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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 ) 
Wording: 2391-10 : -&-s Exams .
Earth Fault Loop Impedance .
-&-s) Test that has to be carried on anEnergized Installation . ( Safety Precautions must be Taken )
➥ Purpose of thetest is to ensure that . in the event of a ( Line – to – Earth Fault ) enough current will flow around the ( Earth Fault LoopImpedance )
Tooperate the protection within a specified time . Example )- Ring Final Circuit – 411.3.2.2.– p/53 . The maximum disconnection timestated in ( 41.1.) shall be applied to Final Circuits NOT exceeding 32A . (Time – 0.4s )
Themeasured value of Loop Impedance is then compared with that given in )- ( 41.1. )
Wording : -&-s Exams . Earth Fault LoopPath .
TheEarth Fault Current Loop ( Line – to – Earth Loop ) comprises the followingparts .
( Starting ) at the point of Fault . → the circuit protective conductor . → the consumer’s earthing terminal &earthing conductor : For .
TN- systems the (Metallic return path ) Or .
TT– systems the earth return path – the path through the earthed neutral point ofthe transformer & the windings . & the Line conductor from thetransformer to the point of Fault . Theearth fault loop impedance ( Zs ) at apoint distant from the origin of the installation is measured with the ( Main protective bonding conductor )conductors in place .
ElectricalSeparation : 612.4.3.p/192.
Test. is usually an Inspection to ensure that certain circuit(s) are separatedfrom one another .
612.4.3. : The separation of the Live parts from those ofother circuit(s) & from earth .
● Thisis – Only the Basics .
-&-s. Q/As . BS-7671:2011. make NOmistake . -&-s will get them IN somewhere .
BS-7671:2011. IS an OPEN BOOK . to -&-s . Yeah . Q/As
Choosethe correct type of Þ ( SPD ) for the Installation& Site in the correct location .
Regulation 534.1 : For the protection of ( A.C. ) power circuits . ( SPDs ) are allocated a Type Number .
Regulation 534.2 : Type ( 1 ) SPDs . are “ ONLY “ usedwhere there is a risk of direct lighting current . ↯↯ ( Origin of theInstallation ) -&-s
( Equipotential bonding . Or . Lighting protection / current . SPD) ▲▲
(Overvoltage ) - Type ( 2 ) SPDs .are “ USED “ at distribution boards. ( Distribution board / Consumer unit ) -&-s . 2392-10 . - ( CU) ⇇
(Overvoltage ) - Type (3 ) SPDs . are “ USED “ nearTerminal equipment . ( Terminal equipment ) -&-s
534.2.1. : Use of ( SPD )
Whererequired by Section 443 . or Otherwise specified . ( SPDs shall be Installed )
i) Near the Origin of an Installation . OR
ii) in the Main distribution assemblynearest the Origin of an Installation .
➥ if we look at . Regulation – 534.1. Scope & Object . Concept describes the Installation of . Type 1 : Type 2: Type 3:
EAWR– 1989 . Regulation ( 5 )
Strength& Capability of electrical equipment which “ States “ NO electrical equipmentshall be put into USE where its strength &capability may be exceeded in such a WAY as may give rise to danger .
CircuitArrangements .
Therequirements of Regulations . 521.8.1. &521.8.2. “ Reference “ Designer & Electricians’ installers ofElectrical Installations .
521.8.1. p/122
Eachpart of a circuit shall be arranged such that the conductor(s) are notdistributed over different muiticore cables . conduit . ducting systems .Trunking systems . or Tray or ladder systems .
521.8.1.
Theconductor(s) of a circuit must NOT be distributed over different multicorecables . Conduits . Ducting . Trunking or Ladder systems . “ However “
Where multicore cables arerun in Parallel . forming ONE circuit this requirement need NOT be met . if multicore cables are installed in Parallel. then each cable must contain ONE conductor of each Line .
521.8.2. The Line & Neutral conductors of each final circuit.
Thisrequirement need NOT be met where a number ofmulticore cables forming ONE circuit areinstalled in Parallel .
Class11 . Lights - NoEarth Required . Double Insulated .
Class11 . lights which can be used where the Earth is missing .
( Lighting can onlyreally be corrected by a re-wire )
Westill have houses without an Earth tothe Lights . 2392-10 . you will find this during Inspections .
