Sizing Cables, Conduit and Trunking

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Sizing Cables, Conduit

and Trunking

Learner Work Book

Name:

Group:

Tutor:

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Foreword

When trying to determine the size of conductors necessary for the safe working of a circuit many factors need to be considered. It is not acceptable to guess or to use cable sizes that are in common usage just because someone may say that you should use 1.5mm² ‘twin and earth’. They may just be wrong! If you get cable sizes wrong, then there may be a fire risk, a load that won’t function properly, or you may be 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. An

aluminium cable would have nearly twice the resistance of a copper cable with the same dimensions. Therefore the energy losses in the aluminium cable will be higher than 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, we have to make it larger. The larger cross sectional area reduces its resistance and brings 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 are each designed to be able to carry up to 500A without the conductor going above 90°C. The copper cable (on the right) is thinner than the aluminium one, because copper is a better conductor. Its cross sectional area is 300mm2 as opposed to 500mm2 for the aluminium.

All cables have electrical resistance, so there must be an energy loss when they carry current. This loss appears as heat and the temperature of the cable rises. 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 of the difference in temperature between the conductor and the surroundings, so as the conductor temperature 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 ability depends on the type of insulation that is used and how the cables are installed into a wiring system. When we consider the current carrying capacity we have to ensure that the cable won’t overheat when the normal current is flowing. If conductors are installed into wiring systems that are incorrectly sized they will not be able to “lose” enough heat 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 and sizing of conduit and trunking.

Cable calculation is the method used to ensure that all the factors of a circuit have been taken into consideration, in particular, the operating current of a cable that is determined by how hot the cable gets. This is affected 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 PowerPoint “Sizing Cables Conduit and Trunking”

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Sizing Cables, Conduit and Trunking Unit Overview

Practical Skills

To 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 with BS7671

Knowledge Requirements

To 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 circuits Select the correct rating of protective device

How to allow and apply factors for: Grouping

Thermal insulation Ambient temperature Protective device type

How to check that the voltage drop is not excessive Establishing circuit disconnections times are met Thermal constraints are satisfied

How to determine the size of conduit and trunking appropriate to the size and number of cables

Methods used of establishing a circuit’s maximum demand after diversity is applied

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The price of cable

Before we look at different cable sizes let us gain some appreciation for the price of cable. As you can imagine the cost of cable varies according to the amount of materials 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 how the price varies from size to size? Also note that the price per metre is more than the calculated amount for 100 metres.

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

4 core 1.5mm² XLPE / SWA 1.11 93.99

4 core 2.5mm² XLPE / SWA 1.52 126.00

4 core 4.0mm² XLPE / SWA 2.15 179.99

4 core 6.0mm² XLPE / SWA 3.00 251.99

4 core 10.0mm² XLPE / SWA 4.40 385.00 4 core 16.0mm² XLPE / SWA 6.10 550.00

Cable prices as of JULY 2008

Price (£) 1. 50M of 4 core 1.5mm² XLPE / SWA in one coil

2. 200M of 4 core 1.5mm² XLPE / SWA in two coils

3. 100M of 4 core 16.0mm² XLPE / SWA in five coils

4. 100M of 4 core 16.0mm² XLPE / SWA in one coil

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

Complete the following exercise.

The following lengths of XLPE/ SWA are required for a job. Your job as the company’s buyer is to process this order form from the site supervisor. The client has requested a price of the materials before they will let the job go ahead. Work out the prices using the table above

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Cable calculation process

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

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

Design current (Ib)

Rating of the protective device (In)

Reference methods

Correction factors (Ca, Ci, Cg, Cc)

Application of correction factors Voltage drop

Shock protection Thermal constraints Diversity.

External influences

If 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 high risk of mechanical damage the cable or installation will have to be sufficiently

protected form danger. Once you have decided on the type of cable suitable for the environmental 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 various current ratings of various sizes of XLPE armoured cable.

The Basics

So that we can understand a full cable calculation we must first understand the basics. Listed below are four terms that describe vital information used in the

calculation process. Try to remember them, as they will appear frequently throughout this course.

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

I b - term used to describe a circuit’s design current in amps i.e. the load.

I n - term used to describe a circuit’s protection size in amps i.e. the fuse size.

