***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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HBC fuse link . BS-1361 . ( House Service &Consumer Unit fuse )
Notpopular for use in Consumer Unit(s) . gives good Short-circuit currentprotection / & does not result in cable de-rating . ranges from . 5A – 100A .

Steel WireArmoured ( Insulation ) XLPE . Cross (X )linked ( L ) poly ( P) ethylene ( E ) :26:

 

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The( X ) in a code simply means that protection isnot specified . [ The code ( IPX2 ) onlyprotection against mechanical objects ] Not Moisture .

2392-10: Inspection – Usually referred to as Visual Inspection . As it is broken into two words. Inspection & Testing.:icon_bs:

 

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2392-10:Total Loop Impedance ( Zs ) What’s the main reason behind this- ??
:icon_bs:This has to be measured in order to ensure that the protective device will operatein the specified time Under Fault Condition(s) . You’ll be coning for your 2391-10 .

 

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2392-10)- Indicates residual current ratings& Uses as Mentioned in BS-7671:2008 –Red Cover .

30mA – RCDs . …. ( a Break down forYou ) :icon_bs:

• Allsockets outlets rated at no more than ( 20A )& for Un-supervised : General Use.
• Mobileequipment rated at NO More than ( 32A ) for use Outdoor(s)
• Allcircuits in a Bath/ Shower room
• Preferredfor all circuits in a TT System
• Allcables installed Less than ( 50mm ) from thesurface of a Wall or Partition [ in a Safe Zone] if the installation is Un-supervised . & also at any depth if theconstruction of the Wall or partitionincludes Metal Parts .
• Inzone ( 0 . 1& 2 ) of Swimming pool locations
• Allcircuits in a location containing Saunas . Etc
•Socket outlet final circuits NOT exceeding ( 32A ) in Agricultural Locations .
• Circuitssupplying ( Class II ) equipment in Restrictive Conductive Locations
•Each socket outlet in caravan parks & marinas & final circuits forhouseboats
•All socket outlet circuits rated NOT MORE than (32A ) for show . Stands . Etc
•All socket outlet circuits rated NOT MORE than (32A ) for construction sites [ Where Reduced Low Voltage . Etc – is NOT USED ]
•All socket outlets supplying equipment Outside mobile or transportable units
•All circuits in Caravans
•All circuits in Circuses . Etc
•A circuit supplying ( Class II ) heating equipment for floor & ceilingheating systems

500mA– RCD
•Any circuit supplying One or More outlets of rating EXCEEDING( 32A ) on a Construction Site

300mA– RCD
•At the ORIGIN of a TEMPORAYSUPPLY to circuses .
•Where there is a ( RISK OF FIRE ) due to storageof combustible materials
•All circuits [ Except Socket Outlets ] in Agricultural Locations

100mA– RCD
•Socket outlets of rating EXCEEDING ( 32A ) in AgriculturalLocations

WhereLoop Impedance values CANNOTbe MET . RCDsof an appropriate rating can be Installed . Their rating can be Determined from( I∆n = 50/Zs )
Where ( I∆n ) is therated operating current of the device – ( 50 )is the Touch Voltage & ( Zs ) is themeasured loop impedance

 

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2392-10)- Learning Curve . for when you go for 2391-10 ( Reasons )

Loop Impedance Test – must be conducted ( First ) to confirm that an ( Earth Path ) exits or the RCD test could Prove Dangerous.
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Apprentices)- Electrical - Calculation(s) Basics
:38:
Alternating Current Circuit
Impedance:

i)in D.C. circuit(s) - The current is limited by Résistance . ◄◄
ii)in A.C. circuit(s) - The current is limited by Impedance ( Z ) – (Résistance& Impedance are measured in Ohm’s. )◄◄

U/Z= I . or voltage ( U ) ÷ Impedance ( Ohm’s ) = current ( Amperes )

• The voltage applied to a circuit with an ( Impedance ) of ( 6Ω is 200 volts ) Current in the Circuit [ U/Z = I ] 200 ÷ 6 = 33.33A ← Round off to4
•The Current in a 230V – single-phase motor is (7.6A ) Impedance of the Circuit . [ U/I = Z ] 230 ÷ 7.6 = 30.26Ω .
•A discharge lamp has an Impedance of ( 265Ω )& current by the lamp is ( 0.4A ) [ Z x I = U ] 265 x 0.4 = 110 Volts .
• Thecurrent through an Impedance of ( 32Ω . is 8A ) Voltage drop across the Impedance . [ U = I x Z] 8 x 32 = 256V

• Thecurrent through an Electric Motor is ( 6.8A )at ( 230V ) The Impedance of the motor .
[U = I x Z ]
Transposefor Z ] Z = U/I . – 230 ÷ 6.8 =33.82Ω . ….. 33.82352941 / Roundoff to → ( 33.82 )

•An coil has an Impedance of ( 430Ω ) Voltage ifthe coil draws a current of ( 0.93A )
[ U= I x Z ]
Transposefor U ] 0.93 x 430 = 400V . ….. round of 399.9 / 400

 

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Apprentices: “ Meaning “ ( PSCC ) 612.11
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(PSCC ) - Prospective ShortCircuit Current –The maximum current which could flow ( Between )Live Conductor(s) ? L/N . for thesake of -&-s . 2 live conductor(s)
(PFC ) - Prospective FaultCurrent – The ( Highest) current which could flow in a Circuit due to a Fault. this will come up on 2392-10 .Exams ◄

UsefulJunk . Yeah
Transpose )- “ Meaning “ Change order to calculate a Value .

