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

[amberleaf]

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 )

 

[amberleaf]

•From a Exam point of View 2392-10 ?
:6:

- AStatutory Document is a Legal Requirement .
- ANon-Statutory Documentis a Recommendation . ◄

UsefulJunk . Appendix 2 – BS-7671 . covers Statutory Regulations .

2391-10)- Answer the Questions in full using correct terminology .

Example. Q / Which is the type of Inspection to be carried out on a New Installation . A / Initial Verification .

2392-10- Certification & Reporting . p/163 ( This is oneyou will get ) Simple – one Q/As
631.3– Where Minor Electrical Installation Work ( DOES -NOT-INCLUDE ◄◄ ) The provision of a ( NEW CIRCUIT) -) The hint is in there !! pleaseread

2391-10- Don’t git mixed up (- 633.1 . Additions & Alterations
ElectricalInstallation Certificate or a – ( Minor Electrical Installation WorkCertificate . shall apply to all the Work of the Additions& Alterations -)

O.S.G.– 9A . 2.5mm[SUP]2[/SUP] / 1.5mm[SUP]2[/SUP]has a résistance of ( 19.51mΩ) per metre .( 58 metres ) of cable will have a résistance of ?
[5.8 x 19.51mΩ = 113.158 ] ÷ 1000 = 0.113]

Root meansquare ( r.m.s. )or effective value of a waveform ( Voltage &Current )

Apprentices.
Test method 1 (- Uses the lineas a return lead . – Otherwise known as the ( R1 + R2 )
Thisis the most common method used for final circuits . Why ?

The Line conductor runs alongside a CPC conductor & this factor is Utilised in theTest .
• Polarityof the circuit is also obtained at the same time .

- YouInitial checking for Continuity .
-So Continuity testing & ( R1 + R2 ) readings .
- (R1 + R2 ) on the Schedule of Test Results .

► Ifwe look at the O.S.G. . 9A . (- Somerows will have the résistance for ( R1 ) &some for just ( R2 ) & also for BOTH .

►2392-10.When using Test Instrument explain the importance of i) Verifying TestInstrument . ii) Regular Calibration . iii) Use of DocumentaryEvidence .

Asa Tester its your reasonably (- You must ensure that your Testing Equipment is Calibrated. Why ? This will indicates that the Instrument is working properly & providingaccurate reading . Downside to this . “ Question“ if you don’t do this . Your Test Results could be Void .

TheBasic Standard Covering the performance &Accuracy of Electrical Test Instruments is ( BS-EN61557 ) which incidentally alsorequires Compliance with the Safety Requirements of ( BS-EN 61010)

 

[amberleaf]

MeasuredEarth Fault LoopImpedance ( Zs)
:39:

Asa Tester point of View . it should be compared with the . Ze + ( R1 + R2 ) Total & if it is ( Higher ) then further Investigation should be carriedout .

2392-10. every little bit helps towards 2391-10 .
The measured value of . Ze + ( R1 + R2 ) will →[ NOT include Parallel Paths ] if carried OUT Correctly .

Parallel Paths . you will getenough on your 2391-10 . “ Very Important “

(R1 + R2 ) This value must be Entered for all circuit(s) unless it is notpossible to measure
(R2 ) Where the measurement of ( R1 + R2) is not possible then the ( End to End ) Résistance of the CPCcan be measured using a Long Lead Method –Wander Lead

Apprentices(- RCD – setting on 30mA / be carful & do not mistake the TrippingCurrent for the current rating of the Device ) Just apoint . Yeah

2392-10– Functional Switching .
Table. 53.2. –p/117 .

Theone you will use a lot . BS-EN 60898 . MCB . ( YES ) / Table 41.3 – ( Zs)
Theone you will use a lot . BS-EN 61009-1 RCBO . ( YES ) / Table 41.3 – ( Zs)

 

[amberleaf]

(-Résistance in a Conductor . :willy_nilly:

2392-10. you will get this on your Exams Principles’

Ifyou Double the Lengthof a Wire . you will Doublethe Résistance of the Wire .
Ifyou Double the Cross-sectional-Area ofthe Wire you will cutits Résistance in HALF.

