Electrical Loads the first step in the electrical designing for any construction is to estimate the electrical loads This enables the electrical engineering.

electrical Loads

• the first step in the electrical designing for any construction is to estimate the electrical loads

• This enables the electrical engineering (designer) to know exactly who should the electrical power feed this construction

An-Najah National UniversityFaculty of Engineering

Electrical Engineering Department

Electrical Loads

• Electrical loads in the buildings can be divided into three types:- Electrical loads used in lighting ( illumination system) , different types of lamps) Electrical loads feed the normal sockets and outlets ( used for small electrical devices , TV, radio , washing machine, etc….) Electrical loads that feed the mechanical devices used in the buildings ( elevators , heating system ,cooling system , pumps ,etc)



estimation of the electrical loads used in lighting

• the estimation of the electrical loads used in lighting is achieved by calculating the sum of the lamps used in the illumination system • there are different types of lamps :- Incandescent Lamp

Gas-discharge lamps



Old method for Electrical Load estimation in lightings

• this methods assumes that all the lighting system consists of either Incandescent Lamp or florescent lamps or a mixed of them . • in the case of using incandescent lamps :- Power(watt/m2) = illuminances (Lux)/2.78 • in the case of using incandescent lamps :- Power(watt/m2) = illuminances (Lux)/11.148 form the above equations it is quite clear that the using of florescent lamps can save power so it is quite important to use them as much as possible.This method is no longer used as it depends on approximations in addition to the fact that there a lot of different types of lamps used these days



example

• An office with an area of 60 m2, and the illuminance equals 500 lux, calculate the power needed in the cases of using the incandescent lamp and florescent lamps.

Answer :- • in the case of using candescent lamps :- P= 60* 500 /2.78709 = 10763 watt• in the case of using florescent lamps : P= 60* 500 /11.148 = 2691 watt power saving using the florescent lamps= 10763-2691= 8072 watt



Unit or Specific Load per Square meter•To over come the disadvantage of the previous method the “Unit or specific load per square meter” which depends on the natural using of the space in the building.•This method give more accurate results than the previous one.



Room Type Unit or specific load(watt/m2)

Living room (not dark colours for walls) 20-30

Living room (dark colours for walls) 40-50

children room (not dark colours for walls) 30-40

children room (dark colours for walls) 60-70

Sleeping room 20-30

kitchen 50-60

Working areas 40-50

Stairs , corridors, 20-30



Room Type Unit or specific load(watt/m2)

theatres 10

banks 21

Hair dressing shops 32

Worship areas 10

clubs 21

hospitals 21

hotels 21

offices 53

schools 32

Restaurants 21

ways for energy consumption saving in lights systems

• using lamps unites with high efficiency which is called “energy saving lamps” which have a high efficiency per watt in addition to the fact that they lasting for long times compared to the normal lamps• using lamps unites with low electrical losses on the form of heat which reduce the need of using the cooling systems for the construction.• reducing the light losses by using lamp units with low light losses coefficient• using lamp units with higher coefficient utilization



Unit Power density procedure

• this method differs than the previous methods in:- 1- it takes into account the shape of the rooms or the space(square,rectangle, etc) 2- it assumes that the lighting system in that room or space is based on the maximum using of the electrical power.

This method depends on dividing the building into rooms depending on the nature of using. And for every room or space there is a tables shows base power unit density(UPD) for that space in watt/m2




• the power needed for lighting a room or a space according to this method is:-

Power= Area* Power unit density(UPD)*room factor* space utilization factor(SUF). Power= A* a1* a2 Power: wattArea (A):m2

Power Unit density(UPD): (F):watt/ m2

Room factor : (a1)Space utilization factor :(a2)




Power Unit density(UPD): obtained from tables(1.2)



Room Type Power unity density

Bank 50.59

hospitals 15.07

offices 34.44

buildings 23.68

hotels 15.07

Libraries 9.69

Schools 23.68

Shops 40.9



Unit Power density procedureRoom factor : (a1)

Room factor (a1) is a number between 1.0 and 2.0 and its important comes from that it shows the effect of the room or space shape on the lighting system . From table (2.2)

Eg 1 :- length of the room is 2.4 . Width is 3.7 and the length is 2.7. the room factor from the table will be 1.8

Eg 2 :- length of the room is 7.3 . Width is 2.4 and the length is 3.4. the room factor from the table will be 1.9

If the shape of the room is not regular we choose a nearest regular shape such that the area of the regular shape is equal to the irregular shape of the room or space



Unit Power density procedureSpace utilization factor :(a2)

The value of the space utilization factor (S.U.F) is between 0.4 and 1.0 .

it is depends on the ration between the area of the room or space that is used and the total area of the space or the room.

Ratio between the used area and the total area space utilization factor (S.U.F)

0.5-1.0 1

0.4-0.49 0.85

0.3-0.39 0.77

0.2-0.29 0.55

Up to 0.19 0.4




• After calculating the electrical power needed for lighting the rooms it means that we calculate the electrical power needed for lighting all the listed areas.

•For the unlisted spaces . First : the area of these spaces are calculated and it equals the area of each flat in the building –the area of the listed spaces.

•The electrical power for the unlisted spaces is then calculated by multiplying its area by 2.15 watt/m2

•the total electrical power for lighting the inner area will be sum of the powers needed for listed and unlisted area.

Unlisted areas




• it consists of 1- the electrical power calculated for the inner space of the project(listed and un listed) 2- the electrical power needed for lighting the outside of the project

a- the electrical power needed for lighting the outside walls of the project (5% of the electrical power needed for the inner lighting ) b- the entrance of the building and the exit (table 1.2)

Unit power density for the inner lighting of the project= power for inner light/ total area of the flats Unit power density for the lighting of the project= power for inner light/ total area of the project

The maximum electrical power for the whole building




• after calculating the total power for the whole building we multiply this power by factor called demand factor which equals 0.85 in the case of lighting.

Unit power density for the inner lighting of the project= power for inner light/ total area of the flats Unit power density for the lighting of the project= power for inner light/ total area of the project

The maximum electrical power for the whole building



Electrical loads needed for feeding the sockets and outlets

• used to feed electrical devices with small power

• if the devices that connected with socket is already known, then the rated power for that device is considered.• if not , then for each socket , we assume that the power for the devices that connected to that socket is between 200 to 250 watt if it is single•If it is double socket then we assume the power for that socket is about 350 watt.



Electrical loads needed for feeding the sockets and outlets

•Table 5.1 shows the number of sockets needed in each room in the building

•Form the table the number of sockets needed in kitchen is 4 in addition to a special socket for the cooker unit.

•The total power needed for feeding the sockets will be then multiply by 0.7 as a demand factor.



Electrical loads needed for feeding the mechanical devices

•Conditiong unit :- each ton refrigeration needs 1500 watt electrical power.•Table 6.1 shows the number of ton refrigerating needed in different types of buildings•For the pumps :- the electrical engineering obtained the mechanical Horsepower (HP) needed for each pumps

•The electrical power =for that pump will be: mechanical Horsepower *746 watt

•For the Escalator and Electrical Stair the power will be given from the catalogue



Electrical lamps

• Electrical lamps that used in lighting is divided into two main type:- 1- Incandescent lamps 2- Discharge lamps

For the incandescent lamps :- the light is emitted as a result of electrical current passing through a filament , which makes it heated to a high temperature then glowing and luminating .

For the discharge lamps:- the light is emitted due to the glowing of the gas atoms between the poles of the lamp.



