***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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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 )

 

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(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 .

 

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(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% )

 

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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 .

 

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(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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(4 ) from a Testing point of View .

“ Motors lying inStorage “

Point to NOTE )- Check that the ( Insulation Résistance) of the Motor is satisfactory beforeSwitching on the Supply .

Why. ?? . Dampnessmay have been picked up During Storage . (( Bang )) :banghead:

 

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(5 ) Importance ☑

MotorTerminal Block :
Terminalarrangements for a (( Six – terminal )) Motor & the correct winding connections .

A1 – A2 :

B1 – B2 :

C1 – C2:

( Motor Windings ) if the connections to any winding are ( Reversed) the ( Magnetic fields ) will work against each other & a serious ( Overload ) willoccur .

Whenthe ( Starter) changes the Motor to the ( Delta ) configuration .

Never allow )- A2 .to be connected to B2 & A1 to C1 .


Importance )- This “ MUST NOT “ be confused with the ( Reversal of any TWO phases of the Supply ) when Motorrotation needs to be ( Reversed ) :banghead:

 

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(6 )

Frequency of Rotor Currents : :banghead:

Asthe rotating field is an ( Alternating One ) the currents intended in theRotor Cage Bars are also alternating . These are . however . NOT the same frequency as the supply . The frequencyof the rotor currents ( ƒs ) is given by .

( ƒs ) = Slip xsupply frequency .

Therefore: = S x ƒ .

S ) is expressed as a per Unit value . “ Example “ ( for 4% Slip )

S = 4 ÷ 100 = 0.04 .

 

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(7 )

An ( Eight – pole Squirrel – cage InductionMotor ) has a Synchronous speed of ( 12.5 revs/second) & a ( Slip of 2% )

Calculatethe frequency of the ( Rotor Currents )

ƒ = n x p .

= 12.5 x 4 = 50Hz .


ƒs = S x ƒ .

= 2 ÷ 100 x 50 . = 1Hz

 

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(8 )

Three– phase Motors may be ( Reversed ) by changing over any ( Two – phases )

 

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Re: ***Cont../ Useful Information for Electricians & Apprentices***
(4 ) from a Testing point of View .

“ Motors lying in Storage “

Point to NOTE )- Check that the ( Insulation Résistance) of the Motor is satisfactory before Switching on the Supply .

Why. ?? . Dampness may have been picked up During Storage . (( Bang )) :banghead:

Also, if Motors that have been in storage for extended periods of time, there is a very real chance that flat spots in baring faces have occurred. Motors or any rotary equipment such as generators etc, need baring supported shafts turned on a regular basis to stop this very real problem (and very costly) occurring...

On a job in Libya, the main contractor had purchased and delivered to site 6 No of 750kVA generators then left them covered up (tarp's) on site for 3 years in temperatures approaching 45C!! Every seal, baring and fiber gasket in those 6 generators needed replacing before we could put them eventually into service!! It cost the main contractor more than the original cost of the generators to put them right. lol!!! Thank god they didn't do the same with the 3MVA 10.800KV generator at the main HQ site ...lol!!!

 

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(9 ) Wound – round type . ( Slip ring motor )

Slip:

Themotor of an induction motor cannot travel at ( Synchronous speed) as there would be NO flux cutting& the machine would NOT work .

The ( rotor )is . then . said to “ Slip “ inspeed behind the ( Synchronous speed )

Slip– ( S ) is usually expressed as a percentage & is given by .

Slip– ( % ) = ( N[SUP] S [/SUP]- N[SUP] r[/SUP][SUP] [/SUP][SUP] [/SUP]) / N[SUP] S [/SUP]x 100 .

Where. ( N[SUP] S [/SUP]) [SUP] [/SUP]isthe ( Synchronousspeed )
N[SUP] r[/SUP][SUP] [/SUP][SUP] [/SUP]) is the rotorspeed .

► Six – pole ) . Cage induction motor runs at ( 4% ) Calculate the motor speed if the supplyfrequency is ( 50Hz )

( Synchronous speed ) N[SUP] S [/SUP]= ƒ / p

= 50 ÷ 3 = 16.666 revs / second

Therefore: 4 = ( 16.66 - N[SUP] r[/SUP][SUP] [/SUP][SUP] [/SUP]) 16.66 x 100 .
Therefore: 4 x 16.66 / 100 = ( 16.66 - N[SUP] r[/SUP][SUP] [/SUP][SUP] [/SUP])
Therefore: N[SUP] r[/SUP][SUP] [/SUP][SUP] [/SUP]) = 16.66 – ( 4 x 16.66 / 100 )

= 16.66 .sub 0.66 = 16 revs/seconds

Six – pole- ( 3 ) ◄


● (Synchronous Induction Motor ) :banghead:

Greatadvantage of the ( Synchronous Inductiontype ) is its ability to sustainheavy mechanical “ Overloads “ suchan “ Overload “ pullsthe motor out of ( Synchronism ) but it continues to run as an ( Induction Motor ) until the “ Overload “ is removed . at which time it pulls back into ( Synchronism again )

 

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● Useful Junk . Old Notes ◄

Nowadays the Instruments doit all for Us .

Determinethe power in Watts . & Volts . Amperes .& the power factor of a single –phase circuit . ( Load )

Calculatethe ( kVA ) & the power factor ofthe Load .

Only Example : 8A : 240V : 1.152kW.

kVA = VA / 1000 .

= 8 x 240 ÷ 1000 . = 1.92 kVA

PF. = kW / kVA . = 1.152 ÷ 1.92 = 0.6

 

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Remember this is Only an Analogy ) What a wasteof a Good Bud . !

The ( Beer Analogy)

Thisis a useful way to explain the ( Power Factor )

Wehave a pint beer glass with the main body of beer & the head . Although theglass is ( Full) part of it is useless . & the true amount of beer is ( Less ) than apint .
A ratioof true to ( Apparentbeer ) would indicate how much head there was . So if this ratio ( Pint Factor ) where (1 ) or ( Unity ) there would be NO head . & a ( PF ) of ( 0.5 ) wouldmean half beer & half head . clearly . it is better to have a ( PF ) close to ( Unity )

Apparent pint . Actual or True beer ◄► Head or Useless beer . Pintfactor = True pint / Apparent pint .

 

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-&-s– BS-7671:2008 ( 2011 )

This is an Open Book Assessment . & Learners will be allowed to take in thefollowing permitted reference material .
IET. Wiring books taken into examinations can containthe following .

• Bookmarks ( e.g. . blank post-it-notes orcorners of pages folded )
• Highlighting of text .
• The Corrigendum . issued by the IET as asupplements to the wiring regulations in July 2008 .

IET wiring regulations . books taken into exams cannot contain the following : :32:

• Sample exam questions . answers or diagrams .
• Any writing in the regulations or accompanyingwritten notes .
• Notes . diagrams or any content that may inanyway advantage the candidate in answering questions within the exam .

Itis the ( Responsibility) of any ( Centre ) to ensure the material in the candidate’s . IET Regulations to Electrical Installationscannot unfairly advantage candidates in any way . if there is any doubt overthe suitability of content the publication should NOT be used & replacedwith “ Clean Regulations “

-&-s : - Candidates will also require a ( Non-programmable calculator )