UsefulJunk .
► Thermosetting )- “ Meaning “ Cable Insulationwhich becomes “ SOFT “ when “ HEATED “& is “ RIGID “ when Cooled Down .

 

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Apprentices: Voltagedrop ( Vd ) . – Amount of voltage “ Lost “ due to a “ Résistance “ ◄ :dots:

 

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Apprentices: This willserve you well in Exams .
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Earth Fault Loop Impedance .
Résistance of the conductors inwhich the current will “ FLOW “ in the event ofan Earth Fault .

Thevalue includes the Supply Cable . Supply Transformer & the Circuit Cable up ( To the point of the Fault )

 

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2392-10: Exams :dots:

Remember )- Circuit Résistance in Parallel: ( I/RT = I/R1 = I/R2= I/R3 ) .. Etc . Electrical Calculations . PS . they will hit you on Parallel Circuit(s) . ◄
Remember )- Circuit Résistance in Series : ( R1 +R2 + R3 ) Etc .

 

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Scenario/ Exercise / Electrical Calculations ( I don’t know if you still get them in 2391-10 . Exams ?? ) ◄◄
:86:
i). Shower circuit has a measured ( Zs ) value of( 2.2Ω )
ii). Previously the circuit ( R1 + R2 ) value was measured as ( 1.8Ω)
iii). The temperature factor is ( 1.2 ) & cablefactor is ( 1.04 )
iv). if the maximum tabulated ( Zs ) value forthis circuit is ( 2.6Ω )

► Show by Calculation if the measured ( Zs ) value is Acceptableor Not .

Zs) = Ze + ( R1+ R2 )
Ze) = Zs + ( R1+ R2 )
Ze) = 2.2 – 1.8 . ◄
Ze) = 0.4Ω ◄
Zs) = Ze + ( R1+ R2 x factors )
Zs) = 0.4Ω + ( 1.8 x 1.2 x 1.04 )
Zs) = 0.4Ω + 2.2464

Zs) = 2.64Ω ….. is this circuit Acceptable. NO

TheWording - ► Show by Calculation ( 2391-10 )

 

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2392-10: Why’s
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Conductor Résistance & Insulation Résistance.

Whencarrying out an Insulation Résistance Test . on a Consumer Unit it is often easyto forget that ( ALL ) circuits basically are connected across Line & Neutral .Whatever they are .

Thismeans that when ( ALL ) of these circuits areconnected together & an Insulation Résistance Test is done .
“The overall reading may well be Unacceptable . “

Back to Basics .
Becausethe circuits ( ALL ) possess Résistance & Resistorsin Parallel have the effect of Reducing theOverall Value .

BS-7671:2008
TheRegulations allow us to ( Sub-divide ) Thesecircuits to get ( Around ) this problem .

AsElectricians’ we must keep our Fundamental Principle in Mind .

p/38- As GN-3 rightly put it .

2392-10– IR . You do not want . a Low Résistance on your readings . ( Which would indicate defective insulation )

GN-3. p/38
InsulationRésistance value of Not Less than ( 1.0MΩ ) complieswith the Regulation(s) where an Insulation Résistance of Less than ( 2MΩ ) is Recorded the possibility of a latent defectExists .

p/39- As GN-3 rightly put it .
where a LOW READING is obtained ( Lessthan 2MΩ ) it may be necessary to test EACHCONDUCTORSEPARAETLTY to EARTH .
(After ensuring that all Equipment is Disconnected)

2392-10. Résistance Readingsobtained should be Not Less than the minimumvalues referred to in . Regulation(s) Table 61 . p/158

TheRegulations are tell us . The Minimum Insulation Résistance ( MΩ .≥ 0.5 / ≥ 1.0MΩ )

 

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General Information about ( Prospective Fault Current ) :icon_bs:

BS-7671:2008. requires the following . Prospective FaultCurrent to be determined for each installation by “ Calculation “ or “ Enquiry “

•The Prospective Short – Circuit Current at the Origin . Regulation 313.1. /refers . &
•The Prospective Short – Circuit & the Prospective Earth Fault current at every point of the installation . Regulation . 434.1. refers

Thiscovers the determination of the Maximum ProspectiveShort – Circuit Current at the Origin of an installation by means of Enquiry toa distributor . Where the installation is supplied at ( 400V three-phase / Similar. for an installation having 230V . single-phase ) supply of up to 100A

 

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►►Example Only . ( kW ) Making you Aware of Diversity :uhoh2:

Conveyorbelt made up of ( Six ) sections . – Eachdriven by a ( 2kW/ Motor )
Materialis Transported along this belt – it is first carried by section ( 1 ) Then each section in succession until the finalsection is reached .