Electrical lamps

• Specification of electrical lamps:-

1- Luminous flux (lumen) 2- Luminous efficiency ( Lumen/watt) 3- Life time 4- Illumination (Lux) or (Lumen/m2) 5- power , voltage (watt ,V) 6- size of the lamp

The luminous efficacy of a lamp



Electrical lamps

• Incandescent lamps

1 – Filament (Tungstun) 2- Bulb 3- Base 4- inert gas or empty volume




Electrical lamps

Incandescent lamp


• the current is fed from the electrical source to the filament using the base• when the current pass through a filament, its temperature is increased and after a short time it is becomes glowing and emitting the light.

• carbon used to be used in the filament , but now a days , Tungstun is used in filament and sometimes some elements such as AL, K, SI are added to Tunstun to improve the hardness of the filament

• luminous efficiency of the incandescent lamps increases as power of that lamps increase . For example, a lamp with power 150 watt give more illumination in a percentage of 34% than using three 50 watt lamps.



Electrical lamps

Incandescent lamp


• the tungstun incandescent lamps are used a lot because of:-

1- it is found in a lot of shapes (table 2.3)

2- the quality of light (suitable for the eye)

3- is very cheap

•The lifetime for the incandescent is about 1000 hour, •The luminous flux of the incandescent lamps is directly related to the applied voltage on the lamp. •Table 3.3 shows the relation between the luminous flux , voltage, and the lifetime .



Electrical lamps

Incandescent lamp


•The end life of the Tungstun lamps

•The melting of the filament is the indication of the end life of the Tungstun incandescent lamps•Due to a defect in a position on the filament , a hot-spot is formed on it, and the temperature of the on that position at beginning will be higher than other position on the filament.• the high temperature makes the Tungstun to evaporate.

•All these process happened at the beginning of the lighting .•The evaporated Tungstun is stuck on the inner surface of the bulb making a black colour to appears on the bulb surface.(blackening).



Electrical lamps

Incandescent lamp


•The end life of the Tungstun lamps•The blackening phenomena appears only in the Tungstun lamps where the bulb is evacuated from the air

• To overcome the blackening phenomena , a Halogen Lamps are used where one of the following halogen is used (Cl, Br, I, F) in addition to the inert gas. This halogen helps to return back the evaporated Tungstun. •they are more efficient than incandescent bulbs using only half the energy to produce the same light output and last twice as long•The halogen lamps are used for specific applications where a low voltage and clear light is needed like , cinema , projectors, theaters, cars lamps , in TVs



Electrical lamps

Discharge lamps


• the principal operation of the discharge lamps is based on an electrical discharge in the atoms of the inert gas or the vapour of metals or a mixed of them.• this results on some visible lights •It is divided into two types:- 1- low pressure discharge lamps (Neon , Florescent) 2- high pressure discharge lamps (Soduim and mercury )



Electrical lamps Florescent lamps

• The current passes through the circuit heating up the filament in each electrode, which are located at both ends of the tube. •The heated electrodes in addition to the high voltage applied to two ends of the lamps cauasing electrical discharge (ionizing the gas(argon)). •The mercury vapor becomes "excited" and it generates radiant energy, mainly in the ultraviolet range. •This energy causes the phosphor coating on the inside of the tube to fluorescent, converting the ultraviolet into visible light.



Electrical lamps Florescent lamps

• in order to obtain a light from florescent lamps, the following things should be happened 1- electrical discharge 2- tranforming the ultraviolet radiated energy to a visible light



Electrical lamps

Power of the lamps(watt)

Incandescent lamps 1- general service lamps (15 to 250 watt) 2- projectors (40-1000) watt 3- halogen Lamps (1000-2000) watt 4- table lamps for specific use (15- 50)watt (positional use)Florescent lamps 1- general (15-80) watt 2- special types (125-200) watt 3- table lamps (4-31) watt (positional use)



Electrical lamps

Mercury Lamps 1- general (80-1000) watt

Soduim Lamps 1- low pressure (35-180) watt

2- high pressure (70-1000) watt



Electrical lamps

Lifetime

Incandescent Lamps :-

1- Genral (1000 ) hour 2- projector (750-1500) hour 3- Halogen (2000) hourFlorescent :- 1- (15-80 ) watt :- 1000 hour 2- (125-200) watt:- 3000 hour 3- (4-13) watt:- 2000 hourMercurry :-7500 hour Soduim :- 2000 hour

CFL :- (5000 -15000)



Electrical lamps

Luminous efficiency ( Lumen/watt)

Lamp type Luminous efficiency

Incandescent 25

Florescent 80

Mercury 60

Soduim (Low pressure) 185

Soduim (high pressure) 140

CFL (sl, pl) (compact florescent lamps) (energy saving lamps)

75-95



Electrical lamps

Energy saving Lamps (CFL)

It is new lamps and comes to replace the traditional lamps (florescent and incandescent)

It has the following characteristics:-

1- Low consumption of energy2- long life time (5000 hour)3- high luminous efficiency (50 lumen/watt)4- have a small size.



Electrical lamps

Comparison between incandescent lamps and CFL lamps

1- Life time 5- Size * incandescent 1000 hour * bigger * CFL 5000 hour * smaller2- power 6- weight * incandescent :75 watt * bigger * CFL : 18 watt * smaller3- luminous flux :- 7- cost * incandescent : 900 lumen * less * CFL : 90 lumen * more4- luminous efficiency * incandescent : 12 lumen/watt * CFL 50 lumen/watt



Electrical lamps

Comparison between incandescent lamps and CFL lamps

Eg:- compare between incandescent lamps (75 watt) and CFL lamp with power (18 ) watt . If The price of incandescent is 0.200 JD and the price of the florescent is 4.5 JD and the price of the kwh is 0.04 JD assuming that both kamps will work 5000 hourNo of incandescent lamps =5000/1000=5No if CFL lamps =1The price of the incandescent lamps =5*0.2 = 1 JDThe price of the CFL lamps = 4.5 JDThe power consumed by incandescent in 5000 hour in Kwh= 5000 *75/1000 = 375 kwhThe price of the kwh = 375 * 0.04 = 15 JD Total for incandescent = 15+1= 16 JDKwh for CFL = 5000*18/1000= 90Price of kwh = 90 *0.04= 3.6 JD

Total price for CFL = 3.6+4.5 = 8.1 JDSaving = 16- 8.1 = 7.9 JD



Lighting calculations

The objective of performing the lighting calculations is to determine the type, power and Distribution the lamps unites

This needs the to know the illumination needed in the sapce depinding on the natural use of the space

the method that we want to use here is called Lumen method




Lumen Method

This method depends on using the utilization factor assuming that the illumination is distributedRegularly in all directions in the space.

N= Em.A/n*Fl*ku*kn

N= no of lamp unites needed to obtain the desired illuminationEm= illumination (obtained from tablesA = arean= no of lamps in the uniteFl= luminous flux for the lamp (lumen)Ku= utilization factorKn= maintenance factor




Lumen Method

Kn= maintenance factor

The maintenance factor depends on the situation of space :-

It is value obtained from the table below:-

Room situation Maintenance factor

Normal 0.8

Dirty 0.7

Very dirty 0.6




Lumen Method

Ku= utilization factor The utilization factor depends on the dimensions of the room .

