Re-take - Useful Information for 2394 :

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Blast from the past )) 16[SUP]th[/SUP] Edition only in this article ◄◄

Accessibility of equipment mg_smile:
RCDs devices is very important if the end user is to be able to operate the trip button on the device , for compliance with BS-7671:

Positioning of RCD
Regulation 513-01-01 which states that every piece of equipment that requires operation or attention by a person shall be so installed that adequate and safe means or access and working space are afforded for such operation or attention

 

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RCDs are used to provide protection
• fault protection
• additional protection
• protection against fire

An RCD is a protective device used to automatically disconnect the electrical supply when an imbalance is detected between live-conductors.
In the case of a single-phase circuit, the device monitors the difference in currents between the line and neutral conductors.

The term ‘ live ’ conductor includes both the line & neutral conductors .. residual current

Additional protection - RCD test (5 x ) 500% fast test

Fault current IΔn .. Sensitivity - instantaneous only (A) 0.03A
RCD .. In) rated current of the contacts, expressed in amperes . e.g.

The testing device then allows a larger fault current (normally 150mA) to flow. This should trip the RCD more quickly, again within a predetermined time for the type of RCD and supply.

An RCD does not provide protection against Overcurrent
Overcurrent protection is provided by a circuit breaker

Residual current operated circuit-breaker (RCBO) with integral overcurrent protection
O.S.G.
• Overload
• Short circuit
• Earth fault

Manufacture details & information.
Residual current operated circuit breaker with Integral Overcurrent Protection ( RCBO)

A residual current operated circuit-breaker designed to perform the functions of protection against :
(Iƒ) fault current
• Short circuit
• Earth fault
• Overload
These devices comply with BS-EN-61009-1

Technology : There are two-types of residual current devices
• Electromagnetic .
• Electronic .

- Electronic devices do not need such a sensitive torroid as electronic circuits within the device amplify the signal to operate the trip relay .
However, these devices often require a safety earth reference lead to ensure that the device will continue to operate in the event of the supply neutral being lost , The power to trip the device is taken from both the fault-current and the mains supply .

(RCBOs) These devices should be disconnected whilst carrying out Insulation résistance to prevent damage to the device and to avoid incorrect test results .

Regulations RCDs
States : The test results should be within the parameters set out in Table 3A .. “ performance criteria ”

Standards’
Only - BS-4293 had a limit of 200mS .. 1983 (1993) replaced by: still out there
(G) BS-EN-61008-1 ( RCCB )
(G) BS-EN-61009-1 (RCBO)

5 x IΔn (500%)
when periodically testing a RCD , it is often advisable to undertake the 5 x IΔn test first .
if the device fails this test , it is a good indication that the quarterly test button check has not been carried out and ? dirt or dust on the contacts may be causing a slower operation than normal or ( just seized up )

Before you reject the RCD , completely, the test should be carried out twice more, as it is quite usual for the device to operate properly after these two-subsequent operations

 

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2394: -&-s
Q) Which test required to confirm the correct operation of a 30mA RCBO installed to provide additional protection .
A) Operation of residual current devices .
Com) The operation of the RCD component of the RCBO must be tested to confirm correct operation . as required in GN-3 .

RCD.s Type of Test’s
½ or 0.5 ) 15mA no trip .. how sensitive is your RCD
performs a no-trip as half the rated current 15mA . / 30mA

Sine wave
~ 0° trip test . The test is always started on a Zero , crossing when the instantaneous voltage is on the rise .
~ 180° trip test . The test is always started on a Zero , crossing when the voltage is on the fall .

30mA - 5 x IΔn . as aptly named . Fast trip ( 5 x 500% times ) 40mS

150mA / 40mS

Q) 30mA BS-EN-61009-1 RCBO is installed on a socket-outlet circuit to provide additional protection , which of the following identifies the maximum test current to be applied and the maximum disconnection time at that test current . A) 150mA & 40mS .. 5 x 30mA = 150mA

The RCBO is provided for additional protection and the 5 x IΔn is the maximum fault current that needs to be applied .
As this device is providing additional protection , the maximum disconnection time is 40mS , as required by BS-7671: regulation 414.1.1.

