35718780 Sizing Cables Conduit and Trunking

7/30/2019 35718780 Sizing Cables Conduit and Trunking

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Sizing Cables, Conduitand Trunking

Learner Work Book

Name:Group:Tutor:


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Sizing Cables Conduit and Trunking REV4.1 2


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Table of Contents

Foreword ........................................................................................................4

Sizing Cables, Conduit and Trunking Unit Overview..................................6Practical Skills.................................................................................................... 6Knowledge Requirements .................................................................................. 6

The price of cable ..........................................................................................7

Cable calculation process.............................................................................8Design Current I b ........................................................................................... 9Over-Current Protective Device I n................................................................ 11Reference Methods.......................................................................................... 12Reference Methods.......................................................................................... 13

Correction Factors.......................................................................................15Ambient temperature - Ca................................................................................ 16Thermal Insulation - Ci ..................................................................................... 17Thermal Insulation - Ci ..................................................................................... 18Grouping circuits - Cg....................................................................................... 20Protection by BS3036 semi-enclosed (re-wireable) fuses - Cc......................... 22

The effects of volt drop ...............................................................................23What is volt drop? ............................................................................................ 23How to establish the value of volt drop ............................................................. 24

Shock protection..........................................................................................27Earth fault loop impedance and fault current .................................................... 27Earth Loop Impedance..................................................................................... 28The earth fault loop path .................................................................................. 29What value is acceptable? ............................................................................... 29How is earth loop impedance calculated? ........................................................ 31Earth fault current............................................................................................. 34Time / current characteristics and disconnection times..................................... 36

Thermal Constraints ....................................................................................38Minimum size of c.p.c....................................................................................... 40

Cable capacities of conduit and trunking..................................................44Conduit Capacities ........................................................................................... 44Trunking capacities .......................................................................................... 46

Maximum demand and diversity.................................................................48Maximum Demand........................................................................................... 48Diversity ........................................................................................................... 49


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Foreword

When trying to determine the size of conductors necessary for the safe working of acircuit many factors need to be considered. It is not acceptable to guess or to usecable sizes that are in common usage just because someone may say that you

should use 1.5mm twin and earth. They may just be wrong! If you get cable sizeswrong, then there may be a fire risk, a load that wont function properly, or you maybe wasting money on excessive material.

As we have already discovered previously copper is a very good electrical conductor.This means that the resistance of a length of copper cable is relatively low. Analuminium cable would have nearly twice the resistance of a copper cable with thesame dimensions. Therefore the energy losses in the aluminium cable will be higherthan in the copper cable. The copper cable is more energy efficient.

To make an aluminium cable with the same energy losses as a copper cable, wehave to make it larger. The larger cross sectional area reduces its resistance andbrings the energy losses down to the same as a narrower copper cable.

Al and Cu Comparison for 500A cable

The two cables in the photograph have similar current-carrying capacity. They areeach designed to be able to carry up to 500A without the conductor going above90C. The copper cable (on the right) is thinner than the aluminium one, becausecopper is a better conductor. Its cross sectional area is 300mm2 as opposed to500mm2 for the aluminium.

All cables have electrical resistance, so there must be an energy loss whenthey carry current. This loss appears as heat and the temperature of the cablerises. As it does so, the heat it loses to its surroundings by conduction,convection and radiation also increases. The rate of heat loss is a function ofthe difference in temperature between the conductor and the surroundings, soas the conductor tem erature rises, so does its rate of heat loss.


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Cables are designed to be able to withstand a certain amount of heat, and this abilitydepends on the type of insulation that is used and how the cables are installed into awiring system. When we consider the current carrying capacity we have to ensurethat the cable wont overheat when the normal current is flowing. If conductors areinstalled into wiring systems that are incorrectly sized they will not be able to lose enoughheat and could have an affect on the insulation properties.

This will be looked at in more detail when we take a look at cable calculations andsizing of conduit and trunking.

Cable calculation is the method used to ensure that all the factors of acircuit have been taken into consideration, in particular, the operatingcurrent of a cable that is determined by how hot the cable gets. This isaffected by a number of variables:

The resistance of the cable - a higher resistance cable will get hotter at

a given current. The insulation on the cable - this will tend to keep it warm like a jacket. The environment of the cable - if it is in a duct with other cables

(especially with no airflow) it will tend to get hotter.

This workbook is to be accompanied by PowerPointSizing Cables Conduit and Trunking


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Sizing Cables, Conduit and Trunking Unit Overview

Practical SkillsTo achieve the learning outcome the candidate must be able to:

Calculate cable sizes for circuits to ensure overload ratings, voltage drop,shock protection and thermal constraints are all met in accordance withBS7671

Knowledge RequirementsTo achieve the learning outcome the candidate must know:

How to select a suitably sized cable including:How to calculating the current demand of single and three line circuitsSelect the correct rating of protective deviceHow to allow and apply factors for:

GroupingThermal insulationAmbient temperatureProtective device type

How to check that the voltage drop is not excessiveEstablishing circuit disconnections times are metThermal constraints are satisfiedHow to determine the size of conduit and trunking appropriate to the size andnumber of cablesMethods used of establishing a circuits maximum demand after diversity isapplied


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The price of cable

Before we look at different cable sizes let us gain some appreciation for the price ofcable. As you can imagine the cost of cable varies according to the amount ofmaterials and manufacturing needed to construct it. Basically the more copper the

more pennies.

Shown below is a table with various prices for one particular type of cable. See howthe price varies from size to size? Also note that the price per metre is more than thecalculated amount for 100 metres.

Size and number of cores - Price per metre - Price per 100 metres

4 core 1.5mm XLPE / SWA 1.11 93.994 core 2.5mm XLPE / SWA 1.52 126.004 core 4.0mm XLPE / SWA 2.15 179.994 core 6.0mm XLPE / SWA 3.00 251.99

4 core 10.0mm XLPE /SWA

4.40 385.00

4 core 16.0mm XLPE /SWA

6.10 550.00

Cable prices as of JULY 2008

Price ()

1. 50M of 4 core 1.5mm XLPE / SWA in one coil

2. 200M of 4 core 1.5mm XLPE / SWA in two coils

3. 100M of 4 core 16.0mm XLPE / SWA in five coils

4. 100M of 4 core 16.0mm XLPE / SWA in one coil

What reason could you give to explain why buying 100 metres costs less than buying fivecoils of twenty metres?

Complete the following exercise.The following lengths of XLPE/ SWA are required for a job. Your job as the companys buyer isto process this order form from the site supervisor. The client has requested a price of thematerials before they will let the job go ahead. Work out the prices using the table above

How might the price be reduced to add a further saving in cost to the job?


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Cable calculation process

Cable selection can be defined as the rules that you must follow when decidingwhich cable to choose for any installation.

This unit will consider the following areas used in cable selection:

External influencesDesign current (Ib)Rating of the protective device(In)Reference methodsCorrection factors (Ca, Ci, Cg,Cc)

Application of correction factorsVoltage dropShock protectionThermal constraintsDiversity.

External influencesIf the circuit is to be installed in a hot or wet environment then the cable has to be

suitably rated and sealed from any affecting factors. If the environment carries a highrisk of mechanical damage the cable or installation will have to be sufficientlyprotected form danger. Once you have decided on the type of cable suitable for theenvironmental conditions, you must choose the size of conductor to be used.

In order to gain appreciation for cable selection will only take a look at the variouscurrent ratings of various sizes of XLPE armoured cable.

The Basics

So that we can understand a full cable calculation we must first understand thebasics. Listed below are four terms that describe vital information used in thecalculation process. Try to remember them, as they will appear frequently throughoutthis course.

The formula above states the underlying principle of the calculation of a circuitscable size. The first factor you need to consider is design current.

I b - term used to describe a circuits design current in amps i.e. the load.I n - term used to describe a circuits protection size in amps i.e. the fuse size.

I z - term used to describe a circuits value, in amps, once all de-rating factors have beenconsidered

I t - term used to describe the tabulated current rating of a cable in amps i.e. the current a cablecan safely carry.

I b I n I z I t


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Design Current I b

The first stage of the design process is to determine how much current will flow in thecircuit. This current is known as the design current and is the full load current of thecircuit. It is calculated using one of the formulae below depending on the type of load.

