claulation for check suitability of ips

7/26/2019 Claulation for check suitability of IPS

1/45

CLIENT:-

Client: Mundra Port & SEZ.

Project:- 220/66/11kV Electrical Distribution. Sales Ref.: IS/WR/

Description:

Amapacity Calculation of 2" IPS AL Tube.

(Schedule 40)

Drawing No.:

Rev G71000-X0009-ED07

Chkd. SO

Date 15.04.2009

Siemens Ltd

KG OfficeRemarks Date

Prep. SGP

Name

Busbar Current Rating and BPI Cantilever Calculation 1/45


2/45

1 Introduction

A. A rigid bus is supported on bus post insulators in an out-door substation. These bus post insulators are

subjected to all types of stresses due to electrical and mechanical forces acting on bus bar.

The principal forces on the bus post insulators supporting a rigid conductor are as follows:

a. Short circuit force on bus conductor

b. Wind force on the bus conductor

c. Wind force on the bus post insulatord. Weight of the bus conductor span supported by the bus post insulators

The resultant of these forces, correlated to the type of 'end Connection' of the rigid conductor at the bus

post insulator terminal, gives the values of the net Electromagnetic Force on the bus post insulator. The

values of forces obtained herewith forms valuable input for the design of bus post insulator support structure

and its foundation design.

In this report along with the force on bus post insulators, the maximum permissible unsupported span length

of the rigid bus conductor connecting the bus post insulators is also determined. The maximum permissible

span is calculated within the limits of 'Vertical Deflection' and permissible 'Fiber Stress' corresponding to the

type of 'End Connection' to the supporting bus post insulator terminals at each end of the span. The

calculation is performed on the method stipulated by IEEE 605 and the minimum of the span length obtained

from the calculation is conservatively considered as the maximum allowable unsupported span for all

influencing conditions.



3/45

1 Continuous Current Carrying Capacity

Bus bar continuous rating required Amps.

As per INDAL handbook current rating (nominal for outdoor) are as below:-

IPS Aluminum Tube

- Inches Amps.

As per INDAL handbook for the maximum operating temperature of the bus to 75 Deg. C, the deratingfactor to be adopted is 0.88; hence the derated current rating (nominal for outdoor) are as below:-

2 Inches IPS Al Tube - Schedule 40 = 1440 = Amps.

From the above it is summrised that, for the required current carrying capacity of 875 Amps.,

2 Inches IPS Al Tube - Schedule 40 is sufficient.

2 Thermal Effects on IPS Aluminium Tube as per IEEE 605

Bus Bar Selected Inch

Ambient temperature in deg. C (T2)

Final temperature in deg. C (T1)

2.1 Radiation Loss (Qr)From Stefan-Boltzman law,

Qr = 36.9 x 10-12

x e x ( T14-T2

4) x A (Clause No. 3.2.4, IEEE 605)

A = Surface area = 12 x d x = 12 x x = In2/Ft

T1 = Temperature of the bar,oK = + 273 =

0K

T2 = Ambient temperature,oK = + 273 =

0K

E = Emissivity factor =

Qr = 36.9 x 10-12

x 0.6 x (4-

4) x = Watts/Ft

2.2 Convective Loss (Qc)

Heat is dissipated from a tubular bus bar by forced convection

Heat transfer is considered at wind speed 2 fps and 1 atmospheric pressure.

Qc = 0.010 x x A (Clause No. 3.2.2, IEEE 605)d

0.4


d = outside diameter of bus bar in inch = Inch

= Temperature difference between conductor surface & ambient air in deg C = T1 - T2 =

= - = deg. C

Qc = 0.010 x x = Watts/Ft

( )0.4

2.3 Heat Gain Due To Solar Radiation (Qs)

Qs = h x s x d x 12

h = co-efficient of absorption of conductor surface = 0.5

s = intensity of solar radiation in Watts/Inch2= Watts/In

2

Assumed - 849.27 Watts/Mtr.2


Qs = 0.5 x 0.5479 x 2.375196 = Watts/Ft

90.0

875

323.0

2.00

50.0

2.38 89.54

Schedule 40 2.00

50.0

1440

12670.88 x

40.0

0.6

363.0 323.0 89.54

25.340

90.0 363.0

2.375

0.5479

2.38 89.54

89.54

90.0 50.0 40.0

2.38

7.81

12.8435

2.38



4/45

2.4 Conductor Resistance

Rt2 = 8.145 x 10-

x 1 + 0.00403 x C' x (T1 - 20) (Clause No. 3.2.9, IEEE 605)

