frac{\Delta \theta }{R^{\text{'}}T}}$

As an example, consider finding the current carrying capacity of a 50 mm² conductor, with XPLE insulation directly buried (with an insulation thermal resistance of 5.88 K.m/W and soil thermal resistance of 2.5 K.m/W)  and at an ambient temperature  of 25 °C

by using the related resources links given at the end of the posts, we are able to find the following:

• the dc resistance of the cable is 0.387 mΩ/m
• the maximum allowable temperature for XLPE insulation is 90 °C

and a total thermal resistance of 5.88+2.5 = 8.38 (insulation, plus soil)

Δθ = 90-25 = 65 K, giving
I = √ [65/(0.000387*8.38)] = 142 A

The Standard in More Detail

Applying the IEC 60287 Standard
(click to enlarge) The reality of any cable installation is more complex than described above.  Insulating materials have dielectric losses, alternating current introduces skin effect, sheath and eddy current losses, several cables are simultaneously producing heat and the surrounding materials are non-homogeneous and have boundary temperature conditions.

While the standard addresses each of these issues, the resulting equations are more complex do take some effort to solve.  Anyone attempting to apply this method should be working directly from a copy of the standard.  As an overview, the standard looks at the following situations:

• differences between alternating and direct current systems in calculating cable capacity
• critical temperatures of soil and possible requirements to avoid drying out the soil
• cables directly exposed to solar radiation
• calculation of the a.c. and d.c. resistance of conductors (including skin effect, proximity effect and operating temperature)
• insulation dielectric losses
• conductor I2R losses
• losses in sheaths and screens (including flat, trefoil and transposed formations)
• circulating current losses (including sheath, armour and pipes)
• thermal resistance (and it's calculation)

Each of these areas is discussed in more detail in the following posts (which together form a comprehensive guide to the standard):

•  
• IEC 60287 Current Capacity of Cables - rated current

Applying the Standard

Within the standard, there are a lot of equations and it can be confusing to persons who are new to the method.  However, a step by step working through it approach will enable the current carrying capacity to be calculated.  The flow chart shows one recommended path for working through a cable sizing exercise in line with the standard.

Given the number of equations which need to be solved, it is tedious to calculate in accordance with the standard by using hand or manual methods.  More practically software applications are used, which allow the sizing of cables to take place quickly. A quick Google search will turn up several software programs capable of performing the calculation.

Tip:  a cable run can move through different installation environments (for example it may start in a cable basement, more through ducts in a wall, be buried for some of the route, suspended under a bridge, buried again, through ducts and into the receiving building).  In this instance the current capacity should be evaluated for each type of installation condition and the worse case taken.

Summary

Within the note the IEC 60287 have been introduces and the problem of finding the current capacity of a cable boiled down to that of a thermal calculation.  The note has given an overview of the contents of the standard, ways to navigate and perform the calculation and provided links to more detailed posts.

Hopefully the note has achieved the objective of providing an introduction to the current capacity sizing methods of IEC 60287.  If you have any comments or something is not clear enough, please post these below.

When sizing cables, the heat generated  by losses within any sheath or armour need to be evaluated. When significant, it becomes a factor to be considered in the sizing of cables.  To understand how sheath and armour losses affect the sizing of cables, you can review the post: IEC 60287 Current Capacity of Cables - Rated Current. 

This note looks at how to obtain the necessary loss factors for use within the IEC 60287 calculation.  In addition, the loss factors quantify the ratio of losses in the sheath to total losses in all conductors and have application outside the IEC 60287 standard.

Calculation of Sheath and Armour Loss

Any cable sheath (or screen) the loss λ1, consists of two components:

 $${\lambda }_1={\lambda }_1{}^{\prime \text{}\prime }+{\lambda }_1{}^{\prime \text{}\prime }$$

• λ1' - losses caused by circulating currents.  These losses only occur in single core cables and for any circulating current to be present, it is necessary to the sheaths of each cable to be bonded at two or more points along its length.
• λ1''  - losses caused by eddy currents.  These are small circulating currents setup in the sheath due to changing magnetic fields.

The loss in armour is considered as only one component, λ₂.

Sheath and armour losses are only applicable to alternating current (a.c.) cables.  The actual formula for calculation of sheath and armour loss depend on the installation and arrangement of cables.  The tables below presents some of the common installation situations and are based on equations given in IEC 60287:

Calculation of sheath or screen loss in - Single Core Cables

For installations bonded only at one point, circulating currents are not possible and the loss is zero. Except in the case of large segmental type conductors (see Some Special Cases below), eddy current loss λ1'', for single core cables can be ignored.

