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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
15
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
Choose No.of Light Fixtures as 2
__________________________________________________________________________
Room Application : Dinning Area (DA)
Lux Level : 200
Room Length : 2.25mt
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
No.of Fixture : (2.25x2.44x200) / (0.5x0.95x1x3350) = 0.69 1
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Choose No.of Light Fixtures as 1
Room Application : TOILET1 ( T1 )
Lux Level : 100
Room Length : 2.4 mt
Room width : 1.2 mt
Height of lighting fixture : 2.5 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
No.of Fixture : (2.4x1.2x100)/(0.5x0.95x1x3350) = 0.18 1
Choose No.of Light Fixtures as 1
__________________________________________________________________________
Room Application : TOILET2 ( T2 )
Lux Level : 100
Room Length : 2.4 mt
Room width : 1.2 mt
Height of lighting fixture : 2.5 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
No.of Fixture : (2.4x1.2x100)/(0.5x0.95x1x3350) = 0.18 1
Choose No.of Light Fixtures as 1
_______________________________________________________________________
Room Application : Master Bed Room (MBR)
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Lux Level : 250
Room Length : 3.64
Room width : 3.19
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 36 W( FL LAMP )
Room Index : (3.64 x 3.19)/ (3.64 + 3.19)x2.5= 0.75
No.of Fixture : (3.64x3.19x250) / (0.5x0.95x1x3350) = 1.824 2
Choose No.of Light Fixtures as 2
__________________________________________________________________________
Room Application : Hall
Lux Level : 300
Room Length : 3.15mt
Room width : 4.89
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 36 W( FL LAMP )
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
Choose No.of Light Fixtures as 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.
Therefore Tonnage = 148 / 120 = 1.23 TR 1 TR AC = 1400 W
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 Length : 3.15mt
Room width : 4.89mt
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 36 W ( FL LAMP )
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
Choose No.of Light Fixtures as 2
__________________________________________________________________________
Room Application : KITCHEN ( K )
Lux Level : 300
Room Length : 3.65 mt
Room width : 2.47mt
Height of lighting fixture : 2 .5mt
Type of Lighting Fixture : 1 x 36 W ( FL LAMP )
Room Index : (3.65 x 2.47)/ (3.65+2.47)x2.5= 0.59
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No.of Fixture : (3.65x2.47x300) / (0.5x0.95x1x3350) = 1.69 2
Choose No.of Light Fixtures as 2
__________________________________________________________________________
Room Application : DINNING AREA ( DA )
Lux Level : 200
Room Length : 2.22mt
Room width : 2.2 mt
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 36 W ( FL LAMP )
Room Index : (2.22 x 2.2) / (2.22 + 2.2 ) x 2.5= 0.44
No.of Fixture : (2.22x2.2x200) / (0.5x0.95x1x3350) = 0.61 1
Choose No.of Light Fixtures as 1
__________________________________________________________________________
Room Application : TOILET 1 (T - 1)
Lux Level : 100
Room Length : 1.2mt
Room width : 2.2mt
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 36 W ( FL LAMP )
Room Index : (1.2 x 2.2 ) / (1.2 + 2.2 ) x 2.5 = 0.31
No.of Fixture : (1.2x2.2x100) / (0.5x0.95x1x3350) = 0.165 1
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Choose No.of Light Fixtures as 1
___________________________________________________________________________
Room Application : MASTER BEDROOM ( MBR )
Lux Level : 250
Room Length : 3.65mt
Room width : 4.25mt
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 36 W ( FL LAMP )
Room Index : (3.65 x 4.25) / (3.65+4.25) x 2.5 = 0.78
No.of Fixture : (3.65x4.25x250) / (0.54x0.95x1x3350) = 2.25 2
Choose No.of Light Fixtures as 2
__________________________________________________________________________
Room Application : BED ROOM ( BR )
Lux Level : 200
Room Length : 3.15mt
Room width : 4.25mt
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 36 W ( FL LAMP )
Room Index : (3.15 x 4.25) / (3.15+4.25) x 2.5 = 0.72
No.of Fixture : (3.15x4.25x200) / (0.54x0.95x1x3350) = 1.68 2
Choose No.of Light Fixtures as 2
Room Application : BALCONY ( BAL)
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Lux Level : 150
Room Length : 1.77