Kr Utilization factor %

0.6 35

0.8 44

1 51

1.25 58

1.5 64

2 72

2.5 77

3 81

4 85

5 89

>5 90




Lumen Method

Ku= utilization factor Where kr is the room index and it is calculated as follows:-

Kr= L*w/Hm(L+w)Where

Kr:- room indexL :- length of the roomw:- width of the roomHm:- the high of the lamp unit from the surface of works in m




Lumen Methodkr

If the height of the room is H , and the height of the woorking surface from the ground of the room is Hp, and the lamp unit is hanged a distance HL from the ceiling then Hm=H-HL-Hp

Usually Hp = 75 cm except it is given a value different than that.




Lumen Methodeg

an office wit dimensions = 8m length , 5 m width and 3 m height . Calculate the number of unit lamps needed for illumination.

Solution:- we use lumilux flourescent lamps from table 16.3 page 111 where luminous flux = Fl=L=3450 lumen .Ilumination =500 lux. Page 275.Maintenance factor=kn= 0.8Utilization factpr kr= L*w/Hm(L+w).HL=0Hp=0.75 mHm=3-0-0.75=2.25 mKr=8*5/(8+5)*2.25=1.37We choose kr=0.58+0.64/2=0.61N=500*8*5/2*3450*0.61*0.8=5.59 = 6 units




Lumen Methodeg

an office wit dimensions = 8m length , 5 m width and 3 m height . Calculate the number of unit lamps needed for illumination.

Solution:- we use lumilux flourescent lamps from table 16.3 page 111 where luminous flux = Fl=L=3450 lumen .Ilumination =500 lux. Page 275.Maintenance factor=kn= 0.8Utilization factpr kr= L*w/Hm(L+w).HL=0Hp=0.75 mHm=3-0-0.75=2.25 mKr=8*5/(8+5)*2.25=1.37We choose kr=0.58+0.64/2=0.61N=500*8*5/2*3450*0.61*0.8=5.59 = 6 units



Electrical Conductors

• the electrical conductors that are used in electrical installation are usually made from AL or CU or a mixed of them.

•The difference between the AL and CU is that the specific resistance for the Cu is less than that in AL

•This means that a cable with bigger radius is needed in the case of AL compared to Cu to conduct the same current.




Conductor resistances and the factors that affect it

•Conductors have a high conductivity due to the existence of a high no of free electronsor in other meanings its electrical resistance is very low :- ρ=1/σWhere ρ is the resistivity and σ is the conductivity of the conductors.•The resisivity is an indication of the quality of the materials as the resistivity is wanted to be as small as possible .




The resistivity of the conductors is affected by the temperature according to the following formulas:-

ρt=ρ0(1+ℓ (T-T0)

Whereρt is the resistivity at temperature Tρ0 is the resistivity at 250

ℓ is the thermal expansion coefficient for the conductors at 250




thermal expansion coefficient for the conductors

Material Thermal expansion coefficient

silver 0.0038

Copper 0.00393

Aluminium 0.00377

Nickel 0.006

Iron 0.006

Platinum 0.0025

NI mixture 0.0001

http://www.4shared.com/file/_qrtIq3h/Pokemon_Platinum__U__English_V.html




The resistance of any cable is given as:-

R= ρL/AWhere R is the resistanceρ resistivityL length of the cableA area of the cableFrom the above equations the resistance depends on:-1-the conductor type (AL ,Cu)2- length of the cable3- area of the cable4- temperatureIt is meant by temperature :-1- the temperature of the space around the conductors2- the temperature of the cable itself which is produced by the current passing in the cable




Ampacity is defined as the maximum amount of electrical current which a cable can carry before sustaining immediate or progressive deterioration

1- its insulation temperature rating;

2- the electrical resistance of the cable material;

3-frequency of the current, in the case of alternating current;

4 - ability to dissipate heat, which depends on cable geometry and its surroundings;

5- ambient temperature.

Ampacity of cables

http://en.wikipedia.org/wiki/Thermal_insulation

http://en.wikipedia.org/wiki/Temperature




How to calculate the cross section area of the cable

1- calculate the designing current in the circuit ore part of the circuit2- according to that , the protection for the circuit is selected3- the choosing of the cross sectional area of the cable will be according to

I cable > I protection > I design




Coloures for electrical conductors

Current IEC

Protective earth (PE) Green/yellow bi-colour

Neutral (N) Blue

Single phase: Live (L)Three phase: L1 Brown

Three phase: L2 Black

Three phase: L3 Grey

http://en.wikipedia.org/wiki/Ground_and_neutral




Area of Conductor

Resistance at 20°C Max. ohm/km Current Rating (Amp.)

0.05 39.00 40.75 26.00 71.0 18.10 111.5 12.10 14

2.5 7.41 19

4.0 4.95 26

6.0 3.30 33

10.0 1.910 45

16.0 1.210 60

25.0 0.780 75

35.0 0.554 95

50.0 0.386 125




Drop voltage calculations

Voltage drop = V1 – V2 =2(I1*r1 +I2*r2+I3*r3+I4*r4)I1,2,3,4 are the currentsr1,2,3,4 are the branches resistances

Let us assume I1=i1+i2+i3+i4 I2=i2+i3+i4 I3=i3+i4 I4= i4




Drop voltage calculations

Let us assume R1=r1+r2+r3+r4 R2=r2+r3+r4 R3=r3+r4 R4= r4

Voltage drop = 2(i1*R1+i2*R2+i3*R3+i4*R4)=2∑iαR α

As R=ρL/AVoltage drop = 2 ρ/A ∑ILA= (2 ρ/voltage drop) ∑IL



Sockets connections

220 V -50 Hz

N P



Switch connections (2 lamps in series with single switch

220 V -50 Hz

N P



Switch connections (2 lamps in parallel with single switch 220 V -50 Hz

N P



Switch connections (2 lamps with double switch

N P

220 V -50 Hz




N P

220 V -50 Hz




N E P

220 V -50 Hz

chandelier



Switch connections (2 lamps with double switch 220 V -50 Hz

E N L



Sockets connections

220 V -50 Hz

N E P



Sockets connections

N E P



lighting 2 lamps from 3 different places220 V -50 Hz

E N L



Bell with push button switch

N E P



Meausring devices

N E PV

A

Kwh

V A



Circuit breakersProtect from :-

1- short circuit 2- over load3- short circuit and over loadCurrent ratings of circuit breakers :

6 A, 10 A, 13 A, 16 A, 20 A, 25 A, 32 A, 40 A, 50 A, 63 A, 80 A and 100 Ainstantaneous tripping current, that is the minimum value of current that causes the circuit-breaker to trip without intentional time delay (i.e., in less than 100 ms), expressed in terms of In

Type Instantaneous tripping current B above 3 In up to and including 5 In

C above 5 In up to and including 10 In D above 10 In up to and including 20 In K above 8 In up to and including 12 In



Circuit breakersTypes of circuit breaker

1- thermal circuit breaker

2- magnetic circuit breaker

3- thermal and magnetic circuit breaker

In choosing circuit breakers we should consider the following

1- ICB>= maximum load current2- VCB> = V supply3- Ibreaking capacity > 1.2 IscMCB :- miniature circuit breaker MCCB :- moulded case circuit breaker (63AMP (min.))