Calculation of sheath or screen loss in - Multi-Core Cables

Due to any sheath or screen surrounding all cores, the possibility of circulating current does not exist, and the λ1' loss can be ignored.  Eddy current loss, λ1'' does need to be considered.

Calculation of armour loss

For armoured cables, the losses are estimated as shown.

Calculating the Parameters

Sheath (Rs) or armour (RA) resistance  - values used above are calculated at their operating temperature.  The operating temperature (in °C) and resistance can be determined from:

$${\theta }_{sc}=\theta -\left(I^{2}R+0.5W_d\right)\times T_1$$   - for any sheath
 $${\theta }_{ar}=\theta -\left\{\left(I^{2}R+0.5W_d\right)\times T_1+\left[I^{2}R\left(1+{\lambda }_1\right)+W_d\right]\times n{T}_2\right\}$$    - for any armour
 $$R_s=R_{s20}\left[1+{\alpha }_{20}\left({\theta }_{sc}-20\right)\right]$$   - for the cable sheath
 $$R_A=R_{A20}\left[1+{\alpha }_{20}\left({\theta }_{ar}-20\right)\right]$$   - for the cable armour

Note: for calculation of the dielectric loss W_d, refer to our Dielectric loss in cables note.

Cable Reactance - for single core cables, where there is significant spacing between conductors, it is necessary to use the reactance in the calculating of circulating current loss.  Accurate values for reactance can be obtained from cable manufacturers or by using software.  Alternatively, the following equations can be used to estimate the reactance (Ω.m-1):

Single core cable reactance estimates (assume bonded at both ends)

 $$X=2\omega {10}^{-7}\ln \left(\frac{2s}{d}\right)$$   - trefoil or flat without transposition
 $$X=2\omega {10}^{-7}\ln \left(2^3\sqrt{2}\frac{s}{d}\right)$$  - flat with transposition
  $$X_m=2\omega {10}^{-7}\ln \left(2\right)$$-  mutual reactance of flat formation cables

Steel tape armour resistance - depending on how steel tape is wound, the resistance can be estimated as follows:

1. tapes laid longitudinally - calculate the resistance as that of an equivalent cylinder (same mass and diameter)
2. tapes laid ≈54° to cable axis - use twice the value obtained from (1)
3. tapes with a very short lay - take resistance as infinite (neglect losses)
4. tapes with a very short lay (double layered) - use twice the value obtained form (1)

Cable Transposition

Transposing of cables (see image) is a technique to reduce the circulating currents within cable sheaths and consequently increase the rating of the cable.

Transposed and cross bonded cable

By cross bonding the sheath the induced currents are in opposite directions, cancelling each other out and significantly improving the current rating of the cable.  Transposing the cables ensures that the reactance balance out and aids in implementation.

At intermediate transposition points, over voltage devices are installed to protect the cable and personnel in the event of voltage build up during faults.

In practice, three minor sections (part between the cross bond) would from a major section (three full transpositions).  It makes sense to do these at each joint point - at each cable drum length.

Transposition and cross bonding are normally carried out in link boxes.

Some Special Situations

Large segmental type conductors

Eddy current losses λ1'', are normally small relative to other losses and can be ignored for single core cables.  This changes for large conductors, which are of a segmented construction.  Under these conditions, the eddy current loss should be considered.

For this condition, the value of λ1'' is derived from the circulating current loss factor λ1' by:

 $${\lambda }_1{}^{\prime \text{}\prime }={\lambda }_1{}^{\prime }\times \frac{4M^2{N}^2+{\left(M+N\right)}^2}{4\left(M^2+1\right)\left(N^2+1\right)}$$

where:
 $$M=N=\frac{R_s}{X}$$   - for cables in trefoil
 $$M=\frac{R_s}{X+X_n}\text{ and }N=\frac{R_s}{X-\frac{X_m}{3}}$$   - for cable in flat formation

Single core cables - variation of route spacing

If the spacing if not maintained the same for the full cable route than the reactance will vary along the route.  In instances such as these, an equivalent overall reactance can be calculated from:

 $$X=\frac{l_a{X}_a+l_b{X}_b+\dots +l_n{X}_n}{l_a+l_b+\dots +l_n}$$

- where l_a, l_b, ... are the section lengths and X_a, X_b, ... are the reactance of each section

Symbols

Cables in troughs pose a particular problem as there tend to be few defined sizing methods. One common approach is to use the empirical method given in IEC  60287-2-1 "Calculation of the current rating - Calculation of thermal resistance".  The approach is to calculate a temperature rise of the air in the trough above it's ambient, with the rating then being calculated as though it were in free air but with the increased temperature. 