I z - term used to describe a circuit’s value, in amps, once all de-rating factors have been considered

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

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Design Current – I b

The first stage of the design process is to determine how much current will flow in the circuit. This current is known as the design current and is the full load current of the circuit. 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 to be 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 Φ = the power factor

Resistive loads

The following formulae apply to single and three line supplies:

Inductive and / or capacitive loads

The following formulae apply to single and three line supplies:

In a.c. circuits, the effects of either highly inductive or highly capacitive loads can produce a poor power factor (cos Ф) (inductive and capacitive loads will be explained later). For now it is satisfactory to know that in circuits where there are inductive and electronic components such as coils and capacitors there are losses. These losses slightly increase the amount of current the equipment uses. You will have to allow for this in such circuits. Note √3 = 1.732

Single-line 230v Inductive and or Capacitive

Φ

×

=

cos

V

P

I

Three line 400v Inductive and or Capacitive

Φ

×

=

cos

3 V

P

I

Single-line 230v Resistive

V

P

I =

Three line 400v Resistive

V

P

I

×

=

3

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Sizing Cables Conduit and Trunking REV4.1 10 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 to add 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 to add the unit (A).

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

V

P

I =

A

I

8 .

70

230 2000

=

Φ

×

=

cos

3 V

P

I

A

I

45 .

58

95 .

0

400

3 30000

=

×

=

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Over-Current Protective Device – I n

Once the design current has been established we must then select an over-current protective device. The function of an over-current protective device is basically as it’s name 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 limited amount of harm or danger, to persons, livestock or property, will exist under fault conditions. The two main factors to consider when selecting a device are shown below.

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 a rise in insulation temperature. High conductor temperatures are of little consequence except that the resistance of the conductor will be increased leading to greater levels of voltage drop.

Insulation cannot tolerate high temperatures since they will lead to deterioration and eventually failure. The most common insulation material is p.v.c. If it becomes too hot it 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 higher current than the design value due to overloaded machines, an error in the

assessment 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 circumstances the temperature of p.v.c. Insulation may rise to 160°C.

Device Selection

The protective device current rating must be equal to or next largest size so that the circuit is sufficiently protected. Take a look at the formula used for the cable calculation 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 the

materials in use will tolerate for a long period of time. The term can be divided into 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 as 300°C. This is why the conductors and or the cutting tool used will melt at

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Sizing Cables Conduit and Trunking REV4.1 12 There are many protective devices in existence and these will be looked at in more detail in Unit 4.4. What is important for now is to realise that these devices all have different ratings or current ratings. Shown below is part of BS7671 that shows the different 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 total power of 6KW

3. A single line, inductive lighting circuit with a total power of 1200 watts and power factor of 0.85

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

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

What reasons can you state for the protective device being equal to or slightly larger than the design 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 Regulations lists the common methods that can be used to install a cable.

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

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

I t

1. A 4.0mm², two core cable, carrying alternating

current that has been clipped to a wall.

2. A 10.0mm², four core cable, carrying A.C current ran on a cable tray

3. A 16.0mm², single line cable supplying a consumer unit, clipped direct

4. A 25.0mm², three line and neutral supply for an uninterruptible power supply unit in an office block clipped direct to a wall.

5. A 25.0mm², 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 basic current ratings (tabulated current ratings I t) of the cables below. Be sure to fully observe the details 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 circuit’s installation types.

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

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Sizing Cables Conduit and Trunking REV4.1 14 Appendix 4 of BS7671 contains tables for each type of conductor type from single core PVC to multi-core paper insulated lead steel wire armoured. Please refer to BS7671 for more information.

C E

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Cc

Cg

Ci

Ca

In

Iz

×

=

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 the protection (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 highest tabulated value (It).

Temperature can have a serious effect on a circuit. It can increase the risk of a fault, can cause a cable’s insulation to melt and can even cause an electrical fire. There are various factors that form a circuit’s temperature and we will look at these in more detail below. It is essential to follow the correct design procedures and apply the correct correction factors to ensure that the many effects of temperature do not affect the normal operation of a circuit. We will look at these factors individually below but remember 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 one calculation shown below

Example

A three line, 32A circuit is to be installed using XLPE SWA. It will run through a boiler house clipped direct to a wall where the ambient temp will be 40ºC. It is not grouped nor does it come into contact with any thermal insulation. Discover the “de-rated” value i.e. the minimum permissible rating of cable.

i) Establish all the correction factors present and obtain the values from the tables 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 and include the value (A)

iii) Select the correct cable table then select the correct column based upon the reference method and select the next highest value (or equal) to Iz. Table 4E4A; column 3; value of 42 amps corresponds to a 4mm² 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.

