Ice Loading Effect - Sag tension

SAG TENSION CALCULATION

CG-309(W)-ED-101

0 FIRST ISSUED JD PKK SKK 17-Feb-14

REV. NO. DESCRIPTIONINITIAL SIGN INITIAL SIGN INITIAL SIGN

DATEPRPD CHKD APVD

NOA. NO. CC-CS/296/NR2/SS-1770/3/G1/NOA-II/4683 DATED 16/05/2013

PROJECT AUGMENTATION OF TRANSFORMERS IN NR PART-A

SUBSTATION 400/220 kV WAGOORA EXTENSION S/S

OWNER

DOCUMENT TITLE :

DOCUMENT NO.

POWER GRID CORPORATION OF INDIA LIMITED (A GOVERNMENT OF INDIA ENTERPRISE)

EPC CONTRACTOR

CROMPTON GREAVES LTD.EPD DIVISION 3rd Floor, Cyber Green, Tower -A

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Construction of 220kV Bays at Wagoora SubstationSAG TENSION CALCULATION FOR 220kV JACK BUS - TWIN ACSR MOOSE (41m Span) at 16.6m height

The Sag-Tension is calculated for the following

i) For Span length 41 Mtr.

Sl.no Swing (m)

1 -15 2931323.28 1750 0.9208 1.5795 1726082.25 1030.4711033 0.8806

2 -10 2915743.43134591 1740.6988 0.9257 1.5879 1722823.89021 1028.52586246 0.8822

3 -5 2900403.73444983 1731.5410 0.9306 1.5963 1719583.82383 1026.59154283 0.8839

4 0 2885298.19838806 1722.5230 0.9355 1.6047 1716361.88107 1024.668043 0.8856

5 5 2870421.03316379 1713.6414 0.9404 1.6130 1713157.89426 1022.75526287 0.8872

6 10 2855766.64408025 1704.8927 0.9452 1.6213 1709971.69786 1020.85310362 0.8889

7 15 2841329.62347543 1696.2738 0.9500 1.6295 1706803.12848 1018.9614677 0.8905

8 20 2827104.74287434 1687.7815 0.9548 1.6377 1703652.02481 1017.08025881 0.8922

9 25 2813086.94553457 1679.4129 0.9595 1.6459 1700518.22757 1015.20938186 0.8938

10 30 2799271.3393622 1671.1650 0.9643 1.6540 1697401.57952 1013.34874297 0.8955

11 35 2785653.19017697 1663.0350 0.9690 1.6621 1694301.92541 1011.49824947 0.8971

12 40 2772227.91530666 1655.0201 0.9737 1.6701 1691219.11194 1009.65780983 0.8987

13 45 2758991.07749223 1647.1177 0.9783 1.6781 1688152.98772 1007.82733367 0.9004

14 50 2745938.37908608 1639.3252 0.9830 1.6861 1685103.40326 1006.00673175 0.9020

15 55 2733065.65652736 1631.6402 0.9876 1.6941 1682070.21095 1004.19591594 0.9036

16 60 2720368.87507877 1624.0602 0.9922 1.7020 1679053.26498 1002.39479919 0.9052

17 65 2707844.12381073 1616.5829 0.9968 1.7098 1676052.42137 1000.60329556 0.9069

18 70 2695487.61081933 1609.2061 1.0014 1.7177 1673067.5379 998.821320128 0.9085

19 75 2683295.65866551 1601.9275 1.0059 1.7255 1670098.47411 997.048789041 0.9101

SUMMARY

1 Maximum working tension per conductor T = 1750

2 Maximum sag of Conductor S = 1.006 M

3 Height of Jack bus level taking H = 16.60 M

4 Height of equipment bus = 9 M

5 Vertical Clearance between lower most conductor & Equipment = 6.59 M

6 Minimum clearance between phase to phase for 220kV as per CBIP Manual = 2.1 M

The max proposed tension is 2000kg/conductor. The maximum ambient temperature is 50°C. However for the calculation of SAG, the stress,tension at still air condition for various temperatures are calculated and tabulated.similarly the stress, tension,deflection and swing for various temperatures at full load wind condition are calculated and tabulated.

Temp °C

Stress (Full wind) (kg/M2)

TensionTfull wind (kg)

SAG AT FULL Wind (M)

Stress (Still air) (kg/M2)

Tension Tstill air (kg)

SAG AT STILL Wind

(M)

Kg at 0⁰ C

Vclr

Since the calculated vertical clearance between Equipment and Lower most conductor is greater than the minimum clearance between phase to phase, The selected height of tower is adequate.

