***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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Continuity of CPC.(1 test ) : r[SUP]1[/SUP] & r[SUP]2 [/SUP].

Dead Test :
Safety Isolation Procedures’ . are required hereprier to Testing ☑
Continuity of CPC.Lighting circuit . having identified the right Circuit Breaker

i) Remove the Lineconductor from the Circuit breaker . Then place it into the Earth bar ( MET ) –Line & Earthtogether
if we look at the Test books some indicate the Link between the top of the Circuit breaker to the Earth bar .
ii) Some indicate the CPCfor that circuit is removed & the Line removed& placed into a connectorblock [ anyone of the will do ] “ Loop “

You must go to the furthest point . & With aTest Instrument low-ohmmeter place it on the Low-ohms scale ( Ω ) The Test isthen taken between the CPC & Line conductor at theCeiling Rose :- in this case a reading of .16Ω that is obtained. proofs continuity of the CPC .
• To proof and theory to the ↔ Consumer Unit breakthe Line conductor go back & take a reading . [ it shows greater than rangeproving continuity was CPC was Testedinitially ]

 

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The PSC { prospective short circuit current }is the maximum available current which can flow in the event of a fault betweenLive conductors { Line – Line } or { Line – Neutral } it iseither directly measuredor if your meterwon’t do direct { Line –Line} then
{ Line – Neutral doubled will suffice }

Line & Neutralcables are ShortedCircuited is called the Prospective Short Circuit Current PSCC .

PSC ↔ ( MCB ) ↔ Protectivedevice must be capable of breaking it Safely ◄ ≈ A.C. Sign .

PFeC { prospective earthfault current } is the maximum fault current between a Liveconductor and earth . { Fault Current is ShortCircuit }
This can measured directly or
by dividing the maximum Line– Earth supply voltage by ( Ze )

PFC {prospective short circuit current } is either the PSCC or PEFC . Whichever isGreater .

By Calculation’s to Obtain PSCC { 230 ÷ by Ze 0.15Ω = 1.53kA } Supply has been connected .
{ 400V ÷ Ze 0.15Ω = 2.6kA }

When we are referring to Design ! Designing for an Installation for Domestic . Design stage – Design purposes.

 

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Broken RingFinal Circuit ? ( As a Tester this is what -&-s arelooking for )

However . We find the Ring to be Broken . the Protective Device couldnot do it’s job as it is rated at 32A & the cable is rated only at 20A .Hence Overloading .

TheRing ? becomes Two Overload Radial . ( From a Tester point of View )

TheRing is a Continuous Loopno matter how it has been designed

p/362)- Appendix 15 ( Informative ↔ telling us )
Ring& Radial circuit arrangements

1) Ring Final Circuit arrangements Regulation– 433.1.
( Informative ↔ telling us )
TheLoad current in any part of the circuit should be ( Unlikely) to exceed for long periods the current carrying capacity of the cable ( Unbroken Ring Final Circuit )
Regulation– 433.1. refers ) This can generally be achieved by .

iv)Taking account of the total floor area being served ( The Regulations aresaying - Rule of Thumb . a Limit of 100m[SUP]2[/SUP][SUP] [/SUP] has been adopted) Ring
Nowif we look at the O.S.G. p/158 . A1 Ring 30 or 32A . 2.5 / 1.5 . – Maximum floorarea served ( m[SUP]2[/SUP] ) 100m[SUP]2 [/SUP]

i)locating socket outlets to provide reasonable sharingof the load around the ring .

 

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Consumer Unit Arrangements . Design Stage .



Part “ M “ requires that reasonable provision be made for people togain Access to a building and use its Facilities .

Prescribes that Switches . Sockets outlets . &“ Other Equipment “needs to be at Appropriate Heights .

Wording → “ Other Equipment“ Consumer Unit . Height of between 0.45m from the floor level Up . 0.⇈ to 0.45m ↳ 0.45m to ↱1.2m from floor level .

 

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2392-10 .

