Short Circuit Force Calculation

ETHS1S/3002074160/ED2.104.001

s

SHORT-CIRCUIT FORCE CALCULATION ON FLEXIBLE CONDUCTOR FOR 400 kV Switchyard at O P Jindal Super Thermal Power Plant, 4X600 MW Units, Tamnar, Raigarh, Chattisgarh

Client : Jindal Power Limited Name

Department

Telephone

Place

Date

Author: Mohd. Sharib

SPEL

4344251

Gurgaon

25.08.10

Approval: Saurabh Jain

SPEL

4344046

Gurgaon

25.08.10

Revised Items

Page

Remarks

Signature

Index of Revisions: Rev:

Date

Name of Reviser

Name of Approver

Siemens Power Engineering Pvt. Ltd., Copying of this document, and giving it to others and the use or communication of the contents thereof, are forbidden without express authority. Offenders are liable to the payment of damages. All rights are reserved in the event of the grant of patent or registration of a utility model or design.

400 kV Switchyard at O P Jindal Super Thermal Power Plant

Page 1 of 11

ETHS2S/3002074160/ED2.104.001

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Page

Contents 1.0

Introduction

.....……………………………………………………………………………

3

2.0

System Data .....……………………………………………………………………………

3

3.0

Conductor Data……………………………………………………………………………..

3

4.0

Installation Data……………………………………………………………………………..

3

5.0

Calculation

.....……………………………………………………………………………

4

6.0

Attachments .....……………………………………………………………………………

4

7.0

Conclusion

.....……………………………………………………………………………

4

8.0

References

.....……………………………………………………………………………

4


Page 2 of 11

ETHS3S/3002074160/ED2.104.001

s 1

Introduction In installations with strained flexible conductors the short circuit current carried by the conductors causes development of stress in the conductors due to the electromagnetic force amongst the conductors in par-allel phases. This conductor force is in addition to the static tension on the conductor (measured from sag-tension calculation) and is sustained by the supporting structures. The objective of this calculation is to determine the maximum force generated on the conductor bundle during a short circuit in order to design the switchyard structures to sustain such forces. Further for line-to-line short circuits, conductor swing out typically results in decreasing phase-to-phase clearance between conductors strung in parallel. In this calculation the 'spacer span' on conductor bun-dle is determined to optimize the tensile forces to values that enables maintaining of the minimum air clearances between phases during horizontal span displacements of conductor bundles at L-L short circuit.

2

System Data Switchyard 1 2 3 4 5 6

3

Nominal System Voltage System Frequency Short Circuit Fault Current at Bus Duration of Fault Current Average (daily ambient) Temperature Maximum Operating Temperature

400 50 50 1

kV Hz kA sec

32

°C °C

75

Conductor Data ACSR MOOSE

4

1

Conductor Diameter

2

Total Cross-sectional Area

3

Conductor Weight

4

Modulus of Elasticity

5

Mass of Spacer Unit

6

Effective distance between sub-conductor in bundle

31.77 597 2.004 6.73E+10 1.94 0.45

mm mm 2 kg/m

N / m2 kg m

Installation Data 1

Phase-to-Phase Spacing

2

Length of Insulator String - TWIN ACSR MOOSE


7 3.843

m m

Page 3 of 11

s 5

ETHS4S/3002074160/ED2.104.001

Calculation The basic methodology of this is as explained in IEC-865 (Part 1) : 1993 and further elaborated in the so-lved examples of IEC-865 (Part 2) : 1994. All the terms , factors and some miscellaneous data used in the calculation are as defined in IEC-865 (Part 1) : 1993. As per the values obtained in the calculation, we have plotted curves of Pinch Force ( F pi ), Drop Force ( F f ) and Tensile Force ( Ft) against the spacer span varying from 1.0 m to 10.0 m through every 0.5 m. The curves give the maximum force in the conductor and the 'critical spacer span' at which it occurs. The actual spacer span selected is a span lower than the critical spacer span. Minimum air clearance is determined for the force optimized with the selected spacer span and with the value of static normal tension as the minimum tension under no wind condition (from sag-tension calculation) for line-to-line fault current.