You need to ensure all units have to be calculated at the same value (i.e. kW have tobe divided by kV; W have to be divided by V)

Where: I = the design current in amps (A)P = the circuit power in watts (W)V = the circuit voltage in volts (V)Cos = the power factor

Resistive loadsThe following formulae apply to single and three line supplies:

Inductive and / or capacitive loadsThe following formulae apply to single and three line supplies:

In a.c. circuits, the effects of either highly inductive or highly capacitive loads canproduce a poor power factor (cos ) (inductive and capacitive loads will be explainedlater). For now it is satisfactory to know that in circuits where there are inductive andelectronic components such as coils and capacitors there are losses. These lossesslightly increase the amount of current the equipment uses. You will have to allow forthis in such circuits. Note 3 = 1.732

Single-line 230vInductive and or

Ca acitive

=

cosV

PI

Three line 400vInductive and or Ca acitive

=

cos3 V

PI

Single-line230v

Resistive

V

PI=

Three line 400vResistive

V

PI

=

3


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Example 1.A single-line lighting circuit has a total power consumption of 2000 watts using 100-watt filament lamps. Calculate the design current.

i) Select the correct formula. (Single line; 230v, resistive)

ii) Input the data into the formula and work it out to two decimal places and be sure toadd the unit (A).

Example 2.A three-line inductive load has a total power consumption of 30,000 watts (30kW)with a power factor of 0.95. Calculate the design current.

i) Select the correct formula. (Three line; 400v, inductive)

ii) Input the data into the formula and work it out to two decimal places and be sure toadd the unit (A).

There are a few examples for you to calculate for yourself shortly.

V

PI=

AI 70.8

230

2000==

=

cos3 VPI

AI 58.4595.04003

30000=

=


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Over-Current Protective Device I n

Once the design current has been established we must then select an over-currentprotective device. The function of an over-current protective device is basically as itsname suggests to protect the circuit from over-current and of course, faults. It is the

weakest part of the circuit and should operate in a given time so that only a limitedamount of harm or danger, to persons, livestock or property, will exist under faultconditions. The two main factors to consider when selecting a device are shownbelow.

1 Amount of Overload current

These are currents higher than those intended to be present in the system. If such

currents persist they will result in an increase in conductor temperature, and hence arise in insulation temperature. High conductor temperatures are of little consequenceexcept that the resistance of the conductor will be increased leading to greater levelsof voltage drop.

Insulation cannot tolerate high temperatures since they will lead to deterioration andeventually failure. The most common insulation material is p.v.c. If it becomes too hotit softens, allowing conductors, which press against it and possibly pass through it.

Overload currents occur in circuits which have no faults but are carrying a highercurrent than the design value due to overloaded machines, an error in theassessment of diversity, and so on.

2 Amount of Fault current

These currents will only occur under fault conditions, and may be very high indeed.As we shall shortly see such currents will open the protective devices very quickly.These currents will not flow for long periods but under such short-term circumstancesthe temperature of p.v.c. Insulation may rise to 160C.

Device Selection

The protective device current rating must be equal to or next largest size so thatthe circuit is sufficiently protected. Take a look at the formula used for the cablecalculation process again below.

I b I n then I z I t

You can see from this that:The over-current protective device must be equal to or greater than the design

current.

'Over current' means what it says - a greater level of current than thematerials in use will tolerate for a long period of time. The term can be dividedinto two types of excess current. 1. Overload current and 2. Fault current

The temperature of two conductors at the point of a fault can be as high as300C. This is why the conductors and or the cutting tool used will melt at

the point of the fault.


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There are many protective devices in existence and these will be looked at in moredetail in Unit 4.4. What is important for now is to realise that these devices all havedifferent ratings or current ratings. Shown below is part of BS7671 that shows thedifferent ratings of BS88 fuses from 6A to 200A.

Fuse sizes of BS88 over-current protective devices (Amps)

6 10 16 20 25 32 40 50 63 80 100 125 160 200

Ib In

1. A single line, resistive lighting circuit

with a total power of 1200 watts

2. A single line heating circuit with a totalpower of 6KW

3. A single line, inductive lighting circuitwith a total power of 1200 watts andpower factor of 0.85

4. A three line heating system with a totalpower of 10KW watts

5. A single-line supply to a 100Aconsumer unit expected to carry 48amps maximum.

What reasons can you state for the protective device being equal to or slightly larger than thedesign current? You may be required to read your reasons to the class.

Using the table above and a calculator see if you can work out the design currents and over-current protection size (BS88) of each of the circuits listed.


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Reference Methods

Table 4A2 from the IEE Wiring Regulationslists the common methods that can beused to install a cable.

You need to decide at this stage in the cable selection process which method ofinstallation to use. This will make sure that the correct cable column is chosen in thelater stages of cable selection.

This choice is also important when you calculate correction factors for thermalinsulation.

I t

1. A 4.0mm, two core cable, carrying alternatingcurrent that has been clipped to a wall.

2. A 10.0mm, four core cable, carrying A.C currentran on a cable tray

3. A 16.0mm, single line cable supplying aconsumer unit, clipped direct

4. A 25.0mm, three line and neutral supply for anuninterruptible power supply unit in an officeblock clipped direct to a wall.

5. A 25.0mm, three line and neutral (TP&N)

supply to a control panel, ran on a cable tray

Using the table (4E4A) from BS7671 on the next page, see if you can determine the basiccurrent ratings (tabulated current ratings I t) of the cables below. Be sure to fully observe thedetails so you choose the correct column.

NB: This exercise displays how different sizes of cables can carry different amounts of current.

Also considered are the circuits installation types.

Consider your findings and try to explain the reasons why we sometimes need to selectdifferent sized cables. You may be required presenting your reasons to the class


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Appendix 4 of BS7671 contains tables for each type of conductor type from singlecore PVC to multi-core paper insulated lead steel wire armoured. Please refer toBS7671 for more information.

C E

E


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CcCgCiCa

InIz

=

Correction Factors

You, as the designer of the installation, need to know the different correction factors,know where they are required and then apply these to the nominal rating of theprotection (In) to obtain a value for Iz. Once Iz is calculated we then refer to the

correct cable tables from BS7671and select the size based upon the next highesttabulated value (It).

Temperature can have a serious effect on a circuit. It can increase the risk of a fault,can cause a cables insulation to melt and can even cause an electrical fire. Thereare various factors that form a circuits temperature and we will look at these in moredetail below. It is essential to follow the correct design procedures and apply thecorrect correction factors to ensure that the many effects of temperature do not affectthe normal operation of a circuit. We will look at these factors individually below butremember there may be more than one factor present in one installation.

What If More Than One Factors Are Present?

If more than one correction factor is present they can be considered in onecalculation shown below

Example

A three line, 32A circuit is to be installed using XLPE SWA. It will run through a boilerhouse clipped direct to a wall where the ambient temp will be 40C. It is not groupednor does it come into contact with any thermal insulation. Discover the de-ratedvalue i.e. the minimum permissible rating of cable.

i) Establish all the correction factors present and obtain the values from thetables in the regulations.

Ca=0.91; Ci=n/a; Cg=n/a; Cc=n/a(Where there is no factor present we assume a value of one).

ii) Input the values into the formula and work out Iz to two decimal places andinclude the value (A)

iii) Select the correct cable table then select the correct column based uponthe reference method and select the next highest value (or equal) to Iz.

Table 4E4A; column 3; value of 42 amps corresponds to a 4mm XLPE SWA cable.

Ca Ambient Temperature (the surrounding temperature the circuit will operate in)Ci Thermal Insulation (the existence and contact of thermal insulation with the cct)Cg Grouping of ccts (whether or not the circuit is bunched with other circuits)Cc Protection type (whether or not the circuit is supplied with a BS3036 fuse or not.

ACcCgCi

Iz 16.3591.0

32=

=


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Ambient temperature - Ca

This is the temperature of the surroundings ofthe cable, often the temperature of the air in aroom or building in which the cable is installed.

When a cable carries current, it gives off heat.Therefore the hotter the surroundings of thecable, the more difficult it is for the cable to getrid of this heat. But if the surroundingtemperature is low, then the heat given offcould be easily let out and the cable could carry more current. Cables must give offthis heat safely or they could be damaged and there is a risk of a fire. You can findthe correction factor for ambient temperaturein Tables 4B1 and 4B2 of the IEEWiring Regulations.

These tables are based on an ambient temperature of 30C. This means that anycables installed in an ambient temperature above this will need the correction factor

applying to them. This is because the cable will not be able to get rid of the heat itgives off safely when carrying current.