Room width : 1.2
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 36 W ( FL LAMP )
Room Index : (1.77 x 1.2) / (1.77 + 1.2) x 2.5 = 0.367
No.of Fixture : (1.77x1.2x150) / (0.5x0.95x1x3350) = 0.2007 1
Choose No.of Light Fixtures as 1
__________________________________________________________________________
Room Application : TOILET 2 (T2)
Lux Level : 100
Room Length : 2.4mt
Room width : 1.2 mt
Height of lighting fixture : 2.5 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
No.of Fixture : (2.4x1.2x100) / (0.54x0.95x1x3350) = 0.189 1
Choose No.of Light Fixtures as 1
Room Application : TOILET 3 (T3)
Lux Level : 100
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Room Length : 2.4mt
Room width : 1.2 mt
Height of lighting fixture : 2.5 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
No.of Fixture : (2.4x1.2x100) / (0.54x0.95x1x3350) = 0.189 1
Choose No.of Light Fixtures as 1
_________________________________________________________________________
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
1 No Water Heater (16A Socket) = 1000 W
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
1 No Water Heater (16A Socket) = 1000 W
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.
Therefore Tonnage = 148 / 120 = 1.23 TR 1 TR AC = 1400 W
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 Length : 16.14mt
Room width : 1.5 mt
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 24 W ( FL LAMP )
Room Index : (16.14 x 1.5) / (16.14 + 1.5) x 2.5 = 0.54
No.of Fixture : (16.14x1.5x150) / (0.5x0.95x1x1750) = 4.36 4
Choose No.of Light Fixtures as 4
____________________________________________________________________________
Room Application : CORRIDOR 2 (COR2)
Lux Level : 150
Room Length : 23.7mt
Room width : 1.2 mt
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 24 W ( FL LAMP )
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Room Index : (23.7 x 1.2) / (23.7 + 1.2) x 2.5 = 0.4
No.of Fixture : (23.7x1.2x150) / (0.54x0.95x1x1750) = 5.13 5
Choose No.of Light Fixtures as 5
__________________________________________________________________________
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 Length : 23.7mt
Room width : 16.14 mt
Height of lighting fixture : 2.5 mt
Type of Lighting Fixture : 1 x 36 W ( FL LAMP )
Room Index : (23.7 x 16.14) / (23.7 + 16.14) x 2.5 = 3.84
No.of Fixture : (23.7x16.14x100) / (0.67x0.95x1x3350) = 17.93 18
Choose No.of Light Fixtures as 18
__________________________________________________________________
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 Length : 23.7mt
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
Step 2 10 KVAR ON / OFF
Step 3 5 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
Load Current (I) = 76.5 A
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
Cable is in Air and Derating Factor is 0.8
Here, Load is 53.39KW( 3) and using the Formulae provided
Load Current (I) = 92.84 A
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
Cable is in Air and Derating Factor is 0.8
Here, Load is 5.32KW (3) and using the Formulae provided
Load Current (I) = 9.25 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
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
Cable is in Air and Derating Factor is 0.8
Here, Load is 38.66KW( 3) and using the Formulae provided
Load Current (I) = 67.22 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
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
Cable is in Air and Derating Factor is 0.8
Here, Load is 25 KVAR ( 3) and using the Formulae provided
Load Current (I) = 57.9 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
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
Cable is in Air and Derating Factor is 0.8
Here, Load is 20KW( 3) and using the Formulae provided
Load Current (I) = 34.77 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
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
Cable is in Air and Derating Factor is 0.8
Here, Load is 7.45KW( 3) and using the Formulae provided
Load Current (I) = 12.9 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
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
Cable is in Air and Derating Factor is 0.8
Here, Load is 11.17KW( 3) and using the Formulae provided
Load Current (I) = 19.24 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
a) For 4 C x 4 SQMM (Cu) Cable = CCC = 40 A x 0.8 = 32 A (Satisfied)
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 .
