building systems wiring
TRANSCRIPT
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Building Wiring Calculations
4-1. Introduction:
Inthis chapter it is required to plan the distribution system of the residentialarea. The planning of this area starts from inside the individual flats of thebuildings.
The various types of loads in the house like lighting, normal sockets (N.S),power sockets (P.S) and load break switches (L.B.S) are estimated accordingto the standard forms. The feeding circuits are defined and the main wiringc.s.a. are calculated.
The mains wiring is generally built using insulated copper cables. Thechoice of conductor material is a compromise among electrical properties,mechanical properties, and price. From the start, copper has been the
material of choice for household branch circuits.
Aluminum is softer than copper and weaker, and a poorer electricalconductor, so is not widely used in small sizes for home wiring. Aluminumcable material is sometimes used (for economical reasons) for thick mainsfeeder cables coming from electrical utility to the mains distribution panel.
The ratings of the sub-circuits' miniature circuit breakers (M.C.B) and themain circuit breaker of the flat or the villa as well as energy meter areselected.
Any house that has been properly wired will have a circuit breaker panel
used to shut circuits off in case they draw too much current. It is the currentcapacity of circuit breaker (in amperes) that determines how much current acircuit can supply. In case of an overload or a short-circuit on that circuit, thebreaker trips and automatically shuts off power to that circuit. Ground faultcircuit breakers offer protection against more than just overloads.
After the load of the flat is being calculated, the diversified estimation of thetotal load of the building is made. The buildings are fed from distribution boxesvia cables of suitable sizes, forming a part of the low voltage distributionnetwork. The distribution boxes are fed from 11 KV/380 V distributiontransformers, preferably in loops, to secure the continuity of supply to thedistribution boxes and hence to the buildings.
The distribution transformers are located in the appropriate sites andconnected in loops to the 11KV Distribution points and the 66/11KVsubstation.
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Detailed calculations and planning of the 380V low voltage distributionnetwork, the 11KV medium voltage network as well as details of the 66/11KVsubstation feeding the area, are presented in the following chapters. Beforethis, the principles of lighting and wring are summarized in the followingsections.
4-2. Lighting background:
(4-2-1 )Importance of light:
Light is the prime factor in the human life as all activities of human beingsultimately depend upon the light. Where there is no natural light, use ofartificial is made. Lighting increases production and reduce accidents.
(4-2-2) Basic Definitions:
Candela
International unit (SI) of luminous intensity; term evolved from considering astandard candle, similar to a plumber's candle, as the basis of evaluating theintensity of other light describe the relative intensity of a source .
Candlepower Distribution CurveA graphic presentation of the distribution of light intensity of a lamp orluminaire.
Illuminance (E)The quantity of light (measured in foot-candles, Lux, etc) at a point on asurface.
Inverse Square LawFormula stating that illumination at a point on a surface varies directly with theintensity of a point source, and inversely as the square of the distancebetween the source and the point; it illustrates how the same quantity of lightflux is distributed over a greater area as the distance from the source to thesurface is increased.
Light Loss FactorThe product of all considered factors that contribute to a lighting system'sdepreciated light output over a period of time, including dirt and lamp lumendepreciation.
LumenThe international unit of luminous flux or quantity of light.
LuminaireA complete lighting unit consisting of a lamp (or lamps) together with the partsdesigned to distribute the light, position and protect the lamps, and connectthem to the power supply. This is sometimes referred to as a "fixture".
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Lamp efficiencyIt is the amount of output lumen per watt.
Lux (lumen/m2)SI (international system) unit of illumination. One lumen uniformly distributed
over an area of one square meter.
Mounting Height
Distance from the bottom of the fixture to either the floor or workplane, depending on usage.
Spacing to Mounting Height RatioRatio of fixture spacing (distance apart) to mounting height above the workplane. Sometimes it is called spacing criterion. A normal range is 11.5
(4-2-3) Requirements of a good lighting scheme:
A good lighting scheme should fulfill the following:1. Provide adequate illumination.2. Provide uniform illumination allover the working plan.3. Provide light of suitable color.4. Avoid glare and hard shadows.
(4-2-4): Factors affecting the illumination and wattage of a certainlamp:
1-Utilization factor (U.F): (0.2 0.6)It is the ratio of the lumen actually received to the total Lumens emitted by
the source, it depends on:1- Room dimensions.2- Color of the walls.3- Type of lighting scheme.
2- Maintenance factor (M.F):It is the ratio between illumination under normal working conditions to the
illumination when every thing is clean. It depends on the rate of cleaning.
M.F. = 0.8 for houses.= 0.3 for streets.= 0.60.7 for schools and shopping centers.
3- Waste factor:The ratio between the resultant illumination due to more than one luminaireto the summation of their illumination when they work individually. Wastefactor is less than unity due to the loss when a place is illuminated by morethan one source due to overlapping.
4- Reflection factor:Due to the fact that light reflected by an angle of incidence when impinged
on a surface.
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5- Room index (k):
It is a factor that depends on the dimension of the room. It equals the ratiobetween the product of length (L) and breadth (W) of the room to the productof the mounting height (H) and the summation of the length and breadth ofthat room.
K=)(*
*
WLH
WL
+
Generally K varies from 0.6 to 5.0
(4-2-5): Designing the lighting system:
To produce a new lighting system in a construction, it must be designed.The designer must determine the desired light levels for tasks that are to beperformed in a given space, then determine the light output that will berequired to meet those objectives consistently, taking into account all the
factors that degrade both light output and light levels over time. Equipmentmust then be chosen and placed in a layout to produce the desired lightdistribution.
A- Types of lamps:The lighting design process in its most basic form entails identifying a task
and then providing a light source that will provide proper quantity and qualityof light for the task. The fixture protects the light source, connects it to thepower source and distributes its light.
The light source is the actual light-producing component of the lightingsystem. It may operate simply as a lamp (incandescent/halogen) or as a lamp
powered by a ballast (fluorescent and high-intensity discharge [HID]).
I-Incandescent Lamps:Incandescent light sources are the cheapest light sources..
Do not require a ballast
It is based on the fact that current is passed through a filament, whichheats until it glows
Less efficacious light source
Shorter service life than other light sources in most cases
Filament is sensitive to vibrations
Bulb can get very hot during operation Must be properly shielded because incandescent lamps can produce
direct glare as a point source
Require proper line voltage as line voltage variations can severelyaffect light output and service life
An example of incandescent lamps is given in figure 4.1Efficiency of incandescent is 14 lumen/watt.