C' x A2 61

Rt2 = Direct current resistance at maximum operating temperature

C' = Conductivity as % IACS =

A2 = Cross sectional area in In

d1 = outside diameter of bus bar in inch = Incht = Thickness of tube = Inch

d2 = inside diameter of bus bar in inch = d1- 2t = - 2 x = Inch

A2 = (/4) x (d1 -d2 ) = (/4) x ( - ) = In

T1 = Final temperature in deg. C = deg. C

Rt2 = 8.145 x 10-

x 1 + 0.00403 x x ( - 20 )

x

Rt2 = /Ft

2.5 Skin Effect Ratio

0.5

f = 0.0636 x F (Clause No. 6, IEEE 605)

r 5280 x Rt2

f = Skin Effect Ratio

r = outer diameter / 2 = / 2 = Inch

F = Frequency In Hz = 50 Hz

Rt2 = Direct current resistance at maximum operating temperature = / Ft

f = x 50 0.5

5280 x

=

2.6 Current Rating / Current Carrying Capability

0.5

I = Qc + Qr - Qs (Clause No. 3.2, IEEE 605)

Rt2x f

0.5

= + - = Amps.

x

61

90.0

90.0

55.0

0.0636

1.72924E-05 1.25

55.0 1.07422

25.340 12.8435

1.72924E-05

2.38

7.81 1183.91

2.067

2.38 2.067 1.07422

2.38 0.154

1.187598

1.72924E-05

1.25

1.1876 1.72924E-05

55.0

2.380.154



5/45

3 Bus Span Calculations For - 2 Inch IPS Aluminium Tube

Short circuit forces maximum span calculation are as per IEEE 605

Bus Bar Selected Inches Al Tube

Bus Bar outer Diameter in Inches

Bus Bar Cross Section Moment of inertia in cm4

(J)

Bus Bar Section modulus in cm3

(S)

Bus Bar ultimate tensile strength in Kg/mm2

(UT)

Modulus of Elasticity of Busbar in Kg/mm2

(E)

Bus bar Unit Self weight in Kgf/mtr (Fc)

Maximum allowable stress in Kg/mm2

(Fa)

Clamping material unit weight in Kgf/mtr (Fd) 7 Kg x 1 Nos. = 0.88

8.0 mtrs

Bus Phase to Phase Spacing in Mtrs (D)

Short Circuit Fault Level in kA (Isc)

Duration of the short-circuit current in seconds Tk

Rated short time in seconds Tkr

Wind Speed In Mtr/sec (V)

Allowable deflection in inches: 1 time dia of tube

3.1 Force Due to Wind (Fw)

Unit force due to wind on conductor is given by

Fw = C Cd Kz Gf V I (Dia + 2xrI) in Lbf/Ft (Equation No. 9, IEEE 605)

C = 2.132 x 10-4

for English units

Cd = Drag coefficient for smooth tubular conductor = 1 (Table 1, IEEE 605)

Kz = Height & Exposure factor = (Clause No. 9.2, for category D, IEEE 605)

Gf = Gust factor = (Clause No. 9.3, for category C&D, IEEE 605)

V = Wind speed in Miles/hr

= 55 x ( 1/ 0.44704) = Miles/hr

Dia = Outer Dia of Conductor in inches = Inch

I = Importance factor = (Clause No. 9.4, IEEE 605)

rI = radial ice thickness in Inches = 0 Inch

Fw = 2.132 x 10-4 x 1 x 1.16 x 0.85 x x x 1.15 x (2.3751968503937 + 2x0)

= Lbf/Ft = x 1.488 = Kgf/mtr

3.2 Short Circuit Force (Fsc)

Short Circuit Force is given by

Fsc = Kf x C x x (Df x 2 x Isc)2

(Equation No. 12, IEEE 605)

D

C = 5.4 x 10-7

for English units

Isc = RMS short circuit current in A kA

D = Phase to Phase Spacing of Busbars in Inch = x 39.37 = Inch

= Constant based on type of short circuit & conductor location =

(Table 2, for 3 phase short circuit on Y phase, IEEE 605)

Kf = Mounting structure flexibility factor = 1

(Figure 4, IEEE 605)

2.00

2.375

27.709

9.186

0.85

2

31.5

1

1

7030.67

1.871

20.5

123.03

2.375

1.15

123.03

55

20.5

1.16

31.5

2 78.74

0.866

123.03

8.69 8.69 12.93294



6/45

Df = Decrement factor = (Equation No. 11a, IEEE 605)

Ta = X 1

R 2 x x fX = System Reactance

R = System Resistance

f = System Frequency = 50 Hz

X =

RTa = x 1 =

2 x x 50

tf = Fault current duration = 1 sec.

