G + 5 ( Calculation Details )

8/10/2019 G + 5 ( Calculation Details )

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FIRSTPOINT CONSTRUCTION SERVICES

(INDIA ) PRIVATE LIMITED

Electrical Design for G+5 Building


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ABSTRCAT

PROJECT TITLE : -GROUND + 5 FLOORS RESIDENTIAL BUILDING.

NOVELTY :- This Project is about G + 5 Floors with 20 Flats. Each flat consisting of Two

Bed Rooms , Provided with SPLIT ACS in each room. OWER is distributed

throughout the building using 315 KVA T/F with IMPEDANCE (Z) = 5 % .

Also Power Backup of 100 KVA DG SET has been Provided for Ground Floor

(SMDBS) .

DESCRIPTION :- Illumination i.e. No.of Light Fixtures required in each room is Calculated by

considering Standard Lux Levels . Based on Lighting , Power , AC Loads

LOAD BALANCING SHEET has been provided to avoid phase difference

between each phases.T/F sizing is done base on calculated TCL & Similarly

D.G set sizing is done based on SMDBS Load. CAPACITOR BANKS have

been provided in grond floor for Inductive Loads.

Various Calculations that have been performed at each Feeder are as Follows

Cable Sizing is done based on Full Load Current (FLC).

Voltage Drop is done based on Resistance of the cable.

S.C is done based on Impedance of the cable.

Circuit Breaker (C.B) sizing is done based on FLC.

Cable Checking , Breaking Capacity and Tripping Time of C.B short

circuit currents (ISC) and No.of Runs of the cable.

Bus Bar Sizing is done at LT Side based on Current Carrying

Capacity (CCC) of the Material.

Earthing is done based on the calculated Max S.C Fault Current at

particular feeder.

Lighting Protection is also provided based on the Avg No.of Strokes

or Thunders of that particular area.

Finally, SLD and WIRING details have been shown using AUTO CAD

software.


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S. NO TOPIC PAGE. NO

1 Civil Layout 4

2

Total Connected Load ( TCL ) OF G + 5 BUILDING

Load Balancing Sheet of Flat 1 & 2

Load Balancing Sheet of Flat 3 & 4

5 - 31

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24

3 Transformer Sizing Calculation 32- 34

4 DG Set Sizing Calculation 3537

5 Capacitor Bank Sizing Calculation 38 , 39

6 Cable Sizing Calculation 40 - 48

7 Voltage Drop Calculation 49 - 60

8 Short Circuit Current Calculation 61 - 76

9 Circuit Breaker Rating 77 - 81

10Cable Checking , Breaking Capacity &

Tripping Time of Circuit Breaker82 - 93

11 LT Side Bus Bar Size Calculation 94 , 95

12 Earthing Calculation 96 - 98

13 Lighting Protection Calculation 99 - 101

14 Bill of Quantity ( BOQ ) 102

15 SUMMARY OF G + 5 PROJECT 103

16 LIST OF AUTO CAD DRAWINGS 104

Lighting Wiring Layout

Power Wiring Layout

AC Wiring LayoutMain Schematic Layout (SLD )

-

-

--


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TOTAL CONNECTED LOAD


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FORMULAES USED:-

ROOM INDEX ( R.I) =

()NOTE : - Given dimensions are converted from Feets to Meters.

No.of Light Fixtures NOTE : -

Coefficient of utilization factor (CUF), manufacturing unit (M.U) and lumens

output values are chosen from catalogue lamp output data.

Type of light fixture value depends on no. of lights in a frame.

Height (H) is taken as 2 mt i.e., from Floor Finishing Level (FFL) .

LIGHTING FIXTURE LOAD :-

Total Fixtures Load = Total No.of Lights Fixtures x Wattage

Ceiling Fan Load = No.of Ceilng Fans x Wattage

Exhaust Fan Load = No.of Exhaust Fans x Wattage

LIGHTING FIXTURE LOAD = Total Fixtures Load + Ceiling Fan

Load +Exhaust Fan Load

POWER LOAD FORMULAE:-

POWERLOAD = ( No.of 6A Sockets x 200 W ) + { No.of 16 A Sockets x

(7501500) W } + { No.of 20 A Socket x ( 2000 - 3000) W }

AC LOAD FORMULAE:-

STEP 1:- Given dimensions are converted to Square Feet (Sq. Ft).

STEP 2 :- As per standards from 10 x 10 ( 100 Sq. Ft ) to 10 x 12 ( 120 Sq. Ft)

choose 1 TR Split AC.


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STEP 3 :- Calculate required TR as Follows,

Room Sq. Ft

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

120 Sq. Ft

STEP 4 :- If calculated value is greater than 1.5 then choose 1.5 TR AC .

NOTE :-

1 TON RADIATION ( TR ) = 1400 Watts .

1.5 TR = 1400 X 1.5 = 2100 W , 2 TR = 1400 X 2 = 2800 W .

AC LOAD = Total No.of ACs X Wattages.

TOTAL CONNECTED LOAD ( TCL ) =

LIGHTING FIXTURE LOAD + POWER LOAD + AC LOAD


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No.of Fixture : (3.15x1.2x100)/(0.5x0.95x1x3350) = 0.356 1

Choose No.of Light Fixtures as 1

__________________________________________________________________________

Room Application : BED ROOM ( BR )

Lux Level : 200

Room Length : 3.15mt

Room width : 4.2 mt

Height of lighting fixture : 2.5mt

Type of Lighting Fixture : 1 x 36W( FL LAMP )

Room Index : (3.15 x 4.2)/ (3.15 +4.2)x 2.5=0.75

No.of Fixture : (3.15x4.2x200) / (0.5x0.95x1x3350) = 1.66 2


__________________________________________________________________________

Room Application : Dinning Area (DA)

Lux Level : 200


Room width : 2.44mt

Height of lighting fixture : 2.5 mt

Type of Lighting Fixture : 1 x 36 W( FL LAMP )

Room Index : (2.25 x 2.44 ) / (2.25 + 2.44 ) x 2.5 = 0.75



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Room Application : TOILET1 ( T1 )

Lux Level : 100

Room Length : 2.4 mt

Room width : 1.2 mt


Type of Lighting Fixture : 1 x 36 W ( FL LAMP )

Room Index : (2.4 x 1.2) / (2.4+1.2) x 2.5 = 0.32



__________________________________________________________________________

Room Application : TOILET2 ( T2 )

Lux Level : 100


Room width : 1.2 mt



Room Index : (2.4 x 1.2) / (2.4+1.2) x 2.5 = 0.32



_______________________________________________________________________

Room Application : Master Bed Room (MBR)


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Lux Level : 250

Room Length : 3.64

Room width : 3.19



Room Index : (3.64 x 3.19)/ (3.64 + 3.19)x2.5= 0.75



__________________________________________________________________________

Room Application : Hall

Lux Level : 300


Room width : 4.89



Room Index : (3.15 x 4.89)/ (3.15 + 4.89)x2.5= 0.76

No.ofFixture : (3.15x4.89x300)/(0.54x0.95x1x3350) = 2.86 3


__________________________________________________________________________

Consider Each Light Point as 36 W

A. Total No.of Fixtures = 13 Nos Lights ( 36 W FL LAMP)

a. Total Fixtures Load = Total No.of Lights x Wattage


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= 13 Nos x 36 W

= 468 Watts = 0.468 KW .

Total Fixtures Load = 468 Watts = 0.468 KW

b. Ceiling Fan Load = 5 Nos Ceiling Fan x 80 W

= 400 Watts

Ceiling Fan Load = 400 Watts = 0.4 KW

c. Exhaust Fan Load = 3 Nos Exhaust Fan x 75 W

= 225 Watts

Exhaust Fan Load = 225 Watts = 0.225 KW

A.LIGHTING FIXTURE LOAD = a + b + c = 468 + 400 + 225 = 1093 W

B. POWER LOAD ( Typical Flat 1 ) :-

i.Balcony : - 2 Nos 6 A Socket ( 200 W each ) = 400 Wii.Kitchen :- 1 No Micro oven (16 A Socket ) = 1000 W

1 No Rice Cooker(16 A Socket ) = 450 W

1 No Feezer Socket (16 A Socket) = 550 W

1 No Mixer (16 A Socket) = 450 W

iii.Bed Room :- 2 Nos 6 A Socket ( 200 W each ) = 400 W1 No Iron Box (16 A Socket) = 450 W

iv. Dinning Area:- 1 No 6 A Socket = 200 W

1 No Washng M/c (16A Socket) = 450 W

v. Toilet1 :- 1 No 6 A Socket = 200W1 No Water Heater (16A Socket) = 1000 W

vi. Toilet2 :- 1 No 6 A Socket = 200W1 No Water Heater (16A Socket) = 1000 W

vii. Master Bed Room :- 4 Nos 6 A Socket = 800 Wviii. Hall :- 4 Nos 6 A Socket = 800 W

TOTAL - 8,350 W

LIGHTING FIXTURE LOAD = 1093 Watts = 1.093 KW


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C. AC LOAD ( Typical Flat 1 ) :-

Note: As per standard 100 to 120 Square Feet area requires 1 Ton refrigeration

of Cooling.

a. Master Bedroom ( MBR ) :-

Room Dimension :- 122 x 108 = (12.16) x (10.66) = 129 Sq Ft.

