Re-take - Useful Information for 2394 :

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2392-10 / 2394. taken for the main-forum . I do thank my fellow-members for this information . ◄◄

2392-10 / 2394. Please go into the , Main-forum and ask questions . :gettree:


Howdo I work out the following question; When measuring the insulation resistancebetween live conductors of five existing circuits in the installation thefollowing results were obtained.

185Mohms, 150Mohms, 20Mohms, 200Mohms,2.5Mohms

State

a) the value of the installation resistance ifthese circuits were measured as a whole show calculations.
b) whether the value obtained in a)isacceptable
c) which further test would be requiredto complete the insulation resistance testing
d) the minimum acceptablevalue from BS 7671:2008 ( 1MΩ ) table - 61
e) any observations to be made to theclient following the test

1/Rt = 1/R1 + 1/R2 + 1/R3 + 1/R4
1/Rt = 1/185 + 1/150 + 1/20+ 1/200 + 1/2.5
1/Rt = 0.005 + 0.007 + 0.05 + 0.005 + 0.4
1/Rt = 0.467
Rt = 1/0.467

Rt = 2.14 Megohms

Value is acceptable.

c) Insulation-résistance-test betweenlive conductors and earth (cpc)
d) 1Mohm
e) the 2.5 value while acceptable isvery low and would require further investigation if it is a new installation ;if it is an old installation it could be a sign that it is soon going to needrewiring.

 

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thetotal résistance with three-resistors.

R[SUP]1[/SUP] = 10Ω, R[SUP]2[/SUP] = 20Ω, R[SUP]3[/SUP] = 30Ω,

R[SUP]1[/SUP] = 1 /( 10Ω ) + 1 / ( 20Ω ) + 1 / ( 30Ω ) = 0.183 ( 1/Ω )

R = 1/ 0.183Ω = 5.46Ω

Parallel-path-résistance.

 

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( Rt) means , total-résistance of the circuit . :gettree:


From collage notes .

Additionof resistances in Parallel .
Three-resistances connected in Parallel across a supply

1 = 1[SUP]1[/SUP] + 1[SUP]2[/SUP] + 1[SUP]3[/SUP] . if we now use Ohms-law to convert , 1 to V/R .

V/R = V/R[SUP]1[/SUP] + V/R[SUP]2[/SUP] + V/R[SUP]3 [/SUP]

Dividing right - through by ( V )

V/V.R = V/ V.R [SUP]1[/SUP] = V/V.R[SUP]2[/SUP] + V/V.R[SUP]3 [/SUP]

( V ) cancels out , Leaving .

1/R = 1/R[SUP]1[/SUP] + 1/R [SUP]2[/SUP]+ 1/R[SUP]3[/SUP] + ….

By calculation . Total-résistance .

1/R = 1/R[SUP]1[/SUP] + 1/R[SUP]2[/SUP] + 1/R[SUP]3[/SUP] . 1/R = 1/4 + 1/8 + 1/2 .

1/R = 0.25 + 1.125 + 0.5 .

1/R = 0.875 : Transporting. R = 1 / 0.875 = 1.143Ω

 

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Test Instruments :

To comply with Regulation- 612.1. test instruments must be chosen according to the relevant-partsof BS-EN-61557 :

Residual-current-devices - Test-instruments . 612.8.1. and 612.10 require the use of an RCD - test-instrument to BS-EN-61557-6 , part - 6
To test the operation RCDs used for , fault-protection and additional-protection , respectively . :carols:

 

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Q) Identify the British-Standard for electrical-test-instruments that are to be used by the test-engineer to obtain-measurement-values on circuits within an installation .

BS-EN-61557.



Q)2392 / 2394: list the Three-stages in a new-installation when inspection & Testing should be carried-out . 610.1. Initial-Verification :gettree:



During-erection.

On-completion

Before being put into-service .

 

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Non-Statutory :

Guidance-Notes , are published by the IET are intended to be “ Explanations “ of the requirements of BS-7671:2008:2011:

Guidance-Notes , are for “ Guidance“ Only , theymust not be used independently of BS-7671:2008:2011: for ensuring-compliance .

