***Cont../ Useful Information for Electricians & Apprentices***

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kW)- True power ( x 1000 )
kVA)- Apparent power ( x 1000 )
kVAr)- Reactive power ( x 1000 )

Load )- The current drawn by ElectricalEquipment connected to an Electrical Circuit .
Prospective Short Circuit Current . – The maximum currentwhich could flow between live conductors . ( PSCC)
Prospective Fault Current - The highest current which could flow in a Circuit due to aFault . ( PFC )

Regulation612.11. Prospective FaultCurrent

(PFC ) → to any of ( PSCC ) or ( PEFC )
The ProspectiveShort Circuit Current & Prospective EarthFault Current shallbe Measured . ETC ( at the Origin ) – Consumer Unit / Tails

(PSCC ) – Line & Neutral
(PEFC ) – Line &Earth

 

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Q)The Instrument to be used for a Polarity test . Before the Supply is connected. ?
A)This test is usually conducted while measuring the value of ( R1+ R2 ) where a Low Reading Ohm Meter is used .
(if conducted as a separate test any from of Continuity Tester is used )

O.S.G.p/77
( I must stress O.S.G. is Guidance only )

Guidanceon Initial Testing of Installations .
ii) Checking that the TestInstrument is made in accordance with the appropriate safety standards such as( BS-EN 61243-3for Two pole voltage detectors & ( BS-EN 61010 or BS-EN 61557 for Instruments .
iv) Observing the safetymeasures & procedures set out in HSE Guidance NoteGS-38 for all Instruments . Leads . Probes & Accessories . ETC .

O.S.G.with the Supply Connected
vi)Check “ Polarity “ of supply . using anApproved Voltage Indicator .

Q)An instrument that may be used for a “ Live “ Polarity Test – after the Supplyis connected .
A)A voltage indicator meeting the requirements of ( GS-38) HSE … This will come up a lot on “ Exams “( GS-38Leads )

Specification)- BS-EN 61243-3 : 1999 –Live working . Voltage detectors . Two-pole low voltage type .
Specification)- BS-EN 61010-1 : 2010 -Safety requirements for electrical equipment for measurement . Control . & Laboratory use ( General Requirements)

IfInstallation(s) are to be Tested to show that they complywith BS-7671 – The following instruments will be Necessary

Testing)- 612.1.
The test of Regulations 612.2. to 612.13.ETC.

Measuringinstruments & monitoring equipment & methods shall be chosen inaccordance with the relevant parts of ( BS-EN 61557 ) if other measuring equipmentis used . it shall provide NO LESS degree of performance & SAFETY .

- Lowrésistance ohmmeters . ( BS-EN 61557- 4 )
- Insulationrésistance ohmmeters . ( BS-EN 61557- 2 )
- Earthfault loop impedance testers ( BS-EN 61557- 3 )
- Earthelectrode résistance testers ( BS-EN 61557- 5 )
- RCDtesters ( BS-EN 61557- 6)

 

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Insulation Résistance ( IR )

2392-10. Dead Test - Line& Neutral to be Linked out for this Test( Megger 1552 - Two lead(s) Lead one L&N )– Lead two Earth
Instrumentto be Used. ? … ( Insulation ResistanceTester ) for Exam purposes -&-s ►► PS NOT MEGGER

ElectricalInspection & Testing .

ContinuityTesting . ( Closed Circuit )
Examplelittle ( r[SUP]1[/SUP] ) end to endLine conductor ( Closed Circuit ) Youwill get a reading Ωs ?? ( r[SUP]1[/SUP]) to ( r[SUP]1[/SUP])

 

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AnInspection & Test which is carried out on a NewInstallation to prove Compliance is called an InitialVerification .

( EIC ) The Electrical Installation Certificatemust be accompanied be a Schedule of Test results & Schedule of Inspection
Withoutthese two documents’ . The Electrical Installation Certificateis not Valid .

Regulation. Chapter 63 p/163 .
632.1. Initial Verification

Followingthe Initial Verification required by Chapter 61. anThe ElectricalInstallation Certificate. together with a Schedule of Inspections & Schedule of Test results .shall be given to the person ordering the work .

 

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Thiscalculation is for BS-EN MCBs

MCBtype B . Time must be achieved when a currentof between ( 3 & 5) times its rating passes through it
MCBtype C . Time must be achieved when a currentof between ( 5 & 10) times its rating passes through it
MCBtype D . Time must be achieved when a currentof between ( 10 & 20) times its rating passes through it (disconnection of the device Ia )

Producetable(s) for Zs values . a specificvalue of current is required & the worst case is Used. ( B – 5 . C – 10 . D - 20 )

Whenthese values are used it is simply a case of ApplyingOhms Law – to the supply voltage 230V
&( Ia ) to obtain a ( Zs ) value

Thiscalculation will satisfy regulation 411.4.5. ( Zs x Ia ≤ Uo )

MCB– B : The ( Zs ) for a 20A type B ?? Calculated .
5 x 20= 100 … ( 230 ÷ 100 = 2.3Ω )

MCB– C : The ( Zs ) for a 20A type C ?? Calculated .
10 x 20 = 200 … ( 230 ÷ 200 = 1.15Ω )

MCB– D : The ( Zs ) for a 20A type D ?? Calculated .
20 x 20 = 400 … ( 230 ÷ 400 = 0.57Ω )

 

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Working my way around the 17[SUP]th[/SUP] Edition.

