Re-take - Useful Information for 2394 :

[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