3200 Busbar Calculation

Client‐ Project:‐ Current Rating of the Bus Duct ( I )

3200 Amp ,415V, TPN, Air insulated‐NSPBD

R.M.S. Value of short circuit current in Amp ( Isc )

50 kA

Duration of short circuit

1 Sec.

Enclosure Size

760

x

420

mm

Conductor Size

200

x

10

mm

Area of conductor in cm

20 cm

Phase conductor configuration

200

x

10 x 2

Neutral conductor configuration

200

x

10 x 1

Total conductor in Enclosure

7 Nos.

Space between the two conductor ( S )

95 mm

Thickness of conductor ( a )

10 mm

Width of conductor ( b )

200 mm

TEMPERATURE RISE CALCULATION 4750 Amp

1) Current carrying capacity of conductor(200 x 10) for Two bars (Refer to table 1) 2) Uprating by Interleaved design (Refer to table 2)

4750 x 1.18 5605 Amp

3) Temperature rise derating factor for 35 ºC rise

5605 x

4) Housing derating Area of conductor x 100 Area of enclosure

200 x 10 x 7 x 100 760 x 420 4.4 %

5) Derating factor due to enclosure size (Refer to table 4) 6) Derating factor due to busbar alloy factor (Refer to table 1)

4568 x 0.75 3426 Amp 3426 x 0.97 3323 Amp

This shows that bars are capable of working in above condition, without Hence Design is safe.

Note:‐

Tables were taken as reference from K.C. Agarwal Book. VOLTAGE DROP CALCULATION RDC at 20ºc

0.815

4568 Amp

over 50ºC ambient (Refer to table 3)

3.133 x 100 Area of Conductor (Sq.cm) (Refer to table 8, Electrical Resistivity) 3.133 x 100 = 20 =

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(RDC |20ºc )

=

15.7 μΩ/mtr

=

0.0157 mΩ/mtr.

=

200 10

=

20

or

Ratio of width to thickness

(From table 5 & interploating, we get value of "K" ) = 1.15 RAC

Since,

RDC we get,

RAC at 20ºc

=

K

=

1.15 x 0.0157

=

0.0181

mΩ/mtr.

Since the operating temprature have been considered to be 85º C RAC at 85ºc

=

(RAC |20ºc ) [1+α20θ (θot ‐ θrt )

α20θ

=

Temperature co‐efficient of resistance

From the Table "9"

0.0033 per ºC at 20ºC

θot

=

Operating temperature i.e.

85 º C

θrt

=

Reference temperature i.e.

20 º C

So, [By putting values & calculation ] = RAC at 85ºc Now by using relation

( .018 ) x [1+ (.0033) x

=

0.022

=

1.26 x S

(85 ‐

20) ]

mΩ/mtr. a

&

a + b

b

here,'S' is space between two bars and 'a' & 'b' are thickness & width of the Busbar respectivily. = 1.26 x 95 10 & 10 + 200 200 =

0.57

&

0.05

From (Table 6 ) we get value of reactance ( a/b and 1.26 x S/a+b) X

Impedance (Z)

=

=

64

μΩ/mtr

0.064

mΩ/mtr.

X² + R²AC | 85ºc

=

(0.064 x 0.064 ) + ( 0.02198245 x 0.02198245)

=

0.068

mΩ/mtr.

(Due to double layer design, value of "Z" become halved) Voltage Drop (Vd)

= I x Z = 3200 x 0.068 = 204.80 mV/mtr. = 0.2048 V/mtr. = 0.1024 V/mtr. ( for Phase splitting and interleaved design,volatge drop get halved) This show's that, the voltage drop is minimum for mtr. Length run This show's that, the voltage drop is minimum for mtr. Length run Power loss,

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PL

Vd x I 0.1024 x 3200 327.68 Watts 0.328 Kw

= = = =

PL PL

SHORT CIRCUIT CALCULATION Minium cross section of conductor required to withstand temp rise due to short circuit current it can be calculate by formula

Isc

=

2.17 x A x 10^4

(θf + 258)

Log10

√t

Isc = Short circuit current A = Area of cross section of conductor in sq.cm t = Duration of short circuit θf = Final temperature after short circuit θi = Intial temperature before short circuit

50 ?? 1 200 85

(θi + 258) kA Sq.cm Sec ºC ºC

Hence, Area is 50 x 1000

=

2.17 x A x 10^4 Log10 ( 200+258) (85 + 258 ) 1

50000

=

50000 = A

=

21700 7689.6227

x A

6.50

sq.mm.

The cross section area of provided conductor is

x

0.3544 x A

sqcm which is above the requirement .

DYNAMIC FORCES (ELECTRODYNAMIC FORCE ) Developed between bus bars due to short circuit current Formula Used

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‐4

Fm = 16 x ISC² x 10 x K N/mt S Where Fm is force between conductor in N/mt Isc = R.M.S value of short circuit current in Amp S = centre space between two conductor K = Space factor for rectangular conductor's which is functio of size & spacing between two conductor. a = Thickness of conductor pair b = Width of conductor

50 kA 95 mm 0.65 10 mm 200 mm

Which is calculated as under ‐ s ‐ a a + b

= =

95 + 10 10 + 200

=

0.50

And =

a b

=

10 200

=

0.05

From (Table 7 ) we get value of K = 0.65 Each phase having two circuit,hence[I SC = 50KA /2] , So, the force developed during short ISC =

circuit is given by:‐ (Putting in above formula) Fm

=

25

( 16 x 25x 25 x 10‐4 x 0.65) 95 x 10000 6500000000

=

950000

(1) 1 Nm = 1/9.807 Kgfm

0.102

=

6842.11

N/m

=

697.68

Kg/m

FRP Bus Bar Support The supports used at every 200 mm apart for this total force acting on each support shall be

=

697.68

=

139.54

x

0.2 ……………………….."A"

STRESS ON BUS BAR SUPPORT So the bearing area of support (Depth x thickness of support x 2)

= = =

Flexural strength of support is

=

Force support can withstand

=

(Bearing area x Flexural Strength)

=

90

x

10

x

2

1800 Sqmm 18

Sq.cm

1350 Kg/cm2 18

x 24300

1350 Kg

……………………….."B"

Force developed at “A” above is very less than the withstand capacity at “B” Hence supports, are capable of withstand these forces.

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