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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 ) :)
 
Workingmy way around the Regulations :

Switch Wires )- Appendix 7 - p/343
Wherea two-core cable with cores coloured Brown& Blue is used as a Switch wire , bothconductors being Line conductors . the Blue conductor should be marked Brown or ( L ) at its terminations .

Intermediate & Two-way Switch Wires )- p/343
Wherea Three-core cable with cores coloured Brown .Black & Gray is used as a Switch wire , All Three conductors’ being Line conductors . The Black & Gray conductors shouldbe marked Brown or ( L ) at their terminations .

2392-10As Designers Are you aware the BS-7671 is British Standards . ina “ Court of Law “ ( Compliance )
TheO.S.G. is only Guidance’s

Appendix3 – ( Informative ) its tell us ◄
Time/CurrentCharacteristics of Overcurrent Protective Dives & ( RCDs )

Time-DelayedResidual Current Devices ( Type S )
Example: Appendix 3. Table 3A p/243 .

Two– RCDs )- Characteristics of the Two devices
itis required to select Suitable devices to EnsureDiscrimination . ( PS - Discrimination. does not Only come with Fuses ) point to Note.

i)The downstream device . is selected to be a ( General Device - BS-EN 61008-1 ) … This is theFIRST Line of DEFENCE. its rated residual operating current (I∆n ) selected 30mA. – The Operating Time . in accordance with the FIRSTLINE

Example :
Ata its rated residual current of ( 30mA – Device will operate within 0.3s ) ↔ ( 60mA – Devicewill operate within 0.15s ) ↔ ( 150mA – Device will operate within 0.04s )

ii)The Upstream Device )- as Designers’ you haveselected it to be ( Type S BS-EN 61009-1) Device in accordance with the second& third lines of the Table
Therated residual operating current ( I∆n ) is selected to be 500mA. The operating time . in accordancewith the second & third lines for such a device .

Ata residual operating current of )- 500mA– 1A – 2.5A .
Thedevice will NOT operate before )- 0.13s – 0.06s –0.05s .
Butwill OPERATE within )- 0.5s – 0.2s – 0.15s .

Itis now required to Establish whether correct Discrimination between the Downstream general device & the Upstream time-delayed device will be Achieved under EarthFault Conditions. Values of Earth Fault Current in the Socket-outlet circuit are considered .

•30mA )- Should an Earth Leakage Fault of 30mA. develop on the Socket outlet circuit . the Downstreamwill Operate within ( 0.3 seconds ) The Upstreamdevice . being a ( 500mA ) Will not “ SEE “this level of EarthFaultCurrent & WILL NOT OPERATE

•150mA )- The Downstream device operates within( 0.04 seconds ) The Upstreamdevice will “ NOT “ operate ( being a500mA device )

•500mA )- The Downstream device operates within ( 0.04seconds ) The Upstream time-delayeddevicewill “ NOT “operate during the first ( 0.13s ) of the Faultthereby allowing the Downstream device to Disconnect .

Herecomes the How’s & Whys –
Note : The Upstream devicewould operate if the Fault Persisted for ( 0.5 second ) DUE tofailure of the Downstream device

In – is the Rated Current ofthe Device .
( RCD ) I∆n - is the RatedResidual Operating Current .

RememberApprentices . You need a MCB – for Overload .
RCD– is for Additional Protection ( RCD alone will NOTdetect OVERLOAD )

Asyoung Designers – Head for Regulation / Definitions p/29
RCBO – is MCB / RCD )-Overload & or Short circuit .
RCCB - )- Overload & orShort circuit .

► Youmust give Discrimination to allow the 30mAdevice a chance to Operate first ( 100mAtime-delayed / 30mA General Device )
 
This bring me to another point !!

Regulation514.12.2 . p/94

Wherean Installation incorporates an RCD a “ Notice “ shallbe fixed in a prominent position at or “Near “ the Origin of the Installation . etc.