Df =

Fsc = 1 x 5.4 x 10-7 x 0.866 x ( x 2 x ) = Lbf/Ft

= x 1.488 = Kgf/mtr

3.3 Gravitational Forces (Fg)

Fg = Fc + Fi + Fd (Equation No. 13, IEEE 605)

Fg = Total bus unit weight in kgf/mtr

Fc = Conductor unit weight in kgf/mtr =

Fi = Ice unit weight in kgf/mtr = 0

Fd = Clamping material unit weight I kgf/mtr =

Fg = 1.871 + 0 + 0. = kgf/mtr = / 1.488 = Lbf/Ft

3.4 Total Force (Ft)

Total force on conductor in Horizontal configuration is given by

Ft = (Equation No. 22, IEEE 605)

= ( + )2

+( )2 = Kgf/mtr

4 Calculation Of Allowable Spans

4.1 For continuous bus, the maximum allowable span length based on vertical deflection is (Ld)

Ld = C x (185 x E x J x Ya)1/4


Fg

C = for English units

E = Modulus of Elasticity in Lbf/In2

= Kgf/mm2

= x 1422.34 = Lbf/in2

J = Cross sectional moment of inertia in In4

= cm4

= x (1 / 41.623) = In4

Ya = Allowable Deflection as vertical dimension of tube = 1D = 1 x = Inch

Fg = Total bus unit weight in kgf/mtr = Kgf/mtr = x (1/1.488)

= Lbf/Ft

Ld = 1.86 x ( 185 x 10000000.03 x 2 )1/4

= In

Ld = / 39.37 = Mtrs.

Hence, maximum allowable span due to vertical deflection is mtrs.

12.35

78.74

15

15 0.0477

1.02

12.35 18.37

1.871

0.875

1.02 31500

2.746 2.746 1.85

12.9329 18.37 2.746 31.43

27.709

27.709 0.67

2.000

1.86

7031

7031 1E+07

0.67 355.52

1.845

355.52 9.03

2.00

2.746 2.746

1.85

9.03

f

a

2t-

Ta

f

T1 + 1 exp

t

2 2

W SC GF F F



7/45

4.2 Allowable Span For Fibre Stress (Ls)

Ls = C x ( (28/3) x Fa x S ) (Equation No. 29c, IEEE 605)

Ft


Fa = Maximum allowable stress in Lbf/In2

= Kg/mm2

= x 1422.33 = Lbf/In2

S = Section modulus in In3

For Inch Al Tube, S = cm3

= / 16.3871 = In3

Ft = Total Force in Lbf/Ft

= Kgf/mtr = / 1.488 = Lbf/ft

Ls = 3.46 x ( (28/3) x x ) = Inch

= / 39.37 = Mtrs.

5 Conclusion

As per IEEE 605, page number 25, Clause No. 11.3, the maximum allowable span is the lower of the spans

as calculated for Ld and Ls.

The maximum allowable bus span is Mtrs. for Inches Aluminium tube.

6.1 Natural Frequency of Conductor Span (fb)

fb = x k2 ( E x J ) (Equation No. 5, IEEE 605)

C x L2

m

C = 24 for English units

L = Span length in Feet = Mtrs. = / 0.305 = Feet


= Kgf/mm2

= x 1422.34 = Lbf/In2


= cm4

= x (1 / 41.623) = In4

m = Mass per unit length = Fc = Kgf/mtr = / 1.488 = Lbf/Ft

k = for two fixed ends

fb = x ( )2

( x ) = Hz

24 x ( )2

Twice the calculated natural frequency of the bus span = 2 x = Hz

6.2 Wind Induced Vibration or Aeolian Force Frequency (fa )

fa = C x V (Equation No. 6, IEEE 605)

d



= 55 x ( 1/ 0.44704) = Miles/hr

d = Conductor diameter in inches = Inches = x 2.54 = cm

fa = x = Hz

As per IEEE 605, page number 10, clause number 7.2.2, if twice the calculated natural frequency of the

bus span is greater than the Aeolian force frequency, then the bus span should be changed or bus should

be damped.

Since, twice the calculated natural frequency of the bus span is Hz, which is less than the

Aeolian force frequency of Hz, the selected bus span of meters is safe from

Aeolian vibrations.

2.0 9.186 9.186 0.56

3.46

20.5 20.5 29157.8

294.1 7.47

31.43 31.43 21.12

29157.8 0.56

7.47 2.00

7.47 7.47

294.1

21.12

27.709

27.709 0.67

1.871 1.871

24.49

7031

7031 1E+07

2.29

1.26

1.51

1.51 1.150.671E+07

3.26

123.03

2 2

24.49 1.26

1.15

5.08

2.29

79.0 7.47

5.08

3.26 123.03 79.0



8/45

7 Insulator Cantilever Strength Calculation As per IEEE 605

Effective Bus Span Length in Mtrs (Le) +

2

= Mtrs.

Height of Insulator Stack in mm (Hi)

Height of Centre Line of Bus over Insulator Height in mm (Hf)

= / 25.4

= Inch

Effective Insulator Diameter in mm (Di) =

7.1 Force Due To Wind On Bus (Fwb)

Fwb = Fw x Le (Equation No. 31, IEEE 605)

Fw = Unit Wind Force On Bus in Lbf/Ft = Lbf/Ft

Le = Effective Bus Span Length in Feet = x (39.37/12) = Feet

Fwb = x = Lbf

7.2 Bus Short Circuit Force (Fsb)

Fsb = Fsc x Le (Equation No. 30, IEEE 605)

Fsc = Unit short circuit force on Bus in Lbf/Ft = Lbf/FtLe = Effective Bus Span Length in Feet = x (39.37/12) = Feet

