Distribution Course

7/22/2019 Distribution Course

1/272

Part 1

Prepared by

Eng. Abd Elmonem Shaban

Eng. Amr Mohamed


2/272

Contents:-

Ch1: AutoCAD

Ch2: Lighting

Ch3: Sockets (Power)

Ch4: Circuit Breaker

Ch5: Cables

Ch6: Air Condition (HVAC)

Ch7: Lefts

Ch8: Bus Duct

Ch9: Feeding Systems

Ch10: Load Estimation

Ch11: Power Factor Correction


3/272


Eng. Amr Mohamed

Ch1: AutoCAd 4A Consultant and Training

4A for Consultant and Training

Ch: 1 AutoCAD

(1) Line L + Enter

AutoCAD: 1- 2-( L + Enter (

( ESC. (: ( F8 )

(F8 .(ortho mode)( :

: ( Format)( Line Weight( )Display Line Weight)

: . : . : .-CLICK L

(2) Arc A+ Enter

Second

Arc) ( 1-. -2 A + Enter

. .

First point

End Point

First point End point

3


4/272


Eng. Amr Mohamed

Power System Distribution 4A Consultant and Training


3- Rectangular rec+ Enterfirst point

End point

1- -2 Rec + Enter 20 cm x 10 cm .Rec + Enter

enter,: .x axis L&Y axis W

.

4-Circle C + Enter C + Enter 10 cm

Enter10 cmC + Enter

5-CopyCo + Enter

-1 -2 Co + Enter

: : : ((selectCo + Enter

.

4


5/272


Eng. Amr Mohamed



6-Move M + Enter

1- -2 M + Enter

: m + Enter

7-Rotate Ro + Enter

-1 -2 Ro + Enter

: 90ORo + Enter Rotate L-Click90O

Enter ( (Manual

8-Scale SC + Enter

-1 -2 SC + Enter

: 3 SC + EnterL-CLICK 3Enter

L-Click

9-TrimTR + Enter-1 ------2 Tr + Enter : 23Tr + Enter

3

2

5


6/272


Eng. Amr Mohamed



10-Extend Ex + Enter

-1 -----2 Ex + Enter

, 3, 2, 14Ex +Enter 13.

11- Stretch S + Enter

-1 -2 S + Enter

S + Enter CLICK-L .

12- Hatch H + Enter

1- -2 H + Enter1

: Hatch1H + EnterPatten Add

Pick PointsL-Click1EnterOk.SolidPatternD-ClickOther Prede fiend

Solid Hatch H + Enter

Gradient. 13- Match ma + Enter

-1 ma + Enter:

ma + Enter. text

2

1

3

4

6


7/272


Eng. Amr Mohamed



14- Offset O + Enter

-1 -2 O + Enter

: . O + Enter

5, 4, 3, 2

15- MirrorMI + Enter

-1 MI + Enter: Mirror( )MI + Enter1 )

( Enter

16- Fillet F + Enter

-1 -2 F + Enter

: 2, 1F + EnterEnter 16- Explode X + Enter

-1 X + Enter.

: X + Enter

17- Area AA + Enter

area + Enter 5,4,3,2,1Enter18- Dimension DI + Enter

DI + Enter 2, 1Enter 12

7


8/272


Eng. Amr Mohamed



19- Dimension

Dimension-1 linear

3-Aligned.

1- (A)-2 T+ Enter

Text D. Click

Al + Enter 1 2.

-1 Br + Enter-2 12

: Br + Enter2,1

-1 B + Enter2-

: BlockB + Enter PickPoint Enter

21- Aline Al + Enter

12

20- Text T + Enter

Br + EnterBreak-22

23- Block B + Enter

8


9/272


Eng. Amr Mohamed



: inserti + EnterPick PointEnter

: Enter+Ap

LoadEnter

: FIRST CORNER

SECOND CORNER

: Dialux 42 2 EnterB + + I

Enter

1-RF + EnterBlock-23- Zero-25-4 26- First CornerEnter7- Second ComerEnter

Enter+ntuCoB Enter

24- Insert Block I + Enter

25-RFRF + Enter

26- B count+ Enter

9


10/272


Eng. Amr Mohamed



- DIV + Enter2-

-------------------------------------------------------------------------------------------------

--- End of Chapter 1 ---

-------------------------------------------------------------------------------------------------


Power System Distribution (Design & Installation) Power System Protection (Basic & Advanced) Distributed Control System DCS (Basic & Advanced) National Instrument LabVIEW (Basic & Advanced) Programmable Logic Controller PLC (Basic & Advanced)

MATLAB (Introduction & SIMULINK Electrical) Classic Control CCNA

:702 02007 211181720/2121118:

E-mail:[email protected]/ [email protected] FB:[email protected]

div+ Enterdivide-27

10
mailto:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]


11/272


Eng. Amr Mohamed

Ch:2 Lighting 4A Consultant and Training


Ch: 2 Lighting

Lighting

Outdoor Lighting Indoor Lighting

1)Sport areas

2)Street lighting

3)Swimming pool

4)Landscape

5)Layouts

Contents:-

1-Luminaire Selection (From Catalogue)

2-Distribution of Luminaire ( Manual & Programs )

3-Design of Distribution Board and wiring system

4-Control of Lighting (Manual & Automatic )

11


12/272


Eng. Amr Mohamed



To distribute any area must be specified the following:-

To know

Lighting Level [LUX] Type of Luminaire

Get from standard code tables Get from Lighting Catalogue

IEC Code

EC CodeNEC Code

To know number of luminaires which achieve the suitable LUX.

N area

Surface Recessed Suspended

[1] Room Function

[2] Room Dimension

[3] Ceiling Type

12


13/272


Eng. Amr Mohamed



Surface mounted ceiling

Recessed mounted ceiling

((60 X 60 cm((30 X 60 cm.

13


14/272


Eng. Amr Mohamed



.

Suspended mounted ceiling

14


15/272


Eng. Amr Mohamed



6mt, 8mt, 12mt, 14mt and so on.

Zero

[4] Work plane height

80 cm

Depends on furniture:-

Take zero if corridor or shops.

Take 0cm if office.

[5] Environment

To specify the index protection

Dust Moisture Vibration

[0 - 6] [08] [010]

15


16/272


Eng. Amr Mohamed



[6] Indoor or Outdoor Lighting

To specify the following:-

Type of luminaire

Maintenance factor ( Indoor take 0.8 & outdoor take 0.6 )

16


17/272


Eng. Amr Mohamed



[7] Direct or Indirect Lighting

Direct lighting (ON/OFF)

Indirect lighting

V Lumen

Lumen = no. of lighting Lines = flux / lamp

Dimming done by using: Resistance, Thyristor or Triack.

17


18/272


Eng. Amr Mohamed



Luminaire selection

Luminaire

Lamp Ballast (Gear) Housing

Type of lamps

Filament Gas Discharge

Incandescent Fluorescent

Halogen (Tungsten) High Pressure sodium

Reflector Low Pressure Sodium

High Pressure Mercury

Metal Halide

Led

18


19/272


Eng. Amr Mohamed



Filament Lamps:-

[1] Incandescent Lamp

Color rendering factor 00%

Color: yellow

Power

(Watt)60 W 75 W 100 W 150 W 200 W 300 W 500 W 1000 W

Lumen 730 960 1350 2220 3150 3500 8400 13800

Lum/W 12.2 12.5 13.5 14.9 15.5 16.7 16.8 18.8

Lumen: Flux/Lamp [Lm]

19


20/272


Eng. Amr Mohamed



[2] Halogen or tungsten lamps

20 lm/w 00000%

Color: Yellow

Used in shops and landscape.

