***Cont../ Useful Information for Electricians & Apprentices***

[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 )

 

[amberleaf]

Re: Continued Useful Information for Electricians’ & Apprentices

“ Meaning “ Instantaneous :

O.S.G.Table 1B – column 5
Water–Heater(s) Instantaneous

Instantaneouswater heaters heat Only the water that is needed .
Thisis done by controlling the flow of water through a small water tank which hasheating elements inside it
Themore restricted the flow of water then the hotter the water becomes . Thetemperature of the water can therefore be Continuously Altered or Stabilisedlocally at whatever temperature is selected .

TheShower-type water heater must be supplied via its OWN MCB in the ConsumerUnit - Double pole isolator “ as a rule“ located near the shower .

Thereare Two-main methods of heating Water Electrically
Eitherheating a large quantity stored in a Tank or Heating only what is required whenit is needed .

•Non-pressure .
Non-pressurewater heaters . which are typically rated at ( Less than 3kW & contain Lessthan 15 litres of water ) Heat water ready for use & are Usually situateddirectly over the sink

WaterHeating Control . ( must have a Functional Switching Device for Convenience )

BS-7671some references
Regulations554.3.3. )- An Immersion Heater must be connected to the supply by Double-polelinked Switch Only . The use of a plug top & socket outlet is Not permitted
Note: No change for 16[SUP]th[/SUP] Edition to the 17[SUP]th[/SUP] Edition 2008 –Red Cover

Regulations554.2.1. )- An Immersion Heater must be Provided with a Thermal Cut-Out toprevent the water from boiling if the thermostat fails .

ImmersionHeaters should be supplied through a Switched Cord-Outlet Connection Unitcomplying with BS-1363-4 .
BS– 1363-4 )- Specification for 13A FusedConnection Units – Switched & Unswitched .

TheRing Final Circuit :
RedCover . BS-7671:2008. makes the following reference to Ring Final Circuits .

RingFinal Circuit must Not supply an Immersion Heater . Storage Heaters etc . 433.1.5.
TheRing Final Circuit Arrangements . p/362 .
(ii)Not supplying ImmersionHeater(s) . comprehensive Electric Space Heatingor Loads of a Similar profile from the Ring Circuit

Regulations314.2 )- All Final Circuits must be wired Separately from all Other Final Circuits

 

[amberleaf]

Re: Continued Useful Information for Electricians’ & Apprentices

Earth Loop Impedance ( EL-Z ) Test

This is a Live Test:

The Main Power has to be Energised to carry outthis procedure . There a certain amount of information required .as a rule thisTest is straightforward
{ Three-prong Test } if was carried out on alighting circuit you would be required to connect to the Line . Neutral . Earth.

Example : were Testing on the Ring Final circuit .What will be measured is called the Maximum ( Zs ) or the Measured ( Zs ) itshould be compared to the Maximum ( Zs ) in BS-7671:2008 .
• so Prior to Testing it is Important to known whichCircuit-breaker is protecting the circuit to be Tested . { in this case it willbe an RCBO to BS-EN61009-1 Rated at 32A }

Look in the Regulations BS-7671 page 49 table 41.3– the maximum ( Zs.s ) you must obtain a reading which is a reading less thanin this case it is 32A - 1.44Ω . it is Important to Note when Testing youshould obtain less than 80% of the figure . Sousing a calculator 1.44Ωs x .8 = we are looking for 1.15Ω

► We are looking for LESS THAN [ 1.15Ω < ] With the Instrumentplugged into the furthest socket ↔ The Loop Scale .
Test is taken & a reading of [ 1.1Ω ] → Less than the Maximum value .

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Symbol \ : Therefore. ◄ Electrical Therefore [\Sometimes used in proofs beforeLogical ] – Therefore : so : hence . Therefore : it follows –that .
Symbol ¥ : What is it ? Infinity – as a Tester you will use this a lot

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Mineral Insulated Copper Conductor “ Pyro “

Orange : General use .
Red : Fire Alarms .
White: Emergency Lighting .

