# IEC 60287 Current Capacity of Cables - An Introduction

By on January 31st, 2013

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 with 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
Θa - ambient temperature, °C
ΔΘ - temperature difference (Θ-Θa), 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:

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

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

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 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):

## 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.

## Other Related Resources

Cable Sizing Tool  - describes the procedure for the sizing of cable to BS 7671 and IEC 60364

Standard Cable & Wire Sizes  - list of standard IEC 60228  wire sized and AWG conversion table

8  Steps to Low Voltage Power Cable Selection and Sizing - general guide to selecting/using LV cables

Cable Insulation Properties - typical properties of various types of cable insulation

## 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.

More interesting Notes:

Steven has over twenty five years experience working on some of the largest construction projects. He has a deep technical understanding of electrical engineering and is keen to share this knowledge. About the author

#### View 3 Comments (old system)

1. nightex says:
2/5/2013 2:47 PM

Thanks for great info. But I have little problem with calculation. in given formula I=(deltaQ/RT)^1/2 thermal reistance T is given by K*m/W and temperature diference deltaQ in celsius (in given example). So how can this formula be true? I was trying to convert C to K (deltaQ) but then I got huge amperage.

• Steven says:
2/6/2013 2:56 AM

Thanks for the comment.

If you are looking at ΔΘ, then it is a change in temperature, so if you use Celsius or Kelvin it's the same. I've added the units on the example calculation so it may help. Also if you do a units analysis, it shows than answer would be in Amperes.

In the second post (I'll write soon), I intend to cover the actual current rating equations in the standard. By simplifying these equations for the given example, they also reduce to the the same formula.

2. Notes says:
2/17/2013 7:31 AM

In the previous note we looked at the approach taken by the standard to the sizing of cables and illustrated this with an example.  We then looked at one method of applying the standard and identified resources enabling the calculation of all the... ...

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