Fsb = x = Lbf

7.3 Force Due To Wind On Insulator (Fwi)

Fwi = C x Cd x Kz x Gf x Vx (Di + 2rI) x Hi (Equation No. 32, IEEE 605)

C = for English units Formula as per IEEE 605, 1987

Cd = Drag coefficient = 1 (Table 1, IEEE 605)


Gf = Gust Factor = (Clause No. 9.3, for category C&D, IEEE 605)


= 55 x ( 1/ 0.44704) = Miles/hr Di = Effective Dia of Insulator in Inch = = / 25.4 = Inch

Hi = Height of insulator in Feet = / (0.305 x 1000) = Feet

rI = Radial ice thickness in Inch = 0 Inch

Fwi = 1.776 x 10-5 x 1 x 1.16 x 0.85x ( )2x( + 2x0 ) x

Fwi = x = Lbf

7.4 Gravitation Force ( Fgb)

Fgb = Fg x Le (Equation No. 34, IEEE 605)

Fg = Total bus unit weight in Lbf/ Ft = Lbf/Ft

Le = Effective Bus Span Length in feet = x (39.37/12) = Feet

Fgb = x = Lbf

80

80

3.15

160

6 6

6

770

8.69

6 19.7

8.69 19.7 171.09

12.356 19.7

12.35 19.7 243.09

160 160 6.30

770 2.52

1.776 x 10-5

1.16

0.85

123.03

19.7

1.85 19.7 36.33

123.03 6.30 2.52

0.00001776 237349 4.21531

1.85

6



9/45

7.5 Insulator Cantilever Load (Fis)


Fis = K1 x Fwi + ( Hi + Hf ) x Fwb + K2 x ( Hi + Hf ) x Fsb

2 Hi Hi

K1 = Overload factor applied to wind forces =

K2 = Overload factor applied to short-circuit current forces = 1

1x + ( + ) x ) + 1 x

( + ) x = Lbf = x 4.44822

= Newton

Hence, Insulators with 4kN cantilever strength is selected.

1

4.22 30.31 3.15 171.09

459.316 459.316

30.31 2043.14

2 30.31

30.31 3.15 243.09



10/45

CLIENT:-


Project:- 220/66/11kV Electrical Distribution. Sales Ref.: IS/WR/

Description:


(Schedule 40)

Drawing No.:

Rev G71000-X0009-ED07Name

Date 15.04.2009

Prep. SGP

Chkd. SO

Siemens Ltd




11/45

1 Introduction




















12/45




IPS Aluminum Tube

- Inches Amps.










Qr = 36.9 x 10-12

x e x ( T14-T2




0K


0K


Qr = 36.9 x 10-12

x 0.6 x (4-

4) x = Watts/Ft





0.4




= - = deg. C


( )0.4


Qs = h x s x d x 12



2



Qs = 0.5 x 0.5479 x 2.375196 = Watts/Ft

12.8435

2.38

2.38

40.0 25.340

90.0 363.0

363.0

7.81

323.0 89.54

2.375

0.5479

2.38 89.54

89.54

90.0 50.0 40.0

Schedule 40 2.00 1440

12670.88 x

0.6

89.54

875

90.0

50.0 323.0

2.00

50.0

2.38



13/45


Rt2 = 8.145 x 10-


C' x A2 61






A2 = (/4) x (d1 -d2 ) = (/4) x ( - ) = In


Rt2 = 8.145 x 10-

x 1 + 0.00403 x x ( - 20 )

x

Rt2 = /Ft


0.5


r 5280 x Rt2





f = x 50 0.5

5280 x

=


0.5


Rt2x f

0.5

= + - = Amps.

x

1.1876 1.72924E-05

55.0

2.380.154

2.067

2.38 2.067 1.07422

2.38 0.154

1.72924E-05

1.187598

1183.91

0.0636

1.72924E-05 1.25

1.25

61

90.0

55.0 1.07422

25.340 12.8435

1.72924E-05

2.38

7.81

90.0

55.0



14/45






(J)


(S)


(UT)


(E)



(Fa)


8.0 mtrs










C = 2.132 x 10-4

for English units





= 55 x ( 1/ 0.44704) = Miles/hr




Fw = 2.132 x 10-4 x 1 x 1.16 x 0.85 x x x 1.15 x (2.3751968503937 + 2x0)






D

C = 5.4 x 10-7

for English units







31.5

2 78.74

0.866

123.03

2.375

1.15

123.03 123.03

8.69 8.69 12.93294

55

20.5

1.16

0.85

2

31.5

1

1

7030.67

1.871

20.5

2.00

2.375

27.709

9.186



15/45


Ta = X 1




X =

RTa = x 1 =

2 x x 50


Df =

Fsc = 1 x 5.4 x 10-7 x 0.866 x ( x 2 x ) = Lbf/Ft

= x 1.488 = Kgf/mtr







Fg = 1.871 + 0 + 0. = kgf/mtr = / 1.488 = Lbf/Ft




= ( + )2

+( )2 = Kgf/mtr



Ld = C x (185 x E x J x Ya)1/4


Fg



= Kgf/mm2

= x 1422.34 = Lbf/in2


= cm4

= x (1 / 41.623) = In4



= Lbf/Ft

Ld = 1.86 x ( 185 x 10000000.03 x 2 )1/4

= In

Ld = / 39.37 = Mtrs.