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

The Basics .

Ib)– term used to describe a circuit’s design current in Amp’s . ( TheLoad )
In)– term used to describe a circuit’s Protective size in Amp’s ( MCB ) orfuses .
Iz)– term used to describe a circuit’s value . in Amp’s once all de-rating factor have been considered .
It)– term used to describe the tabulated current rating of a cable in Amp’s .( The Current a cable can Safety Carry )

[ Ib ≤ In ≤ Iz ≤ It ]

The formula above states the Underlying principle of the calculation of a circuitscable size .
The first factor you need to consider is Design Current. ( Ib )

 

[wcampbell]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Excellent.

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Inspection& Testing – 2391-10 / Designer

TheCertificate couldbe used in a Court of Lawto prove the competence of the Electrical Tester should anything happen at alatter date . if we were to certify an Electrical Installation that would laterresult in damage or harm to persons or property we would Require Proof that we carried outa Full Inspection & Test in accordance with BS-7671:2008 which would satisfythe Electricity at Work Act . The legalities ofour responsibilities are that we are guilty until proven innocent . So havingcorrect paper work & Test Records could save your neck .

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Remember . if any of the Initial Verification Checks require you to Remove Covers then you will need to Carry Out Safe Isolation .
Otherwise you will CONTRAVENE the ELECTRICITY at WORK ACT

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Keeping Circuits inOrder .

BS-7671:2008.
Regulations314.1. (i) & (ii) states that every Installation shall be divided intoCircuits to avoid Danger & minimise Inconvenience & to facilitate safeoperations . Inspection . Testing & Maintenance .

( T ) Regulation 514.1.2.
Statesas far as reasonably practicable . wiring shall be arranged or marked that itcan be Indentified for Inspection . Testing . Repair or Alteration of theInstallation .

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

BS-7671:2008.
Makes the following reference to Ring Final Circuits .

Regulation132.12 )- All Electrical Equipment must be )- Accessible for Operation . Inspection & Testing . Fault detection. Maintenance & repair .
Regulation 411.3.3. )- Unlessspecifically labelled or suitably Indentified . All 13A socket outletsmust be 30mA – RCD protected .
Regulation 543.2.9. )- Except where the Circuit Protective Conductor is formed by a MetalCovering or Enclosure containing all of the Conductors of the Ring .The Circuitprotective conductor of every Ring Final Circuit shall also be run in the formof a Ring having BothEnds connected to the Earthing Terminal atthe Origin of the Circuit.

Apprentices. Top Page Note : “ Wording “ Informative . ◄◄
Apprentices. Read the Wording ) p/362 . Ring FinalCircuit Arrangements . Regulation 433.1.5.
TheLoad current in any part of the circuit should be unlikely to EXCEED for LONG PERIODS the Current-Carrying-Capacity of the cable .
Regulation433.1.5 – refers )- This can generallybe Achieved BY

Regulation 433.1.5. )- Appendix 15 p/362 . Note : Informative . Cookerrated more than 2kW .

ARing Final Circuit must not serve a floor Area Greater than ( 100m[SUP]2[/SUP]) p/362 .. PS Ring – 100m[SUP]2 -[/SUP]32A
Regulation 433.1.5. p/362 )- A Ring Final Circuit mustStart & Finish in the Consumer Unit & be connected to ( 30A / 32A ) MCB etc .
Regulation 433.1.5. p/362 )- The minimum C.S.A. of theLine & Neutral conductor is 2.5mm[SUP]2 [/SUP] . The minimum C.S.A. of the CPC is 1.5mm[SUP]2 [/SUP]
Regulation 433.1.5. p/362 )- An Unfused Sur can feedeither One Single or One Double outlet only & can be taken from the MCB inthe Consumer Unit .
Regulation 433.1.5. p/362 )- The number of Spurssupplied from a Fused Spur connected unit ( a Switched Fused Spur ) depends on the size of the Fuse in the Fused connection unit .
Regulation314.4. )- All FinalCircuits must be wired separately from all other Final Circuits .
Note : O.S.G. IB )- Top of the Page. Purpose of the Final Circuit fed from theconductors . ETC.