The required temperature adjustment, ΔΘ_tr is given by:

$\Delta {\theta }_{tr}=\frac{W_{TOT}}{3p}$

When calculating p, only the effective heat dissipation parts of the trough are included.  Any side exposed to sunlight is not to be included.  In practice this typically becomes the length of the base and two sides of the trough.

Symbols

ΔΘ_tr           - temperature rise above ambient adjustment, °C
W_TOT         - total power dissipated in trough per metre length, W/m
p                - part of trough perimeter effective for heat dissipation, m

Part 2-2 of the IEC 60287 "Calculation of current rating" standard, gives various approaches to obtaining reduction (derating) factors for cables in free air and protected from solar radiation. These are principally designed for multicore cables or single core cables in trefoil but can be extended to other arrangements.

Symbols

Reduction factors for cables with existing ratings

Where the sustained current for an isolated cable or circuit is known, the rating of the cable installed within a group of similar cables is given by:

$I_g=F_g{I}_t$

The reduction (derating) factor F_g for the group is calculated from:

$F_g=\sqrt{\frac{1}{1-k_1+k_1\left({\displaystyle \raisebox{1ex}{$T_{4g}$}\!\left/ \!\raisebox{-1ex}{$T_{4i}$}\right.}\right)}}$

where the surface temperature rise k₁ is provided by:

$k_1=\frac{W{T}_{4i}}{{\theta }_c-{\theta }_a}$

and the term (T_4g/T_4i) is derived iteratively from the ratio (hi/h_g) by:

${\left(\raisebox{1ex}{$T_{4g}$}\!\left/ \!\raisebox{-1ex}{$T_{4i}$}\right.\right)}_{n+1}=\left(\raisebox{1ex}{$h_i$}\!\left/ \!\raisebox{-1ex}{$h_g$}\right.\right){\frac{1-k_1}{\left({\displaystyle \raisebox{1ex}{$T_{4g}$}\!\left/ \!\raisebox{-1ex}{$T_{4i}$}\right.}\right)}+k_1}^{0.25}$

starting with ${\left(\raisebox{1ex}{$T_{4g}$}\!\left/ \!\raisebox{-1ex}{$T_{4i}$}\right.\right)}_{n+1}=\left(\raisebox{1ex}{$h_i$}\!\left/ \!\raisebox{-1ex}{$h_g$}\right.\right)$

with values for (h_i/h_g) are given in the data section below.

Reduction factor for IEC 60287 calculated ratings

When using IEC 60287 itself to calculate the sustained current ratings of cables, part 2-2.1 gives methods for the calculation of thermal resistance. To apply group derating for cables, the heat emission coefficient h is substituted by the group heat emission coefficient h_g, with h_g being given by:

$h_g=\frac{h}{\left({\displaystyle \raisebox{1ex}{$h_i$}\!\left/ \!\raisebox{-1ex}{$h_g$}\right.}\right)}$

and values for (h_i/h_g) as provided in the data section below.

Clearance values to avid a reduction in rating

Data for calculating reduction coefficients (below) give the necessary conditions which need to be achieved to avoid the need to apply any group derating.

Data for calculating reduction coefficients

Image reproduced from IEC 60287-2-2, Table 1

Where cables are arranged in both the horizontal and vertical planes, the sustained current capacity shall be derived using values of h_i/h_g for the vertical plane. 

A common question is how to size an armoured single-core cable buried in the ground to BS7671. This is not possible, but there are alternatives.  This post will examine one alternative, namely ERA 69-30.

 You cannot size single-core unarmoured buried cables to BS7671 (although the standard allowed these cables to be used) because Appendix four does not contain the necessary data. The appendix contains data for multicore buried cables but not single-core. Alternatively, you can use ERA 69-30, which has installation methods and sizing data for buried single-core cables.  

The following video shows how this works:

Cable exposed to direct solar radiation will be subject to greater thermal loads and additional heating. This extra heating needs to be considered in the calculation of sustained current capacities and a potential increase in the resistance of the cable.

IEC 60287

Where cables are subject to solar radiation (or other infra-red radiation), the current capacity can be derived using methods given in IEC 60287.  This is accomplished by reducing the allowable temperature rise, Δθ of the rating equations (1.4.4.1, 4.4.4.1).  The amount of reduction in Δθ is given by:

$\sigma D_e^{*}H{T}_4^{*}$

σ         - absorption coefficient of solar radiation for the cable surface (table 4 of the standard)
H        - intensity of solar radiation W.m2
$T_4^{*}$      - external thermal resistance (adjusted for solar radiation) K.m.W-1

In the calculation of current rating, T4 is the thermal resistance of the surrounding medium.  Part 2 of the IEC 60287 standard gives guidance on how to calculated T4, both considering and not-considering solar radiation. 

