study of galloping mitigation: tug boats and power lines · pdf filestudy of galloping...
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Learning Objectives
► Myths
► Weather Cases
► Mitigation
► Events
► Mechanics
► Tug boats?
► Metric
► CIGRE 322
2
Galloping Myth Busting ► Galloping is a Force
Equilibrium
► Galloping is a Single Span
problem
► Flashovers are the primary
problem
𝑓𝐿
𝐺𝑀𝑊𝑖𝑟𝑒
𝑓𝐷
4
Observed Galloping on
Operating Lines
► Single loop
galloping
observed
on spans
greater
than 200 m
(600-700
ft)
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Galloping Mitigation
Options ► Increase the phase
spacing
► Interphase spacers
► Aerodynamic devices
► Tuning devices
► Ice removal “The Ice
Buster”
8
Mitigation: Increased
Phase Spacing per RUS
Voltage 115 kV 138 kV 230 kV 345 kV 500 kV
Phase-Phase 0.46 m
(1.5 ft)
0.46 m
(1.5 ft)
0.76 m
(2.5 ft)
1.07 m
(3.5 ft)
1.83 m
(6.0 ft)
Phase-Ground 0.30 m
(1.0 ft)
0.30 m
(1.0 ft)
.61 m
(2.0 ft)
.76 m
(2.5 ft)
1.22 m
(4.0 ft)
𝑀(𝑓𝑡) = 1.25 𝑆𝑎𝑔 .5𝑖𝑛 𝑖𝑐𝑒/32 𝐷𝑒𝑔 𝐹 + 1.0
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► Single Loop Galloping
Mitigation:
Interphase Spacers ► Most widely
used device
on all
voltages
and bundles
► Prevents
flashovers
► Galloping /
Dynamic
loading
persist
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Mitigation:
Aerodynamic Devices ► T2/VR2 Conductor
• One Full Twist over ~3m (8-10 ft) for T2 Drake (2 x 795 KCMIL 26/7
ACSR)
• See ASTM B911: Standard Specification for ACSR Twisted Pair
Conductor
• Vertical bundle of controls galloping
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Mitigation:
Aerodynamic Devices ► Air flow spoilers
• Limits: Single conductors
• 25% percent of the span is wrapped in two groups
• Galloping prevention / amplitude reduction
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Galloping Mitigation:
Aerodynamic Devices ► AR Twister:
• Eccentric weight mounted about the conductor
• Creates a smooth ice profile
• Two per span (minimum)
14
Galloping Events and
the Results
► 345 kV, Tubular Steel
Line Retrofit with
Interphase Spacers, MN
Winter 2014
► 345 kV, Lattice Tower
Failures, SD Winter
2010
► 345 kV, H-Frame Tubular
Steel Cascade Failure,
IA Winter 1990
ICE
WIND
Flashover &
Structural Damage
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Galloping Mechanics:
Modes of Failure
► 1 Degree of Freedom:
Vertical
► Cable System: Damped
Simple Harmonic
Motion
► 2 Degrees of Freedom:
Vertical and
Torsional