Cable is in Air and Derating Factor is 0.8
Here, Load is 12.24KW( 3) and using the Formulae provided
Load Current (I) = 21.28 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
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.
Cable is in Air and Derating Factor is 0.8
Here, Load is 12.24KW( 3) and using the Formulae provided
Load Current (I) = 21.28 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
a) For 4 C x 4 SQMM (Cu) Cable = CCC = 40 A x 0.8 = 32 A (Satisfied)
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
Cable is in Air and Derating Factor is 0.8
Here, Load is 14.45KW( 3) and using the Formulae provided
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Load Current (I) = 25.12 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
a) For 4 C x 4 SQMM (Cu) Cable = CCC = 40 A x 0.8 = 32 A (Satisfied)
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
Cable is in Air and Derating Factor is 0.8
Here, Load is 100 KVA ( 3) and using the Formulae provided
Load Current (I) = 139.12 A
Now, Check with Nearest CCC of various Cable Size from LT Catalogue.
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)
10 1R X 3C X 25 SQMM (Al)
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11 1R X 3C X 25 SQMM (Al)
12 1R X 4C X 6 SQMM (Cu)
13 1R X 4C X 4 SQMM (Cu)
14 1R X 4C X 4 SQMM (Cu)
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
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 0.2655 V
%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
(R) = 1.113 / KM d. No.of Runs = 1 e. Load Current = 76.5 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 0.442 V
%VD=
= 0.106639
%VD = 0.1066
Feeder 4(SMDB1F) ,Feeder 5(SMDB2F) , Feeder 6(SMDB3F) ,Feeder 7(SMDB - 4F),
Feeder 8 (SMDB5F)
Given Data :-
a. Cable Size = 3C x 35 SQMM (Al) b. Length of Cable (L) = 3 mt c. Resistance of Cable
(R) = 1.113 / KM d. No.of Runs = 1 e. Load Current = 92.84 A
Voltage Drop is calculated using the Formulae Provided,
VOLTAGR DROP ( VD ) = 0.536 V
%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
(R) = 3.95 / KM d. No.of Runs = 1 e. Load Current = 9.25 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 0126 V
%VD=
= 0.0303
%VD = 0.0303
Feeder 10MCC PANEL
Given Data :-
a. Cable Size = 3C x 25 SQMM (Al) b. Length of Cable (L) = 2 mt c. Resistance of Cable
(R) = 1.539 / KM d. No.of Runs = 1 e. Load Current = 67.22 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 0.358 V
%VD=
= 0.0836
%VD = 0.0863
Feeder 11 CAPACITOR BANK
Given Data :-
a. Cable Size = 3C x 25 SQMM (Al) b. Length of Cable (L) = 3 mt c. Resistance of Cable
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(R) = 1.539 / KM d. No.of Runs = 1 e. Load Current = 57.9 A
Voltage Drop is calculated using the Formulae Provide,
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
(R) = 3.95 / KM d. No.of Runs = 1 e. Load Current = 34.77 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 7.136 V
%VD=
= 1.719
%VD = 1.719
Feeder 13 WATER SERVICE MOTOR (W.S)
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Length of Cable (L) = 25mt c. Resistance of Cable
(R) = 5.91 / KM d. No.ofRuns = 1 e. Load Current = 12.9 A
Voltage Drop is calculated using the Formulae Provide,
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VOLTAGR DROP ( VD ) = 3.301V
%VD=
= 0.795
%VD = 0.795
Feeder 14FIRE FIGHTING MOTOR (FF)
Given Data :-
a. Cable Size = 4C x 4 SQMM (Al) b. Length of Cable (L) = 30 mt c. Resistance of Cable
(R) = 5.91 / KM d. No.of Runs = 1 e. Load Current = 19.42 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 5.96 V
%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 :-
a. Cable Size = 4C x 6 SQMM (Cu) b. Length of Cable (L) = 3 mt c. Resistance of Cable
(R) = 3.95 / KM d. No.of Runs = 1 e. Load Current = 21.28 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 0.436 V
%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.