Who is responsible for Inspection & Testing .? The person carrying out the Inspection( YOU )

Remember ? Inspection& Testing . are two different words ( Inspection ) ( Testing )

 

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2392-10 )- Work Near Electricity “ Assessor “

You are doing a ►► Risk Assessment

Introduction to Short Electrical Training :
2391-10 – 2392-10 –

All Dead Testing should have Safety IsolationProcedures’ prier to Testing .
Safety Isolation Procedures’ . are required hereprier to all Testing ☑

All Dead Testingshould have Safety Isolation Procedures’ prier to Testing .

i) Using GS-38 Prove Test Leads .
ii) first thing to do is Isolate the Supply . That is going to beTested . - Take the GS-38 test leads to a known sourceto ensure that theyoperate correctly
“ if they are workingaright “ they go tothe supply that is to be Tested – then Test between Line to Neutral - Line to Earth - Neutral to Earth .
iii) Then the Test Leads are replied to a known source to ensure that they are still Operatingcorrectly .

iv) At this stage you are placing a Locking OFF- device – followed by Signs in the area that is tobe Tested .

 

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Execution:

SinglePole ( 1P )
SinglePole & Neutral ( 1P + N )
DoublePole ( 2P )
ThreePole ( 3P )
ThreePole & Neutral ( 3P+N )
FourPole ( 4P )

ElectricalEndurance : MCB
≥32A .. Over .
≤32A .. Under .
AmbientWorking Temperature ( °C )

 

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Part( 1 ) Sorry am having problemdownloading Amber .

DomesticInstallation )- 2392-10 . “ Electrical Calculations’ “

100A supply . ( Scenario )

ElectricShower 50A : ( Power Circuit )
Cooker– 32A :
RingFinal Circuit ( 1 ) 32A :
RingFinal Circuit ( 2 ) 32A :
ImmersionHeater 16A :
UpstairsLighting 6A :
DownstairsLighting 6A :

Add )- up the requirements ofEach Circuit ( 50A + 32A + 32A + 32A + 16A+ 6A = Total Demand 174A )
►►This is because the CIRCUITS are RARELY LOADED to theirFULL CAPACITY . ◄◄ at the One Time

Apprentices. You are Using the Principle of Diversity
Note )- NOT Total Demand 174A

Shower )- Classed as Instantaneous Water Heater )- O.S.G. Table 1B – p97
Row– ( 5 ) NO DEVIRSITY . Plain English 100%

Apprentices. Only the rated LOAD of the Appliance should BE USED …….. ( NOT the RATING of the PROTECTIVE DEVICE ) ◄◄◄ MCB

Shower[ 10.000 ÷ 230V = 43.47A ] Power Circuit.

Cooker)- As a designer . Would you apply Diversity ( YES )

1x 4 kW – Oven :
2x 1.5kW – Hob/rings :
1x 2kW Hob/ring :
1x 1kW Hob/ring :
1x 2kW – Grill :

Totalpower required = 12kW … 12.000 )
TotalLoad ( Amps ) - 12.000 ÷ 230V = 52A

Itis NOT likely that all of the elements of the Cookerwill be Used at the same time .
Dueto the required cooking times of the food the Oven will be On first . & itwill have reached its required temperature before the Hobs are Used .
[The Cooker will have ( Thermostats & SimmerStats build into it which will Switch ON & OFF to maintain the Cooking Temperatures ]

Asform a Designer point of View ? For aShort periods of time an Overload could Occur ) if all Loads on at the SameTime . Xmas Time .
Regulation133.1.1. – Selection of Electrical Equipment . Every item of equipment shall COMPLY with the appropriate BritishStandard .

RegulationBS-7671:2008 . Suggest that a small Overload isbetween ( 1.25 & 1.45A )
RegulationBS-7671:2008 . Protective Devices have toCOMPLY with certain requirements .

(Table 41.3 – p/49 . MCB / RCBOs ) … RCBOs are – MCB/RCDs in One ) Overload & / or Short Circuit :
BS-EN60898 & BS-EN 61009-1 … Protective Devices must CARRY a CURRRENT of atLEAST ( 1.13 times the device rating for 1 hourwithout operating -
Theymust OPERATE within 1 hour at a Maximumof ( 1.45 times their Rating )

 

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Part( 2 ) Sorry am having problemdownloading


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Apprentices, The current which causes Operation of the PROTECTIVEDEVICE in the conventional timeis Shown by the Symbol ( I[SUP]2[/SUP] )



Diversitycan now be applied .

