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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 ) :)
 
TestingContinuity of the Ring Final Circuit Conductor(s) .

Short& Sweet . -&-s

Thereare two main reasons for Conducting this Test .??

i)To establish that Interconnections in a RingFinal Circuit Do Not Exist .
ii)To Ensure that the CPC is continuous . & indicate the Value of ( R1 + R2 )for the Ring Final Circuit .
 
SAFETYat ALL TIMES .

ExternalLoop Impedance ( Ze )

Thevalue of ( Ze ) is measured at the Intake Position on the Supply Side ( Withall Main Equipotential Bonding Disconnected )

Thisis what -&-s are looking for ▼▼▼

PointTo Note : ( Unless the Installation can be Isolatedfrom THE SUPPLY ) ◄◄
ThisTest should NOT ◄◄◄be CARRIED OUT . as a POTENTIALShock Risk WILL EXIST with the Supply ON & the MainBonding Disconnected.

So2392-10 – You have be Warned ……… Bigtime . -&-s
 
Fused Connection Unit :

• FromGN-3Guidance to British StandardsEtc. BS-7671:2008

i)Correct rating & Fuse . Regulation – 533.1.
ii)Complies with BS-1363-4 - Regulation – 559.6.1.1. – ( vii )

BritishStandards . Connection to the Fixed Wiring

Lighting Point(s)

• FromGN-3 – Guidance to British Standards Etc. BS-7671:2008

- Correctlyterminated in a suitable accessory or fitting . Regulation – 559.6.1.1.
- Ceilingrose complies with BS-67 . Regulation – 559.6.1.1.
- Notmore than one flex unless designed for multiple pendants . Regulation –559.6.1.3.
- Flexsupport devices used . Regulation – 559.6.1.5.
- Switchlines identified . Regulation – 514.3.2.
2392-10)- Holes in ceiling above rose made good toprevent spread of Fire . Regulation – 527.2.1. ( One is not digging for Gold here )
- Notconnected to a supply exceeding 250V . Regulation– 559.6.1.2.
- Suitablefor the Mass Suspended . Regulation – 559.6.1.5.
- Lampholdersto BS-EN 60598 . Regulation – 559.6.1.1.l
- Luminairecouplers comply with BS-6972 or BS-7001 . Regulation – 559.6.1.1.
- Tracksystems comply with BS-EN 60570 . Regulation – 559.4.4.

Relationshipwith Statutory Regulations )- 114.1. IET.Wiring Regulations Seventeenth Edition . BritishStandard .

-&-s point of View )- Non-statutory . BS-7671:2008 . - British Standard . Compliances
Theymay . however . be Used in a Court of Law .in Evidence to claim Compliance with a Statutory Requirement . Etc .
 
Socket-outlets.

- Complieswith BS-196 . BS-546 . BS-1363 . BS-EN 60309-2 . Regulation553.1.3. & shuttered for Household & similar installation . Regulation553.1.4.
- Mountingheight above the floor or working surface suitable . Regulation 553.1.6.
- Correctpolarity . Regulation 612.6.
- Ifinstalled in a location containing a bath or shower . installed beyond 3mhorizontally of the bath or shower unless shaver supply unit or SELV . Regulation701.512.3.
- Protectedwhere mounted in a floor . Section 522 .
- Notused to supply a water heater having uninsulated elements . Regulation 554.3.3.
-Circuit protective conductor connected directly to the earthing terminal of thesocket outlet . on a sheathed wiring installation . Regulation 543.2.7.
- Earthingtail from the earthed metal box . on a conduit installation to the earthingterminal of the socket outlet . Regulation 543.2.7.
 
2392-10.

InitialVerification – ( New Installation )

Purposeof Initial Verification .
InitialVerification . in the context of Regulation . 610.1. is intended to confirmthat the installation complies with the design & been designed &constructed in accordance with BS-7671

Regulation. 610.1.
Regulation. 611.2.
Regulation. 612.1.

TheWhy’s & How’s.

