***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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Sequenceof Tests : 2392-10
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Point to Note )- The Danger is compounded if Tests are NOT carried out in the CorrectSequence .
Why???

SomeTests require the Supply to be On . -&-s . will come up on this Matter .

BS-7671.States the Order weshould carry out the Tests .

 

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Approved Document P . Electrical Safety – Dwellings .

Note : Lackof Knowledge of a Regulation is No Defence in Law .

Electricians’ Guide to the Building Regulations
ApprovedDocument P .

p/16)- ( P1 ) Design Installation .
Reasonableprovision shall be made in the Design & Installation of electrical installationin order to protect person operating . maintaining or altering the installationfrom Fire or Injury .

BuildingRegulations - ( BR ) p/7 .
ApprovedDocuments are intended to prove guidance for some of the more common buildingsituations . However there may well be alternative ways of achieving compliancewith the requirements .

Thusthere is No Obligation to adopt any particular solution contained in an ApprovedDocument if you prefer to meetthe relevant requirement in some other way .

Asfor -&-s )- ADwelling is a place of Work when work is Undertaken . p/9
This( ACT ◄◄◄) Empowers the “ SECRETARYof STATE “ ( -&-s) to make Regulations . Etc.
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2392-10: Just an Insight . to some Testingmethods :waving:

EarthLoop Impedance ( Zs )

612.11. Prospective Fault Current . ( PFC )
Whendesigning an Installation . 2392-10 – it is theDesigner’s ( You) responsibility to ensure that . Should a Line –to Earth Fault develop .
TheProtective Device will operate safety & withinthe time Specified by BS-7671 .

Itis ” NOT “ until the Installation is complete& “ Live “ that the calculations can be “ Checked “ ( Energized )

What -&-s are looking for)- it is Necessary therefore to “ Determine “the Earth Fault Loop Impedance ( Zs )
Atthe “ Furthest Point “ in each circuit & to “ Compare the Reading “ obtained with calculated valuesTabulated in BS-7671 or the O.S.G.

2392-10)- With an Earth Loop Impedance Tester ◄◄ ( For the purpose of -&-s ) You will beUsing a Megger multifunction tester –MFT 1552 ??
Thevalue of Earth Fault Loop Impedance ( Zs ) may be“ Determined by ?

)- Direct ►Measurement of ( Zs ) at the furthestpoint in the Circuit .

MaximumPermissible Measured Earth Fault LoopImpedance .

“ Wording“ Note .
O.S.G.– p/103 )- Table 2D gives the Maximum Measured ◄( Zs ) for circuits protected by MCBs to BS-EN 60898 & BS-EN 61009-1 ( At Ambient Temperature ) ◄ ( 0.1 to 5s . disconnectiontimes )
Thetabulated values apply only when the Nominal Voltage to Earth ( Uo ) is 230V . scenario only . PFC – 230 /Uo ÷ 1.16 = 198A . – ( B – 32A – 1.16 )
(PFC ) - Although the Designer can calculate this in theory .
Regulation41.3. p/49 )- Maximum ◄ Earth Fault Impedance( Zs ) Compliance with ( 0.4s – 5s )

)- Direct ►Measurement of ( Ze ) for-&-s ( At the Origin of the Circuit ) &adding value of ( R1 + R2) measured during ( Continuity Tests )

ElectricitySupplier )- Obtaining the value of ( Ze ) from the DNO. & adding to this value of ( R1 + R2 ) … Zs = Ze + ( R1+ R2 )

(Ze ) this is a Live Test . ( So –Ze is the Loop Impedance value External to the Source of the Circuit(s) Supply)
(Ze ) Live at the Incoming Terminals ) -&-s are looking for ( Safe Isolation ) You are only testing the Top of the MainsTails Live . The bottom has been ( Safe Isolation)

► DirectMeasurement of ( Zs)
Earth Fault Loop Impedance isachieved by use of an Earth Fault Impedance Tester . This Instrument operatesfrom the “ Main Supply “ & therefore canonly be used on a “ Live Installation “ ( Care must beTaken Here )
Testersare connected directly to the “ Furthest Point “ of the Circuit – it must be Noted that Parallel Paths may be present whichwill affect the True Circuits Earth Loop Impedance Reading .

