Re-take - Useful Information for 2394 :

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2395 Practical assessment

There are also 2 faults to identify out of a possible 7 that the assessor can put on the rig. This could be something like a low insulation résistance, to an open ring-final-circuit. There are not difficult faults trying to trick you, but rather faults you should pick up if you do the tests and interpret the test results correctly.

 

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You are pressed for time in the practical part and have quite a few things to remember. Sometimes a small thing like forgetting to prove dead in the correct manner could cost you the whole test. If you do not isolate, lock off and prove dead in accordance with GS-38 then you will fail.

 

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A very large number of candidates were unable to demonstrate an understanding of voltage drop and its determination during a periodic inspection and test. This is a common and recurring situation across the 2395-302 series of examinations.

A large number of candidates appeared to be aware of the determination of voltage drop during the design of an electrical installation and then attempt to apply that to a periodic inspection. The main incorrect responses included:

• Being unable to explain why voltage drop cannot be determined by direct measurements at the origin and furthest point of the circuit.
• Stating voltage drop can only be determined using the design calculation mV/A/m x Ib x L ÷ 1000 even when this information is not available for an existing installation.
• No correction for conductor operating temperature where resistance is measured at 20°C

Where the method of measurement of conductors was correctly identified a number of candidates lost marks by incorrectly stating measuring R[SUP]1[/SUP] + R[SUP]2 [/SUP]and not ( R[SUP]1[/SUP] & R[SUP]N[/SUP] ) as required for voltage drop. ◄◄◄◄ :6:

( Vd ) Volt drop using test readings .

“ learning curve only ”

Always try to answer the questions in full using the correct terminology

Using ( R[SUP]1[/SUP] + R[SUP]2 [/SUP])
Often the R[SUP]2[/SUP] ( CPC ) will be a conductor with a smaller CSA than the live conductor(s), this of course will result in the calculation showing a higher voltage drop than there would be in reality ( because the current flow in R[SUP]1[/SUP] & R[SUP]N[/SUP] ) ... live conductor(s) L / N

Providing the R[SUP]1[/SUP] + R[SUP]2 [/SUP]calculation give a result of less than the permitted value of 3% ( 6.9V ) or 5% ( 11.5V ) depending on the type of circuit, then all is good . Remember this is only a check , But where the R[SUP]2 [/SUP]calculation for a smaller CPC gives a higher than permitted volt drop it will be worth tying again using R[SUP]N [/SUP]instead .

As an example let’s take a circuit is wired using a 2.5mm[SUP]2[/SUP] / 1.5mm[SUP]2[/SUP] T&E , it has an R[SUP]1[/SUP] + R[SUP]2 [/SUP]value of ( 0.6Ω ) and the circuit is protected by a 20A circuit-breaker

Using the R[SUP]1[/SUP] + R[SUP]2 [/SUP]value of ( 0.6Ω ) you can now calculate the voltage drop for the circuit . (( 0.6 x 20 x 1.20 = 14.4V )) This is to high, if you use R[SUP]1[/SUP] + R[SUP]N [/SUP]you may end up with an acceptable result

How to calculate R[SUP]N [/SUP]
The problem is you do not have a reading for ( R[SUP]N [/SUP]) but a simple calculation will give you all of the information you need
CSA line x ( R[SUP]1[/SUP] + R[SUP]2 [/SUP]) = R[SUP]2 [/SUP]
CSA line + CSA cpc

To put figures to this :
2.5 = 0.625 x 0.6 = 0.375Ω
2.5 + 1.5mm[SUP]2 [/SUP]

0.375Ω is the résistance of R[SUP]2 [/SUP], therefore if you subtract this value from the R[SUP]1[/SUP] + R[SUP]2 [/SUP] value you will have the value of R[SUP]2 [/SUP]0.6Ω – 0.375Ω = 0.225Ω

The résistance of the 2.5mm[SUP]2[/SUP] line-conductor is 0.225Ω
Therefore the (( 2.5mm[SUP]2[/SUP] Neutral-conductor will be the same )) if you now double this value you will have (( R[SUP]1[/SUP] + R[SUP]N [/SUP] )) 0.225 + 0.225 = 0.45Ω

Now you can carry out the voltage drop calculation using 0.45Ω as the résistance value . 0.45 x 20 1.20 = 10.8V

3% ( 6.9V ) or 5% ( 11.5V )

 

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value . 0.45 x 20 x 1.20 = 10.8V

