***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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Electric Shock : FaultProtection

The term “ Fault Protection “ is used in BS-7671:2008 .to refer to protection of persons & livestock against Electric Shock as aresult of making contact with Exposed-conductive-parts that have become LiveUnder Fault Conditions . ( mainly with regard to failure of basic insulation )

Definitions – p/24. FaultProtection : Protection against electric shock UnderSingle-Fault Conditions .

• Preventing a current resulting from a fault frompassing through the body of any person
• Limiting the current which can pass through the bodyto a Non- hazardous value .
• Limiting the duration of a current resulting from aFault which can pass through the body to a Non- hazardous time period .

Regulation . 131.2.2. ( goes onto Confirm that )

In connection with Faultprotection . The application of the method of ProtectiveEquipotential Bonding is one of the ImportantPrinciples of Safety . ◄◄

:thinking2:

SP&N . single-pole & neutral .
TP&N . triple-pole & neutral .

 

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2392-10 : Design .

Section 133 of BS-7671:2008 Containing ( 10 Regulations) . requires equipment to be suitably selected . ?

Erection )- Initial Verification of ElectricalInstallations / & Periodic Inspection & Testing .
The requirements for erection & Initial Verificationof electrical installations & those for Periodic Inspection & Testing arecontained in Sections 134 & 135 . Respectively

BS-7671 . The requirement contain in Regulation134.1.1. For Good Workmanship & Proper Materials to be Used .Etc.

BS-7671 : Summary of Section 133of Electrical Equipment .

Regulation . 133 . General – Dealing With ?

133.1.1 . Equipment to ( Comply) with standards unless Agreed Otherwise.
133.1.2.
133.1.3.

133.2. Characteristics’ .
Equipment to have suitable Characteristics to suit theValues & Conditions on which the Design of the Installation is based.

133.2.1. Voltage . These FourRegulations require equipment to be selected to SUIT.Voltage : Current : Frequency & Power :
133.2.2.
133.2.3.
133.2.4.

133.3. Conditions of Installation : Equipment to be selected to be suitable for the Environmental Conditions. or otherwise Protected .

133.4. Prevention of Harmful Effects : ElectricalEquipment not to cause harmful effects on other Equipmentor the Supply .

 

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2392-10 : Design .
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► Important Regulation )- 2392-10 . The requirementfor Electrical Equipment to be Installed in Accordance with the ( Instruction(s) Provided by the Manufacturer )
134.1.2.)- Goes on to Require that Equipment must NOTbe IMPARIRED by the INSTALLATION PROCESS .

 

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2392-10 : Design .

Summary of Section 134 – Erection & InitialVerification of Electrical Installations . & Section 135 – Periodic Inspection& Testing .

Regulation .

134.1.1. – Good Workmanship & proper Materials .
134.1.2. – Characteristics of equipment Not to be Impaired.

134.1.3. – Conductors to beIndentified .
134.1.4. – Electrical joints& connections to beAdequate .
134.1.5. – Design temperatures NOTto be Exceeded .
134.1.6. – Equipment causing high temperatures or Arcsto be suitably located or guarded .

( T ) 134.1.7. –Suitable warning sigs & / or notices to be provided where necessary forSafety Purposes.
( T ) 134.2.1. – Initial Verification )- Appropriate Inspection & Testing to beperformed during Erection & upon completion of an Installation or an Additionor Alteration . & Appropriate Certification to beIssued .
( T ) 135.1. – PeriodicInspection & Testing )- Recommendation for subsequent Periodic Inspection& Testing required .

 

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2392-10 : Design .
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Whilst “ Maximum Demand “ & Diversity “ are notdefined in BS-7671:2008 .

Following Meanings ..

• Maximum Demand )- The Maximum anticipatedinstallation “ Load “ plus allowance for ( Future LoadingDesigning )◄◄ Potential Load
• Connected Load )- The totalof all Electrical Loads . ≈
• Diversity )- The ratio of Maximum Demand to the ( Connected Load )

Diversity . Therefore can be represented by .

Diversity = Maximum Demand ---- Connected Load.

Diversity )- is never going to be more than ( Unity or 100% )percentage terms .