Reduced– voltage transformer ( Centre Trapped ) ◄◄

(Zs ) values for Circuit Breakers ( Table 41.6 – p/54 )
( Calculatedin the same way as the Circuit Breakers ( Table 41.3 – p/49 )

The only difference is the Voltage ( 55V or 63.5V ) Table 41.6 : We arenot Calculating 230V ◄◄

32Atype B – MCB . ( 32 x 5 = 160 )
(55 ÷ 160 = 0.34Ω )

Maximum Earth Fault Loop Impedance ( Zs ) for 5 sec Etc
&Uo of ( 55V ) Single Phase .
&( 63.5V ) Three Phase

 

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MCB– BS-EN 60898 & BS-EN 61009-1 protective devices . Usually have two short circuit capacitiesmarked on them .

(Icn ) Which is the rated Short Circuit Capacity – is the Maximum Current whichthe device could interrupt without causing ant damage to other Equipmentsurrounding it .
( Ics) Which is the ( In-service Rated Short Circuit Capacity ) is the Maximum Fault Current which the device could interrupt safetywhilst remaining safe to Use .

Itis better to select a device with an ( Ics )which is higher than the ( PFC ) as the devicewill still be fine for continued use .
( Markedon the MCB - Icn ?? 6000 or 10.000 )

BS-EN 61009-1 / RCBOs

 

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Insulation Résistance : ( 3 Test ) Regulations . [ Testing IR at theMain Switch ]

Dead Test .
Safety Isolation Procedures’ . are required hereprier to Testing

This can be carried out at the “ Main Switch “ of the Consumer Unit being Tested . You mustmake sure that the Main Switch is on the ONposition with all the Circuit Breakersfor all the Circuits to be tested . in doing this you must ensure that “ Lamps . Dimmer Switches . & Appliancesare removed Prior to Testing .

• The Test is carried out between the Line & Neutral - using the Insulation Résistance Tester [ Megger ] Using the 500MΩ Volts scale
500V is applied – Your reading should be Greaterthan 999MΩ is Obtained .
• The Test is carried out between the Line & Earth -using the [ Megger 1552 ]
500V is applied – Your reading should be Greaterthan 999MΩ is Obtained .
• The Test is carried out between the Neutral & Earth -using the [ Megger 1552 ]
500V is applied – Your reading should be Greaterthan 999MΩ is Obtained .

Reason(s) :
This proofs that the Line& Neutral are not touching :
This proofs that the Neutralto Earth are not touching :
This proofs that the Line&Earth arenot touching : Throughout the Circuits’ .

 

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Polarity ( 4 ) : This Test is the Forth Test . “ DeadTesting “ “ Wiring Regulations “

Safety Isolation Procedures’ . are required hereprier to Testing ☑
This Test is to ensure that all Switches Fuses& Circuit breaker are in the line conductorOnly & Socket-outlets are wired correctly . 612.6 ☑
• In this case am only Testing the Lighting Circuitonly . ↔ Polaritywould have been Tested through the Continuity Ring Final Circuit Test .
Very similar to Continuityof CPC. Although a read is not required . So the ( R1 plusR2 – Need Not be Recorded )

Consumer Unit – Inside . Having removed the Line Conductor from the circuit to be Tested . it isplaced into the Earth bar ( Met )
The Test Instrument to be used is then brought toeach point in the Lighting Circuit to ensure that Polarity is correct . while areading is not required this can be put on the Buzzer or Bell set mode . ↔ The Test is carried out between the Switch live & Earthat the Ceiling Roseor Batten Holder → with the Switch on the OFF position “ No Buzz “

Switch - the Switch ON it Indicates a Buzzing“ The polarityis correct . if this was carried out using a Low-résistance Ohmmeter thesame Test would show the meter reading from greater from scale ( 999MΩ ) down the a lowrésistance 0.15Ω in the same in the case of ( R1 plus R2 ) forContinuity of CPC

 

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“ Useful Junk “
( 999MΩ ) megohms is a Near-perfect résistancereading for a Wire/Cable

The Megohmmeter . or Megger 1552 – multi function tester . is a measuring device thatTests high Electrical Résistance .
Typically these measurements aremade on Electrical Wires & Motor Winding to test the Insulation value ofthe wires . The prefix “ Meg ” describes a numerical value of 1.000.000. in thecase of an Ohmmeter that value is also described in Ohms .

An Electrical/ Electric terminal positive polarity 612.6 ☑

 

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Continuity of CPC.(1 test ) : r[SUP]1[/SUP] & r[SUP]2 [/SUP].

Dead Test :
Safety Isolation Procedures’ . are required hereprier to Testing ☑
Continuity of CPC.Lighting circuit . having identified the right Circuit Breaker

i) Remove the Lineconductor from the Circuit breaker . Then place it into the Earth bar ( MET ) –Line & Earthtogether
if we look at the Test books some indicate the Link between the top of the Circuit breaker to the Earth bar .
ii) Some indicate the CPCfor that circuit is removed & the Line removed& placed into a connectorblock [ anyone of the will do ] “ Loop “

You must go to the furthest point . & With aTest Instrument low-ohmmeter place it on the Low-ohms scale ( Ω ) The Test isthen taken between the CPC & Line conductor at theCeiling Rose :- in this case a reading of .16Ω that is obtained. proofs continuity of the CPC .
• To proof and theory to the ↔ Consumer Unit breakthe Line conductor go back & take a reading . [ it shows greater than rangeproving continuity was CPC was Testedinitially ]

 

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The PSC { prospective short circuit current }is the maximum available current which can flow in the event of a fault betweenLive conductors { Line – Line } or { Line – Neutral } it iseither directly measuredor if your meterwon’t do direct { Line –Line} then
{ Line – Neutral doubled will suffice }

Line & Neutralcables are ShortedCircuited is called the Prospective Short Circuit Current PSCC .