Feeder 17(SMDB1F TO DB F3) , Feeder 18(SMDB1F TO DB F4)
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Length of Cable (L) = 4 mt c. Resistance of Cable
(R) = 5.91 / KM d. No.of Runs = 1 e. Load Current = 25.12 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 1.028 V
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%VD=
= 0.247
%VD = 0.247
Feeder 19(SMDB2F TO DB F1) , Feeder 20(SMDB2F TO DB F2)
Given Data :-
a. Cable Size = 4C x 6 SQMM (Cu) b. Length of Cable (L) = 5 mt c. Resistance of Cable
(R) = 3.95 / KM d. No.of Runs = 1 e. Load Current = 21.28 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 0.727 V
%VD=
= 0.175
%VD = 0.175
Feeder 21(SMDB2F TO DB F3) , Feeder 22(SMDB2F TO DB F4)
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Length of Cable (L) = 6mt c. Resistance of Cable
(R) = 5.91 / KM d. No.of Runs = 1 e. Load Current = 25.12 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 1.542V
%VD=
0.371
Feeder 23(SMDB3F TO DB F1) , Feeder 24(SMDB3F TO DB F2)
%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
(R) = 9.5 / KM d. No.of Runs = 1 e. Load Current = 21.28 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 2.45V
%VD=
= 0.5905
%VD = 0.5905
Feeder 25(SMDB3F TO DB F3) , Feeder 26(SMDB3F TO DB F4)
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Length of Cable (L) = 8mt c. Resistance of Cable
(R) = 5.91 / KM d. No.of Runs = 1 e. Load Current = 25.12 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 2.507 V
%VD=
= 0.495
Feeder 27(SMDB4F TO DB F1) , Feeder 28(SMDB4F TO DB F2)
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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Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 3.51 V
%VD= = 0.759
Feeder 29(SMDB4F TO DB F3) , Feeder 30(SMDB4F TO DB F4)
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Length of Cable (L) = 10 mt c. Resistance of Cable
(R) = 5.91 / KM d. No.of Runs = 1 e. Load Current = 25.12 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 2.571 V
%VD= = 0.169
Feeder 31(SMDB5F TO DB F1) , Feeder 32(SMDB5F TO DB F2)
Given Data :-
a. Cable Size = 4C x 2.5 SQMM (Cu) b. Length of Cable (L) = 11mt c. Resistance of Cable
(R) = 9.5 / KM d. No.of Runs = 1 e. Load Current = 21.28 A
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 3.851 V
%VD= = 0.928
%VD = 0.759
%VD = 0.619
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%VD = 0.928
Feeder 33(SMDB5F TO DB F3) , Feeder 34(SMDB5F TO DB F4)
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Length of Cable (L) = 12 mt c. Resistance of Cable
(R) = 5.91 / KM d. No.of Runs = 1 e. Load Current = 25.12 A
Voltage Drop is calculated using the Formulae Provide,
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
Voltage Drop is calculated using the Formulae Provide,
VOLTAGR DROP ( VD ) = 0.273 V
%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
Pu Impedance based on System Voltage =
() = 1.59529685Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
= 16.16505247+ 1.59529685=17.76034932Pu
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MAX FAULT CURRENT UPTO C.B (2) =
= 5.630519 MVA
TOTAL MAX FAULT CURRENT =
= 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
Pu Impedance based on System Voltage =
() = 1.822035128Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
= 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
Pu Impedance of 3mt Cable =
= 0.00738Pu
Pu Impedance based on System Voltage =
() = 4.285092176 Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
= 19.58238445+ 4.285092176= 23.86747663Pu
MAX FAULT CURRENT UPTO C.B (9) =
= 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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a. Cable Size = 3C x 25 SQMM (Al) b. Length of Cable (L) = 3mt c. Cable Impedance
= () () = 1.44 / KM d. No.of Runs = 1
Pu Impedance of 3mt Cable =
= 0.00432Pu