► Measurement of ( Ze )
Thevalue of ( Ze ) can be Measured using an Earth Fault Loop Impedance Test at the“ ORIGIN “ of the Installation .

Megger multi function tester – MFT 1552 ??
Theinstrument is connected using approved Leads between the “ Line “ terminal of the Supply & the means ofEarthing . ( with the Main switch closed OFF /Isolated )

2392-10)- -&-s . In order to remove the possibility of Parallel Paths . The meansof Earthing must be Disconnected from the Main Protective Bonding conductors -&-sfor the duration of the TEST.

► Measurement of Prospective Fault Current ( PFC &PSCC )
ThisTest is made with an Earth Fault Loop Impedance Tester set to Amps
TheProspective Fault Current . Which is how much current will flow in the event ofa Fault . is divided into two-parts .
ProspectiveEarth Fault Current ( Line to Earth Fault )
The( PFC ) test is measured at the furthest point or can be calculated using the (Zs ) value
The( PFC - Test ) is made at the Origin of the Installation & its value .Obtained in ( kA ) is then compared to the Fault Current ratings of theProtective Devices .

(PFC ) & ( PSCC ) tests are actually both carried out at the relevant point& the Higher of the Two-Values is recorded in particulars of the Origin ofthe Installation .

2392-10.
Insulation Résistance . Tests should be made on a Dead Test .

Competenceof the Inspector .
A finalconsideration when carrying out Inspection & Tests is the competence of theInspector . ( it is the Responsibility of the Inspector to )

132.1– ( i ) • Ensure no Danger occurs to People . Property & Livestock .
• Confirmthat Test & Inspection results comply with the requirements of BS-7671:2008. & the designers requirements .
• Expressan Opinion as to the condition of the Installation & recommend remedialwork(s) .
• Makeimmediate recommendations . in the event of a dangerous situation . to the Client to Isolate the Defective Part(s) -&-s

 

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InspectionRequirements : 2392-10

Requirementsfor both Basic & FaultProtection .

ElectricalSeparation . Which means NO ElectricalConnection between the SELV Circuit & HIGHER VOLTAGE SYSTEMS .

 

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2392-101: The Customer .

FollowingInitial Verification . BS-7671 requires that an Electrical InstallationCertificate .
Togetherwith a Schedule of Test Results & an Inspection Schedule .

Shouldbe given to the person ordering the Work .

Initial Verification 632.1. ..2392-10 Exam ◄◄ . -&-s Should be given to theperson ordering the Work
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Unlessthis has be Done “ The Regulations have Not been MET.

Calculation – gives you theFigure that it should be .
Measurement – give you theFigure that it is .

 

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Where an Earth Electrode is to be used considerationmust be given to the Earth Electrode .
Inmost cases . we accept that the Electrode Résistance must be no Greater than ( 200Ω )

Wherean Electrode is used for any RCD with a rating Greater than ( 200mA ) theformula ( Ra ≤ 50 / I∆n ) is applied .

Wherea ( 300mA ) RCD is to be used for FireProtection . ( 50 ÷ 0.3 = 167Ω )

Thiswould be the maximum permitted résistance of the Electrode & the protectiveconductor connecting it to any Exposed Conductive Parts .

 

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Table11A ) Rating Factor(s) for triple harmonic currents in four-core &five-core cables
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Balancedthree phase load is to be installed which has adesign current of ( 26A )

Manufacturerhas stated that the equipment being installed has a harmonic distortion of ( 30% )
Therating factor of ( 0.86 ) Table 11A – BS-7671can be applied to calculate the minimum permissible rating for the conductor ( 26A ÷ 0.86 = 30.23A )

Table11A – BS-7671 . p/356 .

Therating factor of ( 0.86 ) can be used only forequipment with Harmonic Distortion of up to ( 33%) p/356
Abovethat the calculation has to include the percentage distortion .

