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O.S.G.. The use of other methods’ of determining Maximum Demand is Not Precludedwhere Specified by the Installation Designer
FirstlyI make no Apologies for the Way am Writing on any Matters . it can be a first day Apprentice or some one Needing aJog of Memory .
Sowe are all in the Same Boat . “ To Learn “
For the Apprentices . The Day we stop learning is the Day we hang Upour Tool-Bag
CookerDesign Current Calculations
Thefirst thing you have to do is get Your Head around the Calculations !!
(From a Design point of View ) 2392-10
DomesticInstallation Oven(s) & Hob(s) are to be Calculated upon their MAXIMUM LOADING
Startwith a simple Calculation ( An Oven has a rating of 2kW ) 2000
(I = P/V ) Formula … I = 2000 ÷ 230V = 8.70A …. Weare Using the Unit Amps
2392-10/ Domestic Installation Oven(s)
Ovenhas 4 Rings ( 2 x 1kW ) & ( 2 x 1.5kW ) & Grill ( 2kW ) & Oven (3kW )
-Controlled via a CookerSwitch with a Socket outlet .
Asa Designer . we’ll have to Apply Diversity ??
Important )- Diversity allowance to be Applied to the FULL LOAD CURRENT for CookingAppliances .
TheO.S.G. is telling us . Purpose of the Final Circuit fed from theConductors )
O.S.G.Table 1B p/97 – column (3) Cooking Appliances → At the Top of the Page Note : Type ofPremises ( 2392-10 → Household Installations ) Domestic Installation(s)
DomesticInstallation(s) Only O.S.G. - 10A + 30% f.l – Full Load ) of connected Cooking Appliances in the Excess of 10A+ 5A if a socket-outlet is incorporated in the Control Unit . ( C.C.U. ) – 45A + 13A Socket Switched with Neon .
Fromyour point of View ( The First 10A ofthe rated current plus 30% of the reminder ( Plus) 5A if the Control Unit incorporates s Socket.
Calculations)- You bank “ Hold OFF“ the first 10 Amps of the Maximum Load Current )
The10A will be used at the End of the Calculations’
-So your Work out the Total Power Rating & then calculate the Full Load Current
Calculations)- Power = ( 2 x 1 ) + ( 2 x 1.5 ) + ( 2+ 3 ) = 10kW
I= 10000 ÷ 230V = 43.48A … round it up to the first four numbers43.47826087 ( 48 ) 43.48A
UsingDiversity allowance stated ↑↑ ( 43.48A sub 10A = 33.48A )
I= 33.48 x 30 ÷ 100 = 10.04A
Youradding the ( 5A ) for Socket outlet . I = 10A + 10.04 + 5A = 25.04A )- Asa Designer this is your Expected Current Demand .
Remember )- Supply Cables Rated to suit DesignCurrent ( Iz ) 
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 )
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
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:
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
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
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 )
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. )
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:
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 )
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 )
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 )
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
Apprentices)- ( MCB ) - Each protective device in a consumer unit is Classified as a “ WAY “ . We have 12 Way – ConsumerUnit ( CU )
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 .
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 .
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 .
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 .
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 .
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
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 .
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