Regulation. 611.2.
Chapter61 of BS-7671 states the requirements for “ InitialVerification “ As far as reasonably practicable . an Inspection shall becarried out to verify . Regulation . 611.2.

i)All installed electrical equipment & material is of the correct type &complies with applicable British Standards or Acceptable Equivalents .
ii)All part of the Fixed installation are correctly selected & erected .
iii)No part of the fixed installation is visibly damaged or otherwise defective .
 
AmbientTemperature – ( Ca )

Thisis the Temperature of the surrounding(s) of the cable . often the Temperatureof the air in a room or building in which the cable is installed .
Whena cable carries current . it gives of Heat . Thehotter the surroundings of the cable . The more difficult it is for the cableto get rid of the Heat .
Temperatureis Low . then the Heat given off could be easilylet out & the cable could carry more current . Cables must give off thisheat safety or thy could be damaged & there is a risk of a fire . You can find the correction factor for Ambient Temperature– ( Ca ) Tables ( 4B1- 30°C ◄ ) & ( 4B2 20°C ◄ ) BS-7671 . p/267

Thesetables are based on an ►► ( Ambient Temperature of 30°C ) ◄◄This means that any cables instead in an Ambient Temperature above this willneed the correction factor applying to them
Thisis because the cable will not be able to get rid of the Heat it gives off safety when carrying current .

Whena cable runs through areas having different Ambient Temperature(s) . correctionfactors should be applied to the highest Temperature(s) Only

&O.S.G.

Themost common of the correction factors are given in Table 4B1 & 4B2 )

4B1– Used cable(s) in “ AIR “ ◄
4B2– Used cables laid in “ Ground “ ◄
 
ThermalInsulation – ( Ci )

Theuse of Thermal Insulation in buildings . in the forms of Cavity wall filling .Roof space blanketing . & so on . ( is now standard )
Sincethe purpose of such materials is to LIMIT thetransfer of Heat . They will clearly affect theability of a cable to Dissipate the Heat build up within it when in contact with them . ThermalInsulation has the EFFECT of wrapping a cable ina fur coat on a Hot Summer(s) Day .
TheHeat produced when the cable carries currentcannot Escape .

LoftInsulation .
The cable rating table(s) of the regulation(s) as theone you have already used . ( Table 4E4A ) ALLOWfor the reduced Heat loss for a cable which is enclosed in an insulating wall& is assumed to be in contact with the insulation on one side .

Inall other cases . The cable should be fixed in a position where it is unlikelyto be completely covered by the insulation . Where this is not possible & acable is buried in thermal insulation for ( 0.5m500mm ) or more . a rating factor of ( 0.5 ) is applied .
Thismeans that the current rating is halved or in other words the ( Iz ) value will be doubled

Ifa cable is Totally Surrounded by Thermal Insulation for only a short length (for . where a cable passes through an insulated wall ) The heating effect on the cable insulation will Not be That Significant . This is because Heat will be conducted away from the short High – Temperature Length through the cable conductor.

Clearly. The longer the length ofcable enclosed in the insulation the greaterwill be the de-rating effect . Table . BS-7671 – Table / 52.2.- p/104
Showsthe de-rating factor for lengths in insulation of up to ( 400mm ) & appliesto cables having Cross-SectionalArea up to ( 10mm[SUP]2[/SUP]) T/52.2.

( Ci ) Rating factor for conductors embedded in ThermalInsulation –
 
Note: That there is no such thing as a Three-Line / Earth Fault .
( Although it is possible )
ForThree Faults to occur on the Three Lines to Earth simultaneously . As far as calculationsfor Fault Current are concerned . The voltage to Earth for Standard UK Suppliesis always ( 230V )

Forboth Single – Line & Three- Line(s) system(s)
Thusthe Tables of Maximum Earth Fault Loop Impedance . Which are given in the Appendices . (Apply both to Single & Three Line systems)
 
2392-10. Testing -
:earmuffs:
ResidualCurrent Devices ( RCD )

Wherea Residual Current Devices ( RCD ) fails to trip out when passing the Integral Test Button .
Thiswould indicate a Fault within the device itself . ( Which should therefore bereplaced )

Wherea Residual Current Devices fails to trip out when being Tested by an RCD Tester?? ( Then it would suggest a Break in the EarthReturn Path )
Whichmust be ( Investigated ) if the RCD does tripout but within the time ( Specified ) then acheck should be made that the ( Test Instrument) is Set Correctly for the Nominal Current of the device UNDER TEST.
 