Ramp test : “ by choice ” I always do a ramp test whenever I test a RCD
5x and x5 are the same thing
150mA or 0.15amps

 

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Checking Voltage Drop
For lighting-circuits the maximum volt drop is 3% (6.9V)
&
For all other circuits the maximum voltage drop is 5% (11.5V) “ power ”

BS-7671 round up for simplicity

Learning curve . Example only

4D5 : Conductor operating temperature at full load for thermoplastic cable ( PVC ) is 70°C
(T&E) Flat cable with protective-conductor .. copper conductors

Note : We all know that 10mm[SUP]2[/SUP] for Shower-circuit . (4.4 ) mV/A/m

Method (C) column ( 6 & 8 )
:icon_bs: 4D5 : (mm[SUP]2[/SUP]) 10.0mm[SUP]2[/SUP] cablewill carry 64A .. clipped direct , So the voltage drop is 4.4
length 15m
Current is 10000 ÷ 230V is = 43.5A

( 4.4 x 43.5 x 15 ) / 1000 = ? .. by calculation 4.4 x 43.5 x 15 ÷ 1000 = 2.9V is it within the limit ??

 

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Insulation resistance tests should be carried out using the appropriate D.C. test voltage specified in Table 61 of BS-7671:
Instruments’ conforming to BS-EN-61557-2 .. ( part 2 )

These tests are to verify that for compliance with BS-7671: :icon_bs:

Insulation résistance test . (IR)
Testing . i) can be carried out on a complete installation .. ( Whole )

2011: GN-3 : The inspector will need to measure the values of insulation résistance for a given distribution board and then take a view based on his / her engineering judgement as to whether the results obtained are acceptable

it should be noted that distribution boards with large numbers of final circuits will generally give a lower installation résistance value than distribution boards with fewer final circuits .

Testing . ii) A section of the installation
Testing . iii) Single circuit ..

612 Testing
firstly :- if any test indicates a fail to comply that test and any preceding test results of which may have been influenced by the fault indicated
Secondly :- shall be repeated after the fault has been rectified .

GN-3 : (IR)
i) The purpose of the insulation résistance test is to verify that the insulation of conductors provides adequate insulation .
ii) is not damaged and that live-conductor(s) or protective-conductor(s) are (( not short-circuited ))

Note GN-3 2008: ◄
Extracts
Although an insulation résistance value of not less than 1.0MΩ complies with the Regulation’s
Where an insulation résistance of less than 2MΩ is recorded the possibility of a latent defects exists .

In these circumstances, each circuit should be tested separately
This will help indentify :
i) whether one particular circuit in the installation has a lower insulation résistance value, possibly indicating a latent defect that should be rectified .
or
ii) whether the low insulation résistance represents, for example, the summation of individual circuit insulation résistance and as such may not be a cause for concern .

 

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Test your batteries firstly on your Meggers before you go on site .

Battery exhausted :mad2: ( O S*** ) can happen to the best of Us .
All testing will be inhibited in the event of a flat battery .

 

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Guidance on how to comply with BS-7671. One such document, Guidance Note 3, Inspection and Testing, gives specific guidance on how inspection and testing should be carried out.
[h=3][/h][h=3]Don't lose sight of what it is you're trying to achieve :yesnod:[/h]Which of these should a Minor Electrical Installation Work Certificate NOT be issued for ?
a) New circuit
b) Moving a light position .
c) Adding a new lighting point to an existing circuit .
d) Adding an extra socket-outlet to an existing circuit .

Minor Electrical Installation Work Certificate . :icon_bs:

Notes : p/394
The Minor Work Certificate is intended to be used for:
Additions and alterations to an installation that do not extend to the provision of a new circuit .◄

Example .
include the addition of socket-outlet(s)
Lighting points to an existing circuit .
the relocation of a light switch .

This Certificate may also be used for the replacement of equipment such as accessories or luminaires .

Note : Not for the replacement of distribution boards or similar items .

 

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Why do we lose valuable Marks : Candidates often identified test instruments using incorrect titles. The titles of Instruments must be in line with those given in GN-3.

Instruments GN-3. the operative knows the limitation and use of the instrument(s) “ in Theory ”

Don’t lose sight of what it is you’re trying to achieve .