When a cable runs through areas having different ambient temperatures, correctionfactors should be applied to the highest temperature only.

The most common of the correction factors are given in the Tables 4B1 and 4B2from BS7671 Correction factors for ambient temperature and are given in yourTables from BS7671 and the Onsite Guide.

Complete the questions on the nest page to understand how ambient temperatureaffects minimum cable ratings.


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Here you will see how ambient temperature can affect the selection of cable. Theperfect situation is 30C therefore Ca = 1.0.

1. A single-line supply to a DB with adesign current of 50A and a MCCBsize of 100A, clipped direct at 30C.

2. A single-line supply to a DB with adesign current of 50A and a MCCBsize of 100A clipped direct and ran

through a boiler house at 50C.

3. A three line supply to a DB with BS88fuses 50A and a maximum demandof 45A, clipped direct and ran near tohot machinery at 40C

4. A three line supply to a DB with BS88fuses 50A and a maximum demandof 45A, clipped direct but re-routed to30C

I n - term used to describe a circuits protection size

in amps i.e. the fuse size.I z - term used to describe a circuits value, in amps, onceall de-rating factors have been considered.

C a term for the correction factor of ambient temperature.CaInIz =

Using the table 4E4A and the Ca correction factors in your essential tables determine theminimum size of XLPE cable for each of the circuits below. Make note of the cable insulationtype before you proceed with your calculations. You must write down:In (fuse size), Ca (correction factor), Iz (de-rated CCC), cable size (mm), It (tabulated CCC)and your working out.

In your own words how does ambient temperature affect the selection of cable? You may berequired to read you answer out to the class.


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Thermal Insulation - Ci

The use of thermal insulation in buildings, in the forms of cavity wall filling, roof spaceblanketing, and so on, is now standard. Since the purpose of such materials is to limitthe transfer of heat, they will clearly affect the ability of a cable to dissipate the heat

build up within it when in contact with them. Thermal insulation has the effect ofwrapping a cable in a fur coat on a hot summers day. The heat produced when thecable carries current cannot escape.

Loft insulation

The cable rating tables of the regulations as the one you have already used (Table4E4A) allow for the reduced heat loss for a cable which is enclosed in an insulatingwall and is assumed to be in contact with the insulation on one side.

In all other cases, the cable should be fixed in a position where it is unlikely to becompletely covered by the insulation. Where this is not possible and a cable is buriedin thermal insulation for 0.5 m (500 mm) or more, a rating factor of 0.5 is applied.

This means that the current rating is halved or in other words the Iz value will bedoubled.

If a cable is totally surrounded by thermal insulation for only a short length (forexample, where a cable passes through an insulated wall), the heating effect on thecable insulation will not be that significant. This is because heat will be conductedaway from the short high-temperature length through the cable conductor.

Clearly, the longer the length of cable enclosed in the insulation the greater will bethe de-rating effect. Table 52.2 (BS7671) shows the de-rating factors for lengths ininsulation of up to 400 mm and applies to cables having cross-sectional area up to 10mm. Table 52.2 from BS7671 De-rating factors for cables up to 10mm in cross-

sectional area buried in thermal insulation. Is given in your Tables from BS7671and the Onsite Guide.

The Regulations use the symbol Ci to represent this correction factor.

Complete the questions on the next page to understand how thermal insulationaffects minimum cable ratings.


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Here you will see how thermal insulation can affect the selection of cable. The perfectsituation is no insulation therefore Ci = 1.0.

1. A single-line supply to a load with adesign current of 13A and a MCBsize of 16A, reference method C withno thermal insulation.

2. A single-line supply to a load with adesign current of 13A and a MCBsize of 16A, reference method C with3000mm thermal insulation.

3. A three-line supply to a DB with BS88fuses 32A and a maximum demandof 27A, reference method E with600mm of thermal insulation.

4. A three-line supply to a DB with BS88fuses 32A and a maximum demandof 27A, reference method E withoutany thermal insulation.

I n - term used to describe a circuits protection sizein amps i.e. the fuse size.

I z - term used to describe a circuits value, in amps, onceall de-rating factors have been considered.

C i term for the correction factor of thermal insulation.Ci

InIz =

Using the table 4E4A and the Ci correction factors in your essential tables determine theminimum size of XLPE cable for each of the circuits below. Make note of the cable insulationtype before you proceed with your calculations. You must write down:In (fuse size), Ci (correction factor), Iz (de-rated CCC), cable size (mm), It (tabulated CCC) andyour working out.

In your own words how does thermal insulation affect the selection of cable? You may berequired to read you answer out to the class


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Grouping circuits - Cg

If a number of cables are installed together and each is carrying current, they will allwarm up. Those which are on the outside of the group will be able to transmit heatoutwards, but will be restricted in losing heat inwards towards other warm cables.

Cables 'buried' in others near the centre of the group may find it impossible to shedheat at all, and will rise further in temperature.

Due to this, cables installed in groups with others (for example, if enclosed in aconduit or trunking) are allowed to carry less current than similar cables clipped to, orlying on, a solid surface that can dissipate heat more easily.

If surface mounted cables are touching the reduction in the current rating is, as wouldbe expected, greater than if they are separated. The picture below illustrates thedifficulty of dissipating heat in a group of cables.

The symbol Cg is used to represent the factor used for de-rating cables to allow for

grouping. Table 4C4 from BS7671 Correction factors for groups of more thanone circuit shows some of the most common values of Cg.

The grouping factors are based on the assumption that all cables in a group arecarrying rated current.

Complete the questions on the next page to understand how grouping affectsminimum cable ratings.

Widelyspacedcablesdissipateheat easily

A closely packedcable cannoteasily dissipateheat and so itstemperaturerises

Note: If a cable is expected to carry no more than 30% of its grouped rated current, itcan be ignored when calculating the group-rating factor. For example, if there arefour circuits in a group but one will be carrying less than 30% of its grouped rating,the group may be calculated on the basis of having only three circuits.


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Here you will see how grouping circuits can affect the selection of cable. The perfectsituation is no circuits grouped therefore Cg = 1.0.

1. A three-line supply to a socket with adesign current of 32A and a MCB size of32A, reference method E not grouped.

2. A three-line supply to a socket with adesign current of 32A and a MCB size of32A, reference method E (touching) andgrouped with 7 circuits.

3. A single-line circuit with a 63A BS88fuse protecting it, reference method Egrouped with one other circuit.

4. A single-line circuit with a 63A BS88fuse protecting it, reference method Eand not grouped with any other circuit.

I n - term used to describe a circuits protection sizein amps i.e. the fuse size.

I z - term used to describe a circuits value, in amps, onceall de-rating factors have been considered.

C g term for the correction factor of grouping of circuits.Cg

InIz =

Using the table 4E4A and the Cg correction factors in your essential tables determine theminimum size of XLPE cable for each of the circuits below. Make note of the cable insulationtype before you proceed with your calculations. You must write down:In (fuse size), Cg (correction factor), Iz (de-rated CCC), cable size (mm), It (tabulated CCC)and your working out.

In your own words how does grouping circuits affect the selection of cable? You may berequired to read you answer out to the class


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Protection by BS3036 semi-enclosed (re-wireable) fuses - Cc

If the circuit concerned is protected by a BS3036 semi-enclosed (re-wireable) fusethe cable size will need to be larger to allow for the fact that such fuses are not socertain in operation as are cartridge fuses or circuit breakers. In other words, in the

event of a fault or an overload they will not disconnect as quickly as the otherprotective devices available would do.

It has been known for a 5 amp BS3036 fuse to carry in excess of twice its ratingwithout any signs of operating!! Therefore the fuse rating must never be greater than0.725 times the current carrying capacity of the lowest-rated conductor protected. Ineffect, this is the same as applying a correction factor of 0.725 to all circuits protectedby semi-enclosed fuses.

Complete the questions below to understand how BS3036 fuses affect minimumcable ratings.

1. A three-line supply to a socket with adesign current of 25A and a BS3036size of 30A, reference method E.

2. A three-line supply to a socket with adesign current of 25A and an MCBsize of 32A, reference method E.

3. A single-line circuit with a 60ABS3036 fuse protecting it, referencemethod C.

4. A single line circuit with an old 60ABS88 fuse protecting it, referencemethod C

In - term used to describe a circuits protection size inamps i.e. the fuse size.

