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Monroe L. Weber-ShirkSchool ofCivil and
Environmental Engineering
When the Steady-
State design fails!
Hydraulic Transients
http://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cornell.edu/http://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/faculty/info.cfm?abbrev=faculty&shorttitle=bio&netid=mw24http://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://ceeserver.cee.cornell.edu/mw24/Default.htm8/4/2019 06 Hydraulic Transients
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Hydraulic Transients: Overview
In all of our flow analysis we have assumed
either _____ _____ operation or ________
______ flow
What about rapidly varied flow?
How does flow from a faucet start?
How about flow startup in a large, longpipeline?
What happens if we suddenly stop the flow of
water through a tunnel leading to a turbine?
steady state gradually
varied
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Hydraulic Transients
Routine transients
change in valve settings starting or stopping of pumps
changes in power demand for
turbines
changes in reservoir elevation
turbine governor hunting
action of reciprocating pumps
lawn sprinkler
Unsteady Pipe Flow: time varying flow and pressure
Catastrophic transients
unstable pump or turbineoperation
pipe breaks
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References
Chaudhry, M. H. 1987. Applied Hydraulic
Transients. New York, Van Nostrand
Reinhold Company.
Wylie, E. B. and V. L. Streeter. 1983. Fluid
Transients. Ann Arbor, FEB Press.
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Analysis of Transients
Gradually varied (Lumped) _________
conduit walls are assumed rigid
fluid assumed incompressible
flow is function of _____ only
Rapidly varied (Distributed) _________
fluid assumed slightly compressible
conduit walls may also be assumed to be elastic
flow is a function of time and ________
ODE
PDE
time
location
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Establishment of Flow:
Final Velocity
2V
EGL
HGL
1
H
g
V
2
22
V2
L Lf hhz
g
Vpz
g
Vp2
2
221
2
11
22
Ken= ____Kexit= ____
g = 9.8 m/s2
H = 100 m
K = ____
f = 0.02L = 1000 m
D = 1 m
1.5
0.5
1.0
How long will it take?
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Final Velocity
Lf hhzzH 212
2
f f 2
VL
h D g g
V
KhL 2
2
2
f
2
V LH K
g D
9.55 m/s2
ff
gHV
LK
D
g = 9.8 m/s2
H = 100 m
K = 1.5
f = 0.02L = 1000 m
D = 1 m
What would V be without losses? _____44 m/s
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Establishment of Flow:
Initial Velocity
dt
dVALHA
before head loss becomes significant
maF mdVFdt
=
gt
L
HV
Vt
dVALdtHA00
ALVHAt AL
HAtV
g = 9.8 m/s2
H = 100 m K = 1.5
f = 0.02
L = 1000 m
D = 1 m01
2
3
4
5
67
8
9
10
0 5 10 15 20 25 30
time (s)
velocity(m/s)
gtL
HV
2
ffgH
VL
KD
F=
m =
pA HAg=
ALr
Navier Stokes?
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________, ________
Flow Establishment:
Full Solution
)(mVdt
dF
020
1 f
2
tV dV
dtgH K
V
L L D
2
f2
VL d ALVA H K
D g dt g
2
0 0 f
2
t VL
dt dV VL
g H K
D g
F = gravity drag 04
lh D
L
gt = -
0F L Dt p=
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a
bV
ab
t 1tanh1
abtb
aV tanh
2 ftanh
f 2
gH gH K V t
L L L DK
D
L
gHa
1 f
2
Kb
L D
1
2 2 20
1tanh
V dV bV
a b V ab a
-
=
-
b
aV if
b
aV
f
Flow Establishment:
tanh!
V < Vf
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Time to reach final velocity
1 11 1tanh tanh
f
bV Vt
ab a ab V
11
0.9
0.91 tanh (0.9)tanh
f
2
f
f
V
f
Vt
ab V gH K
L L D
47.1)9.0(tanh 1
b
aVf
Time to reach 0.9Vfincreases as:
L increases
H decreases
1
0.9
2
tanh (0.9)
f2
fVt
gH LK
L D
Head loss decreases
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Flow Establishment
g = 9.8 m/s2
H = 100 m
K = 1.5
f= 0.02
L = 1000 m
D = 1 m
s34.149.0 fVt
2 ftanh
f 2
gH gH K V t
L L L DK
D
0
2
4
6
8
10
12
0 10 20 30 40
time (s)
velocity
(m/s)
Was f constant?
Re VDn
= 107
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Household plumbing example
Have you observed the gradual increase in flow
when you turn on the faucet at a sink?
50 psi - 350 kPa - 35 m of headK = 10 (estimate based on significant losses in faucet)
f = 0.02
L = 5 m (distance to larger supply pipe where velocity
change is less significant)D = 0.5 - 0.013 m
time to reach 90% of final velocity? T0.9Vf= 0.13 s
No? Good!
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V > Vf?
ifa
Vb
>
0
0
0
1ln
2V
a bVt
ab a bV
+=
-
( )oV
aV ctnh ab t t
b = +
1
2 2 2
1 1ln
2
V dV bV a bV t ctnh
a b V ab a ab a bV
-
+
= = =
- -
( )( )
sinh(2 )
cosh 2 1
xctnh x
x=
-
0
5
10
15
20
0 5 10 15 20
time (s)
velocity(m/s)If V0=( )aV ctnh abt
b=
Why does velocity approach final velocity so rapidly?
