Francis turbine part load resonance risk analytical assessment
Transcript of Francis turbine part load resonance risk analytical assessment
Francis turbine part load resonance risk analytical assessment
C. Nicolet, C. Landry, S. Alligné, A. Béguin
SHF/AFM 2019, Sion, Switzerland, November 06-07,2019
Power Vision Engineering sàrl1, ch. des Champs-CourbesCH-1024 EcublensSwitzerlandhttp://www.powervision-eng.ch
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Contents• Context and key goals
• Eigenvalue computation
✓Numerical equations
✓Analytical equations
• Examples of application
✓Hydraulic system 1
✓Hydraulic system 2
✓Hydraulic system 3
• Conclusions / Take away message
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Contents• Context and key goals
• Eigenvalue computation
✓Numerical equations
✓Analytical equations
• Examples of application
✓Hydraulic system 1
✓Hydraulic system 2
✓Hydraulic system 3
• Conclusions / Take away message
Context and Key goals
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• Massive penetration of alternative renewable energies (Solar, wind power)
• Stochastic nature of the renewable energy production
Mean annual growth rates of electricity production 2002-2012
• To maintain balanced production:✓ Sufficient reserve capacity
✓ Primary and secondary control capabilities
→ Hydropower plants
→ Off-design operation
Context and Key goals
• Frequent power transients:✓ Require a wide operating range
✓ May induce high levels of vibration and large pressure fluctuations
Part load condition
[0.2 – 0.4] • Turbine rotational speed
* Favrel et al., 2013
HYPERBOLE (ERC/FP7-ENERGY-2013-1-Grant 608532)
Full load condition
* Müller et al. ,2013
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Contents• Context and key goals
• Eigenvalue computation
✓Numerical equations
✓Analytical equations
• Examples of application
✓Hydraulic system 1
✓Hydraulic system 2
✓Hydraulic system 3
• Conclusions / Take away message
SIMSEN software
Modelingfrom
water to wire 8
• SIMSEN software✓ Hydraulic circuit
✓ Electrical installations
✓ Rotating inertias
✓ Control system
Electrical Analogy• Mass and momentum equations:
• Electrical analogy:✓ (Bergeron, 1950; Paynter, 1953 )
R/2Lh/2Lh/2
C
Rh
R/2
Hi
Hci
Qi QiHi+1
2
0H a Q
t gA x
+ =
02
12
=+
+
Q
gDA
Q
t
Q
gAx
H
0
01
=+
+
=
+
IRt
IL
x
U
x
I
Ct
U
ee
e
Storage
Losses
Inertia
Datum.
Q1Q2
D, a, dx
H1 H2
• Assumptions:✓ Uniform flow
✓ 1-D approach
✓ Convective terms neglected
✓ Vertical displacements neglected
Eigenvalue computation
10
• Set of differential equations
( ) ( )XdX
A B X X Cdt
+ =
• Small perturbation
0X X X= +
( )( )
0
0
d X Xf X X
dt
+= +
1det 0
l lI s A B
−
+ =
• Eigenfrequency
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Contents• Context and key goals
• Eigenvalue computation
✓Numerical equations
✓Analytical equations
• Examples of application
✓Hydraulic system 1
✓Hydraulic system 2
✓Hydraulic system 3
• Conclusions / Take away message
Natural frequencies
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• Hydraulic system modelled by an equivalent pipe
1
n
tot ii
l l=
=
1
totequ n
i
ii
la
la
=
=
2
equ equk
totk
a af k
l= =
Total length Equivalent wave speed Natural frequency (Order k)
Standing wave
First natural frequency of the cavitating draft tube
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• Simplified hydroacoustic model of the frictionless cavitating draft tube
2DT DT
DTDT
l g AC
a
= DT
DT DT
lL
g A=
1 1 12 2
DTo
DTDT DT
af
lL C = =
lDT = DT length [m]ADT = DT cross section area [m2]aDT = DT wave speed [m/s]
* Jacob, 1993
Draft tube
Cavitation compliance Draft tube inductance First natural frequency
Sim
ilar
to
a s
pri
ng–
mas
s sy
stem
𝑘 ≈1
𝐶𝐷𝑇𝑚 ≈ 𝐿𝐷𝑇 𝑓0 =
1
2π
𝑘
𝑚≈
1
2π
1
𝐶𝐷𝑇𝐿𝐷𝑇
𝑘
𝑚
1 1 12 2
DTo
TR DT DTDT TR
TR
af
L C Al l
A
= =
First natural frequency of the cavitating draft tube
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• Simplified hydroacoustic model of the frictionless cavitating draft tube
lTR = TR pipe length [m]ATR = TR pipe cross section area [m2]