4 OF 16

-15 -10 -5 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75980

990

1000

1010

1020

1030

1040

0.865

0.87

0.875

0.88

0.885

0.89

0.895

0.9

0.905

0.91

0.915

Still Air ConditionTemp Vs Tension & Sag

Tension

Sag

Temp Deg C

Ten

sio

n k

g

Sag

m

-15 -10 -5 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 751500

1550

1600

1650

1700

1750

1800

0.86

0.88

0.9

0.92

0.94

0.96

0.98

1

1.02

Full Wind ConditionTemp Vs Tension & Sag

Tension

Sag

Temp Deg C

Ten

sio

n k

g

Sag

m

5 OF 16

6 OF 16

Construction of 220kV Bays at Wagoora Substation

REFERENCE :

1) TNEB Power Engineers Handbook

2)

3) IS802( Partl/Sec 1 ) :1995 - Use Of Structural Steel In Overhead Transmission Line Towers-Code Of Practise-Third Revision

4) IS 5613 (Part 1/ Section 1): 1995

A. SITE INFORMATION

WIND ZONE 2

BASIC WIND SPEED = 39 M/Sec

TERRAIN CATEGORY 2

RELIABILITY LEVEL 1

HEIGHT OF STRINGING CONDUCTOR 16.6 M

B CALCULATION FOR WIND PRESSURE

DESIGN WIND PRESSURE =

=

= M/Sec

= 28.36 M/Sec

Risk Coefficient = 1

Terrain Roughness Co-efficient = 1

"=> = 28.36 M/Sec

Hence, Design Wind Pressure = 482.70

= 49.20

C Basic Information

Distance Between Two Girders L = 41 M

Girders Width = 1.5 M

Span Length Excluding Grider Width a = 39.5 M

Length Of Long Rod Insulator = 2.175 M

Length Of Stringing Hardware = 0.8 M

Length Of ACSR Conductor = 33.55 M

No. of Conductor Per Phase N = 2

Maximum Temperature Considered = 75

Minimum Temperature Considered = -15

Temperature Range Taken -15 to 75

Maximum Initial Tension (As Per Technical Specification, Annex. - D) = 1750

D Conductor Data

Type of Conductor ACSR MOOSE

Diameter of Conductor = 31.77 mm

Weight of Conductor = 2.004 Kg/M

Area of Cross Section of Conductor = 597

Modulus Elasticity of Conductor = 6860

Co-efficient of Thermal Expansion = 0.0000193

ANNEXURE-A SAG TENSION CALCULATION FOR 220kV JACK BUS - TWIN ACSR MOOSE (41m Span) at 16.6m height

IS:875 (Part 3) - 1987-Code of practice or design loads (other than earthquake) for buildings and structures, Part-3 Windloads (Second Revision)

Vb

Pd 0.6 x Vd N/M2

Where, Vd is Design Wind Speed Vd Vr x K1 x K2

Vr is Refrence Wind Speed Vr Vb / Ko

Where, Ko = 1.375 Vr

K1

K2

Vd

Pd N/M2

Pd Kg/M2

Gw

L1

L2

L3

T1 ⁰C

T0 ⁰C

⁰C

IT Kg at 0⁰ C

Dc

Wc

Ac mm2

Es Kg/mm2

Et Per ⁰C

7 OF 16

E CALCULATION FOR NO. OF SPACERS

Spacer Span = 2.5 M

Nos. of Spacers = Nos.

= 13 Nos.

Weight of Each Spacer = 2 Kg

Spacing Between Sub-Conductor SS = 0.25 M

F Loading on Conductor

Weight of Conductor including Spacers per Unit Length =

= 2.404 Kg/M

Effect of Ice Loading on Conductor

Radius of Conductor r = 0.0159 M

Thickness of Ice (As per Section Project, Cl. 4.2.2, Pg. 20) t = 0.0150 M

Volume of Ice Per unit Length = = 0.0022

Density of Ice = 0.917

= 917

Weight of ice Per Unit Length of Conductor = Density of ice x Vol. of Ice Per unit Length

= 2.0204 Kg/M

Weight of ice on Conductor cosidering ice loading effect = 135.57 Kg

Effect of Ice Loading on Spacer

Diameter of Axial Length in Spacer = 32 mm

Radius of Axial Length in Spacer r = 0.0160 M




= 917

Weight of ice Per Unit Length on Spacer = Density of ice x Vol. of Ice Per unit Length

= 2.0303 Kg/M

Axial Length of Spacer is approx. same as Sub Conductor Spacing

Hence Axial Length of Spacer = 0.25 M

No. of Spacers 13 Nos.

Total Axial Length = 3.355 Mtr.

