***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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Isolation& Switching : :49:

537.4.2.8.
Plugs& Sockets are NOT suitable for EmergencySwitching .

 

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MainElements of Circuit design for conventional Circuits . :juggle2:

•Determine Design Current . actual flow of current through the cables ( Ib )less diversity where appropriate .
•Select type & size of protectivedevice ( In )
•Apply any correction factors to reduce self heating of cables ( It )
•Select type & size of cable fromAppendix 4 ( Iz )
•Determine whether volt drop is satisfactory for Length . Loading & routingof cables .

Note (- 2392-10
Inmost practical applications where an Installation already exists you may havethe choice of Type but only Size .

ApplyingElectrical Design . Cable Sizing .

LoadCurrent ( I ) & design current ( Ib )
Effectivecontinuous current flow through a circuit
-Basic Calculation : P = V . I or I = P/V

Whencalculating Design Current with inductive loading & where there isinsufficient information available apply a factor ( 1.8 ) to the power consumption . App 1 – Table1A O.S.G. note 2 - I = P x1.8 / V

Alternatively. if the powerfactor is known then apply ( P = V x I x Cos Ø or P = √3 x V[SUP]L [/SUP]x I[SUP] L [/SUP] x Cos Ø )

ApplyingElectrical Design . Cable Sizing .

DesignCurrent ( Ib )
Maximum Demand 311.1. ( MD )

Diversitymay be taken into account when determining MaximumDemand
Diversity– O.S.G. Appendix 1 . Table 1A . 1B .

Point to Note : Not all circuits run at their full powerrating .
CookerCircuit - Minus first ( 10A ) then 30% of the rest add( 10A ) to the Total + 5A for Socket / Outlet .

LightingCircuits :
Domesticlighting ( 66% ) is assumed to be on Continuously .

PowerCircuits :
1[SUP]st[/SUP] Circuit = Full Rating . 2[SUP]nd[/SUP] Circuit 40% & all Other Circuits . ◄◄

Diversity applies to groups of circuits or / & cables supplying groups of circuitssuch as a Sub-mains distribution boards .

 

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560.7. Circuits for SafetyServices shall be Independent of Other Circuit . :vanish:

Does NOT include Smoke Alarms & Emergency Lightingthat are ( Battery Powered ) & supplied by a ( Charging Circuit ) i.e. Lights .

 

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AppropriateBritish Standard or do a “ Risk Assessment“ :35:

• SafetyServices to be Installed to allow adequate duration equipment to be suitable tolast for the duration of Escape .
•560.5.3. – Continual ( IT ) insulation monitoring devices that give an Audibleor Visual indication of the first fault in the System .

Selection& Erection (- Safety Services . :rolleyes4:

Classification of a SafetySystem .

i)Non-automatic supply .
ii)Automatic – Self Starting .

AutomaticSupply is Classified by the following time it takes to get ( Fully Operational )

- No breaks – UPS .
- Very short breaks - < 0.15s
- Short breaks - > 0.15 < 0.5s
- Lighting breaks - > 0.5s < 5s .
- Medium breaks - > 5s < 15s
- Long breaks - > 15s .

 

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Ido fell the Humble MCB is in theBackground now . The RCD has taking Front Row . :aureola:

TimeDiscrimination :

Isachieved by delaying the Opening of the ( RCD or MCB ) upstream ( RCD or MCB ) untilthe downstream ( RCD or MCB ) has Opened & cleared the Fault .
Thetotal clearing time of the downstream ( RCD or MCB ) MUSTbe LESS than the time setting of the upstream ( RCD or MCB ) & the upstreammust be able to withstand the Fault Current for the time setting period .

Thereforethe upstream ( RCD or MCB ) must be acategory which has been designed & tested for this purpose .

MCBs- To determine timediscrimination it is only necessary to compare the Time / Currentcharacteristic curves of the Two Devices to ensurethat NOoverlap Occurs .

RCBOs – have taking a primaryrole in Industry now - RCD/MCB . ( B & C . curves ) C forLighting . B forPower .

 

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2392-10: Your first role is a Designing . second is Electrician . :aureola:

Forthe Protection of Lighting Circuits the “ Designer “ must select the CircuitBreaker with the Lowest Instantaneous trip current compatible with the inrushcurrents likely to develop in the circuit

 

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Whatis an Insulation RésistanceTest ( IR ) :aureola:

AnInsulation Résistance Test measures the total Résistance of a products’ ( Cable) Insulation System(s)
Itis also useful test to perform when doing preventative maintenance on ( Cable ) overtime .

Whatis an Insulation RésistanceTest ( IR )
Measuring& Recording long term Stability of Insulating Materials over time .

InsulationRésistance Testing . as a Tester . – The Test results can provide usefulinformation about a ( Cable Integrity ) & can yield clues when itsInsulation should be Replaced .

 

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InsulationRésistance (- 2392-10 . from -&-s point of View . :aureola:
InsulationRésistance measurement is preformed inorder to ( Assure Safety ) against ElectricShock through Insulation .