 

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Harmonicdistortion for the equipment ( 45% ) …. 26A x 45% ÷ 100 x ( 3) = 35.14 .
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3) it is a three-phase supply .

three-phasecircuit(s) . triple harmonic neutral currents [ 3[SUP]rd[/SUP] . 9[SUP]th[/SUP] . 15[SUP]th[/SUP] . ] etc
causethe largest concern as they add instead of cancelling . since they aremultiples of three times the – Fundamental power frequency & are spacedapart by ( 120 Electricaldegrees )
basedon the fundamental frequency . triple harmonic currents of each phase are inphase with each other & so add in the neutralcircuit . Under the worst case conditions . the neutral current can be ( 1.73 times ) the phase current .

greatdeal of electronic equipment requires what is known as ( Clean Power Earth ) unfortunately .
itis the high-frequency harmonic currents due to the switch mode power supplies whichcause the distortions in the first place .
thesedistortions can also have an adverse effect on ( the Earthingsystem of Installation(s) particular lot of PCs in use . These distortionsare often the cause of noise bars on PCs . monitors. in these cases consideration must be given to providing a copper Earth mat which is connected to a very good Earth& connecting each piece of ( IT equipment separately )

 

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Designing:
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Forall cables over 16.0mm[SUP]2[/SUP]there are Three-values given for Volt –drop [ r . x . & y . ]
Thereis a complex calculation in which these values can be used to calculate areduced voltage drop within a cable depending on the conductor temperature .

Anyoneusing cables greater than 16.0mm[SUP]2[/SUP] would be to use the Value ( r) in the voltage drop calculation as this will always give a value which is onthe right side of SAFE

 

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Apprentices. RCDs

TheTerm “ Additional Protection “ is usedthroughout the “ 17[SUP]th[/SUP] Edition “ The use of RCDsare recognised as a means of providing additional protection in the event ofFailure of the provision for “ Basic Protection “

Asan “ Additional Means ” of “ Fault Protection “& to protect againstCarelessness by USERS.

Fromyou point of View ??
Ifan RCD is Used to provide “ Additional Protection“ it then meet the requirements of Regulation 415.1.1.Thatis . The RCD must have a rated Residual Operating Current not Exceeding ( 30mA ) & an operating time NOT Exceeding ( 40mS ) at ( 5 x therated current ) . Yeah

Thisone you will get in 2392-10 Exams . “ AdditionalProtection “ :earmuffs:

 

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Therequirements of Regulation 314.1.. Division of Installation . :6:

Needto be taken into account when designing & installing the circuit protectivearrangements . 314.1. State(s) that Every Installationshall be divided into circuits as necessary:

i)Avoid hazards & minimize inconvenience in the event of a Fault .
iii)Take account of Danger that may arise from the Failure of a Single Circuitsuch as a Lighting Circuit .
iv)Reduce the possibility of Unwanted Tripping of RCDsdue to excessive protective conductor current produced by equipment in Normal Operation .

Apprentices )- This means a Single RCD cannot be Used toProtect the Whole Installation . As in the Event of a Fault . Power will be Lostto all Circuit potentiallycausing Hazards . ( Inconvenience & Danger )

So)- Therefore Multiple Devices should be Installed to protect the Outgoingcircuits & should also be ( SPLITacross the RCD )

 

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Apprentices. RCDs :39:

Scenario)- Shower – PowerCircuit . ( Remember – No Diversity on a Shower . 100% )
(RCBO ) 40A upto 50A
RCBO. Type ( B ) SP / 50A – 30mA – ( 6kA ) ….. 50A . is Breaking Capacity / 30mA . isFault Prospective Current . ( it meets the criteria of the 17[SUP]th[/SUP] Edition – ADS)

IndividualProtection for Designated ( Single ) Circuit . ?
(RCBO ) is a ( MCB / RCD ) Functions ofprotection against ( Overload & / or Short Circuit ) Regulations . Definitions – p/29

( MCB) from same Origin ( Overcurrent )
50A- B . Type – This means it will tripbetween ( 3 & 5) time Full Load Current
B . Type– is Used on Domestic Installation(s)

Asa Designer . the ( RCBO is about £ 50 + Vat )
Asa Designer . the ( MCB is about £ 3 pounds + Vat )
Asa Designer . It all comes down to COST ./ Client

So the Designer may not have the last say on this Matter .

