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

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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 )

 

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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 .

 

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2391-101/ 2392-10 . Designing

Howis Earth Loop Impedance calculated ??

Earth Loop Impedance can becalculated by ??

(Zs = Ze + ( R1 + R2 )
Where–
(R1 + R2 = Table – 9A ---- 1000 x L x Table – 9B or 9C )

Zs)- Used to describe a circuits EarthFault Loop Impedance – ( Ohm’s )
Ze)- Used to describe part of the Earth Loop Impedance ( External to the Installation )
R1)- Used to describe the Impedance of the Line conductor ( Ohm’s – Table 9A )
R2)- Used to describe the Impedance of the circuitprotective conductor( Ohm’s Table 9A )
Length)- Length of circuit from supply to FURTHESTpoint . ( in Metre(s) )
O.S.G.– Table 9B or 9C multiplier (- Factor applied to allow for Expected Ambient Temperatureor conductor résistance at Maximum Operating Temperature ( respectively )

( Zs = Ze + ( R1 + R2 )

Theactual ( Zs ) is the sum of all the Impedance that are present in a Circuit(s) Earth Fault Path .

 

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2391-10/ 2392-10 . Designing

Multipliersare used by the Designer & are required to allow for one of the following .??

O.S.G.- Table 9B So the Designer can give valueof résistance at the Ambient Temperature expected during the Test ( 20°C is classedas 1 )
O.S.G.- Table 9C So the Designer can give valueof résistance at the conductors Maximum Operating Temperatures .

Scenario:
Acircuit supplying a ( D/B ) where a muilti core armoured cable is clippeddirect using 50 ( Metres ) of militia core – 25.0mm[SUP]2[/SUP]/ 70°C armoured ThermoplasticInsulated Cable . The bunched CPC conductor size is ( 16.0mm[SUP]2[/SUP] ) The Ze – is ( 0.5Ω ) Calculate the Earth Loop Impedance in Ohm’sat the Maximum Operating Temperature . ??

i) Write down the formulas & obtain thevalues for each part .

Zs = Ze + ( R1 + R2 ) -------- ( R1 + R2 = Table 9A – 1000 x L x Table / 9B or 9C )

Ze= 0.5Ω )- R1 + R2 must be calculated ( L = 50 m )

ii)Obtain the value for ( R1 + R2 ) in Ohm’s

UsingTable 9A we can see that the Résistance . in mili-ohms per metre . of 25.0mm[SUP]2 [/SUP]& 16.0mm[SUP]2[/SUP] is ( ►►1.877mΩ/m )

iii)Obtain the multiplier from 9C .

TheLine & Earth conductors are part of a Thermoplastic multicore cable so are classedas incorporated in a cable or bunched giving us a value of ( 1.20 ) Input all values into the ( R1 + R2 ) formula

R1+ R2 = 1.877 --- 1000 x 50 x 1.20 = 0.11Ω

iv)Input the values into the main ( Zs ) formula & Calculate ( Zs ) at theMaximum Conductor Operating Temperature

Zs= 0.5 + 0.11 = 0.61Ω

Note)- if the ( R1 + R2 ) values is Measured this can be added to ( Ze ) to givethe Total ( Zs )

 

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Apprentices( Dead Testing ) Ring / Radial . Basic(s)

Measuring( R1 + R2 ) for the Ring Final Circuit is straight forward . Unit in Ohm’s
ByWay of Testing between BOTH ENDS of the Line & likewise for the CPC.

MeasuringRadial Circuit .
TheMeasurement can be determined by Testing ► ACROSS◄ Line & CPC

Line to CPC . Why . if we Test withoutthem being connected – We would have an Open Circuit . Yeah .
( Line & CPC on &Connector Block ) Megger 1552 – One lead on the Line .& the other lead on the CPC . We have a Closed Circuit - LOOP )

PS.We are making a Loop . “ Radial “ Only for Testing Purposes .
Notlike the Ring a Continuous Loop ( “ Ring“ ) final circuit

-&-s. Ring - 4 Liveconductors . 2 . CPC / Earthing ( 6 –wires in a Ring )
2x Line . 2 x Neutral = 4 . Live conductors.

Beforeyou thing old Amber is of his Head .? Neutralis classed as Live conductor .

 

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Apprentices. Basic(s) DeadTests

Key Principle(s) of theInspection & Testing of Electrical Installation 2392-10 / Apprentices

Thefollowing Tests . Where relevant . are to be Tested “ Preferably“ in the Sequence indicated .

612.2.1.- Continuity of protective conductors . including Main & SupplementaryEquipotential bonding . ( Do we have a Earthing)
612.2.2.- Continuity of Ring Final Circuit Conductors .( Do we have correct Polarity / Continuity Method ) L/L . N/N . CPC/CPC . ( So the Earthing is True )

Beforewe hit any more Tests )- 612.2.1. / 612.2.2.You have Tested / Checked – We have an Earthing .Continuity – Polarity– The Ring is complete .

Atthis Stage ?? As a Tester 2392-10
Oneof the main reasons the 17[SUP]th[/SUP] Edition . Harp on about Testing. Where relevant . are to be Tested “ Preferably“in the Sequence . Etc
BasicProtection ( ADS ) → need an EARTH .

PS. This is just one important point. Am making.

Firstthing I do Testing ( PEFC ) Live Test : Externalto the Building . Do we have an Earth

Earth must have a LowImpedance . The possibility exist that a Fault Current will cause a High current to flow into theEarth ( ADS ) RCD will Trip .

Scenario: The Importance of Earthing . Washing machines – Class 1 . !
Protectagainst any Electrical Shock “ Dangerous Voltage “ on Metal Parts in our DomesticInstallation(s) ? from Line to Earth Fault must be Quickly removed by opening the CircuitOvercurrent Protection device . ( The RCD will trip before the MCB ) if we havea Overload . The 13A fuse will be thingabout - RCD

Regulation(- p/27 – Overcurrent )
Acurrent exceeding the RATED value . For conductors the RATED value is thecurrent-carrying-capacity . ( C.C.C. )

Overload Current (- An Overcurrent occurring in a circuit whichis Electrically Sound )- “ Short Circuit “ Occurring in a Health Circuit(s) “

LowImpedance Fault Path
Aswe can see . The Impedance of the Fault Current plays a CRITIAL & VITAL role in REMOVING Dangerous Voltages .

TheImpedance of the Fault Current part must be Permanent & ElectricallyContinuous .
Capableof Safety carrying the MAXIMUM FAULT likely to be IMPOSED on it . & it musthave SUFFICIETLY Low Impedance to facilitate the Operation of OvercurrentDevices under Fault Conditions .

TheOnly time an Earth is Live is ( UnderFault Conditions )

 

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ContinuityTests – A Continuity test is a Test of circuit Integrityto determine if a circuit is COMPLETE between TWO POINTS . (No Open Circuits ) A to B .