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
-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
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 & Lineconductor 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 ]
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.
Broken RingFinal Circuit ? ( As a Tester this is what -&-s arelooking for )
However . We find the Ring to be Broken . the ProtectiveDevice 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
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 .
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 .
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 )
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 )
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
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 )
Electricians’.
:19: Lighting Protective System .Should be considered with regard to Bonding . Theperson carrying out the WORK has a full Understanding of ( BS-EN 62305 )
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 .
UsefulJunk ( A ) is NOT used asIDENTIFICATION as it could be CONFUSEDwith the CURRENT RATING of the DEVICE .
Whatmakes Insulation GO bad ? Just one Example .Yeah
:45: 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 )
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