short circuit force calculation

s ETHS1S/3002074160/ED2.104.001

Client : Jindal Power Limited

Name Department Telephone Place Date Signature

Author: Mohd. Sharib SPEL 4344251 Gurgaon 25.08.10

Approval: Saurabh Jain SPEL 4344046 Gurgaon 25.08.10

Index of Revisions:

Rev: Date Revised Items Page RemarksName of Reviser

Name of Approver

Siemens Power Engineering Pvt. Ltd.,

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

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 of 11

s ETHS2S/3002074160/ED2.104.001

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


s ETHS3S/3002074160/ED2.104.001

1 Introduction

2 System Data

Switchyard

1 kV

2 Hz

3 kA

4 sec

5

6

3 Conductor Data

4 Installation Data

7m

5

2 Length of Insulator String - TWIN ACSR MOOSE3.843

m

6Effective distance between sub-conductor in bundle 0.45

m

1.94

design the switchyard structures to sustain such forces.

clearance between conductors strung in parallel. In this calculation the 'spacer span' on conductor bun-

kg

4 Modulus of Elasticity6.73E+10

circuit.

2 Total Cross-sectional Area597

3 Conductor Weight2.004

kg/m

1 Phase-to-Phase Spacing

Mass of Spacer Unit

50

32

75

501

Average (daily ambient) Temperature

Maximum Operating Temperature

-dle is determined to optimize the tensile forces to values that enables maintaining of the minimum airclearances between phases during horizontal span displacements of conductor bundles at L-L short

1 Conductor Diameter31.77

ACSR MOOSE

Duration of Fault Current

400System FrequencyShort Circuit Fault Current at Bus

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

In installations with strained flexible conductors the short circuit current carried by the conductors causes

Further for line-to-line short circuits, conductor swing out typically results in decreasing phase-to-phase

sag-tension calculation) and is sustained by the supporting structures. The objective of this calculationis to determine the maximum force generated on the conductor bundle during a short circuit in order to

Nominal System Voltage

C°C°

mm

2mm

2/mN


s ETHS4S/3002074160/ED2.104.001

5 Calculation

6 Attachments

7 Conclusion

8 References

The basic methodology of this is as explained in IEC-865 (Part 1) : 1993 and further elaborated in the so-

1. For design of support structure the maximum value of short circuit force to be considered is to be

fault current.

( ) and Tensile Force ( ) 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

-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 ( ), Drop Force

5. Technical Specification for 400 kV Switchyard

decided from Attachment 3. This includes the tensile forces owing to dead load of conductor and windload on conductor, and has been expressed on 'per phase' basis. This force shall be applied to twoPhases of the girder supporting them whereas the third phase shall have normal static tension fromsagtension calculation. This is as per note no. 2 under clause no. 2.4.2 of IEC-865 (Part 1).

1. IEC-865 (Part 1) : 1993 Short Circuits Currents - Calculation of Effects.

2. IEC-865 (Part 2) : 1994 Short Circuits Currents - Calculation of Effects.

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

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 greaterthan 2 m.

3. IEC-61936 (Part 1) : 2002 Power installations exceeding 1 kV a.c. - Common Rules

4. Technical Data sheets for ACSR Moose Conductor Manufacturer.

Attachment - 1: Detailed Calculation for TWIN ACSR MOOSE conductor of 137 m.

tension as the minimum tension under no wind condition (from sag-tension calculation) for line-to-line

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.

piFfF tF


s Attachment - 1 ETHS1S/3002074160/ED2.104.001

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.

a : =

7m

{ref.: Layout Plan}

as : Effective distance between sub-conductors = 4.50E-01 m {assumed}As : Cross-section of one sub-conductor = 5.97E+02 mm2

{ref.: Vendor Data}

= 5.97E-04 m2

ds : Diameter of the sub-conductor in bundle = 31.77 mm {ref.: Vendor Data}= 0.03177 m

E : = 6.73E+10 N/m2 {ref.: Vendor Data}

= 6.73E+10 N/m2

Fst (max): Max Initial static tensile force in main conductor = 137428.96 N{No.of sub-conductor x 7005 x 9.81}Fst (min): Min Initial static tensile force in main conductor = 60068.40 N{No.of sub-conductor x 3062 x 9.81}f : System frequency = 50 Hz {ref.: Tech Spec.}gn : acceleration due to gravity = 9.81 m/s2