Therefore Tonnage = 129 / 120 = 1.08 TR 1 TR AC = 1400 W

b. Bedroom ( BR ) :-

Room Dimension :-106x 142 = (10.5) x (14.16)= 148.68Sq Ft.


Total AC Load = 1400 + 1400 = 2800 W

TOTAL CONNECTED LOAD (TCL) = LIGHTING LOAD + POWER LOAD + AC LOAD

= 1093 + 8,350 + 2800

= 12,243W

After considering a Maximum Demand of 60 % Load

TCL = 12,243 x 60 % = 7,345.8 W = 7.34 KW .

POWER LOAD = 8,350 Watts = 8.35 KW

AC LOAD = 2,800 Watts = 2.8 KW

TCL OF FLAT - 1 = 12,243 Watts = 12.24 KW


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Now Similarly,

TCL for Flat 2 = 12.24 KW

TCL for Flat 1 + TCL for Flat 2 = 12.24 + 12 .42

= 24.48 KW

TOTAL CONNECTED LOAD OF FLAT 1 & 2 = 24,486W = 24.48 KW


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LOAD BALANCING SHEET FOR TYPICAL FLAT 1 & FLAT 2

( 3 - 8 WAY DISTRIBUTION BOARD)

CKT NO. CABLE SIZE MCB LOAD & LOCATION R Y B

R1 2.5 sq mm 5A KITCHEN + BAL + BR 415

Y1 2.5 sq mm 5A SPARE 0

B1 2.5 sq mm 5A DA + T 1 + T 2 + MBR 338

R2 2.5 sq mm 5A SPARE 0

Y2 2.5 sq mm 5A MBR + HALL 340

B2 2.5 sq mm 5A SPARE 0

R3 4sqmm 10A K(2 16A SKT-MIXER & RICE CKR) + BAL (1 6A SKT) 1100

Y3 4sqmm 10A SPARE 0

B3 4sqmm 16AK(1 16A SKT-FRZ) + BR(2 6A & 1 16A IRON BOX) +

BAL(1 6A SKT) 1600

R4 4sqmm 10A K(1 16A MICROOVEN SKT) 1000

Y4 4sqmmDA(1 6A&1 16 A WASH M/C SKT) + T 1 (1 16A

WATER HTR SKT) 1650

B4 4sqmm 10A SPARE 0

R5 4sqmm 10A SPARE 0

Y5 4sqmm 10A T 1 (1 6A SKT) + T 2 (1 6A SKT) + MBR (1 6A SKT) 600

B5 4sqmm 10A T 2 (1 16A WATER HTR SKT ) 1000

`

R6 4sqmm 10A MBR (3 6A SKT) + HALL (1 6A SKT) 800

Y6 4sqmm 10A HALL (3 6A SKT) 600

B6 4sqmm 10A SPARE 0

R7 6sqmm 16A BR ( 1 TR AC ) 426.6

Y7 6sqmm 16A BR ( 1 TR AC ) 426.6

B7 6sqmm 16A BR ( 1 TR AC ) 546.6

R8 6sqmm 16A MBR ( 1 TR AC ) 426.6

Y8 6sqmm 16A MBR ( 1 TR AC ) 426.6

B8 6sqmm 16A MBR ( 1 TR AC ) 546.6

4168.2 4043.2 4031.2

TOTAL 12242.6

Note- ADD DIVERSITY FACTOR OF 0.6 FOR RESIDENTIAL PURPOSE = 12.24 KW X 0.6 =7354.8 W =7.35KW


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Load Calculations of FLAT 3 4 :-

a.LIGHTING LOAD CALCULATION

a.

LIGHTING FIXTURE LOAD:

Room Application : HALL( H )

Lux Level : 300


Room width : 4.89mt



Room Index : (3.15 x 4.89) / (3.15 + 4.89) x 2.5 = 0.76

No.of Fixture : (3.15 x 4.89 x 300) /( 0.54 x 0.95 x 1 x 3350) = 2.68 2


__________________________________________________________________________

Room Application : KITCHEN ( K )

Lux Level : 300


Room width : 2.47mt

Height of lighting fixture : 2 .5mt


Room Index : (3.65 x 2.47)/ (3.65+2.47)x2.5= 0.59


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__________________________________________________________________________

Room Application : DINNING AREA ( DA )

Lux Level : 200


Room width : 2.2 mt



Room Index : (2.22 x 2.2) / (2.22 + 2.2 ) x 2.5= 0.44



__________________________________________________________________________

Room Application : TOILET 1 (T - 1)

Lux Level : 100

Room Length : 1.2mt

Room width : 2.2mt



Room Index : (1.2 x 2.2 ) / (1.2 + 2.2 ) x 2.5 = 0.31



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___________________________________________________________________________

Room Application : MASTER BEDROOM ( MBR )

Lux Level : 250


Room width : 4.25mt



Room Index : (3.65 x 4.25) / (3.65+4.25) x 2.5 = 0.78



__________________________________________________________________________

Room Application : BED ROOM ( BR )

Lux Level : 200


Room width : 4.25mt



Room Index : (3.15 x 4.25) / (3.15+4.25) x 2.5 = 0.72



Room Application : BALCONY ( BAL)


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Lux Level : 150

Room Length : 1.77

Room width : 1.2



Room Index : (1.77 x 1.2) / (1.77 + 1.2) x 2.5 = 0.367



__________________________________________________________________________

Room Application : TOILET 2 (T2)

Lux Level : 100

Room Length : 2.4mt

Room width : 1.2 mt



Room Index : (2.4 x 1.2) / (2.4 + 1.2) x 2.5 = 0.32



Room Application : TOILET 3 (T3)

Lux Level : 100


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Room Length : 2.4mt

Room width : 1.2 mt



Room Index : (2.4 x 1.2) / (2.4 + 1.2) x 2.5 = 0.32



_________________________________________________________________________

Consider Each Light Point as 36 W

Total No.of Fixtures = 14 Nos Lights ( 36 W FL LAMP)

a. Total Fixtures Load = Total No.of Lights x Wattage

= 14 Nos x 36 W= 504 Watts = 0.504 KW .

Total Fixtures Load = 504 Watts = 0.504 KW

b. Ceiling Fan Load = 5 Nos Ceiling Fan x 80 W

= 400 Watts

Ceiling Fan Load = 400 Watts = 0.4 KW

c. Exhaust Fan Load = 4 Nos Exhaust Fan x 75 W

= 300 Watts

Exhaust Fan Load = 300 Watts = 0.3 KW

A. LIGHTING FIXTURE LOAD = a + b + c = 504 + 400 + 300 = 1204 W


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A. LIGHTING FIXTURE LOAD = 1204 Watts = 1.204 KW

B. POWER LOAD ( Typical Flat 3 ) :-

I. Balcony : - 2 Nos 6 A Socket ( 200 W each ) = 400 W

II. Kitchen :- 1 No Micro oven (16 A Socket ) = 1000 W

1 No Rice Cooker (16 A Socket ) = 450 W1 No Freezer Socket (16 A Socket) = 550 W1 No Mixer (16 A Socket) = 450 W1 No 6 A Socket = 200 W

III. Bed Room :- 2 Nos 6 A Socket ( 200 W each ) = 400 W

1 No Iron Box (16 A Socket) = 450 W

IV. Dinning Area:- 1 No 6 A Socket = 200 W

1No Washng M/c (16A Socket) = 450 W

V. Toilet1 :- 1 No 6 A Socket = 200W

1 No Water Heater (16A Socket) = 1000 W

VI. Toilet2 :- 1 No 6 A Socket = 200W


VII. Master Bed Room :- 4 Nos 6 A Socket = 800 W

VIII. Hall :- 4 Nos 6 A Socket = 800 W

IX. Toilet3 :- 1 No 6 A Socket = 200W


TOTAL - 9,750 W

B.POWER LOAD = 9,750 W = 9,75 KW


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C. AC LOAD ( Typical Flat 3 ) :-

Note:- As per standard 100 to 120 Square Feet area requires

1 Ton refrigeration of Cooling.

a. Master Bedroom ( MBR ) :-

Room Dimension :- 122 x 142 = (12.66) x (140.166) = 179.35Sq

Therefore Tonnage = 179.35 / 120 = 1.49 TR 1.5 TR AC

= 1.5 x 1400 W= 2100 W

b. Bedroom ( BR ) :-

Room Dimension :- 106 x 142 = (10.5) x (14.16) = 148.75Sq Ft.


Total AC Load = 2100 + 1400 = 3500 W

C. AC LOAD = 3,500 W = 3.5 KW

TOTAL CONNECTED LOAD( TCL) = LIGHTING LOAD + POWER LOAD + AC LOAD

= 1204 + 9,750 + 3500

= 14,454 W

TCL OF FLAT 3 = 14,454 W = 14.45 KW

After considering a Maximum Demand of 60 % Load

TCL = 14,454 x 60 % = 8,672.4W = 8.67KW .