Circuit breakers

No of units lamps in each feeder should be less than 9No of sockets in each feeder should be less than 5 if it is 2 A and less than 3 if it is 5 A socket

the circuit breaker for the lamp units feeder is 10 A and the cross section of wire is 1.5 mm2

The circuit breaker for the socket feeder is 16 A and the cross section of the wire is 2.5 mm2

There should be feeder for the kitchen and at least spare feeder



RCD single phase (residual-current device)

2* 40 A, 0.03 A

http://en.wikipedia.org/wiki/File:FISkizze.svg



RCD 3 phases (residual-current device)

4* 40 A, 0.03 A

http://en.wikipedia.org/wiki/File:Fi-rele2.gif



Many flats in building (main distribution board

KwH KwH KwH

3 phase supply




3 phase supply

KwHBasement 2

Basement 1

Ground floor

first floor

second floor

Third floor

fourth floor

DB-B2

DB-B1

DB-GF

DB-F1

DB-F2

DB-F3

DB-F4

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe



Main DSB

3 * 160 A

KwH

DB-B2

5*16mm

2

3 * 63 A

DB-B15*16m

m2

3 * 63 A

DB-GF

5*16mm

2

3 * 63 A

DB-F1

5*16mm

2

3 * 63 A

DB-F2

5*16mm

2

3 * 63 A

DB-F3

5*16mm

2

3 * 63 A

DB-F4

5*16mm

2

3 * 63 A

From mains

RCD 160 A 30 mA



Basement 2

3 * 63 A

from DB-B2

light

3*1.5mm

2

10 A

light3*1.5m

m2

10 A

light

3*1.5mm

2

10 A

light

3*1.5mm

2

10 A

socket

3*2.5mm

2

16 A

socket

3*2.5mm

2

16 A

socket

3*2.5mm

2

16 A

spare

3*2.5mm

2

16 A

RCD 63 A 30 mA




3 phase supply

Basement 2

Basement 1

Ground floor

first floor

second floor

Third floor

fourth floor

DB-B2

DB-B1

DB-GF

DB-F1

DB-F2

DB-F3

DB-F4

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

KwH

KwH

KwH

KwH

KwH

KwH

KwH




3 phase supply

KwH

Basement 2

Basement 1

Ground floor

first floor

second floor

Third floor

fourth floor

DB-B2

DB-B1

DB-GF

DB-F1

DB-F2

DB-F3

DB-F4

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

KwH KwH KwH KwH KwHKwH



Main DSB

3 * 160 A

KwH

DB-B2

3*16mm

2

1 * 63 A

DB-B13*16m

m2

1 * 63 A

DB-GF

3*16mm

2

1 * 63 A

DB-F1

3*16mm

2

1 * 63 A

DB-F2

3*16mm

2

1 * 63 A

DB-F3

3*16mm

2

1 * 63 A

DB-F4

3*16mm

2

1 * 63 A

From mains

RCD 160 A 30 mA



Basement 2

1 * 63 A

from DB-B2

light

3*1.5mm

2

10 A

light3*1.5m

m2

10 A

light

3*1.5mm

2

10 A

light

3*1.5mm

2

10 A

socket

3*2.5mm

2

16 A

socket

3*2.5mm

2

16 A

socket

3*2.5mm

2

16 A

spare

3*2.5mm

2

16 A

RCD 63 A 30 mA



Electrical installation In factories

the electrical load in the factories depends mainly on the technological process used in the factoryAs the industrial process is determined in the planning design for the factory , it is possible to use the electrical information about the loads directly from the company that designs the machinesThe electrical installation in the factories consists of the following elements:-

1- main distribution board2- feeders3- auxiliary distribution boards and circuit breakers




connected with the main supply coming from the electrical distribution company and distributed from it the feeders to auxiliary distribution boards it consists of :-1- mean switch (circuit breaker) :- disconnect the distribution board from the electrical power and for protection it can be either air switch or oil siwtch2- bus-bars :- one bus-bar for each phase ,N ,E3- auxiliary switches4- instrumentation devices (power , A , V, f)

Main distribution boards



Electrical installation In factories Auxiliary distribution boards




there are different types of distribution boards either single phase or 3 phase and it can be designed to be used outside the factory or inside it in chosing the distribution boards . It is better to chose a place easy to be reached and in the middle of the electrical loads as this will reduce the material used in the installation and reduce the voltage dropsIt better that the electrical installation in the factories to be clear to make the maintenance easyThe cables and the sockets should be designed to carry the short circuit current for short time complete information for safty procedures and the electricla loads on eachD

General Notes about electrical installation in factories



Electrical distribution in factories

radial distribution

ring system voltage drop is less , more reliable

Both radial and Ring system



Electrical distribution in factories

it is recommended to choose the substation to be near the centre of electrical loads in order to reduce the cost of the electrical netwrok and the no of circuit breaker and protection devices.

substations in factories

y(m)

x(m)

load 2 load 1

load 3

Xo= sum(pi *xi)/sum(pi)yo= sum(pi *yi)/sum(pi)




consist of : 1- core 2- windings 3- cooling equipment a- radiators b- cooling fans c- oil 4- main tank :- a metal tank full of oil in which the windings and core exist 5- Tap changers :- to change the voltages of the transformers to meets the change in the voltages and loads in the electrical network

off load:- the transformer needs to disconnect from the network to change the connection of the tap changer oil load:-no need to disconnect the transformer from the network usually in the on load a special devices (relays) is used to regulate the voltage automatically

6- auxiliary devices a- measurement instruments b- protection devices

Transformers




conditions to operate 2 transformers in parallel :- 1- have the same voltage ratio 2- have the same voltage impedance 3- same vector groups star-star , delta-delta, star-delat, , delta-star,

Group I - (0°) - delta/delta, star/starGroup II - (180°) - delta/delta, star/starGroup III - (-30°) - star/delta, delta/starGroup IV - (+30°) - star/delta, delta/star(Minus indicates LV lagging HV, plus indicates LV leading HV)

Transformers in parallel




Sn> Scal/KoverSn : the nominal capcity of the transformerScal :calculated capcity of the loadKover : over load factor at each transformer

Where Kover =function (Kl,tmax)

Kl- load curve factorTmax- max time overload operation transformerKl=S average/S max= P average/Pmax

Selection of Transformers

A

P max

timePaverage= A/24KoverK load

11.05

1.10.85

1.2

1.150.9

0.95

1

time



Earthing system

earting is the electrical connection between the metal parts that are not carrying electrical current in normal conditions with the earth.It is important in earthing system to look at the earthing system at the source of the electrical power and at the consumer There is a special symbols to refer to the earthing system in both sides

the first letter refers to the earthing system at the electrical source and there are two letters used for that:- a- T:- refers that one point or more than one point are connected directly to the earth (Terre) b- I :- refers to that the electrical power source is either not connected to the earth or connected to the earth through fault limiting impedance (isolated).



Earthing system

the second letter refers to the earthing system at the consumer and there are two letters used for that:- a- T:- all the metal parts that are not carrying electrical current in normal condition are connected directly to the local earth b- N :- all the metal parts that are not carrying electrical current in normal condition are connected directly to the earthing system at the source

The third and fourth letters refers to the earthing system connections at the source a- S :- N and and E are sperated b- C:- E and N are one cable



Earthing system

the electrical source is connected directly with earth and the metal parts that are not carrying electrical current in normal conditions at the consumer are connected to the earthing system at the source.