► Dynamic Analysis of a
Cable Section
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Analysis of Vertical
Conductor Motion ► The motion will be (+) or (-) from a neutral
position
► Likewise the conductor velocity will also
change signs
► Analysis of a conductor in motion
Y=-MAX, ⅆ𝑦/ ⅆt =0
Y=0, (+/-) Max = ⅆ𝑦/ ⅆt
Y=+MAX, ⅆ𝑦/ ⅆt =0
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Physics: NASA Drag
Coefficients
WIND (UO)
Drag Force (𝑓𝑑) =1
2𝜌𝐴𝐶𝑑𝑈𝑂
2
𝐶𝑑 −𝑠𝑝ℎ𝑒𝑟𝑒 = .07 − .5
𝐶𝑑 − 𝑓𝑙𝑎𝑡 𝑝𝑙𝑎𝑡𝑒 = 1.28
𝐶𝑑 − 𝑏𝑢𝑙𝑙𝑒𝑡 = 0.295
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Physics: Aerodynamic
Forces
► Asymmetrical Cable Shapes and the Lift Force
WIND (UO)
+𝑓𝐿
Lift (𝑓𝐿) =1
2𝜌𝑎𝑖𝑟ⅆ𝑣𝑟
2𝐶𝐿 𝜑 ANGLE OF ATTACK (𝜑)
(- 𝜑)
(+𝜑) −𝑓𝐿
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Aerodynamic Forces on
an Iced Conductor
► Lift and Drag Coefficient vs Angle of Attack
+LIFT 𝑓𝐿
+DRAG 𝑓𝐷
(+ 𝜑) (- 𝜑)
-π π π/2 -π/2
-LIFT 𝑓𝐿
+DRAG 𝑓𝐷
90 45
-45 90
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Analysis of Vertical
Conductor Motion
JJQ
► Change in Angle of Attack due to conductor
motion
► Relative Wind at Neutral Position
(+) ⅆ𝑦/ ⅆt
− ⅆ𝑦/ ⅆt
WIND (UO)
WIND (UO) -α
(+ 𝜑) ANGLE OF ATTACK
(- 𝜑) ANGLE OF ATTACK
α
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► Determine if the Aerodynamic Forces are
damping or destabIilizing the vertical
motion
► Case 1: Initial Angle of Attack = 90 deg, π/2
Analysis of Vertical
Conductor Motion
dy/dt =(+)
Δ𝑓𝐿=(-) dy/dt =(-)
Δ𝑓𝐿=(+)
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𝑓𝐿 = (ⅆ𝑓𝐿ⅆ𝜑)𝛼
Review of Lift and
Drag: Damped ► For a small ∆𝜑 (𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝛼), the lift force is linear:
+LIFT 𝑓𝐿
+DRAG 𝑓𝐷
(+ 𝜑) (- 𝜑)
-π π π/2 -π/2
-LIFT 𝑓𝐿
+DRAG 𝑓𝐷
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► Determine if the Aerodynamic Forces are
damping or destabIilizing the vertical
motion
► Case 1: Initial Angle of Attack = 0 deg, 0
Analysis of Vertical
Conductor Motion
dy/dt =(+) Δ𝑓𝐿=(+)
dy/dt =(-) Δ𝑓𝐿=(-)
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Review of Lift and
Drag: Destabilized ► For a small ∆𝜑 (𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝛼), the lift force is linear:
+LIFT 𝑓𝐿
+DRAG 𝑓𝐷
(+ 𝜑) (- 𝜑)
-π π π/2
-LIFT 𝑓𝐿
+DRAG 𝑓𝐷
𝑓𝐿 = (ⅆ𝑓𝐿ⅆ𝜑)𝛼
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Galloping: An
Aerodynamic
Instability
𝑓𝐿 cos 𝛼 > 𝑓𝐷 sin 𝛼, SYSTEM UNSTABLE
tan−1 𝛼 = (−ⅆ𝑦/ⅆ𝑡)/𝑈𝑂
− ⅆ𝑦/ ⅆt
WIND (UO)
α
For small 𝛼, in Radians