Now Similarly,

TCL for Flat 4 = 14.45 KW


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Therefore,

TCL for Flat 3 + TCL for Flat 4 = 14.45 + 14 .45

= 28.90 KW

TCL OF FLAT 3 & FLAT 4 = 14,454 W = 14.45 KW

TCL OF FOUR FLATS= TCL OF FLAT 1 & 2 + TCL OF FLAT 3 & 4

= 24,486 + 28,908

= 53,394 W

TCL OF 4 FLATS = 53,394 W = 53.394 KW

Note : - Each Floor 4 Flats , here total 5 Floors

TCL of Flats Load = 5 x TCL OF FOUR FLATS

= 5 x 53,394

= 266,970 W

TTCCLLOOFF FFLLAATTSSLLOOAADD==226666,,997700WW==226666..9977KKWW


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LOAD BALANCING SHEET FOR TYPICAL FLAT 3 & FLAT 4

( 3 - 8 WAY DISTRIBUTION BOARD)CKT NO. CABLE SIZE MCB RATING LOAD & LOCATION R Y B

R1 2.5 sq mm 5A HALL + KITCHEN 415

Y1 2.5 sq mm 5A SPARE 0

B1 2.5 sq mm 5A DA + T 1 + MBR 379

R2 2.5 sq mm 5A SPARE 0

Y2 2.5 sq mm 5A BR + BAL + T 2 + T3 410

B2 2.5 sq mm 5A SPARE 0

R3 4sqmm 10A HALL ( 4 6A SKT ) 800

Y3 4sqmm 10A K(1 16A SKT-FRZ & 1 6 A SKT) 750

B3 4sqmm 10A K(MICROOVE 16A SKT) 1000

R4 4sqmm 10A K(2 16A RICE CKR & MIXER SKT) 900

Y4 4sqmm 16A

DA(1 6A&1 16 A WASH M/C SKT)

+ T 1 (1 16A WATER HTR SKT) 1650

B4 4sqmm 10A

T 1 ( 1 6A SKT ) + MBR ( 4 6A SKT

) 1000

R5 4sqmm 10A

T 2 (1 16A WATER HTR SKT & 1

6A SKT ) + T 3(1 6A SKT) 1400

Y5 4sqmm 10A T 3 (1 16A WATER HTR SKT ) 1000

B5 4sqmm 10A

BR (1 6A SKT & 1 16A IRON BOX

SKT) 650

`

R6 4sqmm 10A SPARE 0

Y6 4sqmm 10A SPARE 0

B6 4sqmm 10A BR (1 6A SKT) + BAL (1 2 6A SKT) 600

R7 6sqmm 16A MBR ( 1.5 TR AC ) 780

Y7 6sqmm 16A MBR ( 1.5 TR AC ) 660

B7 6sqmm 16A MBR ( 1.5 TR AC ) 660

R8 6sqmm 16A BR ( 1 TR AC ) 426.6

Y8 6sqmm 16A BR ( 1 TR AC ) 426.6

B8 6sqmm 16A BR ( 1 TR AC ) 546.6

4721.6 4896.6 4835.6

TOTAL 14453.8Note :- ADD DIVERSITY FACTOR OF 0.6 FOR RESIDENTIAL PURPOSE = 14.45 KW X 0.6 =

8672.8 W = 8.67 KW


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2.LOAD CALCULATIONS OF DISTRIBUTIN BOARDSERVICE ( DBS)

DBS = CORRIDOR LIGHTING LOAD + GROUND FLOOR

PARKING LIGHTING LOAD +GROUND FLOOR PARKING

POWER LOAD

A. CORRIDOR LIGHTING LOAD :-

Room Application : CORRIDOR 1 (CORR1)

Lux Level : 150


Room width : 1.5 mt



Room Index : (16.14 x 1.5) / (16.14 + 1.5) x 2.5 = 0.54



____________________________________________________________________________

Room Application : CORRIDOR 2 (COR2)

Lux Level : 150


Room width : 1.2 mt




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Room Index : (23.7 x 1.2) / (23.7 + 1.2) x 2.5 = 0.4



__________________________________________________________________________

Now, Total lights required in Corridor of First Floor = 9 Nos Light

Here, Total 5 Floors i.e. 9 x 5 = 45 Lights

A. Corridor Lighting Load = 45 Nos x 24 W (CFL LAMP)

= 1080 W

A .Corridor Lighting Load = 1080 W = 0.108 KW

B. GROUND FLOOR PARKING LIGHTING LOAD :-

Room Application : GROUND FLOOR PARKING (GF - PARKING)

Lux Level : 100


Room width : 16.14 mt



Room Index : (23.7 x 16.14) / (23.7 + 16.14) x 2.5 = 3.84



__________________________________________________________________

B.GF Parking Lighting Load= 18 Nos x 36 W = 648 W


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B. GF Parking Lighting Load = 648 W

C.GROUND FLOOR PARKING POWER LOAD :-

FORMULAE FOR NO.OFSOCKETS:-

2 ( L + W )NO.OF SOCKETS = ----------------------

15

Room Application : GROUND FLOOR PARKING (GF - PARKING)


Room width : 16.14 mt

NO.OF SOCKETS = 2 ( 23.7 + 16.14) / 15 = 5.3 6 SOCKETS

3 Nos 6A Socket 200 Wand 3 Nos 16A Socket 1000 W

GFPARKING POWER LOAD = 3 X 200 W + 3 X 1000 W= 600 + 30000 = 3,600 W

C. GFPARKING POWER LOAD = 3,600 W = 3.6 KW

DISTRIBUTIN BOARDSERVICE ( DBS) LOAD

= A + B + C

= 1080 + 648 + 3,600 = 5,328 W

DBS LOAD = 5,328 W = 5.32 KW

MCC PANEL LOAD:-

MCC LOAD = LIFT LOAD + WATER SUPPLY (W.S) + FIRE FIGHTING LOAD (FF)

= 20 KW + 7.45 KW + 11.17 KW


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= 38.62 KW

3. MCC LOAD = 38.62 KW

TOTAL CONNECTED LOAD OF THE G + 5 BUILDING:-

TCL OF G+5 BUILDING =FLATS LOAD + DBS LOAD + MCC LOAD

= 266.97 KW + 5.32 KW + 38.66 KW

= 310.95 KW

TTCCLLOOFFGG++55BBUUIILLDDIINNGGoorrMMDDBBLLOOAADD==331100..9955KKWW

MAIN DISTRIBUTION BOARD ( MDB LOAD ) :-

SMDB-S + SMDB-1F+ SMDB-2F + SMDB-3F + SMDB-4F + SMDB-5F

= 43.88 KW + 53.39 + 53.39 + 53.39 + 53.39 + 53.939

= 310.95 KW

MDB LOAD = 310.95 KW


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TRANSFORMER SIZING CALCULATIONS


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TRANSFORMER CALCULATIONS

SIZING :-

TCL OF BUILDING = 310.95 KW

Add 10 % Extra for future purpose ,

TCL = 310.95 + 31.095 = 342.045 KW

Considering Maximum Demand of 60 % Load (Diversity Factor),

TCL = 342.045 x 0.6 = 205.22 KW

Convert KW Rating to KVA Rating using Formulae shown below

KVA= =

= 256.33 KVA

Efficiency of Transformer is 90 % .So, take Derating Factor as 0.9 toimprove Eff .

= 284.81 KVA

So, Proposed Transformer size is 315 KVAand is available in Market

CHECKING OF T/F WITH HIGHEST MOTOR RATING :-

Here Highest Motor Rating is Lift Load = 20 KW

KVA= =

=25 KVA

Add Derating Factor as 0.9 for Motor Voltage Fluctuations,

= 27.77 KVA

At the time of Starting Motor take 4 times the Full Load Current (FLC),

27.44 x 4 = 111.11 KVA


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We considered Normal Motor current while calculating TCL, so Deduct and add 4 times

FLC Load.

I.e. 284.8127.77 + 111.11 = 366.22 KVA ------ Eqn 1

T/F Tolerance Factor is taken as 1.5 ,

284.81 x 1.5 = 427.21 KVA ------ Eqn 2

Eqn 2 > Eqn 1, that implies Proposed T/F is Safe.

ProposedTransformer size is 315 KVA (Impedamce) Z = 5 %


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DG SET SIZING CALCULATIONS


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DG SET CALCULATIONS

Proposed DG SETis only for Ground Floor. ( SMDBS )

SIZING :-

TCL OF SMDB - S = 43.98 KW

Add 10 % Extra for future purpose ,

TCL = 43.98 + 4.398 = 48.37 KW

Considering Maximum Demand of 60 % Load (Diversity Factor),

TCL = 48.37 x 0.6 = 29.02 KW

Convert KW Rating to KVA Rating using Formulae shown below

KVA= =

= 36.27 KVA

Efficiency of Transformer is 90 % .So, take Derating Factor as 0.9 toimprove Eff .