I"k3 : = 50000 A {ref.: Tech Spec.}

l : = 137.00 m {ref.: Layout Plan}

lc : = 127.315 m

li : Length of one insulator string = 3.8425 m {ref.: Vendor Data}lb: Beam Depth = 1.00 mm' : Mass per unit length of one sub-conductor = 2.004 kg / m {ref.: Vendor Data}

ms' : = 2.38 kg / m {ref.: IEC-865-Part 2}

=

n : = 2 TWIN

S : = 1.00E+05 N/m

ns : Nos. of spacers in the span = 50

mz : = 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 Sec

μ0 : Magnetic constant, permeability of vacuum = 1.257E-06 N A-2

κ : = 1.8

τ : Time constant of the network = 0.042

γ := 1.49

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 :

No. of sub-conductor in a main conductor bundleResultant spring constant of both supports of one span

Factor for the calculation of peak short circuit current

Factor for the relevant natural frequency estimation

ACSR MOOSEConductor Type

{ref.: IEC-865-Part 1}

Mass of one set of connecting pieces (spacer)

Centre line distance between supports (main conductor span) Cord length of the main conductor in the span. (l - 2lb-2li)

Resultant mass per unit length of one sub-conductor after considering the mass of spacers within the span = m' + {(ns x mz)/(n x lc)}

Detailed Calculation for TWIN ACSR MOOSE conductor of 137 m

Centre line distance between conductors (phase to phase spacing)

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

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

400 kV Switchyard at O P Jindal Super Thermal Power Plan of 11



F' = (μ0/2π) x (0.75) x (I"k32/a) x (lc/l) N/m {ref.: Eqn. 19 of IEC-865-I}

= 49.78 N/m

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) {ref.: Eqn. 20 of IEC-865-I} = 1.06

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

δ1 = tan-1 r degrees {ref.: Eqn. 21 of IEC-865-I} = 46.77 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) m {ref.: Eqn. 22 of IEC-865-I} = 0.80 m

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

T = 2π x {0.8 x (bc / gn)}1/2

sec {ref.: Eqn. 23 of IEC-865-I} = 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}] sec {ref.: Eqn. 24 of IEC-865-I} = 1.38 sec

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

Es = E[0.3+0.7 x sin{Fst /(n x As) x (900 /σfin)}] N/m2 for : Fst/nAs ≤ σfin {ref.: Eqn. 26

= E N/m2 for : Fst/nAs > σfin IEC-865-I}

where, σfin = 5.00E+07 N/m2 {ref.: Eqn. 27 of IEC-865-I}

=> Es = 5.00E+07 N/m2

2.8 The stiffness norm is given by:

N = {1/(S x l)} + {1/(n x Es x As)} 1/N {ref.: Eqn. 25 of IEC-865-I} = 1.68E-05 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) {ref.: Eqn. 28 of IEC-865-I}

= 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:

δk = δ1[1-cos(3600 xTk1/Tres)] degrees for :0 ≤ (Τk1/Τres) ≤ 0.5 {ref.: Eqn. 29 IEC:865-1}

or δk = 2δ1 degrees for : (Tk1/Tres) > 0.5

where, Duration of the first short circuit current flow is given by:Tk1 = 0.64 sec

now, Tk1/Tres= 0.46 sec

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




=> δk = 92.30 degrees

2.11 Maximum swing out angle during or after short circuit current flow, δm, is given by:

δm = 1.25 cos-1χ degrees for : 0.766 ≤ χ ≤ 1 {ref.: Eqn. 31 of IEC-865-I}

or δm = 100 + cos-1χ degrees for : -0.985 ≤ χ < 0.766

or δm = 1800degrees for : χ < -0.985

2.11.1 where, Quantity for the maximum swing-out angle, χ, is given as:χ = 1-r sinδk for : 0 ≤ δk ≤ 900

{ref.: Eqn. 30 of IEC-865-I}or χ = 1-r for : δk > 900

=> χ = -0.06therefore

δm = 103.67 degrees

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

ϕ = 3[ √(1+r2) - 1] for : Tk1≥ Tres/4 {ref.: Eqn. 32 of IEC-865-I}or ϕ = 3[r x sin(δk) + cos(δk) -1] for : Tk1< Tres/4