Thejob I was On . I asked the Lady when was the Last time you Tripped your RCD . (What RCD ?? ) – Lets put it another way . when did you get the CU . 8 Years ago
The RCD Incomer had Seized Up . The MCBs had to take the Fault

Sowe’ll take this as a Learning Cure .
Megger1552 . 3 x Tests + 1 – Ramp Test

Weare Working on ( General Purpose ◄ / They areNon-delayed RCDs )
( ½ ) ↔ ( 50% ) is 0.5 x 30mA .No Trip
( 100% ) is ( x 1 –This must cause the RCD device to Trip ( 300mS )◄◄ Note :BS-EN ◄◄ ( -&-s BS/ Old British Standards must go Out in 200mS )
( 500% ) is ( x 5 if we are using a 30mA Thismust go out within 40mS

Example. The

Regulation411.3.3. . requires that Socket outlets . Not exceeding 20A . That are for useby Ordinary persons intended for General Use .
AreProtected by an RCD rated at ( 30mA ) . With anoperating time NOT exceeding ( 40mS ) at a residual current of ( x 5 I∆n ) I∆n - is theRated Residual Operating Current .

If -&-s asked Q/As you about 100mARCD : in Exam - what would you Test an100mA RCD on ( 0.5 x . x 1 . x 5 ) Whatis x 1 / 100%
 
As a Tester :
As a Tester :

Insulation Résistance Testing

Removeall Lamps from Fitting where Accessible ?

Greater care must betaken. As during this Test ( 500V D.C. ) This could damage the Equipment .

- If we leftequipment which is left connected ?? ( Any accessories with indicator Lamps areswitched OFF . “ Neon’s

Neon Indicator Lamps. Will recognized as a Load by the Test Instrument & will give a Very LowInsulation Value.

All that is requiredis for the Switch on the accessory to be Turned OFF.
 
Brackets“ Apprentices

These should be used to carry out calculation within a calculation .

32--------------------- ( 0.8 x 0.65 x0.94 ) = 65.46

Enter in to Calculator )- 32 ÷ ( 0.8 x 0.65 x 0.94 ) =

( 0.8x 0.65 x 0.94 ) = 0.4888 . )- 32 ÷ 0.4888 = 65.46
 
kW)- True power ( x 1000 )
kVA)- Apparent power ( x 1000 )
kVAr)- Reactive power ( x 1000 )

Load )- The current drawn by ElectricalEquipment connected to an Electrical Circuit .
Prospective Short Circuit Current . – The maximum currentwhich could flow between live conductors . ( PSCC)
Prospective Fault Current - The highest current which could flow in a Circuit due to aFault . ( PFC )

Regulation612.11. Prospective FaultCurrent

(PFC ) → to any of ( PSCC ) or ( PEFC )
The ProspectiveShort Circuit Current & Prospective EarthFault Current shallbe Measured . ETC ( at the Origin ) – Consumer Unit / Tails

(PSCC ) – Line & Neutral
(PEFC ) – Line &Earth
 
Q)The Instrument to be used for a Polarity test . Before the Supply is connected. ?
A)This test is usually conducted while measuring the value of ( R1+ R2 ) where a Low Reading Ohm Meter is used .
(if conducted as a separate test any from of Continuity Tester is used )

O.S.G.p/77
( I must stress O.S.G. is Guidance only )

Guidanceon Initial Testing of Installations .
ii) Checking that the TestInstrument is made in accordance with the appropriate safety standards such as( BS-EN 61243-3for Two pole voltage detectors & ( BS-EN 61010 or BS-EN 61557 for Instruments .
iv) Observing the safetymeasures & procedures set out in HSE Guidance NoteGS-38 for all Instruments . Leads . Probes & Accessories . ETC .

O.S.G.with the Supply Connected
vi)Check “ Polarity “ of supply . using anApproved Voltage Indicator .