I z - term used to describe a circuits value, in amps, once allde-rating factors have been considered.

C c term for the correction factor for the use of BS3036fuses.

Cc

InIz =

Using the table 4E4A in your essential tables determine the minimum size of XLPE cable foreach of the circuits below. You must write down your formulas. Note that this only applies tocircuits that use BS3036 fuses. You must write down:In (fuse size), Iz (de-rated CCC), cable size (mm), It (tabulated CCC) and your working out.


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The effects of volt drop

What is volt drop?

All cables have resistance, and when current flows in them it results in a volt drop.

Hence, the voltage at the load is lower than the supply voltage by the amount of thisvolt drop.

BS7671 states that the voltage at any load must never fall so low as to impair thesafe working of that load, or fall below the level indicated by the relevant BritishStandard where one applies.

BS7671 also indicates that these requirements will be met if the voltage drop doesnot exceed a certain % of the declared supply voltage. (See table 12A below).

Table 12A (BS 7671)Maximum value of voltage drop

Lighting Other uses(i) Low voltage installationssupplied directly from apublic low voltagedistribution system

3% 5%

(ii) Low voltage installationsupplied from a private LVsupply

6% 8%

Public supplies are those that are supplied by the local authority (from the NationalGrid) where the consumer pays a bill for energy used. Private supplies are thosewhere the consumer generates their own electricity (E.g. An onsite power generationplant such as a combined heating and power plant).

Public Supplies

For lightingIf the supply is single-line at the usual level of 230 V, this means a maximumvolt drop of 3% of 230 V, which is 6.9 V. This means the voltage at the load is aslow as 223.1 V. For a 400 V three-line system, allowable volt drop will be 12 Vwith a line load voltage as low as 388 V.

For other uses (power, motors etc)If the supply is single-line at the usual level of 230 V, this means a maximumvolt drop of 5% of 230 V, which is 11.5 V. This means the voltage at the load isas low as 218.5 V. For a 400 V three-line system, allowable volt drop will be 20 Vwith a line load voltage as low as 380 V.


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How to establish the value of volt drop

:

ExampleA 4 mm p.v.c. sheathed circuit feeds a 6 kW shower and has a length of run of 16m.Find the total voltage drop.

i) Work out the design current.

ii) Obtain the mV/A/m from Appendix 4

From Table 4D5A the volt drop figure for 4 mm two-core cable is 11 mV/A/m.

iii) Input all the values into the formula and work out the volt drop to twodecimal places and add the value (V).

Since the permissible volt drop in this instance is 5% of 230 V, which is 11.5 V, thecable in question meets volt drop requirements.

AV

PI 08.26

230

6000===

vVoltdrop 59.41000

161108.26=

=

Each cable rating in the Tables of Appendix 4 of BS7671 has acorresponding volt drop figure in milli-volts per ampere per metre of run(mV/A/m). To calculate the cable volt drop

Where:Ib = the design current in amps

mV/A/m = the milli volts per amp per metre droppedL= the circuit length in metres

1000 = converts the millivolts into volts

( )

1000

// LmAmVIbVoltdrop

=


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Application of volt drop

It is important to appreciate that the allowable volt drop applies to the whole of aninstallation from its source to the furthest point on the final circuit. If an installationhas mains, sub-mains and final circuits, for instance, the volt drop in each must be

calculated and added to give the total volt drop as indicated below.

Ways that too much volt-drop can affect equipment can be seen as follows.

While for a light bulb a large voltage drop will result in a harmless condition of slightlyless bright light being produced, incorrect voltages supplied onto delicate circuitry (asfor example in a DVD player, computer, and so forth) may quite easily result in anelectrically damaging condition. It is quite easy to have a circuit well within the

tabulated (I t) guidelines for its wiring, but whose voltage drop is too large.

For these reasons we are required to size wiring not only for the total current to bedrawn, but also to ensure that the total voltage drop shall not exceed the maximumpercentage (%). This is particularly the case when running long lengths of cable fromone end of a large building to another.

A consumer unit at 60 meters from a main distribution board to which is intended tosupply 30 amps should not necessarily be supplied with a 30 A cable. It may befound that the resistance per metre of the cable is of such a value that the loadvoltage would be below the required level.

Complete the questions on the next page to understand how voltage drop affectsminimum cable ratings.

Bigger cables Smaller cables

DB1

DB2

DB3

LOAD

1.5 volts 2.0 volts 4.4 volts

Overall voltage dropped = 7.9 V


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LEARNER WORK BOOK


1. 20 metres of 1.5mm cable supplying aline line machine which carries 14amps using Multi core 70C armouredthermoplastic (p.v.c) insulated cable

2. 4.0mm three line supply, 40 metres in

length to a 32A socket used forresistive loads only using Multi core90C armoured thermosetting (x.l.p.e)insulated cables

3. A single line lighting circuit to supply 10x 100w lamps with 60 metres of

1.0mm using 70 thermoplastic (p.v.c)insulated and sheathed flat cable withprotective conductor.

4. 80 metres of 1.5mm cable for a 10KW,three line load that has a power factorof 0.90 using Multi core 90C armouredthermosetting (x.l.p.e) insulated cables

Using the tables in your Tables from the regulations and On-site Guide notes calculate the voltdrops of the various circuits below. Make note of the cable insulation type before you proceedwith your calculations. You must write down your formulas. You must record the table number,show your calculations and state whether the cable passes or fails the voltage dropre uirement.

In your own words how does voltage drop affect the selection of cable? Look at the It valuesand compare them with the Ib / In values. Analyse your findings. You may be required to readyou answer out to the class


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LEARNER WORK BOOK


Shock protection

Protection against shock is a massively important factor to consider when designinga circuit. A person in contact with a supply voltage for any length of time can be veryharmful, as we have previously seen. We as designers need to ensure that this

potential is limited to a very small amount of time by ensuring a faulty circuitdisconnects automatically.

Earth fault loop impedance and fault current

The path followed by fault current as the result of low impedance occurring betweenthe line conductor and earthed metal or circuit protective conductor is called the earthfault loop. Fault current is driven through the loop impedance by the supply voltage.

The over-riding requirement is that sufficient fault current must flow in the event of anearth fault to ensure that the protective device cuts off the supply before dangerousshock can occur.

It must be appreciated that the longest disconnection times for protective devices,leading to the longest shock times and the greatest danger, will be associated withthe lowest levels of fault current, and not, as is commonly believed, the highestlevels. Why is this?

Note that there is no such thing as a three-line line/earth fault, although it is possiblefor three faults to occur on the three lines to earth simultaneously. As far ascalculations for fault current are concerned, the voltage to earth for standard UKsupplies is always 230 V, for both single-line and three-line systems. Thus the tablesof maximum earth-fault loop impedance, which are given in the appendices, applyboth to single- and to three-line systems.

For normal 230 V TN systems, there are two different levels of maximumdisconnection time. These are:

Any final circuit not exceeding 32A must disconnect within 0.4sSo any circuit rated at 32A or less must disconnect within 0.4 seconds. A distribution circuit or circuit exceeding 32A must disconnect

within 5sA distribution circuit is a db supply or sub-mains feeder. So anydistribution circuit or circuit rated higher than 32A must disconnect

Why are low levels of fault current more dangerous than higher levels?


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Earth Loop Impedance

Resistance (measured in ohms) is the property of a conductor to limit the flow ofcurrent through it when a voltage is applied. The larger the conductor is the lessresistance it has. The smaller the conductor is the more resistance it has.

Thus, a voltage of one volt applied to one ohm resistance results in a current of oneampere.

1. A 230v supply is connected to aresistance of 20 ohms. What is thecurrent?

2. A 230v supply with 3 amps of current

flows in a circuit. What is theresistance?

3. A current of 12.5 amps flows through aresistance of 18 ohms. What is thesupply voltage?

4. A 230v supply is connected to aresistance of 50 ohms. What is thecurrent?

5. A 230v supply with 5 amps of current

flows in a circuit. What is theresistance?

Q. Why is it called impedance when it clearly is just resistance?

A. It is termed impedance because part of the circuit is the transformer orgenerator winding, which is inductive. This inductance, along with theresistance of the cables to and from the fault, makes up the impedance.

Refresh your memory with some Ohms law calcs.


