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