20


21/272


Eng. Amr Mohamed



[3] Reflected Lamps

:

:- Filament Lamps

.

21


22/272


Eng. Amr Mohamed



Gas Discharge Lamps:-

[1] Fluorescent Lamps

Tube Compact

TL-D (T8) Lamps Integrated

TL-5 Lamps Non Integrated

TL-D Lamps:-

18 Watt 60 Cm 1300 Lumen



For Recessed mounted ceiling:

If L= 60cm & W= 60cm must be use luminaire 4x18 watt.

22


23/272


Eng. Amr Mohamed



If L= 30cm & W= 60cm must be use luminaire 2x18 watt.

23


24/272


Eng. Amr Mohamed



For Surface mounted ceiling:

Use luminaire 4x18 watt

Use luminaire 2x36 for saving

18 W 1300 Lumen

36 W 3350 Lumen

Also 4x18 is more expensive than 2x36 luminaire.

24


25/272


Eng. Amr Mohamed



58 wattlamps used in open areasuch as: Supermarket, Factories andHyper malls.

Because 18 W 1300 Lumen

58 W 5400 Lumen

Ex: for Hyper

Required Lux = 700 lux Area = 53x30

25


26/272


Eng. Amr Mohamed



From calculations No. of required luminaires [N] = 330 luminaire.

No. of luminaires / tray = 3/1.6= 33Luminaire

No. of trays = (total required of luminaires ) / (No. of luminaires per tray)

No. of trays = 330 / 33 = 10 Tray

26


27/272


Eng. Amr Mohamed



27


28/272


Eng. Amr Mohamed



TL-5 Lamps:-




For Recessed mounted celling: according to the area of the celling

For Surface mounted celling: the same as in TL-D type.

35Watt: lamps used in open area.

28


29/272


Eng. Amr Mohamed



Type TL-5 TL-D

Lumen High Lm/W Low Lm/W

Size Diameter: 16mm Diameter: 35mm

Life time 30,000 hours 10,000 hours

Watt Low wattage High power

cost High Low

Compact Fluorescent Lamps:- ( Saving Lamp )

A)Integrated type

Low Lumen/watt

18w = 900 lumen

29


30/272


Eng. Amr Mohamed



B)Non Integrated type

High Lumen/watt

18w =1350 lum

Types: 13W, 18W, 26W, 36W

Used in spots: 2x18 - 2x26 - 2x36 Watt

Power Factor = 0.890.9

30


31/272


Eng. Amr Mohamed



Spot for Compact Type

31


32/272


Eng. Amr Mohamed



[2] High Pressure Sodium [HPS]

Color Rendering 20%

Color: yellow

High cost

-:

[3] Low Pressure Sodium [LPS]

32


33/272


Eng. Amr Mohamed



Color Rendering 45%

Color: yellow

Low cost

[4] Low Pressure Mercury [LPM]

Color Rendering 40%

Color: white

33


34/272


Eng. Amr Mohamed



[5] Metal Halide

Color Rendering 7090 %

Color: white

Power: 35Watt ----------------- up to --------------------- 2000 Watt

.

34


35/272


Eng. Amr Mohamed



Ballast

Types of Ballast:-

1)Magnetic Ballast

PF [ 0.4- 0.6]

2)Electronic Ballast

PF [ 0.9- 0.97]

Ex:-

For a luminaire 4x18W with magnetic ballast(PF= 0.5)

The consumed power is S= P/PF = (4x18)/0.5 = 144 VA

For a luminaire 4x18W with electronic ballast(PF= 0.95)

The consumed power is S= P/PF = (4x18)/0.95 = 75.78 VA

35


36/272


Eng. Amr Mohamed



Housing

Cover Mirror

Prismatic diffuser Opal Diffuser Aluminum

Used against dust used against moisture used to distribute the light

To select any luminaire according to:

1-Type of ceiling2-Lamp type3-Ballast4-Cover (Opal or Prismatic )5-Mirror type6-IP [ Index Protection ]

36


37/272


Eng. Amr Mohamed



)(

0 60:

0300

00:

0 00: 300

: - 300

- 00

0

300 300 00

000

: 300

60: 000

300:

300 300

6

37


38/272


Eng. Amr Mohamed



)(

0300

: 000

: 300

003000

: 300 00

000 :

00000

300000

00000

: 00

000-000

00000:

60300

:

30000

00

38


39/272


Eng. Amr Mohamed



)(

00 60 00

000

: 0

3003000

3000000

00300

000

: 300

0:

00 3000

: 00

0000

: 000003000

39


40/272


Eng. Amr Mohamed



)(

: 300

: 00

000300

: 30000

00 000

00:

30000

0000

00000

000:

30000

000

00 000

000000

000 3000

40


41/272


Eng. Amr Mohamed



)(

: 60300

6000

: - 300- 0

41


42/272


Eng. Amr Mohamed



Lighting Distribution

Manual distribution:-

N =

Where

N: Number of Luminaires

n: number of lamps per luminaire

Lux: Lighting level, get from standard table ( IEC, EC and NEC )

a: Room width

b: Room length

U.F: Utilization factor M.F: Maintenance factor

1)How to calculate utilization factor (U.F):-

First calcite room index [k] =

()

Then by knowing the color of walls, celling and work plane we can get the

U.F from the table

42


43/272


Eng. Amr Mohamed



REF

K

0.7 ---> (walls)

0.5 ---> (celling)0.2 ---> (work plane)

0.5

0.75

1 (U.F)

1.25

R: Reflection of walls, celling and work plane.

K: Room index.

2)Maintenance Factor:-

Indoor lighting M.F = 0.8

Outdoor lighting M.F = 0.4 - 0.6

3)Lux [ Lighting Level ]:-

Take the value from standard tables codes ( EC or IEC or NEC )

4)Lumen get from specification of lamp in catalogues:-

For example:

18 W 1300 Lum (TL-D)

36 W 3200 Lum (TL-D)

43


44/272


Eng. Amr Mohamed



58 W 5400 Lum (TL-D)

14 W 1200 Lum (TL-5)

26 W 1800 Lum (Compact)

From calculation if the total number of luminaire in the room = [N], then to

get the number of luminaire in length and width calculate the following:-

Number of luminaire in length =

Number of luminaire in width = Where

W: Width L: Length N: Total number of luminaires

Example

If we have office room with the following dimensions:-

a= 8 mt b= 8 mt h= 3mt

Lux= 600 Lux use (4X18 Watt) ----->lum= 1300

M.F = 0.8 U.F = 0.6

N=

So, No. of luminaire in width = = = 4No. of luminaire in length = = = 4

44


45/272


Eng. Amr Mohamed



Distance between two luminaire must be equal double distance between

wall and luminaire.