Weight of Spacer cosidering ice loading effect = 6.81 Kg

Net Weight of ice on Conductor including Spacers considering Ice Loading Effect = 142.38 Kg

Weight of ICE ON Conductor including Spacers considering Ice Loading Effect per Unit Length = 2.0209 Kg/M

Effect of Wind Loading on Conductor

Projected Area per unit Length on which Wind is Acting =(r+t).1 Sq.M = 0.031

Wind Pressure On Conductor =

= 1 ; = 2.021 {As Per IS 802 - Table-7} = 99.46

Wind Loading Per Unit Length = 6.1667 Kg/M

Resultant Load on Conductor =

= 7.590 Kg/M

SINθ = 1.715

G Loading on Insulator

Length of Long Rod Insulator = 2175 mm

Diameter of Long Rod Insulator = 135 mm

Ls

Ns (L3 / Ls )

Ws

W1 Wc +((Ns x Ws ) / (L3 x N))

Dice gm / Cm3

Kg / M3

Wice

Wice-conductor

Dice gm / Cm3

Kg / M3

Wice

Wice-spacer

W2

M2

Pc Pd * Cd * Gc

Cd Gc Pc Kg/M2

W3

W4 Sqrt{(W1 +W2)2+W32}

W4

L4

Dins

π {(r+ t )2̂−r 2̂}∗1

π {(r+ t )2̂−r 2̂}∗1

8 OF 16

Weight of Long Rod Insulator = 9 Kg

No. Of Long Rod Insulator per String = 2 Nos.

Total Weight of Insulator = 18 Kg

Effect of Ice Loading on Insulator String

Radius of Long Rod Insulator r = 0.0675 M




= 917

Weight of ice Per Unit Length of Insulator = Density of ice x Vol. of Ice Per unit Length

= 6.4798 Kg/M

Weight of ice on Insulator cosidering ice loading effect = 28.19 Kg

Effect of Wind Loading on Insulator String

Wind Pressure On Insulator String =

= 1.2 ; = 2.105 {As Per IS 802 - Table 6} = 124.28

Wind on Insulator String =

= 36.49 Kg

Resultant Load on Insulator String =

= 58.863 Kg/M

H Loading on Stringing Hardware

Weight of Stringing Hardware = 29 Kg

Effect of Ice Loading on String Hardware

Radius / thickness of Stringing Hardware r = 0.0100 M




= 917

Weight of ice Per Unit Length of Stringing Harware

= Density of ice x Vol. of Ice Per unit Length

= 1.5119 Kg/M

Length Of Stringing Hardware = 0.8 M

Weight of ice on Stringing Hardware cosidering ice loading effect = 1.21 Kg

Effect of Wind Loading on Stringing Hardware

Wind Pressure On Stringing Hardware =

= 1.2 ; = 2.105 {As Per IS 802 - Table 6} = 124.28

Width of Hardware = 0.25 M

Wind on Stringing Hardware =

= 24.86 Kg

Resultant Load on Stringing Hardware =

= 39.121 Kg/M

I) FULL WIND LOAD CONDITION-LOAD DISTRIBUTION

LOAD DISTRIBUTION

Wins

Nins

W5

Dice gm / Cm3

Kg / M3

Wice

Wice-insulator

W6

Pins Pd * Cd * Gc

Cd Gc Pins Kg/M2

W7 0.5*Pins*Lins*Dins*Nins

W7

W8 Sqrt{(W5 +W6)2+W72}

W8

W9

Dice gm / Cm3

Kg / M3

Wice

Wice-Stringing

L2

W10

PHS Pd * Cd * Gc

Cd Gc PHS Kg/M2

WH

W11 PHS*L2*WH

W11

W12 Sqrt{(W9 +W10)2+W112}

W12

π {(r+ t )2̂−r 2̂}∗1

π {(r+ t )2̂−r 2̂}∗1

9 OF 16

SHEAR FORCE DIAGRAM

REACTION AT EACH END R

225.30 Kg

TOTAL CROSS FORCE AREA UPTO MAX SAG

245.02 Kg-M

247.58 Kg-M

50.93 Kg-M

1067.91 Kg-M

TOTAL 1611.44 Kg-M

0.9208 M

W8 +W12 +1/2(L3 *W4)

I1 R*L1/2

I2 (R-W8)*(L1/2 + L2/2)

I3 (R-W8-W12)*(L2/2)

I4 (R-W8-W12)*(1/2 * L3/2)