= 45.33 KVA

So, Proposed DG Set size is 50 KVAand is available in Market

CHECKING OF DG SET WITH HIGHEST MOTOR RATING :-

Here Highest Motor Rating is Lift Load = 20 KW

KVA= =

= 25 KVA

Add Derating Factor as 0.9 for Motor Voltage Fluctuations,

= 27.77 KVA

At the time of Starting Motor take 4 times the Full Load Current (FLC)

27.44 x 4 = 111.11 KVA


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We considered Normal Motor current while calculating TCL, so Deduct and add 4 times

FLC Load.

50 27.77 + 111.11 = 133.34 KVA ------ Eqn 1

DG SET Tolerance Factor is taken as 1.5 ,

50 x 1.5 = 75 KVA ------ Eqn 2

Eqn 2 < Eqn 1, that implies Proposed DG SET not is Safe.

Choose 100 KVA DG SET which is Next Size available in the market.

DG SET Tolerance Factor is taken as 1.5 ,

1000 x 1.5 = 150 KVA ------ Eqn 3

Eqn3>Eqn 1, that implies Proposed DG SET is Safe.

Proposed DG SET size for Ground Floor is 100 KVAand is available in Market.


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CAPACITOR BANK SIZING CALCULATIONS

Capacitor Bank is used only for Motor Loads.

Here, Total MCC Load = 38. 66 KW

Formulae for Calculating Capacitor Size :-

Capacitor Bank = KW { Tan Cos -10.8Tan Cos -1( Req P.F ) }

Here , Req P.F is 0.96

Therefore , Capacitor Size = 17.74 KVAR 18 KVAR

Proposed , Capacitor Bank Size is 25 KVARand is available in Market.

Note:-

In order to minimize the cost we use Automatic Power Factor Relay Correction Panel (APFCP)

and here Choose 3 steps .

Step 1 10 KVAR ON / OFF




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CABLE SIZING CALCULATIONS


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CABLE SIZING CALCULATIONS

NOTE :-

Cable Size mainly depends on Load Current .

Current Carrying Capacity ( CCC ) Ranges as Follows

1 Sqmm Copper Cable = 2.2 A to 2.4 A

1 Sqmm Aluminium Cable = 1.2 A to 1.4 A

From 1 Sqmm to 16 Sqmm 4 Core Copper Cable is Chosen.

From 25 Sqmm and above 3 Core or 3.5 Core Aluminium Cable is Chosen.

For Underground Cable Derating Factor= 0.7.

Factors to be considered for Underground Cable are

a.)Soil Resistivity b) Temperature of Cable c.) Cable Trenches

d) Space between the Cables and e) grouping factor

For Cable being laid in Air Derating Factor = 0.8

For LT Cable ie, From Transformer Secondary Side to MDB always Choose

Underground Cable.

Formulaes Used:-

Load Currentis calculated using below equations

Case 1:- If Load is in KWthen P = 3 V I Cos (3) (Above 4 KW Load)

P = V I Cos (1)

Case 2:- If Load is in KVAthen P = 3 V I (3 )

P = V I (1 )

Case 3:- IfLoad is in KVARthen P = 3 V I Sin (3)

Where, = 36.8in all cases.


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In case Load Current amps is very High and not available in Catalogue

then No.of Runs need to be calculated using below equation.

()

Note:-

Diversity Factor is chosen based on Cable Lying i e. Air or Ground.

Chosen cable size must have less No.of Runs in order to reduce cost and Space

consumption .

_________________________________________________________________

HT SIDE CABLE (1)Standard cable size is considered.

Proposed cableis 1R X 3C X 240 SQMM ( Al )

___________________________________________________________________

LT SIDE CABLE (2)FROM T/F SECONDARY TO MDB

Consider Cable as Underground and Derating Factor is 0.7

Here, Load is 310.95 KW( 3) and using the Formulae provided

Load Current (I) = 540.76 A

Since, Load Current is high as mentioned above No.of Runs need to be calculated

based on Formulae provided.

.For 3.5 C x 400 SQMM (Al) Cable = CCC = 426 A

()

a. For 3.5 C x 300 SQMM (Al) Cable = CCC = 375 A No.of Runs = 2


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Similarly,

b. For 3.5 C x 300 SQMM (Al) Cable = CCC = 375 A No.of Runs = 2

c. For 3.5 C x 240 SQMM (Al) Cable = CCC = 333 A No.of Runs = 2

d. For 3.5 C x 185 SQMM (Al) Cable = CCC = 287 A No.of Runs = 3

Conclusion: - 240 SQMM Cable is chosen because its most Economicaland less cost.

Therefore, Proposed Cable is 2Rx 3.5C x 240 SQMM (Al).

_______________________________________________________________

SMDB - S CABLE (3)SUB MAIN DB SERVICE CABLE

Cable is in Air and Derating Factor is 0.8

Here, Load is 44KW( 3) and using the Formulae provided


Now, Check with Nearest CCC of various Cable Size from LT Catalogue

a.For 3 C x 25 SQMM (Al) Cable = CCC = 95 A x 0.8 = 76 A (Not Satisfied)b. For 3 C x 35 SQMM (Al) Cable = CCC = 117 A x 0.8= 93.8 (Satisfied)

So, Proposed SMDBSCable is 1R X 3C X 35 SQMM (Al)

_______________________________________________________________

SMDB1F (4),SMDB2F (5),SMDB3F (6),SMDB4F (7),SMDB5F(8)

CABLE OF 1st, 2

nd, 3

rd, 4

thand 5

thFloor


Here, Load is 53.39KW( 3) and using the Formulae provided


Now, Check with Nearest CCC of various Cable Size from LT Catalogue.

a.For 3 C x 25 SQMM (Al) Cable = CCC = 95 A x 0.8 = 76 A (Not Satisfied)

b.For 3 C x 35 SQMM (Al) Cable = CCC = 117 A x 0.8= 93.8 (Satisfied)

So, Proposed SMDBCABLE OF 1st, 2

nd, 3

rd, 4

thand 5

thFloor

1R X 3C X 35 SQMM (Al)


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DB - S CABLE (9)SMDBS to DB SERVICE CABLE


Here, Load is 5.32KW (3) and using the Formulae provided



a)For 4 C x 1.5 SQMM (Cu) Cable = CCC = 22 A x 0.8 = 17.6 A (Satisfied)

Note:-

Though, CCC of 1.5 SQMM Cu cable is Satisfied but it cannot withstand its

System Fault Current so, we Propose 6 SQMM Cu Cable

So, Proposed DBSCable is 1R X 4C X 6 SQMM (Cu) .

_______________________________________________________________

MCC PANEL CABLE (10)SMDBS to MCC PANEL CABLE





a) For 3 C x 16 SQMM (Al) Cable = CCC = 70 A x 0.8 = 56 A (Not Satisfied)

b) For 3 C x 25 SQMM (Al) Cable = CCC = 95 A x 0.8 = 76 A (Satisfied)

So, Proposed MCC PANEL CABLE is 1R X 3C X25 SQMM (Al) .

_______________________________________________________________

CAPACITOR BANK CABLE (11)MCC PANEL TO CAPACITOR BANK


Here, Load is 25 KVAR ( 3) and using the Formulae provided



a) For 3 C x 16 SQMM (Al) Cable = CCC = 70 A x 0.8 = 56 A (Not Satisfied)

b) For 3 C x 25 SQMM (Al) Cable = CCC = 95 A x 0.8 = 76 A (Satisfied)


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So, Proposed CAPACITOR BANK CABLE 1R X 3C X 25 SQMM (Al) .

_______________________________________________________________

LIFT MOTOR CABLE (12)MCC PANEL CABLE TO LIFT MOTOR


Here, Load is 20KW( 3) and using the Formulae provided



a) For 4 C x 6 SQMM (Cu) Cable = CCC = 51 A x 0.8 = 40.8 A (Satisfied)

So, ProposedLIFT MOTOR CABLE 1R X 4C X 6 SQMM (Cu) .

_______________________________________________________________

WATER SERVICE (W.S) MOTOR CABLE (13)MCC TO W.S MOTOR





a) For 4 C x 4 SQMM (Cu) Cable = CCC = 40 A x 0.8 = 32 A (Satisfied)

So, Proposed W.S MOTOR CABLE is1R X 4C X4 SQMM (Cu) .

_______________________________________________________________

FIRE FIGHTING (F.F) MOTOR CABLE (14)MCC TO F.F MOTOR






So, Proposed F.F MOTOR CABLE is 1R X 4C X 4 SQMM (Cu) .


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DBF1(15), DBF2 (16) OF SMDB 1FAND DBF1(19), DBF2 (20) OF

SMDB 2F CABLE .





a) For 4 C x 2.5 SQMM (Cu) Cable = CCC = 30A x 0.8 = 28 A (Satisfied)

Note:-

Though, CCC of 1.5 SQMM Cu cable is Satisfied but it cannot withstand its System Fault

Current so, we Propose 6 SQMM Cu Cable

So, Proposed Cable(15) , (16) , (19) , (20)is 1R X 4C X 6 SQMM (Cu) .

___________________________________________________________

DBF1(23), DBF2 (24) OF SMDB 3F,DBF1(27), DBF2 (28) OF SMDB

4FAND DBF1(31), DBF2 (32) OF SMDB 5F CABLE.






