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IEC 60287 Current Capacity of Cables - Introduction

Last updated on 2023-03-31 5 mins. to read

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:

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

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

$Δ θ / T$

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

$I = \sqrt{\frac{Δ θ}{R^{'} T}}$

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.

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Cable Sheath and Armour Loss

Last updated on 2023-03-31 13 mins. to read

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:

$λ_{1} = λ_{1}^{' '} + λ_{1}^{' '}$

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.

	Sheath Circulating Current Loss, λ1'
Single core cables - trefoil, bonded at both ends	$λ_{1}^{'} = \frac{R_{S}}{R} \frac{1}{1+ {(\frac{R_{s}}{X})}^{2}}$
Single core cables - flat, with transposition, bonded at both ends	$λ_{1}^{'} = \frac{R_{S}}{R} \frac{1}{1+ {(\frac{R_{s}}{X})}^{2}}$
Single core cables - flat, without transposition, bonded at both ends	λ₁₁′ - loss factor for the outer cable with the greater losses $λ_{11}^{'} = \frac{R_{s}}{R} [\frac{0.75 P^{2}}{R_{s}^{2} + P^{2}} + \frac{0.25 Q^{2}}{R_{s}^{2} + Q^{2}} + \frac{2 R_{s} P Q X_{m}}{\sqrt{3} (R_{s}^{2} + P^{2}) (R_{s}^{2} + P^{2})}]$ λ₁₂′ - loss factor for the outer cable with the least losses $λ_{12}^{'} = \frac{R_{s}}{R} [\frac{0.75 P^{2}}{R_{s}^{2} + P^{2}} + \frac{0.25 Q^{2}}{R_{s}^{2} + Q^{2}} - \frac{2 R_{s} P Q X_{m}}{\sqrt{3} (R_{s}^{2} + P^{2}) (R_{s}^{2} + P^{2})}]$ λ_1m′ - loss factor for the middle cable $λ_{1 m}^{'} = \frac{R_{s}}{R} \frac{Q^{2}}{R_{s}^{2} + Q^{2}}$ where: $P = X + X_{m}$ $Q = X + \frac{X_{m}}{3}$

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.

	Sheath Eddy Current Loss, λ1''
Two core cable - common sheath, unarmoured	- for round or oval conductors $λ_{1}^{' '} = \frac{16 ω^{2} 10^{- 14}}{R R_{s}} (\frac{c}{d}) [1 + {(\frac{c}{d})}^{2}]$ - for sector shaped conductors $λ_{1}^{' '} = \frac{10.8 ω^{2} 10^{- 16}}{R R_{s}} (\frac{1.48 r_{1} + t}{d}) [12.2 + {(\frac{1.48 r_{1} + t}{d})}^{2}]$
Three core cable - common sheath, unarmoured	- round or oval conductors, R_s ≤ 100 µΩ.m^-1 $λ_{1}^{' '} = \frac{3 R_{s}}{R} [{(\frac{2 c}{d})}^{2} \frac{1}{1 + {(\frac{R_{s}}{ω} 10^{7})}^{2}} + {(\frac{2 c}{d})}^{4} \frac{1}{1 + 4 {(\frac{R_{s}}{ω} 10^{7})}^{2}}]$ - round or oval conductors, R_s >100 µΩ.m^-1 $λ_{1}^{' '} = \frac{3.2 ω^{2}}{R R_{s}} {(\frac{2 c}{d})}^{2} 10^{- 14}$ - for sector shaped conductors (any R_s) $λ_{1}^{' '} = 0.94 \frac{R_{s}}{R} {(\frac{2 r_{1} + t}{d})}^{2} \frac{1}{1 + {(\frac{R_{s}}{ω} 10^{7})}^{2}}$
Two or three core cable - steel tape armour	Multiple the unarmoured cable factor by: ${[1 + {(\frac{d}{d_{A}})}^{2} \frac{1}{1 + \frac{d_{A}}{μ δ}}]}^{2}$
Cables with each core in a separate sheath or pipe-type cables	${λ_{1}}^{' '} = \frac{R_{s}}{R} \frac{1.5}{1 + {(\frac{R_{s}}{X})}^{2}}$ where: $X_{S L} = 2 ω 10^{- 7} \ln (\frac{2 c}{d})$

Calculation of armour loss

For armoured cables, the losses are estimated as shown.

	Armour Loss, λ2
Non-magnetic armour	Use equation for λ1'', substituting: parallel combination of sheath and armour resistance for R_s root mean square of sheath and armour diameter for d
Single core cables - steel wire armour	General advice is not to use magnetic armour for single core cables. If required, then the guidelines given in IEC 60287 on estimating losses should be followed.
Two core cable - steel wire armour	$λ_{2} = \frac{0.62 ω^{2} 10^{- 14}}{R R_{A}} + \frac{3.82 A ω 10^{- 5}}{R} {[\frac{1.48 r_{1} + t}{d_{A}^{2} + 95.7 A}]}^{2}$
Three core cable - steel wire armour	- round conductor $λ_{2} = 1.23 \frac{R_{A}}{R} {(\frac{2 c}{d_{A}})}^{2} \frac{1}{{(\frac{2.77 R_{A} 10^{6}}{ω})}^{2} + 1}$ - sector shaped conductor $λ_{2} = 0.358 \frac{R_{A}}{R} {(\frac{2 r_{1}}{d_{A}})}^{2} \frac{1}{{(\frac{2.77 R_{A} 10^{6}}{ω})}^{2} + 1}$