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LEARNER WORK BOOK


The earth fault loop path

The earth loop impedance we are concerned with is the worst-case scenario. Thismeans that we must ensure that the protective device will operate within the specifieddisconnection time at the furthest point on the final circuit i.e. the furthest point on the

circuit from the protective device. This will then account for the highest impedancepath because each metre of cable has resistance and it will calculate the value at thefurthest length of the circuits conductors.

See the diagram below for the actual earth loop path of a fault.

What value is acceptable?

A circuit is deemed to be in compliance if the value of earth loop impedance isequal to or less then the maximum allowable for the device type and rating. Eachdevice type and rating has its own limits. Complete the questions on the next page tobecome familiar with these limits and how they apply to a circuit.

1. The circuitprotective conductor

2. The mainearthing

conductorand the

consumersearthingterminal

3. The suppliersreturn path,

either

combined,separate or the

general mass ofearth

4. The earthed neutral ofthe supply transformer

5. The supplytransformerwinding

Faulty appliance

Live to earth fault

6. The lineconductorsupply from thetransformer to

the consumerunit

7. The finalcircuit liveconductor

So that we use the correct table to verify the calculated Zs value we need to knowthe maximum disconnection time for the circuit being designed. Remember:

Any circuit rated at 32A or less must disconnect within 0.4 seconds

A distribution circuit or circuit exceeding 32A must disconnect within 5s


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LEARNER WORK BOOK


Maximum values of earth loop impedance for various over-current protective devicesare shown in Table 41.2, 41.3 & Table 41.4 in your Tables from the regulations andOn-site Guide. Once the Zs has been established these tables are referred toensure the designed circuit is in compliance. Turn to these tables now to get familiarwith them.

1. A calculated value of earth loopimpedance (Zs) equating to 8.5 for acircuit supplying portable equipment.With BS88 circuit protection rated at6A.

2. A calculated Zs value of 1.2 for acircuit supplying portable equipment.With BS3036 circuit protection rated at30A

3. A calculated calculated Zs value of 3.2

for a circuit supplying fixedequipment. With BS88 circuit protectionrated at 16A

4. A calculated Zs value of 0.5 for acircuit supplying fixed equipment. WithBS88 circuit protection rated at 100A

5. A calculated Zs value of 5.0 for acircuit supplying portable equipment.With BSEN60898 Type B circuit

protection rated at 6A

6. A calculated Zs value of 5.0 for acircuit supplying portable equipment.With BSEN60898 Type C circuitprotection rated at 6A

7. A calculated Zs value of 0.36 for acircuit supplying fixed equipment. WithBSEN60898 Type D circuit protectionrated at 32A

8. A calculated Zs value of 1.25 for acircuit supplying fixed equipment. WithBSEN60898 Type B circuit protectionrated at 40A

What is the maximum Zs allowable for a 6 amp BS EN60898 Type D device?

Complete the following exercise to determine whether the Zs values of the circuits listed complywith BS7671. You will need your Tables from BS7671 and the on-site guide appendices. Youmust state the maximum disconnection time for the circuit; record the maximum Zs and the tablenumber; and state whether the cable passes or fails the shock protection requirement


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LEARNER WORK BOOK


How is earth loop impedance calculated?

Actual earth loop impedance can be calculated as follows:

The actual Zs is the sum of all the impedances that are present in a circuits earthfault path.

The R1 and R2 values are calculated in the following way. Table 9A from the On-siteguide lists the resistances in milli-ohms per metre of all sizes of cables up to 25mm.Note that the values are in milli-ohms per metre. Table 9A values need to beconverted into ohms (by dividing by 1000) so you can add it to Ze.

Table 9B and 9C are multipliers that take into account the expected ambienttemperature at the time of test or maximum operating temperature of conductors.

where

Zs - term used to describe a circuits earth fault loop impedance, in ohms.Ze - term used to describe part of the earth loop impedance that is external to theinstallation.R1 term used to describe the impedance of the line conductor, in ohms found in Table 9AR2 term used to describe the impedance of the circuit protective conductor, in ohmsfound in Table 9ALength length of circuit from supply to furthest point, in metresTable 9B or 9C Multiplier factor applied to allow for expected ambient temperature orconductor resistance at maximum o eratin tem erature res ectivel

( )21 RRZeZs ++=

R1

R2

Ze

Zs = Ze + (R1 + R2)

The Ze is something that we can either measure or obtain from our electricityprovider and is expressed in ohms. Typical maximum values are: TN-C-S (PME)

system 0.35 ohms, TN-S (cable sheath) 0.8 ohms, TT system 21 ohms

( )CorBTableLATableRR 991000

921

=+


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LEARNER WORK BOOK


Example

A circuit supplying a DB where a multi core armoured cable is clipped direct using 50metres of multi core 25.0mm 70C armoured thermoplastic insulated cable. Thebunched CPC conductor size is 16.0mm. The Ze is 0.5. Calculate the earth loopimpedance in Ohms at the maximum operating temperature.

i) Write down the formulas and obtain the values for each part.

Ze = 0.5; R1+R2 must be calculated; L = 50m

ii) Obtain the value for R1 and R2 in Ohms

Using Table 9A we can see that the resistance, in milli-ohms per metre, of 25.0mmand 16.0mm is 1.877m/m.

iii) Obtain the multiplier value from table 9C

The line and earth conductors are part of a thermoplastic multicore cable so areclassed as incorporated in a cable or bunched giving us a value of 1.20. Input allvalues into the R1+R2 formula

iv) Input the values into the main Zs formula and calculate Zs at the maximumconductor operating temperature

Turn to Table 9A of your Tables from BS7671 and Onsite Guide and complete thequestions on the next page to gain some understanding of earth loop impedancecalculations.

==+ 11.020.1501000

877.121 RR

=+= 61.011.05.0Zs

Multipliers are used by the designer and are required to allow for one of thefollowing:

Table 9B Used so the designer can give values of resistance at the ambienttemperature expected during the tests (200C is classed as 1)

Table 9C Used so the designer can give values of resistance at the conductorsmaximum operating temperatures

( )21 RRZeZs ++= ( )CorBTableLATableRR 991000

921

=+

Note: If the R1 and R2 value is measured this can be added to Ze to give the total Zs


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LEARNER WORK BOOK


1. A circuit supplying a DB using50 metres of multi core 70Carmoured thermoplasticinsulated cable. The line andearth conductors are both25.0mm. What is the expectedZs at the maximum operatingtemperature?

2. A motor circuit where singlecore conductors are installed inp.v.c conduit using 40 metres of90C thermosetting single core.The line and earth conductorsare both 4.0mm. What is theexpected Zs at the maximumoperating temperature?

3. A cooker circuit where 10m ofcable is installed in buildingfabric using 70 thermoplastic(p.v.c) insulated and sheathedflat cable with protectiveconductor. The line and earthconductors are 6.0mm and4.0mm respectively. What willbe the expected Zs at 10oC?

4. A power circuit where a 2.5mm cable is installed on a trayusing multi core 90C armouredthermosetting insulated cable.The circuit length is 80 metres.What will be the expected Zs at5oC?

Complete the following exercise to determine the earth loop impedance (Zs) of the circuits.Assume in all cases that the Ze = 0.3 . You will need your Tables from BS7671 and the on-site guide appendices. You must show all working out and state the R1 and R2 values andshow what Zs is for each circuit.


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LEARNER WORK BOOK


Earth fault current

Once the Zs has been calculated we then calculate the earth fault current (If) usingthe calculation below and then ensure that the device will disconnect within the giventime using the time / current trip curves in the tables from BS7671. If the time is less

than the maximum allowable for the circuit (i.e. At 230v, 0.4 or 5 seconds) we can besure that it will disconnect in time so that it provides protection from indirect contact.We can also ensure that the protective devices short circuit fault current capacity hasnot been exceeded.

ExampleA 230 V circuit is protected by a 15 A semi-enclosed (BS3036) fuse and has anearth-fault loop impedance of 1.6 Ohms. What will be the maximum earth faultcurrent?

This level of earth-fault current will cause the fuse to operate quickly. From the time /current trip curves in BS7671, Fig 3.2A (more on these next) the time taken for thefuse to operate will be about 0.15 s. Any load current in the circuit will be additional tothe fault current and will cause the fuse to operate slightly more quickly.

However, such load current must not be taken into account when decidingdisconnection time, because it is possible that the load may not be connected whenthe fault occurs. Therefore if the earth loop impedance is higher this will restrict theflow of fault current meaning the protective device will take longer to operate.