Depending on the neutral and earth connections , there are three types of TN earthign system:- 1- TN-S :- the earthing system at the source is connected directly with earth (T). The metal parts at the consumer are connected to the earthing system at the source (N) . The neutral and the earth are sperated . Nablus

TN earthing system

http://en.wikipedia.org/wiki/File:TN-S-earthing.svg



Earthing system

1- TN-C-S :- the earthing system at the source is connected directly with earth (T). The metal parts at the consumer are connected to the earthing system at the source (N) . The neutral and the earth are connected at the source and separated at the consumer

TN earthing system

http://en.wikipedia.org/wiki/File:TN-C-S-earthing.svg



Earthing system

1- TN-C :- the earthing system at the source is connected directly with earth (T). The metal parts at the consumer are connected to the earthing system at the source (N) . The neutral and the earth are connected

TN earthing system

http://en.wikipedia.org/wiki/File:TN-C-earthing.svg



Earthing system

the earthing system at the source is connected directly with earth (T). The metal parts at the consumer are connected directly to the earth at the consumer . The neutral and the earth are connected

TT earthing system (Jerusalem)

http://en.wikipedia.org/wiki/File:TT-earthing.svg



Earthing system

the earthing system at the source is not connected directly with earth (T) but connected through isolation (I). The metal parts at the consumer are connected directly to the earth at the consumer . The neutral and the earth are connected

IT earthing system

http://en.wikipedia.org/wiki/File:IT-earthing.svg



Earthing system

TN networks save the cost of a low-impedance earth connection at the site of each consumer. Such a connection (a buried metal structure) is required to provide protective earth in IT and TT systems.TN-C networks save the cost of an additional conductor needed for separate N and PE connections. However, to mitigate the risk of broken neutrals, special cable types and lots of connections to earth are needed.TT networks require proper RCD protection.

http://en.wikipedia.org/wiki/Residual_current_device



Earthing system

Fault path impedanceIf the fault path between accidentally energized objects and the supply connection has low impedance, the fault current will be so large that the circuit overcurrent protection device (fuse or circuit breaker) will open to clear the ground fault. Where the earthing system does not provide a low-impedance metallic conductor between equipment enclosures and supply return (such as in a TT separately earthed system), fault currents are smaller, and will not necessarily operate the overcurrent protection device. In such case a residual current detector is installed to detect the current leaking to ground and interrupt the circuit.

http://en.wikipedia.org/wiki/Residual-current_device



Earthing system

Earthing system components:-1- electrodes (under gorund)

2- cable connecting the electrode with the earthing bus-bar3- earthing bus-bar4- conductors connecting the earthing bus-bar with the metal parts that are not carrying any currents in normal operation

Ground

electrode

Earth bus-bar

Conductor connecting earth bus-bar with electrode

Conductor connecting earth bus-bar with metal parts not carrying any currents



Earthing system

resistivity for the soil the electro physical characteristics for the soil depends on the resistivity of the soil the resistivity of the soil is the resistance of one cube of that soil whose dimensions are 1m x 1m x 1mThis resitivity equals :- ρ=r F/L (ohm.m)r:- the resistance of a specific volume of soil whose area is F m2 and whose length is L m



Earthing system

resistivity for the soil the resistivity of the soil depends on the followings :-1- the wetness in the soil:- the resistivity of the soil is reduced significantly with the increase of the wetness of the soil2- the physical structure of the soil:- the differences in physical structure of the soil leads to differences in the resitivities of the soils : clay soil has less resitivty than that of the dry sand and rocks3- the chemical structure of the soil :- whether the soil contains metals or salts . The resitivty of the soil is reduced if the salts in the soil is increases.4- the thermal degree of the soil :-the rising of the thermal degrees of the soil will increase the resitivity of the soil



Earthing system

charcteritics that should be found in the soil to be a good earthing system:-1- low electrical resistivity2- good resistance against the rust3- the ability to carry high electrical currents for many times4- the ability to sustain the above characteristics fro at least 30 years

To obtain a good earthing , the following factors should be found:-1- low resistivity for the soil2- a wetness soil3- the temperature of the soil is higher than the freezing temperature



Earthing system Electrodes:-It metal rods buried under the ground . It is used to connect the metal parts that are not carrying any currents in normal conditions with the ground.it should have the following characteristics1- has a conical shape in order to make it easy to be buried under the ground 2- has a high mechanical solidity3- high resistance to rust 4- the ability to high a high current at fault conditions length of the electrodes between ( 2.5 – 5 m) it usually made of

1- copper clad steel2- solid copper3- galvanized steel4- stainless steel



Earthing system Vertical Electrodes:- the resistance of the vertical electrodes are calculated through the following formula:- r =ρ /l Where ρ is the resistivity of the soil ( ohm.m) and L is the length of the electrodeHorizontal bar:-It usually connects the vertical electrodes together and it is either thin plates or circular conductors The resistance of this horizontal plates does not depend on the their width or thickness but only depends on the length of this plates and the depth at which they are buried at . The depth of these horizontal plates is 0.6m and 0.7m.The resistance of this bars are calculated through :- r= (ρ/4)*sqrt(π/A)Ρ:- resistivity of the soil in ohm.cmA :- area of the bar(cm2)

If the resitance of the bar is high then many bars are connected in parallel to reduce the total resitance of the horizantal bars



Earthing system

water installation In the past the water pipes were used as electrodes for the earthing system The resistance of the electrical pipes is between 0.2 to 1 ohm for a soil has resitivity of 10 ohm. MNowadyas , the water pipes are not allowed to be used as the only source for the earthing system as a result of increasing use of the plastic pipes instead of metal ones it is preferred to connect the earthing system with the water installation in the construction



Earthing system utilization factor of the electrodes Usually the earthing system is consist of more than one vertical electrode connected with each otherEach electrode will affect the other if the distance between the electrodes is not enoughThis will affect the total resistance of the electrodes if the distance between the electrode is big enougth then the total resistance of electrodes will be:-Re= Ro/n Re :- earth resitanceRo:- the resitance of the electrodeIn parctical we cannot make the desitance between the electrode very big as this requires a big areasBecaue of that we should consider the utilization factor when calculating the earth resitance.The utilization factor is a no to consider the effect of electrodes on each others. ſ= Ro/Re and it value is less than 1.

Re= Rh/ ſh for horizantal electrodesRe =Rv/ ſv.n for vertical electrodes



Earthing system utilization factor of the electrodes Usually the earthing system is consist of more than one vertical electrode connected with each otherEach electrode will affect the other if the distance between the electrodes is not enoughThis will affect the total resistance of the electrodes if the distance between the electrode is big enougth then the total resistance of electrodes will be:-Re= Ro/n Re :- earth resitanceRo:- the resitance of the electrodeIn parctical we cannot make the desitance between the electrode very big as this requires a big areasBecaue of that we should consider the utilization factor when calculating the earth resitance.The utilization factor is a no to consider the effect of electrodes on each others. ſ= Ro/Re and it value is less than 1.

Re= Rh/ ſh for horizantal electrodesRe =Rv/ ſv.n for vertical electrodes



Earthing system reducing the earth of the soil there is two ways to reduce the resistivity of the soil 1- by using electrodes buried deeply in the ground 2- processing the soil

Processing the soil is done through adding one of the following salts :- 1- magnesium sulphate2- copper sulphate3- carbon or coal4- Iron filling

Eg adding 10 kg from salt to the soil will reduce its resistivity is 80%

Usually it needs about 30-40 kg from salt for processing the soil around the vertical electrode and this will reduce the resistivity of the soil by 2 to 6 times.