O.S.G.p/18 )- Socket Outlets must be spaced at least 150mm away from Gas-pipes unlessthere is a plane of Non-combustible Insulating Material Separating them .
O.S.G.p/54 )- Lengths of Unfused Spurs off a Ring Final Circuit should not generallyExceed ( 1/8 ) the cable length from the Spur to the furthest part of theRing
O.S.G.p/54 )- The Wording : As a Rule of Thumb for Rings . Unfused Spur lengths ETC. ( 1/8 )

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Conduit Capacities . Basics

To calculate the number of Cables that may be draw into a Conduit .

O.S.G.You make use of the FOUR Tables

Table5A – Cable factors – conduit for short straightruns up to ( 3m )
Table5B – Cable factors – Use in short straight runs up to ( 3m )
Table5C – Cable factors – for long short straight over ( 3m ) or runs of any lengthincorporating bends .
Table5D– Cable factors – conduit factors for conduit incorporating bends & long straightruns

AllCables are Stranded . ( there are Two types of Cables in use Solid or Stranded )

Example:
Conduit25 metres long without bends with ( 3 x 1.5mm[SUP]2[/SUP]& 3 x 2.5mm[SUP]2 [/SUP]Stranded conductors )

a) Select the correct table for the cable factors & conduit ( 5A )
b) Obtain the cable factors for each size of cable ( 1.5mm[SUP]2 [/SUP] = 31 & 2.5 mm[SUP]2 [/SUP] = 43 ) → ( 5A )
c) Multiply the number of cables by the factor ( 3 x 31 ) +( 3 x 43 ) = 222
d) Select the correct table for the conduit factor & length of run ( 5B )
e) Obtain the factor which is greater than the sum of cable factor(s) ( 20mm = 460)
f) The conduit size is 20mm .

O.S.G.Trunking Capacities : & Tables from “ BS-7671 “

Minimum trunking size is calculate in much the same way conduit although trunking hasits own Table .
O.S.G.– No consideration is paid to the Length of Run .

Firstlyf or Exam - &- maximum allowable 45%space factor .

Table 5E O.S.G. - Cable factors for Trunking .
Table 5F O.S.G. – Factors for Trunking .

Trunking with 100 x 1.5mm[SUP]2[/SUP] & 100 x2.5mm[SUP]2[/SUP]

a) Obtain the cable factors for each size & type of cable ( 1.5mm[SUP]2[/SUP] = 8.6 & 2.5mm[SUP]2[/SUP] = 12.6 … ( P.V.C. Cable factor ) → Stranded Cable
b) Multiply the number of cables by the factor ( 100 x 8.6 ) + ( 100 x 12.6 ) = 2120
c) Obtain the factor which is greater than the sum of cable factor(s) ( 75 x 75 =2371 )
d) The Trunking size is ( 75mm x 75mm )

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

O.S.G.
Apart from Indicating that Diversity & Maximum Demand must be assessed . The Regulations themselves gives little help on calculations’ .

In BS-7671:2008 ( 17th Edition ) Regulation 311.1 gives requirements for the assessment of requirements of Maximum Demand . etc. determining the maximum demand have to consider Diversity .