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Fig. 4.1 Incandescent lamps
ii-Fluorescent Lamps:These lamps rely on the gaseous discharge method.
Require a ballast
Low surface brightness compared to point sources
More efficacious compared to incandescent
Ambient temperatures and convection currents can affect light outputand life
Options for starting methods and lamp current loadings
Requires compatibility with ballast Low temperatures can affect starting unless"cold weather" ballast is
specified.
An example of fluorescent lamps is given in figure 4.2Efficiency of fluorescent lamps is 46 lumen/watt
Fig. 4.2 Different types of Fluorescent lamps
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iii- Compact Fluorescent Lamps (CFL): It is a new and advanced lighting technology More efficient than incandescent lamps CFL use 70 - 75% less energy than their incandescent equivalents. When
replacing a 100 watt incandescent lamp a 28 watt CFL is used. CFL last approximately 10,000 hours, which is 10 to 13 times the life of an
incandescent lamp (expected life approximately 750 hours).
Compact fluorescents are most cost-effective when used at least 2-3hours per day.
Although CFL may appear different than the common incandescent, theyfit most standard fixtures found in homes today. The screw-in base is thesame on both lamps.
The typical incandescent lamp wastes 90% of the energy it uses,producing heat rather than light.
CFL will provide the same amount of light (or lumens) at a fraction of theelectricity used.
An example of CFL lamps is given in figure 4.3
Fig. 4.3Types of CFL lamps
B- Lighting Schemes:
i- Direct LightingIn this type of lighting the light from the source falls directly on the object or
the surface to be illuminated.
ii- Semi-direct Lighting
Only 60-90 % of the light reaches the working plan directly while the rest isreflected to illuminate the ceiling and walls.
iii- Indirect Lighting:In this form of lighting, light doesn't reach the surface directly from the
source but indirectly as the ceiling reflects the light. 90% of the light is directedupwards and only 10% reaches the working plane.
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iv-Semi-Indirect LightingLike the pervious type except that 75% of the light is directed upwards and
25% of it reaches the working plane.
C- Requirements needed to specify right choice ofperfect luminaire:
First, the engineer must fully understand the demands of the application andconditions in the space that will affect the operation of the lighting system:
Tasks to be performed in the space
Desired light levels based on the tasks performed in the space
Room size and dimensions
Layout of furniture and obstructions such as partitions
Special concerns such as safety and security
Room and object surface colors and reflectances
Hours of operation
Cleanliness of the area during operation availability of daylight
(4-2-6) Lighting loads calculations:
Installed wattage =efficiencyFMFU
areaillumenace
*.*.
*
Common figures of M.F and U.F are 0.8 and 0.4 respectivelyApproximate equation to calculate the required wattage for certain area andillumination are given by:
For incandescent lamp: Required installed wattage =0.2 x lux x area. Watt.
Rating: 25 , 40 , 50 , 60 , 80 , 100 ,120,200 Watt
Power factor = 1
For fluorescent lamp:
Required installed wattage = 0.068 x lux x area. Watt.
Rating: 20 , 40 Watt
Power factor = 0.6
N.B The recommended lux in every place is given in table 1 in the appendix ofthis chapter.
The total light load on a line is given by:
0.66(light loads on this line)
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According to the standards, a minimum of two lighting lines are to be foundin the flat. The Cu wire used for wiring of lighting circuits must be of a c.s.a notless than 2 mm2. We will use the 2 mm2for flats and the 3 mm2for villas.
4-3. Socket Loads:
4.3.1 Normal sockets (N.S):
They have different ratings, which can be used such as 3 Amp., 5A,10A theratings of 3A,5A, can be used for bedrooms, entrance, balcony, whichrequires low electrical sets as T.V, radio and small electric fans...etc. Ingeneral we are going to use only the 5A sockets in all the rooms since this ismore practical.
4.3.2 Power sockets (P.S):
Sometimes we need some sockets to be used for special purposes like: full
automatic washing cloth machines, air conditions, water heaters, dishwashers, electric ovens and toasters. Such sockets are called power socketsand they require higher current rating and taking into consideration thestarting period which increases the delivered current to a value higher thannormal operation.
To estimate the socket load for certain domestic units the following are to beconsidered:
1- Generally there are 2-5 sockets in the room.2- Generally there are 5-8 normal sockets on a line.3- Referring to the IEC standard specification, the ratings of sockets are:
*M.C.B. rating for normal socket = 10 A*M.C.B. rating for power sockets = 16 A. or 26 A.
4- Calculate the normal socket loads on a line is according to theformula:
Socket load on a line = 100% of largest normal socket rating on theline + (20% to 40%) of ratings of other normal sockets.
5- Each power socket has its own line.6- Total socket load is given by the formula:
Total socket load = 100% of largest M.C.B. Rating of sockets + (20%to 40%) of Rating of other M.C.B.
7- To make calculations more exact, we should expect the loads to be
used and their power like: Radio cassette : 40w, 0.182 A
T.V set: 65 w, 0.3 A
Video: 30w, 0.137 A
Vacuum cleaner: 800w, a p.f of 0.85 , 4.7 A
fans :200w, a p.f of 0.85, 1.069 A
Shaving Machine: 150w, 0.7 A
Hair dryer: 600w, a p.f of 0.85, 3.2 A
Small fridge to be placed in the bedroom: 80w, a p.f of 0.85, 0.43 A
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Fridge :160w, p.f of 0.85, 0.86 A
Kitchen machine :600w, p.f of 0.85, 3.2 A
Heater:1000w, 4.6 A
Normal washing machines: 400w, p.f of 0.85, 2.14 A.
Iron: 1000w, 4.64 A.
Sound system: 800w, 3.64 A
4-4. Calculations of the riser :
The riser is cable, which passes upward in each building for transmitting theelectric power from the coffree of the building to each unit of this building, inother words, it starts from the fuse at the bottom of the house to the highestflat.
It is a three phase cable made usually of copper and has a number ofoutputs equals to the number of floors, the output of riser is connected to thefuses which feeds this floor.
Riser may be one cable or double cable depending on the height of thehouse, the number of flats and on the load of each flat.
When choosing the riser we follow the next steps:
1- Calculating the KVA of the flat before diversification and use todetermine the suitable diversification curve.