A

Cc

Cg

Ci

Iz

35 .

16

91 .

0

32 =

×

=

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Ambient temperature - Ca

This is the temperature of the surroundings of the cable, often the temperature of the air in a room or building in which the cable is installed. When a cable carries current, it gives off heat. Therefore the hotter the surroundings of the cable, the more difficult it is for the cable to get rid of this heat. But if the surrounding

temperature is low, then the heat given off

could be easily let out and the cable could carry more current. Cables must give off this heat safely or they could be damaged and there is a risk of a fire. You can find the correction factor for ambient temperature in Tables 4B1 and 4B2 of the IEE Wiring Regulations.

These tables are based on an ambient temperature of 30°C. This means that any cables 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 it gives off safely when carrying current.

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

The most common of the correction factors are given in the Tables 4B1 and 4B2 from BS7671 Correction factors for ambient temperature and are given in your “Tables from BS7671 and the Onsite Guide”.

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

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

1. A single-line supply to a DB with a design current of 50A and a MCCB size of 100A, clipped direct at 30°C.

2. A single-line supply to a DB with a design current of 50A and a MCCB size of 100A clipped direct and ran through a boiler house at 50°C.

3. A three line supply to a DB with BS88 fuses 50A and a maximum demand of 45A, clipped direct and ran near to hot machinery at 40°C

4. A three line supply to a DB with BS88 fuses 50A and a maximum demand of 45A, clipped direct but re-routed to 30°C

I n - term used to describe a circuit’s protection size in amps i.e. the fuse size.

I z - term used to describe a circuit’s value, in amps, once all de-rating factors have been considered.

C a – term for the correction factor of ambient temperature.

Ca

In

Iz =

Using the table 4E4A and the Ca correction factors in your essential tables determine the minimum size of XLPE cable for each of the circuits below. Make note of the cable insulation type before you proceed with your calculations. You must write down:

In (fuse size), Ca (correction factor), Iz (de-rated CCC), cable size (mm²), It (tabulated CCC) and your working out.

In your own words how does ambient temperature affect the selection of cable? You may be required 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 space blanketing, and so on, is now standard. Since the purpose of such materials is to limit the 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 of wrapping a cable in a fur coat on a hot summer’s day. The heat produced when the cable carries current cannot escape.

Loft insulation

The cable rating tables of the regulations as the one you have already used (Table 4E4A) allow for the reduced heat loss for a cable which is enclosed in an insulating wall 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 be completely covered by the insulation. Where this is not possible and a cable is buried in 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 be doubled.

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

Clearly, the longer the length of cable enclosed in the insulation the greater will be the de-rating effect. Table 52.2 (BS7671) shows the de-rating factors for lengths in insulation of up to 400 mm and applies to cables having cross-sectional area up to 10 mm². Table 52.2 from BS7671 De-rating factors for cables up to 10mm² in cross-sectional area buried in thermal insulation. Is given in your “Tables from BS7671 and 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 insulation affects minimum cable ratings.

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

1. A single-line supply to a load with a design current of 13A and a MCB size of 16A, reference method C with no thermal insulation.

2. A single-line supply to a load with a design current of 13A and a MCB size of 16A, reference method C with 3000mm thermal insulation.

3. A three-line supply to a DB with BS88 fuses 32A and a maximum demand of 27A, reference method E with 600mm of thermal insulation.

4. A three-line supply to a DB with BS88 fuses 32A and a maximum demand of 27A, reference method E without any thermal insulation.

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

Ci

In

Iz =

Using the table 4E4A and the Ci correction factors in your essential tables determine the minimum size of XLPE cable for each of the circuits below. Make note of the cable insulation type before you proceed with your calculations. You must write down:

In (fuse size), Ci (correction factor), Iz (de-rated CCC), cable size (mm²), It (tabulated CCC) and your working out.

In your own words how does thermal insulation affect the selection of cable? You may be required 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 all warm up. Those which are on the outside of the group will be able to transmit heat outwards, 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 shed heat at all, and will rise further in temperature.