6

Attachments Attachment - 1: Detailed Calculation for TWIN ACSR MOOSE conductor of 137 m. Attachment - 2: Characteristic curves showing behavior of Short Circuit forces against spacer span for arrangement of conductors in different spans.

Attachment - 3: Summary of Results.

7

Conclusion

1. For design of support structure the maximum value of short circuit force to be considered is to be decided from Attachment 3. This includes the tensile forces owing to dead load of conductor and wind load on conductor, and has been expressed on 'per phase' basis. This force shall be applied to two Phases of the girder supporting them whereas the third phase shall have normal static tension from sagtension calculation. This is as per note no. 2 under clause no. 2.4.2 of IEC-865 (Part 1). 2. As per IEC-61936 (Part-1, Clause 5.4.3), the minimum air clearance between phase conductors at maximum swing out position due to short circuit shall be at least 50 % of the mandatory phase-to-phase clearance. 400 kV outdoor switchyard with rated Lightning Impulse Withstand voltage of 1425 kV, the mandatory phase-to-phase clearance is 4 m. Thus the minimum air clearance shall be greater than 2 m.

8

References 1. IEC-865 (Part 1) : 1993 Short Circuits Currents - Calculation of Effects. 2. IEC-865 (Part 2) : 1994 Short Circuits Currents - Calculation of Effects. 3. IEC-61936 (Part 1) : 2002 Power installations exceeding 1 kV a.c. - Common Rules 4. Technical Data sheets for ACSR Moose Conductor Manufacturer. 5. Technical Specification for 400 kV Switchyard 6

400kV ELECTRICAL LAYOUT PLAN - DRAWING NO. (0)-G71770-AC152-L152-001

7.

400kV ELECTRICAL LAYOUT SECTION - DRAWING NO. (0)-G71770-AD152-L153-001

8.

400kV CLEARANCE DIAGRAM PLAN & SECTION- DRAWING NO. (0)-G71770-AD152-L156-001


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Attachment - 1

s

ETHS1S/3002074160/ED2.104.001

Detailed Calculation for TWIN ACSR MOOSE conductor of 137 m 1.0 Input Data Following are the input data for the calculation. Other terms, factors and data used in the calculation are as defined in IEC-865, Part-I.

as : As :

Centre line distance between conductors (phase to phase spacing) Effective distance between sub-conductors Cross-section of one sub-conductor

ds :

Diameter of the sub-conductor in bundle

E:

Final value of Young's modulus of elacticity of the sub-conductor

a:

7

= = = = = =

m m mm 2 m2 mm m

4.50E-01 5.97E+02 5.97E-04 31.77 0.03177

=

6.73E+10 N/m2 2

{ref.: Layout Plan} {assumed} {ref.: Vendor Data} {ref.: Vendor Data} {ref.: Vendor Data}

Fst (max): Fst (min): f: gn :

Max Initial static tensile force in main conductor Min Initial static tensile force in main conductor System frequency acceleration due to gravity

= = = = =

6.73E+10 137428.96 60068.40 50 9.81

I"k3 :

3φ initial symmetrical short circuit current (r.m.s)

=

50000

A

{ref.: Tech Spec.}

=

137.00

m

{ref.: Layout Plan}

=

127.315 m

= = =

3.8425 1.00 2.004

m m kg / m

{ref.: Vendor Data}

=

2.38

kg / m

{ref.: IEC-865-Part 2}

=

ACSR MOOSE

l: lc : li : lb: m' : ms' :

Centre line distance between supports (main conductor span) Cord length of the main conductor in the span. (l - 2lb-2li) Length of one insulator string Beam Depth Mass per unit length of one sub-conductor Resultant mass per unit length of one subconductor after considering the mass of spacers within the span = m' + {(ns x mz)/(n x lc)}

