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Technical Paper 

Armour Currents in Single Core Cables 

by John.V.H. Sanderson CEng, FIEE

Introduction

Use of armoured single core cables in HV installations may be the designer's choice - easy toinstall and terminate and they contribute to good earthing, but there can be seriousconsequences if the flow of armour current is neglected. The same is true for single core LVcables and for any multi core cables where currents flow but do not return via cores in the samecable. All current carrying conductors produce magnetic fields and these induce armour currentswhen gland terminations are made to metalwork at both ends of the cable. The magnitude of 

armour current induced can be comparable with conductor current.

The diagram shows just two armoured cables and the path of current in the armour.The current that flows depends on the physical arrangement of cables, the resistance of thearmour wire, the gland connection at the cable ends, the cable core currents and the presenceand connection of cable screens.It is usual for armour to be connected to brass glands which terminate the armour wire neatlyand provide electrical connection to the armour. The glands should be fixed to a non-ferrousgland plate(s) and they then support the cable. The cable screen is usually brought out to a

substantial terminal lug which can be fixed to an earth terminal at one end of the cable andeither connected or left unconnected at the other end of the cable. If the cable screen isconnected via terminal lugs at both ends of a cable the effect is similar to having armour connected via glands at both ends of the cable. Reference to armour currents in this paper means both screen currents and armour currents.

 Any armour current induced produces its own magnetic field and this partly cancels the

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magnetic field due to the power conductors. If the armour resistance is near zero, the inducedcurrent in armour and screen together will be close in magnitude to the main conductor current.

If there was just one strand of armour on each of the two cables and if this was located in theposition shown in the figure, then it would be correct to use the flux shown to calculate inducedvoltages and the induced current.In practice, there are many armour strands and they are wound in a spiral fashion with arepetition rate of, typically, 1m. So, in a 100m run of armoured cable, each strand occupies theposition closest to the adjacent conductor, 100 times.

The individual armour strands do not make good electrical contact with each other and can betreated as insulated, so that the armour as a whole does not behave as a solid cylindricalconductor. The amount of flux linkage to be used for the armour as a whole should therefore bean average amount that represents the different linkages with each of the individual strands.The proper mathematical analysis should be based on establishing the magnetic flux densityvector, B(Ik), due to each of the main currents, I k, and to perform surface integrals to derive fluxlinkage. The integration limits define the amount of flux linkage to be used.

For three single cables carrying three phase load current, the magnetic field produced is quiteinteresting but impossible to draw on a simple diagram.

The main conductor currents may not be identical in amplitude and they are displaced in time

or, as we say, separated from each other by phase angles. In HV power systems the threecurrents add to zero at all times except during faults. The magnetic field between conductorsdepends on all three currents, and the resulting armour currents depend on the geometry andseparation of all the cables. Hand calculation is very difficult and a computerised method isneeded. When there are several conductors per phase, all cables must be taken into accountand the physical layout matters too. The computerised method made available here can dealwith up to 10 cables and so caters for up to 3 cables per phase plus one cable to represent

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earth. The earth conductor arrangement is an important factor, this is dealt with in thecomputerised method and is discussed at some length below. Although originally intended to beused for calculation of armour currents due to normally loaded cables, the method will dealequally well with fault currents. Similarly, although intended to be used for single core cables,the method works quite well for multi core cables if care is taken to specify the data.

Case Study

 A generator switchboard fed a main HV board via 3 aluminium wire armoured single core 185mm2 cables per phase. The armour was connected to the generator using brass cable glandsfixed to an aluminium gland plate. The connection at the main switchboard, 150 m away, wasterminated by similar brass glands fixed to an aluminium gland plate. Circulating currents of about 100A flowed in the armour of each of the nine cables. One of the brass glands was badlyfitted. It was too big and the aluminium armour made a poor connection. The bad glanddeteriorated and the heat generated caused deterioration and eventually damage to several of the cables as they entered the glands. Plant operators noticed smoke and they wisely shutdown the generator immediately. Our investigation showed that the induced voltage driving thearmour currents was about 12V and the armour currents were about 100A. The nine cables

were fixed along the cable route originally as shown below.

We re-glanded both ends of all cables using the correct sized fittings and re-arranged the cablesas shown below, with the cables in each group laid in trefoil as far as possible. We then

measured armour currents of about 20A and much reduced armour voltages.

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 As a three phase group, preferably in trefoil, the exterior magnetic field is much reducedcompared with the separated phase arrangement and so the induced current are reduced too.

Measuring the voltage when the glands are insulated from the earthed metal gland plate at oneend was useful, and shows that connecting the glands to an insulated board completely stopsthe armour currents. The voltages developed across the insulated glands can be measured andcalculated. These give indications of potential currents should the glands be conventionally fixedat both ends, rather than insulated at one end.

Our calculation sheet is available free of charge and it should be useful to designers who wishto arrange cable in such a way as to minimise induced currents

Computerised worksheet.

The worksheet is designed to deal with parallel cable arranged in a plane such as on a singlecable tray, or fixed individually to a wall. Even if conductors are arranged in a different way, for example on several cable trays, as in the Case study above, the worksheet will still givereasonable estimates of armour voltages and currents.