PSC ↔ ( MCB ) ↔ Protectivedevice must be capable of breaking it Safely ◄ ≈ A.C. Sign .

PFeC { prospective earthfault current } is the maximum fault current between a Liveconductor and earth . { Fault Current is ShortCircuit }
This can measured directly or
by dividing the maximum Line– Earth supply voltage by ( Ze )

PFC {prospective short circuit current } is either the PSCC or PEFC . Whichever isGreater .

By Calculation’s to Obtain PSCC { 230 ÷ by Ze 0.15Ω = 1.53kA } Supply has been connected .
{ 400V ÷ Ze 0.15Ω = 2.6kA }

When we are referring to Design ! Designing for an Installation for Domestic . Design stage – Design purposes.

 

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Broken RingFinal Circuit ? ( As a Tester this is what -&-s arelooking for )

However . We find the Ring to be Broken . the Protective Device couldnot do it’s job as it is rated at 32A & the cable is rated only at 20A .Hence Overloading .

TheRing ? becomes Two Overload Radial . ( From a Tester point of View )

TheRing is a Continuous Loopno matter how it has been designed

p/362)- Appendix 15 ( Informative ↔ telling us )
Ring& Radial circuit arrangements

1) Ring Final Circuit arrangements Regulation– 433.1.
( Informative ↔ telling us )
TheLoad current in any part of the circuit should be ( Unlikely) to exceed for long periods the current carrying capacity of the cable ( Unbroken Ring Final Circuit )
Regulation– 433.1. refers ) This can generally be achieved by .

iv)Taking account of the total floor area being served ( The Regulations aresaying - Rule of Thumb . a Limit of 100m[SUP]2[/SUP][SUP] [/SUP] has been adopted) Ring
Nowif we look at the O.S.G. p/158 . A1 Ring 30 or 32A . 2.5 / 1.5 . – Maximum floorarea served ( m[SUP]2[/SUP] ) 100m[SUP]2 [/SUP]

i)locating socket outlets to provide reasonable sharingof the load around the ring .

 

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Consumer Unit Arrangements . Design Stage .



Part “ M “ requires that reasonable provision be made for people togain Access to a building and use its Facilities .

Prescribes that Switches . Sockets outlets . &“ Other Equipment “needs to be at Appropriate Heights .

Wording → “ Other Equipment“ Consumer Unit . Height of between 0.45m from the floor level Up . 0.⇈ to 0.45m ↳ 0.45m to ↱1.2m from floor level .

 

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2392-10 .

Who is responsible for Inspection & Testing .? The person carrying out the Inspection( YOU )

Remember ? Inspection& Testing . are two different words ( Inspection ) ( Testing )

 

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2392-10 )- Work Near Electricity “ Assessor “

You are doing a ►► Risk Assessment

Introduction to Short Electrical Training :
2391-10 – 2392-10 –

All Dead Testing should have Safety IsolationProcedures’ prier to Testing .
Safety Isolation Procedures’ . are required hereprier to all Testing ☑

All Dead Testingshould have Safety Isolation Procedures’ prier to Testing .

i) Using GS-38 Prove Test Leads .
ii) first thing to do is Isolate the Supply . That is going to beTested . - Take the GS-38 test leads to a known sourceto ensure that theyoperate correctly
“ if they are workingaright “ they go tothe supply that is to be Tested – then Test between Line to Neutral - Line to Earth - Neutral to Earth .
iii) Then the Test Leads are replied to a known source to ensure that they are still Operatingcorrectly .

iv) At this stage you are placing a Locking OFF- device – followed by Signs in the area that is tobe Tested .

 

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Execution:

SinglePole ( 1P )
SinglePole & Neutral ( 1P + N )
DoublePole ( 2P )
ThreePole ( 3P )
ThreePole & Neutral ( 3P+N )
FourPole ( 4P )

ElectricalEndurance : MCB
≥32A .. Over .
≤32A .. Under .
AmbientWorking Temperature ( °C )

 

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Part( 1 ) Sorry am having problemdownloading Amber .

DomesticInstallation )- 2392-10 . “ Electrical Calculations’ “

100A supply . ( Scenario )

ElectricShower 50A : ( Power Circuit )
Cooker– 32A :
RingFinal Circuit ( 1 ) 32A :
RingFinal Circuit ( 2 ) 32A :
ImmersionHeater 16A :
UpstairsLighting 6A :
DownstairsLighting 6A :

Add )- up the requirements ofEach Circuit ( 50A + 32A + 32A + 32A + 16A+ 6A = Total Demand 174A )
►►This is because the CIRCUITS are RARELY LOADED to theirFULL CAPACITY . ◄◄ at the One Time

Apprentices. You are Using the Principle of Diversity
Note )- NOT Total Demand 174A

Shower )- Classed as Instantaneous Water Heater )- O.S.G. Table 1B – p97
Row– ( 5 ) NO DEVIRSITY . Plain English 100%

Apprentices. Only the rated LOAD of the Appliance should BE USED …….. ( NOT the RATING of the PROTECTIVE DEVICE ) ◄◄◄ MCB

Shower[ 10.000 ÷ 230V = 43.47A ] Power Circuit.