2- We have two methods to get the diversified KVA of the flat:a- Using the total number of flats in the building to get the
diversified KVA of the flat. Multiply this diversified KVA by thenumber of flats in the building to get the total KVA of the
building. Dividing this KVA by 3 380 we get the current that
flows in the riser.b- Using the total number of flats on each phase to get the
diversified KVA of the flat. Multiply this diversified KVA by thenumber of flats on the phase to get the total KVA per phase.Dividing this KVA by 220 we get the current that flows in theriser.
3- Assuming that the riser must never be loaded by more than 70% of itscurrent ampacity, we can get the current ampacity of the riser bydividing the current obtained in the last step by 0.7
4- By knowing the value of the current ampacity and using the tables ofcables attached in the appendix we can get the c.s.a of the riser andalso the rating of the fuse used for protection. In general 3-ph risersthat are used are of the following sizes: 10 mm2, 16mm2, 25mm2,50mm2and 70mm2.
5- Other services loads like water pumps, elevators (for buildings morethan 6 floors) and stairs lighting are to be considered in ourcalculations.
6- A fuse is added for protection.
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4-5. General points to be considered in the design:
1- In distribution of loads among light circuits or socket circuits we shouldconnect the rooms that are next to each other on the same subcircuit toavoid crossing between connections. Also it is recommended that thecircuits of the same type are equally loaded.
2- Diversity factor between the sockets on the same line depends on somefactors like the area of the flat, the larger the area the smaller thediversity factor used.
3- In calculating the required amount of light for the shaving mirror in thebathroom we consider the recommended lux to be half of that requiredfor the bathroom yet the area is the same area of the bathroom. We useincandescent lamps for the shaving mirror.
4- Flats of area less than 90 m2 are considered as youth housing thussingle phase energy meters are used in them.
5- The distribution of flats among riser phases is done in a way to makevoltage drop on each phase exactly equal to other phases.
6- Single phase energy meters are of ratings 20A and 40A. Three phase
energy meters are 320A, 325A, 340A and 380A (a catalog ofElsweedy Electrometeris attached in the appendix).
7- For the c.s.a of the neutral conductor, we follow the Egyptian ElectricCode (EEC) which states " If the c.s.a of the phase conductor is lessthan or equal 16 mm2then the neutral conductor is of the same c.s.a asthe phase conductor. If the phase conductor is of c.s.a less than 35 mm2then the neutral conductor is of c.s.a equal to the one preceding theconcerned phase conductor. If the c.s.a of the phase conductor is morethan or equal 50 mm2 then the c.s.a of the neutral conductor is half ofthe concerned phase conductor.
8- In general all our distribution of loads among the lines or the phases wemust care that the loads are almost balanced as much as we can to
avoid the unbalanced operation.9- Low voltage fuses ratings are as follows:
2,4,6,8,10,16,20,25,32,35,40,50,63,80,100 and 125 according to ABBpocket book(switchgear manual), 8thedition.
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4-6. Types of buildings:
The present calculations are carried out for residential buildings of 4 types:
a- Flat type (F)
b- Flat type (J)
c- Villa type (C)
d- Villa type (E)
It was assumed that we have 4,000 persons to be distributed in this
residential district such that an average of 5 persons live in a flat while an
average of 8 persons live in a villa. The percentages and number of flats
required is given in the following table:
Villa
Type
Required
Percentage
Total no.
of
persons
required
No. of
units
Unit per
floor
No. of
floors
per
building
No. of
units
per
building
Total no.
of
buildings
Flat F 60 % 2400 480 3 5 15 32
Flat J 20 % 800 160 2 5 10 16
Villa C 10 % 400 50 1 1 1 50
Villa E 10 % 400 50 1 1 1 50
Remark: 1 unit =1 villa or 1 flat.
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4.6.1 Calculations for flat type (F):
A- Lighting calculations:
The following rooms are found in flat (F)
Roo
m no. Functionof room
Area(m2)
lux lampused
Req.wattage
No of lamps Power(w)
Actual
current(A)
1 Bedroom1 3.094.01 150 F 126.39 340 W 120 0.91
2 Bedroom2 3.094.01 150 F 126.39 340 W 120 0.91
3Living
room5.083.00 150 I 457.20
660+160+160W
480 2.18
4 Balcony 1.920.96 50 I 18.43 1 25 W 25 0.11
5 Kitchen 2.753.00 300 F 168.30 440 W 160 1.21
6Bedroom3(main
bedroom)
4.263.38 150 I 431.96 4100 W 400 1.82
7 Bathroom 1.923.5 100 F 45.70 140 W 40 0.30
Shaving
mirror50 I 67.20 240 W 80 0.36
8 Hall 4.261.5 150 I 191.70 2100 W 200 0.91
9 Balcony 5.08*1.5 50 I 76.20 1100 W 100 0.45
Remarks: F means fluorescent, while I means incandescent.
Calculations of lighting circuits loads:
Lighting loads will be divided in 2 circuits
Circuit (1): L1This circuit containing loads of rooms 1, 2, 3 and 9
L1 = 0.66 actual currents of the rooms in the circuit)= 0.66 (0.91+0.91+2.182+0.455)= 2.94 Amps.
And for safety and good design; we take a safety factor of 1.6
L1 = 4.7 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 2 mm2
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Circuit (2): L2This circuit containing loads of rooms 4,5,6,7 and 8
L2 = 0.66 actual currents of the rooms in the circuit)= 0.66 (0.114+1.212+1.82+0.303+0.364+0.91)= 3.118 Amps.
And for safety and good design; we take a safety factor of 1.6
L2 = 5.0 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 2 mm2
B- Sockets calculations:
- Since the area of the flat is less than 90 m2 and the flat is of youthhousing we don't expect to have any power sockets in this flat.
- The following loads are expected in the flat:
>> Bedroom1:This is the kids' room, we expect the use of 1 fan of 200 W.
>> Bedroom 2:We expect the use of 1 cassette of 40 W.
>>Living room:We expect 1 TV set of 65 W and 1 fan of 200 W
>>BalconySince this balcony is very small and accompanied to the main
bedroom so no need for sockets in it.
>> KitchenWe expect the use of a fridge of 160 w and a kitchen machine of 600 w.
>>Bedroom 3 (main bedroom):we expect the use of a fan of 200 w and a T.V of 65 w.