45


46/272


Eng. Amr Mohamed



Programing Distribution using (DIALux program):-

DIALUX

66 164

OK

46


47/272


Eng. Amr Mohamed



Luminariesselection

:lu in

47


48/272


Eng. Amr Mohamed



indoorfamily name

((CENTURA

TCS 098/4184186

ADD COLSE

48


49/272


Eng. Amr Mohamed



(INSERT LUMINAIRE FIELD)

MOUNTING 455

PASTE

49


50/272


Eng. Amr Mohamed



OUTPUTSUMMARY

U0>0.5 555((514

50


51/272


Eng. Amr Mohamed



Wiring system

51


52/272


Eng. Amr Mohamed



52


53/272


Eng. Amr Mohamed



From standard: one circuit () ---------------------------1200VA

VA/LINE C.B CABLE

1200VA 10 Amp or16Amp 3X3mm2

L

N E

53


54/272


Eng. Amr Mohamed



For indoor lighting

No of luminaire/line from (12-------16) outlet on circuit

( (30

)(( 0%-

30 )%

30

830

54


55/272


Eng. Amr Mohamed



55


56/272


Eng. Amr Mohamed



Control of lighting

56


57/272


Eng. Amr Mohamed



Project Application

1-selection of luminaire

Type: (A)

Type: (B)

57


58/272


Eng. Amr Mohamed



Type: (A)

Technical Data

Recessed mounting Luminaire.:DescriptionHousing of sheet steel with internal and external

electrostatic white paint:Luminaire housingAluminum sheet electrostatic white paint reflector

High mirror diffuser.:Reflector

Electronic ballast.

:

Control Gear

Fluorescent lamps 4*14 w:LampsIP20:Degree of protection

58


59/272


Eng. Amr Mohamed



Type: (B)

Technical Data

Indirect Luminaire.:DescriptionHousing of sheet steel with internal and external

electrostatic white paint:Luminaire housingAluminum sheet electrostatic white paint

reflector&opal diffuser.:ReflectorElectronic ballast.:Control GearFluorescent lamps 2*55 W:LampsIP43:Degree of protection

59


60/272


Eng. Amr Mohamed



Type (C )

Technical DataFixed recessed down lights with automatic

fixing clips for compact fluorescent lamps.:Description

Polycarbonate closed gear box.Luminaire housing

High brilliant metallized reflector.:RefelectorIntegral control gear in the housing..:Control GearCompact fluorescent 2x18 W:LampsElectronic BallastControl GearIP 44.:Degree of protection

60


61/272


Eng. Amr Mohamed



Type: (D)

Technical Data

Prismatic type clear polycarbonate injected.:DiffuserReinforced polyester with fiberglass (GRP):Luminaire housingSteel plate.:ReflectorElectronicballast:Control GearPolyurethane gasket, elastic washers and rubber

seals, .:Tightness2*36W:LampsIP65:Degree of protection

61


62/272


Eng. Amr Mohamed



Type: (E)

Technical Data

Energy saving and slim lined luminaire.:DescriptionHeat resistant thermoplastic housing.:Luminaire housingAcrylic diffuser.:ReflectorElectronic ballast.:Control GearCompact fluorescent 2x9 W.:LampsIP20:Degree of protection

62


63/272


Eng. Amr Mohamed



Type (F)

Technical DataEnergy saving and particle universal luminaire.:DescriptionHeatresistant thermoplastic housing.:Luminaire housingAcrylic diffuser reflector.:ReflectorElectronic Ballast:Control GearCompact fluorescent, Dulux F36W:LampsIP44:Degree of protection

63


64/272


Eng. Amr Mohamed



TOT

AL

VA

NO

.

VA/L

UMI

NAI

RE

4X3

60VA

A

RECESSED

500LUX

3 6 9

1X3

60VA

A

RECESSED

250LUX

3 4 7

4X3

60VA

A

RE

CESSED

5

00LUX

3 6 9

4X3

60VA

A

RECESSED

500LUX

3 6 9

3X3

60VA

A

RECESSED

500LUX

3 6 7

3X3

60VA

A

RECESSED

500LUX

3 6 7

3X3

60VA

A

RECESSED

500LUX

3 6 7

4X

3

60V

A

A

RECESSED

500LUX

3 6 9

2X4

60VA

A

RECESSED

500LUX

34.5

7

4X3

60VA

ARECES

SED

500L

UX 3 6 9

64


65/272


Eng. Amr Mohamed



1X2

60VA

D

SURFACE

200LUX

3.8

34.5

2X4

60VA

A

RECESSED

500LUX

3 4 9

2X3

120VA

B

RECESSED

500LUX

3 5 6

3X4

120V

A

A

RECESS

ED

500LUX

3 10 6

3X4

120VA

A

RECESSED

500LUX

3 10 5

1X25

40VA

C

RECESSED

150LUX

3 46 2

12

40VA

F

SU

RFACE

100LUX

3.8

11

40VA

E

RECESSED

100LUX

3

65


66/272


Eng. Amr Mohamed



66


67/272


Eng. Amr Mohamed



67


68/272


Eng. Amr Mohamed



68


69/272


Eng. Amr Mohamed



: - - -3 -8 -

-

- 60

--8

-0

Street lighting

69


70/272


Eng. Amr Mohamed



- -6 -6 6-3 36-8 -3

D=

Where:-

Lumen- flux per lamp

U.F= utilization factor

M.F=maintenance factor

E=LUX (LIGHTING LEVEL)

D=distance between towers in m)

W=street width in )m)

70


71/272


Eng. Amr Mohamed



A3B03C51

D51E53

71


72/272


Eng. Amr Mohamed



For example

For high way street with 500m length and 10 m width.

Calculate the distance between two towers.

-from the table as high way street lux=30

-we select high pressure sodium with 250watt

-From catalogue 250 watts----------------33200 lumen

-height of tower about 10m

-width /height =w/h=1

-from curve get U.F=.34

D =

D =

=23

No of poles = 500/23=22pole

72


73/272


Eng. Amr Mohamed



stadiums Lighting

By using calculux program

Open the program then Press file then choose new project

73


74/272


Eng. Amr Mohamed



Choose data then project options correct the maintenance factor

74


75/272


Eng. Amr Mohamed



75


76/272


Eng. Amr Mohamed



Press on data then choose application fields then press on add choose the

type then ok and close .

76


77/272


Eng. Amr Mohamed



77


78/272


Eng. Amr Mohamed



Choose data then project luminaries then press on add and choose file

dialux Then press add then close then close

78


79/272


Eng. Amr Mohamed



79


80/272


Eng. Amr Mohamed



80


81/272


Eng. Amr Mohamed



Choose data then individual luminaries and then press on new

.new

81


82/272


Eng. Amr Mohamed



x & y- . . z--rot 081 .

tilt 90: the best angle is 53 or 54

then press ok

Press on report then choose setup follow the pictures to make the complete

report and pick what you want from its data.

82


83/272


Eng. Amr Mohamed



83


84/272


Eng. Amr Mohamed



Press on calculation then choose presentation mark at all then

Press show results.

84


85/272


Eng. Amr Mohamed



85


86/272


Eng. Amr Mohamed



From the results see:

Graphical table and see if its ok or needs to get modified (individualluminaries &calculate again)

86


87/272


Eng. Amr Mohamed



To save the results as pdf file a program like snagit 8 is needed and

choose file then print report .

87


88/272


Eng. Amr Mohamed



-------------------------------------------------------------------------------------------------


-------------------------------------------------------------------------------------------------


:706 611181760/616112182:026007 E-mail:[email protected]/ [email protected] FB:[email protected]

88
mailto:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]


89/272


Eng. Amr Mohamed

Power System Distribution 4A for Consultant and Training


Contents:-

1.Types of sockets.2.Distribution of sockets.3.Design of distribution board and wiring system.

Types of Sockets1 General used sockets or Single sockets.

Standard rating for single socket:-

V = 250 volt ; I = 10 A or I = 16 A

S = according to codes

IEC EC

180 VA 250 VA

Ch3: Sockets (Power)

89


90/272


Eng. Amr Mohamed

Ch: 3 Sockets 4A for Consultant and Training


2 Double socket

Standard rating for double socket:-

V = 250 volt ; I = 10 A or16 A

S = according to codes

IEC EC

360 VA 500 VA

90


91/272


Eng. Amr Mohamed



3 Power socket

Standard rating for power socket:-

V = 250 volt ; I = 20 A or 32 A

For (S=500 ~ 2000 VA) Take 20A. For (S=2000 ~ 5000 VA) Take 32A.