TCFA

DEFLECTION at 0⁰ C AT WIND LOAD CONDITION DWL TCFA / IT

10 OF 16

II) STILL WIND LOAD CONDITION-LOAD DISTRIBUTION

LOAD DISTRIBUTION

SHEAR FORCE DIAGRAM

REACTION AT EACH END ( STILL AIR CONDITION)

112.43 Kg

TOTAL CROSS FORCE AREA UPTO MAX SAG

122.26 Kg-M

132.88 Kg-M

29.69 Kg-M

622.58 Kg-M

TOTAL 907.41 Kg-M

R0((W5+W6)/2)+((W9+W10)/2)+

((1/2)*(W1+W2)*L3)

I01 R0*L1/2

I02 (R0 - (W5+W6)/2)*(L1/2+L2/2)

I03 (R0 - (W5+W6)/2) - ((W9+W10)/2))*(L2/2)

I04(R0 - (W5+W6)/2) -

((W9+W10)/2))*(1/2)*(L3/2)

T0CFA

11 OF 16

TENSION CALCULATION AT Full Wind CONDITION + ICE LOADING EFFECT

A Conductor Data






= 6860000000







Temperature Difference = 5


Wind pressure at zero Deg celcius = 2.93132328

= 2931323.28

Weight of Conductor including Spacers per Unit Length = 7.590 Kg/M

δ is density of Conductor material δ = (w/A) Constant δ = 12713.46

Weight of Conductor including Spacers per Unit Length considering Ice Loading & Wind Loading = 7.590 Kg/M

Loading Factor = (W/w) = 3.157

Loading Factor For Full Wind = 3.157

K = -5.74E+07

α Coefficient of linear expansion per deg C α = 0.0000193

= -15

= -10

t = 5

αtE = 6.62E+05

N = K - αtE N = -5.81E+07

M = 5.18E+20

Where,

a = 1

b = -5.81E+07

c = 0

d = 5.18E+20

= 2915743.43135

Dc

Wc

Ac mm2

Es Kg/mm2

Kg/M2

Et Per ⁰C

L3

T1 ⁰C

T0 ⁰C

⁰C

Δt ⁰C

IT Kg at 0⁰ C

f1 Kg/mm2

f1 Kg/M2

W1

Kg/M3

W4

q1 q1

q2

Per ⁰C

t1 ⁰C

t2 ⁰C

Temperature difference = ( t2 - t1 ) ⁰C

Evaluation of f2 from the cubic equation

On Solving the above cubic equation we get f 2

f 2 Kg/M2

K=f 1−l2δ2 q1

2E

24 f 12

M=l2δ2q2

2E

24

f 23−f 2

2N−M=0

12 OF 16

13 OF 16

INITIAL STRESS CALCULATION AT Stand Still Air CONDITION + ICE LOADING EFFECT

A Conductor Data






= 6860000000







= 2931323.28





Loading Factor ICE only = 1.840

K = -5.74E+07


t = 0

αtE = 0.00E+00

N = K - αtE N = -5.74E+07

M = 1.76E+20

Where,

a = 1

b = -5.74E+07

c = 0

d = 1.76E+20

= 1726082.25009

Dc

Wc

Ac mm2

Es Kg/mm2

Kg/M2

Et Per ⁰C

L3

Δt ⁰C

IT Kg at 0⁰ C

f1 Kg/mm2

f1 Kg/M2

W1

Kg/M3

W4

q1 q1

q2

Per ⁰C




f 2 Kg/M2

K=f 1−l2δ2 q1

2E

24 f 12

M=l2δ2q2

2E

24

f 23−f 2

2N−M=0

14 OF 16

15 OF 16

TENSION CALCULATION AT Stand Still Air CONDITION + ICE LOADING EFFECT

A Conductor Data






= 6860000000








Maximum Initial Tension (As Per Technical Specification, Annex. - D) =


= 1726082.25





Loading Factor For Stand Still Air = 3.157

K = -1.72E+08


= -15

= -10

t = 5

αtE = 6.62E+05

N = K - αtE N = -1.73E+08

M = 5.18E+20

Where,

a = 1

b = -1.73E+08

c = 0

d = 5.18E+20

= 1722823.89021

Dc

Wc

Ac mm2

Es Kg/mm2

Kg/M2

Et Per ⁰C

L3

T1 ⁰C

T0 ⁰C

⁰C

Δt ⁰C

IT

f1 Kg/mm2

f1 Kg/M2

W1

Kg/M3

W4

q1 q1

q2

Per ⁰C

t1 ⁰C

t2 ⁰C




f 2 Kg/M2

K=f 1−l2δ2 q1

2E

24 f 12

M=l2δ2q2

2E

24

f 23−f 2

2N−M=0

16 OF 16

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Ice Loading Effect - Sag tension

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