Guidance-Note - 3 , is based upon the , requirements of Part- 6 in BS-7671:2008:2011:

As part of the competence-requirement , the Inspector must have a sound-working-knowledge of the, test-methods as described in the Guidance-Note .

It provides details of the methods to beused for carrying out the tests as required by BS-7671:2008:2011: for both :

New-installations . ( EIC ) or ( MEIWC )
Existing-installations . ( EICR ) Condition-Report

( EIC , is to be used only for the ,initial-certification of a new-installation . 2392-10 / 2394 :
) for an addition or alteration to an Existing-installation , where new circuits have been introduced .

The “ Verification “ of Electrical-Installations.
All measures by means of which compliance of theelectrical-installation with the relevant-requirements of BS-7671:2008:2011: are confirmed.

A “ Duty ofCare “ responsibility makes the Inspector a “ DutyHolder “ for the duration of the inspection andtesting-process .
The “ Titlein Law “ placedupon the Inspector is that of “ Duty Holder “

Inspector :
BS-7671:2008:2011: requires the inspection andtesting shall be carried out by a competent-person.

Regards to “ Verification “

- Sufficient-technical-knowledge. fully-versed in the inspection andtesting-requirements contained in BS-7671:2008:2011:

Relevant-practical-skills :
- have sufficient-practical-skill to carry out the ,inspection and test-procedures outlinedin Guidance-Note - 3

Experience with regards to the type of electrical-work undertaken : and employ

Appropriate testing-equipment :
Instrument(s) used in the testing-process , where the values are to be recorded in the schedule of test-results , must comply with the appropriate-parts of BS-EN-61557 .

611.1. Precautions for working on equipment made-dead .

Isolation : requires thatinspection shall be carried out beforethe testing-process-commences and shallnormally be done with the part of theinstallation being inspected-isolated from thesupply .

611.1. Installation is securely and securely-isolated prior to this-process being-undertaken

- Indentify the point of Isolation .
- Confirm with the client that it is safe to Isolate .

Regulation - 537.2.2.4.. p.151. requires .
That provision should be made for securing OFF Load-isolating-devices against , inadvertent or unintentional-opening . Etc.

Where Circuit-breakers are to be used as the ( Isolating-device ) then dedicated ( Locking OFF Devices ) with unique keys or combinations should be used . :christmaswreath:

 

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Instrument Calibration: Calibration is not a warranty or a statement of how-long the instrument will remain-accurate . :gettree:

Instrument(s) regarding the Inspector in BS-7671:2008:2011: isthat you must employ adequate-testing-equipment . !!
Instrument(s) generally be split up into five-separate-categories.

Indicate thepresence of voltage .
Measure - voltage .
Measure - current .
Measure - résistance .
Measure - time .

 

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Effects of length upon conductor-résistance . Basics . :gettree:

The résistance of a conductor is directly-proportional to its length . short piece of cable - Length - résistance . 5Ω
2 x Length : if length is doubled ( 2 x length ) then the résistance will increase by the same-factor , résistance = 10Ω

Résistance of a 60m length of cable is ( 4Ω) if the length of the cable wasreduced by ( 25% determine the résistance )

New - Length : 75% of 60m = 45m
New - résistance : 4 x new - length / original - length , = 45 / 60 = 3Ω . ( 4 x 45 ÷ 60 = 3Ω )

Cable -Length :

Résistance ofa 50m length of conductor is 0.1Ω . determine the résistance of 200m of the same-conductor ?
New - résistance : 0.1 x 200 / 50 = 0.4Ω

The effects of (C.S.A. ) upon conductor-résistance.
The résistance of a conductor is inversely-proportional to its (C.S.A. ) 25mm[SUP]2[/SUP]

If the cross-sectional-area is , Doubled then the résistance is ( Halved)

50mm[SUP]2[/SUP] Unknown-cable-résistance , = themeasured-résistance , x Original - C.S.A. / New - C.S.A.
Résistance of a length of 4.0mm[SUP]2[/SUP] cable is 0.6Ω , whatwould be the résistance of a 6.0mm[SUP]2[/SUP] cable if the length remained the same ?