So, Proposed Cable(23),(24),(27),(28),(31),(32)is 1R X 4C X 2.5 SQMM (Cu) .

_______________________________________________________________

DBF3(17), DBF4 (18) OF SMDB 1F,DBF3(21), DBF4(22) OF SMDB 2F,

DBF3(25),DBF4(26) OF SMDB 3F,DBF3(29), DBF4 (30) OF SMDB 4F,

DBF3(33), DBF4 (34) OF SMDB 5F CABLE DB-F3, DB-F4 of SMDB1F to

DB-F3,DB-F4OF SMDB5F CABLE




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So, ProposedCable(17),(18),(21),(22),(25),(26),(29),(30),(33),(34) is

1R X 4C X 2.5 SQMM (Cu) .

_______________________________________________________________

DG SET CABLE (35)SMDB - S TO DG SET


Here, Load is 100 KVA ( 3) and using the Formulae provided



a) For 3 C x 70 SQMM (Al) Cable = CCC = 176 A x 0.8 = 140.8 A (Satisfied)

So, Proposed DG SET CABLE is1R X 3C X 70 SQMM (Al)

LIST OF PROPOSED CABLE SIZES

FEEDERN0 PROPOSED CABLE SIZE

1 1R X 3C X 240 SQMM ( Al )

2 2R x 3.5C x 240 SQMM (Al)

3 1R X 3C X 35 SQMM (Al)

4,5,6,7,8 1R X 3C X 35 SQMM (Al)

9 1R X 4C X 6 SQMM (Cu)



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15,16,19,20 1R X 4C X 6 SQMM (Cu)

23,24,27,28,31,32 1R X 4C X 2.5 SQMM (Cu)

17,18,21,22,25,26,29,30,33,34 1R X 4C X 2.5 SQMM (Cu)

351R X 3C X 70 SQMM (Al)


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VOLTAGE DROP SIZING CALCULATIONS


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VOLTAGE DROP SIZING CALCULATIONS

Voltage drop plays a very important role in operation of any Equipment.

Ex- If a system works with a minimum voltage of 207 V and in case 206 V is supplied thenParticular system doesnt works.

As per NBC Standards Voltage drop for Lighting Loop should be Less than 3 % to 5 % for

Each loop.

As per NBC Standards Voltage drop for Power Loop should be Less than 10 % to 15 % for

each loop.

In case , after adding the voltage drops for all feeders in Lighting (or) Power Loops & total

value of %VDis greater than 3 % or 10 %then Choose the Feeder with highest value

of%VD& change the Size of the Cable.

Formulaesused to Calculate %VD:-

VOLTAGR DROP ( VD )

%VD=


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Feeder 1HT Side

Given Data :-

a) Cable Size = 3c x 240 SQMM (Al)

b) Length of Cable (L) = 45 mt

c) Resistance of Cable (R) = 0.167 / KM

d) No.of Runs = 1e. Load Current = 16.53 A

Voltage Drop is calculated using the Formulae Provide,

VOLTAGR DROP ( VD ) = 0.2061 V

%VD=

= 0.00187

%VD= 0.00187

Feeder 2 LT Side

Given Data :-

a. Cable Size = 3.5C x 240 SQMM (Al) b. Length of Cable (L) = 35 mt c. Resistance of Cable

(R) = 0.167 / KM d. No.of Runs = 2 e. Load Current = 540.76 A



%VD=

= 0.639

%VD= 0.639


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Feeder 3 SMDB - S

Given Data :-

a. Cable Size = 3C x 35 SQMM (Al) b. Length of Cable (L) = 3 mt c. Resistance of Cable




%VD=

= 0.106639

%VD = 0.1066

Feeder 4(SMDB1F) ,Feeder 5(SMDB2F) , Feeder 6(SMDB3F) ,Feeder 7(SMDB - 4F),

Feeder 8 (SMDB5F)

Given Data :-



Voltage Drop is calculated using the Formulae Provided,


%VD=

= 0.129

%VD = 0.129


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Feeder 9 DB - S

Given Data :-

a. Cable Size = 4C x 6 SQMM (Cu) b. Length of Cable (L) = 2mt c. Resistance of Cable



VOLTAGR DROP ( VD ) = 0126 V

%VD=

= 0.0303

%VD = 0.0303

Feeder 10MCC PANEL

Given Data :-





%VD=

= 0.0836

%VD = 0.0863

Feeder 11 CAPACITOR BANK

Given Data :-



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VOLTAGR DROP ( VD )

= 0.463 V

%VD=

= 0.111415

%VD = 0.111

Feeder 12 LIFT MOTOR LOAD

Given Data :-

a. Cable Size = 4C x 6 SQMM (Cu) b. Length of Cable (L) = 30 mt c. Resistance of Cable




%VD=

= 1.719

%VD = 1.719

Feeder 13 WATER SERVICE MOTOR (W.S)

Given Data :-


(R) = 5.91 / KM d. No.ofRuns = 1 e. Load Current = 12.9 A



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VOLTAGR DROP ( VD ) = 3.301V

%VD=

= 0.795

%VD = 0.795

Feeder 14FIRE FIGHTING MOTOR (FF)

Given Data :-





%VD=

= 1.437

%VD = 1.437

NOTE:- For Power Loop i.e. ,

Feeder 1 + Feeder 2 + Feeder 10 + Feeder 9 + Feeder 11

= 0.00187 + 0.639 + 0.1066 + 0.0863 + 1.719 = 2.552 V

TOTAL %VDof one Power Loop = 2.552 V

Therefore, As per Standards % VD is less than 10 % and is satisfied.


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Feeder 15(SMDB1F TO DB F1) , Feeder 16(SMDB1F TO DB F2)

Given Data :-





%VD=

= 0.105

%VD = 0.105

NOTE :- For Lighting Loop i.e.,

Feeder 1 + Feeder 2 + Feeder 4 + Feeder 15

= 0.00187 + 0.639 + 0.1066 + 0.129 + 0.105 = 1.022 V

TOTAL %VDof one Lighting Loop = 1.022 V

Therefore, As per Standards % VD is less than 3% and is satisfied.


Given Data :-






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%VD=

= 0.247

%VD = 0.247


Given Data :-





%VD=

= 0.175

%VD = 0.175


Given Data :-





%VD=

0.371


%VD = 0.371


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Given Data :-

a) Cable Size = 4C x 2.5 SQMM (Cu) b. Length of Cable (L) = 7 mt c. Resistance of Cable




%VD=

= 0.5905

%VD = 0.5905


Given Data :-





%VD=

= 0.495


Given Data :-

a) Cable Size = 4C x 2.5 SQMM (Cu) b. Length of Cable (L) = 9 mt c. Resistance of Cable

(R) = 9.5 / KM d. No.ofRuns = 1 e. Load Current = 21.28 A

%VD = 0.495


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%VD= = 0.759


Given Data :-





%VD= = 0.169


Given Data :-

a. Cable Size = 4C x 2.5 SQMM (Cu) b. Length of Cable (L) = 11mt c. Resistance of Cable




%VD= = 0.928

%VD = 0.759

%VD = 0.619


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%VD = 0.928


Given Data :-




VOLTAGR DROP ( VD )

= 3.085 V

%VD=

= 0.743

Feeder 35 DG SET

Given Data :-

a) Cable Size = 3C x 70 SQMM (Al) b. Length of Cable (L) = 2 mt c. Resistance of Cable

(R)= 0.568 / KM d. No.ofRuns = 1 e. Load Current = 139.12 A



%VD=

= 0.0659

%VD = 0.0659

%VD = 0.743


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SHORT CIRCUIT ( S.C) CALCULATIONS

S.C Calculation is done to know the amount of fault current passing through a cable.

S.C Current Values For Different C.B :-

a.Miniature Circuit Breaker ( MCB ) - 10 KA to 16 kA( 0.5 A to 63 A )

b.Moulded Case Circuit Breaker ( MCCB ) - 25 KA to 35 kA

( 63 A to 800 A )

c.VCB / ACB / SF6C.B - 50 KA to 65 kA

(800 A to 6300 A )

FORMULAES USED TO CALCULATE S.C CURRENT

Impedance of cable

Pu Impedance of T/F

Pu Imp of Cable =

Pu Impedance based on System Voltage=

( )

Total Pu Imp =Previous Pu Imp + Present Pu Imp

HT Side PreviousPu Imp is POS

Pu Impedance based at POS=

Note:-While Changing a Loop PrevPu Imp has to be taken carefully


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Max Fault Current Upto C.B=

Total Max Fault Current = _________________________________________________________________

Feeder 1:- HT SIDE

Given Data :-

a. Cable Size = 3C x 240 SQMM (Al) b. Length of Cable (L) = 45mt c. Cable Impedance

= () ()= 0.157 / KM d. No.of Runs = 1e) Maximum Fault Current = 350 MVA (Taken from nearest Substation) and

f) Minimum Fault Current or Base MVA = 100 MVA (Taken from nearest Substation)

Calculate S.C Current by using FormulaesProvided,

Pu Impedance of 45 mtCable = = 0.007065Pu

Pu Impedance based on System Voltage =

() = 0.00632231405Pu

Total Pu Imp = Previous Pu Imp + Present Pu Imp

HT Side PreviousPu Imp is at POS

Pu Impedance at Point of Supply (POS) = = 0.285714285Pu

Total Pu Imp = 0.285714285 + 0.00632231405 = 0.292036599Pu


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MAX FAULT CURRENT UPTO C.B (1) =

= 342.422834 MVA

Note : - 342.422834 MVA > 350 MVA ( Given Max Fault Current )

TOTAL MAX FAULT CURRENT =

= 17.973KA 18 KA

S.C Current at HT Side = 18 KA

Only For T/F:

Pu Impedance of T/F =

= 15.87301587Pu

Total PuImp of T/F = Previous Pu Imp + Present Pu Imp

= 0.292036599 + 15.87301587= 16.16505247Pu

Feeder 2 :-LT SIDEAlways Choose Underground Cable .