 

[amberleaf]

Apprentices(- “ Plugs “ Résistance . Why . a Larger EarthPin .
:39:

Becauseof Safety Reasons .
AThick Earth Pin provides a ( Less Résistance )path for the Current to be Earthed.

ForSafety so the ( Leakage Current ) is any ( Will take the Less resistant Path ) to Earth pin. to Earth rather than a person operating the Appliance/ Equipment

Plusyour Ring Final Circuit . is a Continuous Ring - Earthing . sowhen you plug into a Socket . The rest speaks Volumes . Yeah

 

[amberleaf]

2392-10– you will hear this in Exams Principles’ . 3/5 day course .Cold Test / Testing .The cable has Not been Energized yet. Yeah . :aureola:
Doyou have ( Parallel Paths ) at this point . No . ( A ↔ B ) Yeah .

Whenwe test the ( r1 to r1 ) we are doing a Cold Test . – Initial Verificationis a NewInstallation .

 

[amberleaf]

Socket(s).

Firstly. you are Complying with BS-7671:2008 . ( Dual EarthTerminals )

Example. MK. Socket - Socket have two IndependentEarth Terminals . ( used for “ Clean Earth“ Installation )

Scenario . Keep in Mind you may have 12 Sockets on Ring Final Circuit. ?? Run off Cable ? 50m ↔ ( No Spur at this point )

Mypoint . Megger 1552 time . Set on Ohm’s. ? Zeroed my Test Leadsfirst . ◄
TwoLead(s) on Socket Earth bar = my Readingis 0.23Ω / 0.55Ω - Radical Reading(s) . CopperBar .
Keepthis in mind . This was not on Backbox . yet ( No Parallel Paths ◄ ) A ↔ B

Justa Point to Note Here .
3x 2.5mm[SUP]2 [/SUP] . ( Adding a Spur ?? )
3x 4.0mm[SUP]2 [/SUP] ( Radial ?
► 6 x 6.0mm[SUP]2 [/SUP] (Stranded . We use Stranded cable for Flexibly .) you will get this on CoP .

Apprentices. Just a Leaning Curve

 

[amberleaf]

Apprentices.

Thistakes me back to a point here .

Example. if we use the principles off the Earth pinon a Plug . “ Bigger – Least Résistance . Bigger – Cable - Least Résistance
Amreferring to the Socket . BS-1363 only here . ifwe look at the Earthing Bar – Bigger – if willhelp to clear an Earth Fault much Faster .Yeah

10.0mm[SUP]2[/SUP] / Bigger . CPC .
16.0mm[SUP]2[/SUP] / Bigger . CPC

Amnot using the full Title . ?? Scenario .Domestic Installation – would a 16.0mm[SUP]2[/SUP]help ( Parallel Paths ) :gettree:

 

[amberleaf]

Apprentices. So
( Parallel Path ) ????? :gettree:

Parallel Paths (- through the Heating Pipes ( Combi boiler) Before you start saying . you didn’t need to Earththe Boiler . is the Main Water Pipe Earthed . Metal . Yes
Dowe have ( Parallel Paths ) Cold Water feeds the ( Combiboiler )

Sockets. has Contact with the Backboxs . ( Parallel Paths ) . ?? - 20 x backboxes . 20 x Sockets

Seemto be another ( Parallel Path ) involved .

