electrical Loads - An-Najah National University

Report
An-Najah National University
Faculty of Engineering
Electrical Engineering Department

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
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
 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
 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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
 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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
 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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
 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
 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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
 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):m2
Power Unit density(UPD): (F):watt/ m2
Room factor : (a1)
Space utilization factor :(a2)
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
 Unit Power density procedure
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
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
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 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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
 Unit Power density procedure
Unlisted areas
• 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.
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
 Unit Power density procedure
The maximum electrical power for the whole building
• 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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
 Unit Power density procedure
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
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.
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.
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
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.
The luminous efficacy of a lamp
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
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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
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.
The luminous efficacy of a lamp
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
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 .
The luminous efficacy of a lamp
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
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).
The luminous efficacy of a lamp
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
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
The luminous efficacy of a lamp
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
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 )
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.
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
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 (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)
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical lamps
Mercury Lamps
1- general (80-1000) watt
Soduim Lamps
1- low pressure (35-180) watt
2- high pressure (70-1000) watt
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical lamps
Lifetime
Incandescent Lamps :1- Genral (1000 ) hour
2- projector (750-1500) hour
3- Halogen (2000) hour
Florescent :1- (15-80 ) watt :- 1000 hour
2- (125-200) watt:- 3000 hour
3- (4-13) watt:- 2000 hour
Mercurry :-7500 hour
Soduim :- 2000 hour
CFL :- (5000 -15000)
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
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 75-95
lamps) (energy saving lamps)
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.
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical lamps
Comparison between incandescent lamps and CFL lamps
1- Life time
* incandescent 1000 hour
* CFL 5000 hour
2- power
* incandescent :75 watt
* CFL : 18 watt
3- luminous flux :* incandescent : 900 lumen
* CFL : 90 lumen
4- luminous efficiency
* incandescent : 12 lumen/watt
* CFL 50 lumen/watt
5- Size
* bigger
* smaller
6- weight
* bigger
* smaller
7- cost
* less
* more
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 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 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 = 3.6+4.5 = 8.1 JD
Saving = 16- 8.1 = 7.9 JD
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
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
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 situation
Maintenance factor
Normal
0.8
Dirty
0.7
Very dirty
0.6
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 .
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
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
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.
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Lighting calculations
eg
Lumen Method
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=3-0-0.75=2.25 m
Kr=8*5/(8+5)*2.25=1.37
We choose kr=0.58+0.64/2=0.61
N=500*8*5/2*3450*0.61*0.8=5.59 = 6 units
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Lighting calculations
eg
Lumen Method
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=3-0-0.75=2.25 m
Kr=8*5/(8+5)*2.25=1.37
We choose kr=0.58+0.64/2=0.61
N=500*8*5/2*3450*0.61*0.8=5.59 = 6 units
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.
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 .
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:ρt=ρ0(1+ℓ (T-T0)
Where
ρt is the resistivity at temperature T
ρ0 is the resistivity at 250
ℓ is the thermal expansion coefficient for the conductors at 250
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical Conductors