>> BathroomWe expect the use of a normal washing machine of 400 w and a shavingmachine of 150 w.
>>HallWe expect the use of any appliance not more than 3 A.
>> BalconyThis is the main balcony and we may use a cassette in it.
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The result of the required sockets is tabulated in the following table:
RoomNo.
1 2 3 4 5 6 7 8 9
RoomFunction
Bedrrom1 Bedroom2Livingroom
balcony kitchen Bedroom3 bathroom Hall balcony
No. ofN.S
1 1 2 --------- 2 2 2 1 1
Calculations of sockets circuits loads:Sockets loads will be divided into to 2 circuits each with 6 normalsockets
Circuit (3): NS1This circuit contains the loads of rooms 1,2,3,8 and 9NS1 = largest rating socket in ampere + (the rest sockets ratings in
ampere of the rooms in the circuit)diversity factor= 5+0.3 5 5 = 12.5 A
The M.C.B. used = 16 AThe copper wire used is of c.s.a 4 mm2
Circuit (4): NS2This circuit contains the loads of rooms 4,5,6,7 and 8NS2 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit) diversity factor
= 5+0.3 5 5 = 12.5 A
The M.C.B. used = 16 AThe copper wire used is of c.s.a 4 mm2
-
C- Calculation of the KVA of the flat:
Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.Bratings in amperes) diversity factor.
= 16 + 0.3 16= 20.8 Amps
Total Lighting Load= L1 + L2 = 5.0 + 4.7 = 9.7 AmpsTotal Load of flat= Total Socket Load + Total Lighting Load
= 30.5 Amps.
The incoming M.C.B. used = 32 AThe copper wire used is of c.s.a 16 mm2Energy meter is 40 A single phase meter.
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KVA for one flat= 220 30.5 10-3= 6.5 KVA.
After using diversity curves (done for 15 units)
Diversified KVA of the flat= 2.75 KVA / flat.
D- Riser Calculations:
Flat KVA = 2.75 KVA after diversity for 15 flats.
Total number of flats in the building = 15 flatsuse one riser
Total riser KVA= 2.75 x 15 = 41.25 KVA
Phase KVA = 41.25/3= 13.75 KVA
Phase current =220
1000x13.75= 62.5 Amps
For safety considerations, the riser is loaded with 70%of its ampacity.
\I phase max= 62.5/0.7 = 89.3 Amps. From the tables attached:
Riser used is 3x50+ 25 mm2Cu & the resistance per phase is0.836x10-3V/m.
Let auxiliary loads be:
a- Water Pump = 1.5 HP, 0.85 p.f. KVA of the pump =85.0
746x5.1= 1.32
KVA.
b- Stairs lighting: if each floor uses one incandescent lamp of 100 W,p.f= 1, KVA of stairs lighting = 5 x 100 x 10-3= 0.5 KVA.
Total KVA of the building= 41.25 + 1.32 + 0.5 = 43.1 KVA
For protection we use a three phase fuse 80A.
Building box is of rating 43.1 KVA.
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4.6.2 Calculations for flat type (J):
A- Lighting calculations:
The following rooms are found in flat (J)
Room
no. Functionof room
Area (m2) lux lampused
Req.wattage
No of lamps Power(w)
Actual
current(A)
1 Kitchen 3.483.63 300 F 257.70 340 +340 W 240 1.82
2 Bathroom 2.731.9 100 F 35.27 140 W 40 0.30
Shaving
mirror50 I 51.87 225 W 50 0.23
3
Bedroom1
(main
bedroom)
3.53.63 150 I 381.15 4100 W 400 1.82
4
Bedroom2
(kids'bedroom)
3.353.62 150 F 123.70 340 W 120 0.91
5 Balcony 3.35*.88 50 I 62.98 160 W 60 0.27
6 Living room 3.35*.87 150 I 388.94 225 +660 W 410 1.86
7 Hall1 1.23.93 130 I 122.62 1100 W 100 0.45
8Dining
room4.83.5 150 I 504.00
660 +160+160W
480 2.18
Remarks: F means fluorescent, while I means incandescent.
Calculations of lighting circuits loads:Lighting loads will be divided in 2 circuits
Circuit (1): L1
This circuit containing loads of rooms 1, 2, 7 and 8L1 = 0.66 actual currents of the rooms in the circuit)
= 0.66 (1.82+0.303+0.455+2.182+0.227)= 3.29 Amps.
And for safety and good design; we take a safety factor of 1.52
L1 = 5 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 2 mm2
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Circuit ( 2 ): L2This circuit containing loads of rooms 3, 4, 5 and 6
L2 = 0.66 actual currents of the rooms in the circuit)= 0.66 (1.82+0.91+0.27+1.86)= 3.21 Amps.
And for safety and good design; we take a safety factor of 1.52
L2 = 4.88 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 2 mm2
B- Sockets calculations:
- Since the area of the flat is less than 90 m2 and the flat is of youth
housing we don't expect to have any power sockets in this flat.- The following loads are expected in the flat:
>> Kitchen:We expect the use of a fridge of 160 w and a kitchen machine of 600 w.
>> BathroomWe expect the use of a normal washing machine of 400 w.
>>Bedroom 1 (main bedroom):We expect the use of a fan of 200 w and a T.V of 65 w.
>> Bedroom 2:This is the kids' room. We expect the use of a fan of 200w.
>>Balcony (of the kids' room)We expect the use of any appliance of rating not more than 3 A.
>>Living room:We expect 1 TV set of 65 W and 1 fan of 200 W
>>HallWe expect the use of any appliance like a vacuum cleaner of 800w.
>> Dinning room.We expect the use of a cassette of 40w and a fan of 200w.
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The result of the required sockets is tabulated in the following table:
RoomNo.
1 2 3 4 5 6 7 8
RoomFunction
Kitchen BathroomBedroom
1Bedroom
2Balcony
Livingroom
HallDinningroom
No. ofN.S
2 1 2 1 1 2 1 2
Calculations of sockets circuits loads:Sockets loads will be divided into to 2 circuits each with 6 normalsockets
Circuit (3): NS1
This circuit contains the loads of rooms 1, 2, 7 and 8.