Given Data :-

a. Cable Size = 3C x 240 SQMM (Al) b.Cable Length (L) = 35 mt c. Cable Impedance

=() ()= 0.157 / Kmd. No.of Runs = 2

Pu Impedance of 35 mt Cable =

= = 0.0027475 Pu


() = 1.59529685Pu


= 16.16505247+ 1.59529685=17.76034932Pu


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= 5.630519 MVA


= 7.833 KA

S.C Current at LT Side = 7.833 KA

Feeder 3(SMDB-S) , Feeder 4(SMDB-1F) , Feeder 5(SMDB-2F) , Feeder 6(SMDB-3F),Feeder 7(SMDB-4F) ,Feeder 8(SMDB-15F)

Given Data :-

a. Cable Size = 3C x 35 SQMM (Al) b. Length of Cable (L) = 3 mt c. Cable Impedance

= () ()= 1.046 / KM d. No.of Runs = 1

Pu Impedance of 3mt Cable =

= 0.003138Pu


() = 1.822035128Pu


= 17.76034932 +1.822035128= 19.58238445Pu

MAX FAULT CURRENT UPTO C.B (3,4,5,6,7.8) =

= 5.106630 MVA

TOTAL MAX FAULT CURRENT = = 7.104KA


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S.C Current at SMDB-S,SMDB-1F,SMDB-2F,SMDB-3F,SMDB-4Fand

SMDB-5F = 7.10 KA

Feeder 9 :- DB -S

Given Data :-

a. Cable Size = 4C x 6 SQMM (Cu) b. Length of Cable (L) = 2mt c. Cable Impedance

= () () = 3.69 / KM d. No.of Runs = 1


= 0.00738Pu


() = 4.285092176 Pu


= 19.58238445+ 4.285092176= 23.86747663Pu


= 4.189801 MVA

TOTAL MAX FAULT CURRENT = = 5.829 KA

S.C Current at DB-S = 5.82 KA

Feeder 10 :- MCC PANEL

Given Data :-


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= () () = 1.44 / KM d. No.of Runs = 1


= 0.00432Pu


() = 2.50834664Pu


= 19.58238445+ 2.50834664 = 22.09073109Pu


= 4.526785 MVA


= 6.297KA

S.C Current at MCC PANEL = 6.29 KA

Feeder 11 : MCC PANEL To Capacitor Bank

Given Data :-


= () () = 1.44 / KM d. No.of Runs = 1


= 0.00432Pu


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() = 2.50834664 Pu


= 22.09073109+ 2.50834664 = 24.59907773Pu


= 4.065193 MVA


= 5.655KA

S.C Current at Capacitor Bank = 5.65 KA

Feeder 12MCC PANEL To Lift Motor

Given Data :-

a. Cable Size = 4C x 6 SQMM (Cu) b. Cable Length (L) = 30mt c. Cable Impedance

= () ()= 4.61 / KM d. No.of Runs = 1

Pu Impedance of 3mt Cable = = 0.1383Pu


() = 80.40193061Pu


= 22.09073109+ 80.40193061= 102.4926617Pu


= 9.75679 MVA

TOTAL MAX FAULT CURRENT = = 1.357 KA


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S.C Current at Lift Motor = 1.35 KA

Feeder 13MCC PANEL To Water Service(W.S) Motor

Given Data :-


= () ()= 5.53 / KM d. No.of Runs = 1


Pu Impedance based on System Voltage = () = 77.36971984Pu


= 22.09073109+ 77.36971984= 99.46045093Pu

MAX FAULT CURRENT UPTO C.B (13) = = 1.00542 MVA


= 1.398 KA

S.C Current at W.S Motor = 1.39 KA

Feeder 14MCC PANEL To Fire Fighting (F.F) Motor

Given Data :-


= () () = 5.53 / KM d. No.of Runs = 1



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() = 92.84366381Pu


= 22.09073109+ 92.84366381= 114.9326098Pu


= 0.870075 MVA


= 1.210 KA

S.C Current at F.F Motor = 1.21 KA

Feeder 15 ( SMDB-1F to DB F1), Feeder 16 ( SMDB-1F to DB F2):-

Given Data :-


= () () = 3.69 / KM d. No.of Runs = 1

Pu Impedance of 3mt Cable =0.01107Pu

Pu Impedance based on System Voltage = 6.427638264Pu


=19.52238445+6.427638264 = 25.95002271Pu

MAX FAULT CURRENT UPTO C.B (15,16) = 3.853561MVA

TOTAL MAX FAULT CURRENT = 5.36KA

S.C Current at Feeder 15,16 = 5.36 KA


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Feeder 17( SMDB-1F to DB F3), Feeder 18 ( SMDB-1F to DB F4) :-

Given Data :-


= () () = 5.53 / KM d. No.of Runs = 1

Pu Impedance of 4mt Cable = 0.022121616Pu

Pu Impedance based on System Voltage = 12.84460212Pu


=19.52238445+12.84460212 = 32.36698657Pu

MAX FAULT CURRENT UPTO C.B (17,18) = 3.089567 MVA

TOTAL MAX FAULT CURRENT = 4.29KA


Feeder 19 ( SMDB-2F to DB F1), Feeder 20( SMDB-2F to DB F2) :-

Given Data :-


= () () = 3.69 / KMd. No.of Runs = 1 Pu Impedance of 5mt Cable =0.01845Pu

Pu Impedance based on System Voltage= 10.71273044Pu


=19.52238445+ 10.71273044=30.23511489Pu



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TOTAL MAX FAULT CURRENT= 4.60 KA



Given Data :-


= () ()= 5.53 / KM d. No.of Runs = 1

Pu Impedance of 6mt Cable = 0.03318Pu



=19.52238445+19.26549572=38.78788017Pu




Feeder 23 ( SMDB-3F to DB F1) , Feeder 24 ( SMDB-3F to DB F2) :-\

Given Data :-

a. Cable Size = 4C x 2.5 SQMM (Cu) b. Cable Length (L) = 7mt c. Cable Impedance

= () () = 8.88 / KM d. No.of Runs = 1 Pu Impedance of 7mt Cable = 0.06216Pu

Pu Impedance based on System Voltage= 36.09232109 Pu



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=19.52238445+ 36.09232109= 55.61470559 Pu





Given Data :-


= () () = 5.53 / KM d. No.of Runs = 1 Pu Impedance of 8mt Cable = 0.04424 Pu



=19.52238445+25.68732762 = 45.20971297 Pu





Given Data :-




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=19.52238445+46.40441283 = 65.92679728Pu




Feeder 29 ( SMDB-4F to DB F3) , Feeder 30 ( SMDB-4F to DB F4):-

Given Data :-


= () ()= 5.53 / KM d. No.of Runs = 1 Pu Impedance of 10mt Cable = 0.0553 Pu



=19.52238445+32.10915953 = 51.63154398 Pu





Given Data :-



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=19.52238445+56.71650457=76.2388902Pu





Given Data :-





=19.52238445+38.53099144 = 58.05337589 Pu




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Feeder 35 ( SMDB-S TO DG SET ) :-

Given Data :-

a. Cable Size = 3C x 70 SQMM (Al) b. Cable Length (L) = 2mt c. Cable Impedance




=19.52238445+0.615473944 =20.13785839Pu

MAX FAULT CURRENT UPTO C.B (35) = 4.965771 MVA


S.C Current at Feeder 35 = 6.90 KA


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CIRCUIT BREAKER (C.B) RATING CALCULATIONS


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CIRCUIT BREAKER(C.B) RATING CALCULATIONS

FormulaesUsed:-

Load Currentis calculated using below equationsCase 1:- If Load is in KWthen P = 3 V I Cos (3 ) (Above 4 KW Load)

P = V I Cos (1 )

Case 2:- If Load is in KVAthen P = 3 V I (3 )

P = V I (1 )

Case 3:- IfLoad is in KVARthen P = 3 V I Sin (3 )

Where, = 36.8in all cases.

C.B Rating = 1.25 x Full Load Current (FLC)

__________________________________________________________________________

HT SIDE :- C.B (1)

Note:-

a. As per Local standardsC.B Rating is 200A VCB / ACB.

b.As per International standardsC.B Rating is 630A VCB / ACB.