 

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Designing )- Trunking : Calculations.
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Trunkingcontains ( 5 – Three-phase & 3 – Single-phase Circuits )
Equivalentnumber of Single Circuits .

5x 1.5 = 7.5.
7.5+ 3 = 10.5 . in this case a rating factor of ( 0.51) could be Used ( 4C1 )

Equivalentnumber of Three-phase circuits .

3x 0.666 = 1.998 – round up ( 2 )
2+ 5 = 7 )- in this case a rating factor of ( 0.54) could be used ( 4C1 )

Whencalculating the cable size use the Three-phase factor for Three-phase circuits. In this case ( 0.54 )

&Single-phase factor for Single-phase circuits . in this case ( 0.51 )

 

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Designing )- Trunking : Calculations. Part ( 5)


Designing )- part ( 4 )
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Ambienttemperature factor ( Ca ) = 0.87
Groupingfactor ( Cg ) = 0.7
Tabulatedcurrent rating of the cable is ( 24A )
Asthe circuit is in Conduit at ( 40°C ) &grouped with two-other circuits both rating factors must be Applied .

24 x 0.87x 0.7 = 14.61A

Asyou can see . This now reduces the current-carrying capacity to ( 14.61A ) Thisis due to the likely build up of temperature in the cable from being at agreater Ambient temperature ( 30°C ) & theimported Heat from the OtherCables .
Whereall the rating factors do not affect the cable(s) . at the same time only the Worst case needs to be Applied – in this case ( Cg – 0.7 ) in other words .
Ifthe conduit passes through an Environment at ( 40°C ) withonly one cable in & then Enters another roomwhich has a temperature of ( 30°C ) but picksup another two circuits in the room . Only the Worst case needs to be Appliedto your cable . This would be ( Cg – 0.7 )

Insome installation(s) . Three-&Single-phase circuit(s)will be contained in the same Enclosure .
Table. ( 4C1 ) only provides factors for ( Uniform Groups of Cables) . if all of the cables are Three-phase then each Three-phase circuit iscounted as ( One Circuit )
Thefactors from table can be used just as they can for ( Single-phase-circuit(s)This is not the case where there are ( Single-&Three-phase circuit(s) mixed together . & slightly differentapproach to ( Grouping ) has to be taken .

Toestimate the grouping factors for Three-phase circuit(s) in a containmentsystem having mixed groups we must ( Multiply theNumber ) of Three-phase circuit(s) by ( 1.5) then add this number to the number of ( Singlecircuits ) This will give us a figure to USEwith table ( 4C1 )

Estimatingthe grouping factor for ( Single-phase circuits) in a mixed containment system . We must ( Multiplythe number of Single-phase circuits by 0.666 ) & then add this to the number of ( Three-phase circuit(s)

 

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Designing )- part ( 3 )
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2.5mm[SUP]2[/SUP] cable can carry ( 24A )before any ( Correction Factors ) are applied .To calculate the amount of current which the cable can carry before it needs tobe counted as ( A Circuit ) as a circuit in ourgrouping calculation the following formula can be Used

24 x 0.3 x 0.7 = 5.04A )

24 )- is the rating factor ofthe cable from ( column 4 ) table ( 4D1A ) ( 0.3 ) is the multiplier for ( 30%) & ( 0.7) is the grouping factor / Example 2 -
Usingthis calculation shows us that the ( 2.5mm[SUP]2 [/SUP])cable was only carrying ( 5.04A ) we could discount it from our groupingcalculation . for the remaining circuits only a grouping factor for two circuits( 0.8 ) need applied .

Theimportant thing to remember is that for a cable to be discounted for groupingit must only be carrying ( 30% ) of its capacity with the grouping factor forall of the circuits applied . Hence in our example the grouping factor forThree-phase circuits was applied to the ( 2.5mm[SUP]2 [/SUP])cable which has been discounted in the calculation .

Ifwe now look at examples ( 1 & 2 ) again . it is possible that the cable beinstalled in a conduit with two-other-circuits & the Ambient Temperature ofthe Area where the conduit is installed (40°C ) in this case . both correction factorsmust be Applied to the circuit to be installed .
Thiswould result in a further reduction of Current-Carrying Capacity for the circuit .