The basis of the worksheet is the evaluation of magnetic flux due to each conductor, linking thearmour of adjacent pairs of cables. Data is needed about each cable and their separationdistances. The worksheet will handle up to ten parallel conductors but it is suggested that one of these, the last one, be reserved for an earth cable that ties together the metalwork at one end of the cables to the metalwork at the other end. An electrical earth connection is always present

and induced currents will flow in this path, as they will in the armour of the power cables.Specifying an earth path also enables armour voltages relative to earth to be calculated. Theworksheet calculates first the voltages that would appear between cable glands on adjacentcables when they are isolated from each other at one end, but connected to each other and toearth at the other end. This array of (n-1) voltages for a total (n) cables is a good guide to theinduced currents that will flow should the isolated glands be connected to each other and toearth.

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Data to be entered are:

1. The cable to cable centre separation distances mm.

2. The outside diameters of each cable in mm. The worksheet assumes that the armour iscovered by 2mm thick PVC outer sheath.

3. The current magnitude in each cable in A, and angle in degrees. The sum of currents in the(n) cables must of course be zero and so the worksheet calculates the current in the n thconductor. The user must then adjust the (n-1) conductor currents to make this n th conductor have zero current if desired. Angles should be in the range -360 to +360 degrees. Note that 180degrees = -180 degrees.

4. The other essential data:

n, the number of cables including the earth conductor is deduced by the worksheet according tothe number of rows of data entered.

L, the length of the cable run, in meters.

f, the frequency of the cable currents in Hz, default is 50Hz. Choosing f = 60 Hz is, of course,OK. Choosing say f = 250Hz would mean addressing the induced voltage & current problem for 5th harmonics on 50Hz. The cable currents in 3 above must be at the specified frequency.

5. The armour resistance/meter of each cable. This data is not needed to calculate inducedvoltages but is needed to calculate induced currents. Suggested values for different size cablesare displayed.

Example data and further advice about specifying earth

Suppose we wish n to be 7. Enter 6 rows of data and look carefully at the 7th cable which is inthe 10th row of the table of data. The value of n = 7 cannot actually be entered because theworksheet itself deduces the value based on the rows of data entered.

The cable centre to centre distances d1 = 100, d2 = 130, d3 = 100 ….. all in mm, see diagrambelow.

Cable outside diameters as measured by callipers on the PVC outer insulation OD1 = 50, OD2= 50 ….all in mm, see diagram below. The data for the last conductor, the 7th one in this case,does not need to be accurate but it needs to be considered further before settling on data to beentered.

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L = 150 m, f = 60 Hz

 All cable current are 1000A balanced three phase.The earthing Method 1 below is chosen and so the input table should be as shown.

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The switchboards at opposite ends of the power cables can be earthed by one of four methodsdescribed below. First there is a preamble. It is not necessary to have an earth conductor but, of course, it is normal to have one. Any of the n cables can be designated as the earth but isexpected that the nth cable will be the one chosen. This is the bottom cable in the diagram andthe bottom row in the data. It is the top cable if the diagram is viewed upside down! It is notessential for the earth cable to carry zero current at the set up of the worksheet, but there maybe some difficulty in interpreting results if a non-zero value is chosen.

Method 1. tied together with an earth conductor directly. A reasonable choice of conductor is 95mm2 copper at a distance of 1 m from the centre of the (n-1)th cable with cable resistance of 0.25 ohm/m and cable OD of 39mm. The length of the earth conductor cannot be different from

the length of the other (n-1) cables.

Method 2. separate earth conductors to a common earth point. This differs from Method 1because of the effective length of the nth cable. A distance of 2 m from the (n-1)th cable and aresistance of double the value chosen for Method 1 is suggested. The choice of distance fromthe (n-1)th cable may seem rough and ready, but the value is not very critical in evaluatinginduced currents. This is because the self and mutual impedances calculated within theworksheet involve logarithms of distance, and logs only change a little for sizable changes indistance. For example, doubling distance means only a 30% increase in log(distance). Making areasonable estimate of the earth arrangement is worthwhile because knowledge about theinduced currents in the earth path is valuable. Try running the worksheet with different values of cable resistance and separation.

Method 3. connected to earth via earth rods driven into the ground or connected to the buildingstructure at a common point at each end of the power cables. This amounts to the same thing.

 A value of 4m for the distance from the (n-1)th cable is suggested and cable resistance shouldbe (RA+RB)/L, where RA and RB are the earth resistances at respective ends. In HV systemsRA and RB are each likely to be less than 1 ohm. Making (RA+RB) too large will lead to anunderestimate of induced currents. Resistance values do not affect mesh voltages.

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Method 4. tied together via the cable armour which is the subject of the induced currentcalculation in this worksheet. There are two ways of doing this. One is to designate say 3 power cables and to give the 4th cable zero current and a high resistance armour, say 10 ohm/m. Theother way is to put in say two of the power cables, in row 1 and row 2, and to give the 10th rowthe dimensions of the 3rd cable and to examine the cable current to make sure it is the valuedesired.

Method 5. A combination of Methods 1 - 4. A complex earthing system is difficult to model andthe worksheet is not intended to deal with such. However, a few points are mentioned here.

1. The worksheet can deal with n<=10 and several of the n cables can be considered as earths.

2. Extraneous earth paths, i.e. those via service pipes and conduit and armoured signal multicores, are usually disregarded in assessing earth performance.

3. Small section cables in earth paths have higher resistance than substantive earth cablesinstalled for earth purposes.

4. Currents share earth paths but not simply according to their resistance. Induced currents alsodepend on magnetic coupling of all power cables and all armour currents and all earth pathcurrents. The idea that currents take the path of least resistance is just nonsense.