N/m N{No.of sub-conductor x 7005 x 9.81} N{No.of sub-conductor x 3062 x 9.81} Hz {ref.: Tech Spec.} m/s 2

{ref.: Vendor Data}

ns :

Conductor Type No. of sub-conductor in a main conductor bundle Resultant spring constant of both supports of one span Nos. of spacers in the span

=

50

mz :

Mass of one set of connecting pieces (spacer)

=

1.94

kg

{ref.: Vendor Data}

ls :

Distance between two adjacent spacer

=

2.5

m

{selected}

Tk1 :

Duration of first short circuit current flow

=

1

μ0 :

Magnetic constant, permeability of vacuum

=

κ:

Factor for the calculation of peak short circuit current

=

1.8

τ:

Time constant of the network

=

0.042

γ:

Factor for the relevant natural frequency estimation

=

1.49

n: S:

= =

2

TWIN

1.00E+05 N/m

{ref.: IEC-865-Part 1}

Sec

1.257E-06 N A

-2

2.0 Detailed Calculation Calculation for Short Circuit Forces 2.1 The characteristic Electromagnetic Load per unit length on flexible main conductor in 3φ system is given by :

400 kV Switchyard at O P Jindal Super Thermal Power Plan

Page 5 of 11

Attachment - 1

s

ETHS2S/3002074160/ED2.104.001

Detailed Calculation for TWIN ACSR MOOSE conductor of 137 m F'

= =

(μ0/2π) x (0.75) x (I"k32/a) x (lc/l) 49.78 N/m

N/m

{ref.: Eqn. 19 of IEC-865-I}

2.2 The ratio of electromagnetic force under short circuit to gravitational force on a conductor is given by : r

= =

F'/(n x m's x gn) 1.06


2.3 The direction of the resulting force exerted on the conductor is given by : δ1

= =

tan-1 r 46.77

degrees degrees


2.4 The equivalent static conductor sag at midspan is given by : bc

= =

(n x m's x gn x l2 ) / (8 x Fst) 0.80 m

m


sec


2.5 The period, T, of conductor oscillations is given by : T

= =

2π x {0.8 x (bc / gn)}1/2 1.60 sec

2.6 The resulting period, Tres, of the conductor oscillation during the short circuit current flow is given by : Tres

= =

T/[{1+r2}1/4 x {1-(π2/64) x (δ1/900)2}] 1.38 sec

sec


2.7 Actual Young's modulus of conductor, E s, is given by: Es

= =

E[0.3+0.7 x sin{Fst /(n x As) x (900 /σfin)}] E

where,

σfin =

5.00E+07

N/m2

=> Es

=

5.00E+07

N/m2

N/m2 N/m2

for : Fst/nAs ≤ σfin for : Fst/nAs > σfin

{ref.: Eqn. 26 IEC-865-I}


2.8 The stiffness norm is given by: N

= =

{1/(S x l)} + {1/(n x Es x As)} 1.68E-05 1/N

1/N


2.9 The stress factor, ζ, of the main conductor is given by: ζ

= =

(n x gn x ms' x l)2 / (24 x Fst3 x N) 0.00


2.10 During or at the end of the short circuit current flow, the span will have oscillated out of the steady state to the angle given by: δ1[1-cos(3600 xTk1/Tres)] degrees for :0 ≤ (Τk1/Τres) ≤ 0.5 {ref.: Eqn. 29 IEC:865-1} δk = 2δ1 for : (Tk1/Tres) > 0.5 or δk = degrees where,

Duration of the first short circuit current flow is given by: Tk1 = 0.64 sec Tk1/Tres= 0.46 sec now, therefore, as per Cl. No. 2.3.2.1 of IEC 865-1, the value of Tk1 (=0.4T) shall be used to calculate δk


