## Armour Currents in Single Core Cables.

- Details
- Written by John V. H. Sanderson CEng, FIEE

Use of armoured single core cables in HV installations may be the designer's choice; easy to install and terminate and they contribute to good earthing, but there can be serious consequences if the flow of armour current is neglected.

The same is true for single core LV cables and for any multicore cables where currents flow, but do not return via cores in the same cable. All current carrying conductors produce magnetic fields and these induce armour currents when 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 current that flows depends on the physical arrangement of cables, the resistance of the armour wire, the gland connection at the cable ends, the cable core currents and the presence and connection of cable screens.

It is usual for armour to be connected to brass glands which terminate the armour wire neatly and provide electrical connection to the armour. The glands should be fixed to a non-ferrous gland 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 and either connected or left unconnected at the other end of the cable. If the cable screen is connected 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 magnetic field due to the power conductors. If the armour resistance is near zero, the induced current 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 the position shown in the figure, then it would be correct to use the flux shown to calculate induced voltages and the induced current.

In practice, there are many armour strands and they are wound in a spiral fashion with a repetition rate of, typically, 1m. So, in a 100m run of armoured cable, each strand occupies the position closest to the adjacent conductor, 100 times.

The individual armour strands do not make good electrical contact with each other and can be treated as insulated, so that the armour as a whole does not behave as a solid cylindrical conductor. The amount of flux linkage to be used for the armour as a whole should therefore be an 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 density vector, B(Ik), due to each of the main currents, Ik, and to perform surface integrals to derive flux linkage. 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 quite interesting 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 three currents add to zero at all times except during faults. The magnetic field between conductors depends on all three currents, and the resulting armour currents depend on the geometry and separation of all the cables. Hand calculation is very difficult and a computerised method is needed. When there are several conductors per phase, all cables must be taken into account and the physical layout matters too. The computerised method made available here can deal with up to 10 cables and so caters for up to 3 cables per phase plus one cable to represent earth. The earth conductor arrangement is an important factor. This is dealt with in the computerised method and is discussed at some length below. Although originally intended to be used for calculation of armour currents due to normally loaded cables, the method will deal equally well with fault currents. Similarly, although intended to be used for single core cables, the method works quite well for multicore 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 185 mm2 cables per phase. The armour was connected to the generator using brass cable glands fixed to an aluminium gland plate. The connection at the main switchboard, 150 m away, was terminated 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 badly fitted. It was too big and the aluminium armour made a poor connection. The bad gland deteriorated 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 shut down the generator immediately. Our investigation showed that the induced voltage driving the armour currents was about 12V and the armour currents were about 100A. The nine cables were fixed along the cable route originally as shown.

We re-glanded both ends of all cables using the correct sized fittings and re-arranged the cables as shown, 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.

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

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

Our calculation sheet is available free of charge and it should be useful to designers who wish to 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 single cable 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 give reasonable estimates of armour voltages and currents.

The basis of the worksheet is the evaluation of magnetic flux due to each conductor, linking the armour of adjacent pairs of cables. Data are needed about each cable and their separation distances. 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. The worksheet calculates first the voltages that would appear between cable glands on adjacent cables when they are isolated from each other at one end, but connected to each other and to earth at the other end. This array of (n-1) voltages for a total (n) cables is a good guide to the induced currents that will flow should the isolated glands be connected to each other and to earth.

Data to be entered are:

- The cable to cable centre separation distances in mm.
- The outside diameters of each cable in mm. The worksheet assumes that the armour is covered by 2mm thick PVC outer sheath.
- 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 th conductor. 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 180 degrees = -180 degrees.
- The other essential data:

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

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

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.

The armour resistance/metre of each cable. This data is not needed to calculate induced voltages but is needed to calculate induced currents. Suggested values for different size cables are 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 in the 10th row of the table of data. The value of n = 7 cannot actually be entered because the worksheet 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 diagram.

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 be entered.

L = 150 m, f = 60 Hz

All cable currents are 1000A balanced three phase.

The earthing Method 1 below is chosen and so the input table should be as shown.

The switchboards at opposite ends of the power cables can be earthed by one of four methods described 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 it is expected that the nth cable will be the one chosen. This is the bottom cable in the diagram and the bottom row in the data. It is the top cable if the diagram is viewed upside down! It is not essential for the earth cable to carry zero current at the set up of the worksheet, but there may be 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 95 mm2 copper at a distance of 1 m from the centre of the (n-1)th cable with cable resistance of 0.25 mohm/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 1 because of the effective length of the nth cable. A distance of 2 m from the (n-1)th cable and a resistance of double the value chosen for Method 1 is suggested. The choice of distance from the (n-1)th cable may seem rough and ready, but the value is not very critical in evaluating induced currents. This is because the self and mutual impedances calculated within the worksheet involve logarithms of distance, and logs only change a little for sizable changes in distance. For example, doubling distance means only a 30% increase in log(distance). Making a reasonable estimate of the earth arrangement is worthwhile because knowledge about the induced 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 building structure 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 should be (RA+RB)/L, where RA and RB are the earth resistances at respective ends. In HV systems RA and RB are each likely to be less than 1 ohm. Making (RA+RB) too large will lead to an underestimate of induced currents. Resistance values do not affect mesh voltages.

**Method 4.** tied together via the cable armour which is the subject of the induced current calculation 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. The other way is to put in, say two of the power cables, in row 1 and row 2, and to give the 10th row the dimensions of the 3rd cable and to examine the cable current to make sure it is the value desired.

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

- The worksheet can deal with n<=10 and several of the n cables can be considered as earths.
- Extraneous earth paths, ie those via service pipes and conduit and armoured signal multicores, are usually disregarded in assessing earth performance.
- Small section cables in earth paths have higher resistance than substantive earth cables installed for earth purposes.
- Currents share earth paths but not simply according to their resistance. Induced currents also depend on magnetic coupling of all power cables and all armour currents and all earth path currents. The idea that currents take the path of least resistance is just nonsense.

### DISCLAIMER

PLEASE NOTE: The information and worksheet on this page are preliminary. Consultation and worksheet testing continues. The results obtained from the worksheet should not be relied upon.

Any persons using this spreadsheet, or results obtained from this spreadsheet, do so at their own risk. PEC cannot accept responsibility for any consequences