Q ) For what purpose should an Electrical Installation Certificate be issued and who to ?

Answers should include :
i) The Electrical Installation Certificate is to be used only for the initial certification of a new installation or for an addition or alteration to an existing installation where new circuits have been introduced .

ii) It is not to be used for a Periodic Inspection , for which an Electrical Installation Condition Report should be used .

iii) It should be issued to the person ordering the work .

Extracts .
Condition Report .
GN-3 . reminds us , This Report is an important and valuable document . etc

This report should only be used for reporting on the condition of an Existing electrical installation .
The purpose of this Condition Report is to confirm , so far as reasonably practicable, whether or not the electrical installation is in a satisfactory condition for continued service .

The Report should identify any damage , deterioration , defects and / or conditions which may give rise to danger

Facts :- Why do we still have to issue BS-7671 electrical installation certificates ?

For reasons of safety

BS-7671 required all electrical work to be inspected, tested and certified long before Part P came into effect .

Facts :- ( EIC )
This Safety Certificate has been issued to confirm that the electrical installation work to which it relates has been . ETC

Knowledge of BS-7671 and Guidance Note 3 .. 2014

One question required the candidates to list the three documents that must be completed and handed to the client on completion of an initial verification of an installation.

A number of candidates were unable to correctly identify the three documents

Electrical Installation Certificate .. (EIC)
Initial verification , BS-7671: requires that an (EIC) together with a
i) Schedule of test results
ii) Schedule of inspection ... (for new installation work only)

This certificate is only valid if accompanied by the Schedule of Inspection and the Schedule(s) of Test Results

be given to the person ordering the work

Schedule of test results

The schedule of test results is a written record of the results obtained when carrying out the electrical test(s) required by Part 6 of BS-7671:
( Generic ) Schedule of test results, The following notes give ( guidance on the compilation ) of the “ Schedule ”

BS-7671: Minor works are defined as :
To be used only for minor electrical work which does not include the provision of a new circuit .

Q) Which test is not normally required for an initial verification ? 612.14. Voltage drop

 

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Lighting calculations’

Current loading : florescent lamp ratings must be multiplied by 1.8 to take into account control gear losses .

florescent :- 22 lamps x 70W x 1.8 ÷ 230V = 12.0A

Extra low voltage ( spotlights ) 17 x 50W ÷ 230V = 3.7A

O.S.G. table A1 p/110 .
Notes 2 .
2 . final circuits for discharge lighting must be arranged so as to be capable of carrying the total steady current, viz. that of the lamp(s) and any associated controlgear and also their harmonic currents.

Where more exact information is not available , the demand in volt-amperes is taken as the rated lamp watts . multiplied by not less than 1.8 .

This multiplier is based upon the assumption that the circuit is corrected to a power factor of not less than 0.85 lagging, and takes into account controlgear losses and harmonic current .

 

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Regulation :icon_bs:; there are tables giving the values of voltage drop for various types and sizes of conductor(s)
These values are given in millivolts (mV) for every ampere (A) that flows along a length of 1 metre (m) .. mVA/m .

So ; you should be able to check , that résistance, and hence voltage drop , reduces with an increase of (CSA)
10.0mm[SUP]2[/SUP] conductor should have ten times less of a voltage drop than a 1.0mm[SUP]2[/SUP] conductor .

p/333 . Table 4D1B , column 3 confirms this ; the millivolts drop for 1.0mm[SUP]2[/SUP] being (( 44 mV )) and that for 10.0mm[SUP]2[/SUP] being (( 4.4 mV))

-&-s Q) Which of the following factors directly affects the conductor résistance of a cable . (( length and CSA ))
Insulation and CSA
b) Length and CSA
Length and insulation
Load current and CSA

Of the four options given (b) is the only one in which both components will directly affect the conductor résistance .
Load current will only have an indirect effect on conductor résistance by raising the conductor temperature .

-&-s Q) Two identical cables are connected in parallel. Which of the following describes the effect on the combined conductor résistance compared to the single cable. (( Halved ))

C) The two cables will have the same résistance, which when connected in parallel will result in their combined résistance being (( half )) that of one single cable .