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

If surface mounted cables are touching the reduction in the current rating is, as would be expected, greater than if they are separated. The picture below illustrates the difficulty 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 than one circuit shows some of the most common values of Cg.

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

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

Widely spaced cables dissipate heat easily A closely packed cable cannot easily dissipate heat and so its temperature rises

Note: If a cable is expected to carry no more than 30% of its grouped rated current, it can be ignored when calculating the group-rating factor. For example, if there are four 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 perfect situation is no circuits grouped therefore Cg = 1.0.

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

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

3. A single-line circuit with a 63A BS88 fuse protecting it, reference method E grouped with one other circuit.

4. A single-line circuit with a 63A BS88 fuse protecting it, reference method E and not grouped with any other circuit.

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

Cg

In

Iz =

Using the table 4E4A and the Cg correction factors in your essential tables determine the minimum size of XLPE cable for each of the circuits below. Make note of the cable insulation type before you proceed with your calculations. You must write down:

In (fuse size), Cg (correction factor), Iz (de-rated CCC), cable size (mm²), It (tabulated CCC) and your working out.

In your own words how does grouping circuits affect the selection of cable? You may be required 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) fuse the cable size will need to be larger to allow for the fact that such fuses are not so certain 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 other protective devices available would do.

It has been known for a 5 amp BS3036 fuse to carry in excess of twice its rating without any signs of operating!! Therefore the fuse rating must never be greater than 0.725 times the current carrying capacity of the lowest-rated conductor protected. In effect, this is the same as applying a correction factor of 0.725 to all circuits protected by semi-enclosed fuses.

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

1. A three-line supply to a socket with a design current of 25A and a BS3036 size of 30A, reference method E.

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

3. A single-line circuit with a 60A BS3036 fuse protecting it, reference method C.

4. A single line circuit with an old 60A BS88 fuse protecting it, reference method C

I n - term used to describe a circuit’s protection size in amps i.e. the fuse size.

I z - term used to describe a circuit’s value, in amps, once all de-rating factors have been considered.

C c – term for the correction factor for the use of BS3036 fuses.

Cc

In

Iz =

Using the table 4E4A in your essential tables determine the minimum size of XLPE cable for each of the circuits below. You must write down your formulas. Note that this only applies to circuits that use BS3036 fuses. You must write down:

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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 this volt drop.

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

BS7671 also indicates that these requirements will be met if the voltage drop does not 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 installations supplied directly from a public low voltage distribution system

3% 5%

(ii) Low voltage installation supplied from a private LV supply

6% 8%

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

Public Supplies For lighting

If the supply is single-line at the usual level of 230 V, this means a maximum volt drop of 3% of 230 V, which is 6.9 V. This means the voltage at the load is as low as 223.1 V. For a 400 V three-line system, allowable volt drop will be 12 V with 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 maximum volt drop of 5% of 230 V, which is 11.5 V. This means the voltage at the load is as low as 218.5 V. For a 400 V three-line system, allowable volt drop will be 20 V with a line load voltage as low as 380 V.

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How to establish the value of volt drop

:

Example

A 4 mm² 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 mm² two-core cable is 11 mV/A/m. iii) Input all the values into the formula and work out the volt drop to two

decimal places and add the value (V).

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

A

V

P

I

26 .

08

230 6000

=

v

Voltdrop

4 .

59 1000

16

11

08 .

26 =

×

=

Each cable rating in the Tables of Appendix 4 of BS7671 has a

corresponding 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 dropped L= the circuit length in metres

1000 = converts the millivolts into volts

(

)

1000

/

A

m

L

mV

Ib

Voltdrop

=

×

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Application of volt drop

It is important to appreciate that the allowable volt drop applies to the whole of an installation from its source to the furthest point on the final circuit. If an installation has 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 slightly less bright light being produced, incorrect voltages supplied onto delicate circuitry (as for example in a DVD player, computer, and so forth) may quite easily result in an electrically 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 be drawn, but also to ensure that the total voltage drop shall not exceed the maximum percentage (%). This is particularly the case when running long lengths of cable from one end of a large building to another.