Page 6 of 11

Attachment - 1

s

ETHS3S/3002074160/ED2.104.001

Detailed Calculation for TWIN ACSR MOOSE conductor of 137 m =>

δk

=

92.30

degrees

2.11 Maximum swing out angle during or after short circuit current flow, δm, is given by: 1.25 cos-1χ for : 0.766 ≤ χ ≤ 1 δm {ref.: Eqn. 31 of IEC-865-I} = degrees 100 + cos-1χ for : -0.985 ≤ χ < 0.766 or δm = degrees 1800 for : χ < -0.985 or δm = degrees 2.11.1 where, Quantity for the maximum swing-out angle, χ, is given as: 1-r sinδk for : 0 ≤ δk ≤ 900 χ = or χ for : δk > 900 = 1-r => χ = -0.06 therefore δm = 103.67 degrees


2.12 The load parameter, ϕ, is obtained as follows:

or

ϕ ϕ

=> ϕ

= = =

3[ √(1+r2) - 1] 3[r x sin(δk) + cos(δk) -1] 1.38

for : Tk1≥ Tres/4 for : Tk1< Tres/4


2.13 Span reaction factor, ψ (ϕ,ζ), is calculated as a real solution of the equation:

ϕ2ψ3 + ϕ(2 + ζ)ψ2 + (1 + 2ζ)ψ − ζ(2 + ϕ) = 0 => ψ

=

0.024

for : 0 ≤ ψ ≤ 1


(approximately)

2.14 The 'short circuit tensile force', F t, is given by: Fst(1 + ϕψ) Ft = 1.1 x Fst(1 + ϕψ) or Ft = => Ft = 156199.57 N 2.15 The 'drop force' is given by: 1.2 x Fst √(1+ 8ζ x δm/1800) Ff = or Ff = non-significant => Ff = 164929.65 N

{ref.: Eqn. 34 of IEC-865-I} for : n = 1, single conductor for : n ≥ 2, bundled conductor

for : r > 0.6, δm ≥ 700 for : r < 0.6, δm < 700


2.16 Short circuit current force between sub-conductors in a bundle is given as: (n-1) x (μ0/2π) x (I"k3/n)2 x (ls/as) x (ν2/ν3) Fν = {ref.: Eqn. 45 of IEC-865-I} where, the factor, v2, is given by figure 8 of IEC:865-1, as a function of factor v 1 which is in turn given as: {f /sin(1800/n)} x [{(as-ds) x m's}/{(μ0/2π) x (I"k3/n)2 x (n-1)/as}]1/2 ν1 2.16.1 = = 3.00 {ref.: Eqn. 46 of IEC-865-I} corresponding to this value of factor ν1, the value of ν2 is observed from curve as: 1- {sin(4πfTpi-2γ) + sin2γ} / 4πfTpi + fτ / fTpi x (1-e-2fTpi/fτ) x sin2γ-8πfτsinγ/{1+(2πfτ)2} x ν2 = [{2πfτ x cos(2πfTpi-γ)/2πfTpi + sin(2πfTpi-γ)/2πfTpi} x e-fTpi/fτ + (sinγ-2πfτcosγ)/2πfTpi] ν2 = 1.68 the factor, v3, is given by : ν3 {(ds/as)/sin(1800/n)} x [{(as/ds)-1}1/2 / {tan-1{(as/ds)-1}1/2}] 2.16.3 = = 0.20 = 5946.95 N => Fν

2.16.2

2.17 Tensile force, Fpi,caused by pinch effect is given as: Fst x {1+ (νe/εst) x ξ} Fpi =

{ref.: AnnexA. A.6 of IEC-865-I} {ref.: AnnexA. A.7 of IEC-865-I}

for : j ≥ 1, i.e. sub-conductors in bundle clash against each other during short circuit. {ref.: Eqn. 50 of IEC-865-I}