Take it in calculation S = 500 ~5000 VA (Depend on load)

Application:

(1) Kitchen (2) Bath Rooms (heater and hand drier)

(3) Laundry (4) Drilling Machines

0.5 KVA ---------------***---------------***---------------***---------------

4 U P S socket

Standard rating forU.P.S Sockets:-

V = 250 volt ; I = 10 A or 16 A

Take it in calculation S = 250 VA

Total power of computer 250 VA

91


92/272


Eng. Amr Mohamed



A separated distribution board for UPS- socket shouldbe design.

92


93/272


Eng. Amr Mohamed



5 Weather proof socketIt is normal socket with cover (IP)

V = 250 VA I = 10 or 16 A take S as in single socket 180 or 250

VA

Normal Weather proof

`

93


94/272


Eng. Amr Mohamed



Application

Corridors Kitchen Bath Room Outdoor Stores Factories

---------------***---------------***---------------***---------------

6 3 socket

Standard rating for 3 sockets:-V = 400 volt;

I = 16A, 20A, 32A, 50A, 63A, 80A, 100A or 125A.

94


95/272


Eng. Amr Mohamed



7 Column sockets

Standard rating for 3 sockets:-V = 250 VA I = 16 A

Outlet: 6, 12, 18, 24 outlet

8 Trunking SocketV = 250 VA I = 16 A

Outlet: 6, 12, 18, 24 outlet

95


96/272


Eng. Amr Mohamed



[1] Wall Mounting

30 cm(offices) or

120 cm(water areas)

[2] Floor mounting

Must be IPZ very high

( IP:67)

[3] Furniture mounting

IP ( IP:65)

Sockets Mounting

96


97/272


Eng. Amr Mohamed



Legend of sockets

Symbol Type No.

Single socket with H = 30 cm

V = 250 Volt, 16 A

Double sockets with 30 cm

V = 250 Volt, 16 A

UPS sockets 30 cm

250 Volt, 16 A

Power sockets 30 cm

250 Volt, 20 A

.Weather proof sockets

Single socket V = 250 Volt, 50 HZ, 16 A

W.PWeather proof sockets

Double socket 250 Volt, 50 HZ, 16 A

.Weather proof sockets

Power socket V = 250 Volt, 16 A

Floor mounting. Double sockets

16 A, 250 V, IP= 67

Floor mounting. UPS sockets

16 A, 250 V, IP= 67

WPWall mounting. Power sockets

Weather proof 120 cm, 20 A, 250 V.

97


98/272


Eng. Amr Mohamed



( : Standard )(Type)

Design of DB and wiring system

There are two types of distribution boards:-

NSDB (Normal Socket Dist. Board)U.P.S.DB (Uninterruptible Power Supply Dist. Board)

Normal socket Distribution board

NSDBfeeds the below types of sockets:-

16 A, 250 Volt 16 A, 250 Volt

20 A, 250 Volt

According to code

IEC EC

Linea C.B Cable Linea C.B Cable2000 VA 16 A 3 x 3 2000 VA 16 A 3 x 3

( ) ()

( )()

98


99/272


Eng. Amr Mohamed



( ) ()

( )()

So, we connect from (610) on one line.

&.( 8-01 (

.21. Power Socket

. UPS - Socket

99


100/272


Eng. Amr Mohamed



- :( ) 11100) ) 111005110340111,9501111,9601111,957 01111,958011109011100101111,950001111,9507511,9035111,85045111,95511,90506111,90511,5070511,508111,8 091051,900111,90111,93 0111,94611,955511,96511,9

100


101/272


Eng. Amr Mohamed

Ch: 4 Circuit Breaker 4A Consultant and Training


Specification of circuit Breaker:-

[1] Operating voltage of C.B[2] Rated current of C.B (Ir or In) Amp.[3] Instantaneous short circuit current (Ics) KA[4] Rated breaking capacity (Icu) KA[5] Types of C.B[6] Types of poles.[7] Earth leakage C.B

Construction of low voltage C.B

Ch4: Circuit Breaker

101


102/272


Eng. Amr Mohamed



Operating Principle of low voltage C.B

Ir Ic.s Ic.u

Ir:rated current of C.B

Ic.s:short circuit current of C.B

Ic.u: max short circuit current or (Rated Breaking capacity)

Note:

Ir:depend on KVA of load

(Discussed in details in this capter)

Icu & Ics:depend on the impedance of (Cables, Bus Bars and Transformers)

(Will be discussed in details in SC calculation chapter)

102


103/272


Eng. Amr Mohamed



Operating voltage of C.B

Low voltage

(1V 1KV)

Medium voltage

(1KV 66KV)

H igh voltage

(66KV 500KV)

1 220

3 380 V

MCB

MCCB - ACB

11 KV, 22KV

6.6KV,3.3 KV

SF6 - Vacuum

132KV, 220KV

500KV

Oil - SF6

Low Voltage C.B

1 C.B

220 Volt

S < 5 KVA

3 C.B

380 Volt

S > 5 KVA

How to select C.B according to Ir ?

Ir of C.B= ?

Feed 1 , 50Hz.Load 4 KVA

As, S 1= VI So,

(Single Phase)

IL= 4.5 * KVA for 1- LoadI load= 4.5 * 4 = 18 A

103


104/272


Eng. Amr Mohamed



IC.B = Safety factor * Iload

Safety factor

E.C

25% over load

IEC

20%over load

NEC

10% over load

IC.B= 1.25 * 18 = 22.5 Amp.

(But there is no C.B with Ir = 22.5A)

So, from C.B standard:-

20A 25A 32A

So, Select C.B = 25 Amp.

22.5A

CB

(A)

630050004000320025002000160012501000800630400250200160125100806350403225201610

Circuir Breaker Ratings

ACBMCB

MCCB

104


105/272


Eng. Amr Mohamed



Ir of C.B = ?

3 , 50Hz

LOAD 50 KVA

As, So,

(Three Phase)

IL= 1.5 * KVA for 3-LoadIL= 1.5 *50 = 75 A

IC.B= 75 * 1.25 = 94 A

From C.B standard

80A 100A 125A

94A

Select C.B = 100 AMP

EX:- 3ph load, 140KVA

IL= 140 * 1.5 = 210 AmP

IC.B= 210 * 1.25 = 262.5 AmP

From C.B standard

250A 250A 400A

262.5

C.B = 250 AMP

105


106/272


Eng. Amr Mohamed



Ex:- 3ph load, 160KVA

Irated= 1.5 * 160 = 240 A IC.B= 1.25 * 240 = 300 A

From Standard 200A 250A 400A

300 A

Select C.B = 400 A / Adjustable

To get 300 A

Ir = 400 * 0.75= 300

Adjust the potentiometer

at 0.75

106


107/272


Eng. Amr Mohamed



How to select C.B according to IC.Sand IC.U?

- We should know short circuit level first, so this part as stated before willbe discussed later at Short Circuitcalculations chapter in this book.