𝛼 = (−ⅆ𝑦/ⅆ𝑡)/𝑈𝑂
sin 𝛼 = 𝛼
cos 𝛼 = 1
𝑓𝐿 1 > 𝑓𝐷(α)
𝑑𝑓𝐿
𝑑𝜑 𝛼 > 𝑓𝐷(α)
𝑓𝐷
𝑓𝐿 = (ⅆ𝑓𝐿ⅆ𝜑)𝛼
𝑓𝐿 cos 𝛼 < 𝑓𝐷 sin 𝛼, SYSTEM DAMPED
𝑑𝑓𝐿
𝑑𝜑 > 𝑓𝐷, SYSTEM UNSTABLE
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Idealized Structure &
Cable System ► Cable system with fixed ends, sag (mass)
► Tension enters the system as a restorative
force (opposite to motion)
𝐹𝑟𝑒𝑠𝑡𝑜𝑟𝑒 = −𝑘𝑦 𝑡 + 𝑚𝑔 − 𝑘(𝑌𝑖𝑛𝑖𝑡𝑖𝑎𝑙)
𝐹𝑜𝑟𝑐𝑒𝑠 =ma=m(ⅆ2𝑦/ⅆ𝑡2)
𝑚 (ⅆ2𝑦/ⅆ𝑡2) + 𝑘𝑦(𝑡) = 0
DIFFERENTIAL EQUATION:
𝑚𝑔 − 𝑘(𝑌𝑖𝑛𝑖𝑡𝑖𝑎𝑙) = 0 WHERE:
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Cable System: Simple
Harmonic Motion ► Hooke’s Law: cable spring constant
𝜎 = 𝐸(𝜀)
GIVEN: 𝜎 = Δ𝑇/𝐴
𝜀 = Δ𝑙/𝐿
𝑅𝑒𝑠𝑡𝑜𝑟𝑎𝑡𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 𝑘𝑥 = Δ𝑇 =𝐸𝐴
𝐿(Δ𝑙)
THEREFORE: 𝑘 =𝐸𝐴
𝐿
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Physics: Sum of the
Forces in SHM ► Key Relationships and Solution
y(t) = 𝑦𝑚𝑎𝑥 cos(𝜔𝑡)
DIFFERENTIAL EQUATION:
WHERE:
𝑚(ⅆ2𝑦/ⅆ𝑡2) = −𝑘𝑦(𝑡)
ⅆ2𝑦/ⅆ𝑡2 = −𝜔2𝑦𝑀𝐴𝑋 cos(𝜔𝑡)
THEREFORE: −𝑚𝜔2𝑦𝑚𝑎𝑥 cos(𝜔𝑡) = 𝑚ⅆ2𝑦/ⅆ𝑡2 = −𝑘𝑦(𝑡)
THEREFORE THESE RELATIONSHIPS EXIST: 𝑚𝜔2 = (2𝜋𝜈)2 = 𝑘
𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝜈𝑁𝐴𝑇) =1
2𝜋
𝑘
𝑚
SOLUTION DESCRIBING LOCATION
VALUE DESCRIBING SYSTEM:
MECHANICAL ENERGY: 𝐸𝑛𝑒𝑟𝑔𝑦 = .5𝑘 𝑦 𝑡 2+ .5𝑚ⅆ𝑦
ⅆ𝑡
2
y(t) = 𝑦𝑚𝑎𝑥 cos(𝜔𝑡)
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Physics: Damped
Harmonic Motion ► Damping vane in a viscous fluid attached to
an oscillating block
𝐹𝑜𝑟𝑐𝑒𝑠 = −𝑘𝑦 − 𝑏𝑣 + (𝑚𝑔 − 𝑘(𝑌𝑖𝑛𝑖𝑡𝑖𝑎𝑙)) =ma
NOW THE SUM OF THE FORCES:
DAMPING FORCE:
𝐹𝐷𝐴𝑀𝑃𝐼𝑁𝐺 = −𝑏 𝑉𝐸𝐿𝑂𝐶𝐼𝑇𝑌 = −𝑏(ⅆ𝑦
ⅆ𝑡)
𝐹𝑜𝑟𝑐𝑒𝑠 = −𝐹𝑟𝑒𝑠𝑡𝑜𝑟𝑒 − 𝐹𝑑𝑎𝑚𝑝𝑖𝑛𝑔 + (𝐹𝑔𝑟𝑎𝑣𝑖𝑡𝑦 − 𝐹𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑠𝑝𝑟𝑖𝑛𝑔 𝑙𝑜𝑎𝑑) =ma
𝑚𝑔 − 𝑘(𝑌𝑖𝑛𝑖𝑡𝑖𝑎𝑙) = 0
m(ⅆ2𝑦
ⅆ𝑡2) + 𝑏
ⅆ𝑦
ⅆ𝑡+ 𝑘𝑦(𝑡) = 0
WHERE:
NEW DIFFERENTIAL EQUATION:
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Physics: Damped
Harmonic Motion
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝐷𝑎𝑚𝑝𝑖𝑛𝑔 (𝑐𝑐) = 2 𝑘𝑚
𝐷𝑎𝑚𝑝𝑖𝑛𝑔 𝑅𝑎𝑡𝑖𝑜 𝜍 =𝐴𝑐𝑡𝑢𝑎𝑙 𝐷𝑎𝑚𝑝𝑖𝑛𝑔 𝑐
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝐷𝑎𝑚𝑝𝑖𝑛𝑔 𝑐𝑐=
𝑏
2 𝑘𝑚
𝐷𝑎𝑚𝑝𝑒ⅆ 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦(𝜈𝐷𝐴𝑀𝑃) = 𝜈𝑁𝐴𝑇 1 − 𝜍2
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 (𝜔) = (𝑘
𝑚−
𝑏
4𝑚2)
y(t) = 𝑦𝑚𝑎𝑥 𝑒−(𝑏𝑡2𝑚)cos(𝜔𝑡)
NEW DIFFERENTIAL EQUATION:
SOLUTION DESCRIBING LOCATION
m(ⅆ2𝑦
ⅆ𝑡2) + 𝑏
ⅆ𝑦
ⅆ𝑡+ 𝑘𝑦(𝑡) = 0
VALUE DESCRIBING SYSTEM:
𝐸𝑛𝑒𝑟𝑔𝑦 𝐹𝐼𝑁𝐴𝐿 + 𝐸𝑛𝑒𝑟𝑔𝑦𝐷𝐼𝑆𝑆𝐼𝑃𝐴𝑇𝐸𝐷 = 𝐸𝑛𝑒𝑟𝑔𝑦 𝐼𝑁𝐼𝑇𝐼𝐴𝐿 MECHANICAL ENERGY:
35
Normal Case: Cable
Motion is Underdamped
SHM
► Review the force equation:
► Air Resistance (Drag) Acts as a Viscous
Damper
► Results: Aerodynamic Stability
𝐹𝑜𝑟𝑐𝑒𝑠 = −𝐹𝑟𝑒𝑠𝑡𝑜𝑟𝑒 − 𝐹𝑑𝑎𝑚𝑝𝑖𝑛𝑔 + (𝐹𝑔𝑟𝑎𝑣𝑖𝑡𝑦 − 𝐹𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑠𝑝𝑟𝑖𝑛𝑔 𝑙𝑜𝑎𝑑) =ma
𝑓𝐿 cos 𝛼 < 𝑓𝐷 sin 𝛼, SYSTEM DAMPED
dy/dt =(+)
Δ𝑓𝐿=(-) dy/dt =(-)
Δ𝑓𝐿=(+)
37
► Review the force equation:
► Aerodynamic Lift Over Comes Drag: Excitation
► Results: Aerodynamic Destabilized, but
Requires Wind Energy
𝐹𝑜𝑟𝑐𝑒𝑠 = −𝐹𝑟𝑒𝑠𝑡𝑜𝑟𝑒 + 𝐹𝑒𝑥𝑐𝑖𝑡𝑖𝑎𝑡𝑖𝑜𝑛 + (𝐹𝑔𝑟𝑎𝑣𝑖𝑡𝑦 − 𝐹𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑠𝑝𝑟𝑖𝑛𝑔 𝑙𝑜𝑎𝑑) =ma
Galloping Case: Cable
Motion is Excited SHM
𝑓𝐿 cos 𝛼 > 𝑓𝐷 sin 𝛼, SYSTEM UNSTABLE
dy/dt =(+) Δ𝑓𝐿=(+)
dy/dt =(-) Δ𝑓𝐿=(-)
38
Torsion: The Second
Degree of Freedom ► Pitching Moment creates the twisting motion
WIND (UO) (+ 𝜑) ANGLE OF ATTACK
-α
+𝜃𝑖𝑐𝑒
+θ
𝜑 = −𝛼 + 𝜃 + 𝜃𝑖𝑐𝑒 PITCHING MOMENT =𝑀𝑊= 1
2𝜌𝑎𝑖𝑟ⅆ
2𝑣𝑟2𝐶𝑀 𝜑
Mw
39
Vertical / Torsional
Motion Phasing ► Determine impacts of motion out of phase
► Exaggerates the Aerodynamic Instability
► Case 1: Max Twist at Y=0
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Vertical / Torsional
Motion Phasing ► Determine impacts of phased oppositely
(Aeroelastic)
► Case 2: Max Twist not at Y=0, “Coupled Flutter”
Occurs
41
Structural Components:
3 Centers of an Iced
Cable ► 2 degrees of freedom complicated by
structural data
AERODYNAMIC CENTER
CENTER OF GRAVITY
CENTER OF CABLE
42
Galloping Standing
Waves in a Section ► Creates Dynamic Loading:
• Dynamic Loads on a Tangent: Vertical Force 2 times the static load
(CIGRE 322)
• Dynamic Loads on a Deadend: Tension Force 2.5 times the static load
(CIGRE 322)
44
Conclusions ► Galloping weather: ice plus high winds
► Galloping mitigation exists
► Single cable due to aerodynamic
instability
► Multi-cable due to aero-elastic
instability
46
Advanced topics of
Galloping Analysis ► Discussion of Cigre, Alberta ESO, and RUS galloping
design
► The dynamic loading
• Effects on the structures due to allowing the cable to act as a spring
• Structure has one fundamental frequency and wire section has a different
frequency
► Multi-Span analysis for fundamental natural
frequency
► Vertical and torsion motion, 2 total degrees of
freedom system: forces and stiffness
► Vertical, torsion, and horizontal motion, 3 total
degrees of freedom: forces and stiffness
► Weather cases, ice shapes, aerodynamic coefficients
► Amplitude of motion (energy balance wind as
excitation)
► Influence of hardware configuration
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