Cooker)- As a designer . Would you apply Diversity ( YES )

1x 4 kW – Oven :
2x 1.5kW – Hob/rings :
1x 2kW Hob/ring :
1x 1kW Hob/ring :
1x 2kW – Grill :

Totalpower required = 12kW … 12.000 )
TotalLoad ( Amps ) - 12.000 ÷ 230V = 52A

Itis NOT likely that all of the elements of the Cookerwill be Used at the same time .
Dueto the required cooking times of the food the Oven will be On first . & itwill have reached its required temperature before the Hobs are Used .
[The Cooker will have ( Thermostats & SimmerStats build into it which will Switch ON & OFF to maintain the Cooking Temperatures ]

Asform a Designer point of View ? For aShort periods of time an Overload could Occur ) if all Loads on at the SameTime . Xmas Time .
Regulation133.1.1. – Selection of Electrical Equipment . Every item of equipment shall COMPLY with the appropriate BritishStandard .

RegulationBS-7671:2008 . Suggest that a small Overload isbetween ( 1.25 & 1.45A )
RegulationBS-7671:2008 . Protective Devices have toCOMPLY with certain requirements .

(Table 41.3 – p/49 . MCB / RCBOs ) … RCBOs are – MCB/RCDs in One ) Overload & / or Short Circuit :
BS-EN60898 & BS-EN 61009-1 … Protective Devices must CARRY a CURRRENT of atLEAST ( 1.13 times the device rating for 1 hourwithout operating -
Theymust OPERATE within 1 hour at a Maximumof ( 1.45 times their Rating )

 

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Part( 2 ) Sorry am having problemdownloading


“ UsefulJunk “

Apprentices, The current which causes Operation of the PROTECTIVEDEVICE in the conventional timeis Shown by the Symbol ( I[SUP]2[/SUP] )



Diversitycan now be applied .

RingCircuit(s) – O.S.G. . Table 1B - row 9.. ( Only Applies to Domestic Installations ) 2392-10



Diversityis Calculated at 100% .. for circuit ( 1 ) & 40% for circuit ( 2 ) – Plain English Largest / 100% - Second / 40%



O.S.G.– Table 1B row 9 . Ring ( 1 ) Largest / 100% - 32A

O.S.G.– Table 1B row 9 . Ring ( 2 ) Second / 40% ( by Calculations 40% is 0.4 = )



Nowthe Calculation is ( 32A x 40% = 12.8A ) 32A x 0.4 = 12.8A

PlainEnglish Largest / Second .. ( 32 + 12.8 = 44.8A ) → Total allowance for Ring =44.8A



O.S.G.the hint is in the Name 1B ? Immersion Heaters & Thermostatically controlled Water Heater )- NO Diversity



Atthis Stage I must point out )- O.S.G. 1B– Individual Household Installations ( 2392-10Domestic Installation(s) Top of Page .Look PS. That’s what you are Applying



Apprentices)- Only the Total Load should be Calculated . Immersion Heater … ( 3000 ÷ 230V= 13A )



Note: Domestic / 100A . Single Phase



Lightingcan be Calculated using ( 66% ) of the Total Demand.

Twolighting circuits at 6A = 12A … [ 12 x 66% = 7.92A ]



(MD ) Maximum Demand for your Installation is [ 43.37A – Shower : 32A – Cooker : Ring Circuits – 42.8A : Immersion heater 13A : Lighting 7.92A :

Apprentices?? What is the MaximumDemand used Now ??? 43.37+ 32 + 42.8 + 13 + 7.92 = 139.09A



DomesticInstallation(s) Ring Final Circuits areNOT fully Loaded

DomesticInstallation(s) Shower(s) are used for avery short time .

DomesticInstallation(s) Lights are NOT all on atthe same time . Yeah

DomesticInstallation(s) Cooker is Why diversity can be Used

 

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ExternalEarth Fault Loop Impedance ( Ze )

Thisis the Résistance between Live Conductors & Earth

Remember)- You have Résistance in every Cable .

p/161- 612.11.ProspectiveFault Current .ETC.
Itcan be Obtained by ( Measurement ) ( Enquiry ) (Calculation )

230V/ 400V . protected at the Supply cut-out by Fuses [ BS-1361 type 2 : BS-88-2 : BS-88-6 : ??
( Enquiry )
TTSystem - 21Ω . → Apprentices . O.S.G.- p/11
TN-S - 0.8Ω
TN-C-S– 0.35Ω

Calculationsfor cable size . can be carried out using these values . it is Important toMeasure these values before allowing the Installation to be put into Service .
PLEASE NOTE )- The Only system for which calculation of ( Ze) would be possible is a ( TN-C-S system )

Thisis because the Fault Path External to the Installation is the same for both (PSCC ) & ( PEFC ) - Prospective EarthFault Current

Forthis Calculation to be carried out ( The value of ►► PSCC between Line & Neutral shouldbe Known .

Ohm’sLaw )- Uo / Ze = PSCC .

Symbols’used in the Regulations - p/36
(Uo ) Nominal A.C. rms Line Voltage to Earth ( V )

(Ze ) That part of the Earth Fault Loop Impedance which is External to the Installation . ( Ωs )

HiBruce .

 

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Electricians’.
:19:
Lighting Protective System .Should be considered with regard to Bonding .
Theperson carrying out the WORK has a full Understanding of ( BS-EN 62305 )

 

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2392-10:

BS-7671:2008 . Only provides us with Informationwhich is relevant to Equipment which we could purchase & fit at the time itwas published .
Regulations)- Table 41.3 . Example – BS-EN 60898MCBs . BS-EN 61009-1 – RCBOs

Why?? for us . To use different ProtectiveDevice(s) to be used on Domestic Installation(s) Why ? We have different types of Levels of OVERLOAD .