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

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

Bigger cables Smaller cables

D

B

1 D

B

2

D B 3 LOAD

1.5 volts 2.0 volts 4.4 volts

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Sizing Cables Conduit and Trunking REV4.1 26 1. 20 metres of 1.5mm² cable supplying a

line line machine which carries 14 amps using Multi core 70°C armoured thermoplastic (p.v.c) insulated cable

2. 4.0mm² three line supply, 40 metres in length to a 32A socket used for

resistive loads only using Multi core 90°C armoured thermosetting (x.l.p.e) insulated cables

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

1.0mm² using 70° thermoplastic (p.v.c) insulated and sheathed flat cable with protective conductor.

4. 80 metres of 1.5mm² cable for a 10KW, three line load that has a power factor of 0.90 using Multi core 90°C armoured thermosetting (x.l.p.e) insulated cables

Using the tables in your “Tables from the regulations and On-site Guide” notes calculate the volt drops of the various circuits below. Make note of the cable insulation type before you proceed with 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 drop

requirement.

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

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Shock protection

Protection against shock is a massively important factor to consider when designing a circuit. A person in contact with a supply voltage for any length of time can be very harmful, 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 circuit disconnects automatically.

Earth fault loop impedance and fault current

The path followed by fault current as the result of low impedance occurring between the line conductor and earthed metal or circuit protective conductor is called the earth fault 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 an earth fault to ensure that the protective device cuts off the supply before dangerous shock 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 with the lowest levels of fault current, and not, as is commonly believed, the highest levels. Why is this?

Note that there is no such thing as a three-line line/earth fault, although it is possible for three faults to occur on the three lines to earth simultaneously. As far as

calculations for fault current are concerned, the voltage to earth for standard UK supplies is always 230 V, for both single-line and three-line systems. Thus the tables of maximum earth-fault loop impedance, which are given in the appendices, apply both to single- and to three-line systems.

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

• Any final circuit not exceeding 32A must disconnect within 0.4s

So any circuit rated at 32A or less must disconnect within 0.4 seconds.

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

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

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Earth Loop Impedance

Resistance (measured in ohms) is the property of a conductor to limit the flow of current through it when a voltage is applied. The larger the conductor is the less resistance 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 one ampere.

1. A 230v supply is connected to a resistance of 20 ohms. What is the current?

2. A 230v supply with 3 amps of current flows in a circuit. What is the

resistance?

3. A current of 12.5 amps flows through a resistance of 18 ohms. What is the supply voltage?

4. A 230v supply is connected to a resistance of 50 ohms. What is the current?

5. A 230v supply with 5 amps of current flows in a circuit. What is the

resistance?

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 or generator winding, which is inductive. This inductance, along with the resistance of the cables to and from the fault, makes up the impedance. Refresh your memory with some Ohms law calcs.

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The earth fault loop path

The earth loop impedance we are concerned with is the worst-case scenario. This means that we must ensure that the protective device will operate within the specified disconnection 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 impedance path because each metre of cable has resistance and it will calculate the value at the furthest length of the circuit’s 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 is equal to or less then the maximum allowable for the device type and rating. Each device type and rating has its own limits. Complete the questions on the next page to become familiar with these limits and how they apply to a circuit.

1. The circuit protective conductor 2. The main earthing conductor and the consumers earthing terminal 3. The suppliers return path, either combined, separate or the general mass of earth 4. The earthed neutral of the supply transformer 5. The supply

transformer winding

Faulty appliance – Live to earth fault 6. The line

conductor supply from the transformer to the consumer unit 7. The final circuit live conductor

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

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

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Sizing Cables Conduit and Trunking REV4.1 30 Maximum values of earth loop impedance for various over-current protective devices are shown in Table 41.2, 41.3 & Table 41.4 in your “Tables from the regulations and On-site Guide”. Once the Zs has been established these tables are referred to ensure the designed circuit is in compliance. Turn to these tables now to get familiar with them.

1. A calculated value of earth loop

impedance (Zs) equating to 8.5 Ω for a circuit supplying portable equipment. With BS88 circuit protection rated at 6A.

