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 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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Presentation on theme: " electrical Loads the first step in the electrical designing for any construction is to estimate the electrical loads This enables the electrical engineering."— Presentation transcript:

1  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 University Faculty of Engineering Electrical Engineering Department

2  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) An-Najah National University Faculty of Engineering Electrical Engineering Department

3  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 An-Najah National University Faculty of Engineering Electrical Engineering Department

4  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/m 2 ) = illuminances (Lux)/2.78 in the case of using incandescent lamps :- Power(watt/m 2 ) = illuminances (Lux)/ 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 An-Najah National University Faculty of Engineering Electrical Engineering Department

5  example An office with an area of 60 m 2, 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 / = watt in the case of using florescent lamps : P= 60* 500 / = 2691 watt power saving using the florescent lamps= = 8072 watt An-Najah National University Faculty of Engineering Electrical Engineering Department

6  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. An-Najah National University Faculty of Engineering Electrical Engineering Department Room TypeUnit or specific load(watt/m 2 ) 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 room20-30 kitchen50-60 Working areas40-50 Stairs, corridors,20-30

7 An-Najah National University Faculty of Engineering Electrical Engineering Department Room TypeUnit or specific load(watt/m 2 ) theatres10 banks21 Hair dressing shops32 Worship areas10 clubs21 hospitals21 hotels21 offices53 schools32 Restaurants21

8  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 An-Najah National University Faculty of Engineering Electrical Engineering Department

9  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/m 2 An-Najah National University Faculty of Engineering Electrical Engineering Department

10  Unit Power density procedure 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: watt Area (A):m 2 Power Unit density(UPD): (F):watt/ m 2 Room factor : (a1) Space utilization factor :(a2) An-Najah National University Faculty of Engineering Electrical Engineering Department

11  Unit Power density procedure Power Unit density(UPD): obtained from tables(1.2) An-Najah National University Faculty of Engineering Electrical Engineering Department Room TypePower unity density Bank50.59 hospitals15.07 offices34.44 buildings23.68 hotels15.07 Libraries9.69 Schools23.68 Shops40.9

12 An-Najah National University Faculty of Engineering Electrical Engineering Department  Unit Power density procedure Room 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

13 An-Najah National University Faculty of Engineering Electrical Engineering Department  Unit Power density procedure Space 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 areaspace utilization factor (S.U.F) Up to

14 An-Najah National University Faculty of Engineering Electrical Engineering Department  Unit Power density procedure 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/m 2 the total electrical power for lighting the inner area will be sum of the powers needed for listed and unlisted area. Unlisted areas

15 An-Najah National University Faculty of Engineering Electrical Engineering Department  Unit Power density procedure 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

16 An-Najah National University Faculty of Engineering Electrical Engineering Department  Unit Power density procedure 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

17 An-Najah National University Faculty of Engineering Electrical Engineering Department  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.

18 An-Najah National University Faculty of Engineering Electrical Engineering Department  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.

19 An-Najah National University Faculty of Engineering Electrical Engineering Department  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

20 An-Najah National University Faculty of Engineering Electrical Engineering Department  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.

21 An-Najah National University Faculty of Engineering Electrical Engineering Department  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

22 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Incandescent lamps 1 – Filament (Tungstun) 2- Bulb 3- Base 4- inert gas or empty volume The luminous efficacy of a lamp

23 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Incandescent lamp The luminous efficacy of a 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.

24 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Incandescent lamp The luminous efficacy of a 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.

25 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Incandescent lamp The luminous efficacy of a 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).

26 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Incandescent lamp The luminous efficacy of a 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

27 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Discharge lamps The luminous efficacy of a lamp 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 )

28 An-Najah National University Faculty of Engineering Electrical Engineering Department  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.

29 An-Najah National University Faculty of Engineering Electrical Engineering Department  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

30 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Power of the lamps(watt) Incandescent lamps 1- general service lamps (15 to 250 watt) 2- projectors ( ) watt 3- halogen Lamps ( ) watt 4- table lamps for specific use (15- 50)watt (positional use) Florescent lamps 1- general (15-80) watt 2- special types ( ) watt 3- table lamps (4-31) watt (positional use)

31 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Mercury Lamps 1- general ( ) watt Soduim Lamps 1- low pressure (35-180) watt 2- high pressure ( ) watt

32 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Lifetime Incandescent Lamps :- 1- Genral (1000 ) hour 2- projector ( ) hour 3- Halogen (2000) hour Florescent :- 1- (15-80 ) watt : hour 2- ( ) watt: hour 3- (4-13) watt: hour Mercurry :-7500 hour Soduim : hour CFL :- ( )

33 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Luminous efficiency ( Lumen/watt) Lamp typeLuminous efficiency Incandescent25 Florescent80 Mercury60 Soduim (Low pressure)185 Soduim (high pressure)140 CFL (sl, pl) (compact florescent lamps) (energy saving lamps) 75-95

34 An-Najah National University Faculty of Engineering Electrical Engineering Department  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 energy 2- long life time (5000 hour) 3- high luminous efficiency (50 lumen/watt) 4- have a small size.