NS1 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit)diversity factor
= 5+0.3 5 5 = 12.5 A
The M.C.B. used = 16 AThe copper wire used is of c.s.a 4 mm2
Circuit (4): NS2
This circuit contains the loads of rooms 3, 4, 5 and 6
NS2 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit) diversity factor
= 5+0.3 5 5 = 12.5 A
The M.C.B. used = 16 AThe copper wire used is of c.s.a 4 mm2
-
C- Calculation of the KVA of the flat:
Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.Bratings in amperes) diversity factor.
= 16 + 0.3 16
= 20.8 AmpsTotal Lighting Load= L1 + L2 = 5.0 + 4.88 = 9.88 AmpsTotal Load of flat= Total Socket Load + Total Lighting Load
= 30.68 Amps.
The incoming M.C.B. used = 32 AThe copper wire used is of c.s.a 16 mm2Energy meter is 40 A single phase meter.
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KVA for one flat= 220 30.68 10-3= 6.75 KVA.
After using diversity curves (done for 10 units)
Diversified KVA of the flat= 3.02 KVA / flat.
D- Riser Calculations:
Flat KVA = 3.02 KVA after diversity for 10 flats.
Total number of flats in the building = 10 flatsuse one riser
Total riser KVA= 3.02 x 10 = 30.2 KVA
Phase KVA = 30.2/3= 10.06 KVA
Phase current =220
1000x10.06= 45.71 Amps
For safety considerations, the riser is loaded with 70%of its ampacity.
\I phase max= 45.71/0.7 = 65.31 Amps. From the tables attached:
Riser used is 3x35+ 25 mm2Cu & the resistance per phase is1.097x10-3V/m.
Let the auxiliary loads be:
a- Water Pump = 1.5 HP, 0.85 p.f. KVA of the pump =85.0
746x5.1= 1.32
KVA.
b- Stairs lighting: if each floor uses one incandescent lamp of 100 W,p.f= 1, KVA of stairs lighting = 5 x 100 x 10-3= 0.5 KVA.
Total KVA of the building= 30.2 + 1.32 + 0.5 = 32.02 KVA.
And since phases R and S have only 3 flats on each while phase T has 4flats, thus to cause relative balance between phases then we feed theauxiliary loads from either phase R or S.
For protection we use a three phase fuse 63A.
Building box is of rating 32.02 KVA.
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4.6.3 Calculations for villa type (C):
I- 1st
floor:
A- Lighting calculations:
The following rooms are found in the first floor of villa (C)Room
no.Function
of roomArea (m2) lux
lamp
used
Req.
wattageNo of lamps
Power
(w)
Actual
current
(A)
1 Kitchen 42.8 300 F 228.48340 + 340
W240 1.82
2 Bathroom 1.82.8 100 F 34.27 140 W 40 0.30
Shaving
mirror50 I 50.40 240 W 80 0.36
3 Living room 4.54 150 I 540.00860 W
+21*60 W600 2.73
4 Dinningroom 3.54 150 I 420.00 660 W +2160 W480 2.18
5 Salon 3.54 150 I 420.00660 W +2160 W
480 2.18
6 Balcony 3.51.5 50 I 52.50 160 W 60 0.27
7 Balcony 4.51.5 50 I 67.50 160 W 60 0.27
8 Balcony 0.82.8 50 I 22.40 125 W 25 0.11
9 Hall1 7.751.5 100 I 232.50 21100 W 200 0.91
10 Stairs 50 I 50.00 160 W 60 0.27
11 Entrance 1.52.8 100 I 84.00 1100 W 100 0.45
Remarks: F means fluorescent, while I means incandescent.
Calculations of lighting circuits loads:
Lighting loads will be divided in 3 circuits
L1
This circuit containing loads of areas 1, 2, 8, 10 and 11L1 = 0.66 actual currents of the rooms in the circuit)
= 0.66 (1.82+0.3+0.36+0.11+0.27+0.45)= 2.1846 Amps.
And for safety and good design; we take a safety factor of 1.6
L1 = 3.5 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 3 mm2
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L2
This circuit containing loads of areas 3, 7 and 9L2 = 0.66 actual currents of the rooms in the circuit)
= 0.66 (2.73+0.27+0.91)= 2.581 Amps.
And for safety and good design; we take a safety factor of 1.6
L2 = 4.1296 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 3 mm2
L3This circuit containing loads of areas 4, 5 and 6
L3 = 0.66 actual currents of the rooms in the circuit)= 0.66 (2.18+2.18+0.27)
= 3.06 Amps.And for safety and good design; we take a safety factor of 1.6
L3 = 4.9 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 3 mm2
B- Sockets calculations:
- Normal Socket Lines:
The following loads are expected in each area.>> Kitchen:We expect the use of a fridge of 160 w, a kitchen machine of 600 w anddeep freezer of 160 w.
>> BathroomWe expect the use of a hair dryer of 600w or a shaving machine of 150w.
>>Living room:We expect a TV set of 65 W, a video of 30w and a cassette of 40w.
>> Dinning room.We expect the use of a cassette of 40w and a fan of 200w.
>>Salon:We expect the use of a T.V of 65w and a sound system of 800w
>>Balconies of the salon, the living room and the bathroom:We expect the use of any appliance of rating not more than 3 A.
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>>Hall 1:Since the hall is long we use at least two normal sockets one at each end.
The result of the required sockets is tabulated in the following table:
Room No. 1 2 3 4 5 6 7 8 9
RoomFunction
Kitchen Bathroom Livingroom
Dinningroom
Salon Balcony Balcony Balcony Hall1
No. of N.S 3 1 3 2 2 1 1 1 2
Calculations of normal sockets circuits
loads:Normal sockets loads will be divided into to 3 circuits
NS1
This circuit contains the loads of rooms 1, 2 and 8.NS1 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit)diversity factor
= 5+0.2 4 5 = 9 A
The M.C.B. used = 10 AThe copper wire used is of c.s.a 4 mm2
NS2
This circuit contains the loads of rooms 3, 7 and 9.
NS2 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit) diversity factor
= 5+0.2 5 5 = 10 A
The M.C.B. used = 16 AThe copper wire used is of c.s.a 4 mm2
NS3
This circuit contains the loads of rooms 4, 5 and 6..
NS3 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit) diversity factor
= 5+0.2 4 5 = 9 A
The M.C.B. used = 10 AThe copper wire used is of c.s.a 4 mm2
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- Power Socket Lines:The following loads are expected in each area.