 

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ContinuityFunctions. :aureola:

Apprentices(- ParallelRésistance Paths - Interferingcurrent in measured circuit will influence the TestResults.

From-&-s point of View (- “ Nulling“ / Test Leads Résistance beforeperforming Continuity Measurement

When measuring Insulation Résistance betweeninstallation conductors all Loads must bedisconnected . ( & all SwitchesClosed ) Mains Voltage Switched OFF

“Functionality “

Regulation 612 – BS-EN 61557.

ElectricalSafety in Low-voltage distribution systems up to 1000V. A.C. & 1500V . D.C. .
Regulationsp/31 ( Low . Exceeding extra-low voltagebut not Exceeding 1000V . A.C. & 1500V . D.C. . between conductors.
Equipmentfor Testing . Measuring or Monitoring of Protective Measures .

Thecircuit under test must be Switched OFF .de-energised & Isolated before test connections are made when carrying outInsulation & Continuity Tests .

612.2. & 612.2.1.
Continuityof Protective Conductor(s) & Main Protective Bonding Conductor of New orModified Installation(s) must be verified before carrying out an Earth FaultLoop Impedance . or RCD test .

 

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Loop Impedance .:aureola:

Isa Live Test. which is carried out on a circuit to check the Résistance of the Line conductor to Earth.

Thisshows if the current created under Fault Conditions will trip the MCB .

Apprentices: Refered to as Earth Fault Loop Impedance .
-&-s . Extremely Important Test to verify the System is Safe.

TheEarth Loop Impedance test measures therésistance of the path that a ( Fault Current ) would take between Line & Protective Earth
Thismust be low enough to allow sufficient current to flow to trip a MCB

 

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2392-10. :aureola:

InterpretingResults & taking Remedial Action .
“Remember “ that it is NOT sufficient just to carry out TESTS & Record the Results .

Knowledge of Regulations . & of how to Interpret results . is alsorequired to ( Ensure the Installations ) Safetycharacteristics are within the prescribed limits .
Example(- An excessive Earth Loop Impedance value should . (Prompt an Investigation ) into its CAUSE

Asa “ Inspector “ Remedial Action should then be carried Out .& the Installation Retested .

Inspectionis the Vital Initial Operation . Testingsupports the Inspection :dupe:

 

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Workingmy way around the 17[SUP]th[/SUP] Edition :aureola:

Making you aware of Residual Current Devices .

RCDs as General Protection . ( “ Meanings “ )

►►► Protection from ( Faults ) Earth Faults currents . Shock protection ◄◄◄ :50:

531.2.1.– A residual current device shall be capable of disconnecting all the Line conductors of thecircuit at substantially the same time .

531.2.2.– The Magnetic circuit of the transformer of a residual current device shallenclose all the Live conductors of the protectedcircuit . The associated conductor shall be outside the Magnetic circuit .

531.2.3.– The residual operating current of theprotective device shall COMPLY with therequirements of Section 411 as appropriate to the type of System Earthing.

531.2.4.- A residual current device shall be so SELECTED & the ELECTRCALCIRCUITS so SUBDIVIDED that any protective conductor current whichmay be expected to OCCUR during normal operationof the Connected Load(s) Will be Unlikely to Cause unnecessary tripping of the Device .

531.2.5.– The USE of a residual current device associatedwith a circuit normally expected to have a protective conductor shall NOT be considered SUFFICIENT for Fault protection if there is NO such conductor . EVENif the rated residual operating current ( I∆n ) of the device does not EXCEED 30mA .

 

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Workingmy way around the 17[SUP]th[/SUP] Edition – Selection & Erection / Part 5

Making you aware of Effects of External Influences - ( HEAT ) :66:

Protectionof Wiring Systems by :-
-Shielding .
- Distance.
-Type of Wiring .
-Changing or Reinforcing type of Insulation .

522.2.2.– Extra insulation – ( Lighting )
522.3.- ( IP )rating for Ingress of Water .
522.4.- ( IP )rating for Ingress of Dust & Particles .
522.5.– ( Corrosion) from liquids . Electrolytic Action from dissimilar metals .

Hi. K

 

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Workingmy way around the 17[SUP]th[/SUP] Edition – Selection & Erection / Part 5

Making you aware of “ Connections “ :aureola:

526.1.Connections
- Durableelectrical continuity .
- Adequatestrength .
- Protection.

2392-10. Domestic Installation(s) Yeah :50:

 

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Workingmy way around the 17[SUP]th[/SUP] Edition – Selection & Erection / Part 5

Making you aware of “ Methods& Selection of Connection(s). :yawn:

“ ElectricalConnections “ :aureola:

- Conductor& Insulator materials .
-Number & Shape of wires forming the Conductor .
-C.S.A. of conductor .
-Number of conductors Connected Together .
-Temperature of Joint for normal Operation .
-Prevent mechanical movement under Vibration .