Types of low voltage Circuit Breakers

(1)Miniature C.B ( 10 ~ 125A)

(2) Moulded Case C.B (16~ 1600A)

107


108/272


Eng. Amr Mohamed



(3) Air C.B (630 ~ 6300A)

A.C.BMCCB

A.C.BM.C.C.B

M.C.B

M.C.C.BM.C. B

6300A 1600A 630A 125A 16A 10A

(1) If C.B Ir= 10 A or 16 A M.C.B

(2) If C.B 125 A < Ir< 360 A M.C.B

(3) If C.B 2500 A < Ir


109/272


Eng. Amr Mohamed



Because:

M.C.B operates in 3 msec. M.C.C.B operates in 9 msec. A.C.B operates in 30 msec.

109


110/272


Eng. Amr Mohamed



MCB MCCB ACB

Rating Increase

Speed Increase

If C.B Incoming Select MCCB

If C.B Outgoing Select MCB

If C.B After Transformer must be A.C.B except only one case, If the

transformer locates in any area contain dust such as outdoor & factories must be

selected MCCB.

Because the MCCB can be maintained, but the ACB is very hard to be

maintained.

Types of Poles of CB

(1)Single Phase - Single pole C.B

Line is protected

Neutral is non-protected

(2)Single Phase - Two Pole C.B

Neutral and line are protected

High cost than single pole

110


111/272


Eng. Amr Mohamed



(3)3 Phase - 3 Pole C.B (4) 3 Phase - 4 Pole C.B

Earth leakage C.B (ELCB or RCCB)There are two types: 1 ELCB and 3 ELCB

111


112/272


Eng. Amr Mohamed



Operating Principle:-

For single phase system the ELCB compare the difference between the life and

neutral phases with the adjusted setting value.

Iin

IoutIin= Iout Normal Operation

IinIout Earth Leakage

So,

Iin = Io + ILeakage

ILekage= Iin- IoutIn case of three phase system the ELCB compare the difference between the

three line phase and the neutral with the adjusted setting value.

IR

Is

IT

IN

IR+ Is+ IT= IN= Zero Normal Operation

IR+ Is+ IT= IN= Value But IR+ Is+ IT+ IN= Zero Unbalance

IR+ Is+ IT= IN+ ILeakageOr IR+ Is+ IT+ IN= ILeakage Earth Leakage

LOAD

E

L

C

B

EL

C

B

LOAD

112


113/272


Eng. Amr Mohamed



Main function of ELCB

(1) To protect Human, we select Isetting= Iin- Iout = 30 mA

(2) To protect machines, we select I setting= 300 mA

Medium Vol tage C.B

Rated Voltages are

22 KV 11KV 6.6 KV 3.3 KV

Rated currents are I rated = 630 ~ 4000A

Note:630A frame size

113


114/272


Eng. Amr Mohamed



Rated Breaking Capacity

(A) 11KV MVAS.C= 500 MVA

(B) 22KV MVAS.C= 750 MVA

(C) 6.6KV MVAS.C= 250 MVA

Types of MV C.B are: oil, Vacuum and SF6

Ir of C.B = ?

Motor 2MVA

11 KV

= 104 A

Select: I CB= 630A Type: SF6 C.B

IS.C at 11KV 500 MVA

= 26 KV

Note: for MV motors there are a contribution of currents from the other

neighbored motors fed fault

The contribution range from (50% - 80%)

ICU= IS.C+ 80% IS.C= 1.8 (26) = 46.8 KA

So, ICU= 50 KA

114


115/272


Eng. Amr Mohamed



Note: Medium Voltage C.B Just operation only without fault detection (i.e.

controlled switch by protective relay).

The Multi-Function protective relay protects the motor from:-

(6) Over load

(7) Over current

(8)Under voltage(9)Over voltage

(10)Phase sequence

(1)Phase failure (open of one phase)(2)Over temp (PT 100)(3)Unbalance phase(4)Over frequency(5)Over speed.

C.T

V.T

Relay

IfFault

C.B

Relay Contact

Aux.

Coil

115


116/272


Eng. Amr Mohamed



FUSES

Types of Fuses

(1) Semi-enclosed Fuse

(2) Cartridge Fuse

Mainly used in Siemens boxes

116


117/272


Eng. Amr Mohamed



(3) High Rupture Capacity Fuse (HRCF)

(4) aM-Type Fuse

117


118/272


Eng. Amr Mohamed



Where:

- Semi enclosed and cartridge used in low voltage.- High Rupture Capacity used in medium voltage- H.R.C.F used to protect transformer from short circuit.- aM fuse used to protect short circuit protection in motors, transformer and

other load with high inrush currents due to the good current limiting

capability and low I2t values.

- Rating of fuses start from 10A, 16A, 20A, 25, 32, 40, 50, 63, 80, 100, 125,160, 200, 250, 320, 400, 630, 800, 1000, 1250A.

Ex.

To calculate the rating of fuse

(HRCF)

I rated= 52A

I FUSE = (Safety factor) *Irated

Safety factor = 1.25

If= 1.25 * 52 = 65 A

Select fuse rating = 80 A

Note:HRCF used only to protect

the transformer from short circuit

-------------------------------------------------------------------------------------------------


-------------------------------------------------------------------------------------------------

118


119/272


Eng. Amr Mohamed

Ch: 5 Cables 4A Consultant and Training


Cable Selection Cable Design

Cable Selection

Cables are selected according to:

Operating voltage Operating frequency Insulation level Conductor type Core number

[1] Operating voltage:-Low voltage cable [ 1 V 1000 V ]Medium voltage cable [ 1 KV 66 KV ]Overhead conductor [ 66 KV 500 KV ]Control Cable

Ch: 5 Cables

119


120/272


Eng. Amr Mohamed



For the same C.S.A medium voltage cable insulation higher than low voltage

cables (VInsulation).

In general:-

V Insulation

I Cross Section Area

[2] Operating Frequency:-

50 Hz 60 Hz

120


121/272


Eng. Amr Mohamed



[3] Conductor type:-

Aluminum (Al) Copper (Cu)

Conductivity of Al= 65% of Cuconductivity. Alis lighter than Cuin weight. Cuis higher cost than Al.

Aluminum (Al)

Cupper (Cu)

All medium voltage cablesare made from Al because of two reasons:-

1)Low current I 2)Underground cable cost

121


122/272


Eng. Amr Mohamed



Low voltage cablesare made from Cuexcept underground cables ofelectrical distribution company for residential area.

[4] Insulation Level:-Type PVC XLPE

Standard normal temperature 70c 90c

Max Temp. at short circuit level 150c 250c

Note:-

All medium voltage cables with XLPE insulation

122


123/272


Eng. Amr Mohamed



Low voltage cables may be PVC or XLPE (PVC for low current & XLPE forhigh current).

Note:-

Sheath is always made from PVC.

Conductor: Cu or Al

Insulation: PVC or XLPE

Sheath: PVC

Medium voltage cablesare always:

AL / XLPE / STA / PVC

As short circuit level very high Sheath

@ 11 KV Network ----- SC= 500MVA

@ 22 KV Network ----- SC= 750MVA

123


124/272


125/272


Eng. Amr Mohamed



[5] Core number:-a)Single core cable:

Application of single core cable:-

If CSA > 300mm2. Residential area. (Riser) Earthling cable.

Earthling Cable

Single Core Cable

b)Two core cable:Application of two core cable:-Used in low voltage in 1 where is there is

no earthing system [ L& N only ].

125


126/272


Eng. Amr Mohamed



c)Three core cable:Application of three core

cable:-

Used in low voltage in1 where is there is

earthing system [ L, N

and E ].

In medium voltagethree phase [ R, S, and T

].

Low Voltage

d)Four core cable:Application of four core cable:-

Used for three phase network in low

voltage system [R, S, T and N].

126


127/272


Eng. Amr Mohamed



================================================================

Cables Formations

Trefoil position is preferred than flat position as:flat position Temp RDerating in cables.