 

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Voltage drop in conductors

To check the suitability of the current carrying capacity it is simply a matter of looking at the installation method , and then checking on the current carrying capacity tables for the cable in Appendix 4 of BS-7671

To ensure that the cable meets the voltage drop requirements is slightly more complex , A simple method is to measure the voltage at the origin of the circuit , and then measure the voltage at the end of the circuit with the load connected and switched on , The difference between the two measurements will be the voltage drop .

if the first method is impractical, then a résistance test should be carried out between the Line and Neutral of the circuit. This test is carried out using the same method as the R[SUP]1[/SUP] + R[SUP]2 [/SUP]test although of the test being between Line and CPC (( it is between the Line and Neutral for the circuit ))
Once the résistance of the R[SUP]1[/SUP] + R[SUP]N [/SUP]circuit has been measured it should be multiplied by the current that will flow in the circuit , This will give you the voltage drop for the circuit .

Example . A circuit is wired in 2.5mm[SUP]2[/SUP] & is 25 metres in length . The current in the circuit is 18 amps
Measured value of résistance is 0.37Ω

Voltage drop = I x R = V .. 18 x 0.37 = 6.66 V

 

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All domestic circuits are now effectively protected by RCD's .. so a limit here

2394

Where the method of measurement of conductors was correctly identified a number of candidates lost marks by incorrectly stating measuring R[SUP]1[/SUP] + R[SUP]2 [/SUP]and not ( R[SUP]1[/SUP] & R[SUP]N[/SUP] ) as required for voltage drop. ◄

R[SUP]1[/SUP] + R[SUP]N[/SUP] .. as required for voltage drop .. Elementary T&E cable L / N same size, Your Q the protective-conductor is smaller in size

for any given circuit cross-sectional area then there is résistance , if you make the circuit longer then résistance increases.

more length equals higher résistance so that will influence voltage drop under load and also the effective circuit impedance that in turn determines if circuit breakers operate in the event of fault.

Re-cap
Continuity of ring-final-circuit-conductor(s)
A Three-step-test is required to verify the continuity of the , Line , Neutral , & protective-conductor(s)

R[SUP]1 [/SUP]+ R[SUP]N [/SUP] .. Step 2 : is done to help confirm polarity

Step 3 ( R[SUP]1[/SUP] + R[SUP]2 [/SUP]) at the sockets. i.e. ( r[SUP]1 [/SUP]+ r[SUP]2 [/SUP]) /4 = R[SUP]1[/SUP] + R[SUP]2 [/SUP]. Remembering that you only record the highest R[SUP]1[/SUP] + R[SUP]2 [/SUP]reading

Live-conductor’s , meaning Line & Neutral .

Radial-circuit , This test is usually repeated for the résistance of both the line & neutral-conductors together ( R[SUP]1[/SUP] + R[SUP]N [/SUP]) is done to help confirm polarity

On a radial circuit, this gets higher the further away you get from the source, as the cables have more résistance as they get longer.

 

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Feedback on candidate performance 2394.

These types of responses indicate that the candidates were either not in possession of suitable knowledge or failed to consider and understand the requirements of the questions.

Candidates should also be aware that where questions carry high marks these require a more detailed response, for example, a three word statement is not going to achieve 10 marks.

When things go pear shape

A large number of candidates were unable to correctly explain the effect on an RCD of a line to neutral fault.
Most candidates believed this would result in the operation of the RCD and very few identified that the RCD would operate in the event of a fault between live conductors and earth

 

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Useful Junk

What’s in a name

MICC / MI / Pyro / MIMS

MICC
Mineral-insulated copper-clad , cable
copper tube and filling the intervening spaces with dry magnesium oxide powder.

colloquially known as pyro (because the original manufacturer and vendor for this product in the UK was a company called Pyrotenax .
MICC cable is made by placing copper rods inside a circular copper tube and filling the intervening spaces with dry magnesium oxide powder.

Joistripper in the barrel each hole is for a different size of cable .
joystripper is used for stripping the cable sheath on popular cable sizes; 2L 1 2L1.5 2L2,5 3L1 3L1.5 4L1 4L1.5

The tool is set up at 2L 1.5 , first number determines (2) how many conductors are inside the cable . Hence (2) conductors inside the cable
The ( L ) is for light gauge .. this means that this cable can handle up to 600V
The ( 1.5 ) is the size of the conductors inside the cable . 1.5mm

Note : When cutting the micc-cable use a junior hacksaw , Do not use pliers as this flatten the cable .