Page 7 of 11

Attachment - 1

s

ETHS4S/3002074160/ED2.104.001

Detailed Calculation for TWIN ACSR MOOSE conductor of 137 m or Fpi

2.17.1

2.17.2 2.17.2.1 2.17.2.2 2.17.3 2.17.3.1 2.17.3.2 2.17.3.3

=

Fst x {1+ (νe/εst) x η2}

for : j < 1, i.e. sub-conductors in bundle reduce their

distance but don't clash during short circuit. {ref.: Eqn. 54 of IEC-865-I} The parameter j, determines the bundle configuration during short circuit current flow and is given as: {εpi / (1 + εst)}1/2 j = {ref.: Eqn. 49 of IEC-865-I} = 0.3582 where the strain factors characterizing the contraction of the bundle shall be calculated from : εst 1.5 x {(Fst x ls2 x N)/(as-ds)2} x {sin(1800/n)}2 = {ref.: Eqn. 47 of IEC-865-I} = 123.92 εpi 0.375 x n x {(Fν x ls3 x N)/(as-ds)3} x {sin(1800/n)}3 = {ref.: Eqn. 48 of IEC-865-I} = 16.03 Now, for : j ≥ 1, the factor νe is given as: 1/2 + [9/8 x n(n-1) x (μ0/2π) x (I"k3/n)2 x N x ν2 x {ls/(as-ds)}4 x [{sin(1800/n)}4 / ξ3] x [1 νe = -{tan-1√ν4} / √ν4]-1/4]1/2 {ref.: Eqn. 52 of IEC-865-I} (as - ds) / ds with, ν4 = {ref.: Eqn. 53 of IEC-865-I} = N.A for : j2/3 ≤ ξ ≤ j and ξ is given by the real solution of ξ3 + εst ξ2 −εpi = 0 => ξ = NA {ref.: Eqn. 51of IEC-865-I} => νe

=

N.A

2.17.4 Now, νe 2.17.4.1

for : j < 1, the factor νe is given as: 1/2 + [9/8 x n(n-1) x (μ0/2π) x (I"k3/n)2 x N x ν2 x {ls/(as-ds)}4 x [{sin(1800/n)}4 / η4] x [1 = -{tan-1√ν4} / √ν4]-1/4]1/2 {ref.: Eqn. 55 of IEC-865-I} η(as - ds) / (as -η(as - ds)) = {ref.: Eqn. 56 of IEC-865-I} 2.17.4.2 with, ν4 = 0.03 3 2.17.4.3 and η is given by the real solution of η +εstη−εpifn=0 1/2 x {as-η(as-ds)} ya = = 2.2E-01 as x (2ya/as) / sin(1800/n) x {(1-2ya/as)/2ya/as}1/2/arctan{(1-2ya/as)/2ya/as}1/2 asw = 0.44 asv3/asw fn = =>

η=

0.03

=> νe = 408.14 thereby the magnitude of Pinch Force, Fpi, is calculated as : Fpi = 137753.16 N Calculation for Horizontal Span Displacement and Minimum Air Clearance Note:

Values of all parameters are at 75 °C (maximum conductor temperature ) and L-L fault current of 43.3 kA 2.18 The elastic expansion is given by : εela N x (Ft - Fst) = where, N Ft Fst


= 1.34E-03 at final conductor Temperature (=75°C ) = = =

8.54E-08 1/N 75703.31 N 60068.40401 N

2.19 The thermal expansion is given by : cth {I"k3/(n x As)}2 x Tres/4 εth = cth {I"k3/(n x As)}2 x Tk1 or εth =


for : Tk1≥ Tres/4 for : Tk1< Tres/4


Page 8 of 11

Attachment - 1

s

ETHS5S/3002074160/ED2.104.001

Detailed Calculation for TWIN ACSR MOOSE conductor of 137 m where, cth

=

I"k3 Tk1 Tres

= = =

=> εth

=

2.7E-19

m4/A2sec

43.30 kA 0.97 sec 2.21 sec

for : Aluminium, aluminium alloy & aluminium/steel conductors with cross-section ratio of Al/St >6 at final conductor Temperature (=75°C ) at final conductor Temperature (=75°C )

1.96E-04

2.20 Dilation factor, CD, allows for sag increase caused by elastic and thermal elongation of the conductor is given by: CD where, bc εela εth