35 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical lamps Comparison between incandescent lamps and CFL lamps 1- Life time 5- Size * incandescent 1000 hour * bigger * CFL 5000 hour * smaller 2- power 6- weight * incandescent :75 watt * bigger * CFL : 18 watt * smaller 3- luminous flux :- 7- cost * incandescent : 900 lumen * less * CFL : 90 lumen * more 4- luminous efficiency * incandescent : 12 lumen/watt * CFL 50 lumen/watt

36 An-Najah National University Faculty of Engineering Electrical Engineering Department  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 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 hour No of incandescent lamps =5000/1000=5 No if CFL lamps =1 The price of the incandescent lamps =5*0.2 = 1 JD The price of the CFL lamps = 4.5 JD The power consumed by incandescent in 5000 hour in Kwh= 5000 *75/1000 = 375 kwh The price of the kwh = 375 * 0.04 = 15 JD Total for incandescent = 15+1= 16 JD Kwh for CFL = 5000*18/1000= 90 Price of kwh = 90 *0.04= 3.6 JD Total price for CFL = = 8.1 JD Saving = = 7.9 JD

37 An-Najah National University Faculty of Engineering Electrical Engineering Department  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

38 An-Najah National University Faculty of Engineering Electrical Engineering Department  Lighting calculations Lumen Method This method depends on using the utilization factor assuming that the illumination is distributed Regularly in all directions in the space. N= Em.A/n*Fl*ku*kn N= no of lamp unites needed to obtain the desired illumination Em= illumination (obtained from tables A = area n= no of lamps in the unite Fl= luminous flux for the lamp (lumen) Ku= utilization factor Kn= maintenance factor

39 An-Najah National University Faculty of Engineering Electrical Engineering Department  Lighting calculations Lumen Method Kn= maintenance factor The maintenance factor depends on the situation of space :- It is value obtained from the table below:- Room situationMaintenance factor Normal0.8 Dirty0.7 Very dirty0.6

40 An-Najah National University Faculty of Engineering Electrical Engineering Department  Lighting calculations Lumen Method Ku= utilization factor The utilization factor depends on the dimensions of the room. KrUtilization factor % >590

41 An-Najah National University Faculty of Engineering Electrical Engineering Department  Lighting calculations 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 index L :- length of the room w:- width of the room Hm:- the high of the lamp unit from the surface of works in m

42 An-Najah National University Faculty of Engineering Electrical Engineering Department  Lighting calculations Lumen Method kr 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.

43 An-Najah National University Faculty of Engineering Electrical Engineering Department  Lighting calculations Lumen Method eg 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= 0.8 Utilization factpr kr= L*w/Hm(L+w). HL=0 Hp=0.75 m Hm= =2.25 m Kr=8*5/(8+5)*2.25=1.37 We choose kr= /2=0.61 N=500*8*5/2*3450*0.61*0.8=5.59 = 6 units

44 An-Najah National University Faculty of Engineering Electrical Engineering Department  Lighting calculations Lumen Method eg 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= 0.8 Utilization factpr kr= L*w/Hm(L+w). HL=0 Hp=0.75 m Hm= =2.25 m Kr=8*5/(8+5)*2.25=1.37 We choose kr= /2=0.61 N=500*8*5/2*3450*0.61*0.8=5.59 = 6 units

45 An-Najah National University Faculty of Engineering Electrical Engineering Department  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.

46 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors Conductor resistances and the factors that affect it Conductors have a high conductivity due to the existence of a high no of free electrons or in other meanings its electrical resistance is very low :- ρ=1/σ Where 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.

47 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors The resistivity of the conductors is affected by the temperature according to the following formulas:- ρ t =ρ 0 (1+ℓ (T-T 0 ) Where ρ t is the resistivity at temperature T ρ 0 is the resistivity at 25 0 ℓ is the thermal expansion coefficient for the conductors at 25 0

48 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors thermal expansion coefficient for the conductors MaterialThermal expansion coefficient silver Copper Aluminium Nickel0.006 Iron0.006 Platinum NI mixture0.0001

49 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors The resistance of any cable is given as:- R= ρL/A Where R is the resistance ρ resistivity L length of the cable A area of the cable From the above equations the resistance depends on:- 1-the conductor type (AL,Cu) 2- length of the cable 3- area of the cable 4- temperature It is meant by temperature :- 1- the temperature of the space around the conductors 2- the temperature of the cable itself which is produced by the current passing in the cable

50 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors 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;insulation 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.temperature Ampacity of cables

51 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors How to calculate the cross section area of the cable 1- calculate the designing current in the circuit ore part of the circuit 2- according to that, the protection for the circuit is selected 3- the choosing of the cross sectional area of the cable will be according to I cable > I protection > I design

52 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors 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: L2Black Three phase: L3Grey

53 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors Area of Conductor Resistance at 20°C Max. ohm/kmCurrent Rating (Amp.)