MCBs- B – C – D .

UsefulJunk
( A ) is NOT used asIDENTIFICATION as it could be CONFUSEDwith the CURRENT RATING of the DEVICE .

 

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No Insulation has ( Infinite Résistance ) is perfect . ( IR ) Insulation Résistance

Youcan get Small Amounts of Electricity flowing along & through Insulation to Ground/ Earth

 

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( IR ) Insulation Résistance

CommonSense tells us that the more Voltage we have .The more Current there’ll be .

TheLower the Résistance of the Cable/ Wire – The more Current for the same Voltage.

 

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Whatmakes Insulation GO bad ? Just one Example .Yeah
:45:
GoodInsulation has High Résistance .
PoorInsulation . relatively Low Résistance .

Dependingupon such factors as the Temperature or Moisture content of the Insulation [ Résistance decreasesin Temperature or Moisture]

Combinedwith the Electrical Stresses that Exist ??
Asa Pin Hole(s) or Crack(s) develop. Moisture &foreign penetrate the surface(s) of the Insulation . ( Providing a Low Résistancepath for Leakage Current )

 

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PowerCables : 2392-10 : Insulation Résistance

Whentesting Cables . ( Fault Finding ) it is Usuallybest to disconnect at Both Ends in Order to Test the Cable by itself .

 

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UsefulJunk . ( IR )

Formany years One – Megohm has been Widelyused as a fair allowable lower limit for Insulation Résistance of OrdinaryIndustrial Electrical Equipment
Ratedup to ( 1000 Volts ) PS – Wording Above 500V - 1000V

Regulation. Table 61 – p/158 . Will Reinforce this Matter .

 

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Apprentices> When a Measurement of ( PFC ) is taken as Close to the Supply Intake asPossible . ( Meaning !! At the Origin )

Apprentices> The 5 xtimes test must Only be carried out on RCDs with trip ratings ( I∆n ) up to 30mA -&-sQ/As ( Or less 30mA )

Apprentices> Maximum ( Zs ) What is the Measured value of ( Zs ) for anAltered Circuit ? A ) That it is Lowerthan the Maximum Permitted Value .

Designing: Supply Characteristics .
Natureof Supply Parameters [ This can be gained by Enquiry or Measurement ]

U )-Is Line to Line orLine .
Uo)- is Lineto Earth

Symbolsused in the Regulation . p/36 )- U . Voltage betweenLines . ( V )
Uo)- is Lineto Earth ( V )

 

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Electricityonly has Current when it is on the move . Yeah .

Formulafor Inductive Reactance : XL = 2πƒL

Thereare 4-factors thatdetermine the Résistance of a material :

1)Type of material :
Therésistance of various types of materials are different . for instance . gold isa better conductor of electricity than copper . & therefore has lessrésistance .

2)Length :
Therésistance of a material is directly proportional to it’s length . The longerthe material is . the more résistance it has . This is because the electronsmust flow through more material . & therefore meets more friction over theentire distance .

3)Cross-Sectional Area :
Therésistance of a material is inversely proportional to the cross sectional areaof the material . This means that the thicker the substance is across . thelower the résistance .
Thisis because the larger the cross sectional area is . the less friction there isover a given length [ picture in your mind . if you will . that a fire hosewill pass more water than a garden hose . because the wider the pipe . the lessrésistance it has ]

4)Temperature :
Invarious types of materials . résistance can vary inversely or directly with thetemperature . This is because of the chemical properties of the material . incarbon . for instance . the résistance decreases as the temperature rises . Sowe say it varies inversely . in copper . however . the opposite is true . withthe rise in temperature . we have a rise in the résistance .

* So not only can résistancechange with heat . but causes heat as well .

 

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Regulation705.512.2. External Influences :

( Agricultural & HorticulturalPremises)

Callsfor all equipment to have a degree of Ingress Protection of at least ( IP44 ) for protection against water splashing from any direction .
Wheremore Onerous Environmental Conditions Apply ?? Higher degree of IngressProtection will be required .

Compliance: Regulation 705.512.2. )- Where equipment of ( IP44) rating is NOT available . it shall be placedin an EnclosureComplyingwith ( IP44 ) ◄◄

IPX5)- Protection against Water Jets from anydirection ( Note : Under stated conditions )
IPX6)- Protection against Powerful Water Jets fromany direction
IPX7)- Protection against the effects of TemporaryImmersion in Water .
IPX8)- Protection against the effects of Continuous Immersionin Water .

arty:

 

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Apprentices& 2392-10 )- Please take the Time to Read
:24:

This Test is Conducted on Domestic Installations .

I do this Test First . Why ??To make sure that I have a Earth returning back to Transformer Windings . External
BeforeI connect up My Earthing to the Consumer Unit.

TestSequence & Descriptions .

Thefollowing Test(s) are carried out with the SupplySwitchedOFF
ExternalEarth Fault Loop Impedance . ( Ze )

TheReason behind this Method )- To establish that a good Earth Exists at the Installation in Order for theremaining Test(s) to go ahead.
ProspectiveFault Current . ( PFC )

612.11.)- Protective EarthFault Current shall be Measured .
Bydoing this Test you have to Remove (Disconnect the Main Protective Bonding Conductor ) Earth – from the ( MET ) in the Consumer Unit . ( MET – MainEarthing Terminal)

TheReason behind this Method )- You are Removing Parallel Earth Paths . ◄◄◄◄
ByTesting the Line & Earth. You have a straight path to back to the Transformer Windings .