Regulations. As far as getting an Earth from somewhere else it ( States ) :Earth Wire must run along the same route as the Live supply & it will to beat least 4mm sq if not part of the cable .
542.1.8.– 543.1.1. – 521.8.
542.1.8. p/158 .
Wherea number of installations have separate Earthing arrangements . any protectiveconductors common to any of these installations shall either be capable ofcarrying the maximum fault current likely to flow through them or ( be Earthed) within one installation only & insulated from the Earthing s
Arrangementsof any other installation . in the latter circumstances , if the protective conductor forms part of a cable. the protective conductor shall be Earthed only in the installation containingthe associated protective device .
521.8. p/122 .
521.8. 1. Each part of a circuit shall be arranged such that theconductors are not distributed over different multicore cables .conduits . ducting systems . trunking systems . or tray or ladder systems . Etc.
Condition Report Inspection Schedule for Domestic & Similar Premises with up to100A Supply . -&-s
● Nice one Bruce . Good Call . :yes:
2392-10 : -&-s : What isthe status of an Electrician who will be carrying an Inspection of anInstallation
Competent Person . 621.5. Testing & Inspection shouldonly be carried by a competent person .
710.411.3.2.1. (-&-s ) in MedicalLocations of ( Group 1) & ( Group 2 ) Where RCDs are required . ONLY type ( A ) or Type ( B ) shall beselected . depending on the possible Fault Current arising .
NOTE : Type ( AC ) RCDs SHALLNOT be USED . this iswhat -&-s are looking . Q/As .2011 .
Note – Type ( A ) RCDS . tripping is ensured for residual ( sinusoidal ) alternating current & for ( pulsatingdirect currents ) D.C.
Note – Type ( AC )RCDS . tripping is ensured for residual ( sinusoidal alternating current ) . A.C.
Justa Reminder .
O.S.G. – p182 . Table 11 .
( 10mm[SUP]2 [/SUP][SUP] [/SUP]- Copper / 1.83 )
Continuityof Protective Conductors .
“ Main Protective Bonding Conductor . “
Asthe résistance of copper conductor is known . is continuous throughout itslength .
10m– length . of 10mm[SUP]2 [/SUP] MainProtective Bonding Conductor . Reading ( should givea reading of approximately . 10 x 1.83 ÷ 1000 = 0.0183Ω )
Résistanceof copper conductors in milliohms per metre ( mΩ/m ) at 20°C . ◄
LightingCircuit(s) . Lampholders in damp situation(s) oraccessible to persons in contact with Earthed Metalor within ( 2.5mm of a Bath or Shower ) should be fitted with “ Home Office Skirts “
TheName has NOT died out Yet . ( H/O – Specifications – Straight ◄ )
BattenLampholder(s) : Straight BattenLampholder . C/W Home Office Skirt . Long Skirt . Conformsto BS-EN-61184 .
Useful Junk .
Inthe . 30”s & 40”s . The factoriesAct was also known as the Home Office . ( H.O. ) Regulations& parts of this are still referred to in ( H.S.E. ) publications .
Appendix3 – of the Memorandum of Guidance on the . Electricity at Work Regulations .1989 .
( H.O. ) Shroud – or Skirt . achieved compliance withthese regulations .. because it prevented the metal cap of the lamp beingtouched in damp situations .
Blastfrom the past . ? IET .
Years ago (in the30's/40's) the Factories Act was also known as the Home Office (H.O.)Regulations and parts of this are still referred to in HSE publications (seeAppendix 3 of the Memorandum of guidance on the Electricity at Work Regulations1989). I was told that the H.O. shroud (or skirt) achieved compliance withthese regulations because it prevented the metal cap of the lamp being touchedin damp situations. There were various other standards around in those daysincluding Ministry of Works and Parker Morris requirements ? when I was anapprentice maintaining old domestic properties, I was told this required one lightingpoint and one socket outlet in each room. That was in the days when the bedroom lightingpoint was not positioned in the centre of the ceiling but offset towards thewindow because this wouldprevent the shadow of anyone undressing in the room being seen on the curtainsoutside! ?apparently, the H.O. shroud has even outlived this requirement.