1. PS1 For water heater in the kitchen:Let the water heater be 2000w thus the rated current will be 9.12A thus weuse a power socket 16A and the MCB will be 16A with a copper wire of c.s.a
4 mm2.
2. PS2 For dish washer in the kitchen:Let the dish washer be 1500w with a power factor of 0.85 thus the ratedcurrent will be 8.5A and a starting current of 250% of the rated current thuswe use a power socket 26A and the MCB will be 25A with a copper wire ofc.s.a 6 mm2.
3. PS3 For A.C in the living room:Let the A.C be 2.25HP with a power factor of 0.85 thus the rated current willbe 9.5A and a starting current of 250% of the rated current thus we use a
Load break switch (L.B.S) 26A and the MCB will be 25A with a copper wireof c.s.a 6 mm2.
4. PS4 For A.C in the salon:Let the A.C be 2.25HP with a power factor of 0.85 thus the rated current willbe 9.5A and a starting current of 250% of the rated current thus we use aLoad break switch (L.B.S) 26A and the MCB will be 25A with a copper wireof c.s.a 6 mm2.
C- Distribution of the circuits among the phases:
Phase LightLines
SocketLines
R L1 (3.5A)NS1 (10A)PS1 (16A)PS2 (25A)
S L2 (4.13A) NS2 (16A)PS3 (25A)
T L3 (4.9A)NS3 (10A)PS4 (25A)
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Calculations of phases current:For phase R:Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.B
ratings in amperes) diversity factor.= PS2+0.2 (NS1+PS1)= 30.2 Amps
Total Lighting Load= L1 = 3.5 AmpsTotal Load of phase R for the 1stfloor= Total Socket Load + Total Lighting
Load= 33.7 Amps.
For phase S:Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.B
ratings in amperes) diversity factor.= PS3+0.2 NS2= 28.2 Amps
Total Lighting Load= L2 = 4.13 AmpsTotal Load of phase S for the 1stfloor= Total Socket Load + Total LightingLoad= 32.33 Amps
For phase T:Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.B
ratings in amperes) diversity factor.= PS4+0.2 NS3= 27.0 Amps
Total Lighting Load= L3 = 4.9 Amps
Total Load of phase T for the 1
st
floor= Total Socket Load + Total LightingLoad= 31.9 Amps
The incoming M.C.B.of the 1stfloor panel board is 40 A (3 p)The copper wire used is of c.s.a 416 mm2
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II- 2nd
floor:
A- Lighting calculations:
The following rooms are found in the 2nd
floor of villa (C)
Room
no.Function
of roomArea (m2) lux
lamp
used
Req.
wattageNo of lamps
Power
(w)
Actual
current
(A)
12 Kitchenette 42.8 200 F 152.32 340 W 120 0.91
13 Bathroom 1.82.8 100 F 34.27 140 W 40 0.30
Shaving
mirror50 I 50.40 240 W 80 0.36
14
Bedroom1
(main
bedroom)
4.54 150 I 540.004100 W+2160 W
520 2.36
15 Bedroom2 3.54 150 I 420.00 4100 W 400 1.82
16 Bedroom 3 3.54 150 F 142.80 440 W 160 1.21
17 Balcony 3.51.5 50 I 52.50 160 W 60 0.27
18 Balcony 4.51.5 50 I 67.50 160 W 60 0.27
19 Balcony 0.82.8 50 I 22.40 125 W 25 0.11
20 Hall2 7.751.5 100 I 232.50 21100 W 200 0.91
Remarks: F means fluorescent, while I means incandescent.
Calculations of lighting circuits loads:
Lighting loads will be divided in 3 circuits
L4
This circuit containing loads of areas 12, 13, 19 and 20.L4 = 0.66 actual currents of the rooms in the circuit)
= 0.66 (0.91+0.3+0.36+0.11+0.91)= 1.71 Amps.
And for safety and good design; we take a safety factor of 1.6
L4 = 2.74 AmpsThe M.C.B used =6 Amps
The Copper wire used is of c.s.a 3 mm2
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L5
This circuit containing loads of areas 14 and 18.L5 = 0.66 actual currents of the rooms in the circuit)
= 0.66 (2.36+0.27)= 1.73 Amps.
And for safety and good design; we take a safety factor of 1.6
L5 = 2.78 AmpsThe M.C.B used =6 Amps
The Copper wire used is of c.s.a 3 mm2
L6
This circuit containing loads of areas 15, 16 and 17.L6 = 0.66 actual currents of the rooms in the circuit)
= 0.66 (1.82+1.21+0.27)= 2.2 Amps.
And for safety and good design; we take a safety factor of 1.6
L6 = 3.48 AmpsThe M.C.B used =6 Amps
The Copper wire used is of c.s.a 3 mm2
B- Sockets calculations:
- Normal Socket Lines:
The following loads are expected in each area.>> Kitchenette:
We expect the use of a small fridge of 80 w and a toaster of 800w
>> BathroomWe expect the use of a hair dryer of 600w or a shaving machine of 150wand an air blower of 200w
>>Bedroom1 (main bedroom):We expect a TV set of 65 W, a cassette of 40w, and a lighting spot of100W.
>> Bedroom2:We expect the use of a cassette of 40w and an iron of 1000w.
>>Bedroom3:We expect the use of a T.V of 65w and a fan of 800w
>>Balcony of bedroom3We expect the use of a cassette of 40w
>>Balcony of bedroom1We expect the use of any appliance of not more than 3A
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>>Balcony next to the bathroom:We expect the use of any appliance of not more than 3A
>>Hall 2:Like we did in hall1, we use at least two normal sockets one at each end.
The result of the required sockets is tabulated in the following table:
Room No. 1 2 3 4 5 6 7 8 9Room
FunctionKitchen
etteBathroom
Bedroom1
Bedroom2
Bedroom3
Balcony Balcony Balcony Hall2
No. of N.S 2 2 3 2 2 1 1 1 2
Calculations of normal sockets circuitsloads:
Normal sockets loads will be divided into to 3 circuits
NS4
This circuit contains the loads of rooms 12, 13 and 19.NS4 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit)diversity factor
= 5+0.2 4 5 = 9 A
The M.C.B. used = 10 AThe copper wire used is of c.s.a 4 mm2
NS5
This circuit contains the loads of rooms 14, 18 and 20.NS5 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit) diversity factor
= 5+0.2 5 5 = 10 A
The M.C.B. used = 16 AThe copper wire used is of c.s.a 4 mm2
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NS6
This circuit contains the loads of rooms 15, 16 and 17.NS6 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit) diversity factor
= 5+0.2 4 5 = 9 A
The M.C.B. used = 10 AThe copper wire used is of c.s.a 4 mm2
- Power Socket Lines:
The following loads are expected in each area.