• Water Heater(s) Thermostatic
( 2 x 6kW & 7 x 3kW ) Table 1B states “ No Diversity Allowable “

Small Shop ??
So the Total load will be .
P = ( 2 x 6kW ) + ( 7 x 3kW ) = 12 + 21 = 33kW → I = 33.000 ÷ 230V = 143.47A ( Unit in Amps )

( 2 x 6kW ) = 12
( 7 x 3kW ) = 21

• Ring Circuit(s) 8 x 32A
O.S.G. Table 1B states “ 100% of the largest & 50% of remained “ The Total Load ??
Table 1B Colum 9 ( Next column across 100% / 50% )

First Ring ( I – 32A ) PS … Not Domestic Installation(s)
Remaining Ring Final Circuit(s) ( I = 7 x 32A = 224A x 50% ÷ 100% = 112A ( Total for Ring Final Circuit(s) = 144A

Table 1B Colum 9 ( 100% / 40% - for Domestic Installation(s) 2392-10 Domestic Installation(s)
Ring Final Circuit – Full Demand on First Circuit → MCB is 32A . ( Then add 40% of remaining ) The two Ring Final Circuits are on 32A MCBs
After Diversity = 32 + ( .4 x 32 = 12.8 ) = 32 + 12.8 = 44.8 Amps . [ 40% for rest of Socket Outlets ]

The Term → ( MD ) Maximum Demand . is self-explanatory . it simply the sum of all the Individual Loads to be connected for a Simultaneous & Significant period .

 

[brucelee]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Good reading amber glad to see you back and sure others and I will benefit greatly from your help and info

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Inspection& Testing “ Overview “ 2392-10

Youwill need to demonstrate “ Assessor “

- Carryout an Initial Inspection of the Installation .
- Selectcorrect Instrument(s) to carry out Tests .
- Completethe Correct Sequence of Tests .
- Record the Test ResultsObtained .
- Carryout Functional Testing of an Installation .
- Fill in recognised Certificatesof Completion .

KnowledgeRequirements – 2392-10

• Howto Carry out an Initial Inspection .
• Howto Correct any Deviationsfound during Inspection .
• Howto use Various Test Instruments .
• The Importance of the Sequence of Tests .
• Howto carry out Functional Testing
• Howto Document Inspection & Testing .

Whatwould you do if you Discover Unsatisfactory Test Results . ??

ThePurpose of Inspection & Testing is to PROVIDE . So far is reasonably practicable.

• Thesafety of Persons & Livestock .
132.1. Design – The Electrical Installationshall be designed to prove .
i)The protection of Persons . Livestock & Property . ETC.
ii)The proper Functioning of the Electrical Installation – for the Intended USE .

•Protection against Damage to Property by Fire &Heat arising from an InstallationDefect .
•Confirmation that the Installation is NOT Damaged or Deteriorated so far as to ImpairSafety .
• TheIdentification of Installation Defects & Non-compliance with the Requirementsof the Regulations which may give Rise to Danger .

Inspection& Testing . 2392-10
During & or on Completionof a New Installation

Withthat in Mind ? 2392-10
Youhave to Satisfy the Requirements of Part P of the Building Regulation(s)

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

If you Know → Need to know. Yeah

Volts÷ Ohm’s → Need to know =Amps
Volts÷ Amps → Need toknow = Ohm’s
Voltsx Amps → Need toknow = Watts
Watts÷ Amps → Need toknow = Volts
Watts÷ Volts → Need toknow = Amps
Ampsx Ohm’s → Need toknow = Volts

Resistanceof the Length of Cable ….. Ohm’s x Current Drawn….. Amps = Volt Drop … Vd

FlexibleCables ?

Ifwe are “ Overloading “ a Cable : Keep inmind that the cable . Has a running current . by using more it will get Hot& eventually Melt the PVC coating . ( Plain English – Use the Right Size )
TheRegulations . my be a pain in the But !!There for a reason some kind person sitting at a desk has done the Work for us

Definitions/ p24 – Flexible Cord . A flexible cable in which the Cross-sectional area ofEach Conductor does Not Exceed 4.0mm[SUP]2[/SUP] ( PS. You will get this on CoP -&- ) 2011

BritishStandard BS-7671 : is set of Electrical Standards published by the IET in the UK .
TheStandards are NOT a LEGAL REQUIREMENT but a set of Standards which will ensurea Safe & Proper Electrical Installation

Aswe Say in the Trade ?? Some Contracts Specify that the Electrical Equipment ofthe “ Installation “ will CONFORM to BS-7671 & so at this Point they become( LEGALLY BINDING on the INSTALLER )

Fora Designer point of View .