Pu Impedance based on System Voltage =
() = 2.50834664Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
= 19.58238445+ 2.50834664 = 22.09073109Pu
MAX FAULT CURRENT UPTO C.B (10) =
= 4.526785 MVA
TOTAL MAX FAULT CURRENT =
= 6.297KA
S.C Current at MCC PANEL = 6.29 KA
Feeder 11 : MCC PANEL To Capacitor Bank
Given Data :-
a. Cable Size = 3C x 25 SQMM (Al) b. Length of Cable (L) = 3mt c. Cable Impedance
= () () = 1.44 / KM d. No.of Runs = 1
Pu Impedance of 3mt Cable =
= 0.00432Pu
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Pu Impedance based on System Voltage =
() = 2.50834664 Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
= 22.09073109+ 2.50834664 = 24.59907773Pu
MAX FAULT CURRENT UPTO C.B (11) =
= 4.065193 MVA
TOTAL MAX FAULT CURRENT =
= 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
Pu Impedance based on System Voltage =
() = 80.40193061Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
= 22.09073109+ 80.40193061= 102.4926617Pu
MAX FAULT CURRENT UPTO C.B (12) =
= 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 :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Cable Length (L) = 25mt c. Cable Impedance
= () ()= 5.53 / KM d. No.of Runs = 1
Pu Impedance of 3mt Cable = = 0.13325Pu
Pu Impedance based on System Voltage = () = 77.36971984Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
= 22.09073109+ 77.36971984= 99.46045093Pu
MAX FAULT CURRENT UPTO C.B (13) = = 1.00542 MVA
TOTAL MAX FAULT CURRENT =
= 1.398 KA
S.C Current at W.S Motor = 1.39 KA
Feeder 14MCC PANEL To Fire Fighting (F.F) Motor
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Cable Length (L) = 30mt c. Cable Impedance
= () () = 5.53 / KM d. No.of Runs = 1
Pu Impedance of 3mt Cable = = 0.1599Pu
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Pu Impedance based on System Voltage =
() = 92.84366381Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
= 22.09073109+ 92.84366381= 114.9326098Pu
MAX FAULT CURRENT UPTO C.B (14) =
= 0.870075 MVA
TOTAL MAX FAULT CURRENT =
= 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 :-
a. Cable Size = 4C x 6 SQMM (Cu) b. Cable Length (L) = 3mt c. Cable Impedance
= () () = 3.69 / KM d. No.of Runs = 1
Pu Impedance of 3mt Cable =0.01107Pu
Pu Impedance based on System Voltage = 6.427638264Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=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 :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Cable Length (L) = 4mt c. Cable Impedance
= () () = 5.53 / KM d. No.of Runs = 1
Pu Impedance of 4mt Cable = 0.022121616Pu
Pu Impedance based on System Voltage = 12.84460212Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=19.52238445+12.84460212 = 32.36698657Pu
MAX FAULT CURRENT UPTO C.B (17,18) = 3.089567 MVA
TOTAL MAX FAULT CURRENT = 4.29KA
S.C Current at Feeder 17,18 = 4.29 KA
Feeder 19 ( SMDB-2F to DB F1), Feeder 20( SMDB-2F to DB F2) :-
Given Data :-
a. Cable Size = 4C x 6 SQMM (Cu) b. Cable Length (L) = 5mt c. Cable Impedance
= () () = 3.69 / KMd. No.of Runs = 1 Pu Impedance of 5mt Cable =0.01845Pu
Pu Impedance based on System Voltage= 10.71273044Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=19.52238445+ 10.71273044=30.23511489Pu
MAX FAULT CURRENT UPTO C.B (19,20) = 3.307412 MVA
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TOTAL MAX FAULT CURRENT= 4.60 KA
S.C Current at Feeder 19,20 = 4.60 KA
Feeder 21 ( SMDB-2F to DB F3), Feeder 22 ( SMDB-2F to DB F4):-
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Cable Length (L) = 6mt c. Cable Impedance
= () ()= 5.53 / KM d. No.of Runs = 1
Pu Impedance of 6mt Cable = 0.03318Pu
Pu Impedance based on System Voltage= 19.26549572Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=19.52238445+19.26549572=38.78788017Pu
MAX FAULT CURRENT UPTO C.B (21,22) = 2.578124 MVA
TOTAL MAX FAULT CURRENT= 3.58 KA
S.C Current at Feeder 21,22 = 3.58 KA