1.845

355.52 9.03

9.03

2.000 2.00

2.746 2.746

1.85

0.67 355.52

1.86

7031

7031 1E+07

27.709

27.709 0.67

1.871

0.875

2.746 2.746 1.85

12.9329 18.37 2.746 31.43

1.02

1.02 31500 12.35

78.74

12.35 18.37

15

15 0.0477

f

a

2t-

Ta

f

T1 + 1 exp

t

2 2

W SC GF F F



16/45



Ft



= Kg/mm2

= x 1422.33 = Lbf/In2



= / 16.3871 = In3


= Kgf/mtr = / 1.488 = Lbf/ft

Ls = 3.46 x ( (28/3) x x ) = Inch

= / 39.37 = Mtrs.

5 Conclusion






C x L2

m




= Kgf/mm2

= x 1422.34 = Lbf/In2


= cm4

= x (1 / 41.623) = In4



fb = x ( )2

( x ) = Hz

24 x ( )2




d



= 55 x ( 1/ 0.44704) = Miles/hr


fa = x = Hz



be damped.



Aeolian vibrations.

5.08

2.29

79.0 7.47

2 2 5.08

3.26 123.03 79.0

24.49 1.26

1.15 2.29

3.26

123.03

1.26

1.51

1.51 1.150.671E+07

7031 1E+07

27.709

27.709 0.67

1.871 1.871

7.47 2.00

7.47 7.47 24.49

7031

29157.8 0.56 294.1

21.12

294.1 7.47

2.0 9.186 9.186 0.56

31.43 31.43 21.12

3.46

20.5 20.5 29157.8



17/45



2

= Mtrs.



= / 25.4

= Inch






Fwb = x = Lbf




Fsb = x = Lbf











Fwi = 1.776 x 10-5 x 1 x 1.16 x 0.85x ( )2x( + 2x0 ) x

Fwi = x = Lbf





Fgb = x = Lbf

0.00001776 237349 4.21531

1.85

1.85 19.7 36.33

6 19.7

6.30

770 2.52

123.03 6.30 2.52

1.776 x 10-5

1.16

0.85

123.03160 160

12.356 19.7

12.35 19.7 243.09

8.69

6 19.7

8.69 19.7 171.09

6

770

80

80

3.15

160

6 6



18/45




2 Hi Hi



1x + ( + ) x ) + 1 x

( + ) x = Lbf = x 4.44822

= Newton


459.316

30.31 2043.14

2 30.31

30.31 3.15 243.09 459.316

1

4.22 30.31 3.15 171.09



19/45

CLIENT:-


Project:- 220/66/11kV Electrical Distribution. Sales Ref.: IS/

Description:


(Schedule 40)

Drawing No.:


Date 15.04.2009

Prep. SGP

Chkd. SO

Siemens Ltd




20/45

1 Introduction




















21/45




IPS Aluminum Tube

- Inches Amps.










Qr = 36.9 x 10-12

x e x ( T14-T2




0K


0K


Qr = 36.9 x 10-12

x 0.6 x (4-

4) x = Watts/Ft





0.4




= - = deg. C


( )0.4


Qs = h x s x d x 12



2



Qs = 0.5 x 0.5479 x 3.5 x 12 = Watts/Ft

18.9258

3.50

3.50

40.0 31.977

90.0 363.0

363.0

11.51

323.0 131.95

3.500

0.5479

3.50 131.95

131.95

90.0 50.0 40.0

Schedule 40 3.00 2350

20680.88 x

0.6

131.95

263

90.0

50.0 323.0

3.00

50.0

3.50



22/45


Rt2 = 8.145 x 10-


C' x A2 61






A2 = (/4) x (d1 -d2 ) = (/4) x ( - ) = In


Rt2 = 8.145 x 10-

x 1 + 0.00403 x x ( - 20 )

x

Rt2 = /Ft


0.5


r 5280 x Rt2





f = x 50 0.5

5280 x

=


0.5


Rt2x f

0.5

= + - = Amps.

x

1.75 8.33058E-06

55.0

3.500.216

3.068

3.50 3.068 2.22984

3.50 0.216

8.33058E-06

1.75

1964.56

0.0636

8.33058E-06 1.23

1.23

61

90.0

55.0 2.22984

31.977 18.9258

8.33058E-06

3.50

11.51

90.0

55.0



23/45






(J)


(S)


(UT)


(E)



(Fa)


12.0 mtrs










C = 2.132 x 10-4

for English units





= 55 x ( 1/ 0.44704) = Miles/hr




Fw = 2.132 x 10-4 x 1 x 1.16 x 0.85 x x x 1.15 x (3.5 + 2x0)