 

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Basic understanding.

Insulation résistance (IR) ( Whole installation’s )

► The more resistances there are in parallel, the lower the overall résistance, and the longer a cable the lower the insulation résistance . ◄ :30:

On large installation’s may give , if measured as a (( Whole )) low value’s , even if there are no faults.

GN-3 : 2008:2011:
re-cap . To perform the test in a complex installation it may need to be (( subdivided ))

 

[Richard Burns]

For Amberleaf
Basic’s of resistance




Long-cable has a bigger resistance than a short one of the same material and diameter.
Thick-cable has a lower resistance than a thin one of the same length and material.

 

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Learning curve . ( Ia / 20A )

BS-7671:2011: supplementary-protective-conductor-bonding . tell us ( Where required ) :book: :icon_bs:
it is not unusual to see supplementary-protective-conductor-bonding in places where it is not really required .

2011: Protective device 20A . BS-3036 rewirable fuse .

To find the current which will automatically operate this device in (5 seconds ) Appendix 3 / 3A2(b)
This shows that the current required to operate the fuse in (5 seconds ) is 60A

The value can be found by using the maximum Zs value for 20A fuse . Table 41.4. - 3.38Ω
Calculation . Ia = 230 / Zs ( Ia = 230V ÷ 3.38Ω = 60A )

This value can now be used to verify if the area requires protective supplementary bonding or not .
Calculation is: R = 50V / Ia ( R = 50 / 60 ) R = 0.83Ω
The maximum permitted value between exposed or extraneous-conductive-parts , in the area is 0.83Ω

if the measured résistance is higher than 0.83Ω then protective-supplementary-bonding will be required .
The résistance values used will be different depending on the type and ratings of the protective device which is being used for protection of the circuit .

Where an RCD is being used for protection .
used to find the maximum résistance permitted between exposed or extraneous-conductive-parts before protective bonding is required .

Calculation is: R = 50V / IΔn .
Trip rating of the RCD is (In) 30mA .. ( R = 50 ÷ 0.03A ) R = 1666.666667Ω . round up - 1667*
Providing the résistance between parts is 1667* or less . protective supplementary bonding would not be required .

Determining if a metal part is extraneous or just a piece of metal .
Very often it is impossible to tell whether a metal part is extraneous or not .

Test , using an (( insulation résistance tester )) . MΩ voltage 500V d.c.
One lead from the tester must be connected to a known ( Earth) and the other lead connected to the metal part .
measure the résistance . if the value is less than 0.02MΩ ( 20.000Ω) then bonding is required as the part would be deemed an extraneous-conductive-part

Where the résistance is found to be above ( 0.02MΩ ) it is just a piece of metal and it will not need to be bonded . “ Ohms law ”
V / R - I . ( 500V ÷ 20000Ω = 0.025A ) What’s in a decibel point. 500V ÷ 20.000Ω = 25.A

This show you that a current of 25mA could flow between conductive-parts . is at 500V . if the fault was on a 230V supply then the current flow would be half . which would be 12.5mA ( 0.012A)

This test should (( Not )) be confused with a Continuity test .

 

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Learning curve .

Testing of protective-bonding-conductor(s)
Main-protective-bonding-conductor .. (( Equipotential Bonding ))

This test is carried out to ensure that the protective bonding conductors are unbroken, and have résistance low enough to satisfy the requirements of BS-7671:

The purpose of the protective-bonding is to ensure that under fault conditions a dangerous potential will not occur between earthed metalwork ( exposed-conductive-parts ) and

Other metalwork ( extraneous-conductive-parts) in a building

Note here : the words used is , ( exposed-conductive-parts ) and ( extraneous-conductive-parts)

it is not the purpose of this test to ensure a good earth path , it is to ensure that in the event of a fault the exposed & extraneous-conductive-parts will rise to the same potential . hence the term “ Equipotential Bonding ”

in order to achieve this it is recommended that the résistance of the bonding-conductors does ( Not ) exceed 0.05Ω

O.S.G. P/39 (4) 2011. :book:
4.3. Main protective bonding of metallic services .
The purpose of protective equipotential bonding is to reduce the voltages between the various ( exposed-conductive-parts ) and ( extraneous-conductive-parts) of an installation, during a fault to earth and in the event of a fault on the DNO network

 

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Learning curve only .