UsefulJunk
CircuitBreaker . BS-3871 . 10A - Replaced ◄◄ BS-EN
Thisvalue of current can be carried indefinitely by the device . & is known asits nominal setting ( In )
Thevalue of the current which will cause operation of the device ( I[SUP]2[/SUP] ) will be larger than ( In ) & will be dependenton the device’s “ fusing factor “ . this is a figure which . when multipliedby the nominal setting ( In ) willindicate the value of operating current ( I[SUP]2[/SUP])
Thisfusing factor is approximately ( 1.45 ) our 10A devicewould not operate until the currentreached ( 1.45 ) x 10A = 14.5A . the Regulationsrequire co-ordination between conductors & protection when an Overloadoccurs .
( 1.45 ) x 10A =14.5A …… 10A– Lowest =14.5A
Takenform my old Notes . blast from the past .
Ratingof fluorescent circuits . “ Banks of Fluorescent tubes “
Fluorescenttubes are rated in Watts .
Forany reason .we are not working to ( Watts ) plant is not rated in ( Watts ) now inVolt amps . ??? ( VA ) it is recommended that . if NO other information is Available . The lamp wattage may be multiplied by ( 1.8 ) in order to determine the ( VA ) rating .
VA) rating of a Fluorescent Unit with an 80W tubeis ( 1.8 x 80 = 144VA )
Whensupplied at 230V the current taken would be : I = VA / A : = 144 ÷ 230V = 0.6A .
● Stroboscopic Effect. the Reason(s) ????
forthe sake of Apprentices .
whilea Fluorescent lampis in operation the light may flicker .
undersome circumstances this may make it appear that ( Rotating Machinery ) has ( Slowed down or Even Stopped ) this is calledthe “ Stroboscopic Effect “
One of the following methods .
Balancingthe Lighting Load – Three – phase .
Ifa large lighting load is installed in a Three – phase . Installation where there is some rotatingMachinery . the “ Stroboscopic Effect “ effect may be overcome by connecting alternategroups of lamps to a different “ Phase “
Advantage of balancing the lighting load . Brown – Neutral - 6 / lights . : Black –Neutral - 6 / lights . : Gray – Neutral - 6 / lights . :
Note :
Some of these questions arenow out of date but some are still usable .
State – 3 circumstances thatwould result in an ( Increase inconductor résistance )
Increase in conductortemperature due to Overload .
Increase in conductortemperature due to Fault Current .
Increase in conductortemperature due to increase in Ambient temperature .
Reduction of C.S.A.
Increase in conductor length.
A Single phase ( TN-C-S )installation has a measured value of external loop impedance of ( 0.015W )
Show bycalculations the expected value of Prospective FaultCurrent . at the Origin .
Ipfc = Uo / Ze . Thus a Prospective Fault Current of circa ( 16kA) can be expected at this Point . ( 2011 – Ipf )
The Continuity of a lightingcircuit protective conductor is to be tested . if the test method uses atemporary link between Line & Earth in the Distributionboard – Consumer Unit . State .
At which points in thecircuit the test should be conducted .
The conductors between whichthe tests should be made .
The significance of the testreading at the end of the circuit .
i) at each point .
ii) Line & CPC .
iii) ( R1 + R2 ) for the circuit .
2392-10 .
the Assessment is in twoparts .
i) Practical assessment . ( On the Boards Testing )
ii) Multi – choice examination .
612.5 – theVerification shall be made by a (Competent Person )
610.6. – On completionof the Verification , according to regulations . 610.1. to 610.5. a certificate or Certificates . shallbe prepared .
· Assessing the Risk. 2392-10 :
Hazard – means anythingwhich can cause harm .
Risk – is the change .great or small . that someone will actually be harmed by the hazard .
2391-10 : the problem is being able to transfer theknowledge from your head onto paper .
never assume that the examiner will know what I mean as you will only gain marksif you can clearly ( Demonstrate ) by your answers . that you are Competent inthe Theory of the 2391:10
Question may contain words such as (State . List . Explain with the Aid of a Diagram . Explainbriefly . Show All Calculations )
Explain Briefly ) would require a brief explanation . usually no more than one ortwo sentences .
State ) maybe a short statement & may even be only a few words or Less .
Show All Calculations ) are self explanatory . buta Correct Answer without showing the calculations would only gain a smallpercentage of the actual marks .available for that question .