Derating Factors

Standards based on derating factors (BS 7671, ERA 69-30 and IEC 60502 for example), do not consider the effect of direct solar radiation.  Typically the standards will refer the user to IEC 60287 in these instances.  While IEC 60287, is the preferred method to determine the sustained current rating, the calculations are complex and just using an additional derating factor for solar radiation is commonly employed. 

The French standard NF C 13-200 suggests a derating factor of 0.85 for cables exposed to solar radiation. Cable manufacturers such as Nexans recommend a factor of 0.8 (black sheath).  As a conservative factor, 0.8 would seem a suitable value to apply for cables exposed to direct solar radiation.

Our recommendation is to increase the ambient temperature for the calculation, such that the derating factor used includes an additional 0.8 for solar radiation.  For example, BS 7671 90 °C thermosetting cables, the derating factor for 25 °C ambient is 1.02. Multiplying by 0.8 gives 0.82 which is the factor for 50 °C ambient (adding 25 °C to the expected temperature, will allow for direct solar heating).

Using an increased temperature more accurately reflects the actual condition and will be reflected in resistance calculations as well as sustained current capacity.  From a review of the derating factors given in the standards, allowing an additional 25 °C is generally conservative but will result in safely designed cables.  For some cables or naturally high ambient temperature environments, the engineer may wish to consider to smaller temperature adjustment, so as not to oversize cables.

Avoidance

In addition to (or as an alternative) to allowing for direct solar radiation effects, consideration can be given to physical measures.  These could include relocating the cables to shaded areas, covering ladder or tray with ventilated cable covers or installing some other shading system.

When selecting a cable, the performance of the cable under fault conditions is an important consideration. It is important that calculations be carried out to ensure that any cable is able to withstand the effects of any potential fault or short circuit.  This note looks at how to do this.

The primary concern with cables under a fault condition is the heat generated, and any potential negative effect this may have on the cable insulation. 

Calculation of fault rating is based on the principle that the protective device will isolate the fault in a time limit such that the permitted temperature rise within the cable will not be exceeded.

In addition to the direct heating effect of fault currents, other considerations include:

• electro-mechanical stress and fault levels large enough to cause cable failure
• performance of joint and terminations under fault conditions

When calculating the fault ratings of a cable, it is generally assumed that the duration is short enough that no heat is dissipated by the cable to the surrounding.  Adopting this approach simplifies the calculation and errs on the safe side.

The normally used equation is the so-called adiabatic equation.  For a given fault of I, which lasts for time t, the minimum required cable cross-sectional area is given by:

 $A=\frac{\sqrt{I^{2}t}}{k}$

where: A - the nominal cross-section area, mm²
I - the fault current in, A
t -  duration of fault current, s
k - a factor dependant on cable type (see below)

Alternatively, given the cable cross-section and fault current, the maximum time allowable for the protective device can be found from:

 $t=\frac{k^2{A}^2}{I^2}$

The factor k is dependant on the cable insulation, allowable temperature rise under fault conditions, conductor resistivity and heat capacity.  Typical values of k are:

*where two values; lower value applied to conductor CSA > 300 mm²
* these values are suitable for durations up to 5 seconds, source: BS 7671, IEC 60364-5-54

Derivation  - Adiabatic Equation and k

The term adiabatic applies to a process where there is no heat transfer.  For cable faults, we are assuming that all the heat generated during the fault is contained within the cable (and not transmitted away).  Obviously, this is not fully true, but it is on the safe side.

From physics, the heat Q, required to rise a material ΔT is given by:

 $Q=cm\Delta T$

where Q - heat added, J
c - specific heat constant of the material, J.g^-1.K^-1
m - mass of the material, g
ΔT - temperature rise, K

The energy into the cable during a fault is given by:

 $Q=I^{2}Rt$

where  R - the resistance of the cable, Ω

From the physical cable properties, we can calculate m and R as:

 $m={\rho }_{c}Al$   and     $R=\frac{{\rho }_{r}l}{A}$

where ρ_c - material density in g.mm^-3
ρ_r - resistivity of the conductor, Ω.mm
l - length of the cable, mm

Combining and substituting we have:

 $I^{2}Rt=cm\Delta T$

 $I^{2}t\frac{{\rho }_{r}l}{A}=c{\rho }_{c}Al\Delta T$

and rearranging for A gives:

 $S=\frac{\sqrt{I^{2}t}}{k}$   by letting    $k=\sqrt{\frac{c{\rho }_c\Delta T}{{\rho }_r}}$

Note: ΔT is the maximum allowable temperature rise for the cable:

 $\Delta T={\theta }_f-{\theta }_i$

where θ_f - final (maximum) cable insulation temperature, °C
θ_i - initial (operating) cable insulation temperature, °C

Units:  are expressed in g (grams) and mm², as opposed to kg and m.  This is widely adopted by cable specifiers.  The equations can easily be redone in kg and m if required.