To gain some appreciation of fault current calculations complete the questions on thenext page.

I f - term used to describe a circuits earth fault current, in amps.U o - term used to describe the nominal voltage to earth, in volts.

Z s term used to describe the earth fault loop impedance, in ohms.

Zs

UoIf =

ampsZs

UoIf 75.143

6.1

230===


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LEARNER WORK BOOK


1. A circuit with a BS3036 overcurrentprotective device rating of 30A hasan R1+R2 value measured at 0.79.

2. A power circuit with a BS88

overcurrent protective device ratingof 32A has an R1+R2 valuemeasured at 0.56.

3. A BS88 63A fuse protects a feed toa three-line socket. The measuredR1+R2 value is 0.2 .

4. A lighting circuit is protected by aBSEN60898 10amp Type BMCB. The measured R1+R2 valueis 1.5.

5. A radial power circuit is protectedby a BSEN60898 32amp Type CMCB. The measured R1+R2 valueis 0.8.

Complete the following exercise to determine the maximum earth fault current (If) of the circuits.Assume in all cases that the Ze = 0.3 . You must show all working out.


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LEARNER WORK BOOK


Time / current characteristics and disconnection times

In appendix 3 of BS7671 there are graph-like tables that represent the time / currentcharacteristics of the main types and rating of circuit protection. They are used todetermine the time it takes a device to operate under a certain amount of fault

current.

If you look at the time/current curve you will find that the scales on both the time(seconds) scale and the prospective current (amperes) scale are logarithmic and thevalue of each subdivision depends on the major division boundaries into which itfalls.

For example, on the current scale, all the subdivisions between 10 and 100 are inquantities of 10, while the subdivisions between 100 and 1000 are in quantities of100 and so on. This also occurs with the time scale, subdivisions between 0.01 and0.1 being in hundredths and the subdivisions between 0.1 and 1 being in tenths, etc.

If you look at Fig.3.2A in your Tables from BS7671 and the on-site guide

appendices you can see that current, in amps, is represented along the bottom (-axis) and along the side (Y axis) is the time in seconds.

Also available on each graph is a quick reference table that displays the maindisconnection times and the required amount of current to achieve those times.

Each line between twopoints represents avalue in the lower ofthe two units

Three lines above the number 10represents the value 40

Remember:

Any circuit rated at 32A or less must disconnect within 0.4 seconds

A distribution circuit or circuit exceeding 32A must disconnect within 5s


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LEARNER WORK BOOK


ExampleFind out the expected disconnection time of a circuit that is protected by a 30ABS3036 fuse when 70 amps of fault current flows.

i) Obtain the correct time/current graph for the selected protective devicefrom Appendix 3 of BS7671.

Look at Fig.3.2A. Find the 30amp BS3036 fuse trip characteristic line.

ii) Identify the fault current value on the axis and follow it to the pointwhere it crosses the selected fuse rating curve.

Identify 70 amps on the -axis. Follow 70 amps upwards until it crosses the 30ampcurve.

iii) At the point where the fault current crosses the fuse rating curve follow theline across to the Y axis. Identify the value of time in seconds.

Follow the point where 70amps crosses the fuse line and track it (left) to the Y axisand obtain the time. The disconnection time is expected to be 20 seconds.

Over-current device andrating

Prospective current(Amps)

Disconnection time(seconds)

BS3036 - 30 amp 100

BS3036 - 20 amp 70

BS88 32 amp 260

BS88 63 amp 450

BSEN60898 10ampType B

300

BSEN60898 32ampType C

300

BSEN60898 32ampType D

300

BS88 100 amp 400

BS88 6 amp 18

You should now have a good grasp of the way we determine the disconnection times

of a few devices based upon fault current levels.

What if the disconnection time ishigher than the maximum allowed?What can the designer do to ensurethat shock protection is afforded?

Complete the exercise below using the Tables from BS7671 and the on-site guide appendicesto determine the operation times of different devices.


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LEARNER WORK BOOK


Thermal Constraints

Now that you have chosen the type and size of cable to suit the conditions of theinstallation, we must look at thermal constraints. This is a check to make sure thatthe size of the c.p.c, the earth conductor, complies with the IEE Wiring Regulations.

If there is a fault on the circuit, which could be a short circuit, or earth fault, a faultcurrent of hundreds or thousands of amperes could flow. Imagine that this is a 1 mmor 2.5 mmcable; if this large amount of current was allowed to flow for a short periodof time, i.e. a few seconds, the cable would melt and a fire would start. The c.s.a ofthe circuit protective conductor (c.p.c) is of great importance since the level ofpossible shock in the event of a fault depends on it.

The cable calculation process at this point has ensured the selected cable will safelycarry the design current and will disconnect in the required time therefore protectingthe circuit from danger. The final stage of the calculation process is to confirm thatthe circuit protective conductor can withstand the fault current for a short space oftime, for example 0.4 or 5 seconds at 230 volts.

In very many cases, calculation of the CPC size will show that a smaller size thanthat detailed in is perfectly adequate. The adiabatic equation is:

By using Ohms law the earth loop impedance value and the supply voltageare used to calculate how much fault current (I) will flow.The disconnection time of protective device is determined by using the faultcurrent value and the correct time/current curve table from Appendix 3 ofBS7671.K is determined by assessing what the CPC material is and how it is installed

in relation to the line conductors.

S - term used to describe the c.s.a of the circuit protective conductor, inmm.

I - term used to describe the earth fault current, in amps.t term used to describe the time earth fault current will be present in the

circuit, in seconds.k term used to describe the factor which takes into account the

resistivity, temperature coefficient and heat capacity of the conductormaterial, and the appropriate initial and final temperatures. (See tables

54.1 to 54.6)

k

tIS

=

2

We need to check that the c.p.c will be large enough to be able to carry thisfault current without causing any heat/fire damage. The formula that is usedto check this situation is the adiabatic equation. The c.p.c will only need tocarry the fault current for a short period of time, until the protective device

operates.


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LEARNER WORK BOOK


ExampleFind out the minimum size of c.p.c in mm so that the 50A BS88 circuit complies withBS7671. The protective conductor is a copper conductor incorporated in a 90CThermosettingconductor cable. The calculated fault current is 300 amps.

i) Obtain all values for the adiabatic equation.I is either worked out as shown previously or declaredt-is established using the graphs in Appendix 3 of BS7671k- is established using the tables from Section 5 of BS7671

I is 300amps; t is 1 second (fig.3.3A); k is 100 (table 54.3)

ii) Work out I x t first; then square root; then divide by the k value

The minimum size of c.p.c. is 3 mm. If the actual selected cable is more than thisthen thermal constraints have been satisfied. If the c.p.c is smaller than this it will nothandle the fault current safely.

Safety could always be assured if we assessed the size using Table 54.7 as a basis.However, this could result in a more expensive installation than necessary because

we would often use protective conductors which are larger than those found to beacceptable by calculation

So you can gain some more understanding complete the exercise on the next pageto calculate the minimum size of C.P.Cs. You will need your Tables from BS7671and the on-site guide appendices.

k

tIS =2

22

3100

1300mmS =

=

If S is smaller than the selected cpc size, what will we need to do so that the cable complieswith thermal constraints?


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LEARNER WORK BOOK


Minimum size of c.p.c

1. The protective conductorinstallation is steel armour inside a70C Thermoplastic (PVC) cablewith an 80A BS88 has a calculatedfault current of 500 amps.

2. A fault current of 1000A iscalculated on a circuit with a 100ABS88 fuse where the CPC is 70CThermoplastic (PVC) not bunchedwith cables

3. The protective conductorinstallation is steel armour inside90C Thermosettingconductorincorporated in a cable with a 32ABSEN 60898 Type B has acalculated fault current of 300amps.

4. A fault current of 150A is calculatedon a circuit with a 30A BS3036 fusewhere the CPC is 70CThermoplastic (PVC) bunched withcables

5. The protective conductorinstallation is Steel conduitcontaining 70C Thermoplastic(PVC) with a 32A BSEN 60898Type C has a calculated faultcurrent of 250 amps.

Complete the following exercise to determine the minimum size of CPC for the circuits using theadiabatic equation. You must show all working out; state the calculated size of CPC and theactual size to be installed.

There is a problem with question 5. Can you see what it is Look at the values in the formula?Once you discover it explain what will have to be done.