Multicore cables are more economic than single core cables. Multicore cables designed as trefoil so more technical than single core

cables.

Multicore cables are preferred than single core cables.

127


128/272


Eng. Amr Mohamed



================================================================

For Neutral Cable:-

C.S.A for neutral = C.S.A for any phase C.S.A (R) = C.S.A (Y) = C.S.A (B) = C.S.A (N)

Because: 1) third harmonic.

2)Unbalanced system.

R

S

T

N

128


129/272


Eng. Amr Mohamed



For Earthing Cable:-

If C.S.A 16 mm2 E = L

1 3

L = N = E R = S = T = N = E

3 x mm2

5 x mm2

L N E R S T N E

Ex:-

3 x 3 mm2 5 x 3mm

2

3 x 4 mm2 5 x 4mm

2

3 x 6mm2 5 x 6mm

2

3 x 10mm2 5 x 10mm

2

3 x 16 mm2 5 x 16mm

129


130/272


Eng. Amr Mohamed



If C.S.A >16 mm2 E = L

For three phase only E = L

4 x +

R S T N

Ex:-

4 x 70 + 35 or 4 x 240 + 120 or 4 x 300 + 150

Single core or Multi core

130


131/272


Eng. Amr Mohamed



Cable Design

Cables are designed according to:

Current carry capacity or thermal rating. Voltage drop. Short circuit level.

[1] Current Carry Capacity:-

C.B = 80 A Cables C.S.A =??!

3 PH, 50HZIrated= 40 x 1.5 = 60 Amp

IC.B = 60 x 1.25 = 75 Amp C.B = 80 Amp

Icable=

LOAD 40 KVA

131


132/272


Eng. Amr Mohamed



So must select C.B before cable. C.B rating depends on (KVA of load). Cable sizing depends on C.B rating.

Types of Derating Factor:-

a)Ambient temperature Derating factorb)Ground temperature Derating factorc)Grouping factord)Burial depth Derating factore)Soil thermal resistivity

CB C/Cs

Cable C/Cs

132


133/272


Eng. Amr Mohamed



133


134/272


Eng. Amr Mohamed



Derating Factor

Air Ground

TempTempGrouping Factor Depth

Soil Thermal

Take TA= 50c Take Tg= 50c

So, for PVCtake 0.82 So, for PVCtake 0.76

For XLPEtake 0.89 For XLPEtake 0.85

(for depth = 80 Cm)

How to calculate Derating Factor for group of cables?

Correction factor for cable laying in cable trays.

If cables are single layer and the distance between two cables is equal to

diameter of cable and the distance between cable and wall equal D/2 this

mean:

134


135/272


Eng. Amr Mohamed



Derating Factor D.F = 1

Ex

Icable=

Temp = 50c PVC D.FT= 0.82

Single cable D.FG= 1

Icable=

= 97 Amp

From Elsewedy Catalogue: chose Cu/PVC/PVC 4 X 25 + 16 mm2

135


136/272


Eng. Amr Mohamed



[2] Voltage Drop calculation:-

V.D = (mv / amp / m) x 10-3x Iactualx L

Where:

Iactual: load rated current.

L: Cable length.

(mv / amp / m ): Factor get from cable catalogue.

Note:-

Accepted voltage drop is V 5 V.D % = (V.D / 380) x 100

EX- Calculate the voltage drop of two motor 50hp and 100hp

136


137/272


Eng. Amr Mohamed



137


138/272


Eng. Amr Mohamed



For motor1 :- p = 50 HP

Irated= 50 x 1.5 = 75 A. IC.B = 75 x 1.25 = 94 A. C.B = 100 A

Icable=

= 125 A. 4x50 + 25 mm

2Cu/PVC/PVC

For motor2:- p = 100 HP

Irated= 100 x 1.5 = 150 A. IC.B = 150 x 1.25 = 187.5 A C.B = 200 A

Icable= = 250 A. 4x120 + 70 mm

2Cu/PVC/PVC

For DB:-

Total KVA = 100 + 50 = 150 KVA

Irated= 150 x 1.5 = 225 A. C.B = 250 A

Icable=

= 312.5 A. 4x185 + 95 mm2

Cu/PVC/PVC

Voltage Drop calculation:-

( From 1 ---- to ---- 2 )

L = 30 mt ; Iactual= 225 A ; C.S.A = 185 mm2

V.D = (mv / amp / m) x 10-3

x Iactual x L

V.D = 0.244 x 10-3

x 225 x 30 = 1.647 Volt

V.D % = (1.647/380) x 100 = 0.433 %

( From 2 ---- to ---- 3 )

138


139/272


Eng. Amr Mohamed



L = 100 mt ;Iactual= 75 A ; C.S.A = 50 mm2

V.D = (mv / amp / m) x 10-3

x Iactual x L

V.D = 0.72 x 10-3

x 75 x 100 = 5.4 Volt

V.D % = (5.4/380) x 100 = 1.42 %

From 1to3 :::::::: Total V.D = 1.42 + 0.433 = 1.835 % {Accepted}

( From 2 ---- to ---- 4 )

L = 300 mt ;Iactual= 150 A ; C.S.A = 120 mm2

V.D = (mv / amp / m) x 10-3

x Iactual x L

V.D = 0.341 x 10-3

x 150 x 300 = 15 Volt

V.D % = (15/380) x 100 = 4 %

From 1to4 :::::::: Total V.D = 4 + 0.433 = 4.435 % {Accepted}

Note:-

If total V.D % > 5 % ( not accepted ) we have to solve this problem.

As V.D = (mv / amp / m) x 10-3x Iactual x L , so if the (mv / A / m) reduced

the V.D will be reduced as well.

So, we select the next higher C.S.A cable.

C.S.A R (mv / A / m) V.D V.D%

139


140/272


Eng. Amr Mohamed



[3]

Short circuit calculation:-

For copper conductor: A = 9.1 IS.C mm2

For aluminum conductor: A = 14.2 IS.C mm2

Where:

A: Cable cross sectional area.

t: Operation time of C.B ( worst case = 1 Sec. ).

Is.c: short circuit current (KA)

V.D ,

Xsc

140


141/272


Eng. Amr Mohamed



Standard of cable trays is:

Height: 5cm, 8cm, 10 cm.

Width: 5cm, 10cm, 15cm, 20cm, 25cm, .., 100cm.

Cable tray in two levels:-

There are 3 and 4 levels, but standard distance between two levels 30cm.

Cable Tray

141


142/272


Eng. Amr Mohamed



Cross section area of pipes

Where

d: Cable diameter.

D: Pipe diameter.= %40

- :

Secondary of current transformer

: Pipes made from PVC.Trays.On ground.

-:

.Tray

142


143/272


Eng. Amr Mohamed



Fire Barrier

, .50

500

.

18040.

Compact 2-10 3 4-20 5- -6 7-15.

143


144/272


Eng. Amr Mohamed



18060

Compact 2-10 3-40 4-

144


145/272


Eng. Amr Mohamed

Ch: 6HVAC 4A Consultant and Training


Ch: 6 AIR CONDITION

Types of Air Condition:-

(1)H.V.A.C (Heating Ventilation Air Condition)(2)Split duct( DX) type(3)Split type.(4)Window type

(1) H.V.A.C:

Chiller0.5 mw

: (1)F.C.U (2)A.HU (3)Pumps (4)Chiller

(2) Split duct( DX) type:

Chiller (

(3) Split Unit:

Split duct. )

145


146/272


Eng. Amr Mohamed



(4) Window:

CompressorFanWindowSplit Compressor Split UnitFanCompressor

.Fan :

(1) Exhaust Fan:

. )(2) Smoke Exhaust Fan:

: 1-. 2-.

