Re-take - Useful Information for 2394 :

[amberleaf]

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.

 

[amberleaf]

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

 

[amberleaf]

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

 

[amberleaf]

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

 

[amberleaf]

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:

 

[amberleaf]

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]

 

[amberleaf]

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

 

[amberleaf]

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 )

 

[amberleaf]

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 .

 

[amberleaf]

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

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

 

[amberleaf]

Voltage drop using tables from BS-7671:2011:

Example:
Circuit is wired using 70°C thermoplastic flat T&E cable

- Copper 2.5mm[SUP]2[/SUP] live-conductor(s) & 1.5mm[SUP]2[/SUP] circuit-protective-conductor
- Circuit is 30 metres long
- Carry a current of 17 amperes .. supply voltage 230V

( Vd ) for this cable can be find using :
:icon_bs: Table 4D5 Appendix 4 of BS-7671:2011: P/340
2011: O.S.G. Table F6 . P/150

Using either of ( Table 4D5 or O.S.G. )
4D5 ) The voltage drop for 2.5mm[SUP]2[/SUP] copper cable is ( 8 ) → 18mV/A/m
O.S.G. ) The voltage drop for 2.5mm[SUP]2[/SUP] copper cable is ( 8 ) → 18mV/A/m

18mV/A/m millivolts x amperes x distant in metres ( as value is in millivolts it must be divided by 1000 to convert to Volts )

Circuit is by calculation : 18mV x 17A x 30m ÷ 1000 = 9.18
Point to note : This volt drop value used in BS-7671: has been rounded up for ease of calculation .

 

[amberleaf]

Ohms law is not printed in BS-7671 but it is certainly used :svengo:

Circuit is wired using 70°C thermoplastic ( T&E )
- Copper 2.5mm[SUP]2[/SUP] live-conductor(s) & 1.5mm[SUP]2[/SUP] circuit-protective-conductor
- Circuit is 30 metres long
- Carry a current of 17 amperes .. supply voltage 230V

2011: O.S.G. Table 11
Values of résistance / metre or ( R[SUP]1[/SUP] + R[SUP]2 [/SUP] ) / metre for copper conductors has a résistance of 7.41mΩ per metre at 20°C

The current flowing in a circuit will be the (( Same )) in the Line & Neutral-conductor(s) :yesnod:
Therefore the résistance of both Live-conductor(s) must be taken into account
O.S.G. Table 11 , Line conductor 7.41mΩ / m
O.S.G. Table 11 , Neutral conductor 7.41mΩ / m .. 7.41 + 7.41 = 14.82

Résistance of ( T&E ) copper cable is 14.82 mΩ / m
Total résistance of this cable , mΩ per metre x length .. 14.82 x 30 = 444.6 mΩ

This value is in milliohms and should now be converted to Ohms: mΩ / 1000 = ohms ( 444.6 ÷ 1000 = 0.444Ω )

When conductors are operating their maximum current rating they can operate at 70°C This will result in the résistance of the conductors increasing
This increased résistance must be used in the calculation for (( volt drop ))

To calculate the total résistance of the cables at their operating temperature a factor ( O.S.G. ) Table 11 should be used , a multiplier of ( 1.2 ) should be used for a conductor rated at 70°C .
To calculate the total résistance of the current carrying conductor(s) Ω x multiplier = total résistance of conductor(s) at 70°C . 0.444 x 1.2 = 0.533Ω

These calculations’ ca be carried out in one single calculation . MΩ x length x multiplier / 1000 = total résistance
14.82 x 30 x 1.2 ÷ 1000 = 0.533

Voltage drop can now be calculated using Ohms law . I x R = U .. 17 x 0.533 = 9.06V

 

[amberleaf]

Table 4Ab – voltage drop P/314

re-cap
The maximum voltage drop in any circuit from the origin of the supply , to the terminals of the current using equipment must not exceed ( 3% & 5% ) of the supply voltage

Supply voltage is 230V the calculation to find 3% & 5%
230V x 3% ÷ 100 = 6.9V
230V x 5% ÷ 100 = 11.5V

 

[amberleaf]

Definition of voltage drop is the voltage difference between any two points of a circuit or conductor , due to the flow of current .

 

[amberleaf]

Watch for Question like this -&-s 2015 onwards

Ways in which electrical wiring or equipment can pose a fire-risk. :svengo:

- High resistance & arcing due to loose connections .. PS ; I won’t mention plastic CCUs .
- Overloaded cables

 

[amberleaf]

Inspection :
Examination of an electrical installation using all the senses as appropriate.

Give two inspections that can be carried out using Touch.
Choose from tug-test to check security of terminations, security of fixings and fittings; tracing individual cables; soundness of fixings, lids and covers; using hands to see if anything feels hot or if there is excessive vibration .