Scenario ??
i)First Fix . I test my ring ( if not cut ) . Cut . by connecting to a connector Block(s) [ r1 – r1 ] – [ n1 – n1 ] [ CPC – CPC ] you reading is True& Low . ( Continuity )
Scenario ??
ii)if I connect Ring to” just “ Socket(s) would my Reading go up . ?? in Ohm’s
Scenario ??
iii)if I connect Ring to Socket(s) & Backboxes . ?? would my Readinggo up . ?? in Ohm’s

Sodo we have . ( Parallel Path ) … Parallel Path(s) takes many forms . Yeah

 

[amberleaf]

2392-10.
Ifyou have Measured the ( Zs) then you do not need to ◄►Calculate it. :89:

Scheduleof Test Results . Certification & Reporting . – Earth Loop Impedance . ( 0.74Ω ?? ) Straight onto the Schedule of Test Results

“ DoCheck your Polarity “ ( if we look ◄◄ atSchedule of Test Results / p341 ( Equipment Vulnerable to Testing - Confirmation ofSupply Polarity )

Scheduleof Test Results . ( PFC ) – Ze .at the Origin .

From. GN-3 – Two Fold . 2391-10 .

( Ze ) - i ) . Toverify that there is an Earth Connection . ( Live Test - Line & Earth ) with( Parallel Paths disconnected )
ii ) . To verify that the . ( Ze ). value is Equal to or Less than the valuedetermined by the Designer & used in the Design Calculation .

Remember : to Null your TestingLeads on your Tester . ( kA )

RingFinal Circuit .
• Whatto look for (- there maybe a Dodgy Connection) Readingson Sockets
• Measuredvalue of ( R1 + R2 ) could have a Spur . ( Higher value of ( R1 + R2 )

 

[amberleaf]

OldAmber is off on One .

Electricityat Work Regulation(s) 1989 . ( Law )

Part1 . Introduction .
ThePurpose of these Regulation(s)

“ Circuit Conductor “
“Conductor “

“ Circuit Conductor “ means AnyConductor in a system which is intended to carry electric current in normal conditions . or to be Energised in Normal Conditions

“ Assessor“ Two-Fold Here.

Electricityat Work Regulation(s) 1989 . Requires proof that an ElectricalSystem is Safe .

Proper Inspection & Testingof a System by ( Competent People ) the rest I have left out . Etc

Includes(- Combined Neutral& Earth conductor(s)

2391-10(- Not Include a “ Conductor “ provided “ Solelyto Perform “ a ProtectiveFunction . to Earth . ◄◄ Hope this helps you along the Way .

Canthe . Line & Neutral& Earth . Become “ Live “ YES .

2392-10(- Certification & Reporting . ( Electrical Installation Certificate ) are we complying with Electricity at WorkRegulation(s) 1989 . PROOF. :89:

 

[amberleaf]

UsefulJunk .
:willy_nilly:
612.2.1.
AContinuity Test shall be made . it is recommended that the Test be carried outwith a supply having a No-load voltage between( 4V & 24V d.c.or a.c. ) & a
ShortCircuit Current of not less than ( 200mA ) ↔ ( 0.2A )

LowRésistance Ohm Meters
Thiscan take the form of a specialised Low-ohm-meter or . Continuity range of anIR/Continuity tester .

Instrumentto BS-EN 61557-4

Measuringrange must cover a span of at least ( 0.2Ω to 2Ω )
Digitalresolution must be ( 0.01Ω )

 

[amberleaf]

InsulationResistance Testers . for any 2391-10 . Written Exam. :89:

AnInsulation Résistance Test is the correct term for this ( Form ) of Testing .

Ina Written Exam. ( NOT MEGGER ) . This is Not theName of the Test :89:

• TheElectricity at Work Regulations require that all Electrical Systems . includes TestInstruments – be Maintained to prevent Danger .

 

[amberleaf]

SomePoints on Testing . :willy_nilly:

Point to Note : Resistor in Parallel would bring down the Readings .
Point to Note : Could we have a Trapped a Wire / when the Fittings are Screwed back . ?
Point to Note : Do we have a Faulty Socket . ? Spur . ?