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
Total Pu Imp = Previous Pu Imp + Present Pu Imp
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=19.52238445+ 36.09232109= 55.61470559 Pu
MAX FAULT CURRENT UPTO C.B (23,24) = 1.798085 MVA
TOTAL MAX FAULT CURRENT= 2.50 KA
S.C Current at Feeder 23,24 = 2.50 KA
Feeder 25 ( SMDB-3F to DB F3), Feeder 26 ( SMDB-3F to DB F4):-
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Cable Length (L) = 8mt c. Cable Impedance
= () () = 5.53 / KM d. No.of Runs = 1 Pu Impedance of 8mt Cable = 0.04424 Pu
Pu Impedance based on System Voltage= 25.68732762 Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=19.52238445+25.68732762 = 45.20971297 Pu
MAX FAULT CURRENT UPTO C.B (25,26) = 2.211914 MVA
TOTAL MAX FAULT CURRENT= 3.07 KA
S.C Current at Feeder 25,26 = 3.07 KA
Feeder 27 ( SMDB-4F to DB F1), Feeder 28 ( SMDB-4F to DB F2):-
Given Data :-
a. Cable Size = 4C x 2.5 SQMM (Cu) b. Cable Length (L) = 9mt c. Cable Impedance
= () () = 8.88 / KM d. No.of Runs = 1 Pu Impedance of 9mt Cable = 0.07992Pu
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Pu Impedance based on System Voltage= 46.40441283Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=19.52238445+46.40441283 = 65.92679728Pu
MAX FAULT CURRENT UPTO C.B (27,28) = 1.516833 MVA
TOTAL MAX FAULT CURRENT= 2.11 KA
S.C Current at Feeder 27,28 = 2.11 KA
Feeder 29 ( SMDB-4F to DB F3) , Feeder 30 ( SMDB-4F to DB F4):-
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Cable Length (L) = 10mt c. Cable Impedance
= () ()= 5.53 / KM d. No.of Runs = 1 Pu Impedance of 10mt Cable = 0.0553 Pu
Pu Impedance based on System Voltage= 32.10915953 Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=19.52238445+32.10915953 = 51.63154398 Pu
MAX FAULT CURRENT UPTO C.B (25,26) = 1.936800 MVA
TOTAL MAX FAULT CURRENT= 2.69 KA
S.C Current at Feeder 29,30 = 2.69 KA
Feeder 31 ( SMDB-5F to DB F1), Feeder 32 ( SMDB-5F to DB F2):-
Given Data :-
a. Cable Size = 4C x 2.5 SQMM (Cu) b. Cable Length (L) = 11mt c. Cable Impedance
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= () () = 8.88 / KM d. No.of Runs = 1 Pu Impedance of 11mt Cable = 0.09768 Pu
Pu Impedance based on System Voltage= 56.71650457Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=19.52238445+56.71650457=76.2388902Pu
MAX FAULT CURRENT UPTO C.B (31,32) = 1.311666 MVA
TOTAL MAX FAULT CURRENT= 1.824 KA
S.C Current at Feeder 31,32 = 1.82 KA
Feeder 33 ( SMDB-5F to DB F3), Feeder 34 ( SMDB-5F to DB F4):-
Given Data :-
a. Cable Size = 4C x 4 SQMM (Cu) b. Cable Length (L) = 12mt c. Cable Impedance
= () () = 5.53 / KM d. No.of Runs = 1 Pu Impedance of 12mt Cable = 0.06636 Pu
Pu Impedance based on System Voltage= 38.53099144 Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=19.52238445+38.53099144 = 58.05337589 Pu
MAX FAULT CURRENT UPTO C.B (33,34) = 1.722552 MVA
TOTAL MAX FAULT CURRENT= 2.39 KA
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S.C Current at Feeder 33,34 = 2.39 KA
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
= () () = 0.53 / KM d. No.of Runs = 1 Pu Impedance of 2mt Cable = 0.00106Pu
Pu Impedance based on System Voltage= 0.615473944Pu
Total Pu Imp = Previous Pu Imp + Present Pu Imp
=19.52238445+0.615473944 =20.13785839Pu
MAX FAULT CURRENT UPTO C.B (35) = 4.965771 MVA
TOTAL MAX FAULT CURRENT= 6.90 KA
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
Load Current is calculated by using Formulaes Provided
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
Load Current is calculated by using Formulaes Provided
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
Load Current is calculated by using Formulaes Provided
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
Load Current is calculated by using Formulaes Provided
i e. FLC = 67.22A ,
C.B Raring = 1.25 x 67.22 = 84.03 A
Therefore, Proposed C.B(10) Rating is 100 A MCCB.