D

C = 5.4 x 10-7

for English units







40

4.25 167.32

0.866

123.03

3.500

1.15

123.03 123.03

12.81 12.81 19.05748

55

20.5

1.16

0.85

4.25

40

1

1

7030.67

3.884

20.5

3.00

3.500

125.606

28.258



24/45


Ta = X 1




X =

RTa = x 1 =

2 x x 50


Df =

Fsc = 1 x 5.4 x 10-7 x 0.866 x ( x 2 x ) = Lbf/Ft

= x 1.488 = Kgf/mtr







Fg = 3.884 + 0 + 0. = kgf/mtr = / 1.488 = Lbf/Ft




= ( + )2

+( )2 = Kgf/mtr



Ld = C x (185 x E x J x Ya)1/4


Fg



= Kgf/mm2

= x 1422.34 = Lbf/in2


= cm4

= x (1 / 41.623) = In4



= Lbf/Ft

Ld = 1.86 x ( 185 x 10000000.03 x 3 )1/4

= In

Ld = / 39.37 = Mtrs.


3.002

508.33 12.91

12.91

3.000 3.00

4.467 4.467

3.00

3.02 508.33

1.86

7031

7031 1E+07

125.606

125.606 3.02

3.884

0.583333

4.467 4.467 3.00

19.0575 13.94 4.467 33.30

1.02

1.02 40000 9.37

167.32

9.37 13.94

15

15 0.0477

f

a

2t-

Ta

f

T1 + 1 exp

t

2 2

W SC GF F F



25/45



Ft



= Kg/mm2

= x 1422.33 = Lbf/In2



= / 16.3871 = In3


= Kgf/mtr = / 1.488 = Lbf/ft

Ls = 3.46 x ( (28/3) x x ) = Inch

= / 39.37 = Mtrs.

6 Conclusion






C x L2

m




= Kgf/mm2

= x 1422.34 = Lbf/In2


= cm4

= x (1 / 41.623) = In4



fb = x ( )2

( x ) = Hz

24 x ( )2




d



= 55 x ( 1/ 0.44704) = Miles/hr


fa = x = Hz



be damped.



Aeolian vibrations.

7.62

1.17

52.6 12.73

3 3 7.62

3.26 123.03 52.6

41.72 2.61

0.58 1.17

3.26

123.03

2.61

1.51

1.51 0.583.021E+07

7031 1E+07

125.606

125.606 3.02

3.884 3.884

12.73 3.00

12.73 12.73 41.72

7031

29157.8 1.72 501.0

22.38

501.0 12.73

3.0 28.258 28.258 1.72

33.30 33.30 22.38

3.46

20.5 20.5 29157.8



26/45



2

= Mtrs.



= / 25.4

= Inch






Fwb = x = Lbf




Fsb = x = Lbf











Fwi = 1.776 x 10-5 x 1 x 1.16 x 0.85x ( )2x( + 2x0 ) x

Fwi = x = Lbf





Fgb = x = Lbf

0.00001776 1307151 23.215

3.00

3.00 37.7 113.27

11.5 37.7

11.61

2300 7.54

123.03 11.61 7.54

1.776 x 10-5

1.16

0.85

123.03295 295

9.3711.5 37.7

9.37 37.7 353.55

12.81

11.5 37.7

12.81 37.7 483.22

11.5

2300

130

130

5.12

295

11.5 11.5



27/45




2 Hi Hi



1x + ( + ) x ) + 1 x

( + ) x = Lbf = x 4.44822

= Newton


895.668

90.55 3984.13

2 90.55

90.55 5.12 353.55 895.668

1

23.22 90.55 5.12 483.22



28/45

CLIENT:-


Project:- 220/66/11kV Electrical Distribution. Sales Ref.: IS/

Description:


(Schedule 40)

Drawing No.:

Rev G71000-X0009-ED07

Chkd. SO

Date 15.04.2009

Siemens Ltd


Prep. SGP

Name



29/45

1 Introduction




















30/45




IPS Aluminum Tube

- Inches Amps.










Qr = 36.9 x 10-12

x e x ( T14-T2




0K


0K


Qr = 36.9 x 10-12

x 0.6 x (4-

4) x = Watts/Ft





0.4




= - = deg. C


( )0.4


Qs = h x s x d x 12



2



Qs = 0.5 x 0.5479 x 4.5 x 12 = Watts/Ft

95.0

1250

318.0

4.00

45.0

4.50 169.65

Schedule 40 4.00

45.0

3050

26840.88 x

50.0

0.6

368.0 318.0 169.65

46.476

95.0 368.0

4.500

0.5479

4.50 169.65

169.65

95.0 45.0 50.0

4.50

14.79

30.4744

4.50



31/45


Rt2 = 8.145 x 10-


C' x A2 61






A2 = (/4) x (d1 -d2 ) = (/4) x ( - ) = In


Rt2 = 8.145 x 10-

x 1 + 0.00403 x x ( - 20 )

x

Rt2 = /Ft


0.5


r 5280 x Rt2





f = x 50 0.5

5280 x

=


0.5


Rt2x f

0.5

= + - = Amps.