 

[amberleaf]

2392-10. Volt drop Calculations ( Start of yourDesign Principles’ ) ▼

Regulation. wise Appendix 12 voltage drop in Consumer Installation / Table 41.3 Maximum EarthFault Loop Impedance ( Zs ) tell us / ( Uo / 230V ) Single-phase

• Cablesshould . where “ Practicable “ Where a cable is partially or completelycovered by “ Thermal Insulation “ Regulation .523.6.6.

 

[amberleaf]

2392-10. ( Design Principles’ you may use One Day) ▼ PS. Not on Consumer Installation(s) – 2391-10 & further down the line .

►►Remember that 2392-10 (- p/48 . Table 41.3.For the Types & Ratingsof Overcurrent Devices listed [ May be Used instead of ►►Calculation] MCBs / RCBOs . Yeah

Firstly(- Other Methods are NOT precluded ◄ :41:

Appendix14 – p/361

Measurement of Earth Fault LoopImpedance . Consideration of theIncrease of the Résistance of Conductors with increase of Temperature . ETC

Therequirements ( 411.4.5. or 411.5.4 ) as appropriate . are considered to be met when the measured value of Earth Fault LoopImpedance ] Satisfies the following Equation

[Zs (m) ≤ 0.8 x Uo/Ia . Uo/230V .. ? 24A= 9.58 → [ Ze/0.8x Zs/9.58 = 7.66 ]

Asthe Regulation . tells us “ Where “

Zs (m) Measured Impedance of the Earth Fault Current Loop up to the most distant point of the relevant circuit fromthe Origin of the Installation . → ( Ω )
Uo . is the Nominal ( A.C. –r.m.s. ) Line Voltage to Earth ( V ) .. Symbols used in Regulations . p/35
Ia . isthe current causing the Automatic Operation ofthe Protective Device within the Time Stated in Table ( 41.1 ) Maximum DisconnectionTimes .

Scenario(- MCB – type B. ( p/249 – BS-EN 6089832A ) .. 32 ÷ 160A = ( Ia / 0.2 ) ↔ ( t )

Or . Within ( 5s ) According to the conditions stated in Regulation( 411.3.2.3. ( A )

i)– Line Conductor to ProtectiveConductor Loop Impedance of the Supply is First Measuredat the Origin of the Installation . ( L / E ) “ Live Test “
ii)– The Résistance of the Line Conductor & Protective Conductor of the Distribution/Circuit(s) areMeasured . [ ExternalLoop Impedance ]
iii)– The Résistance of the Line Conductor & Protective Conductor of the FinalCircuit are Measured .
iv)– The values of Résistance measured in Accordance with ( ii ) & (iii ) areIncreased on the basis of the Increaseof conductor Temperature . Taking into considerationthe ( Design Current – Ib )
v)– The values of Résistance increased in Accordance with ( iv) are Finally Addedto the value measured at ( i ) to obtain a realistic value of ( Zs ) Under Earth FaultConditions .

Remember that 2392-10 (- Formula . Calculating – Earth Loop Impedance. - Zs = Ze + ( R1 + R2 ) – for Apprentices Internal/ External - ( R1 + R2) Simple . Yeah


Foodfor Through – Regulation 433.1.5. ( Wording ) Can generally be Achieved by it / Doesnot say it is Achieved BY ?? :34:

Scenario(- 2392-10 .

( Ib ) in each Leg of cable is not to be Exceed ( 20A ) Current will not bethe Same All Around ( Ring FinalCircuit )

Assumedto be ( 20A ) at far End ( Additional / 12A ) Evenly Distributed. ◄◄◄

(32A + 20 = 52A ÷ 2 = 26A . ( Lvd) 4 x11.5 x 1000 ÷ 26 ÷ 18 = 98.4m )

( In ) 32A + 20A = 52A ÷ 2 = 26A

(-Other Methods are NOT precluded ◄ :41:

( Lvd) 2 x 11.5x 1000 ÷ 20A ÷ 18 = 63.8m .. Point toNote . ( 2 )
( Lvd) 4 x 11.5x 1000 ÷ 20A ÷ 18 = 127.7m .. Point toNote . ( 4 )