New - résistance = 0.6 x original-cable-size / new-cable-size , 0.6 x 4.0mm[SUP]2[/SUP] / 6.0mm[SUP]2[/SUP] = 0.4Ω [ 0.6 x 0.4 ÷ 0.6 = 0.4Ω ]

The résistance of alength of 1.0mm[SUP]2[/SUP] is 0.2Ω. what would be the résistance of a 4.0mm[SUP]2[/SUP] conductor of the same-length ?
2 x original-cable-size / new-cable-size , = 0.2 x 1.0mm[SUP]2[/SUP] - 4.0mm[SUP]2[/SUP] = 0.05Ω [ 0.2 x 0.1 = 0.02 ÷ 0.4 = 0.05Ω ]

 

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Basics . Some - facts

What does not changewith the increase in , temperature isits Length .

Consider - 1.0mm[SUP]2[/SUP] ( 20°C / 100m )

The résistance /metre , for 1.0mm[SUP]2[/SUP] PVC -insulated-copper-conductors at 20°C . is 18.10 milli-ohms / metre :

O.S.G. Table - 11 . p.182 . 1.0mm[SUP]2[/SUP] - is 18.10 milli-ohms / metre

Therefore : thetotal-résistance of the cable shown , = 100 x 18.1 / 1000 = 1.81Ω

The résistance / metre for 1.0mm[SUP]2[/SUP] PVC -insulated-copper-conductors at 25°C is 18.1 milli-ohms/ metre x 1.02 .

Therefore : thetotal-résistance of the cable at 25°C = 100 x 18.1 x 1.02 / 1000 = 1.846Ω

The table -provided in the O.S.G. gives the résistance / metre for PVC - insulated-copper-conductors, & of ( R[SUP]1[/SUP]+ R[SUP]2[/SUP] ) per / metre at 20°C in milli - ohms / metre . Where ( R[SUP]1[/SUP] ) is the line-conductor , & ( R[SUP]2[/SUP] ) is the circuit-protective-conductor . Etc.

 

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Basics .

The effects of Temperature , upon conductor-résistance .
Copper : has a positive-temperature-coefficient, the warmer the cable the higher its - résistance . :gettree:

The résistance :tables for conductor-sizes are based upon an ( Ambient temperature ) of ( 20°C ) and the résistance ofcopper-increases by ( 0.004Ω ) per - degree-rise .

If a coil ofcables-temperature-increases by ( 5°C ) the corresponding-increase in résistance will be ( 0.02Ω ) 5 x 0.004 = 0.02

Value of résistance / metre for PVC - insulated-copper-conductors & of ( R[SUP]1[/SUP] + R[SUP]2[/SUP] ) per/ metre at ( 20°C ) in milli - ohms / metre . O.S.G. - Table - 11 . p.182

Where the résistance of a cable is measured at an ( Ambient temperature ) otherthan of ( 20°C ) and the length is required , a correction/factor must be used .

It must be ensured that when thecircuit-0length . etc needs to beobtained , the values used in the calculations must be at the same-temperature .i.e. either .

Change the values in the résistance table at ( 20°C ) to whatever temperature the résistance-measurement was taken at . O.S.G. table 12 . p.183

GN-3 . P.121 . Table- B2
Expected - ambient-temperature ( °C ) Correction - factor *

5°C - 0.94
10°C - 0.96
15°C - 0.98
20°C - 1.00
25°C - 1.02
30°C - 1.04
35°C - 1.06
40°C - 1.08

Or - change the résistance-measurement , from whatever - temperature it wasmeasured at , to (20°C )

 

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Basics .

Continuity Measurement :
Calculate ( Assuming the ambient - temperature to be ( 20°C ) the résistanceof the measured - lengths of cable in table- 11 - O.S.G. p.182 .

Scenario : Cable on Coil . !!

1.0mm[SUP]2[/SUP] . Measured-length ( M ) 24 , Cable-résistance mΩ/M - 18.1 , Cable-résistance Ω/M - 0.0181 x 24 , Cable-résistance - Ω 0.434 .
1.5mm[SUP]2[/SUP] . Measured-length ( M ) 36 , Cable-résistance mΩ/M - 12.1 , Cable-résistance Ω/M - 0.0121 x 36 , Cable-résistance- Ω 0.435 .
2.5mm[SUP]2[/SUP] . Measured-length ( M ) 44 , Cable-résistance mΩ/M - 7.41 , Cable-résistance Ω/M - 0.00741 x 44 , Cable-résistance - Ω 0.326 .
4.0mm[SUP]2[/SUP] . Measured-length ( M ) 52 , Cable-résistance mΩ/M - 4.6 , Cable-résistance Ω/M - 0.00461 x 52 , Cable-résistance - Ω 0.239 .