(3) Pressure Fan:

(4) Heater

A) Bath Room B) Kitchen Room

(5) Hand Driver

.

146


147/272


Eng. Amr Mohamed



(6) Cooling Tower . Note:

1-. 2-.

.

(1) H.V.A.C (Heating Ventilation Air Condition)(1)Chiller

147


148/272


149/272


Eng. Amr Mohamed



KVA(1) 1 Ton 1.5 HP

(2) 12000 BTU 1 Ton

(3) HP KVA

Chiller.

2) Pump(

Used with chiller to make supply and return to water from FCU and AHU to Chiller

No of pumps = no of chiller

Pump is closed to chiller

Pump motor(kw)

)Chiller & PumpKw .

Chiller Pump

149


150/272


Eng. Amr Mohamed



150


151/272


Eng. Amr Mohamed



l Unit)o(Fan Co(3) FCU

Cold water come from chiller and then returns to chiller as hot water

Operation:

(1)In Summer:

.

151


152/272


Eng. Amr Mohamed



Note:

Fan coil unit as Fan

KW or HP(2)In winter

Fan coil unit as heater and fan (Fan + Heater)

Fan


153/272


Eng. Amr Mohamed



In summer:Fan work only

In winterFan + Heater work

calculationLoad

In summer:

(1) Chiller (2) Pumps (3) Fans (F.C.U, A.H.U)

In winter:

(1) Fans (F.C.U, A.H.U) (2) Heaters (FCU, AHU)

Summer load > Winter Load

So,we sizing the transformer on summer load and sizing main distribution board on

winter loads

Transformer summer loads

Distribution boards Winter loads

Cables Winter loads

Riser Bus Bar Summer load

Note:

Maximum load absorbed from transformer when

1. Chiller 2. Pump 3. FanOperate (summer load)

Maximum load absorbed from distribution board when

1. Fans 2. Heaters Operate (winter load)

153


154/272


155/272


Eng. Amr Mohamed



projectExample on

:

1.Chiller

2.Air handling unit (AHU)

3.Fans coil (FCU)

4.Exhausted Fans (EF)

5.Water Pump

6.Sewerage Pump (SP)

7.Smoke Exhaust Fans (SEF)

(1) Basement Floor:

Quantity Cooling Load

Load (K.w)

Fan Heater

A.H U-05 1 30 TR 25 30

A.H U-06 1 30 TR 25 30

155


156/272


Eng. Amr Mohamed



(2) Ground Floor:

Quantity Cooling Load

Load (K.w)

Fan Heater

A.H U-01 1 27 TR 25 30

F.C U-02 4 15 TR 0.25 2

F.C U-04 8 2.5 TR 0.25 3

F.C U-05 6 3 TR 0.25 3

E.F 05 1 - 4

E.F 02 1 - 4

E.F 01 1 - 4

(3) First Floor:

Quantity Cooling LoadLoad (K.w)

Fan Heater

A.H U-01 1 30TR 25 30

F.C U-02 14 1.5 TR 0.25 2

F.C U-04 5 2TR 0.25 2

E.F 01 1 - 4

E.F 02 1 - 4

E.F 03 1 - 4

E.F 04 3 - 4

156


157/272


Eng. Amr Mohamed



(4) First Floor:

Quantity Cooling LoadLoad (K.w)

Fan Heater

A.H U-03 1 30TR 25 30

F.C U-02 17 1.5 TR 0.25 2

F.C U-04 2 2 TR 0.25 2

E.F 01 1 - 4

E.F 02 1 - 4

E.F 03 1 - 4

E.F 04 - 4

(5) Roof:

Quantity Cooling Load Load (K.w)

Chiller 2 150 TR 250

Pumps 3 - 30

SE FAN-01 1 - 30

SE FAN-02 1 - 30

SP ZF-01 1 - 10

SP ZF-02 1 - 10

157


158/272


Eng. Amr Mohamed



Winter load calculation

Design the distribution board

(1) Basement Floor:

A) AHU-05 (Fan = 25 Kw, heater = 30 Kw)

ILoad= 1.5 * 61.25 = 92 a

IC.B = 1.25 * ILoad= 115 A

CB = 125 A

from cables catalogs the suitable cable

4 X 70 + 35 Cu/ PVC/ PVC

Total load for AH-05

Total KWA = (1) (61,26) = 61.25 KWA

158


159/272


160/272


Eng. Amr Mohamed



Ground Floor

(1) AHU-01 (Fan = 25 Kw& Heater = 30 Kw)

C.B = 125 amp

Cable = 4 * 70 + 35 mm2Cu / PVC / PVC

(2) FCU-02 Q= 4

160


161/272


162/272


Eng. Amr Mohamed



Irated= 4.5 * 3.3 = 14.85A

IC.B = 10.35 * 1.25 = 13 A

C.B = 26 A MCB

Cables = 3 * 4 mm2Cu / PVC / PVC

Total Load = 8 * 3.3 = 26.4 KVA

(5) FCU-05 Q= 6

Fan = 0.25 Kw& Heater = 3 Kw

KVA = 3.3 KVA

Cables = 3 * 4 mm2Cu / PVC / PVC

Total Load = 6 * 3.3 = 20 KVA

(6) EXHAUS _ Fan 05,02,01

Fan = 4 Kw

ILoad= 4.5 * 5 = 22.5 A

IC.B = 28 A

C.B = 32 A Cable = 3 X 10mm2

Total Load = 15 KVA Cu / PVC / PVC

Main C.B:

Total Load of C.B = 61.25 + 9.2 + 18.4 + 26.4 + 20 + 15 = 150 KVA

ILoad= 4.5 * 150 = 225 A

C.B = 250 Amp

Cable = 4 * 185 + 95 Cu / PVC / PVC

162


163/272


Eng. Amr Mohamed



3. First Floor

A) AHU 0.2 Q = 1 (Fan 25 + Heater 30)

ILoad= 1.5 * 61.5 = 93 Amp

IC.B = 1.25 * 93 = 115 Amp

163


164/272


Eng. Amr Mohamed



We selected the suitable C.B = 125 Amp

So we select cable Cu / PVC / PVC 4 X 70 + 35 mm2

Total power (KVA) = 61.5 KVA

B) F.Cu-02 Q = 14 ( Fan 0.25 + Heater 2)

ILoad= 4.5 * 2.3 = 10.4 Amp

IC.B = 1.25 * 10.4 = 13 Amp

C.B = 16 Amp

Cable 3 X 3 mm2

Total power (KVA) = 14 * 2.3 = 32.2 KVA

(3) F.C.U-04 Q = 5 ( Fan 0.25 + Heater 2)

C.B = 16 Amp MCB

Cable = 3 X 3 mm2

(4) EF-01, 02, 03, 04 (4KW) Q=6

ILoad= 4.5 * 5 = 22.5 Amp

IC.B = 22.5 * 1.25 = 28 Amp

C.B 32 Amp MCB

164


165/272


Eng. Amr Mohamed



3 X 10 mm2

Cu / PVC / PVC

Total power (KVA) = 6 * 5 = 30 KVA

For main C.B and main cable

Total KVA = 61.5 + 32.2 +11.5 + 30

Total MA = 135.2 MA

Irated= 1.5 * 135.2 MA

IC.B = 1.25 * 202.8 = 253.5 Amp

Select C.B = 250 Amp MCCB

Main cables

4 * 185 + 95 mm2

Cu / PVC / PVC

165


166/272


167/272


168/272


169/272


170/272


Eng. Amr Mohamed



(3)WindowsConsist of :

(1)Grill(2)Compressor(3)Fan

Rating of window type

1.5 HP & 2.25 HP & 3 HP &4 HP & 5 HP

Calculation of Power

According to code

100 VA/mt2

EX-

FORROOM

5X6m

the suitable unit

HP KVA

Rating of window = 3 HP

170


171/272


172/272


Eng. Amr Mohamed

Ch:7 Lifts 4A Consultant and Training


Ch: 7 Lefts

1- 2- 3- -4

:-1-2-34--5 6-

172


173/272


174/272


Eng. Amr Mohamed



. /125,.075 ( (variable(variable speedvoltage(vvvf

:

174


175/272


176/272


Eng. Amr Mohamed



10001.5m/s .