x

61

95.0

95.0

55.0

0.0636

5.93699E-06 1.13

55.0 3.17415

46.476 30.4744

5.93699E-06

4.50

14.79 3045.32

4.026

4.50 4.026 3.17415

4.50 0.237

2.25

5.93699E-06

1.13

2.25 5.93699E-06

55.0

4.500.237



32/45






(J)


(S)


(UT)


(E)



(Fa)


12.0 mtrs










C = 2.132 x 10-4

for English units





= 55 x ( 1/ 0.44704) = Miles/hr




Fw = 2.132 x 10-4 x 1 x 1.16 x 0.85 x x x 1.15 x (4.5 + 2x0)






D

C = 5.4 x 10-7

for English units







4.00

4.500

301.039

52.675

0.85

3

40

3

3

7030.67

5.529

20.5

123.03

4.500

1.15

123.03

55

20.5

1.16

40

3 118.11

0.866

123.03

16.47 16.47 24.50248



33/45


Ta = X 1




X =

RTa = x 1 =

2 x x 50


Df =

Fsc = 1 x 5.4 x 10-7 x 0.866 x ( x 2 x ) = Lbf/Ft

= x 1.488 = Kgf/mtr







Fg = 5.529 + 0 + 0. = kgf/mtr = / 1.488 = Lbf/Ft




= ( + )2

+( )2 = Kgf/mtr



Ld = C x (185 x E x J x Ya)1/4


Fg



= Kgf/mm2

= x 1422.34 = Lbf/in2


= cm4

= x (1 / 41.623) = In4



= Lbf/Ft

Ld = 1.86 x ( 185 x 10000000.03 x 4 )1/4

= In

Ld = / 39.37 = Mtrs.


12.87

118.11

15

15 0.0477

1.01

12.87 19.15

5.529

0.583333

1.01 40000

6.112 6.112 4.11

24.5025 19.15 6.112 44.08

301.039

301.039 7.23

4.000

1.86

7031

7031 1E+07

7.23 628.41

4.108

628.41 15.96

4.00

6.112 6.112

4.11

15.96

f

a

2t-

Ta

f

T1 + 1 exp

t

2 2

W SC GF F F



34/45



Ft



= Kg/mm2

= x 1422.33 = Lbf/In2



= / 16.3871 = In3


= Kgf/mtr = / 1.488 = Lbf/ft

Ls = 3.46 x ( (28/3) x x ) = Inch

= / 39.37 = Mtrs.

6 Conclusion






C x L2

m




= Kgf/mm2

= x 1422.34 = Lbf/In2


= cm4

= x (1 / 41.623) = In4



fb = x ( )2

( x ) = Hz

24 x ( )2




d



= 55 x ( 1/ 0.44704) = Miles/hr


fa = x = Hz



be damped.



Aeolian vibrations.

4.0 52.675 52.675 3.21

3.46

20.5 20.5 29157.8

594.6 15.10

44.08 44.08 29.62

29157.8 3.21

15.10 4.00

15.10 15.10

594.6

29.62

301.039

301.039 7.23

5.529 5.529

49.51

7031

7031 1E+07

1.07

3.72

1.51

1.51 0.547.231E+07

3.26

123.03

4 4

49.51 3.72

0.54

10.16

1.07

39.5 15.10

10.16

3.26 123.03 39.5



35/45



2

= Mtrs.



= / 25.4

= Inch






Fwb = x = Lbf




Fsb = x = Lbf











Fwi = 1.776 x 10-5 x 1 x 1.16 x 0.85x ( )2x( + 2x0 ) x

Fwi = x = Lbf





Fgb = x = Lbf

172

172

6.77

182

12 12

12

1472

16.47

12 39.4

16.47 39.4 648.29

12.8712 39.4

12.87 39.4 506.75

182 182 7.17

1472 4.83

1.776 x 10-5

1.16

0.85

123.03

39.4

4.11 39.4 161.72

123.03 7.17 4.83

0.00001776 516125 9.16639

4.11

12



36/45




2 Hi Hi



1x + ( + ) x ) + 1 x

( + ) x = Lbf = x 4.44822

= Newton


1

9.17 57.95 6.77 648.29

1294.6 1294.6

57.95 5758.66

2 57.95

57.95 6.77 506.75



37/45

CLIENT:-


Project:- 220/66/11kV Electrical Distribution. Sales Ref.: IS/WR

Description:

Amapacity Calculation of 4.5" IPS AL Tube.

(Schedule 40)

Drawing No.:


Date 15.04.2009

Prep. SGP

Chkd. SO

Siemens Ltd




38/45

1 Introduction




















39/45




IPS Aluminum Tube

- Inches Amps.