54 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors Drop voltage calculations Voltage drop = V1 – V2 =2(I1*r1 +I2*r2+I3*r3+I4*r4) I1,2,3,4 are the currents r1,2,3,4 are the branches resistances Let us assume I1=i1+i2+i3+i4 I2=i2+i3+i4 I3=i3+i4 I4= i4

55 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical Conductors 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∑i α R α As R=ρL/A Voltage drop = 2 ρ/A ∑IL A= (2 ρ/voltage drop) ∑IL

56 An-Najah National University Faculty of Engineering Electrical Engineering Department  Sockets connections 220 V -50 Hz N P

57 An-Najah National University Faculty of Engineering Electrical Engineering Department  Switch connections (2 lamps in series with single switch 220 V -50 Hz N P

58 An-Najah National University Faculty of Engineering Electrical Engineering Department  Switch connections (2 lamps in parallel with single switch 220 V -50 Hz N P

59 An-Najah National University Faculty of Engineering Electrical Engineering Department  Switch connections (2 lamps with double switch N P 220 V -50 Hz

60 An-Najah National University Faculty of Engineering Electrical Engineering Department  Switch connections (2 lamps with double switch N P 220 V -50 Hz

61 An-Najah National University Faculty of Engineering Electrical Engineering Department  Switch connections (2 lamps with double switch N E P 220 V -50 Hz chandelier

62 An-Najah National University Faculty of Engineering Electrical Engineering Department  Switch connections (2 lamps with double switch 220 V -50 Hz E N L

63 An-Najah National University Faculty of Engineering Electrical Engineering Department  Sockets connections 220 V -50 Hz N E P

64 An-Najah National University Faculty of Engineering Electrical Engineering Department  Sockets connections N E P

65 An-Najah National University Faculty of Engineering Electrical Engineering Department  lighting 2 lamps from 3 different places 220 V -50 Hz E N L

66 An-Najah National University Faculty of Engineering Electrical Engineering Department  Bell with push button switch N E P

67 An-Najah National University Faculty of Engineering Electrical Engineering Department  Meausring devices N E P V A Kwh V A

68 An-Najah National University Faculty of Engineering Electrical Engineering Department  Circuit breakers Protect from :- 1- short circuit 2- over load 3- short circuit and over load Current 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 A instantaneous 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 I n Type Instantaneous tripping current B above 3 I n up to and including 5 I n C above 5 I n up to and including 10 I n D above 10 I n up to and including 20 I n K above 8 I n up to and including 12 I n

69 An-Najah National University Faculty of Engineering Electrical Engineering Department  Circuit breakers Types 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 current 2- VCB> = V supply 3- Ibreaking capacity > 1.2 Isc MCB :- miniature circuit breaker MCCB :- moulded case circuit breaker (63AMP (min.))

70 An-Najah National University Faculty of Engineering Electrical Engineering Department  Circuit breakers No of units lamps in each feeder should be less than 9 No 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

71 An-Najah National University Faculty of Engineering Electrical Engineering Department  RCD single phase (residual-current device) 2* 40 A, 0.03 A

72 An-Najah National University Faculty of Engineering Electrical Engineering Department  RCD 3 phases (residual-current device) 4* 40 A, 0.03 A

73 An-Najah National University Faculty of Engineering Electrical Engineering Department  Many flats in building (main distribution board KwH 3 phase supply

74 An-Najah National University Faculty of Engineering Electrical Engineering Department  Many flats in building (main distribution board 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 5*16mm 2 xlpe

75 An-Najah National University Faculty of Engineering Electrical Engineering Department  Main DSB 3 * 160 A KwH DB-B2 5*16mm 2 3 * 63 A DB-B1 5*16mm 2 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

76 An-Najah National University Faculty of Engineering Electrical Engineering Department  Basement 2 3 * 63 A from DB-B2 light 3*1.5mm 2 10 A light 3*1.5mm 2 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

77 An-Najah National University Faculty of Engineering Electrical Engineering Department  Many flats in building (main distribution board 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*16mm 2 xlpe KwH

78 An-Najah National University Faculty of Engineering Electrical Engineering Department  Many flats in building (main distribution board 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*16mm 2 xlpe KwH

79 An-Najah National University Faculty of Engineering Electrical Engineering Department  Main DSB 3 * 160 A KwH DB-B2 3*16mm 2 1 * 63 A DB-B1 3*16mm 2 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

80 An-Najah National University Faculty of Engineering Electrical Engineering Department  Basement 2 1 * 63 A from DB-B2 light 3*1.5mm 2 10 A light 3*1.5mm 2 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

81 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical installation In factories  the electrical load in the factories depends mainly on the technological process used in the factory  As 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 machines  The electrical installation in the factories consists of the following elements:- 1- main distribution board 2- feeders 3- auxiliary distribution boards and circuit breakers

82 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical installation In factories  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 siwtch 2- bus-bars :- one bus-bar for each phase,N,E 3- auxiliary switches 4- instrumentation devices (power, A, V, f) Main distribution boards

83 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical installation In factories Auxiliary distribution boards

84 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical installation In factories  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 drops  It better that the electrical installation in the factories to be clear to make the maintenance easy  The 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

85 An-Najah National University Faculty of Engineering Electrical Engineering Department  Electrical distribution in factories  radial distribution  ring system voltage drop is less, more reliable  Both radial and Ring system

86 An-Najah National University Faculty of Engineering Electrical Engineering Department  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)


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