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
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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical Conductors
Ampacity of cables
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.
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
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)
Brown
Three phase: L1
Three phase: L2
Black
Three phase: L3
Grey
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical Conductors
Area of
Conductor
0.05
0.75
1.0
1.5
Resistance at 20°C Max. ohm/km
Current Rating (Amp.)
39.00
26.00
18.10
12.10
4
7
11
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
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
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∑iαR α
As R=ρL/A
Voltage drop = 2 ρ/A ∑IL
A= (2 ρ/voltage drop) ∑IL
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Sockets connections
220 V -50 Hz
N
P
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
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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Switch connections (2 lamps with double switch
220 V -50 Hz
N
P
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Switch connections (2 lamps with double switch
220 V -50 Hz
N
P
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Switch connections (2 lamps with double switch
220 V -50 Hz
chandelier
NEP
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Switch connections (2 lamps with double switch
220 V -50 Hz
L
E
N
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Sockets connections
220 V -50 Hz
N
E
P
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Sockets connections
N
E
P
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
 lighting 2 lamps from 3 different places
220 V -50 Hz
L
E
N
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Bell with push button switch
N
E
P
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Meausring devices
A
V
Kwh
N
V
A
E
P
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 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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Types of circuit breaker
Circuit breakers
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.))
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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
RCD single phase (residual-current device)
2* 40 A, 0.03 A
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
RCD 3 phases (residual-current device)
4* 40 A, 0.03 A
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Many flats in building (main distribution boar
KwH
3 phase supply
KwH
KwH
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Many flats in building (main distribution boar
5*16mm2 xlpe
fourth floor
5*16mm2 xlpe
Third floor
5*16mm2 xlpe
second floor
first floor
5*16mm2 xlpe
Ground floor
5*16mm2 xlpe
5*16mm2 xlpe
Basement 1
Basement 2
KwH
5*16mm2 xlpe
3 phase supply
DB-F4
DB-F3
DB-F2
DB-F1
DB-GF
DB-B1
DB-B2
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Main DSB
From mains
KwH
3 * 160 A
RCD 160 A
30 mA
3 * 63 A
3 * 63 A
3 * 63 A
3 * 63 A
3 * 63 A
3 * 63 A
3 * 63 A
5*16mm2
5*16mm2
5*16mm2
5*16mm2
5*16mm2
5*16mm2
5*16mm2
DB-B2
DB-B1
DB-GF
DB-F1
DB-F2
DB-F3
DB-F4
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Basement 2
from DB-B2
3 * 63 A
RCD 63 A
30 mA
10 A
10 A
10 A
10 A
16 A
16 A
16 A
16 A
3*1.5mm2
3*1.5mm2
3*1.5mm2
3*1.5mm2
3*2.5mm2
3*2.5mm2
3*2.5mm2
3*2.5mm2
light
light
light
light
socket
socket
socket
spare
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Many flats in building (main distribution boar
3*16mm2 xlpe
fourth floor
KwH
3*16mm2 xlpe
Third floor
KwH
3*16mm2 xlpe
second floor
KwH
first floor
3*16mm2 xlpe
Ground floor
3*16mm2 xlpe
KwH
3*16mm2 xlpe
Basement 1
DB-F4
DB-F3
DB-F2
DB-F1
DB-GF
KwH
KwH
DB-B1
KwH
Basement 2
3 phase supply
3*16mm2 xlpe
DB-B2
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Many flats in building (main distribution boar
3*16mm2 xlpe
fourth floor
DB-F4
3*16mm2 xlpe
Third floor
DB-F3
3*16mm2 xlpe
second floor
first floor
3*16mm2 xlpe
Ground floor
3*16mm2 xlpe
DB-F2
DB-F1
DB-GF
3*16mm2 xlpe
Basement 1
DB-B1
Basement 2
KwH
3 phase supply
KwH
KwH
KwH
KwH
KwH
KwH
DB-B2
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Main DSB
From mains
KwH
3 * 160 A
RCD 160 A
30 mA
1 * 63 A
1 * 63 A
1 * 63 A
1 * 63 A
1 * 63 A
1 * 63 A
1 * 63 A
3*16mm2
3*16mm2
3*16mm2
3*16mm2
3*16mm2
3*16mm2
3*16mm2
DB-B2
DB-B1
DB-GF
DB-F1
DB-F2
DB-F3
DB-F4
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Basement 2
from DB-B2
1 * 63 A
RCD 63 A
30 mA
10 A
10 A
10 A
10 A
16 A
16 A
16 A
16 A
3*1.5mm2
3*1.5mm2
3*1.5mm2
3*1.5mm2
3*2.5mm2
3*2.5mm2
3*2.5mm2
3*2.5mm2
light
light
light
light
socket
socket
socket
spare
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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical installation In factories
Main distribution boards
 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)
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical installation In factories
Auxiliary distribution boards
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical installation In factories
General Notes about 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
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
An-Najah National University
Faculty of Engineering
Electrical Engineering Department
Electrical distribution in factories
 substations 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.
y(m)
load 2
load 1
load 3
x(m)
Xo= sum(pi *xi)/sum(pi)
yo= sum(pi *yi)/sum(pi)

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