MCC TO CAPACITOR BANK C.B (11) P = 25 KVAR
Load Current is calculated by using Formulaes Provided
ie. FLC = 57.96A ,
C.B Raring = 1.25 x 57.96 = 72.45 A
Therefore, Proposed C.B(11) Rating is 80 A MCCB.
MCC TO LIFT MOTOR C.B (12) P = 20 KW
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Load Current is calculated by using Formulaes Provided
i e. FLC = 34.77A ,
C.B Raring = 1.25 x 34.77 = 43.77 A
Therefore, Proposed C.B(12) Rating is 60 A MCB.
MCC TO WATER SERVICE (W.S) C.B (13) P = 10 HP = 10 X0.745 = 7.45 KW
Load Current is calculated by using Formulaes Provided
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
Load Current is calculated by using Formulaes Provided
i e. FLC = 19.43A ,
C.B Raring = 1.25 x 19.43 = 24.2 A
Therefore, Proposed C.B(14) Rating is 25 A MCB.
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
Load Current is calculated by using Formulaes Provided
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
Load Current is calculated by using Formulaes Provided
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
Load Current is calculated by using Formulaes Provided
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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CABLE CHECKING,C.B BREAKING CAPACITY AND
TRIPPING TIME CALCULATIONS
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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 ..)
Cable withstanding capacity =
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 :-
Cable withstanding capacity =
= 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
Using Formulaes Provided :-
Cable withstanding capacity =
= 36.36 KA
Full Data of C.B
Cable withstanding capacity@ 1 sec = 36.36 KA----- eqn 1
System Fault Current I e. Calculated S.C Current at HT Side is
= 7.85 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
____________________________________________________________________________
Feeder N0 3 SMDB-S
Cable 1R x 3C x 35 Sq mm (Al) Isc= 2.65 @ t = 0.1 Sec
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
200 A VCB
18 KA 3 Sec
800 A (ACB)
7.85 KA 1 Sec
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Cable withstanding capacity =
= 8.38 KACable withstanding capacity@ 0.1 sec = 8.38 KA----- eqn 1 Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 7.12 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
_____________________________________________________________________________
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
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 8.38 KACable withstanding capacity@ 0.1 sec = 8.38 KA----- eqn 1 Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 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
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 6.9 KACable withstanding capacity@ 0.01 sec = 6.9 KA----- eqn 1 Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
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
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
_____________________________________________________________________________
Feeder N0 10 SMDBS to MCC PANEL 1R x 3c x 25 Sqmm (Al) Isc= 1.89 @ t = 0.08s
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 6.68 KA
Cable withstanding capacity@ 0.08 sec = 6.68 KA----- eqn 1 Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 6.31 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
_____________________________________________________________________________
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
Using Formulaes Provided :-
Cable withstanding capacity =
= 5.97 KA
Cable withstanding capacity@ 0.1 sec = 5.97 KA----- eqn 1 Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 5.65 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
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)
Using Formulaes Provided :-
Cable withstanding capacity =
= 1.54 KA
Cable withstanding capacity@ 0.2 sec = 1.54 KA----- eqn 1 Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 1.35 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
____________________________________________________________________________
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
Isc= S.C Circuit Current Rating = (Taken from Gems Cab S.C Rating Catalogue)
Using Formulaes Provided :-
Cable withstanding capacity =
= 1.45 KACable withstanding capacity@ 0.1 sec = 1.45 KA----- eqn 1 Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 1.39 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
_____________________________________________________________________________
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
Isc= S.C Circuit Current Rating = (Taken from Gems Cab S.C Rating Catalogue)
60 A MCB
1.35 KA 0.2 Sec
25 A MCB
1.39 KA 0.1 Sec
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Using Formulaes Provided :-
Cable withstanding capacity =
= 1.45 KA
Cable withstanding capacity@ 0.1 sec = 1.45 KA----- eqn 1 Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 1.21 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
_____________________________________________________________________________
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
Isc= S.C Circuit Current Rating = (Taken from Gems Cab S.C Rating Catalogue)
Using FormulaesProvided :-
Cable withstanding capacity =
= 6.9 KA
Cable withstanding capacity@ 0.01 sec = 6.9 KA----- eqn 1 Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 5.36 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
____________________________________________________________________________
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
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
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
System Fault Current I e. Calculated S.C Current at HT Side is
= 4.59 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
____________________________________________________________________________
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
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 4.6 KACable withstanding capacity @ 0.01 sec = 4.6 KA----- eqn 1
Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 4.29 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe.