TRTR
TR
lL
g A=
* Dörfler, 2013
Draft tube
2DT DT
DTDT
l g AC
a
=
Cavitation compliance Tailrace pipe inductance First natural frequency
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DTo
DTDT TR
TR
af
Al l
A
=
Analytical equations
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• Summary
1st order1
2DT
oDT
af
l=
2equ equ
ktotk
a af k
l= =
2equ equ
ktotk
a af k
l= =
2equ equ
ktotk
a af k
l= =
1 2equ equ
totk
a af
l= =
2nd-6th order
Equivalent pipe
Draft tube without TR
Draft tube with TR
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Contents• Context and key goals
• Eigenvalue computation
✓Numerical equations
✓Analytical equations
• Examples of application
✓Hydraulic system 1
✓Hydraulic system 2
✓Hydraulic system 3
• Conclusions / Take away message
Example of applications
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• Hydraulic system 1
PenstockL = 300 mD = 1.2 ma = 1’250 m/sλ = 0.012 -
Draft tubeL = 10 mD = 1.2 ma = [50-100] m/sλ = 0.012 -
TurbinePn = 5 MWQn = 5.0 m3/sHn = 100 mWCNn = 750 rpmDref = 0.846 mNq = 53 -
Nn = 750 rpm = 12.5 Hz➔fexcitation = [0.2 – 0.4]·fn
➔fexcitation = [2.5 – 5 Hz]
Example of applications
• Hydraulic system 1
1st Pressure mode 1st Discharge modeElastic mode shape
Elastic mode shape✓ Non-Linear amplitude variation of
pressure in the TR as function of the length
✓ Non-constant amplitude variation of discharge in the TR as function of the length
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equ equ
tot
a af
l= =
1
2DT
oDT
af
l=
Analytical
calculation
Eigen value
calculation
a DT (min) a DT (min) Relative error
[m/s] [m/s] [%]
f0 [Hz] 0.8 1.24 -35.48
f1 [Hz] 1.14 1.24 -8.06
f2 [Hz] 2.27 2.09 8.61
f3 [Hz] 3.41 3.67 -7.08
f4 [Hz] 4.55 4.18 8.85
f5 [Hz] 5.68 5.99 -5.18
f6 [Hz] 6.82 6.15 10.89
System 1
Example of applications
• Hydraulic system 1
fexcitation = [2.5 – 5 Hz]
• 1st order: Better agreement with f1
• 2nd-6th order: Rather good agreement for natural frequencies Maximum error of 14%.
• Risk of resonance with the draft tube in red.
a =
50
m/s
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12
equ equ
tot
a af
l= =
1
2DT
oDT
af
l=
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Contents• Context and key goals
• Eigenvalue computation
✓Numerical equations
✓Analytical equations
• Examples of application
✓Hydraulic system 1
✓Hydraulic system 2
✓Hydraulic system 3
• Conclusions / Take away message
Example of applications
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• Hydraulic system 2
PenstockL = 300 mD = 1.2 ma = 1’250 m/sλ = 0.012 -
Tailrace pipeL = 100 mD = 1.2 ma = 1’250 m/sλ = 0.012 -
TurbinePn = 5 MWQn = 5.0 m3/sHn = 100 mWCNn = 750 rpmDref = 0.846 mNq = 53 -
Draft tubeL = 10 mD = 1.2 ma = [50-100] m/sλ = 0.012 -
Nn = 750 rpm = 12.5 Hz➔fexcitation = [0.2 – 0.4]·fn
➔fexcitation = [2.5 – 5 Hz]
Example of applications
• Hydraulic system 2
1st Pressure mode 1st Discharge modeRigid column mode shape
Rigid column mode shape✓ Linear amplitude variation of
pressure in the TR
✓ Constant amplitude variation of discharge in the TR
→ Similar to surge tank mass oscillation between TR pipe and DT compliance
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12
equ equ
tot
a af
l= =
12
DTo
DTDT TR
TR
af
Al l
A
=
Analytical
calculation
Eigen value
calculation
a DT (max) a DT (max) Relative error
[m/s] [m/s] [%]
f0 [Hz] 0.5 0.55 -9.09
f1 [Hz] 1.19 0.55 116.36
f2 [Hz] 2.38 2.1 13.33
f3 [Hz] 3.57 4.05 -11.85
f4 [Hz] 4.76 4.88 -2.46
f5 [Hz] 5.95 6.18 -3.72
f6 [Hz] 7.14 6.36 12.26
System 2
Analytical
calculation
Eigen value
calculation
a DT (min) a DT (min) Relative error
[m/s] [m/s] [%]
f0 [Hz] 0.25 0.27 -7.41
f1 [Hz] 0.96 0.27 255.56
f2 [Hz] 1.92 2.04 -5.88
f3 [Hz] 2.88 2.53 13.83
f4 [Hz] 3.85 4.12 -6.55
f5 [Hz] 4.81 4.87 -1.23
f6 [Hz] 5.77 5.16 11.82
System 2
Example of applications
• Hydraulic system 2
fexcitation = [2.5 – 5 Hz]
• 1st order: Good agreement with f0
• 2nd-6th order: Rather good agreement for natural frequencies Maximum error of 14%.