Once the cable calculation process has reached this stage the entire process is completeand the requirements of BS7671 have been met


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LEARNER WORK BOOK


Now complete the questions below

1. Write down the formula to establish the total design current I b of in two types ofcircuit.

2. What is I n and how is it established?

3. What four correction factors need to be taken into account when calculating theminimum size of a circuit conductor? Write down their symbols and descriptions

4. Once all these factors have been taken into consideration and we obtain I z whatneeds to be established next and what do we use to achieve it?

5. Once the circuit conductor has been selected we need to ensure that is complies withvolt drop? Write down the formula and explain its parts.

6. Explain what shock protection is and state how we ensure it is adhered to. Writedown the formula

7. Explain what thermal constraints are and state how we ensure it is adhered to. Writedown the formula.


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LEARNER WORK BOOK


Full cable calculation

Now complete a cable calculation and decide what cable size should be installedbased upon a complete installation decided by the group. Make sure you consider allfactors before you begin and write them in the boxes below. Then use the spacebelow to carry out the calculations.

Load detailsType:Rating:Voltage:

Cable Type Installation details

Supply detailsDB type:Ze value

Circuit detailsLength of run:Correction factors:


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LEARNER WORK BOOK


This page is for you the continuation of your calculation Cable calculation process

1. Work out the design current IbUsing a the correct formula

2. Select the protective device sizeIb In

3. Obtain all correction factorsFind the values so they can be used inthe next calcualion

Ca, Cc, Cg, Ci

4. Calculate IzIn divided by all factors

5. Select conductor sizes

With Iz use the tables in Appendix 4 andchoose.Iz It

6. Calculate volt dropUse design current, mV/A/m value fromcable table, the circuit length then 1000

7. Calculate R1+R2Using table 9A, 9B and 9C find the m/mvalue 1000 then apply multipliers

8. Calculate Zs

9. Verify Zs value with BS7671Using the tables in Section 41 ensurecalculated value is less than thetabulated value to ensure shockprotection

10. Calculate the fault currentUsing nominal voltage and Zs

11. Establish the disconnection timeUsing the fault current, device type andtime / current graphs in Appendix 3obtain the expected disconnection time

12. Calculate minimum size of CPCUsing the adiabatic formula, input allvalues (including k factor) and work outmin. size of cpc. Then compare with ourchosen cpc size

13. Once all above has been verified theprocess is complete

V

PI=

CcCgCiCa

InIz

=

( )

1000

// LmAmVIbVoltdrop

=

==+ CBA

RR 99

1000

921

=++= 21 RRZeZs

AZs

UoI ==

k

tIS

=

2


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LEARNER WORK BOOK


Cable capacities of conduit and trunking

Not only must it be possible to draw cables into completed conduit and trunkingsystems, but also neither the cables nor their enclosures must be damaged in the

process. If too many cables are packed into the space available, there will be agreater increase in temperature during operation than if they were given more space.It is important to appreciate that grouping factors still apply to cables enclosed inconduit or trunking.

Conduit Capacities

To calculate the number of cables that may be drawn into a conduit, we make use offour tables below. These have been adapted from the on-site guide to show the twomost common conduit sizes used today. The conduit terms are shown in yourTables from BS7671 and the Onsite Guide.

The number of cables that can be drawn into or laid in any enclosure of a wiringsystem must be such that no damage can occur to the cables or the enclosure duringinstallation. The number of cables that can be used is the overall sum of the cables

cross-sectional area (c.s.a) compared to the overall c.s.a of the trunking. This isexpressed as a percentage and should not exceed 45 per cent.

To calculate the required size of conduit we must first establish the amount ofconductors to be installed. We must then obtain the individual terms for individualconductors and multiply them together. If we have 5 cables of one size we multiplythe term for that cable by 5. We treat each size of conductor as an individual sum.Once we have obtained all the calculated terms we add them all together to give onetotal. We then refer to the relevant table and discover which conduit term iscompatible. Bear in mind that the conduit term shown is the upper limit of spacingand should not be exceeded under any circumstances.

Table 5A (Onsite guide) - Cable factors for conduit in short straight runs up to 3mTable 5B (Onsite guide) - Conduit factors for use in short straight runs up to 3mTable 5C (Onsite guide) - Cable factors for long straight runs over 3m, or runs of

any length incorporating bends.Table 5D (Onsite guide) - Conduit factors for conduit incorporating bends and

lon strai ht runs

Cables in conduit


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LEARNER WORK BOOK


E.g Conduit 2.5 metres long without bends with 3 x 1.5mm and 3 x 2.5mm strandedconductorsi) Select the correct table for the cable factors and conduit (5A)ii) Obtain the cable factors for each size of cable (1.5mm =31 and 2.5mm =43)iii) Multiply the number of cables by the factors (3x31)+(3x43)=222iv) Select the correct table for the conduit factors and length of run (5B)v) Obtain the factor which is greater than the sum of cable factors (20mm=420)vi) The conduit size is 20mm

1. Conduit 2.5 metres long without bendswith 6 x 1.5mm and 6 x 2.5mm

2. Conduit 3.5 metres long without bendswith 5 x 6.0mm

3. Conduit 6.0 metres long with one bendwith 5 x 1.5mm and 3 x 4.0mm

4. Conduit 2.0 metres long with twobends with 3 x 16.0mm

5. Conduit 10 metres long one bend with6 x 1.5mm and 6 x 2.5mm

6. Conduit 3 metres long three bends with5 x 6.0mm

7. Conduit 8.0 metres long with twobends with 5 x 1.5mm and 1 x 4.0mm

8. Conduit 2.0 metres long with one bendwith one ring main and a single line 6.0mm cooker supply

Complete the following conduit sizing exercise using the four tables listed above. You mustrecord the cable factors, show your calculations and select the correct size conduit. All cablesare stranded. See if you can make a guess before you calculate the answers.


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100x100

Trunking capacities

Minimum trunking size is calculated in much the same way as conduit althoughtrunking has its own table of terms for single core cable. No consideration though, ispaid to the length of run.

The sizes should ensure an easy pull, with low riskof damage to cables and enclosures. The electricaleffects of grouping are not taken into account.Therefore, as the number of circuits increases, thecurrent-carrying capacity of the cables willdecrease. Cable sizes would have to be increasedwith a consequent increase in cost of cable andtrunking. It may therefore be more economical todivide the circuits concerned between two or moreenclosures.

The tables show the factors for trunking having taken into consideration themaximum allowable 45% space factor. Sizes in bold are the most common trunkingsizes available.

The trunking terms are shown in your Tables from BS7671 and the Onsite Guide.

E.g. Trunking with 100 x 1.5mm and 100 x 2.5mmi) Obtain the cable factors for each size and type of cable (1.5mm =8.6 and

2.5mm =12.6)ii) Multiply the number of cables by the factors (100x8.6)+(100x12.6)=2120iii) Obtain the factor which is greater than the sum of cable factors (75x75=2371)iv) The trunking size is 75mmx75mm

Table 5E (Onsite guide) - Cable factors for trunkingTable 5F (Onsite guide) - Factors for trunking


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1. A trunking installation with6 x 1.5mm,6 x 2.5mm,3 x 4.0mm and3 x 6.0mm conductors.

2. A trunking installation with12 x 1.5mm,

12 x 2.5mm,9 x 4.0mm, 3 x 6.0mm ,3 x 10.0mm , 1 x 16.0mm and2 x 25.0mm conductors.

3. A trunking installation with18 x 1.5mm, 24 x 2.5mm,

12 x 4.0mm, 20 x 6.0mm ,12 x 10.0mm , 20 x 16.0mm and15 x 25.0mm conductors.

4. A trunking installation with10 single-line lighting circuits,

10 ring mains,8 x A2 radial circuits (32A),10 x 6.0mm SP&N circuits,5 x 10.0mm TP&N circuits,8 x 16.0mm TP&N circuits,5 x TP&N DB supplies with 25.0mmconductors1 x 25mm earthing conductor and6 x 16.0mm Main equipotentialbonding conductors

Complete the following trunking sizing exercise using the two tables above. You must record thecable factors, show your calculations and select the correct size trunking. All cables arestranded to BS6004.


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Maximum demand and diversity

In this section we will look at Maximum Demand and Diversity. The current demandfor a final circuit is determined by adding up the current demands of all points of

utilisation and equipment in the circuit and, where appropriate, making an allowancefor diversity.

