3200a Busduct Busbar Calculation.pdf

APPROVED DATE: SIGN:

Clint:- M/s The Adani Dharma Port company Limited NAME: Project:- Thyssenkrupp I. I Pvt. Ltd. Rating

(760 x 420 ) mm

Conductor Size

(200 x 10 )mm

Phase conductor configuration

(200 x 10) x 2

Neutral conductor configuration

(200 x 10) x 1

TEMPERATURE RISE CALCULATION 1) Current carrying capacity of conductor(200 x 10) (refer to table 1) for Two bars

4750 Amp

2) Uprating by Interleaved design (refer to table 2)

4750 x 1.18 5605Amp

2) Temperature rise derating factor for 35 ºC rise over 50ºC ambient (refer to table 3)

5605 x 0.815 4568 Amp

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

= 200 x 10 x 7 x 100 760 x 420

=

Derating factor due to enclosure size (refer to table 4)

4.3%

4568 x .75 3426 Amp

4) Derating factor due to busbar alloy factor (refer to table 1)

3426 x .97 3323 Amp

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

Tables were taken as reference from K.C. Agarwal Book.

Tables were taken as reference from K.C. Agarwal Book.

VOLTAGE DROP CALCULATION RDC at 20ºc

3.133 x 100 Area of conductor (sq.cm) (refer to table 8,electrical resitivity) (RDC |20ºc ) = 15.6 μΩ/mtr or = 0.0156 mΩ/mtr. Ratio of width to thickness

………………………..……………..…………. ………………………..……………..………….

3200Amp,415V,TPN,Air insulated-N

Enclosure Size

Note:-

………………………..……………..………….

=

=

200

=

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20

APPROVED DATE:

07 March 2017

Pramod Wankhade CONCEPT ENGINEERING PROJECT PVT. LTD.

=

=

10

20

From the Table "5" & interploating, we get value of "K" Since,

RAC RDC

=

we get,

RAC (at 20 degree)

= =

=

1.15

K 1.15 x 0.0156 0.0179 mΩ/mtr.

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

=

(RAC |20ºc ) [1+α20θ (θot - θrt )

α20θ From the Table "9"

=

Temperature co-efficient of resistance 0.0033 per ºC at 20ºC

θot θrt

= =

operating temperature [i.e. 85º C] reference temperature [i.e. 20º C]

Note:-

Tables were taken as reference from K.C. Agarwal Book.

So,

[By putting values &calculation ] RAC at 85ºc

=

Now by using relation

=

0.022 mΩ/mtr.

1.26. S a+b

a b

&

here,'S' is space between two bars and 'a' & 'b' are thickness & width of the bar respectivily. =

1.26 x 95 10 + 200

=

0.57

& &

10 200 0.05

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

Note:-

Tables were taken as reference from K.C. Agarwal Book. X

Impedance (Z)

Voltage Drop

=

64 0.064

μΩ/mtr mΩ/mtr.

=

X² + R²AC | 85ºc

=

0.06

mΩ/mtr. mΩ/mtr. (Due to double layer design, value of "Z" become halved) = IxZ = 3200 x 0.06 = 192 mV/mtr.

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= 0.192 V/mtr. = 0.096 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,

PL PL PL

= = = =

Vd x I 0.096 x 3200 307 watts 0.307 KW

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 t

Log10 θf + 258 θi + 258

Isc = Short circuit current (i.e. 50KA) A = Area of cross section of conductor in sq.cm t = Duration of short circuit ( i.e. 1 second) θf = Final temperature after short circuit ( i.e.200ºC) θi = Intial temperature before short circuit (i.e. 85ºC) Hence, Area is A =

A

50 x 1000 x 1 2.17 x 10^4 Log 10 200 + 258 85 + 258

=

6.76 Sq.cm

The cross section area of provided conductor is 40 sqcm which is above the requirement .

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

=

16 x ISC² x 10-4 x K S

N/mt

Where Fm is force between conductor in N/mt Isc = R.M.S value of short circuit current in Amp (50KA) S = centre space between two conductor

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K = Space factor for rectangular conductor's which is function of size & spacing between two conductor. a = Thickness of conductor pair( i.e.10mm) b = Width of conductor (i.e. 200 mm) Which is calculated as under S-a a+b

=

95 - 10 10+ 200

=

0.4

And a b

=

10 200

= 0.05

The value of K is 0.65 (refer to table 7)

Note:-

Tables were taken as reference from K.C. Agarwal Book.

Each phase having two circuit,hence[ISC = 50KA /2] , So, the force developed during short circuit is given (Putting in above formula) Fm = 16 x 25 x 25 x 1000 x 1000 x 0.65 95 x 10000

or

Fm Fm

= =

6842.1 N/m 697.6 Kg/m

(1) 1 Nm = 1/9.807 Kgfm

FRP Bus Bar Support The supports used at every 200 mm apart for this total force acting on each support shall be = 697.6 x 0.2 = 139.5 ……………………………….."A" STRESS ON BUS BAR SUPPORT So the bearing area of support (Depth x thickness of support x 2)

Flexural strength of support is

= 90 x 10 x 2 = 1800Sq.mm / 18 Sq.cm = 1350 Kg/cm2

Force support can withstand (Bearing area x Flexural Strength) = 18 x 1350 = 24300 Kg ………………….. "B"

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