• Risk of resonance with the draft tube in red.
a =
50
m/s
a =
10
0 m
/s
11
12
equ equ
tot
a af
l= =
12
DTo
DTDT TR
TR
af
Al l
A
=
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Contents• Context and key goals
• Eigenvalue computation
✓Numerical equations
✓Analytical equations
• Examples of application
✓Hydraulic system 1
✓Hydraulic system 2
✓Hydraulic system 3
• Conclusions / Take away message
Example of applications
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• Hydraulic system 3
PenstockL = 300 mD = 1.2 ma = 1’250 m/sλ = 0.012 -
Tailrace pipeL = 100 mD = 2 ma = 1’250 m/sλ = 0.012 -
TurbinePn = 5 MWQn = 5.0 m3/sHn = 100 mWCNn = 750 rpmDref = 0.846 mNq = 53 -
Draft tubeL = 10 mD = 1.2 ma = [50-100] m/sλ = 0.012 -
fexcitation = [2.5 – 5 Hz]
Nn = 750 rpm = 12.5 Hz➔fexcitation = [0.2 – 0.4]·fn
➔fexcitation = [2.5 – 5 Hz]
Example of applications
• Hydraulic system 3
1st Pressure mode 1st Discharge modeRigid column mode shape
Rigid column mode shape✓ Linear amplitude variation of
pressure in the TR
✓ Constant amplitude variation of discharge in the TR
→ Similar to surge tank mass oscillation between TR pipe and DT compliance
11
12
equ equ
tot
a af
l= =
12
DTo
DTDT TR
TR
af
Al l
A
=
Analytical
calculation
Eigen value
calculation
a DT (max) a DT (max) Relative error
[m/s] [m/s] [%]
f0 [Hz] 0.84 0.81 3.70
f1 [Hz] 1.19 0.81 46.91
f2 [Hz] 2.38 2.1 13.33
f3 [Hz] 3.57 4.03 -11.41
f4 [Hz] 4.76 4.72 0.85
f5 [Hz] 5.95 6.14 -3.09
f6 [Hz] 7.14 6.55 9.01
System 3
Analytical
calculation
Eigen value
calculation
a DT (min) a DT (min) Relative error
[m/s] [m/s] [%]
f0 [Hz] 0.42 0.42 0.00
f1 [Hz] 0.96 0.42 128.57
f2 [Hz] 1.92 2.04 -5.88
f3 [Hz] 2.88 2.55 12.94
f4 [Hz] 3.85 4.12 -6.55
f5 [Hz] 4.81 4.84 -0.62
f6 [Hz] 5.77 6.16 -6.33
System 3
Example of applications
• Hydraulic system 3
fexcitation = [2.5 – 5 Hz]
• 1st order: Good agreement with f0
• 2nd-6th order: Rather good agreement for natural frequencies Maximum error of 14%.
• Risk of resonance with the draft tube in red.
a =
50
m/s
a =
10
0 m
/s
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Contents• Context and key goals
• Eigenvalue computation
✓Numerical equations
✓Analytical equations
• Examples of application
✓Hydraulic system 1
✓Hydraulic system 2
✓Hydraulic system 3
• Conclusions / Take away message
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Conclusions• Analytical approach to determine the possible risk of resonance with the draft tube vortex rope
excitation (2.5-5Hz)
• 1st order: Better agreement with f1 for the hydraulic system without a tailrace pipe (hydraulic system 1) → Elastic mode shape
• 1st order: Good agreement with f0 for the hydraulic system with a tailrace pipe (hydraulic system 2 & 3) → rigid column mode shape
• 2nd-6th order: Rather good agreement for natural frequencies (Maximum error of 14%).
• Limitations of the methodology✓ Parallel branches:
• Modelling by a single branch with equivalent parameters to obtain a first order of magnitude.• Real system will feature much more complex and numerous eigenvalues (hydraulic system asymmetry, diameters,
bifurcations)
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Take away Message• This analytical method is included as ANNEXE E.2 of the new IEC
Technical Specification 62882 ED1 (to be issued in 2020)
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DTo
DTDT TR
TR
af
Al l
A
=
2equ equ
ktotk
a af k
l= =
Without a tailrace pipe
With a tailrace pipe
2equ equ
ktotk
a af k
l= =
2equ equ
ktotk
a af k
l= =
1st order
1st order
2nd–6th order
2nd–6th order
Hydraulic machines – IEC Technical specification for Francis turbine pressure fluctuation
Thank you for your attention!
Power Vision Engineering Sàrl1, ch. Des Champs-CourbesCH-1024 [email protected]
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Example of applications