SimpleExample
Only( One ) section of conveyer is carryingmaterial at any point in time .
Therefore( Five ) Motors are Only handling – No-Load mechanicallosses ( ? 1 kW )

Keepingthe belts moving whilst ( One Motor ) is handling the Load. ( ? 1kW) The Demand presented by ( Each Motor ) when it is carrying it’s Load is ( 1 kW )

Demand Load(s) – is ( 6kW ) but the Maximum Loadpresented by the System at any time is Only ( 1.5 kW )

TheDiversity factor for this system is ??
TheDemand Load(s) – 6kW
Maximum Demand – 1.5kW

[= 6kW ÷ 1.5kW = 4 ]

Maximum Demand – 1.5kW
Connected Load - 12kW …………… 1.5kW ÷ 12kW = 0.125

 

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Why? do we Issue Certificates :39:

ElectricalTest Certificates are used to record what has been done & confirm that theinstallation meets the required standard .

2392-10:
Withinthis Standard . Regulation - 610.1. State that .
Everyinstallation shall . during erection & completion before being put intoservice be Inspected & Tested to verify . so far as reasonably practicable. that the requirements of the regulations have been met .

2391-10 PIR )
Regulation. 621.1. State that .
Whererequired . Periodic Inspection & Testing of every electrical installation shall be carried out in accordance withRegulations . 621.2. to 621.5. in order to determine as far as is reasonablypracticable . whether the installation is in asatisfactory condition for continued service .

Design& Installation . from ( Part P ) with respect
Workshould Comply with BS-7671:2008 . Electricalwiring Regulation(s)

( Part P ) with respect
Inspection& Testing before taking into Service .
p/79. 6.2.1. General – Every installation must be inspected During Erection asnecessary & on Completion & before being put into Service to provide a Visual Check that theinstallation including the ( Installed equipment complies with the requirementsof BS-7671:2008 )

( Part P ) BS-7671:2008Installation Certificates .
Compliancewith Part P can be demonstrated by the Issue of a correct Electrical InstallationCertificate

2392-10:
Compliancewith Building Regulations is a ( Legal Requirement ) electrical work carried out in DomesticSector is now in the Building Regulations .
Itis a Criminal Offence “ NOT “ to “ Comply “ withBuilding Regulations .

Justa gentle reminder . 2392-10 .
BS-7671:2008 . does NOT recognize a “ Kitchen asa Special Location “ ( Document P ) p/60 – 5.2. ▼

Kitchen is defined in the Building Regulation(s) as a “ Room or part of a Room “ which contains asink & food preparation facilities .

BS-7671.requires . ?
2392-10: under Part P . The Electrical Installation Certificate. may be used together with Schedule of Inspections. & Schedule of Test Results .
Hint . ►► this will come up inyou -&-s EXAMS / 2392-10 . ( 1 x 1 x1 ) ??

Remember . Approved Document P– Electrical Safety .
AsPart P . reminds us . This Safety Certificate has been issued to confirm thatthe Electrical Installation Work to which it relates has been Designed .Constructed & Inspected & Tested in accordance with British Standard7671

Ifan Electrical InstallationCertificate is required . then the Earthing arrangements must be Upgraded to complywith the current Edition of BS-7671 .

UsefulJunk :
Asits not in the Regulation(s) -Table 61 .Insulation Résistance. it may never come up . -&-s . Q/As – FELV. circuit(s) are tested as ( LV ) circuit(s) at ( 500V . d.c. )

2392-10. Q/As -&-s . Your testing your self . PS. This one is still on the books . ▼▼

Asa Tester ( 100mA) general purpose . RCD to BS-EN 61008 is to be Tested to ensurethat it operates within Speciation . ( You should bedoing Handstands at this Stage .
Yeah )

- TTsystem should preferably use a Slit – Consumer Unit . with RCD rated at ( 100mA) for equipment . & ( 30mA ) for socketoutlets – O.S.G. p/24. fig 3.2.
-Where a circuit has ( High Loop Impedance ) a 100mA RCD issuitable providing it is NOT for socket outletsthat could be used outside & that the circuit is NOTin a special location .

2392-10– as a Tester . S Type 100mA RCD does not normally trip ( Without Reason )

2392-10– Regulation . p/340 . Schedule ofInspections .

( Firstly ) → TheInspections Schedule provides CONFIRMATION that a VISUALINSPECTION has been carried outas required by Part 6 of BS-7671 .

2392-10. don’t worry You will not get this in Exams . 2391-10 / may
Thesum of the Résistance of the Line conductor &the Protective conductor ( R1 + R2 ) should be recorded & . after appropriate correction for temperaturerise .
Maybe used to determine Zs – ( Zs = Ze + ( R1 + R2 )

Where Test method 2 is used .The maximum Value of ( R2 ) should be recorded.