RingCircuit(s) – O.S.G. . Table 1B - row 9.. ( Only Applies to Domestic Installations ) 2392-10



Diversityis Calculated at 100% .. for circuit ( 1 ) & 40% for circuit ( 2 ) – Plain English Largest / 100% - Second / 40%



O.S.G.– Table 1B row 9 . Ring ( 1 ) Largest / 100% - 32A

O.S.G.– Table 1B row 9 . Ring ( 2 ) Second / 40% ( by Calculations 40% is 0.4 = )



Nowthe Calculation is ( 32A x 40% = 12.8A ) 32A x 0.4 = 12.8A

PlainEnglish Largest / Second .. ( 32 + 12.8 = 44.8A ) → Total allowance for Ring =44.8A



O.S.G.the hint is in the Name 1B ? Immersion Heaters & Thermostatically controlled Water Heater )- NO Diversity



Atthis Stage I must point out )- O.S.G. 1B– Individual Household Installations ( 2392-10Domestic Installation(s) Top of Page .Look PS. That’s what you are Applying



Apprentices)- Only the Total Load should be Calculated . Immersion Heater … ( 3000 ÷ 230V= 13A )



Note: Domestic / 100A . Single Phase



Lightingcan be Calculated using ( 66% ) of the Total Demand.

Twolighting circuits at 6A = 12A … [ 12 x 66% = 7.92A ]



(MD ) Maximum Demand for your Installation is [ 43.37A – Shower : 32A – Cooker : Ring Circuits – 42.8A : Immersion heater 13A : Lighting 7.92A :

Apprentices?? What is the MaximumDemand used Now ??? 43.37+ 32 + 42.8 + 13 + 7.92 = 139.09A



DomesticInstallation(s) Ring Final Circuits areNOT fully Loaded

DomesticInstallation(s) Shower(s) are used for avery short time .

DomesticInstallation(s) Lights are NOT all on atthe same time . Yeah

DomesticInstallation(s) Cooker is Why diversity can be Used

 

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ExternalEarth Fault Loop Impedance ( Ze )

Thisis the Résistance between Live Conductors & Earth

Remember)- You have Résistance in every Cable .

p/161- 612.11.ProspectiveFault Current .ETC.
Itcan be Obtained by ( Measurement ) ( Enquiry ) (Calculation )

230V/ 400V . protected at the Supply cut-out by Fuses [ BS-1361 type 2 : BS-88-2 : BS-88-6 : ??
( Enquiry )
TTSystem - 21Ω . → Apprentices . O.S.G.- p/11
TN-S - 0.8Ω
TN-C-S– 0.35Ω

Calculationsfor cable size . can be carried out using these values . it is Important toMeasure these values before allowing the Installation to be put into Service .
PLEASE NOTE )- The Only system for which calculation of ( Ze) would be possible is a ( TN-C-S system )

Thisis because the Fault Path External to the Installation is the same for both (PSCC ) & ( PEFC ) - Prospective EarthFault Current

Forthis Calculation to be carried out ( The value of ►► PSCC between Line & Neutral shouldbe Known .

Ohm’sLaw )- Uo / Ze = PSCC .

Symbols’used in the Regulations - p/36
(Uo ) Nominal A.C. rms Line Voltage to Earth ( V )

(Ze ) That part of the Earth Fault Loop Impedance which is External to the Installation . ( Ωs )

HiBruce .

 

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Electricians’.
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Lighting Protective System .Should be considered with regard to Bonding .
Theperson carrying out the WORK has a full Understanding of ( BS-EN 62305 )

 

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2392-10:

BS-7671:2008 . Only provides us with Informationwhich is relevant to Equipment which we could purchase & fit at the time itwas published .
Regulations)- Table 41.3 . Example – BS-EN 60898MCBs . BS-EN 61009-1 – RCBOs

Why?? for us . To use different ProtectiveDevice(s) to be used on Domestic Installation(s) Why ? We have different types of Levels of OVERLOAD .

MCBs- B – C – D .

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( A ) is NOT used asIDENTIFICATION as it could be CONFUSEDwith the CURRENT RATING of the DEVICE .

 

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No Insulation has ( Infinite Résistance ) is perfect . ( IR ) Insulation Résistance

Youcan get Small Amounts of Electricity flowing along & through Insulation to Ground/ Earth

 

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( IR ) Insulation Résistance

CommonSense tells us that the more Voltage we have .The more Current there’ll be .

TheLower the Résistance of the Cable/ Wire – The more Current for the same Voltage.

 

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Whatmakes Insulation GO bad ? Just one Example .Yeah
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GoodInsulation has High Résistance .
PoorInsulation . relatively Low Résistance .

Dependingupon such factors as the Temperature or Moisture content of the Insulation [ Résistance decreasesin Temperature or Moisture]

Combinedwith the Electrical Stresses that Exist ??
Asa Pin Hole(s) or Crack(s) develop. Moisture &foreign penetrate the surface(s) of the Insulation . ( Providing a Low Résistancepath for Leakage Current )