1. PS5 for washing machine on the bathroom:
Let the washing machine be 1500w with a power factor of 0.85 thus therated current is 8.5 A and the starting current is 250% of the rated currentthus we use a power socket of 26 A and the M.C.B is 25A with a copper wireof c.s.a 6 mm2
2. PS6 for an A.C in bedroom1:Let the A.C be 2.25 HP with a power factor of 0.85 thus the rated current willbe 9.5A and a starting current of 250% of the rated current thus we use aLoad break switch (L.B.S) 26A and the MCB will be 25A with a copper wireof c.s.a 6 mm2.
3. PS7 For an A.C in bedroom2:Let the A.C be 2.25HP with a power factor of 0.85 thus the rated current willbe 9.5A and a starting current of 250% of the rated current thus we use a
Load break switch (L.B.S) 26A and the MCB will be 25A with a copper wireof c.s.a 6 mm2.
C- Distribution of the circuits among the phases:
Phase LightLines
SocketLines
R L4 (2.74A)
NS4(10A)
PS5(25A)
S L5(2.78 A)NS6(10A)PS6(25A)
T L6(3.48A)NS5(16A)PS7(25A)
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Remark: in the 1stfloor, load of phase R> load of phase S> load of phase T;thus in the 2ndfloor we care that the opposite happens to cause a balance onthe phase as a whole.
Calculations of phases current:
For phase R:
Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.Bratings in amperes) diversity factor.
= PS5+0.2 NS4= 27 Amps
Total Lighting Load= L4 = 2.74 AmpsTotal Load of phase R for the 2ndfloor= Total Socket Load + Total Lighting
Load= 29.74 Amps.
For phase S:Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.B
ratings in amperes) diversity factor.= PS6+0.2 NS6= 27 Amps
Total Lighting Load= L5 = 2.78 AmpsTotal Load of phase S for the 2ndfloor= Total Socket Load + Total Lighting
Load= 29.78 Amps
For phase T:Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.B
ratings in amperes) diversity factor.= PS7+0.2 NS5= 28. Amps
Total Lighting Load= L6 = 3.48 AmpsTotal Load of phase T for the 2ndfloor= Total Socket Load + Total Lighting
Load= 31.9 Amps
The incoming M.C.B. of the 2ndfloor panel board 40A (3p)
The copper wire used is 416 mm2
D- Calculation of the KVA of the villa:
For the 1stfloor; KVA of this floor is calculated as follows:KVAbottom= 220 (33.7+32.33+31.9) 10
-3= 21.54 KVA.
For the 2nd floor; KVA of this floor is calculated as follows:
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KVAupper = 220 (29.74+29.78+31.68) 10-3
= 20.06 KVA.Thus the KVA of the two floors after making the appropriate diversificationbetween the loads of the two floors is 41 KVA.Let the garden lighting be 1.5 KVA. It will be connected to the phase S since itis the least loaded phase.
Total KVA of the building= 41 + 1.5 = 42.5 KVA.
E- Riser Calculations:
Villa KVA without the garden lighting is 41 KVA.
Total riser KVA= 41 KVA
Phase KVA = 41/3= 13.67 KVA
Phase current = 220
1000x13.67
= 62.12 Amps
For safety considerations, the riser is loaded with 70%of its ampacity.
\I phase max= 62.12/0.7 = 88.74 Amps. From the tables attached:
the riser used is 3x50 + 25 mm2Cu & the resistance per phaseis 0.836x10-3V/m.
We use a three phase energy meter 3 80 A.
For protection we use a three phase fuse 80A.
Building box is of rating 42.5 KVA.
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4.6.4 Calculations for villa type (E):
A- Lighting calculations:
The following rooms are found in villa (E)
Room
no.Function
Of room
Area
(m2)lux
lamp
used
Req.
wattageNo of lamps
Power
(w)
Actual
current
(A)
1
Bedroom1
(main
bedroom)
55 120 I 600 4100+21100 600 2.73
2 Bedroom2 55 150 F 255 340 W+340 W 240 1.82
3 Salon 55 150 I 750 1060 W+2160 W 720 3.27
4Dinning
room55 150 I 750 1060 W+2160 W 720 3.27
5 Hall1 66.5 100 I 780 1060 W+4160 W 840 3.82
6 Kitchen 45 300 F 359.04 440+440 W 320 2.42
7 Bathroom 23 100 F 40.8 140 W 40 0.30
Shaving
mirror50 I 60 225 W 50 0.23
8 Hall2 22 100 I 80 1100 W 100 0.45
9 W.C. 21.2 100 F 16.32 125 W 25 0.19
10 Hall3 51.5 100 I 150 160+160 W 120 0.55
11 Hall4 51.5 100 I 150 160+160 W 120 0.55
Remarks: F means fluorescent, while I means incandescent.
Calculations of lighting circuits loads:
Lighting loads will be divided in 3 circuits
L1
This circuit containing loads of areas 1, 3 and 11L1 = 0.66 actual currents of the rooms in the circuit)
= 0.66 (2.73+3.27+0.55)
= 4.323 Amps.And for safety and good design; we take a safety factor of 1.2
L1 = 5.1876 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 3 mm2
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L2
This circuit containing loads of areas 2, 5, 7 and 8L2 = 0.66 actual currents of the rooms in the circuit)
= 0.66 (1.82+3.82+0.303+0.23+0.45)= 4.37 Amps.
And for safety and good design; we take a safety factor of 1.2
L2 = 5.244 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 3 mm2
L3
This circuit containing loads of areas 4, 6, 9 and 10L3 = 0.66 actual currents of the rooms in the circuit)
= 0.66 (3.27+2.42+0.189+0.55)= 3.06 Amps.
And for safety and good design; we take a safety factor of 1.2
L3 = 5.088 AmpsThe M.C.B used =10 Amps
The Copper wire used is of c.s.a 3 mm2
B- Sockets calculations:
- Normal Socket Lines:
The following loads are expected in each area.