4.5 Inches IPS Al Tube - Schedule 40 = 3420 = Amps.


4.5 Inches IPS Al Tube - Schedule 40 is sufficient.






Qr = 36.9 x 10-12

x e x ( T14-T2




0K


0K


Qr = 36.9 x 10-12

x 0.6 x (4-

4) x = Watts/Ft





0.4




= - = deg. C


( )0.4


Qs = h x s x d x 12



2



Qs = 0.5 x 0.5479 x 5 x 12 = Watts/Ft

27.0368

5.00

5.00

40.0 39.607

90.0 363.0

363.0

16.44

323.0 188.50

5.000

0.5479

5.00 188.50

188.50

90.0 50.0 40.0

Schedule 40 4.50 3420

30100.88 x

0.6

188.50

1250

90.0

50.0 323.0

4.50

50.0

5.00



40/45


Rt2 = 8.145 x 10-


C' x A2 61






A2 = (/4) x (d1 -d2 ) = (/4) x ( - ) = In


Rt2 = 8.145 x 10-

x 1 + 0.00403 x x ( - 20 )

x

Rt2 = /Ft


0.5


r 5280 x Rt2





f = x 50 0.5

5280 x

=


0.5


Rt2x f

0.5

= + - = Amps.

x

2.5 5.03945E-06

55.0

5.000.247

4.506

5.00 4.506 3.68608

5.00 0.247

5.03945E-06

2.5

3005.68

0.0636

5.03945E-06 1.10

1.10

61

90.0

55.0 3.68608

39.607 27.0368

5.03945E-06

5.00

16.44

90.0

55.0



41/45


42/45


Ta = X 1




X =

RTa = x 1 =

2 x x 50


Df =

Fsc = 1 x 5.4 x 10-7 x 0.866 x ( x 2 x ) = Lbf/Ft

= x 1.488 = Kgf/mtr







Fg = 6.423 + 0 + 0. = kgf/mtr = / 1.488 = Lbf/Ft




= ( + )2

+( )2 = Kgf/mtr



Ld = C x (185 x E x J x Ya)1/4


Fg



= Kgf/mm2

= x 1422.34 = Lbf/in2


= cm4

= x (1 / 41.623) = In4



= Lbf/Ft

Ld = 1.86 x ( 185 x 10000000.03 x 4.5 )1/4

= In

Ld = / 39.37 = Mtrs.


4.887

679.30 17.25

17.25

4.500 4.50

7.271 7.271

4.89

10.44 679.30

1.86

7031

7031 1E+07

434.683

434.683 10.44

6.423

0.848485

7.271 7.271 4.89

27.225 13.94 7.271 41.81

1.02

1.02 40000 9.37

167.32

9.37 13.94

15

15 0.0477

f

a

2t-

Ta

f

T1 + 1 exp

t

2 2

W SC GF F F



43/45



Ft



= Kg/mm2

= x 1422.33 = Lbf/In2



= / 16.3871 = In3


= Kgf/mtr = / 1.488 = Lbf/ft

Ls = 3.46 x ( (28/3) x x ) = Inch

= / 39.37 = Mtrs.

6 Conclusion






C x L2

m




= Kgf/mm2

= x 1422.34 = Lbf/In2


= cm4

= x (1 / 41.623) = In4



fb = x ( )2

( x ) = Hz

24 x ( )2




d



= 55 x ( 1/ 0.44704) = Miles/hr


fa = x = Hz



be damped.



Aeolian vibrations.

11.43

0.92

35.1 17.25

4.5 4.5 11.43

3.26 123.03 35.1

56.57 4.32

0.46 0.92

3.26

123.03

4.32

1.51

1.51 0.4610.441E+07

7031 1E+07

434.683

434.683 10.44

6.423 6.423

17.25 4.50

17.25 17.25 56.57

7031

29157.8 4.18 696.0

28.10

696.0 17.68

4.5 68.454 68.454 4.18

41.81 41.81 28.10

3.46

20.5 20.5 29157.8



44/45



2

= Mtrs.



= / 25.4

= Inch






Fwb = x = Lbf




Fsb = x = Lbf











Fwi = 1.776 x 10-5 x 1 x 1.16 x 0.85x ( )2x( + 2x0 ) x

Fwi = x = Lbf





Fgb = x = Lbf

0.00001776 1307151 23.215

4.89

4.89 54.1 264.54

16.5 54.1

11.61

2300 7.54

123.03 11.61 7.54

1.776 x 10-5

1.16

0.85

123.03295 295

9.3716.5 54.1

9.37 54.1 507.26

18.30

16.5 54.1

18.30 54.1 990.45

16.5

2300

130

130

5.12

295

16.5 16.5



45/45




2 Hi Hi



1x + ( + ) x ) + 1 x

( + ) x = Lbf = x 4.44822

= Newton


1593.97

90.55 7090.34

2 90.55

90.55 5.12 507.26 1593.97

1

23.22 90.55 5.12 990.45

claulation for check suitability of ips

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