____________________________________________________________________________
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
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 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
System Fault Current I e. Calculated S.C Current at HT Side is
= 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
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 2.87 KACable withstanding capacity @ 0.01 sec = 2.87 KA----- eqn 1
Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 2.50 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe
____________________________________________________________________________
Feeder N0 27,28DBF1(27)& DBF2(28) OF SMDB4F
Here For Cable 1R x 4C x 2.5 Sq mm (Cu) Isc= 0.287 @ t = 0.01 Sec
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 2.87 KA
Cable withstanding capacity @ 0.01 sec = 2.87 KA----- eqn 1
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
System Fault Current I e. Calculated S.C Current at HT Side is
= 2.11 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe
____________________________________________________________________________
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
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 4.6 KACable withstanding capacity @ 0.01 sec = 4.6 KA----- eqn 1
Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 3.07 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe
____________________________________________________________________________
Feeder N0 29,30DBF3 (29)& DBF4(30) OF SMDB4F t = (0.01 to 1 ) Sec
Here For Cable 1R x 4c x 4 Sqmm (Cu) Isc= 0.46 @ t = 0.02Sec
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 3.25 KA
Cable withstanding capacity @ 0.02 sec = 3.25 KA----- eqn 1
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
System Fault Current I e. Calculated S.C Current at HT Side is
= 2.69 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe
____________________________________________________________________________
Feeder N0 31,32DBF1(31)& DBF2(32) OF SMDB5F t = (0.01 to 1 ) Sec
Here For Cable 1R x 4C x 2.5 Sq mm (Cu) Isc= 0.287 @ t = 0.02 Sec
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 2.02 KA
Cable withstanding capacity @ 0.02 sec = 2.02 KA----- eqn 1
Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 1.82 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe
____________________________________________________________________________
Feeder N0 33,34DBF3 (33)& DBF4(34) OF SMDB3F t = (0.01 to 1 ) Sec
Here For Cable 1R x 4c x 4 Sqmm (Cu) Isc= 0.46 @ t = 0.02Sec
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 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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Cable withstanding capacity @ 0.02 sec = 3.25 KA----- eqn 1
Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 2.39 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe
____________________________________________________________________________
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
Isc= Short Circuit Current Rating = Taken from Gems Cab S.C Rating Catalogue
Using Formulaes Provided :-
Cable withstanding capacity =
= 7.49 KA
Cable withstanding capacity @ 0.5 sec = 7.49 KA----- eqn 1
Full Data of C.B
System Fault Current I e. Calculated S.C Current at HT Side is
= 6.90 KA ------ eqn 2
Comparing Eqn 1 & 2 I e. Eqn 1 > Eqn 2 So, Proposed Cable is Safe
32 A MCB
2.39 KA 0.02 Sec
180 A MCCB
6.90 KA 0.5 Sec
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LT SIDE - BUSBAR SIZING CALCULATION
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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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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