***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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Flexiblecord on which the lampholder is suspended performs two - functions ‘ it carries the electric currentto the lamp . & supports the weight of the holders . lamp & shade . “ its physical strength “ Theconnection is made by means of a ceiling rose .

Thevast majority of light fittings are installed above head height . so clearly itis important that the fittings are secure . BS-7671-2011 . requires that thefixing for light as a pendant set has to be capable of supporting a Mass of ( 5kg )

559.6.1.5. – p/178
Fixingof Luminaries .

Luminaries. must be fixed securely . & the means of fixing must be able to supportthe weight of the Luminaries . this fixing must be able to carry a mass of notless than ( 5kg )

16[SUP]th[/SUP] Edition . ◄► p/96. blast form the past . ◄ :icon_bs:

522-08-06 .
A flexible wiring system shall be installed so that excessivetensile & torsional stresses to the conductors & connections areavoided .

Flex & Cables . ( pendentLight Fittings )

Theweight of the light fitting may affectyour choice of flex size .
Current – rating . 0.5mm[SUP]2[/SUP] flex takesfittings . up to ( 2kg ) 16[SUP]th[/SUP] Edition . ◄ …….. (1) 100W - ( 3 ) ► x 230 = 690W = ( 2kg ) Load Limit . ( Just a Point)
0.75mm[SUP]2[/SUP] flex takes up to ( 3kg ) inweight . 16[SUP]th[/SUP]Edition . ◄
1.0mm[SUP]2[/SUP] flex will support fittings up to( 5kg ) 16[SUP]th[/SUP]Edition . ◄

StandardCeiling Rose should not be used for light fittings weighing more than ( 2.5kg )for very heavy fittings ( more than 5kg ) a supporting chain is necessary .

559.6.1.3. – p/178
CeilingRoses should only be used for ONE outgoingflexible cord “ Unless “ they aredesigned for multiple pendent’s .

UsefulJunk – 2392-10 . Example .

Lewden Plug in Ceiling Rose.
Thesafe & easy way to remove a light fitting from a Wall or Ceiling .Detachable Ceiling Rose enables . removal of light fittings without disturbing the fixedwiring 6A rated . Maximum Working Load for Hook ( 5kg )

Enclosures .

Theenclosure of any equipment “ Serves “ to keep out dirt . dust . moisture & Prying Fingers.

 

[galinkahaha]

real thanks for putting all the information it helped me with my ongoing course

 

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2392-10 . Fundament requirements for selecting allEquipment .

BS-7671:2011 .addresses equipment so far as ( Selection & Equipment ) within the Installation is concerned –Regulation . 113.1. p/15

 

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2392-10 .Just a Reminder.

Compliance withStandards . 511.1. p/113

□ Every item of equipmentshall comply with :-
i) Currenteditions or agreed amendment dates of its :
ii) Britishstandard .
iii)Harmonized standard .

Or

Compliance with aforeign standard based on IEC standard as long as the degree of safety isequivalent to ( BS- )

□ NOT covered by a BS- EN- orHD- then the designer or person responsiblefor ( Specification ) must establish that the same degree of safety as affordedby the regulations . 511.2.

 

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State )- thedifference between . & the requirements for ( Isolation )

□ Switching offor ( Mechanical Maintenance )
□ Emergencyswitching / Functional switching .

2392-10 .

Testing . One off the most Important things -&-s are looking for is )- Example .

State )- theuses of Monitoring .
□ Insulation .
□ ResidualCurrent .

 

[NrVolts]

some things i really hate about the regs is that you have to use logic to try and understand what is been stated..its a bit like a double negative gives a positive answer ..i give the following example:

559.6.1.5. – p/178
Fixingof Luminaries .

Luminaries. must be fixed securely . & the means of fixing must be able to supportthe weight of the Luminaries . this fixing must be able to carry a mass of notless than ( 5kg )

the underlined bit means the fixing should be able support a load of 5kg and more but not below 5kg.why the hell they dont just say that is beyond me.

 

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BS-7671:2011 :

does not specify a maximum permissible value for voltage drop in installation(s) it does however require that undernormal services conditions the voltage at the terminals of any fixed currentusing equipment shall be Greater than the Lower limitcorresponding to the British Standard relevantto the equipment .

the requirements are deemed to be satisfied for a supply given in accordancewith the Electrical Safety . Quality & Continuity Regulations ( ESQCR ) ifthe voltage drop between the Origin of the Installation & the loadterminals is not Greater than ( 3% ) for lighting & ( 5% ) for ourequipment

drops in excess of these recommendations may be acceptable if itis Verified that the voltage Variations are within the limits specified in therelevant British Standards or in the absence of a standard the Manufacturesrecommendations .