In most cases main or sub-main cables will supply a number and/or variety of finalcircuits. Use of the various loads must now be considered, otherwise if all the loadsare totalled, a larger cable than necessary will be selected at considerable extra cost.Therefore a method of assessing the load must be used. This method is calledapplying diversity.

Maximum Demand

Maximum demand is the largest current normally carried by circuits, switches andprotective devices under normal operating conditions. Assessment of maximumdemand is sometimes straightforward. For example, the maximum demand of a 230Vsingle-line 8 kW shower heater can be calculated by dividing the power (8 kW) by thevoltage (230 V) to give a current of 34.78 A.

There are times, however, when assessment of maximum demand is less obvious.For example, if a ring circuit with 15 x 13A sockets, the maximum demand clearlyshould not be 195A (15 x 13A). Some 13A sockets may feed table lamps with 60 Wlamps fitted, whilst others may feed 3 kW washing machines; others again may notbe loaded at all.

Lighting circuits pose a special problem when determining maximum demand. Eachlamp-holder must be assumed to carry the current of 0.43A per lamp (100 W perlamp holder). Discharge lamps are particularly difficult to assess, and current cannotbe calculated simply by dividing lamp power by supply voltage. We have to multiplythe total lamp power by 1.8 to allow for some losses. The reasons for this are:

Control gear losses result in additional current,The power factor is usually less than unity so current is greater

When assessing maximum demand, account must he taken of the possible growth indemand during the life of the installation. As a rule of thumb this is generally 20%.

The maximum demand is the expected current that a circuit will carry.

Typical demands to be used for this summation are given in Table 1A of theOn-Site Guide.

When calculating the maximum demand of an installation of three ring mains what do youexpect the value of current to be? Explain the reasoning behind your answer


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Diversity

Diversity is defined as making a realistic estimate as to how much the maximumdemand will be under an installations normal operating conditions.

Appendix 1, Table 1B, of the IEE On-Site Guidecontains the recommended diversityallowances, which allows diversity to be applied depending upon the type of load andinstallation premises.

The individual circuit / load figures are added together to determine the totalAssumed Current Demand for the installation. This value can then be used as thestarting point to determine the rating of a suitable protective device and the size ofcable, considering any influencing factors in a similar manner to that applied to finalcircuits.

A domestic ring circuit feeds a large number of 13A sockets and is protected by a32A fuse. If all sockets were feeding 13A loads, more than two of them in use at thesame time would overload the circuit and its protective device would disconnect it.

In practice, the chance of all domestic ring sockets feeding loads taking 13 A is small.Most sockets feed small loads such as table lamps, vacuum cleaner, television oraudio machines and so on. The chances of all the sockets being used simultaneouslyare remote. The consideration that only a few sockets will be in use at the same timeis called diversity.

The sensible application of diversity to the design of an installation calls forexperience and a detailed knowledge of the intended use of the installation. Futurepossible increase in load should also be taken into account. Diversity relies on anumber of factors that can only be properly assessed by having a detailed knowledgeof the type of installation and the habits and practices of the users. Not knowing howan installation is to be used could mean that installation conductors and equipment isunder or over-rated

We will now take a look at how diversity is implemented.

Please remember that the calculation of maximum demand is not an exactscience and a suitably qualified electrical engineer may use other methods

of calculating maximum demand.

By making allowance diversity, the number of circuits and their rating canbe reduced, with a consequential financial saving. However, if diversity isover-estimated, the normal current demands will exceed the ratings of the

protective devices, which will disconnect the circuits. Overheating mayalso result from overloading which exceeds the rating of the protective

devices.


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Point of utilization or current usingequipment

Current demand to be assumed

2 amp sockets At least 0.5A

Lighting outlet Connected load, minimum 100 WShaver outlet, bell transformer or anyequipment rated 5 watts of less

May be neglected

Household cooking appliance The first 10 A of the rated current plus30% of the remainder plus 5A if thecontrol unit incorporates a socket

All other stationary equipment Rated currentTable 1A On site guideCurrent Demand to be assumed for points of utilization and current usingequipment

Note 1 See Table 1B in your Tables from BS7671 and Onsite Guide for the design ofother socket outlets.Note 2 Final circuits for discharge lighting take into to account harmonic currents andcontrol gear losses. Where the exact manufacturers information for gear losses isnot available the maximum demand shall be assumed to be 1.8 x the lamp rating.

Current usingequipment

Number ofpoints /

Rating

Maximumdemand without

diversity

Maximum demand withdiversity

2A sockets 20

Lighting outlet 12 x 100 wattseach

Householdcooker

6kW withsocket on mainswitch

Apart from indicating that diversity and maximum demand must be assessed, theRegulations themselves give little help. Suggestions of values for the allowances fordiversity are given in Table 1B in your Tables from BS7671 and Onsite Guide.

Work out the following current demands with and without diversity using table 1A above. Then inpairs discuss the implications of diversity both when done correctly and incorrectly.


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Example diversity calculationWe will now take a look at diversity on a larger scale by considering the examplebelow.

A shop has the following single-line loads, which are balanced as evenly as possibleacross the 400 V three-line supply.

Q. Calculate the total demand of the system, assuming that diversity can be applied.Calculations will be based on Table 1B. The single-line voltage for a 400V three-linesystem is 400 /3 = 230 V.

All loads with the exception of the discharge lighting can be assumed to be at unitypower factor, so current may be calculated from:

Where: I = the design current in ampsP = the circuit power in wattsV = the circuit voltage in volts

1. Water heaters (thermostatic) 2 x 6 kW and 7 x 3kw

Table 1B states No diversity is allowable, so the total load will be:

2. Water heaters (instantaneous) 2 x 3 kW

Table 1B states 100% of the largest and 100% of 2nd largest. So the total load willbe:

V

PI=

( ) ( ) kWkWP 3321123762 =+=+=

AI 47.143

230

000,33==

kWP 632 ==

AI 08.26230

6000==

Example Small Shop

1. 2 x 6 kW and 7 x 3kw thermostatically controlled water heaters2. 2 x 3 kW instantaneous water heaters3. 2 x 6 kW and 1 x 4 kW cookers4. 12 kW of discharge lighting (Sum of tube ratings)5. 8 x 32 A ring circuits feeding 13 A sockets.


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3. Cookers 2 x 6 kW and 1 x 4 kW

Table 1B states 100% of the largest and 80% of 2nd largest. 60% of remainder Sothe total load will be:

First cooker

Second cooker

Third cooker

Total for Cookers =57.37A

4. Discharge lighting 12 kW

Table 1B states 90% of total current demand So the total load will be:

5. Ring circuits 8 x 32 A

Table 1B states 100% of the largest and 50% of remainder So the total load will be:

First ring main

Remaining Ring mains

Total for Ring mains =144A

AI 08.262306000 ==

AAI 86.20100

8008.26

230

6000===

AAI 43.10

100

6039.17

230

4000===

AAI 52.84

100

9091.93

230

8.112000==

=

AI 32=

AAI 112100

50224327 ===


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Total current demand after diversity is applied.

Thermostatically controlled water heaters = 143.47 A

Instantaneous water heaters = 26.08 A

Cookers = 57.37 A

Discharge lighting = 84.52 A

Ring circuits feeding 13A sockets = 144 A

TOTAL = 455.44 A

Then we divide by three to balance across three lines A81.1513

44.455=

Therefore the switchgear we select can be rated to take 151.81 amps. Inthis instance we would select the next largest size of switchgear, which is

160 Amps.

If diversity hadnt been applied explain what would happen to the maximum demand and statethe implications?


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Now answer the question on diversity below

A small hotel has the following single-line loads, which are balanced as evenly aspossible across the 400 V three-line supply.

EXAMPLE - SMALL HOTEL

a. 10 kW of discharge lighting (Sum of tube ratings)b. 6 kW of low voltage lighting (cos = 0.95)c. 5 x 3kW thermostatically controlled water heatersd. 2 x 3 kW instantaneous water heaterse. 5 x 8.5 kW shower unitsf. 3 x 6 kW and 1 x 3 kW cookersg. 6 x 32A ring circuits feeding 13A sockets.h. 1 x 8 kW Under-floor heating

Calculate the total demand of the installation, assuming that diversity can be applied and workout the approximate balanced load across three lines. Calculations will be based on Table 1B.Use this and the next page for your working out.


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Now complete the questions below.

1. What is another name f

Documents

35718780 Sizing Cables Conduit and Trunking