176


177/272


Eng. Amr Mohamed



/10001.5 12Kw.

= 3*12=36Kw50% (

0.77./ 80% 36*0.77/0.8=34.65Kw

177


178/272


Eng. Amr Mohamed

Ch: 8 Bus Duct 4A Consultant and Training


Ch: 8 Bus Duct

Bus Duct Bus Duct

. Bus Duct

( ).

Bus Duct

Taps Plug-In Units .Bus Duct

) BusDuct Bus Duct

(ContinuousReliability BusDuct

.

178


179/272


Eng. Amr Mohamed



TrunkingBarBus

Specification of Bus Bar

(1) Type of Bus Bar trunking according to :

A) Forms D) Arranged

B) Conductor E) Loading

E) Feeding.

(2) Voltage operation.

(3) Cross section area and weight.

(4) Short circuit current rating.

(5) Voltage Drop.

Types of Bus Bar trunking

A) According to forms

(1) Air type (2) Sandwich type

179


180/272


181/272


182/272


Eng. Amr Mohamed



Total MA = 360 AmpWe must select C.B fvstly

I = 360 * 1.5 = 540 Amp

C.B = 630 Amp M.C.C.B

Note: B.B Rating = C. B rating

Because de-rating factor Y Bus Bar 1

D -according to arranged

(1) Half mentral

(R + S + T + FN + E) (R + S + T + FN + AL)

182


183/272


Eng. Amr Mohamed



E) According to Feeding

(1) Feeder Type

Bus bar used for fed one load only at the end of Bus Bar.

183


184/272


Eng. Amr Mohamed



Irated = 2000 * 1.5 = 3000 Amp

C.B = 3200 Amp safety factor = 1

As oil type transformer

B.B rating = 3200 Amp

(2) Plug in type

Bus Bar used to fed several loads by putting plug in (Top Off)

may by.. [Load Break switch (L.B.S), FUSE< Circuit + breaker]

184


185/272


186/272


187/272


Eng. Amr Mohamed

Ch: 9 Feeding System 4A Consultant and Training


Ch: 9 Feeding SystemsAccording to Cairo Electric Distribution Company (CEDC) there are five types

of feeding systems according to the load rating divided as follow:-

1)1sttype Used for loads less than 200 KVA

2)2ndtype

Used for 200 < loads < 1 MVA

3)3rd

type Used for 1 MVA < Loads < 1.25 MVA

4)4th

type Used for 1.25 < loads < 5 NVA

5)5th

type Used for loads > 5 MVA

Difference between the five types:-

1sttype

The feeding is from one source in radial system.

2ndtype

The feeding is from one source in ring system.

3rd type

The feeding is from two independent feeding sources and there are a

joint between them by using bus coupler located in the low voltage

network after the transformer.

187


188/272


Eng. Amr Mohamed



4thtype

The feeding is also from two independent feeding sources and there are ajoint between them by using bus coupler located in the medium voltage

and low voltage networks.

5thtype

Used if total load of project higher than 5 MVA. The feeding is from

distributer which may be 66/11KV or 66/22 KV.

Distributer

Main component of each distributer:-

There are two types of distributer:

[1] 16 cells [2] 14 cells

16 Cells type consists of:-

10 cells for outgoing.

4 cells for incoming.

2 cells for bus coupler.

188


189/272


190/272


191/272


Eng. Amr Mohamed



Part of distributor (16 Cells)

There are two Bus Bars each one contains 2 transformers incoming these

transformers may be with ratio of 66/11 or 66/22 KV.

The Bus Coupler is 2 out of 3 (2/3) and is used to connect two bus bars if

one of the incoming being out of service to insure power sustainability to

the loads.

191


192/272


193/272


194/272


Eng. Amr Mohamed



Main Components of Feeding System

1)Ring Main Unit (RMU)

194


195/272


196/272


Eng. Amr Mohamed



Note:

Rating of HRCF depend on transformer rating

For example: if a transformer with rating of 1MVA the current in the

MV side will be 52A.

I Fuse= safety factor * I rated = 1.25 * 52 = 65 A

So, select a Fuse = 80 A

Standard of fuse rating20, 25, 32, 40, 50, 63, 80, 100, 125, 160, 200, 250 and 400 A

The same as C.B Rating

-: .

3)Medium Voltage Cable:-

- M.V Cable before thetransformer is always

selected to be 3 * 240 mm2

AL / XLPE / STA / PVC

Because any M.V Cabledesign according to:

1.Current carry capacity2.Short circuit level.

196


197/272


Eng. Amr Mohamed



Example:

At 11 KVS.C MVA = 500 MVAAt 22 KVS.C MVA = 750 MVA

and at t = 1 Sec.

So, select C.S.A = 240 mm

4)TransformerUsed to convert the medium

voltage (11KV or 22KV) to

the low voltage (380V).

Types of transformer:-

There are two types of the

distribution transformers

A) Oil type transformer.

B) Dry type transformer.

The main difference

between the two types are summarized in the below table.

197


198/272


Eng. Amr Mohamed



Oil type Dry type

Operate at normal operation at 80% ofloading capacity

Operate at normal operation at 100% ofloading capacity

During over load operate at 100%May be operate at over load at 110% or

120%

Suitable location at outdoor Suitable location at indoor (Basement)

Low lowses

high

High losses

Oil>dryLow cost compared with dry type High cost compared with oil type

How to select the suitable transformer?

For example if you have the following loads:-

1) Lighting load = 140 KVA

2) HVAC load = 1800 KVA

3) Sockets Load = 50 KVA

4) Lifts load = 70 KVA

5) Ups load = 75 KVA

6) Fire pump = 50 KVA

7) Water pump = 5 KVA

So, the total connected load = 140 + 1800 + 50 + 70 + 75 + 5

Total load = 2140 KVA

198


199/272


200/272


201/272


202/272


203/272


204/272


205/272


206/272


Eng. Amr Mohamed



1stType

Used if load < 200 KVA

So we used 1st

type to fed The Residential Areas.

Feeding Residential area:- Each distributor contains 13 C.B

( C.B )13+ 1There are two bars each bar has 2 incoming equally loaded for high

reliability, and 4 outgoing and there are B.C to connect two Bus Bars each

out go to group of small RMUs

RMU LBS

. CB

. :

CB

.

206


207/272


208/272


209/272


Eng. Amr Mohamed



209


210/272


211/272


Eng. Amr Mohamed



211


212/272


213/272


214/272


215/272


216/272


Eng. Amr Mohamed



45 .

- (Coordination)

.

- 1500( Buss-Risers )

. - )(

( Plug-In 3220/380( + + 50 .

( IEC.)

. ) (

-( Diversity

Documents

Distribution Course