IEC 60364-5-54 gives the following formula for the calculation of the factor k:

$k=\sqrt{\frac{Q_c(\beta +20)}{{\rho }_{20}}\ln \left(\frac{\beta +{\theta }_f}{\beta +{\theta }_i}\right)}$

where Q_c = volumetric heat capacity of conductor at 20°C, J.K^-1.mm^-3
ß = reciprocal of temperature coefficient of resistivity at 0 °C 
ρ₂₀ =  electrical resistivity of conductor material at 20 °C, Ω.mm
ϴ_i = initial temperature of the conductor, °C
ϴ_f = final temperature of the conductor, °C

Note: strictly speaking ß is specified at 0 °C.  Within myCableEngineering calculations, ß is determined by taking the reciprocal of the 20 °C temperature coefficient.  This introduces a small, but negligible error. 

Typical Application

For manual calculations, the following will give realistic results:

Substituting the above values and rearranging the IEC equation slightly, gives:

$k=226\sqrt{\ln \left(1+\frac{{\theta }_f-{\theta }_i}{234.5+{\theta }_i}\right)}$- copper conductors

$k=148\sqrt{\ln \left(1+\frac{{\theta }_f-{\theta }_i}{228+{\theta }_i}\right)}$- aluminium conductors

$k=78\sqrt{\ln \left(1+\frac{{\theta }_f-{\theta }_i}{202+{\theta }_i}\right)}$- steel

For details on thermal withstand and I2t, see I2t Thermal Withstand. Energy limiting devices reduce the I2t let-through in the event of a fault.  The values are typically specified by the manufacturer:

Typical MCB Energy Limiting Characteristic

Using typical data given above, for a 10A, 10kA fault rated MCB would have an Ip/A of 1000 (10³), give an energy let-through of approximates 2,000 A²s.

The thermal withstand of a cable can be determined by comparing the maximum energy at the fault, with the maximum energy the cable can absorb.  The equation for this is:

$I^{2}t\le k^2{A}^2$

I²t is proportional to the energy let-through of the protective device
       I - fault current, A
       t - fault duration, S

k²A² is proportional to the energy withstand of the cable
        k - adiabatic constant
        A - cable cross-sectional area, mm²

To better understand the theory behind the above, please see The adiabatic equation.

As mentioned, the adiabatic equation assumes no heat is dissipated from the cable during a fault.  While putting the calculation on the safe side, in some situations, particularly for longer fault duration there is the potential to be able to get away with a smaller cross section.  In these instances, it is possible to do a more accurate calculation.

Considering non-adiabatic effects is more complex.  Unless there is some driver, using the adiabatic equations is just easier.  Software is available to consider non adiabatic effects, however, there is a cost, time and complexity associated with this.

The IEC also publish a standard which deals with non-adiabatic equations:

• IEC 60949 "Calculation of thermally permissible short-circuit current, taking into account non-adiabatic heating effects".

The method adopted by IEC 60949 is to use the adiabatic equation and apply a factor to cater for the non-adiabatic effects:

 $I=\epsilon I_{AD}$

where I - permissible short circuit current, A (or kA)
I_AD - adiabatic calculated permissible short circuit current, A (or kA)
ε - factor to allow for heat dissipation from cable

The bulk of the IEC 60949 standard is concerned with the calculation of ε.

Voltage drop is calculated in accordance with CENELEC technical report CLC/TR 50480 "Determination of cross-sectional area of conductors and selection of protective devices", dated February 2011. 

The voltage drop (as a percentage) is given by:

$\Delta U=\frac{b\left(R\cos \phi +X\sin \phi \right)I_b}{U_o}\cdot 100$

where
        ΔU         = cable voltage drop, %
        U_o             = nominal  line to neutral voltage, V
        R           = cable resistance, Ω
        X           = cable reactance, Ω
        I_b           = cable design current, A
        b            = circuit factor (=2 for d.c. and single-phase, =1 for three-phase)

Note: within myCableEngineering, we use complex arithmetic and the above is evaluated as:

$\Delta U=\frac{b\left(R+jX\right)I_b}{U_o}\cdot 100$

Note: R and X are per line conductor.  For example, the resistance of a single-phase two core circuit (live and neutral) would be 2R (assuming the live and neutral circuits are of equal resistance).  For parallel cables, R and X are the per line resultant values. For example, n_ph cables in parallel, with the resistance of one line conductor is R_c0ph then the R = R_c0ph/n_ph.