 

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2392-10 :
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( T )Verification )- Regulation. p/31
All measurements by means of which Compliance ofElectrical Installation with the reverent requirements of BS-7671 . are “Checked “ comprising of Inspection . Testing & Certification

 

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Protection against Electric Shock . ( Making you Aware of it )
:41:
Fundamental Requirements . ( Some times it’s rightunder our Noses )

Part 1 – Chapter 13 . 131.2.1. / 131.2.2.
Protective Measures – Part 4 . Chapter 41 – ( All )
Devices for Fault Protection . – Part 5 . Chapter 53 . –Section . 531 . ( All )
Earthing & Protective Conductors . Part 5 . –chapter 54 – ( All )

Part 7 . – Special installations or locations (Particular Requirements )

Chapter 41 (- Appropriate measures for Protection against Electric Shock ( divided into nine sections ) ◄◄

Section – 410 : Introduction .
Section – 411 : Protective measure : Automatic Disconnectionof Supply . ( ADS )
Section – 412 : Protective measure : Double or Reinforced Insulation.
Section – 413 : Protective measure : Electrical Separation .
Section – 414 : Protective measure : Extra – Low –Voltage . provided by ( SELV or PELV )
Section – 415 : Additional Protection.
Section – 416 : Provisions for BasicProtection .
Section – 417 : Obstacles & Placing out of Reach .

Take you Pick .

 

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Protective Measures that are Generally Permitted .
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Protective Measure :
411 )- ADS
412 )- Double orReinforced Insulation .
413 )- Electrical Separation for the Supply to One Itemof Current – Using – Equipment .
414 )- Extra – Low – Voltage – ( SELV or PELV )

 

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Basic Insulation of Live Parts .
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2392-10 )-
i) T&E – PVC . ( Insulation of Live Conductors )

Basic Insulation of Barriers & Enclosures .
i) Consumer Unit . – Barrier ( Protection from Contactwith Busbar )
ii) Consumer Unit . – it has an Enclosure

Why ?? itprevents Physical & Electrical Contact with LiveParts .

Just a Point to Note here :
Regulation 416.1. calls for Liveparts to be covered by Insulation ( That canOnly be Removed by Destruction ) ◄◄

Would you . ?? Use Sleeveor Tape of Insulating Material is Inadequate ifonly because ( itmay be Removed without Destruction ) ◄◄

 

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2392-10 : Basics .

Ring Final Circuit .

The normal requirement relating to Protection againstOverload are relaxed in the case of RingFinal Circuit .
Regulation . 433.1.5.provides a “ deemed to comply “ Status for Protection against Overload wherecertain conditions are a prerequisite .

• The Current – Carrying – Capacity ( Iz ) of the cable must be not less than ( 20A )
• The LoadCurrent in any part of the circuit must be UNLIKELY to Exceed for long Periods the Current – Carrying– Capacity ( Iz ) of the cable

 

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Apprentices: Something(s) to thing about . Yeah .
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Protectionagainst Fault Current .

Fault Current is an Overcurrentcaused by either a Short-Circuit .

- BetweenLine Conductors ) or an EarthFault .
- BetweenLine Conductor & an Exposed –Conductive-Part. or Protective Conductor .
- Faults occur as a result of Insulation Failure .
-Bridging of Conductive-Parts by some Conducting-material

Numberof possible Fault Conditions which needsConsideration .

-Short Circuit Faults between Line & Neutral .
-Short Circuit Faults between Line & Line .
- Earth Faults between Line& Exposed-conductive-parts or Protective Conductor.

 

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ForApprentices . Learning Curve . Only – It startsyou working the Regulations . Designing . ◄
:waving:

Wehave a ( 2.5mm[SUP]2[/SUP]) Twin &Earth thermoplastic ( 70°C ) cable installed inConduit with two-other circuit(s)
TheConduit is clipped direct to a masonry wall but it passes through (100mm ) of thermal insulation at an AmbientTemperature of ( 35°C )
Itis protected by a ( 20A ) BS-EN 60898/ MCB .