1.0mm[SUP]2[/SUP] . = 18.1 mΩ/M . 0.018 Ω/M

Total - résistance . = 24 x 0.018 = 0.432Ω :carolers:

 

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Q) from the measurement taken of the 1.5mm[SUP]2[/SUP] / 1.0mm[SUP]2[/SUP] T&E cable .
Determine :

The résistance of the Neutral-conductor .
The résistance of the Circuit-protective-conductor .
The maximum ( R[SUP]1[/SUP]+ R[SUP]2[/SUP] ) value of the circuit .
The Length ofthe cable-connecting theoutlet-point to the consumer-unit .

Megger - 1552 . reading off , Socket-outlet . 1.2Ω / L & N

By - Calculations
1.5mm[SUP]2[/SUP] = 12.1 mΩ/M
1.0mm[SUP]2[/SUP] = 18.1 mΩ/M

Line - loop = 12.1 mΩ/M
Neutral -loop = 12.1 mΩ/M
L + N = 24.2 mΩ/M ,
= 0.024 Ω/M

 

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Unknown -cable -résistance . = the measured-résistance x original C.S.A. / new C.S.A. .

Solution .

The résistance of the neutral-conductor = 1.2 / 2 = 0.6Ω
The résistance of the circuit=protective-conductor - ( R[SUP]2[/SUP] ) = 0.6 x 1.5 / 1 = 0.9Ω

The maximum ( R[SUP]1[/SUP]+ R[SUP]2[/SUP] ) of thecircuit = 0.6 + 0.9 = 1.5Ω

Cable - length . = measured-résistance of the Line and Neutral - loop / the résistance in Ω/M of the L & N - conductors . = 1.2 / 0.024 = 50M :gettree:

 

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From old notes , ► 2008 . Useful-junk .

( Line & Neutral - conductor - 1.5mm[SUP]2 [/SUP] / 1.5mm[SUP]2 [/SUP] = 24.20 ) O.S.G. 24.2


Three-phase-board

Determine whether the circuit-complies with thevoltage-drop requirements in Appendix - 12, ► moved Appendix - 4 sec 6.4. 421

Circuit- 4L1 - 6A , lighting-circuit supplyinga large-number of lighting-points is supplied via a , 1.5mm[SUP]2 [/SUP] T&E & CPC .
Line / Neutral , loop=measurement at the ceiling-rose = 0.82Ω

( Ipsc ) on the brown-phase = 1.15kA

Thevoltage-drop on a lighting-circuit = 3% of 230V = 6.9V .
Three-factors need to be taken into account .

i). The mV/A/M of the cable .

i.e.. 29 mV/A/M from tables in Appendix - 4 .
ii). the design-current ( Ib )
can be calculated or . the rating of the protective-device may be used .

regulation- 433.1.1. requires that the rating of the protective-device ( In) p.73 . p.81. must not be less than that of the design-current ( Ib) by substituting ( In) for ( Ib ) in the calculation, the maximum-voltage-drop can be calculated .

if this is in compliance then the actual-voltage-drop must also be in compliance .

iii). the circuit-length ( L )
the prospective-short-circuit-current is given as 1.15kA , ( 1150A )
the résistance of the , L / N loop-external to the installation , = 230V / 1150A = 0.2Ω

therésistance of the , L / N -conductors from the supply to the remote-end of the circuit , = 0.82 sub 0.2 = 0.62Ω

Now from , O.S.G. p.182 - Table - 11 . 24.2 milli-ohms per / metre = 0.024 ohms-per-metre .
Therefore:- Circuit-length = 0.62 / 0.024 = 25.8M

The circuits maximum-voltage = ( mV/A/L) x Ib x L / 1000 , = 29 x 6 x 25.8 / 1000 = 4.49V

Q). does the circuit complies . :christmastree1:

 

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From my old-notes . ◄◄ Useful-junk . PS its only a leaning-curve . by calculations Only .