Q)An instrument that may be used for a “ Live “ Polarity Test – after the Supplyis connected .
A)A voltage indicator meeting the requirements of ( GS-38) HSE … This will come up a lot on “ Exams “( GS-38Leads )

Specification)- BS-EN 61243-3 : 1999 –Live working . Voltage detectors . Two-pole low voltage type .
Specification)- BS-EN 61010-1 : 2010 -Safety requirements for electrical equipment for measurement . Control . & Laboratory use ( General Requirements)

IfInstallation(s) are to be Tested to show that they complywith BS-7671 – The following instruments will be Necessary

Testing)- 612.1.
The test of Regulations 612.2. to 612.13.ETC.

Measuringinstruments & monitoring equipment & methods shall be chosen inaccordance with the relevant parts of ( BS-EN 61557 ) if other measuring equipmentis used . it shall provide NO LESS degree of performance & SAFETY .

- Lowrésistance ohmmeters . ( BS-EN 61557- 4 )
- Insulationrésistance ohmmeters . ( BS-EN 61557- 2 )
- Earthfault loop impedance testers ( BS-EN 61557- 3 )
- Earthelectrode résistance testers ( BS-EN 61557- 5 )
- RCDtesters ( BS-EN 61557- 6)
 
Insulation Résistance ( IR )

2392-10. Dead Test - Line& Neutral to be Linked out for this Test( Megger 1552 - Two lead(s) Lead one L&N )– Lead two Earth
Instrumentto be Used. ? … ( Insulation ResistanceTester ) for Exam purposes -&-s ►► PS NOT MEGGER

ElectricalInspection & Testing .

ContinuityTesting . ( Closed Circuit )
Examplelittle ( r[SUP]1[/SUP] ) end to endLine conductor ( Closed Circuit ) Youwill get a reading Ωs ?? ( r[SUP]1[/SUP]) to ( r[SUP]1[/SUP])
 
AnInspection & Test which is carried out on a NewInstallation to prove Compliance is called an InitialVerification .

( EIC ) The Electrical Installation Certificatemust be accompanied be a Schedule of Test results & Schedule of Inspection
Withoutthese two documents’ . The Electrical Installation Certificateis not Valid .

Regulation. Chapter 63 p/163 .
632.1. Initial Verification

Followingthe Initial Verification required by Chapter 61. anThe ElectricalInstallation Certificate. together with a Schedule of Inspections & Schedule of Test results .shall be given to the person ordering the work .
 

Thiscalculation is for BS-EN MCBs

MCBtype B . Time must be achieved when a currentof between ( 3 & 5) times its rating passes through it
MCBtype C . Time must be achieved when a currentof between ( 5 & 10) times its rating passes through it
MCBtype D . Time must be achieved when a currentof between ( 10 & 20) times its rating passes through it (disconnection of the device Ia )

Producetable(s) for Zs values . a specificvalue of current is required & the worst case is Used. ( B – 5 . C – 10 . D - 20 )

Whenthese values are used it is simply a case of ApplyingOhms Law – to the supply voltage 230V
&( Ia ) to obtain a ( Zs ) value

Thiscalculation will satisfy regulation 411.4.5. ( Zs x Ia ≤ Uo )

MCB– B : The ( Zs ) for a 20A type B ?? Calculated .
5 x 20= 100 ( 230 ÷ 100 = 2.3Ω )

MCB– C : The ( Zs ) for a 20A type C ?? Calculated .
10 x 20 = 200 … ( 230 ÷ 200 = 1.15Ω )

MCB– D : The ( Zs ) for a 20A type D ?? Calculated .
20 x 20 = 400 … ( 230 ÷ 400 = 0.57Ω )



 
Working my way around the 17[SUP]th[/SUP] Edition.

Reduced– voltage transformer ( Centre Trapped ) ◄◄

(Zs ) values for Circuit Breakers ( Table 41.6 – p/54 )
( Calculatedin the same way as the Circuit Breakers ( Table 41.3 – p/49 )

The only difference is the Voltage ( 55V or 63.5V ) Table 41.6 : We arenot Calculating 230V ◄◄

32Atype B – MCB . ( 32 x 5 = 160 )
(55 ÷ 160 = 0.34Ω )

Maximum Earth Fault Loop Impedance ( Zs ) for 5 sec Etc
&Uo of ( 55V ) Single Phase .
&( 63.5V ) Three Phase
 
MCB– BS-EN 60898 & BS-EN 61009-1 protective devices . Usually have two short circuit capacitiesmarked on them .