Here, C.B Size is chosen according to Local Standards

Therefore, Proposed C.B(1) Rating is 200 A VCB.

LT SIDE :- C.B (2) P = 310.95 KW

Load Current is calculated by using Formulaes Provided

i e. FLC = 540.76 A .

C.B Raring = 1.25 x 540.76 = 675.95 A

Therefore, Proposed C.B(2) Rating is 800 A ACB.

SMDB-S :- C.B (3) P = 43.98 KW


i e. FLC = 76.48A ,


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C.B Raring = 1.25 x 76.48 = 95.6 A

Therefore, Proposed C.B(3) Rating is 100 A MCCB.

SMDB -1F C.B (4), SMDB - 2F C.B (5),SMDB - 3F C.B (6),SMDB - 4F C.B (7),

SMDB - 5F C.B (8) have same Rating P = 53.39 KW


i e. FLC = 92.84A ,

C.B Raring = 1.25 x 92.84 = 116.05 A

Therefore, Proposed C.B(4),C.B(5),C.B(6),C.B(7),C.B(8) Rating is 125 A MCCB

DB-S C.B (9) P = 5.32 KW


i e. FLC = 9.25A ,

C.B Raring = 1.25 x 9.25 = 11.56 A

Therefore, Proposed C.B(9) Rating is 16 A MCB.

MCC PANEL C.B (10) P = 38.66 KW


i e. FLC = 67.22A ,

C.B Raring = 1.25 x 67.22 = 84.03 A


MCC TO CAPACITOR BANK C.B (11) P = 25 KVAR


ie. FLC = 57.96A ,

C.B Raring = 1.25 x 57.96 = 72.45 A


MCC TO LIFT MOTOR C.B (12) P = 20 KW


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i e. FLC = 34.77A ,

C.B Raring = 1.25 x 34.77 = 43.77 A


MCC TO WATER SERVICE (W.S) C.B (13) P = 10 HP = 10 X0.745 = 7.45 KW


i e. FLC = 12.95A ,

C.B Raring = 1.25 x 12.95 = 16.19 A

Therefore, Proposed C.B(13) Rating is25 A MCB.

MCC TO FIRE FIGHTING (FF) C.B (14) P = 15 KW = 15 X 0.745 = 11.17 KW


i e. FLC = 19.43A ,

C.B Raring = 1.25 x 19.43 = 24.2 A


DB F1 AND DB F2OF SMDB 1F,SMDB 2F,SMDB 3F,SMDB 4F,SMDB 5F

C.B 15,16,19,20,23,24,27,28,31,32 P = 12.24 KW


i e. FLC = 21.28A ,

C.B Raring = 1.25 x 21.28 = 26.60 A

Therefore, Proposed C.B15,16,19,20,23,24,27,28,31,32 Rating is 32 A MCB.

DB F3 AND DB F4OF SMDB 1F,SMDB 2F,SMDB 3F,SMDB 4F,SMDB 5F

C.B 17,18,21,22,25,26,29,30,33,34 P = 14.45 KW


i e. FLC = 25.12A ,


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C.B Raring = 1.25 x 25.12 = 31.41 A

Therefore, Proposed C.B 17,18,21,22,25,26,29,30,33,34 Rating is 32 A MCB.

SMDB - S TO DG SET C.B (35) P = 100 KVA


i e. FLC = 139.12A ,

C.B Raring = 1.25 x 139.12 = 173.89 A

Therefore, Proposed C.B (35) Rating is 180 A MCCB.


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82

CABLE CHECKING,C.B BREAKING CAPACITY AND

TRIPPING TIME CALCULATIONS


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83

CABLE CHECKING, C.B BREAKING CAPACITY AND TRIPPING TIME

Full Data of C.B ----- -

RATING OF C.B is calculated based on FLC

BREAKING CAPACITY IS S.C CURRENT I e.SystemFault Current.

Tripping Time is Duration of Fault Current

List of Formulaes used :-

1. HT Side :- t = 3 sec ( Constant )

Cable withstanding capacity =

Where, K = Constant value for material , A =Area of Cable (I e. Sqmm of cable)& t = Tripping

Timeof C.B

Here K = From HT cable Catalogue For Al = 0.143

2. LT side : - t= 0.01 sec to 1 sec (I e. Check for t = 0.9,0.8,07 ..)


ISC` = Short Circuit Current taken from Gems Cab S.C Rating Catalogue

,Proposed Cable is Safe when

{WITHSTANDING CAPACITY OF CABLE}> {SYSTEM FAULT CURRENT}

S.C Current of Particular Feeder

RATING OF C.B (Amps)

BREAKING

CAPACITY

(KA)

TRIPPING

TIME

(SEC)


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Feeder N0 1HT SIDE t = 3 Sec ( Constant ) , K = 0.143 , A = 1R x 3C x 240 Sq mm(Al)

Using Formulaes Provided :-


= 19.81 KA

Full Data of C.B

Cable withstanding capacity@ 3 sec = 19.81 KA ----- eqn 1

System Fault Current I e. Calculated S.C Current at HT Side is

= 17.97 KA ------ eqn 2

Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.

____________________________________________________________________________

Feeder N0 2LT SIDE Cable 2R x 3C x 240 Sq mm (Al) Isc= 18.18 @ t = 1 Sec

Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue



= 36.36 KA

Full Data of C.B

Cable withstanding capacity@ 1 sec = 36.36 KA----- eqn 1


= 7.85 KA ------ eqn 2


____________________________________________________________________________

Feeder N0 3 SMDB-S

Cable 1R x 3C x 35 Sq mm (Al) Isc= 2.65 @ t = 0.1 Sec



200 A VCB

18 KA 3 Sec

800 A (ACB)

7.85 KA 1 Sec


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= 8.38 KACable withstanding capacity@ 0.1 sec = 8.38 KA----- eqn 1 Full Data of C.B


= 7.12 KA ------ eqn 2


_____________________________________________________________________________

Feeder N0 4,5,6,7 & 8 SMDB 1F to SMDB 5F

Cable 1R x 3C x 35 Sq mm (Al) Isc= 2.65 @ t = 0.1 Sec






= 7.12 KA ------ eqn 2

Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe._____________________________________________________________________________

Feeder N0 9 SMDBS to DB - S 1R x 3C x 6 Sq mm (Cu) Isc= 0.69 @ t = 0.01 Sec






100 A MCCB

7.12 KA 0.1 Sec

125 A MCCB

7.12 KA 0.1 Sec

16 A MCB

5.85 KA 0.01 Sec


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= 5.85 KA ------ eqn 2


_____________________________________________________________________________

Feeder N0 10 SMDBS to MCC PANEL 1R x 3c x 25 Sqmm (Al) Isc= 1.89 @ t = 0.08s




= 6.68 KA

Cable withstanding capacity@ 0.08 sec = 6.68 KA----- eqn 1 Full Data of C.B


= 6.31 KA ------ eqn 2


_____________________________________________________________________________

Feeder N0 11 MCC PANEL to CAPACITOR BANK

Cable 1R x 3c x 25 Sqmm (Al) Isc= Short Circuit Current Rating = 1.89 (Taken from

Gems Cab S.C Rating Catalogue) @ t = 0.1 Sec



= 5.97 KA



= 5.65 KA ------ eqn 2


Feeder N0 12 MCC PANEL to Lift Motor 1R x 4c x 6 Sqmm (Cu) Isc= 0.69 @ t = 0.2Sec

100 A MCB

6.31 KA 0.08 Sec

80 A MCCB

5.65 KA 0.1 Sec


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Isc= S.C Circuit Current Rating = (Taken from Gems Cab S.C Rating Catalogue)



= 1.54 KA



= 1.35 KA ------ eqn 2


____________________________________________________________________________

Feeder N0 13 MCC PANEL to Water Service (W.S) Motor

Cable 1R x 4c x 4 Sqmm (Cu) Isc= 0.46 @ t = 0.1 Sec






= 1.39 KA ------ eqn 2


_____________________________________________________________________________

Feeder N0 14 MCC PANEL to Fire Fighting (F.F) Motor

Cable 1R x 4c x 4 Sqmm (Cu) Isc= 0.46 @ t = 0.1 Sec


60 A MCB

1.35 KA 0.2 Sec

25 A MCB

1.39 KA 0.1 Sec


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= 1.45 KA



= 1.21 KA ------ eqn 2


_____________________________________________________________________________

Feeder N0 15 ,16DBF1(15) , DBF2(16) OF SMDB1F

Here For Cable 1R x 4C x 6 Sqmm (Cu) Isc= 0.69 @ t = 0.01 Sec


Using FormulaesProvided :-


= 6.9 KA



= 5.36 KA ------ eqn 2


____________________________________________________________________________

Feeder N0 19,20DBF1(19) , DBF2(20) OF SMDB2F

Here For Cable 1R x 4C x 6 Sqmm (Cu) Isc= 0.69 @ t = 0.01 Sec



Cable withstanding capacity = = 6.9 KA

25 A MCB

1.21 KA 0.1 Sec

32 A MCB

5.36 KA 0.01 Sec


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Cable withstanding capacity@ 0.01 sec = 6.9 KA----- eqn 1