The above equation), has the relative voltage drop expressed as a percentage of Uo.  Multiplying by U_o will give the actual voltage drop for d.c. and single-phase circuits and for three-phase circuits, this needs to be multiplied by √3.    

Adjusting the CENELEC equation to take into account referring the voltage drop to line-line voltage for three-phase systems, setting R, X in ohms, and using complex forms gives:

	ΔU, V/m	ΔU, % per m
d.c.  systems
a.c.  systems, single-phase
a.c systems, three-phase
-  the resistance of a single conductor, Ω/m           - the reaction of a single conductor,  Ω/m           - the impedance of single conductor , Ω/m          - cable design current, A        - the nominal line to neutral/earth voltage (for single-phase a.c. or d.c.), V        - the nominal line to line voltage (for three-phase systems), V

    

BS 7671  Voltage Drop Tables

BS 7671 , Appendix 4 (Informative), gives a procedure for calculating the voltage drop in low voltage cables.  This procedure is based on looking up resistive and reactive voltage drops in a table and determining the cable voltage drop from these values.

Voltage Drop Calculation

The tables in BS 7671 give the voltage drops for different cables in mV/A/m (or effectively mΩ/m) for various cable types and installation arrangements.  Voltage drop figures are related to the cable's maximum operating temperature and the line to neutral voltage for single phase circuits and line to line voltage for three phase circuits. Values for both resistive (R), reactive (X) and total impedance (Z) voltage drops are given.

Typical BS 71 Voltage Drop Table

Note: the figures in the tables apply to frequencies from 49 to 61 Hz for a.c. cables and assume armour is bonded at both ends for single core armoured cables. For conductors less than 16 mm², the tables ignore inductance and only resistive voltage drops are given.

Typically the cable voltage drop is found by multiply the figures given in the table by the intended cable operating current and its length.  This approach leads to significant inaccuracies where the cable is operating substantially below its rated value due to the cable having a lower operating temperature.  If power factor is not correctly considered, errors in calculated voltage drop will also be apparent.  

Only the resistance of the cable is affected by temperature.  To correct the for the operating temperature of the cable, we can use:

$C_t=\frac{230+t_p-\left(C_a^2{C}_g^2{C}_s^2{C}_d^2-{\displaystyle \frac{I_b^2}{I_t^2}}\right)\left(t_p-30\right)}{230+t_p}$

t_p        - the maximum permitted normal operating temperature of the cable
I_b        - design current of the circuit (current intended to be carried by the cable)
I_t         - value to tabulated current (BS 7671 appendix 4, tables)
C_a       - rating factor for ambient temperature
C_g       - rating factor for grouping
C_s       - rating factor for soil thermal resistivity
C_d       - rating factor for depth of burial

Combining the temperature correction factor, with the power factor,$\cos \left(\varphi \right)$ the voltage drop (in mV) is given by:

$vd =I_b*L*C_t \cos \left(\varphi \right) \left(\mathrm{tabluated}\right(mV/A/m{)}_r) + \sin (\varphi ) (\mathrm{tabulated}(mV/A/m{)}_x)$

vd             - voltage drop across the cable in mV (divide by 1000 for V)
$\cos \left(\varphi \right)$   - the power factor of the circuit
L               - length of the cable, m

Note: power factor is 1 and the tabulated (mV/A/m)_x is 0 for a d.c. circuit. For cable CSA less than 16 mm², BS 7671 ignores inductance and (mV/A/m)_x can be taken as zero. For three phase circuits, the voltage drop is related to the line-line voltage - to relate to the line-neutral voltage divide by √3.

To obtain the voltage drop as a percentage, divide by the nominal voltage for d.c. circuits, the line-neutral voltage for single phase circuits and the line-line voltage for three phase circuits.

From BS 7671 Appendix 4, Table 4Ab, the voltage drop between the origin of any installation and load point should not be greater than:

Note: for installation runs longer than 100 m, the above can be increased by 0.005% per metre beyond 100m, without any increase being greater than 0.5%.