Wenow need to reference the rating factors . Grouping ( Cg) is the first [ this can be found intable ( 4C1 / p/268 ) Three circuits Enclosed is ( 0.7.) 0.70

(100mm )► Next is thermal insulation ( Ci ) which is found in table ( p/104 - 52.2. & is 0.78 ) & finally we need ( Ca )
( 35°C ) ► ( Ca )Ambient Temperature . Which is found in table ( p/267 - 4B1 & is 0.94. )
Ourcorrection factors are :

Cg – 0.7.
Ci – 0.78 .
Ca – 0.94 .

4D1A )- Example : you areusing . We know the minimum current rating of the cable has to be ( 20A ) rating of the protective device . & we are protecting the circuit fromOverload .

Asit is enclosed in Conduit but clipped direct wecan use the Method ( C we can use this method because we are using separate factors for Insulation .

Tocalculate the current which a cable can carry under these conditions ( Iz ) we need to select a cable .

ForApprentices / Your trying ( 4.0mm[SUP]2[/SUP]) from - column ( 6) – ( 37A ) x 0.7 x 0.78 x 0.94 = 18.98A – As wehave a ( 20A ) protective device this cable willbe Unsuitable .
ForApprentices / Your trying ( 6.0mm[SUP]2[/SUP]) from - column ( 6) – ( 47A )- which clipped direct carry ( 47A ) before any factors are Applied .

( 47A ) x 0.7 x 0.78 x 0.94 = 24.12A -( Iz ) Apprentices So will this Cable be Fine .

 

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Workingmy way around the Regulations . “ Apprentices “

Classification of External Influence . ( AD1– Negligible / IPX0)

Firstly )- Definition – p/24 / Part 2 .
ExternalInfluence (- Any Influence external to an electrical installation which affectsthe design&safe operation of the installation.

TheseExternal Influence are shown by the use of Codewhich has two capital Letters & Numbers .

i)-Your first Letter shows the category of - ExternalInfluence
ii)- External Influence relates to the Numberindicates the Level of the External Influence .The higher the ( Worse Influence )

•the first Letter indicates which category the External Influence .
• Thesecond Letter shows what the External Influence relates to the Number indicates theLevel of the External Influence . The higher theNumber the Worse the Influence .

TheFirst indicates which category the External Influence fallsunder

A ) is Environment ( Wet . Dry. Humid . Dust . Vibration . )
B ) is Utilisation ( How isthe building being Used ? . – What is the capability of the person using thebuilding ?. – Are they ordinary personsor perhaps are they Disabled ? .

-How difficult would be if Required
-Does the building contain materials which could Explode .

C ) is Construction ofbuilding . – ( What is the building constructed from ? ) Will it be likely tomove ?

TheSecond Letter indicates what the External Influence couldbe.
IfYou refer to Appendix 5 . p/318 . Look in theEnvironment section ( Category – A ) You cansee that if the Second Letter was ( K ) itwould refer to Flora ( Plant . Moss & the like )

TheLast digit is a Number & indicates theLevel of the Problem which the External Influence islikely to be .

Example: if we look at an External Influence given aCode of ( AD1 ◄ ) We can see that the Letter ( A ) relates to the Environment. & the Letter ( D ) relates to Water & the final digit ( 1 ) Tells us that the problem likely to be caused by Water is negligible .

Whenwe require further information with regard to this ExternalInfluence . We need to look further into Appendix 5 . When we find ( AD1 ) We can see that it relates to Areas where Weather Protection is Notrequired & that the ( IP rating ) forthe equipment is ( IPX0 ) As the External Influence relates to Waternothing is specified for Dust ( X ) & theLevel of protection against protection against Water is ( 0 ) Which means No Protectionrequired . ►► p/323 – ( AD1 – Negligible / IPX0 )

Youcan see that ( AD8 ) is where the Equipment islikely to be totally Submersed . such as in a ( SwimmingPool ) The IP rating given as protectionis ( IPX8 ) . X . shows that nothing is specified for Dust& ( 8 ) shows that it must be protectedagainst the effects of ContinuousImmersion in Water.

Indexof Protection Codes is Used throughoutBS-7671:2008 . To identify the Minimum Level of Protection .