=> ϕ = 1.38

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 for : 0 ≤ ψ ≤ 1 {ref.: Eqn. 33 of IEC-865-I} => ψ = 0.024 (approximately)

2.14 The 'short circuit tensile force', Ft, is given by: {ref.: Eqn. 34 of IEC-865-I}

Ft = Fst(1 + ϕψ) for : n = 1, single conductor

or Ft = 1.1 x Fst(1 + ϕψ) for : n ≥ 2, bundled conductor

=> Ft = 156199.57 N

2.15 The 'drop force' is given by:Ff = 1.2 x Fst √(1+ 8ζ x δm/1800) for : r > 0.6, δm ≥ 700

{ref.: Eqn. 35 of IEC-865-I}

or Ff = non-significant for : r < 0.6, δm < 700

=> Ff = 164929.65 N

2.16 Short circuit current force between sub-conductors in a bundle is given as:Fν = (n-1) x (μ0/2π) x (I"k3/n)2 x (ls/as) x (ν2/ν3) {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 v1 which is in turn given as:

2.16.1 ν1 = {f /sin(1800/n)} x [{(as-ds) x m's}/{(μ0/2π) x (I"k3/n)2 x (n-1)/as}]1/2

= 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:

ν2 = 1- {sin(4πfTpi-2γ) + sin2γ} / 4πfTpi + fτ / fTpi x (1-e-2fTpi/fτ) x sin2γ-8πfτsinγ/{1+(2πfτ)2} x

[{2πfτ x cos(2πfTpi-γ)/2πfTpi + sin(2πfTpi-γ)/2πfTpi} x e-fTpi/fτ + (sinγ-2πfτcosγ)/2πfTpi]

2.16.2 ν2 = 1.68 {ref.: AnnexA. A.6 of IEC-865-I}

the factor, v3, is given by :

2.16.3 ν3 = {(ds/as)/sin(1800/n)} x [{(as/ds)-1}1/2 / {tan-1{(as/ds)-1}1/2}] {ref.: AnnexA. A.7 of IEC-865-I} = 0.20

=> Fν = 5946.95 N

2.17 Tensile force, Fpi,caused by pinch effect is given as:

Fpi = Fst x {1+ (νe/εst) x ξ} for : j ≥ 1, i.e. sub-conductors in bundle clash against

each other during short circuit. {ref.: Eqn. 50 of IEC-865-I}




or Fpi = 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}

2.17.1 The parameter j, determines the bundle configuration during short circuit current flow and is given as:j = {εpi / (1 + εst)}

1/2 {ref.: Eqn. 49 of IEC-865-I}

= 0.35822.17.2 where the strain factors characterizing the contraction of the bundle shall be calculated from :

2.17.2.1 εst = 1.5 x {(Fst x ls2 x N)/(as-ds)

2} x {sin(1800/n)}2 {ref.: Eqn. 47 of IEC-865-I}

= 123.922.17.2.2 εpi = 0.375 x n x {(Fν x ls

3 x N)/(as-ds)3} x {sin(1800/n)}3

{ref.: Eqn. 48 of IEC-865-I} = 16.03

2.17.3 Now, for : j ≥ 1, the factor νe is given as:

2.17.3.1 νe = 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

-{tan-1√ν4} / √ν4]-1/4]1/2 {ref.: Eqn. 52 of IEC-865-I}

2.17.3.2 with, ν4 = (as - ds) / ds {ref.: Eqn. 53 of IEC-865-I} = N.A

2.17.3.3 and ξ is given by the real solution of ξ3 + εst ξ2 −εpi = 0 for : j2/3 ≤ ξ ≤ j => ξ = NA {ref.: Eqn. 51of IEC-865-I}

=> νe = N.A

2.17.4 Now, for : j < 1, the factor νe is given as:

2.17.4.1 νe = 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}

2.17.4.2 with, ν4 = η(as - ds) / (as -η(as - ds)) {ref.: Eqn. 56 of IEC-865-I} = 0.03

2.17.4.3 and η is given by the real solution of η3+εstη−εpifn=0ya = 1/2 x {as-η(as-ds)}