Dry magnesium oxide powder
MICC insulated cables can , Absorb moisture if not correctly terminated resulting in reduced insulation which will cause the protective-device to trip as a certain amount of outgoing line current will return through the MICC cable sheath

 

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What is the purpose of Insulation résistance tests

provide an indication of any condition of any insulation used to provide basic protection against electric shock , and prevent short-circuits and earth faults

Insulation testing should identify the problem

The most effective way of testing for line faults in any wiring , is by measuring the insulation résistance
The most effective way of testing for neutral faults in any wiring , is by measuring the insulation résistance
The most effective way of testing for earth faults in any wiring , is by measuring the insulation résistance

 

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Note that the term ‘live’ conductor includes both the line and neutral conductors. - residual current , meaning two

RCDs detect differences in electrical flow between the line and neutral conductor(s) in an electrical circuit and open (disconnect the supply) when the imbalance is too big.

A basic principle of any electrical circuit, or appliance, is ‘ what goes in must come out ’; electrical flow in equals electrical flow out.
circuit-protective-conductor connects to different parts of an appliance, or circuit, to ensure that they do not become ‘live’.

RCD checks the difference in electrical flow between the line and neutral conductor(s) the flow should be the same , what goes in must come out.

Neutral to earth fault
floorboard(s) nail, screw, picture hooks driven between neutral & earth conductors creates a neutral to earth fault
Requirements to protect cables from impact and penetration are given in regulation 522.6.100 , 522.6.101. (iv) :icon_bs:

 

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2394 My reason is -&-s have been asking the Qs on Voltage drop

Voltage drop in consumer installation’s

Example for a Private Network Supply:

A 10KW single phase load requires a minimum of 220 volts to operate correctly. The final circuit is a 63 amp protected circuit supplying the load via a 2 core 10mm2 PVC/PVC XLPE SWA armoured cable. The final circuit length is 30 metres and the constant load current is 52.17 amps. The Vd/A/m figure is 4.7 (Table 4E2B of BS-7671:2008). :icon_bs:

Maximum voltage drop for the final circuit is 5% (from (i) of the Table above). The note below the table says you must use Public Network figures on Private Network final circuits.

Voltage drop on final circuit:

4.7 x 52.17 x 30/1000 = 7.36 volts. This equates to 3.2% of the nominal voltage, which is below the maximum permitted 11.5 volts (5%).

The load only requires 220 volts to operate, so the minimum voltage we require at the distribution board is 220 + 7.36 = 227.36 volts.

If the Private Network transformer has a single phase open circuit voltage of 245 volts, we have available 17.64 volts for use on the distribution circuit(s) design. This equates to 7.6% of the nominal voltage (230v), which makes the total voltage drop 10.9%. This is below the 14% figure given above, which takes into account the permissible tolerances on the DNO supply.

It can be seen from this, the lower the open circuit transformer voltage, the less the designer has available to him for calculating circuit voltage drop in his design.

[h=3]Key Factsheet - Voltage Drop in Consumer Installations :50:[/h]

 

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it has been down loaded for learning curve . better understanding only .

Voltage drop in a consumer’s installation can be a contentious issue. Nevertheless, it is an important aspect of installation design because, if it is too high, certain equipment will either not function correctly or not function at all.

BS-7671 Requirements: :icon_bs:

525.1 In the absence of any other consideration, under normal service conditions the voltage at the terminals of any fixed current-using equipment shall be greater than the lower limit corresponding in the product standard relevant to the equipment.

525.100 Where fixed current-using equipment is not the subject of a product standard the voltage at the terminals shall be such as not to impair the safe functioning of that equipment.

525.101 The above requirements are deemed to be satisfied if the voltage drop between the origin of the installation (usually the supply terminals) and a socket-outlet or the terminals of fixed current-using equipment, does not exceed that stated in Appendix 4 Section 6.4.

It is important to note that the main criteria in 525.100 is the safe functioning of the equipment which means that, providing the equipment can operate safely and function correctly at its supply voltage, there is no limit on the voltage drop in the system. This is also important where voltage optimisation equipment is utilised.

The designer should be aware that Appendix 4 (referred to in Regulation 525.101) provides one method of complying with BS 7671 requirements. However, other methods that take into account permissible system tolerances are equally valid.