2. A calculated Zs value of 1.2 Ω for a circuit supplying portable equipment. With BS3036 circuit protection rated at 30A

3. A calculated calculated Zs value of 3.2 Ω for a circuit supplying fixed

equipment. With BS88 circuit protection rated at 16A

4. A calculated Zs value of 0.5 Ω for a circuit supplying fixed equipment. With BS88 circuit protection rated at 100A 5. A calculated Zs value of 5.0 Ω for a

circuit supplying portable equipment. With BSEN60898 Type B circuit protection rated at 6A

6. A calculated Zs value of 5.0 Ω for a circuit supplying portable equipment. With BSEN60898 Type C circuit protection rated at 6A

7. A calculated Zs value of 0.36 Ω for a circuit supplying fixed equipment. With BSEN60898 Type D circuit protection rated at 32A

8. A calculated Zs value of 1.25 Ω for a circuit supplying fixed equipment. With BSEN60898 Type B circuit protection rated 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 comply with BS7671. You will need your “Tables from BS7671 and the on-site guide” appendices. You must state the maximum disconnection time for the circuit; record the maximum Zs and the table number; and state whether the cable passes or fails the shock protection requirement

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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 circuit’s earth fault path.

The R1 and R2 values are calculated in the following way. Table 9A from the On-site guide lists the resistances in milli-ohms per metre of all sizes of cables up to 25mm². Note that the values are in milli-ohms per metre. Table 9A values need to be

converted 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 ambient temperature at the time of test or maximum operating temperature of conductors.

where

Zs - term used to describe a circuit’s earth fault loop impedance, in ohms.

Ze - term used to describe part of the earth loop impedance that is external to the installation.

R1 – term used to describe the impedance of the line conductor, in ohms found in Table 9A R2 – term used to describe the impedance of the circuit protective conductor, in ohms found in Table 9A

Length – length of circuit from supply to furthest point, in metres

Table 9B or 9C Multiplier – factor applied to allow for expected ambient temperature or conductor resistance at maximum operating temperature respectively

(

R

1 R

2 )

Ze

Zs

=

+

R1

R2

Ze

Zs = Ze + (

R1

+ R2)

The Ze is something that we can either measure or obtain from our electricity provider 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

(

R

Table

A

L

Table

9 B

or

9 C

)

1000

9

2

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Sizing Cables Conduit and Trunking REV4.1 32 Example

A circuit supplying a DB where a multi core armoured cable is clipped direct using 50 metres of multi core 25.0mm² 70°C armoured thermoplastic insulated cable. The bunched CPC conductor size is 16.0mm². The Ze is 0.5Ω. Calculate the earth loop impedance in Ohms at the maximum operating temperature.

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

Ze = 0.5Ω; 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 25.0mm² and 16.0mm² is 1.877mΩ/m.

iii) Obtain the multiplier value from table 9C

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

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

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

Ω

=

×

=

+

50

1 .

20

0 .

11 1000

877 .

1

2

1 R

R

Ω

=

+

=

0 .

5

0 .

11

0 .

61 Zs

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

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

Table 9C – Used so the designer can give values of resistance at the conductor’s maximum operating temperatures

(

R

1 R

2 )

Ze

Zs

=

+

₍

R

Table

A

L

Table

9 B

or

9 C

₎

1000

9

2

1 +

=

⋅

×

⋅

(33)

1. A circuit supplying a DB using 50 metres of multi core 70°C armoured thermoplastic insulated cable. The line and earth conductors are both 25.0mm². What is the expected Zs at the maximum operating temperature?

2. A motor circuit where single core conductors are installed in p.v.c conduit using 40 metres of 90°C thermosetting single core. The line and earth conductors are both 4.0mm². What is the expected Zs at the maximum operating temperature?

3. A cooker circuit where 10m of cable is installed in building fabric using 70° thermoplastic (p.v.c) insulated and sheathed flat cable with protective conductor. The line and earth conductors are 6.0mm² and 4.0mm² respectively. What will be the expected Zs at 10oC?

4. A power circuit where a 2.5 mm² cable is installed on a tray using multi core 90°C armoured thermosetting insulated cable. The circuit length is 80 metres. What will be the expected Zs at 5oC?

Complete the following exercise to determine the earth loop impedance (Zs) of the circuits. Assume in all cases that the Ze = 0.3 Ω. 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 and show what Zs is for each circuit.

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Earth fault current

Once the Zs has been calculated we then calculate the earth fault current (If) using the calculation below and then ensure that the device will disconnect within the given time 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 be sure that it will disconnect in time so that it provides protection from indirect contact. We can also ensure that the protective device’s short circuit fault current capacity has not been exceeded.

Example

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

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 the fuse to operate will be about 0.15 s. Any load current in the circuit will be additional to the fault current and will cause the fuse to operate slightly more quickly.