Testing Insulation Résistance ( Insulation Résistance value with circuit(s) tested in Parallel )
the more résistance there are in parallel . thelower the overall resistance .& in consequence . the longer a cable thelower the insulation résistance . add to the fact that almost all installation circuits are also wired inparallel .
test on large installationsmay give . if measured as a whole . pessimistically low values . even if thereare no faults .
from a Inspector point of View ( floor by floor . periodic testing . tominimize disruption )
(1 ) . 2011
AlternatingCurrent Motors . ( A.C. )
Thereare many different types of . A.C. motor operating from either Three-phase or Single-phase . A.C. supplies .
TheThree-phase motor depends on the rotation of a ( Magnetic field) for it’s movement .
Coilsor Poles are ( Arranged 120° apart ) &connected as to an Alternating ( Three-phase supply ) Each pole willbecome fully energized at a different time in relation to the Others .
TheIron core of eachcoil becomes ( Magnetized ) as the coil is energized . the arrangement gives the effect of a ( Magnetic field ) rotating around the coils .
-&-s : Thespeed of rotation of the ( Magnetic field) is called the ( Synchronous Speed ) & is dependent on the ( frequency ) ofthe supply & the number of ( Pairs of Poles ) . Hence .
ƒ = N[SUP] S [/SUP] . P
ƒ ) is the supply frequency in Hertz . ( Hz )
N[SUP] S [/SUP] ) is the Synchronous Speed in ( revs / second .
P ) is the number of ( Pairs of Poles )
Calculatethe : ( SynchronousSpeed ) of a ( Four – pole machine ) if the supply ( frequency ) is ( 50Hz )
ƒ = N[SUP] S [/SUP] . P
Therefore: N[SUP] S [/SUP] = ƒ / p . = 50 ÷ 2 = 25 . revs / second – Or 1500 revs / min .
p ) 4 .poles are 2 .
(2 ) Eight – pole Induction Motor . 2011 .
Eight– pole Induction Motor runs at ( 12 revs/second ) &is supplied from a ( 50Hz ) supply . ◄► Calculate the ( Percentage Slip ) – ( % )
ƒ = N[SUP] S [/SUP] . P
50 ÷ 4 = 12.5 revs/second )
S( % ) = ( N[SUP] S [/SUP]- N[SUP] [/SUP][SUP]r [/SUP][SUP] [/SUP]) / N[SUP] S [/SUP]x 100
=( 12.5 [SUP] [/SUP]- 12 ) / 12.5 x 100 .
= 0.5 x 100 / 12.5 [SUP] . [/SUP]= ( 4% )
FunctionalTests .
Circuitbreakers .
Youshould operate the manual mechanism of each circuit breaker to make sure thedevice opens & closes as it should do .
Recordin your Report any of the following items found while Testing or Inspecting .circuit breakers .
i) Doubts about the reliability or effectivenessof the device .
ii) Damage .
iii) Parts that have worn away unexpectedly .
iv)Evidence that a device has failed to operate as it should .
(3 )
“ Three-phase . four – wire system “ !!
Three– phase Circuits .
Three– phase supply comprises ( Three – Wave forms ) each separated by ( 120° ) &the resultant waveform is ( ZERO )
Two main ways ofconnecting Three – phase equipment. ( Star ) & ( Delta )
( Star ) → . Star or Neutral point .
( Delta ) → .Usually - Motor Windings .
( Star Connected Load ) → . Natural Conductor )- “ Generator → Load . “
Fedfrom a Star connected supply . The addition of the conductor between the ( Star points ) converts the system into what is Known as a “ Three-phase . four – wire system “
Currentssupplied by “ Generator “ flow along the Lines . thought the Load .& return via the ( Neutral Conductor)
Currentsin a ( Balanced Three– phase supplyadd up to ZERO ) Therefore - IB = IBLK + IG = In = 0 . thecurrent flowing in the Neutral is ZERO .
Since NO currentflows between the Star points . they MUST both be atthe same (Potential ) which is also ZERO . the Star point of a Transformer is ( Earthed ) is alsoat ( ZERO VOLTS )
i) - One reason for the connection of the ( Neutral Conductor) is to provide a path for currents if the system became ( Unbalanced )
ii) – Another is that it enables ( Single-phase Loads )to be connected to a ( Three – phase supply )
the ( Windings )of most ( Three – phase Motors )are connected in ( Delta ) as the ( Line – Windings ) areperfectly ( Balanced &No Neutral is Needed )
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