Ring final circuit which is wired in 70°C thermoplastic T&E cable . 2.5mm[SUP]2[/SUP] live conductor(s) 1.5mm[SUP]2[/SUP] circuit-protective-conductor .
47 metres long , end-to-end

( Note : T&E cable Live-conductor(s) are the same size 0.34Ω )

O.S.G. Table 11 .
2.5mm[SUP]2 [/SUP]copper conductor has a résistance of 7.41mΩ per metre .
1.5mm[SUP]2 [/SUP]copper conductor has a résistance of 12.10mΩ per metre .
the cable is 47 metres long .

Step 1 :- to measure the résistance of each conductor end-to-end .. ( complete loop )

Line / line - 7.41 x 47m / 1000 = 0.34Ω ←
Neutral / neutral - 7.41 x 47m / 1000 = 0.34Ω ←
Circuit-protective-conductor / CPC – 12.10 x 47m / 1000 = 0.56Ω ←

These are the value(s) which you could expect providing all of the conductor(s) are connected properly and form a complete ( Loop )

Step 2: cross-connected
The open ends of the line & neutral-conductor(s) are then connected together so that the outgoing line-conductor is connected to the returning neutral-conductor . (( vice-versa ))

GN-3. 2.2b - Connections for testing .
• The résistance between line & neutral-conductor(s) is measured at each socket-outlet
• The readings at each of the socket(s) wired into the ring should be substantially the same and the value will be approximately one-quarter of the résistance of the line plus the neutral loop résistance ( r[SUP]1 [/SUP]+ r[SUP]N[/SUP] )

Note:- re-cap . (( See mathematical explanation in Fig 2.3. )) “ drawing’s ”

measure between line & neutral-conductor at each socket-outlet and the résistance value should be half of résistance of one of the live-conductor end-to-end measurement . The value you would expect . 0.34 ÷ 2 = 0.17Ω

This is because that by interconnecting the end of the ring you have halved the length and doubled the cross-sectional area .

if you have one conductor which has a résistance of 0.34Ω and join another conductor with the same résistance to it , the length will now have a résistance of 0.68Ω

Note :- Conductor(s) in Series . 0.34Ω → 0.68Ω ← 0.34Ω .
if you join other ends together you have halved the length of the conductor
This means that the résistance must be halved . 0.68 / 2 = 0.34Ω

Note :- Conductor(s) in parallel . 0.34Ω 0.34Ω
if you measure the loop from end-to-end you are also measuring double the ( CSA) which of course will result in the value halving again
0.34 ÷ 2 = 0.17Ω

the simple calculation to find the expected ( R[SUP]1[/SUP] + R[SUP]N[/SUP] ) value at each socket-outlet is . ( 0.34Ω + 0.34Ω = 0.68 ) ÷ 4 = 0.17Ω

Step 3.

The same principle applies when you need to calculate the R[SUP]1[/SUP] + R[SUP]2 [/SUP]value at each socket-outlet .

From Step 1 :- to measure the résistance of each conductor end-to-end .. ( complete loop )

Line / line - 7.41 x 47m / 1000 = 0.34Ω
Neutral / neutral - 7.41 x 47m / 1000 = 0.34Ω
Circuit-protective-conductor / CPC – 12.10 x 47m / 1000 = 0.56Ω

The only difference is that in this example the (CSA) of the conductor(s) are different as you have a 2.5mm[SUP]2[/SUP] line-conductor & 1.5mm[SUP]2[/SUP] circuit-protective-conductor can be used . ( 0.34Ω + 0.56Ω = 0.9 ) ÷ 4 = 0.225Ω .

This will be the value of R[SUP]1[/SUP] + R[SUP]2 [/SUP]for the circuit .

► if there is Spur(s) on the circuit that the will be higher at the Spur .

BS-7671: Generic Schedule of Test Results .
* Where there are NO spurs connected to a ring final circuit this value is also the ( R[SUP]1[/SUP] + R[SUP]2[/SUP] ) of the circuit .