Earthing system Earthing conductors it is the conductors connected between the electrodes and the earth terminal (busbar)Usually the copper or coated steel conductors are used as the earthing conductors copper 16 mm2 (protected against rust)Steel 16 mm2 (protected against rust)

copper 25 mm2 (not protected against rust)Steel 50 mm2 (not protected against rust)



Earthing system

protective earth

Lowest area of the protective earth in mm2

Area of the phase conductor(s)

s S<= 16

16 16<S<=35

s/2 S>35



Earthing system earthing calculation the meaning of the earth calculation is to know the no of electrodes and its type and the way its distributed in order to get the desired earth resistance .The procedures are :-1- calcuated the single phase short circuit and then calculate the earth resistance.2- measure the earth resisitnce in the place 3- provisionaly specify the earth system . Like to be consist of no of vertical eelctrodes distributed in one line and connected through a horizantal earth busbar . Or in aclosed shape .4- calculate the resitance of the vertical electrode using r=resitivty /l

No of electrodes = Rv/n5- calculate the length of the horizantal plate using r= resistivity*sqrt(pi/A)/4



Earthing system

the value of the earthing resistance:-The fig shows the occuring of short circuit of one phase with ground in TT earthing system

The short circuit current depends on the voltage and the resistances ro and re with addition to the resistance of the cablesAssume ro=re=4 ohmIe=220/8=27.5 AThe voltage on the motor = 110 VIf re=4 ohm and ro=1.5 ohm then the voltage will be (220x4)/5.5= 150 V which can kill the person



Earthing system the value of the earth resitance:-

The regulations states that the maximum voltage that should appears on the metal parts that are not carrying any currents in normal condition should not more than 42 V in fault condition

V phase x re)/(rs+ro) <= 42 V

re<= (42*ro)/ (V phase-42) ohm

If V phase =220 then re <=0.236 ro.

The earth resistance of the consumer should be 4 times less than the earth resistance of the supplier.

The regulation states that the earth resistance of the consumer should be less than 4 ohms So re<= 0.236*4 = 0.944 ohm

Regulations states that the earth resistance of the consumer shouldn’t exceed 2 ohms



Earthing system

In general it is possible to obtain enough protection in the case of fault in one of the following ways:-

1- if the voltage appears on the metal part that are not carrying any current in normal condition is very small (less than 42 V) and this is done if the earth resistance is very small.

2- if the faukt current is very small and enough to make the protection device to work ( either the fuse or the CB)



Earthing system The earthing system in electrical companies:-

1- earthing the distribution networks

Every meduim voltage tower or pole (33 kv) is earthed through copper wire connecting with an electrode buried under the pole. The earth resistance should be less than 10 ohm.

2- the low voltage distribution newtworks (0.4 kv) is earthed through the poles in away that every pole from five or six poles is earthed thorugh electrode.

The earthing in low voltage networks (0.4 V) follow the following earthing systems:-1- earthing system (TN-S) :- the low voltage distribution netwrok consist of ( 3 phase , 1 N , 1 E)

2- earthing system (TN-C-S) :- 3 phase + 1N + 1 E



Earthing system The earthing system in electrical companies:-

1- earthing the distribution networks

Every meduim voltage tower or pole (33 kv) is earthed through copper wire connecting with an electrode buried under the pole. The earth resistance should be less than 10 ohm.

2- the low voltage distribution newtworks (0.4 kv) is earthed through the poles in away that every pole from five or six poles is earthed thorugh electrode.

The earthing in low voltage networks (0.4 V) follow the following earthing systems:-1- earthing system (TN-S) :- the low voltage distribution netwrok consist of ( 3 phase , 1 N , 1 E)

2- earthing system (TN-C-S) :- 3 phase + 1N + 1 E



Earthing system

The earthing system in electrical companies:-

1- earthing the distribution transformers:-

the distribution transformer 33/0.4 kv is connected in delta from the high side and in star in the low voltage sideThe neutral is connected directly with the earth

the distribution transformer 132/33 kv is connected in star in high voltage and the neutral is connected directly with the earthThe low voltage is connected in delta



Earthing system

The earthing system in electrical companies:-

4- earthing the main cables:-

all the ends of the cables 11kv ,33 kv with three cores . The ends of the cable is connected with the earthing network to prevent voltage from appearing between the two ends of the cable in the faulty condition



Lightning protection

about 8 million light flashes occurs every day or 100 light flashes ever second about 18.2 % of the fires in buildings are due to the lightning phenomena




the water vapour is condensed in the air and then water drops are formed

These water drops are frozen due to the low temperature

The temperature inside the water drop is higher than that at the surface

the positive Ions is directed to the surface under the effect of the difference in the temperature and charged it with positive charge The inner side of the water drop is become charged with negative charge




the electrical charges after the broken of the cover is trasfered to the top and so the the tope side of the cloude will be ositively charged and the lower side will be negatively charged.

As the concentrate of the electrical charges is increased in the cloud , the voltage of electrical field of the cloud will increase

When it reaches 20-25 kv .cm which depend on the hieght of the cloud , the air will strat to be ionized and then the air become conducting .




the electrical discharge is started when the cloude is started to send negative charges to the ground .These neagtive charges moving to the earth with avergae speed of 200000 m/s .When these charges moves 45m it stops. And start a new route.The time needed for this step is 30-50 msAs much these charges become closer to the ground, the time will be reduced , and the current will be 50-100 A




the electrical discharge is started when the cloude is started to send negative charges to the ground .These neagtive charges moving to the earth with avergae speed of 200000 m/s .When these charges moves 45m it stops. And start a new route.The time needed for this step is 30-50 msAs much these charges become closer to the ground, the time will be reduced , and the current will be 50-100 A



Fire alarm system

it is important to save life and buildings from firesIt is used to detect the fire and determine its source in the early stagesThen it sends alarming signal in order to take safety procedures

Alarming system goes in stages:-

Detecting the fire (time is important)Response stage (determine the source of the fire and its place)

Extinguishing stage



Fire alarm system

fire alarm consist of

Control unitFire detectorManual stationsPower supplySupplementary equipment



Fire alarm system

detectors

Smoking detectorHeat detectorRadiation detector



Fire alarm system

Manual stations : it is a unites through which the the alarming circuits is closed to give a manual alarm for a fireBells and horns : it is units gives a specific sounds for alarming Control unit:- 1- it is consist of a digital and electronic circuits 2- it is receive the signal comes from the detector3- send a signal to the bells and horns 4- it consists of lamps to show the zone in which the fire exist5- there is switch in it that returns the alarm to the normal condition6- it is supplied with electrical power from main power source and a standbb source7- the main source come from the D.B with sperate circuit breaker and feeder8- the standby power suppll consist of rechargable batteries or UPS



Fire alarm system

spacing of detectorsIt is important to determine the distance covered by the detector such as all the needed ares are covered bb the detectorsThe spacing of detectors rules:- 1- each detector covers area of 83 m2 2- the spacing between two detectors vertically and horizantally not more than 9m3- the distance between any centre of fire and any detector not exceed 6.4 m



Fire alarm system

electrical installation of the fire alarm systems

There are two types of alarm system installation 1- open circuit 2- closed circuit



Fire alarm system

open circuit

In this type the alarm points ( detectors , manual station) is connected in parallel and its switch are normally opened . When a fire is occured, this switches is closed and start passing currents which gives a signal to the control unit and as a result the horns and bells start working.



Fire alarm system

closed circuit

In this type the alarm points ( detectors , manual station) is connected in series with the coil of the relay which is closed when no current is passing in it and open when is passing. The alarming points are closed in normal condition and there is a current passing in the relay coil so the breaker near the bells are openWhen there is a fire one of the alarming point become open circuit and so no current will bas in the oil of the relays and so the circuit breaker near the bells will be closed.

It is more trustable than the previous method . Any fualt in the relay or any cutting in the wire will make the system alaring.