However, such load current must not be taken into account when deciding

disconnection time, because it is possible that the load may not be connected when the fault occurs. Therefore if the earth loop impedance is higher this will restrict the flow of fault current meaning the protective device will take longer to operate.

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

I f - term used to describe a circuit’s 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

Uo

If =

amps

Zs

Uo

If

143 .

75

6 .

1

230 =

=

(35)

1. A circuit with a BS3036 overcurrent protective device rating of 30A has an R1+R2 value measured at 0.79 Ω.

2. A power circuit with a BS88

overcurrent protective device rating of 32A has an R1+R2 value

measured at 0.56Ω.

3. A BS88 63A fuse protects a feed to a three-line socket. The measured R1+R2 value is 0.2 Ω.

4. A lighting circuit is protected by a BSEN60898 – 10amp Type B MCB. The measured R1+R2 value is 1.5Ω.

5. A radial power circuit is protected by a BSEN60898 – 32amp Type C MCB. The measured R1+R2 value is 0.8Ω.

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

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Time / current characteristics and disconnection times

In appendix 3 of BS7671 there are graph-like tables that represent the time / current characteristics of the main types and rating of circuit protection. They are used to determine 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 the value of each subdivision depends on the major division boundaries into which it falls.

For example, on the current scale, all the subdivisions between 10 and 100 are in quantities of 10, while the subdivisions between 100 and 1000 are in quantities of 100 and so on. This also occurs with the time scale, subdivisions between 0.01 and 0.1 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 (Χ-axis) and along the side (Y (Χ-axis) is the time in seconds.

Also available on each graph is a quick reference table that displays the main disconnection times and the required amount of current to achieve those times. Each line between two

points represents a value in the lower of the two units

Three lines above the number 10 represents the value 40…… Remember:

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

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Example

Find out the expected disconnection time of a circuit that is protected by a 30A BS3036 fuse when 70 amps of fault current flows.

i) Obtain the correct time/current graph for the selected protective device from 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 Χ axis and follow it to the point where it crosses the selected fuse rating curve.

Identify 70 amps on the Χ-axis. Follow 70 amps upwards until it crosses the 30amp curve.

iii) At the point where the fault current crosses the fuse rating curve follow the line 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 axis and obtain the time. The disconnection time is expected to be 20 seconds.

Over-current device and rating Prospective current (Amps) Disconnection time (seconds) BS3036 - 30 amp 100 BS3036 - 20 amp 70 BS88 – 32 amp 260 BS88 – 63 amp 450 BSEN60898 – 10amp Type B 300 BSEN60898 – 32amp Type C 300 BSEN60898 – 32amp Type 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 is higher than the maximum allowed? What can the designer do to ensure that shock protection is afforded?

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

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Thermal Constraints

Now that you have chosen the type and size of cable to suit the conditions of the installation, we must look at ‘thermal constraints’. This is a check to make sure that the 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 fault current of hundreds or thousands of amperes could flow. Imagine that this is a 1 mm²

or 2.5 mm²cable; if this large amount of current was allowed to flow for a short period of time, i.e. a few seconds, the cable would melt and a fire would start. The c.s.a of the circuit protective conductor (c.p.c) is of great importance since the level of possible shock in the event of a fault depends on it.

The cable calculation process at this point has ensured the selected cable will safely carry the design current and will disconnect in the required time therefore protecting the circuit from danger. The final stage of the calculation process is to confirm that the circuit protective conductor can withstand the fault current for a short space of time, 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 than that detailed in is perfectly adequate. The adiabatic equation is:

By using Ohms law the earth loop impedance value and the supply voltage are used to calculate how much fault current (I) will flow.

The disconnection time of protective device is determined by using the fault current value and the correct time/current curve table from Appendix 3 of BS7671.

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, in mm².