“Apprentices “ To Covert back to “ Black “ ◄◄

Scrolldown all – by holding the Left Key down on the Mouse . & let go . it willbe all in Black Highlighted
Goto your top browser right hand corner ( A - Hit the Arrow ▼ . ◄ hit the Black Button– Automatic )

 

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Protectionof Conductor(s) in Parallel against Overcurrent .

Toinstall conductor(s) in Parallel . most commonoccurrence is . 2392-10 . Installing Ring Final Circuit .
Byusing Two-conductor(s) which are joined at the Far-End . Your circuit canprovide socket outlet(s) over a greater Area .

Thecurrent can be shared between Two-conductor(s)

Regulation(s)– Appendix 15 p/362 . Informative .

i)Locating sockets outlet(s) to provide reasonable sharing of the Load around the Ring .
ii)Taking account of the Total Floor Area being served – ( 100m[SUP]2[/SUP] )

2392-10. As from your point of View . This will result in a Lower Earth Fault Loop Impedance & Voltage drop. ( Vd )

Regulation(s) 612.14 –Red / Cover BS-7671:2008 .
NOTE )- Verification of voltagedrop is NOT normally required during Initial Verification )

Verificationof voltage drop – GN-3 . ( 3 ) P/57
NOTE )- Verification of voltagedrop is NOT normally required during Initial Verification )

Regs/ Appendix 12 - Informative . p/358
Apprentices– gives Maximum values of voltage drop for either Lighting or other Usesdepending upon whether the installation is supplied directly from ?
Apprentices– Wording ) LV distribution system ( Domestic Installation(s)◄ . OR from a private LV supply ◄ .

RingFinal Circuit (- protected by a 32A device wired in ( 2.5mm[SUP]2[/SUP]/ 1.5mm[SUP]2 [/SUP]) 70°C thermoplastic . Twin-&-Earth ( 4D5 ) Designing point of View .

2392-10- What’s going for the “ Ring “ → Cables is in Parallel . Your halving theLength & Doubling the Cross Sectional Area . ( C.S.A. )

Example Only : ◄ for calculation purposes only

Wehave a Radial Circuit – Protected by a ( 20A device ) – ( 2.5mm[SUP]2[/SUP]/ 1.5mm[SUP]2[/SUP] )
Oneway to look at this !! Maximum Length of the circuit will be limited by voltage drop which is Limited to a Maximumof ( 5% - 11.5V )◄

Calculatethe Maximum Length of the cable . Transpose / Formula – We normally use forvoltage drop . ( Vd )

mVx A x L --- 1000 = Vd .
Transpose/ this for Length . Vd x 1000 --- A x mV = L
Replacing/ Symbols / Figure . - sum in two parts - ( 11.5 x 1000 = 11500 ) ( 20A x 18mV = 360 ) → 11500 ÷ 360 =31.94m . ( L )

Apprentices- Voltage drop islimited to a Maximum of ( 5% - 11.5V )
5%of 230V = 11.5V … ( 5 x 230V ÷ 100 = 11.5V ) Your Calculation for Power / as theRegs say – Other Uses .

AtRadom - 2.5mm[SUP]2[/SUP] / 4D5-Flat . PVC . cables – 18mV

 

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►Very Basic(s) we’ll leave it at that .

Apprentices. you’ll never get this on Exams .
:waving:
Effectof harmonic currents on a balanced Three-phasesystem . ( The Whys )
Harmonicsare caused by electronic equipment which converts ( 50Hz . a.c. to d.c. ) & then backto . a.c. . at a different frequency .

electronicequipment does not usually have constant Impedance . This is because theimpedance changes due to the constant switching of the electronic componentswithin the equipment .

thisswitching On & Off during part of the waveform results in current beingIntroduced back into the distribution system .

thisis because the Currents are at frequencies other than ( 50 Hz ) it is assumedthat a third harmonic is at ( 150 Hz ) the effect of this is Increased Currentin the Neutral .

whichin certain circumstances will be Higher than the Current in the LineConductor(s)

- Uninterrupted Power Supplies – ( UPS ) **
- Computer Power Unit(s) – ( CPU ) **
-Electronic ballast(s)
- Discharge Lighting . ( florescent . mercury .sodium . ) Etc .

TheCalculation(s) has been sent down on one of the pages.