Type of switches

push button switch :

ON:- it is used to pass the current to the coil when it is pressed

OF :- it is used to open the circuit and so no current will pass in the coil



Type of switches

Limit switch

these switches have many auxiliary normally closed and normally open switches and the situation of these switches will change when any thing collapse with it

It is used to conduct or cut the current from the coil when the load reaches a specific distance or height

It is found in elevators

There are special types of them called the photo electronic detector , and the auxiliary switches will be changed in it whenever something passes in front of it at a specific distance without need of mechanical touch.



Type of switches Pressure switch

A pressure switch is a form of switch that makes electrical contact when a certain set pressure has been reached on its input. This is used to provide on/off switching from a pneumatic or hydraulic source. The switch may be designed to make contact either on pressure rise or on pressure fall.

PneumaticUses of pneumatic pressure switches include:switching off an electrically driven gas compressor when a set pressure is achievedSwitching off a gas compressor, whenever there is no feed.

HydraulicHydraulic pressure switches have various uses in automobiles, for example:to switch on a warning light if engine oil pressure falls below a safe levelto switch on brake lights automatically by detecting a rise in pressure in hydraulic brake pipesIn dust control systems (bag filter), a pressure switch is mounted on the header which will raise an alarm when air pressure in the header is less than necessary to gain or decline energy beyond the set value

http://en.wikipedia.org/wiki/Switch

http://en.wikipedia.org/wiki/Pressure

http://en.wikipedia.org/wiki/Pneumatic

http://en.wikipedia.org/wiki/Hydraulic

http://en.wikipedia.org/wiki/Gas_compressor

http://en.wikipedia.org/wiki/Automobile

http://en.wikipedia.org/wiki/Oil_pressure_(internal_combustion_engine)

http://en.wikipedia.org/wiki/Brake_light

http://en.wikipedia.org/wiki/Hydraulic_brake



Type of switches

Float switch:- it is used for controlling the rising or falling of the level of liquids The auxiliary switches state are changed when the level of a liquid is rising above

or falling below a specific level.



Type of switches

Timer :- it is used to conduct the current to the coil or cdisconnect it automatically after a specific time at which the time is adjusted.



Type of switches

Power and control circuits any control unite in any machine has a motor or more is divided into 2 separate circuits, power circuit and contrpl circuit1- power circuit:- it is the circuit used to conduct the current from source to the motor and consist of

A- fuses or CBB– contactors(3)C-heat coils for overload (3)D-motor terminalE-wires

All the components in this circuit should be able to carry the rated current of the motor used



Type of switches

2- control circuit:- it used to conduct the current to the contactor coil and consist of:-

A- Fuse or CBB- on and off switches Contactor auxiliary switches Normally closed auxiliary switches for the heat breaker

All these components should carry current of the coil which is very small current



Type of switches






Type of switches






Type of switches

Power electric circuit consist of:-

1- current sources L1,L2,L32- 3 fuses3- 3 contacts points 4- 3 over load contacts 5- 3 motor terminals



Type of switches

Control circuit

1- fuse2- contacts point for overload coilOn switchOff switchCoil

The voltage across the control circuit equals the voltage that the coil can carry



Type of switches

Control circuitConnecting the lamp units in the control circuit

Rm over load lamp point LM light when the motor rotatesLF over load disconnection



How to choose the terminals of control circuit

the terminals of the control should have a voltage difference equals the rating voltage of the coilIn some circuits , the rating voltage of the coil is the same of the voltages between the three phasesAnd in this case , any two phases are used in control circuitIn some circuits , the rating of the coil equals the volage between any phase an the neutral, and in this case any phase from the three phases is used in control circuitIn many circuits , the rating voltage of the coil will be small , 24,48,110 V, and in this case , a transformer is used to obtain the desired voltage.



How to choose the terminals of control circuit

example : control circuit to control a motor start and stop from two different places note:-

in the case of switching on the same coil from different places , the start switching should be connected in parallel with an auxiliary open switch

In the case of switching off the coil from different places, all the stop switches should be connected in series In the dangerous machines, the control circuit is designed in away that the machine will not work till a two switches are pressed simulatneously. And the two switches are spaced in away that the worker should use both of his hands to press the two switches



Power and control circuit for single phase motor

the contactor and overload that are used in controlling a 3 phase motors can be used in controlling a single phase motor



Controlling the motor in opposite direction

The change of direction of rotation of a 3 phase motor by changing the connection of the two phases from three .

Note:- if the three phase are changed, then the direction of rotation will be in the same direction



Controlling the motor in opposite direction

Power circuit :- when closing the contactor C1 , the electrical current will reach the motor in the sequence R S T……….. W U VWhen the closing the contactor C2 , the electrical current will reach the motor in the sequence R S T…………….W V UThe electrical current in both cases will be the same



Star delta motor control

The external connection of the motor can be either star or delta and in some cases the motor start with star connection and then change to be connected in delta connectionThere are two methods to connect the motor, and each method determine the voltage difference at which the motor will work



Star delta motor controlThe voltage difference in star connection is bigger than that in delta connection

Vstar = sqrt(3) * V delta

The current in star is less than that in delta I star= I delta/ sqrt(3)

In the high power motor , it works at delta,. At the beginning of rotation it will be connected in star, then when it reaches the final speed it will change to delta The six terminals of the motor should be freeThe voltage difference of the source should be same with the voltage of the motor when it works in delta connection




The voltage difference in star connection is bigger than that in delta connection

Vstar = sqrt(3) * V delta

The current in star is less than that in delta I star= I delta/ sqrt(3)

In the high power motor , it works at delta,. At the beginning of rotation it will be connected in star, then when it reaches the final speed it will change to delta




Power circuit:- C3 is used to connect the current to the terminal of the motor and it has a timerWhen the current passes coil 3 through the start button , the auxiliary switch C3 will be closed and the current will pass in coil 1. and the motor will be connected in starAfter some time, the timer will disconnect the current from C1 and connect it to coil 2 and the motor will work in delta



Timer

It used in the cases that needed a start or stop of the motor after a specific timeThere are many types of them1- the timer has a samll motor, when the current passes through it, it will start rotating gears and after a specific time, the gear will close a switch and open another switch

When closing C , B lamp will light and after a specific time, B light will switch off and A lamp will light



Control circuit of 3 motors with timer

The circuit is used to start 3 motors , the first motor will start working using a start switch , after a specific time, the second swithc will start will start working and the first motor will stop, and after a specific time, the third motor will start working and the second motor will stop.



Series resistors

In the motors with high power, another method used to protect the coils of the motors from high current at the start of start of rotation of the motor through connecting it with resistors in series



Series resistors

At the beginning of rotation, a series of motors will be connected in series with the motor , then after a specific time the resistor will be switched off



Control circuit for main supply and auxiliary supply

In some constructions , it is not allowed for the power supply to be switched off even for short timeIn such cases , the construction should have a mian power supply and a standby sourceIn the case of switching of the power , the standby system will supply the constrcution with powerTill the power returns back




A: contactor for the main sourceR: auxiliary contactor with the main sourceT1: timer connected with the contactor A. this timer changes its contacts if a current passing hrough coil A and stays like that. And if the current stop passing in the coil A for a specific time its contacts return back to the normal situation B : contactor for the standby source




If there is current in the main source , the current will pass through coil A and coil R and in this case the timer will colse its contacts but the current will not passs through coil B because the normally closed auxiliary switches A R will be opened. in the case of cutting of the source from the main , the normally closed auxiliary switches A and R will be changed to be closed and so the current will bas through coil b and the stand by source will be the source.When the current returns back in the main source, the current will pass through coil R and the switch will open . And so no current will pass through coil B, then the current will pass through coil A



Control circuit for motor with 2 speeds



UPS Un interruptable power supplyUPS types

Standby Line Interactive Standby-Ferro Double Conversion On-LineDelta Conversion On-Line



UPS Un interruptable power supplyStandby UPS

The Standby UPS is the most common type used for Personal Computers



UPS Un interruptable power supplyStandby UPS

the transfer switch is set to choose the filtered AC input as the primary power source (solid line path), and switches to the battery / inverter as the backup source should the primary source fail. When that happens, the transfer switch must operate to switch the load over to the battery / inverter backup power source (dashed path). The inverter only starts when the power fails, hence the name "StandbyHigh efficiency, small size, and low cost are the main benefits of this design



UPS Un interruptable power supplyThe Line Interactive UPS

is the most common design used for small business, Web, and departmental servers




In this design, the battery-to-AC power converter (inverter) is always connectedto the output of the UPS. Operating the inverter in reverse during times when the input AC power is normal provides battery charging.