Definition)- MechanicalProtection

522.6.5.(v) A cable installed under a floor or above a ceiling shall be run in such a positionthat it is NOT liable to be Damaged by contact with the floor or ceiling ortheir fixings .
PlainEnglish ) Be mechanically protectedagainst Damage . Prevent penetration of the Cable by Nails & Screws . & Like ( Keep Cables Out of Harms Way )

2392-10. We will use this on Part P . also

iii)&& (v) require Conduit & Trunking to be Earthed

• Whatis Mechanical Protection !!

Example : FP-200 is Fire Résistance? say 30 mins

Pyro: Mechanical Protection & Fire Résistance (2/1 )
i)Protection of cables against Mechanical Protection
ii)Protection of cables against Damage by Fire …( Long history of performing extremelywell in Fire(s) - Beyond the duration necessary to ensure theObjectives !!

MostDesigners specify either MICS → ( Pyro ) orProprietary Fire-resisting Cable(s)
Asa young Designer . ?? What would you Use .

?Mechanical Protection – running the cables on Cable Tray .

CableTypes ??
Twomost Important thing about Cable(s)

i)Insulation Properties’ are One of the Most Important things to consider whenselecting a Cable for a Circuit .
(if the Insulation is not rated for the Environment the circuit is installed intit can have a number of Detrimental Effects on its operation & Safety )

ii)Mechanical Protection . for cables is probably as Equally as Important as itsInsulation if the circuit with limited Mechanical Strength .
( Installedin an Environment where there is a risk of it coming into contact with Heavy Objects. as for our point of view ? )

522.6.5.(v) A cable installed under a floor or above a ceiling shall be run in such a positionthat it is NOT liable to be Damaged by contact with the floor or ceiling ortheir fixings .
PlainEnglish ) Be mechanically protectedagainst Damage . Prevent penetration of the Cable by Nails& Screws . & Like ( Keep Cables Out of Harms Way )

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

SeparatedExtra Low Voltage ( SELV )
Isthe most common method of providing protection . ElectricalSeparation ( Which means NO Electricalconnection between the SELV circuit &higher voltage systems )

* BS-7671 . 2392-10 some questions -&- “ Initial Verification “
Asa Tester .

* Regulation 610.1 - Theseare Hand & Glove with Q/As -&-s
Everyinstallation shall during Erection & on Completion before being put into Service. be Inspected & Tested to Verify . So far is reasonably practicable . Thatthe regulations have been met .

( Precautions’shall be taken to avoid Danger to persons . Livestock & to avoid damage toproperty & Installed Equipment during Inspection & Testing )

Certification& Reporting –

* 632.4 . Defects or omissions revealedduring Inspection & Testing of the Installation Work covered by theCertificate shall be made good before the Certificate is issued .

YoungApprentices !! InsulationRésistance Testing . ( IR)

Why we REMOVE LAMPS .

TheCOIL that is that is the LAMP FILAMENT is Effectively creating a Short Circuit between the Line& Neutral conductors

Let’slook at it in an other way - The bulb is only Designed to operate with 230V ( by applying 500V? would we do Damage )

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

All cable(s) have Electrical Résistance. So !

Therewill be Heat loss ( Energy loss ) when they Carry Current . This loss appearsas Heat & the Temperatureof the cable rises.
i)The rate of Heat loss is a function of the difference in Temperature between the Conductor & thesurroundings .
ii)“ Making it “ As the conductor Temperature rises – so does its rate of Heat Loss

Backon to Cable(s) .
Cablesare designed to be able to withstand a certain amount of Heat
Copperis a very good Electrical Conductor ( 300mm[SUP]2[/SUP])
Thismeans that the Résistance of a length of Copper Cable is relatively low .