(Ze ) Measured at the Supply Position .

Onelead on ( The Disconnect the Main Protective BondingConductor ) Earth
&the Other Lead on ( At the Top of Main Switch ) Line

Younow have a Closed Circuit . ( Protective Earth Fault Current shall be Measured )

Itis one off the Easiest of all Test(s)


Line & Earth cable Only which has been Removed from ( MET ) Simple -&-s will pick your Brains on this ◄ 2392-10

-&-s)- The following Test(s) are carried out with the SupplySwitchedOFF ◄

Apprentices( What is Ze I hear you Say ??? )
Symbolsused in the Regulation . p/36 .
Thatpart of the Earth Fault Loop Impedance whichis External to the Installation ( Ωs )

 

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Apprentices: Basics
Rememberthat the : 17[SUP]th[/SUP] Edition – is British Standard :iagree:

Meaning)- ( R1 +R2 ) …. Line / R1

Meaning)- ( R1 + R2) …. CPC / R2

Putthe two together . You have ( Line / CPC ) ↔ ( Line - R1 + CPC - R2 )

Inany Exam . The Regulations will get you out of Trouble .

Toconfirm This .

Symbolsused in the Regulations p/36
( R1 ) – Résistance of the Line Conductorof a Distribution or Final Circuit (Unit Ω )
TheRegulations are pointing you to . Appx 4 – Schedule of Test Results .

Symbolsused in the Regulations p/36
( R2 ) – Résistance of the CircuitProtective Conductor→( CPC )of a Distribution or Final Circuit ( Unit Ω )
TheRegulations are pointing you to . Appx 4 – Schedule of Test Results .

(Ze ) External to the Installation p/36
Meaning )- Outside of theBuilding → Ze ] ↔ [ Zs Inside of Building.

(Ze ) You are checking that you have agood Earth from the Transformers Winding
(Zs ) You are checking That your Installationis Earthed →(Zs ) 612.9 – The Earth Fault Loop Impedance . is required to be determined forthe furthest point of each circuit

-&-s You will be asked the Qs ) Formula Calculating Earth Loop Impedance . itwill be the Option of 4 Qs ) mixed up .
( Zs = Ze + R1 + R2) ◄◄ This Qs is the start of Testing Calculation(s) for 2392-10

The colour has been added for Clarity

 

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ApprenticesBasics
LearningCurve

Regulation(s)p/333

( EIC ) Electrical Installation Certificate

Weare looking under ( Main Protective Conductors )

EarthingConductor : - Material( Copper ) C.S.A ( 10mm[SUP]2[/SUP] ) … Connection Verified Tick inbox
MainProtective Bonding Conductors : - Material (Copper ) C.S.A ( 16mm[SUP]2[/SUP] ) … Connection Verified Tick inbox

Youare Confirming )- To incoming Water & / or Gas service … ConnectionVerified Tick inbox

TheEarthing is very Important in any Installation/ or Jobs

Pointto Note : Please mark your Earthing in ConsumerUnit ?? Gas / Water ( White InsulatingTape . Why ?? if you have to come backfor any Reason you can Indentify for Testing reason . Testing comes under Common Sense . Yeah

Remember that the Certificationis you Proof in a court of Law i) Your complied with British Standards . ii)As a Competent Person . Tester .
:35:

 

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Testing. Yes this will come up on 2392-10 -&-s Q/As

Initial Verification – New Installation ( Certificate )
Regulation632.4.
All)- defects or omissions revealed during Inspection & Testing of theInstallation work covered by the ( Certificate ) shallbe made good before the Certificate is issued

PlainEnglish : The regulation(s) are telling you to Cover you But .
Electricians’are not Perfect 99.9% . mistakes can bemade accidently . The Reason for Testingin Order from the Regulation(s) is that every test has a reason(s) ??

TheRegulation(s) prescribed Tests . Etc

612.2.1.Continuity of ProtectiveConductors ( Dead Test) Do we have Earthing ?? ETC .
612.2.2.Continuity of Ring Final Circuit Conductor( Dead Test ) is the Ring a Complete Loop?? ETC . Do we have reverse Polarity
612.3.2.Insulation Résistance ( Dead Test ) has the cable been damaged ?? L/N – E . Megger 1552 two leads .

500V≥ 1.0MΩ ( Mains ) 2 x 230V = 460 / 500V ( your fire up doublethe Voltage ) Testing the Cable only
SELV: = 50V .A.C. = 250V D.C. ≥ 0.5MΩ
(They are only three Qs on this matter ( Table 61 – p/158 – ( 250V ≥ 0.5MΩ ) ( 500V ≥ 1.0MΩ ) ( above 500V – 1000 ≥ 1.0MΩ )
Look at the Wording . SELV / (250V ≥ 0.5MΩ )

2392-10)- Testing Q/As -&-s Yes ◄
612.3.3.Where the circuit includes ELECTRONICE DEVICESwhich are LIKELY to INFLUENCEthe results or be DAMAGED.
Onlya MEASUREMENT between the LINE conductors connected together & the EARTHING ARRANGEMENT shall be MADE . ( L& E ) Only

Nowthe Regulations are TELLING us ????? ( CoverourButs )
NOTE 1 )- AdditionalPrecautions . such as ( DISCONNECTION ) may be NECESSARY to avoid damage to ELECTRONICEDEVICES

Nowwe can say Informative

 

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Apprentice’sBasics .