It should also be noted that BS 7671 appendices provide guidance and are non-regulatory.

It is important when designing an installation, to assess the characteristics of the equipment being installed. In particular, the designer should identify the equipment manufacturers’ recommended operating voltages and ensure that they can be

achieved. Circuit cable conductor sizes are then calculated and selected to ensure that the total voltage drop from the origin of the installation is such that, under full load conditions, the lower voltage limits recommended by the equipment manufacturers are maintained. In the event that the minimum voltage cannot be achieved it may be necessary to provide protection against under-voltage or voltage fluctuations.

The following does not take account of any spare capacity that may be required within the total voltage drop assessment process. The designer should discuss such requirements with the client before the assessment is undertaken.

The Origin of the Installation: For installations supplied from the Distribution Network Operator (DNO) low-voltage Public Network, the origin is normally the point at which electricity is supplied to the premises; e.g. the service cable at the intake cut out and metering point.

refer to . Link above

 

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it has been down loaded for learning curve . better understanding only .

Voltage drop 2008: :icon_bs:

The requirements concerning voltage drop, set out in Regulations 525.1 / 525.3. relate only to the (( Safety issues of electrical equipment performance ))

They do not address the other operational requirements, which may include . example

Efficiency, it may be that compliance with BS-7671 in this respect is not the only consideration in assessing voltage drop : some equipment may operate at less than its optimum efficiency at voltages permitted by BS-7671: some form of lighting sources may have considerably reduced lifespan and / or efficiency at voltages other than those prescribed by the manufacture . In some cases, the operational considerations may place more stringent limits on voltage drop than the specified requirements. Some cases have been reported where a reduction of 5% of nominal voltage reduces the efficiency of particular lamps by as much as 20%, When one considers that a distributor is permitted a tolerance of + 6% to 10% of the nominal declared voltage , it can be seen that the designer may find difficulty in providing supplies to such circuits if optimum efficiency is to be achieved. From the safety standpoint, the basic requirements, embodied in Regulation 525.1. & 525.2. is the voltage supplied to the current0using equipment provides for safe operation ( Cont )

 

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Appendix 12 ( informative )
Voltage drop in consumer installations .. Deleted by BS-7671:2008: Amendment No 1 . content moved to Appendix 4 sec 6.4.

it has been down loaded for learning curve . better understanding only .

Voltage drop 2008: :icon_bs Cont )

This may be ascertained by reference to the relevant standard where that standard has been addressed the safe functioning requirements , E.g. BS-EN-60335 ) Regulation 525.3.
[h=4][/h][h=4]What is BS EN 60335-1:2012+A11:2014[/h]BS EN 60335-1:2012+A11:2014 gives general requirements to ensure the safety of electrical household appliances – providing their rated voltage is not more than 250 V for single-phase and 480 V for other appliances. These best practice recommendations for electrical safety look at common hazards of household equipment or electrically operated devices that could cause injury to persons in and around the house. The use of appliances by unsupervised children, or young children playing with electrical household equipment, is not covered in this standard.
[h=4][/h][h=4]How does it work?[/h]BS EN 60335-1 looks at the general requirements and conditions to test the domestic safety of electric household appliances. It also defines the classification and marking of electrical equipment, and demonstrates how to ensure protection against live parts. The standard explains heating, void as well as leakage currents and electric strength at operating temperatures. Other topics include moisture resistance, stability and mechanical hazards, internal wiring and connections.

( Cont ) 525.3. provides a deemed to comply “ status provided the voltage drop from the origin of the installation ( supply point ) to the terminals of all current using equipment or to socket-outlets is not greater than stated in Appendix 12 of BS07671: for LV installation supplied directly from a public LV distribution system , the limiting voltage drop stated in Appendix 12 expressed with respect to the nominal voltage , is 3% for lighting equipment and 5% for other uses .

At these limits , the voltage at the current using equipment and 5% for other uses , At these limits, the voltage at the current using equipment may be 91% or 89% of the nominal voltage respectively , i.e. 230V – ( 6 + 3 ) % = 209.3V or 230V = ( 6 + 5 )% = 204.7V
As permitted by Regulation 525.4. a greater voltage drop than prescribed in Appendix 12 may be permitted in the case of motor under starting conditions and other current using equipment having high inrush current .

 

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Section and Erection of wiring systems P/121 (2011)

For domestic and similar installations, The origin of the installation is clearly the supply terminals .