(Icn ) Which is the rated Short Circuit Capacity – is the Maximum Current whichthe device could interrupt without causing ant damage to other Equipmentsurrounding it .
( Ics) Which is the ( In-service Rated Short Circuit Capacity ) is the Maximum Fault Current which the device could interrupt safetywhilst remaining safe to Use .

Itis better to select a device with an ( Ics )which is higher than the ( PFC ) as the devicewill still be fine for continued use .
( Markedon the MCB - Icn ?? 6000 or 10.000 )

BS-EN 61009-1 / RCBOs









 
Insulation Résistance : ( 3 Test ) Regulations . [ Testing IR at theMain Switch ]

Dead Test .
Safety Isolation Procedures’ . are required hereprier to Testing

This can be carried out at the “ Main Switch “ of the Consumer Unit being Tested . You mustmake sure that the Main Switch is on the ONposition with all the Circuit Breakersfor all the Circuits to be tested . in doing this you must ensure that “ Lamps . Dimmer Switches . & Appliancesare removed Prior to Testing .

The Test is carried out between the Line & Neutral - using the Insulation Résistance Tester [ Megger ] Using the 500MΩ Volts scale
500V is applied – Your reading should be Greaterthan 999MΩ is Obtained .
The Test is carried out between the Line & Earth -using the [ Megger 1552 ]
500V is applied – Your reading should be Greaterthan 999MΩ is Obtained .
The Test is carried out between the Neutral & Earth -using the [ Megger 1552 ]
500V is applied – Your reading should be Greaterthan 999MΩ is Obtained .

Reason(s) :
This proofs that the Line& Neutral are not touching :
This proofs that the Neutralto Earth are not touching :
This proofs that the Line&Earth arenot touching : Throughout the Circuits’ .
 
Polarity ( 4 ) : This Test is the Forth Test . “ DeadTesting “ “ Wiring Regulations “

Safety Isolation Procedures’ . are required hereprier to Testing
This Test is to ensure that all Switches Fuses& Circuit breaker are in the line conductorOnly & Socket-outlets are wired correctly . 612.6
In this case am only Testing the Lighting Circuitonly . ↔ Polaritywould have been Tested through the Continuity Ring Final Circuit Test .
Very similar to Continuityof CPC. Although a read is not required . So the ( R1 plusR2 – Need Not be Recorded )

Consumer Unit – Inside . Having removed the Line Conductor from the circuit to be Tested . it isplaced into the Earth bar ( Met )
The Test Instrument to be used is then brought toeach point in the Lighting Circuit to ensure that Polarity is correct . while areading is not required this can be put on the Buzzer or Bell set mode . ↔ The Test is carried out between the Switch live & Earthat the Ceiling Roseor Batten Holder → with the Switch on the OFF position “ No Buzz “

Switch - the Switch ON it Indicates a Buzzing“ The polarityis correct . if this was carried out using a Low-résistance Ohmmeter thesame Test would show the meter reading from greater from scale ( 999MΩ ) down the a lowrésistance 0.15Ω in the same in the case of ( R1 plus R2 ) forContinuity of CPC
 
“ Useful Junk “
( 999MΩ ) megohms is a Near-perfect résistancereading for a Wire/Cable

The Megohmmeter . or Megger 1552 – multi function tester . is a measuring device thatTests high Electrical Résistance .
Typically these measurements aremade on Electrical Wires & Motor Winding to test the Insulation value ofthe wires . The prefix “ Meg ” describes a numerical value of 1.000.000. in thecase of an Ohmmeter that value is also described in Ohms .

An Electrical/ Electric terminal positive polarity 612.6

 

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