2392-10. Testing -

Insulation Résistance . ( IR )

Asa Tester )- A reading below ( 1.0MΩ ) Wouldsuggest a weakness in the Installation .

PVCInsulated Cables are NOT generally subject to a deterioration of Insulation Résistance due to Dampness
( Unless the Insulation or Sheath is Damaged )
 
Continuity( Polarity ) of CPC & ( R1 + R2 )

1)-Isolate Supply .
2)-At the Consumer Unit( CU ) → Line ( R1 ) & Earth ( R2 ) “ Linked “ Connector Block .
3)-Light Switch ( ON ) …. EarthingTerminal in Switch
4)-Megger 1552 - ZeroLeads Résistance / set to Ohm’s. ……… ( The Megger will show a symbolwhen Zeroed ) the two clips leads together & hit the button " Yellow . Zero
5)-Place Megger lead Line & Earth at each Accessory◄.
6)-Take readings & confirm Continuity
7)-Operate Switch :- with Megger connected to confirmPolarity . **

Apprentices’)- you’ll hear this a lot Continuity ?? Polarity **
 
Apprentices: Basic(s).

Continuityof Ring Final Circuit Conductors .

Whydo we do this Test ??

A Testis required to verify the ( Continuity ) of eachconductor including the circuit .

Protectiveconductor ( CPC ) of every ring final circuit.
TheTest results should establish that the ring is ( Complete) & has no ( Interconnection / -&-s )
TheTest will also establish that the ring is ( Not Broken) & that the ( Polarity is correct )

( Polarity byContinuity Methods )

Howdo we do this Test ??
Youhave ( Three Steps for completing this Test ) O.S.G. p/80 -

Step-1 ) Initial check for Continuity at Ends of Ring: … Line –to-Line : Neutral – to Neutral : CPC –to CPC :

TheBook(s) are telling us [ For this Test you need a Low ReadingOhmmeter ] I use my Megger1552 - Set on Ωs ( Zeroed Leads )

Step1 )- Simple . End to End … ( Continues Loop ) A to B …. If they are no broken wires – you willget a Readingon the Megger. 0.03 ??
Step1 )- Simple . End to End … ( Continues Loop ) A to B …. If there is a broken wire – you willget an Open Circuit . … ( Houston We have a Problem)
Step1 .
TheLine . Neutral& Protective Conductors are Indentified& their measured ( Separately ) A finitereading confirms that there is ( No Open Circuit) on the ring final circuit conductors Under Test .

TheRésistance values obtained should be the same ( within 0.05Ω ) if theconductors are the same size C.S.A. . The Résistance of the Protective Loopwill be proportionally higher (Typically 1.67 x higher ) Than that ofthe Line or Neutral Loop . if these relationships’ are not achieved then eitherthe conductor(s) are incorrectly identified or there is a loose connection atone of the Accessories – “ Socket “

Step2 ) Line & NeutralCross-coupled : O.S.G. p/81

Apprentices)- Figure of Eight . Why ?? Draw out aring . L1 / N1 / -L2 / N2
L1 – Linked to N2 .
L2 – Linked toN1 . ……… ( Line & Neutral Coupled)

TheLine & Neutralconductors are then connected together so that the outgoing Line conductor is returning Neutralconductor & Vice Versa . The Résistance between Line & Neutral conductorsis then measured at the Consumer Unit & then at each socket outlet .
Thereadings obtained from those sockets wired into the Ring will be substantiallythe same & the value will be Approximately half the résistance of the Line or Neutral loop résistance. Any sockets wired as Spur will have aproportionally higher résistance value corresponding to the length of the Spur Cable .

Step3 ) O.S.G. p/82
Line & Earth Crossed Coupled .

Step2 )- is then repeated but with the Line & CPC Crossed Coupled .
TheRésistance between Line & Earth is thenmeasured at each Socket . The Highest Value (Recorded Represents the Maximum ( R1 + R2 ) of the circuit & is Recorded on the TestSchedule . Etc .

Ido hope this will guide you in the right way . Amber .
 

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