Aluminiumcable would have Nearly Twice the Résistance of Copper Cables with the sameDimensions . ( 500mm[SUP]2[/SUP])
That’sthe reason why we stop using them / partly Cost

- Therésistance of the cable )- A higher résistancecable will get hotter at a given Current .
-The Insulation on the cable )- This will tend to keep it warm like a quilt .
- TheEnvironment of the cable )- were it is situated .

DesignCurrent ( Ib )

i)The first stage of the Design Current is to determine how much current will flowin the Circuit .
ii)This Current is known as the Design Current & is the FULL LOAD CURRENT of the CIRCUIT.
iii)Depending on Load ????

Singlephase . 230V resistive .
I= P/V . at this stage you areCalculating the Design Current )- I = 2000 ÷ 230V = 8.70A

Three-phaseInductive Load . power consumption of 30.000Watts – 30kW
Powerfactor – 0.95

atthis stage you are Calculating the Design Current )- I = P --- √3 x V cos Ø

I= 30000 --- √3 x 400 x 0.95 = 45.58A

[ √3x 400 x 0.95 ] = 658.1793069
30000 ÷ 658.1793069 = 45.58028441
Roundoff . = 45.58A

UsefulJunk . Ambient Temperature( Ci )
RememberCables must give off Heat … ( Surroundings’ ofthe cable . the more difficult it is for the cable to get rid of Heat )

Tables 4B1 & 4B2 – 17[SUP]th[/SUP] Edition . Note 30°C– 4B1 … Note 20°C – 4B2. & The O.S.G.

Wording C.C.C. - We will see itas sometimes as ?? Tabulated C.C.C. / Iz– de-rated C.C.C. ( Current Carrying Capacity )

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Polarity:
Correct polarity is achieved by the correct termination of conductors’ to the terminals of all Equipment .

Polarity:
Is either Correct or Incorrect– There nothing in between .

 

[amberleaf]

Re: ***Cont../ Useful Information for Electricians’ & Apprentices***

Example. “ it still forms part of Designing “

Cable Calculations ( The Whys & How’s ) it’s a Start

- Workout the design current ( Ib ) … I = P/V
-Select the Protective Device size ( Ib ≤ In )
- if we need to . Obtain all correction factors . Ca .Cc . Cg . Ci .
-Calculate ( Iz ) … Iz = In ----- Ca x Ci x Cg x Cc
-Select Conductor sizes . With ( Iz ) use the tables Appendix 4 & choose . ( Iz ≤ It )
-Calculate Volt-drop . Use design current ( mV/A/m ) value from cable table .The circuit length then ÷ 1000 [ Volt-drop Ib x ( mV / A / m ) x L --- 1000
-Calculate ( R1 + R2 ) Using table ( 9A / 9B & 9C ) find the mΩ/m value ÷1000 then apply multipliers . ▼
(R1 + R2 ) = 9A --- 1000 x 9B x 9C = Ω
- Calculate( Zs ) ……. ( Zs = Ze + R1 + R2 ) = Ω
-Verify ( Zs ) values with BS-7671 – using the tables in section 41 ensure calculatedvalues is less than the Tabulated values . ETC
- Calculatethe Fault Current .. Using nominal voltage & ( Zs ) . I = Uo/Zs = A . PS – Usefuljunk Regulations p/35 . Symbols (If ) fault current ( A ) 442.2.1
-Establish the disconnection time . Usingthe fault current . device type & time / current graphs in Appendix 3 obtainthe Expected Disconnection Time .
- Calculateminimum size of CPC . Using the Adiabatic formula . Input all values ( Including k factor ) & work out minimum size ofCPC . Then compare with our chosen CPC size .

HiBruce