Example Calculation

To illustrate the above, consider a single core, 50 mm² copper XLPE armoured cable, supplying a 400 V three phase load of 152 A with pf=0.87.  Assume the cable is installed in trefoil on a ladder with an ambient of 35 °C, not grouped with any other cables and 80 m in length. 

To view the detailed calculation for the cable, please see BS 7671 Voltage Drop Calculation.  This calculation shows that the absolute voltage drop is 10.06 V (referenced to the line-line voltage).  As a percentage the voltage drop is 2.52%.

myCableEngineering.com - sizing the same cable with myCableEngineering, using a generic cable gives the same voltage drop (10.07 V and 2.52%). Using a BS 6724 cable gives 2.46%.  These types of variation are due to BS 7671 using a one-size-fits-all approach, whereas in myCableEngineering voltage drop is calculated based on each cables actual physical characteristics.

Deriving Impedance & CENELEC 50480

Voltage drop in cables is always related to the impedance of the cable.  Consequently, it is possible to drive an equivalent impedance for the mV/A/m values given in the BS 7671 tables. To convert the three phase table values so that they relate to the input resistance (or reactance) required by CENELEC 50480 it is necessary to divide the BS 7671 three phase values by √3 (the square root of three).  No adjustment is required for single phase values.  

Given values of voltage drop for three phase balanced systems are related to the line voltage.  

$R = (mV/A/m{)}_r / (2*1000)$ and $X = 0$
        - for a d.c. circuit

$R = (mV/A/m{)}_r / (2*1000)$ and $X = (mV/A/m{)}_x/(2*1000)$
        - for a singe phase a.c. circuit

$R = (mV/A/m{)}_r / (\sqrt{3}*1000)$ and $X = (mV/A/m{)}_x/(\sqrt{3}*1000)$
        - for a singe phase a.c. circuit

(mV/A/m)_r        - tabulated resistive mV/A/m from BS 7671
(mV/A/m)_x        - tabulated reactive mV/A/m from BS 7671
R                      - resistance per conductor in  Ω/m
X                      - reactance per conductor in Ω/m

Note: the values given in BS 7671 for resistive voltage drop are at the cables maximum operating temperature (i.e. 90 °C for thermosetting cables).  These can be adjusted to the required temperature by standard formulae.  See Conductor Resistance for details on how to adjust for temperature.

Understanding of how cables (and busbar) perform is at heart a thermal problem. 

Heat is generated in the cable due to current flowing.  This generated heat then interacts with the environment and is dissipated. Thermal modelling of a cable installation establishes that the steady state temperatures are below the safe operating conditions for the materials and personnel.

Note: for a list of symbols, see the bottom of the post.

Heat Flow

The defining equation of heat flow is governed by Fourier's law of heat conduction:

        $q_x=-kA\frac{\partial T}{\partial x}$    

Conduction, Convection & Radiation

For cable modelling, we typically assume constant thermal conductivity and the generalised heat flow equation (PDE) is given by:

        $\rho C\frac{\partial T}{\partial t}-\nabla ·\left(k\nabla T\right)=\dot{q}$    - parabolic form (transient)

        $-\nabla ·\left(k\nabla T\right)=\dot{q}$    -   elliptic form (steady state)

Note: the above equation is for three dimensions.  Often in cable problems, we are only concerned with a cable section and will use partial derivatives in the 'x' and 'y' plane only.

The above equations govern heat flow by conduction.  In practical situations, we often also have to consider Convection and Radiation.

Power (heat generated)

The power dissipated in the cable (or conductors) is calculated I²R.  Where required sheath and dielectric losses can be estimated using IEC 60287. For further information, see:

• conductor loss: Power Loss
• sheath loss: [work in progress]
• dielectric loss: Dielectric loss in cables

It should be noted that heat generated Q, is in W/m³. Any calculated I²R power needs to be converted to W/m³, by dividing the value obtained by the volume over which the power is dissipated. 

Solving the problem

To solve a cable installation problem, the following steps are carried out:

1. the installation is modelled as a series of heat flow PDE
2. boundary conditions are determined
3. the system of equations and boundary conditions are solved to give heat flow and temperatures

Boundary Conditions

Depending on the complexity of the installation, various boundary conditions may be relevant.  For cable installation, we typically come across the following boundary conditions.

Finite Element Analysis (FEA)

To solve the cables analytically, would be extremely difficult if not impossible.  We there for use finite element analysis (FEA).  FEA involves breaking the geometry into a mesh of small solvable grids (such as tetrahedrons).  By solving all the small grids we are able to solve the complete cable installation.  