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 conductor material, and the appropriate initial and final temperatures. (See tables

54.1 to 54.6)

k

t

I

S

=

×

2

We need to check that the c.p.c will be large enough to be able to carry this fault current without causing any heat/fire damage. The formula that is used

to check this situation is the adiabatic equation. The c.p.c will only need to carry the fault current for a short period of time, until the protective device

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Example

Find out the minimum size of c.p.c in mm² so that the 50A BS88 circuit complies with BS7671. The protective conductor is a copper conductor incorporated in a 90°C Thermosetting conductor 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 declared t-is established using the graphs in Appendix 3 of BS7671 k- 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 I² x t first; then square root; then divide by the k value

The minimum size of c.p.c. is 3 mm². If the actual selected cable is more than this then thermal constraints have been satisfied. If the c.p.c is smaller than this it will not handle 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 be acceptable by calculation

So you can gain some more understanding complete the exercise on the next page to calculate the minimum size of C.P.Cs. You will need your “Tables from BS7671 and the on-site guide” appendices.

k

t

I

S

=

×

2 2 2

3

100

1

300 mm

S

=

×

=

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

(40)

Minimum size of c.p.c

1. The protective conductor

installation is steel armour inside a 70°C Thermoplastic (PVC) cable with an 80A BS88 has a calculated fault current of 500 amps.

2. A fault current of 1000A is

calculated on a circuit with a 100A BS88 fuse where the CPC is 70°C Thermoplastic (PVC) not bunched with cables

3. The protective conductor

installation is steel armour inside 90°C Thermosetting conductor incorporated in a cable with a 32A BSEN 60898 Type B has a

calculated fault current of 300 amps.

4. A fault current of 150A is calculated on a circuit with a 30A BS3036 fuse where the CPC is 70°C

Thermoplastic (PVC) bunched with cables

5. The protective conductor installation is Steel conduit containing 70°C Thermoplastic (PVC) with a 32A BSEN 60898 Type C has a calculated fault current of 250 amps.

Complete the following exercise to determine the minimum size of CPC for the circuits using the adiabatic equation. You must show all working out; state the calculated size of CPC and the actual 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 complete and the requirements of BS7671 have been met

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Now complete the questions below

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

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

3. What four correction factors need to be taken into account when calculating the minimum 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 what needs 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 with volt drop? Write down the formula and explain its parts.

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

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

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Sizing Cables Conduit and Trunking REV4.1 42 Full cable calculation

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

Load details Type:

Rating: Voltage:

Cable Type Installation details

Supply details DB type: Ze value Circuit details Length of run: Correction factors:

(43)

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

1. Work out the design current Ib

Using a the correct formula

2. Select the protective device size

Ib ≤ In

3. Obtain all correction factors

Find the values so they can be used in the next calcualion

Ca, Cc, Cg, Ci

4. Calculate Iz

In divided by all factors

5. Select conductor sizes

With Iz use the tables in Appendix 4 and choose.

Iz ≤ It

6. Calculate volt drop

Use design current, mV/A/m value from cable table, the circuit length then ÷1000

7. Calculate R1+R2

Using table 9A, 9B and 9C find the mΩ/m value ÷ 1000 then apply multipliers

8. Calculate Zs

9. Verify Zs value with BS7671

Using the tables in Section 41 ensure calculated value is less than the tabulated value to ensure shock protection

10. Calculate the fault current

Using nominal voltage and Zs

11. Establish the disconnection time

Using the fault current, device type and time / current graphs in Appendix 3 obtain the expected disconnection time

12. Calculate minimum size of CPC

Using the adiabatic formula, input all values (including k factor) and work out min. size of cpc. Then compare with our chosen cpc size

13. Once all above has been verified the process is complete V P I = Cc Cg Ci Ca In Iz × × × =

(

)

1000 / /A m L mV Ib Voltdrop= × × Ω = × × = +R A B C R 9 9 1000 9 2 1 Ω = + + =Ze R1 R2 Zs A Zs Uo I= = k t I S= × 2

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Cable capacities of conduit and trunking

Not only must it be possible to draw cables into completed conduit and trunking systems, 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 a greater 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 in conduit or trunking.

Conduit Capacities

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

The number of cables that can be drawn into or laid in any enclosure of a wiring system must be such that no damage can occur to the cables or the enclosure during installation. 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 is expressed as a percentage and should not exceed 45 per cent.

To calculate the required size of conduit we must first establish the amount of conductors to be installed. We must then obtain the individual terms for individual conductors and multiply them together. If we have 5 cables of one size we multiply the 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 one total. We then refer to the relevant table and discover which conduit term is

compatible. Bear in mind that the conduit term shown is the upper limit of spacing and should not be exceeded under any circumstances.

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

Sizing Cables, Conduit and Trunking