When the input power fails, the transfer switch opens and the power flows from the battery to the UPS output. With the inverter always on and connected to the output, this design provides additional filtering and yields reduced switching transients when compared with the Standby UPS topology.In addition, the Line Interactive design usually incorporates a tap-changing transformer. This adds voltage regulation by adjusting transformer taps as the input voltage varies.High efficiency, small size, low cost and high reliability coupled with the ability to correct low or high line voltage conditions make this the dominant type of UPS in the 0.5-5kVA power range.



UPS Un interruptable power supplyThe Standby-Ferro UPS

The Standby-Ferro UPS was once the dominant form of UPS in the 3-15kVA range.




This design depends on a special saturating transformer that has three windings (power connections). The primary power path is from AC input, through a transfer switch, through the transformer, and to the output. In the case of a power failure, the transfer switch is opened, and the inverter picks up the output load. In the Standby-Ferro design, the inverter is in the standby mode, and is energized when the input power fails and the transfer switch is opened. The transformer has a special "Ferro-resonant" capability, which provides limited voltage regulation and output waveform "shaping". The isolation from AC power transients provided by the Ferro transformer is as good as or better than any filter available. But the Ferro transformer itself creates severe output voltage distortion and transients, which can be worse than a poor AC connection.Even though it is a standby U PS by design, the Standby-Ferro generates a great deal of heat because theFerro-resonant transformer is inherently inefficient. These transformers are also large relative to regular isolation transformers; so standby-Ferro UPS are generally quite large and heavy.In addition, the Line Interactive design usually incorporates a tap-changing transformer. This adds voltage regulation by adjusting transformer taps as the input voltage varies.



UPS Un interruptable power supplyThe Double Conversion On-Line UPS

This is the most common type of UPS above 10kVA




is the same as the Standby, except that the primary power path is the inverterinstead of the AC main.In the Double Conversion On-Line design, failure of the input AC does not cause activation of the transfer switch, because the input AC is charging the backup battery source which provides power to the output inverter. Therefore, during an input AC power failure, on-line operation results in no transfer time.



UPS Un interruptable power supplyThe Delta Conversion On-Line UPS

10 year old technology introduced to eliminate the drawbacks of the Double Conversion On-Line design and is available in sizes ranging from 5kVA to 1.6MW.




Similar to the Double Conversion On-Line design, the Delta Conversion On-Line UPS always has the inverter supplying the load voltagethe additional Delta Converter also contributes power to the inverter output. Under conditions of AC failure or disturbances, this design exhibits behavior identical to the Double Conversion On-Line.



UPS Un interruptable power supply

Summary of UPS types






UPS Un interruptible power supplyUse of UPS types in the industry



UPS Un interruptable power supplyUse of UPS types in the industry

Ultracapacitors have distinct advantages including:> storing high quantities of energy> accepting many charge/discharge cycles without suffering any wear> acceptable service life> satisfactory operation over wide temperatures range> For high power ratings and backup times of≥several seconds required by UPSs, cost is high– ≈€10,000 for 50 kW Useful primarily for peak load shaving due to veryfast charge/discharge cycle Relatively small in size, and are a promisingtechnology for UPS applications Used to supply extra power (e.g. improveperformance of electrical vehicles) or supply lowc urrent,long-duration loads (e.g. computermemory backup applications).).



UPS Un interruptable power supplyUse of UPS types in the industry



UPS Un interruptable power supplyFlywheels – Low speed Operation:> "no-break" system, metal flywheel connected to genset motor & generator> AC-input is available : machine operates in motor mode as a synchronous conditioner> AC-input unavailable : synchronous machine operates as AC generator using flywheels kinetic energy to supply output voltage.> Regulation system maintains output voltage in spite of drop in flywheel speed. Advantages include:> Long backup times when used with engine generator set> Rotating at a relatively low speed (≈1,500 rpm); makes it possible to usestandard bearings Challenges:> Requires regular and costly maintenance with system shutdown> “Environmental costs ?”> Low efficiency, uses <5% of the stored kinetic energy> Systems as a whole are large & noisy



UPS Un interruptable power supplyFlywheels – Low speed Operation:> Compact systems with high energy density thanks to new lightweight composite materials & electronics developments> Currently being paired with UPSs for special applications.> High speeds : 30,000 to 100,000 rpm– kinetic energy proportional to the square of rotating speed> Motor-driven rotor maintained by magnetic bearings in a levitating position in a vacuum Advantages:> compact size> excellent efficiency> low noise levels Disadvantages:> risks due to structural faults in the rotor> Very high cost of the materials> “Environmental costs ?”



UPS Un interruptable power supplyFlywheels – Low speed Operation:> Compact systems with high energy density thanks to new lightweight composite materials & electronics developments> Currently being paired with UPSs for special applications.> High speeds : 30,000 to 100,000 rpm– kinetic energy proportional to the square of rotating speed> Motor-driven rotor maintained by magnetic bearings in a levitating position in a vacuum Advantages:> compact size> excellent efficiency> low noise levels Disadvantages:> risks due to structural faults in the rotor> Very high cost of the materials> “Environmental costs ?”



UPS Un interruptable power supplyHow to determine the correct UPS size

1- Make a list of the equipment that needs protection. Include all individual items thathave their own power cords, such as CPU, monitor, modem, etc

2. Obtain the wattage rating of each device

4-Determine the maximum back-up time needed to shut down critical equipment only, should an outage occur

3 Add the total wattage ratings for all devices to be protected

























Under normal operating conditions (i.e. when the mains voltage is within the accepted range), the synchronous motor draws a compensating current so as to maintain a constant load voltage level. Also during normal operation, kinetic energy is stored in the flywheel (or induction coupling)

When the mains fail, the stored flywheel energy is extracted to support the load.Flywheel inertia (or induction coupling speed range) is sized to support the full load operation for a specific period (typically a few seconds), during which time the backup diesel generators are started.



Passive standby



UPS Un interruptable power supplyPassive Stand-ByIn this topology, the system monitoring and control unit continuously monitors the utility voltage and frequency to determine if they are within the specified limits. If the utility source parameters are within +/-10%, the load is directly fed from the utility mains through the fast acting static switch. During this condition (standby mode) the power inverter remains energized but does not supply any power to the load. The battery charger maintains the voltage level of the batteries during this time. In the event of any power failure at the mains or if the mains power parameters go out of the specifiedlimit, the system control turns off the fast acting static switch and turns on the inverter. Under this condition (Stored energy mode) the inverter supplies the load power by converting energy from the battery.

Electrical Loads the first step in the electrical designing for any construction is to estimate the electrical loads This enables the electrical engineering.

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