building systems wiring

8/12/2019 Building Systems Wiring

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Chapter -4- Building Wiring Calculations

40

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.




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


-

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):


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


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.


L1 = 5 AmpsThe M.C.B used =10 Amps



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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.





- 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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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


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


-

C- Calculation of the KVA of the flat:


= 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



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.




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4.6.3 Calculations for villa type (C):

I- 1st

floor:


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




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.





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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.




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




- 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.


>>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.


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


NS2

This circuit contains the loads of rooms 3, 7 and 9.


= 5+0.2 5 5 = 10 A


NS3

This circuit contains the loads of rooms 4, 5 and 6..


= 5+0.2 4 5 = 9 A



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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


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:


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




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.





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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.




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.






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.


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


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



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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


- 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.


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:


= 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.


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



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.




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4.6.4 Calculations for villa type (E):


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




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




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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.




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.






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.


>>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.


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



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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


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


- 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


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


Total Lighting Load= L1 = 5.1876 AmpsTotal Load of phase R for the 1stfloor= Total Socket Load + Total Lighting

Load= 33.3876 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


\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.


Building box is of rating 24 KVA.


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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

building systems wiring

Documents