Continuityof Protective Conductors . 612.2.1.

Method 1 )- ( R1 + R2 ) Line / CPC → This also proves Polarity
Method 2)- ( Measurement of R2 → or Bonding ) R2 / Wander Lead

Point to Note )- Test Voltages & Minimum Insulation Résistance .
Table 61 – p/158 - 500V to 1000V ( Above )

613.3.1. Insulation Résistance ( Domestic Installation - 500V between Live conductor(s)where appropriate during this measurement . Line & Neutral may be connected together . L/N –E separate

 

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RingFinal Circuit ( DeadTest ) Loop Testing

Thereason behind this Test !!
RingLoop Testing ensures that any Ring Circuits wired in the Installation you havecreated are actually a ( RING CIRCUIT ) & not forming a ( Radial )
“ Figure of Eight loop “

Testing must include the relevantTests from the following Checklist ( 612. )

O.S.G.(- When a Test shows a failure to comply . The installation must be corrected .The test must then be Repeated . As must any earlier test could have beeninfluenced by the failure .

Itthen chases onto the Regulation(s) 612.1.
TheTests of regulation(s) 612.2. to 612.13. where relevant . Shall be carried out &the results compared with relevant Criteria .
TheRegulation are telling us in Order . each test has its Reason

Apprentices. O.S.G. p/80 – 2392-10 will use this system

Continuityof Ring Final Circuit Conductors . p/80 – A Three-step Test is required to Verify the Continuity of the Line . Neutral & Protective conductors & the correct wiring ofthe Ring Final Circuit . Etc

Step1 )- p/81 / End to End ( MeasuredSeparately ) L/N/CPC
Step2 )- p/81
Step3 )- p/82

Youwill see ?? ( Little r[SUP]1[/SUP]+ r[SUP]2[/SUP] ) don’t get confused with ( R1 + R2 )

 

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2392-10: NEW
Purposeof Initial Verification .

InitialVerification . in the context of Regulation 610.1.is intended to confirm that the Installation complies with the Design & hasbeen Designed & constructed in Accordance .
610.1. ) 611.2. ) 612.1. )

 

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2392-10: ( T )
Certificates

Appendix6 of BS-7671 contains Three model forms for the Initial Certification of a ( NEW ) Installation or for an Alteration or Addition toan existing installation as follows.

Ambringing you to the Point )- Single Signature ( ElectricalInstallation Certificate)- That is you 2392-10
WhereDesign . Construction . Inspection & Testing are the RESPONSIBLY of ONE PERSON )- That is you 2392-10

ACertificate with a Single Signature MAY REPLACEthe MULTIPLE SIGNATURE form ◄◄◄
:iagree:

 

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2392-10: This will come up -&-s
:sweatdrop:
Regulationp/339 . ( PIR)

Periodic Inspection Report .
Regulations )- row 5 . The “ Extent& Limitations “ box should fullyIndentify the elements of the Installation that are covered by the report &those that are not . This aspect having been agreed with the CLIENT & other interested parties before the Inspection & Testingis CARRIED OUT .

Tester. GN-3 p/19 .
Scope)- it is essential that the Inspector & the Person ordering the Inspectionknown the “ Extent “ of the Installation to beInspected & any Criteria regarding the “ Limit“ of the Inspection . ( BOTH should be AGREED & RECORDEDon the CERTIFICATE )

Regulation)- Schedule of Inspections P/340
Bottomof Page ??

NOTES : ( LIM ) ◄◄ -&-s/ Big time . PS trip up Q .
ToIndicate that . exceptionally . a “ LIMITATION“ agreed with the person ordering the work preventedthe Inspection being carried out ( Applicable for a PeriodicInspection Only )

 

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2392-10: ( ADS )
:iagree:
TestingRCDs
Wherean RCD has a rated value of ( I∆n ) not exceeding ( 30mA ) & is used toprovide Additional Protection to BASIC PROTECTION the operating time of the device mustnot exceed ( 40mS ) when subject to a Test Current of ( 5 x I∆n )

 

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2392-10:

Example Only . -&- ( 2 Seconds time ) GN-3 . p/56
Generalpurpose RCDs to BS 4293 .
Becauseof the variability of the time delay it is not possibly to specify a maximumtest time . it is therefore imperative that the circuit protective conductordoes not rise more that 50V above Earth potential ( Zs I∆n ≤ 50V ) it issuggested that in practice a ( 2 second )maximum test time is SUFFICENT . -&-s now you know

2392-10: Get your self a copy of ( GN-3 )
FunctionalTesting .
Operationof residual current devices .
Whilethe following tests are NOT a specific requirement of BS-7671 it is recommendedthat they are carried out . GN-3 . p/56

Q)Prior to these RCD tests it is Essential for Safety Reasons , that the EARTH LOOP IMPEDANCE is tested to check the requirements’have been met . WHY ??

( Wehave a Earth ) .. Common sense . RCD must havean Earth ( 612.9↔ 612.10 )

 

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2392-10: 612.6.