>>Bedroom1 (main bedroom):We expect a TV set of 65 W, a small fridge of 80w, and a cassette of 40w.
>> Bedroom2:We expect the use of a cassette of 40w and an iron of 1000w.
>>Salon:
We expect the use of a T.V of 65w, a video of 30w and a sound system of800w.
>> Dinning room.We expect the use of a cassette of 40w and a fan of 200w.
>>Hall 1:Since the area is big (6.5 6m2) thus we must spread at least two or threesockets in this area so that the user can use the required appliance
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wherever he wishes in the hall. We expect the use of a cassette of 40w, avacuum cleaner of 800w and any other appliance not exceeding 3A.>> Kitchen:We expect the use of a fridge of 160 w, a kitchen machine of 600 w anddeep freezer of 160 w.
>> BathroomWe expect the use of a hair dryer of 600w or a shaving machine of 150w
>>Hall 2:We expect the use of any appliance of 3A.
>> W.CWe expect the use of a hair dryer of 600w or a shaving machine of 150w
>> Hall3:We expect the use of any appliance of 3A.
>> Hall4We expect the use of any appliance of 3A.
The result of the required sockets is tabulated in the following table:
RoomNo.
1 2 3 4 5 6 7 8 9 10 11
RoomBedroom
1bedroom
2salon
Dinningroom
Hall1
kitchen bathroomHall
2W.C
Hall3
Hall4
No. of
N.S
3 2 3 2 3 3 1 1 1 1 1
Calculations of normal sockets circuits
loads:
Normal sockets loads will be divided into to 3 circuits with 7 sockets oneach circuit.
NS1
This circuit contains the loads of rooms 1, 3 and 11NS1 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit)diversity factor
= 5+0.2 6 5 = 11 A
The M.C.B. used = 16 AThe copper wire used is of c.s.a 4 mm2
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NS2
This circuit contains the loads of rooms 2, 5, 7 and 8 .NS2 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit) diversity factor
= 5+0.2 6 5 = 11 A
The M.C.B. used = 16 AThe copper wire used is of c.s.a 4 mm2
NS3
This circuit contains the loads of rooms 4, 6, 9 and 10.NS3 = largest rating socket in ampere + (the rest sockets ratings in ampereof the rooms in the circuit) diversity factor
= 5+0.2 6 5 = 11 A
The M.C.B. used = 16 AThe copper wire used is of c.s.a 4 mm2
- Power Socket Lines:The following loads are expected in each area.
4-1. PS1 For A.C in bedroom1:Let the A.C be 2.25HP with a power factor of 0.85 thus the rated current will
be 9.5A and a starting current of 250% of the rated current thus we use aLoad break switch (L.B.S) 26A and the MCB will be 25A with a copper wireof c.s.a 6 mm2.
4-2. PS2 For A.C in bedroom2:Let the A.C be 2.25HP with a power factor of 0.85 thus the rated current willbe 9.5A and a starting current of 250% of the rated current thus we use aLoad break switch (L.B.S) 26A and the MCB will be 25A with a copper wireof c.s.a 6 mm2.
4-3. PS3 For A.C in the salon:
Let the A.C be 2.25HP with a power factor of 0.85 thus the rated current willbe 9.5A and a starting current of 250% of the rated current thus we use aLoad break switch (L.B.S) 26A and the MCB will be 25A with a copper wireof c.s.a 6 mm2.
4-4. PS4 For water heater in the kitchen:Let the water heater be 1500w thus the rated current will be 6.82A thus weuse a power socket 10A and the MCB will be 10A with a copper wire of c.s.a4 mm2.
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4-5. PS5 for washing machine on the bathroom:Let the washing machine be 1500w with a power factor of 0.85 thus therated current is 8.5 A and the starting current is 250% of the rated currentthus we use a power socket of 26 A and the M.C.B is 25A with a copper wireof c.s.a 6 mm2
C- Distribution of the circuits among the phases:
Phase Light Lines SocketLines
R L1 (5.1876A)NS1 (16A)PS1 (25A)
S L2 (5.244A)
NS2 (16A)
PS2 (25A)
T L3 (5.088A)NS3 (16A)PS5(25A)PS4 (10A)
Calculations of phases current:
For phase R:Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.B
ratings in amperes) diversity factor.= PS1+0.2 NS1= 28.2 Amps
Total Lighting Load= L1 = 5.1876 AmpsTotal Load of phase R for the 1stfloor= Total Socket Load + Total Lighting
Load= 33.3876 Amps.
For phase S:Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.B
ratings in amperes) diversity factor.= PS2+0.2 NS2= 28.2 Amps
Total Lighting Load= L2 = 5.244 Amps
Total Load of phase S for the 1stfloor= Total Socket Load + Total LightingLoad
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= 33.44 AmpsFor phase T:Total Socket Load= largest rating M.C.B in amperes + (the rest M.C.B
ratings in amperes) diversity factor.= PS5+0.2 (NS3 + PS4)= 30.2 Amps
Total Lighting Load= L3 = 5.088 AmpsTotal Load of phase T for the 1stfloor= Total Socket Load + Total Lighting
Load= 35.288 Amps
The incoming M.C.B. 40A (3p)The copper wire used is 416 mm2
D- Calculation of the KVA of the villa:
KVAvilla= 220 (33.3876 + 33.44 + 35.288) 10-3
= 22.5 KVA.
Let the garden lighting be 1.5 KVA. It will be connected to the phase R since itis the least loaded phase.Total KVA of the building= 22.5 + 1.5 = 24 KVA.
E- Riser Calculations:
Villa KVA without the garden lighting is 22.5 KVA.
Total riser KVA= 22.5 KVA
Phase KVA = 22.5/3= 7.5 KVA
Phase current =220
1000x7.5= 34.1 Amps
For safety considerations, the riser is loaded with 70%of its ampacity.
\I phase max= 34.1/0.7 =48.7 Amps. From the tables attached:
The riser used is 3x25 + 16 mm2Cu
We use a three phase energy meter 3 40A.
For protection we use a three phase fuse 50A.
Building box is of rating 24 KVA.
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Chapter -4- Building Wiring Calculations
4-7. Summary of results
The load and the number of each building type are summarized in the
following table:
Building typeLoad of the building
(KVA)
No. of buildings
required
Flat type (F) 43.1 32
Flat type (J) 32.02 16
Villa type (C) 42.5 50
Villa type (E) 24 50