Inspection checks that can be carried out using the sense of hearing
Listening for chattering, vibration or crackling sounds .

 

[amberleaf]

Impedance is much more complicated than résistance, taking into consideration that it relates to AC.

Every material has resistance. Copper has a low résistance

Résistance is something that “oppose or resist’s ” the flow of current.

First, the total length of the cables will affect the amount of résistance. The longer the wire, the more resistance that there will be.
Second, the cross-sectional area of the wires will affect the amount of résistance. Wider ( cables ) wires have a greater cross-sectional area

(( Water will flow through a wider pipe at a higher rate than it will flow through a narrow pipe ))
This can be attributed to the lower amount of resistance that is present in the wider pipe
In the same manner, the wider the ( cables ) wire, the less résistance that there will be

Basic’s
Mathematical nature of Résistance .
Cable is directly proportional to the length of the wire and inversely proportional to the cross-sectional area of the cables
Where ( L ) represents the length of the cable (( in meters )) CSA , cross-sectional area of the cable ( in meters[SUP]2 [/SUP])

What is Résistance (( Electrical )) .. “ resist ”
When electrons flow through a bulb or another conductor, the conductor does offers some “ oppose or resist ” to the current. This obstruction is called résistance.
- The longer the conductor higher the resistance
- The smaller its area the higher its resistance

Every material has an electrical résistance and it is the reason that the conductor give out heat when the current passes through it.
Plastic does not conduct electricity. It has a high electrical resistance, Class II

Symbols used in the Regulation’s

Résistance
represented by the uppercase letter (R) The standard unit of résistance is the ohm (Ω) , sometimes written out as a word

BS-7671:
Both résistance and impedance are expressed in unit ohms. Mathematically, however, they are denoted differently. Impedance is often denoted with symbol (Z) while résistance is (R).

(( Circuit loop impedance ))
Following a test of earth fault loop impedance (Zs) the results are compared with the values given in BS-7671:
-&-s 2394 , Which of the following describes the purpose of this comparison ?

purpose of this comparison , You’re Q , To determine whether disconnection times will be achieved under (( earth fault conditions ))

You will have to consult Sherlock Ohms .
The operation of the protective device under earth-fault-conditions relies upon a sufficiently large current flowing .
From Ohms law the supply voltage and impedance of the system will determine the fault-current.
The (Zs value) is the variable and can therefore be used to determine whether disconnection will be achieved within the required time .

 

[amberleaf]

-&-s 2394 . Which of the following factors affect the insulation résistance of a cable . (( Cable length ))

Of the four options given ,
Conductor ( CSA )
Cable length .. is the only one which will have an effect on the Insulation-résistance of the cable .
Installation method
Load current

• Fault-current: current which flows across a given-point of fault resulting from an insulation fault .

 

[amberleaf]

The Load refers to the item that requires the supply in order to function .

Circuits must be designed that are fit for purpose and suitable for the load they are intended to supply. They should be correctly designed in accordance with BS-7671.

Annealed copper
Cable conductors are made of annealed copper, this is copper that has been heat treated to make it tougher

-&-s 2394 : The question they are asking you .
Which of the following factors directly affects the conductor résistance of a cable .

Insulation and CSA
length and CSA .. the only one 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 .
Length and insulation
Load current and CSA

 

[Richard Burns]

For Amberleaf

Resistance ( R ) Suppose you've two water tanks, one with a narrow pipe and one with a wide pipe.

It stands to reason that we can’t fit as much volume through a narrow pipe than a wider one at the same pressure. This is resistance. The narrow pipe “resists” the flow of water through it even though the water is at the same pressure (voltage) as the tank with the wider pipe.

In electrical terms, this is represented by two circuits with equal voltages and different resistances. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.

Ohm. Ohm defines the unit of resistance of “ 1 Ohm ” as the resistance between two-points in a conductor where the application of 1 volt will push 1 ampere, This value is usually represented in schematics with the Greek letter “Ω”, which is called omega, and pronounced “ohm”.


Ohm's Law
Combining the elements of voltage, current, and resistance, Ohm developed the formula: V = I . R
V) Voltage in volts
I) Current in amps
R) Resistance in ohms


This is called Ohm’s law. Let’s say, for example, that we have a circuit with the potential of 1 volt, a current of 1 amp, and resistance of 1 ohm. Using Ohm’s Law we can say: 1V = 1A . 1Ω


Let’s say this represents your tank with a wide pipe. The amount of water in the tank is defined as 1 volt and the “narrowness” (resistance to flow) of the hose is defined as 1 ohm. Using Ohms Law, this gives us a flow (current) of 1 amp.