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:

$θ_{s c} = θ - (I^{2} R + 0.5 W_{d}) \times T_{1}$    - for any sheath
$θ_{a r} = θ - {(I^{2} R + 0.5 W_{d}) \times T_{1} + [I^{2} R (1 + λ_{1}) + W_{d}] \times n T_{2}}$     - for any armour
$R_{s} = R_{s 20} [1 + α_{20} (θ_{s c} - 20)]$    - for the cable sheath
$R_{A} = R_{A 20} [1 + α_{20} (θ_{a r} - 20)]$    - 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)

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

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

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

$λ_{1}^{' '} = λ_{1}^{'} \times \frac{4 M^{2} N^{2} + {(M + N)}^{2}}{4 (M^{2} + 1) (N^{2} + 1)}$

where:
$M = N = \frac{R_{s}}{X}$ - for cables in trefoil
$M = \frac{R_{s}}{X + X_{n}} and N = \frac{R_{s}}{X - \frac{X_{m}}{3}}$ - 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

A	- armour cross sectional area, mm²
R	- conductor a.c. resistance, Ω.m^-1
R_A	- armour resistance maximum at operating temperature, Ω.m^-1
R_A20	- armour resistance at 20 °C, Ω.m^-1
R_s	- sheath or screen resistance at maximum operating temperature, Ω.m^-1
R_s20	- sheath or screen resistance at 20 °C, Ω.m^-1
X	- sheath or screen reactance, Ω.m^-1
X_m	- mutual reactance (sheath one cable to conductors of other cables), Ω.m^-1

λ₁	- ratio of sheath loss to total conductor loss
λ₂	- ratio of armour loss to total conductor loss

λ₁'	- sheath loss caused by circulating currents
λ₁''	- sheath loss caused by eddy currents

c	- distance between axis of conductors, mm
d	- mean diameter of sheath or screen, mm
d_A	- mean diameter of armour, mm
r₁	- circumscribing radius of sector shaped conductors, mm
s	- axial separation of conductors, mm
t	- insulation thickness between conductors, mm
T₁	- thermal resistance between conductor and sheath, K.m.W^-1
T₂	- thermal resistance between sheath and armour, K.m.W^-1

θ	- maximum conductor temperature, °C
θ_ar	- maximum operating temperature of armour, °C
θ_sc	- maximum operating temperature of screen, °C
ω	- angular frequency (2πf)
µ	- relative magnetic permeability of armour
δ	- equivalent thickness of armour, mm

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Nov 28, 2018 2:59 PM

"Cables with each core in a separate sheath or pipe type cables". it should be "Losses in screen and sheaths of pipe-type cables", and (Rs/XSL)^2

Steven McFadyen Nov 29, 2018 11:59 AM

Thanks, your right it the term should be squared. Have updated the article.

Cable Troughs

Last updated on 2023-03-31 1 mins. to read

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:

$Δ θ_{t r} = \frac{W_{T O T}}{3 p}$

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

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Derating Factors - cables grouped in air

Last updated on 2023-03-31 3 mins. to read

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

D_e	external diameter of multicore cable or one single core in trefoil	mm
e	clearance between adjacent cables in a group (between surfaces)	mm
F_g	group reduction factor
h	heat dissipation coefficient, calculation of thermal resistance	W.m^-2.K^-5/4
h_g	heat dissipation coefficient of the hottest cable or circuit in a group	W.m^-2.K^-5/4
h_i	heat dissipation coefficient of isolated cable or circuit	W.m^-2.K^-5/4
I_g	rating of the hottest cable in a group	A
I_t	rating of one cable or circuit assumed isolated	A
k₁	surface temperature rise factor
T_4g	external thermal resistance of the hottest cable in a group	K.m.w^-1
T_4i	external thermal resistance of one cable assumed to be isolated when carrying I_t	K.m.w^-1
W	power loss from isolated cable or circuit in trefoil, carrying I_t	W.m^-1
Θ_c	conductor temperature used for calculating I_t	°C
Θ_a	ambient temperature used for calculating I_t	°C

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:

$F_{g} = \sqrt{\frac{1}{1 - k_{1} + k_{1} (\frac{T_{4 g}}{T_{4 i}})}}$

where the surface temperature rise k₁ is provided by:

$k_{1} = \frac{W T_{4 i}}{θ_{c} - θ_{a}}$

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

${(\frac{T_{4 g}}{T_{4 i}})}_{n + 1} = (\frac{h_{i}}{h_{g}}) {\frac{1 - k_{1}}{(\frac{T_{4 g}}{T_{4 i}})} + k_{1}}^{0.25}$

starting with ${(\frac{T_{4 g}}{T_{4 i}})}_{n + 1} = (\frac{h_{i}}{h_{g}})$

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:

$h_{g} = \frac{h}{(\frac{h_{i}}{h_{g}})}$

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.

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