Selecting the rated-residual-operating-current of an RCD .

Q). if the measured-earth-loop-impedance on a TT - system was , 150Ω

Determine the maximum-residual-operating-current of the RCD to be installed at thesupply-intake .

Regulation 411.5.3 . Ra x IΔn ≤ 50 : IΔn ≤ 50 /Ra : = 50 ÷ 150 = 0.33A = 330mA .

The RCDs rating must not-exceed 330mA .

Therefore: an RCD with a rated-residual-operating-current of 300mA would be acceptable . ( used only as leaning-curve )

 

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Inspection : Examination of an electrical-installation using all the senses as appropriate . :icon_bs:
Q/As
Function-testing: Senses Used .
Visual,
Touch,
Hearing.

Q) An RCD is to be used in aninstallation for protection against the risk of fire , State the maximum-rating of the RCD , 300mA .

State : themaximum-disconnection-time of the device.

Within 300mS , if the RCD is to BS-EN-61008-1 or BS-EN-61009-1
Or ,
Within 200mS if the RCD is to BS-4293 .

 

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415.1.1. - for Additional-protection only . just showing you the regulation number.

Q). State : themaximum-rating of the RCD ,
The maximum-disconnection-time and the applied-test-current as required by Regulation - 415.1.1.

Maximum-rating 30mA .
Maximum-disconnection-time . 40mS
Maximum-test-current up to five-times the maximum-rating .

 

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Q) Describe how you would carry out aninsulation-resistance-test on a circuit with a Neon-indicator-present . Couldthis come up , Written-exams :gettree:

Limitation : No-insulation-résistance-tests were carried-out between Live-conductors after the point of local-isolation .

Neon-indicator - Switch-closed, Line & circuit-protective-conductor , Neutral & circuit-protective-conductor ,


Neon-indicator - ( Whatnot to do ) Switch-open , Line & circuit-protective-conductor ,

Neon-indicator will adversely affect the result of the test ,

Thepresence of the Neon-indicator can influence the measured-insulation-résistance. typical-résistance of an indicator is approximately 200kΩ or 0.2MΩ


Basic Theory : the effectsof circuit-length on insulation-résistance
The insulation-résistance of a cable ( Decreases ) as its ( Length-increases)

Thelength is doubled the insulation-résistance ( Decreasesby half ) the change in insulation-résistance is inversely-proportional to change in the ( Length)

Basic Theory :
Total-combined-résistance of the circuits within a ? Distribution-board or Consumer-unit ,( Decreases ) asthe total-number of circuits-increases .

The résistance does not increase or decrease-proportionally to the number of circuits . installed-cables are of different-types and differing-cross-sectional-areas .
Circuit-lengths will vary as will the number ofoutlet-point per-circuit . ( All these-factors when combined affect the insulation-résistance )

Basic Theory :
Whetherthe installation is single or three-phase, only Two-insulation-résistance-tests are required
Between Live-conductors . - Neutral is a live-conductor . L / N
Between Live-conductors and Earth. L / N & E .

 

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Toavoid things going pear-shape .

Insulation-résistance.

Where the circuit-includes electronic-devices such as :
Smoke-detectors . whichcould-damage , Neon-indictors , that could influence thetest-results . Only a measurement between Live-conductors, connected together to Earth needbe made .

Re-cap.

612.3.3.
Where the circuit includes electronic-devices whichare likely to influence the results or be damaged , only a measurement between the , Live-conductors connected together and the earthing-arrangemernt shall be made .

The Regulations have stated the-facts

NOTE : Additional-precautions, such as disconnection , may be necessary to avoid-damage to electronic-devices . :17:

 

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Old-Notes.

Insulation-résistance :

Main-switch in the On-position and all circuit-breakers - Closed .
Three-phase .

Step- 1 :
Test between L1 and Neutral .
Test between L2 and Neutral .
Test between L3 and Neutral .

Step- 2 :
Now test between
L1 and L3
And then
L2 and L3

Step- 3 :
Finally test between L1 and L2.

Record the Lowest-measured-value .