Typical steps in FEA consist of:

1. defining the geometry of the cable installation
2. setting up the heat flow PDE(s) for the components of the installation
3. applying boundary (and initial) conditions
4. meshing the geometry
5. solving the installation system
6. plotting and visualizing the results

Symbols

A        - area, m²
c        - specific heat of material, J/kg.°C
k        - thermal conductivity, W/m.°C
T        - temperature, °C
T_a        - surface temperature of enclosure

q         - heat flow, W (or Joules/second)
$\dot{q}$        - heat generated per unit volume, W/m³

ρ        - density, kg/m³
t         - time, s   

Boundary conditions:
g       - heat flux, W/m² 
h       - weighting coefficient
r        - temperature, °C
q        - heat transfer coefficient

For cable modelling, we typically assume constant thermal conductivity, see Cable Thermal Analysis.

To consider the effect of convection (natural or forced in a gas or liquid), Newton's law of cooling is used:

$q=hA\left(T-T_a\right)$

for air (a common cable installation medium), the convection heat transfer coefficient h, flowing at velocity v, can be obtained from:

$h=7.371+6.43v^{0.75}$

To set a convection boundary conditions, we can use a Neumann condition with  q = desired convection coefficient and g = temperature of environment times the convection coefficient.

Enclosed Spaces

Convection is a complex topic, requiring a detailed understanding of heat flow, fluid flow and the behaviour of gases as they flow across surfaces.  For cable sizing, to achieve this level of technical detail is not practical nor achievable in many situations. One effective technique when solving equations for enclosed spaces is the use of an effective thermal conductivity, k_e.  Using an effective thermal conductivity enables the air to be treated from a conduction view, eliminating much of the complexity in implements full connectivity solutions.

For an enclosed spaced, the effective thermal conductivity is given by:

$k_e=k{N}_u$ The Grashof number is given by: $G_{r\delta }=\frac{g\beta \left(T_1-T_2\right){\delta }^3}{{\nu }^2}$ with: $\beta =\frac{1}{T_f}$ and $T_f=\frac{T_1+T_2}{2}$ Note: Tf is in Kelvin (not °C). Experimental results for free convection in an enclosure give: $N_{u_{\delta }}=C{\left(G_{r_{\delta }}{P}_r\right)}^n{\left(\frac{L}{\delta }\right)}^m$ Typically we can take g = 9.80665 and for air: Tf, °C ν, 10-5 m2.s Pr 0 1.343 0.720 20 1.568 0.708 80 2.056 0.697 120 2.591 0.689 Note: for differing temperatures, exact factors can easily be looked up on the internet

Note: δ is the hydro-dynamic boundary layer thickness.  

Values of C, n and m can be obtained from the following table using the Grashof-Prandtl number product:

Symbols

A                - area, m²
g                -  acceleration of gravity, m/s²
h                - convection heat transfer coefficient, W/m².°C
k                - thermal conductivity, W/m.°C
k_e               - effective thermal conductivity, W/m.°C 
T, T₁, T₂, T_a        - temperature, K
v                - velocity of air, m/s 
ν                 - kinematic viscosity, m²/s

q         - heat flow, W (or Joules/second)
$\dot{q}$        - heat generated per unit volume, W/m³

β        -  coefficient of thermal expansion
δ        - enclosure dimension, m

G_r         - Grashof number
N_u        - Nusselt number
P_r        - Prandtl number
R_a        - Rayleigh number 

Radiation is the transfer of heat by electromagnetic radiation. 

Luckily in cable installations, we rarely encounter the need to consider radiation.  One exception is that of bare conductors in an enclosure, where the effect of radiation can be taken into account by:

$q=\epsilon \sigma A\left(T^4-{T_a}^4\right)$

with ε and A properties of the busbar, and T_a the temperature of the enclosure

Busbar & Enclosures

Radiation heat flow between two plates (busbar surface to enclosure for example), can be calculated from:

$q=\frac{\delta A\left({T_1}^4-{T_2}^2\right)}{{\displaystyle \frac{1}{{\epsilon }_1}+\frac{1}{{\epsilon }_2}-1}}$

q is the heat flow in W between plates of area A
ε₁, ε₂ are the emissivities of the two surfaces
T₁, T₂ are in K (not °C)

The effect of radiation is included in any solution as a suitable boundary condition.  An alternative approach would be to adjust the heat flow for the volume of the busbar and subtract this directly from the heat generated within the busbar.

Symbols

A         - area, m²
T₁       -  temperature of surface (hottest), K
T₂       -  temperature of enclosure

q         - heat flow, W (or Joules/second)
$\dot{q}$        - heat generated per unit volume, W/m³

ε         - emissivity of a surface