= 2.2E-01asw = as x (2ya/as) / sin(1800/n) x {(1-2ya/as)/2ya/as}

1/2/arctan{(1-2ya/as)/2ya/as}1/2

0.44fn = asv3/asw

=> 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 (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) {ref.: Eqn. 36 of IEC-865-I}

= 1.34E-03

where, at final conductor Temperature (=75 )

N = 8.54E-08 1/NFt = 75703.31 N

Fst = 60068.40401 N

2.19 The thermal expansion is given by :

εth = cth {I"k3/(n x As)}2 x Tres/4 for : Tk1≥ Tres/4 {ref.: Eqn. 37 of IEC-865-I}

or εth = cth {I"k3/(n x As)}2 x Tk1 for : Tk1< Tres/4

C°

C°




where,

cth = 2.7E-19 m4/A2sec for : Aluminium, aluminium alloy & aluminium/steel

conductors with cross-section ratio of Al/St >6

I"k3 = 43.30 kA

Tk1 = 0.97 sec at final conductor Temperature (=75 )

Tres = 2.21 sec at final conductor Temperature (=75 )

=> εth = 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 = {1 + 3/8 x (l/bc)2 x (εela +εth)}

1/2 {ref.: Eqn. 38 of IEC-865-I}

= 2.06where, at final conductor Temperature (=75 )

bc = 1.83 m

εela = 1.34E-03

εth = 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:

CF = 1.05 for : r ≤ 0.8 {ref.: Eqn. 39 of IEC-865-I}

or CF = 0.97 +0.1r for : 0.8 < r < 1.8

or CF = 1.15 for : r ≥ 1.8

=> CF = 1.05

2.22 The maximum horizontal displacement within a span, bh, due to short circuit in spans with strained conductors

connected to potrals with tension insulator strings is given by:

bh = CF CD bc sinδ1 for : δm ≥ δ1 {ref.: Eqn. 41 of IEC-865-I}

or bh = CF CD bc sinδm for : δm < δ1

=> bh = 2.46 m

where, at final conductor Temperature (=75 )δm = 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:

amin = a - 2bh {ref.: Eqn. 42 of IEC-865-I}

amin = 2.08 m

C°C°

C°

C°



Sl. No. Spacer Span Tensile Force (Ft) Drop Force (Ff) Pinch Force (Fpi)

(m) (N) (N) (N)1 1.0 154540.27 167450.01 139415.112 1.5 155467.93 167670.69 139675.503 2.0 155919.98 167766.64 141092.884 2.5 156199.57 167822.58 142050.245 3.0 156371.57 167856.7 143116.656 3.5 156500.57 167882.09 144260.107 4.0 156608.08 167903.09 145483.208 4.5 156672.59 167915.62 146715.849 5.0 156737.12 167928.08 148028.73

10 5.5 156780.14 167936.36 149383.4611 6.0 156823.17 167944.61 150814.5712 6.5 156866.2 167952.83 152350.6913 7.0 156887.72 167956.93 154012.3914 7.5 156930.77 167965.1 189262.1215 8.0 156952.3 167969.18 187356.1116 8.5 156973.83 167973.24 185635.2917 9.0 156973.83 167973.24 184034.4818 9.5 156995.36 167977.3 182609.8519 10.0 157016.9 167981.35 181316.99

Characteristic curves showing behavior of Short Circuit forces against Spacer Span for the arrangment of TWIN ACSR MOOSE conductor for Span Length 137 m

Short Circuit Forces Vs Spacer Span

0

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

9.0

9.5

10

.0

Spacer Span (m)

Sh

ort

Cir

cuit

Fo

rces

(N

)

Tensile Force (N)

Drop Force (N)

Pinch Force (N)



No. of sub conductors Critical Spacer Chosen Spacer Initial Static Tension at Maximum Minimum air clearanceper phase Span Span temperature and Full Wind amin for L-L fault

(m) (m) (m) (N) (m) (N) (kg)

l n ls Fst Ft Ff Fpi Fst amin

1 137 2 7 2.5 137429 156199.6 167822.6 142050.2 60068.4 2.076 167822.58 17107.30

Serial No. Span Short Circuit Force at minimum

SUMMARY OF RESULTS

temperature and Full Wind at 3-Ph fault per phase for design structures

(N)

Maximum Short Circuit Force


short circuit force calculation

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