Using this analogy, let’s now look at the tank with the narrow hose. Because the hose is narrower, its resistance to flow is higher. Let’s define this resistance as 2 ohms. The amount of water in the tank is the same as the other tank, so, using Ohm’s Law, our equation for the tank with the narrow hose is 1V = ? . 2Ω


But what is the current ?
Because the resistance is greater and the voltage is the same, this gives us a current value of 0.5 amps: 1V = 0.5A . 2Ω


Basic understanding (Ohms law)
So, the current is lower in the tank with higher resistance. Now we can see that if we know two of the values for Ohm’s law you can solve for the third

 

[amberleaf]

Thermal insulation
in Theory , This has a similar effect to wrapping a cable in a fur coat on a hot summer’s day and heat produced cannot escape .
contact of cables with thermal insulation is considered as two separate scenarios

• Totally surrounded
• Contact on one side only
​

 

[amberleaf]

Just remember, there’s is a right way and a wrong way to do everything -&-s Wrong way is to keep trying to make everybody else do it right

Which of the following instruments must be used to measure the earth-electrode résistance for a (( standby generator )) ?

• Earth fault loop impedance
• Insulation résistance tester
• Low résistance ohmmeter
• Earth electrode résistance tester

Comments
There is no return part for the test current when testing the earth-electrode for a transformer or generator .
Therefore the test must be undertaking using an earth-electrode résistance tester and not an earth-fault-loop-impedance tester .
The other options are not suitable . :mad2:

 

[amberleaf]

Ensure that the safety of yourself and others is always the priority :svengo:

Earth fault loop impedance .. “ by direct measurement ”

-&-s Direct measurement
This indicates that a test is required and the results are not to be established by using a calculation .

GN-3 . Instruments conforming to BS-EN-61557-3

With (Ze) you are (( Verifying )) that there is an earth-connection into your installation . “ full stop ”

Live tests

There are two tests .
i) measurement of external loop impedance (Ze)
ii) measurement of total earth fault loop impedance (Zs)

(Ze) the installation is isolated from the supply and the ( earthing-conductor ) disconnected to avoid parallel paths .

2394: -&-s
Explain why the earthing-conductor in an installation must be disconnected from the MET when measuring Ze
A) To remove parallel earth paths so that the intended fault path can be confirmed to be reliable .

Com) if the parallel earth paths are not removed during the test then the reliability of the test result is in doubt.

Measurement of (Zs) is made on a live-installation and for safety and practical reasons, ( neither ) the connection with earth nor bonding-conductors are disconnected.

 

[amberleaf]

GN-3: 2008 ◄ Out but not forgotten
Your Q , Direct measurement using an earth-fault-loop-impedance tester

Using the words ; Direct measurement -&-s
This indicates that a test is required and the results are not to be established by using a calculation .

“ Measurement ” 2.7.14.
The reasons that Ze is required to be measured are twofold :

1) to verify that there is an earth-connection . :yesnod:

 

[amberleaf]

((Condition report )) Existing
Doing any Electrical installation condition report is that you need to check that there are at least a couple of grey hairs growing on your head before you start.

Electrical installation condition report is to assess that “ as far as is reasonably practical ” the installation is safe

Initial Verification ..

GN-3 & Regulation’s , Asks us to record the ( little rS) value as Evidence

(r[SUP]1[/SUP]) value as Evidence
(r[SUP]2[/SUP] ) value as Evidence
(r[SUP]N[/SUP]) value as Evidence .. logic applies put food on your table leave saving the world to captain America

The principle reason for dead testing using (R[SUP]1 [/SUP]+ R[SUP]2[/SUP] ) or ( R[SUP]2[/SUP]) is to verify the continuity of the circuit-protective-conductor (CPC)

Unless tested you cannot be sure that any circuit-protective-conductor(s) is continuous

GN-3 continuity (Dead testing) ok verification, “ to prove ” three-steps to take for continuity

Testing continuity of conductor(s) requires us to do
This is to establish and ensure that there are no breaks in continuity in all three conductors ( r[SUP]1 [/SUP], r[SUP] N[/SUP] , r[SUP]2[/SUP] )

Prospective fault current ( PFC ) “ by direct measurement ”

Two tests are required: .. Live Tests

i) Prospective earth fault current (PEFC)
ii) Prospective short circuit current (PSCC)

Regulation 612.11. requires that the (PFC) under both “ short-circuit ” & “ earth-fault-conditions ” be measured at the Origin

:mad2: GN-3 . Which is the greater of the (PSCC) and the (PEFC) obtained should be recorded on the Schedule of Test Results

 

[amberleaf]

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

 

[amberleaf]

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

 

[amberleaf]

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

 

[amberleaf]

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 ??

 

[amberleaf]

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 .

 

[amberleaf]

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 .

 

[amberleaf]

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 .

 

[amberleaf]

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

 

[amberleaf]

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 .

 

[amberleaf]

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 .

 

[amberleaf]

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.

 

[amberleaf]

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 .

 

[amberleaf]

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

 

[amberleaf]

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 .