Voltage Drop . ( mV / A / m) x L x Ib - 1000

5% of 230V = 11.5V .

Just a Reminder . – 2392-10

Regulation – 523.9. . Cable in Thermal Insulation .

A cable should preferably NOT be installedin a location where it is liable to be covered by Thermal Insulation . ETC .

 

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Industrialplugs have a keyway which prevents atool from one voltage being connected to a socket outlet of a different voltage.

400V– Red .
230V– Blue .
110V– Yellow .
50V– white .
25V– Violet .

 

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2392-10.

Thepurpose of Inspection & Testing the repaired circuit . system or equipment isthe confirm the ( Electrical Integrity ) of the system before it is ( Re-Energized )

 

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TheThree effects of an Electric Current .

Whenan electric current flows in a circuit it can have one or more of the followingthree effects . Heating. Magnetic or Chemical .

 

[Graeme Harrold]

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

Just a quick simple illustration for those looking for a basic relationship on Ohms Law...

 

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Isit necessary to Verify voltage drop during Initial Verification ?

□ Verification of voltage drop is ( NOT ) normally requiredunless there is considered to be a voltage drop problem . Regulation 612.14. p/194 .

Regulation 612.14 . Verification of voltage drop .
Where required to verify compliance with Section 525 , thefollowing options ( MAY ) be Used

(i) The voltage drop may be evaluated by measuring the circuitimpedance .
(ii) The voltage drop may be evaluated by using calculations . forexample . by diagrams or graphs showing maximum cable length v load current fordifferent conductor cross-sectional areas with different percentage voltage drops’for specific nominal voltages . conductor temperatures & wiring systems .

NOTE :- Verification ofvoltage drop is ( NOT ) normally required during InitialVerification . BS-2011 . – 2392-10

Foryour Assessors point of View . 2392-10

□ Considering Regulations . 134.1.1. – p/21. & 510.3. p/113 .
Which require equipment to be installed in accordance with ( Instructions by the Manufacturer ) areinstallers now required to ( Check Torque Settings ) for connections Tightness at ConsumerUnits where these are ( Manufacturer’s Instruction’s)

Regulation . 510.3 . Everyitem of equipment shall be selected & erected so as to allow compliancewith the regulation’s stated in this chapter & the relevant regulations in otherparts of . BS-7671 & shall take account of ( Manufacturer’s Instruction’s )

 

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Calculatethe speed of a Three-phase ( 50Hz ) Synchronous motor having (Np . 8 – poles )

Solution. we know .

Synchronousspeed . ( Ns) = ƒ x 120 - p

50 x 120 ÷ 8 = 750 rpm .

□ Fire Alarm System . These are two methods you will here on Site .

( Addressable )
Thebasic principle here is the same as for a conventional system .
Thedifference being that , by using modern technology , the control panel canidentify exactly which device initiated the alarm .

Thesesystems have their detection circuits wired as ( Loops ) with each device then having an “ Address “built in .

( Analogue )
Analoguesystems incorporate more features than eitherconventional or addressable systems . The detectors may include their own mini –computer . & this evaluates the ( Environment) around the detector & is therefore able to let the ( Control Panel ) know whether there is a fire .

A changein circumstance likely to lead to a fire . a fault . or even if the detectorhead needs cleaning .
As a rule . These systems are useful in preventing theoccurrence of False Alarms .

 

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VoltageDrop . ( Vd )

Cablesin a circuit are similar to ( resistors ) in that the ( Longer the conductor) the higher its Résistance becomes & thus the greater the voltage drop .

Regulation 525 . states that the voltage drop betweenthe Origin of the installation ( Usuallythe supply terminals ) & a socket – outlet ( or the terminals of the fixed current using equipment . shall NOTexceed three per cent ( 3% ) of the nominal voltage of the supply forlighting & (5% ) for other circuits.

 

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Taken from old notes . ◄◄

UsefulJunk .

Whyis a spur off a Ring Circuit called a “ Non-fused spur .

□ Because a spur connected inthis way breaks one of the golden rule’s of circuitdesign . i.e.
Thefuse size ( 30/32A ) should always be smaller than the conductorcurrent carrying capacity ( 27A for2.5mm[SUP]2 [/SUP] ) clipped direct) which it isn’t
Thereforeit is called unfused because the circuit fuse is too big to protect the spurunit .