Full Data of C.B


= 4.59 KA ------ eqn 2


____________________________________________________________________________

Feeder N0 17,18DBF3 (17)& DBF4(18) OF SMDB1F

Here For Cable 1R x 4c x 4 Sq mm (Cu) Isc= 0.46 @ t = 0.01 Sec




= 4.6 KACable withstanding capacity @ 0.01 sec = 4.6 KA----- eqn 1

Full Data of C.B


= 4.29 KA ------ eqn 2


____________________________________________________________________________

Feeder N0 21,22DBF3 (21)& DBF4(22) OF SMDB2F

Here For Cable 1R x 4c x 4 Sq mm (Cu) Isc= 0.46 @ t = 0.01 Sec




= 4.6 KA

Cable withstanding capacity @ 0.01 sec = 4.6 KA----- eqn 1

32 A MCB

4.59 KA 0.01 Sec

32 A MCB

4.29 KA 0.01 Sec


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Full Data of C.B


= 3.58 KA ------ eqn 2

Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe

____________________________________________________________________________

Feeder N0 23,24DBF1(23)& DBF2(24) OF SMDB3F

Here For Cable 1R x 4C x 2.5 Sq mm (Cu) Isc= 0.287 @ t = 0.01 Sec





Full Data of C.B


= 2.50 KA ------ eqn 2


____________________________________________________________________________

Feeder N0 27,28DBF1(27)& DBF2(28) OF SMDB4F





= 2.87 KA


32 A MCB

3.58 KA 0.01 Sec

32 A MCB

2.50 KA 0.01 Sec


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Full Data of C.B


= 2.11 KA ------ eqn 2


____________________________________________________________________________

Feeder N0 25,26DBF3 (25)& DBF4(26) OF SMDB3F t = (0.01 to 1 ) Sec

Here For Cable 1R x 4c x 4 Sqmm (Cu) Isc= 0.46 @ t = 0.01 Sec





Full Data of C.B


= 3.07 KA ------ eqn 2


____________________________________________________________________________


Here For Cable 1R x 4c x 4 Sqmm (Cu) Isc= 0.46 @ t = 0.02Sec




= 3.25 KA


32 A MCB

2.11 KA 0.01 Sec

32 A MCB

3.07 KA 0.01 Sec


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Full Data of C.B


= 2.69 KA ------ eqn 2


____________________________________________________________________________

Feeder N0 31,32DBF1(31)& DBF2(32) OF SMDB5F t = (0.01 to 1 ) Sec





= 2.02 KA


Full Data of C.B


= 1.82 KA ------ eqn 2


____________________________________________________________________________


Here For Cable 1R x 4c x 4 Sqmm (Cu) Isc= 0.46 @ t = 0.02Sec




= 3.25 KA

32 A MCB

2.69 KA 0.02 Sec

32 A MCB

1.82 KA 0.02 Sec


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Full Data of C.B


= 2.39 KA ------ eqn 2


____________________________________________________________________________

Feeder N0 35 SMDBS to DG SET t = (0.01 to 1 ) Sec

Here For Cable 1R x 3c x 70 Sqmm (Al) Isc= 5.30 @ t = 0.5 Sec




= 7.49 KA


Full Data of C.B


= 6.90 KA ------ eqn 2


32 A MCB

2.39 KA 0.02 Sec

180 A MCCB

6.90 KA 0.5 Sec


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94

LT SIDE - BUSBAR SIZING CALCULATION


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95

LT SIDE - BUSBAR SIZING CALCULATION

As per Standard for 1 SQMM Cu Current Carrying Capacity (CCC) is 1.6 A

As per Standard for 1 SQMM Al Current Carrying Capacity (CCC) is 0.8 A

LT SIDEHere Cu material is used for Bus Bar I e . CCC = 1.6 A / SQMM

Full Load Current (FLC) = 540.76 A

SIZE OF BUS BAR = ( )

( )

=

= 337.97 SQMM

Therefore, Proposed Copper Bus Bar Sizeavailable in Market is,

SIZE OF BUS BAR = 75 x 6 (Width x Thickness)


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EARTHING CALCULATION


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97

EARTHING CALCULATION

Earthling is calculated based on Maximum S.C Current Value.

Here Max Fault Current is at HT SIDE: - 18 KA

Earthing resistance should always be less than 1 .

LIST OF FORMULAES USED :-

EARTH STRIP SIZE =

ISC= Max Fault Current , t = Duration of TimeK = Constant Value of Conductivity (Cu = 118, Al = 90, GI = 80 )

Here K = 118 (Cu) ISC= 18000 A and t =3 Sec

= 264.2

Nearest Earth Strip Sizeavailable in Market = 75 x 6 (Width x Thickness)

EARTH PIT RESISTANCE ( Rrod)= ( )

r= Resistivity of Soil = 0.2 / km (Taken from soil Test Eng )L= Length of rod = 3 mt

d= Diameter of rod ( 40mm or 65mm or 75mm) = 65 mm = 0.065 mt

Earth Pit Resistance ( Rrod)= ( )

Earth Resistance should be less than 1 .So increase N0.of Pits by using Formulae

Earth Resistance=

Case 1:- If No.of Pits are 2 then Earth Resistance =

= 1.2 (Not Sufficient)

Case 2:- If No.of Pits are 3 then Earth Resistance = = 0.8 (Sufficient) for safety

purpose go for another pit .


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Case 3:- If No.of Pits are 3 then Earth Resistance = = 0.6 (Sufficient and Safe)

Therefore, Proposed No.of Pits = 4 and 1 Rod in Each Pit

EARTH STRIP RESISTANCE () ()

r = Resistivity of Soil= 0.2 / km (Taken from soil Test Eng )

L = Length of the Strip = Assume as 30 mt (Distance from PanelTaken from Site Eng)

W = Depth Level of the Strip (1Feet to 2 Feet)=Mostly 2Feet = 2 x 0.3 = 0.6 mt .

S = Size of Strip = 75 mm = 0.075 mt.

() ()

For 1 Rod =

0.68

OVERAL RESISTANCE =

= = 0.31

Conclusion:-

Overall Resistance is less than 1 , the Earthing Designed for Building is Safe.

Rrod= 0.6

()


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LIGHTING PROTECTION CALCULATION


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LIGHTING PROTECTION CALCULATION

Given Data :-

a) Height of the Building (H) = 18 mt b) Width of the Building (W) = 17.94 mt

c) Length of the Building (L) = 23.7 mt. d) Location of Building = Kurnool , India

Flashes / km2 as per year ( Ng ) :-

Ng= 0.7 (Collected from IS 2309 or NFDA70 2000 Codes )

Where, Ng = Yearly Avg Flux Density in a Region .

Effective Collection Area ( Ac ) :-

Ac= (W x L) + 2 (W x H) + 2 (L x H) + ()= (17.94 x 23.7) + 2 (17.94 x 18) + 2 (23.7 x 18) + ( )Therefore, Total Effected Area of Building ( Ac) = 2,238.37 Sq. mt .

Now, Over all Area of Building = L x W X H = 23.7 x 17.94 x 18 = 7,653.20 Sq.mt.

That implies 2,238.37 sq.mt area of building needs to be protected from Flashes or

lighting strokes 0ut of 7,653.20 Sq.mt ( Total Area of Building) .

Propability of Stroke (P) :-

Here, P = 2,238.37 x 0.7 x 10-6= 1.56 x 10-3

Overall Weighting Factor (W):-

a = 1.7 Use of Structure (Value Varies Depending on Building)

b = 0.8 Type of Construction (Varies)

c = 1.7 Content or Consequential Effect (Constant)

d = 2.0 Degree of Isolation (Constant)

P = Acx Ngx 10-

W = a x b x c x d x e


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BILL OF QUANTITY (BOQ)

ITEM NAME

G.F

+

COR

1ST

FLOOR 2ND

FLOOR 3RD

FLOOR 4TH

FLOOR 5th

FLOOR

F1 F2 F3 F4 F1 F2 F3 F4 F1 F2 F3 F4 F1 F2 F3 F F1 F2 F3 F4 TOT

36 W LAMP 18 13 13 14 14 13 13 14 14 13 13 14 14 13 13 14 14 13 13 14 14 288

24 W LAMP 45 - - - - - - - - - - - - - - - - - - - - 45

CEILING

FANS- 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 100

EXHAUST

FANS- 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 60

6A SOCKET 3 14 14 17 17 14 14 17 17 14 14 17 17 14 14 17 17 14 14 17 17 313

16A

SOCKET3 8 8 9 9 8 8 9 9 8 8 9 9 8 8 9 9 8 8 9 9 173

1 TR AC - 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 1 30

1.5 TR AC - - - 1 1 - - 1 1 - - 1 1 - - 1 1 - - 1 1 10

315 KVA

T/F- - - - - - - - - - - - - - - - - - - - - - - 1

100 KVA

DG1 - - - - - - - - - - - - - - - - - - - - 1

2

Date post:	02-Jun-2018
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