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
-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
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 )
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)- 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 .
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 .
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 . ??
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 .
(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 .
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 )
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 .
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 )
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 .
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