=

{1 + 3/8 x (l/bc)2 x (εela +εth)}1/2


= 2.06 at final conductor Temperature (=75°C ) = 1.83 m = 1.34E-03 = 1.96E-04

2.21 Form factor, CF, allows for a possible increase in the dynamic sag of the conductor caused by a change in shape of the conductor curve and is given by: for : r ≤ 0.8 CF = 1.05 {ref.: Eqn. 39 of IEC-865-I} or CF = 0.97 +0.1r for : 0.8 < r < 1.8 for : r ≥ 1.8 or CF = 1.15 => CF

=

1.05

2.22 The maximum horizontal displacement within a span, b h, due to short circuit in spans with strained conductors connected to potrals with tension insulator strings is given by: CF CD bc sinδ1 bh for : δm ≥ δ1 = {ref.: Eqn. 41 of IEC-865-I} CF CD bc sinδm for : δm < δ1 or bh = => bh = 2.46 m where, δm

at final conductor Temperature (=75°C ) = 86.65 degrees

2.23 The distance between the midpoints of the two main conductors during a line-to-line two-phase short circuit is in the worst case given by: a - 2bh amin = {ref.: Eqn. 42 of IEC-865-I} amin = 2.08 m


Page 9 of 11

s

Attachment - 2

ETHS1S/3002074160/ED2.104.001

n a p S r o f r o t c u d n o c E S O O M R S C A N I W T f o t n e m g n a r r a e h t r o f n a p S r e c a p S t s n i a g a s e c r o f t i u c r i C t r o h S f o r o i v a h e b g n i w o h s s e v r u c c i t s i r e t c a r a h C m 7 3 1 h t g n e L


Short Circuit Forces Vs Spacer Span

9.5

10.0

9.0

8.5

8.0

7.5

7.0

6.5

6.0

5.5

5.0

4.5

4.0

3.5

3.0

2.5

2.0

1.5

200000 180000 160000 140000 120000 100000 80000 60000 40000 20000 0 1.0

Short Circuit Forces (N)

Sl. No. Spacer Span Tensile Force (Ft) Drop Force (Ff) Pinch Force (Fpi) (m) (N) (N) (N) 1 1.0 154540.27 167450.01 139415.11 2 1.5 155467.93 167670.69 139675.50 3 2.0 155919.98 167766.64 141092.88 4 2.5 156199.57 167822.58 142050.24 5 3.0 156371.57 167856.7 143116.65 6 3.5 156500.57 167882.09 144260.10 7 4.0 156608.08 167903.09 145483.20 8 4.5 156672.59 167915.62 146715.84 9 5.0 156737.12 167928.08 148028.73 10 5.5 156780.14 167936.36 149383.46 11 6.0 156823.17 167944.61 150814.57 12 6.5 156866.2 167952.83 152350.69 13 7.0 156887.72 167956.93 154012.39 14 7.5 156930.77 167965.1 189262.12 15 8.0 156952.3 167969.18 187356.11 16 8.5 156973.83 167973.24 185635.29 17 9.0 156973.83 167973.24 184034.48 18 9.5 156995.36 167977.3 182609.85 19 10.0 157016.9 167981.35 181316.99

Tensile Force (N) Drop Force (N)

Spacer Span (m)

Pinch Force (N)

Page 10 of 11

Attachment - 3

s

ETHS1S/3002074160/ED2.104.001

SUMMARY OF RESULTS Serial No.

1

Span (m) l 137

No. of sub conductors Critical Spacer Chosen Spacer per phase Span Span (m) (m) ls n 2

7


2.5

Short Circuit Force at minimum temperature and Full Wind at 3-Ph fault (N) Fst Ft Ff Fpi 137429

156199.6 167822.6 142050.2

Initial Static Tension at Maximum temperature and Full Wind (N) Fst

Minimum air clearance amin for L-L fault

60068.4

2.076

(m) amin

Maximum Short Circuit Force per phase for design structures (N) (kg) 167822.58

17107.30

Page 11 of 11

Short Circuit Force Calculation

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