4 different scenarios . requirea Polarity Test .

i)- Unit used ( Ω ) All single pole devices & circuit breakers are connected inthe Line conductor Only .
ii)- Unit used ( Ω ) The line conductor must be connected to the centre terminal ofan Edison Screw Lamp holder ( with the Exception of E14& E27 lampholders to BS-EN 60238 )These are European . They have known to be usedon exams .
iii)- Unit used ( Ω ) All polarities of Socket outlets ( Ring & Radial must be Verified )
O.S.G. – p/78 ↓
iv)- Unit used ( Ω ) The Polarity of the mains supply must be correct . [ Using anapproved voltage tester ] with the supplyconnected

 

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Apprentices)- ( MCB ) - Each protective device in a consumer unit is Classified as a “ WAY “ . We have 12 Way – ConsumerUnit ( CU )

 

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Apprentices)- Continuity of Main Protective BondingConductor . ( PIR )

Alot off you will have a Multi Function Tester ( At this point you have zero the Megger to the test leads .
(for the sake of -&-s . if you don’t have a Megger ? you do it the old fashion way “ Subtract Leads“ Leads 0.02 – reading 0.4 = 0.38Ω ) you may get asked this one ??

i) Isolate Main Supply . ( Lock OFF )

( R2 ) - 612.2.1.

ii)Disconnect Bonding Conductor from Main Earth Terminal ( MET )
iii)Place One lead on conductor & other on the Earth Clampconnection ( you Using Wandering Lead )

Bonding Conductor → to Water pipe ( you have A to B – continues Loop )

iv)Zero lead Résistance & set to Ω

v)Take reading ensuring Less than ( 0.05Ω )

( Remember to replace Disconnect Bonding Conductor fromMain Earth Terminal ( MET ) BACKwhen Testing is Done )

UsefulJunk .
UnderFault conditions the Fault current should flow in a controlled manner through aCPC.

Oneother way of looking at this . A “Dedicated “ CPC would carry most of the fault current as it provides the mostdirect path & lowest Impedance back to the source of energy .

 

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InitialVerification : Apprentices’

610.1. Wording .

EveryInstallation shall . DURING ERECTION ◄◄◄ ( And )on Completion BEFORE BEING PUT INTO SERVICE ◄◄◄ .be Inspected & Tested to VERIFY .Etc

PlainEnglish : Test it whilst your working on the Job .

 

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Apprentices’: Basics . Yeah Reg – P/35

Youmust understand the basics .
Listedbelow are 4 terms that describe vital information in the Calculation process .

Thefirst factor you need to consider is ( DESIGN CURRENT)

17[SUP]th[/SUP]Edition - ( Ib) Design current of circuit
PlainEnglish ( Ib) Term used to describe a circuits DESIGN current in AMPS (The Load )
( In ) Rated current or current setting of PROTECTIVE device AMPS
17[SUP]th[/SUP]Edition - ( Iz) Current-carrying capacity of a cable for continuous service under theparticular installation conditions concerned .
PlainEnglish ( Iz) Term used to describe a circuits value in AMPS . once all de-rating factorshave been considered .
17[SUP]th[/SUP]Edition - ( It )Tabulated current-carrying capacity of a cable .
PlainEnglish ( It) Term used to describe the Tabulated current rating of a cable in AMPS ( The CURRENT a cable can SEFELYCARRY )

(Ib ≤ In ≤ Iz ≤ It )

Thisformula states the underlying principle of the calculation of a circuits CableSize .

 

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Theprotective device current rating must be ( Equal to or Next Largest Size ) so that the circuit issufficiently protected

Formula)- used for cable calculation .

(Ib ≤ In then Iz ≤ It )

TheOver current protective device must be ( Equal to or Greater ) than the designcurrent .

 

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Howto Establish the value of Volt Drop ( Vd ) Red Book BS-7671:2008

Eachcable rating in the Tables of Appendix 4 of BS-7671:2008
Hasa corresponding volt drop figure in mili-volts per metre of run ( mV/A/m )

Tocalculate the cable volt-drop ??

Volt-drop= Ib x ( mV/A/m ) x L ÷ 1000

Where.

Ib)- Design current in amps .
mV/A/m)- The mili-volts per amp per metre dropped
L)- The circuit length in metres .
► 1000 )- Converts the mili-volts into Volts ◄

Scenario : ◄◄
4.0mm[SUP]2[/SUP]PVC sheathed circuit feeds a ( 6kWshower ) & has a length of run of ( 16m ) Find the total Voltage Drop . ( Vd )

i)-Work out the Design Current . ( I = P/V = 6000 ÷230V = 26.08A ) ◄ Round up to 4– 26.08695652
ii)Obtain the . mV/A/m from Appendix 4
Table( 4D5 ) p/282 .
Thevolt drop figure for ( 4.0mm[SUP]2[/SUP] T&E is 11 mV/A/m )
iii)Input all the values into the Formula & work out the Volt drop to ( Twodecimal places & add the value V )

Voltdrop = 26.08 x 11 x 16 ÷ 1000 = 4.59V

Sincethe Permissible volt drop in this Instance is ( 5% of 230V ) which is ( 11.5V ) Now does the cable in Question meet volt droprequirements . ??

( 4D5 ) p/282
FlatPVC Cables .

 

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Earth Loop Impedance . 2392-10

Résistance ( Measured in Ohm’s )
Isthe property of a conductor to limit the flow of current through it when avoltage is APPLIIED . The larger the conductor is the less résistance it has .
Thesmaller the conductor is the more résistance it has . .

(Thus . a voltage of one volt applied to one ohm résistance results in a currentof one ampere ) Yeah

 

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2391-10 / 2392-10
Multipliersare used by the Designer & are required toallow for one of the following .

O.S.G.Table 9B )-
Usedso the Designer can give values of Résistance at the Ambient Temperatureexpected during the Test(s) ( 20°C is classed as1 )
O.S.G.Table 9C )-
Usedso the Designer can give values of Résistance at the Conductors Maximum OperatingTemperatures .