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Intake Pipe, with
flow Q and cross
sectional area Apipe
Wet Pit,
with plan
view area
Atank
Lake Source Cooling Intake
Schematic
Lake Water Surface
?
Steel Pipe
100 m
Pump inlet
length of intake pipeline is 3200 m
1 m
Motor
What happens during startup?
What happens if pump is turned off?
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Transient with varying driving
force
)( vmdt
dF
g
LVA
dt
dhHA
pipe
lpipe
2
2f 2l
pipe
L Q
h K D A g
thH
L
gAQ
l
pipe
dQdthH
L
gAl
pipe H = ______________________________Lake elevation - wet pit water level
f(Q)
Finite Difference Solution!
Q
where
wetpit
wetpit
dz Q
dt A=What is z=f(Q)?
Is f constant?
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Wet Pit Water Level and Flow
Oscillations
constantsWhat is happening on the vertical lines?
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 200 400 600 800 1000 1200
time (s)
Q(m3/s)
-4
-3
-2
-1
0
1
2
3
4
z(m)
Q z
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Wet Pit with Area Equal to Pipe
Area
Pipe collapse
Water Column Separation-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 200 400 600 800 1000 1200
time (s)
Q(m3/s)
-20
-15
-10
-5
0
5
10
15
20
z(m)
Q z
Why is this unrealistic?
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Overflow Weir at 1 m
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 200 400 600 800 1000 1200
time (s)
Q(m3/s)
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
z(m)
Q z
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Period of Oscillation:
Frictionless Case
dQdthHL
gAl
pipe
zL
gA
dt
dQ pipe Qdt
dzAwetpit
zL
gAdt
zdA pipewetpit 2
2
dt
dQ
dt
zdAwetpit 2
2
z = -H
02
2
z
LA
gA
dt
zd
wetpit
pipe
wetpit
pipe
wetpit
pipe
LA
gAtC
LA
gAtCz sincos 21
Wet pit mass balance
z = 0 at lake surface
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Period of Oscillations
pA
A
g
LT
pitwet2
2
2
27.1
24
/81.9
31702
m
m
sm
mT
plan view area of wet pit (m2) 24
pipeline length (m) 3170
inner diameter of pipe (m) 1.47
gravity (m/s2) 9.81
T = 424 s
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 200 400 600 800 1000 1200time (s)
Q(m3/s)
-4
-3
-2
-1
0
1
2
3
4
z(m)
Q z
Pendulum Period?
2L
T
g
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Transients
In previous example we assumed that the
velocity was the same everywhere in the
pipeWe did not consider compressibility of
water or elasticity of the pipe
In the next example water compressibilityand pipe elasticity will be central
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VV2
Valve Closure in Pipeline
Sudden valve closure at t = 0 causes changein discharge at the valve
What will make the fluid slow down?____
Instantaneous change would require__________
Impossible to stop all the fluid
instantaneously
infinite force
What do you think happens?
p at valve
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Transients: Distributed System
Tools
Conservation of mass
Conservation of momentum
Conservation of energy
Wed like to know
pressure change
rigid wallselastic walls
propagation speed of pressure wave
time history of transient
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Pressure change due to velocity
change
velocity
density
pressure
unsteady flow steady flow
P0
0
V0 VV 0
P0
P
0
P0
0
P0
P
0
a
V0
V0
V
HGL
V0 a V0 V a
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Momentum Equation
2121
ppxx
FFMMx
1
2
111AVM x 2
2
222AVM x
221112111
ApApVVAV
a
V0
V0
V
HGL
222111AVAV
1 2
Mass conservation
A1A2
p = p2 - p1pVV 11
sspp FFFWMM 21
21
Neglect head loss!
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Magnitude of Pressure Wave
pVV 11
a
V0
V0
V
1 2
1
V aV 0
Vap
a VH
g
- DD =
0Va
p HgD = D
Decrease in V causes a(n) _______ in HGL.increase
i d
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Propagation Speed:
Rigid Walls
Conservation of mass
a
V0
V0
V
00
1)(
0
0
0
aVV
0
0 )( aVV
Solve for V
))(()( 0000 VaVAaVA
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Propagation Speed:
Rigid Walls
a
V0
V0
V
00
momentumVaVp )( 00
aV 0 0
2ap
0
0 )( aVV mass
0
2
00 )( aVp
Need a relationship between pressure and density!
P i S d
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Propagation Speed:
Rigid Walls
pK
pa2
Ka
definition of bulk modulus of elasticity
Example:
Find the speed of a pressure wave in a water pipeline
assuming rigid walls.
GPa2.2K
3Kg/m1000
m/s14801000
10x2.29
a
speed of sound in water
(for water)
P i S d
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Propagation Speed:
Elastic Walls
a
V0
V0
V
00
0
Ka D
t = thickness of thin walled pipe
E = bulk modulus of elasticity for pipe
Additional parameters
D = diameter of pipe
t
D
E
K
Ka
1
0 effect of water compressibility
effect of pipe elasticity
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solution
Propagation Speed:
Elastic Walls
Example: How long does it take for a
pressure wave to travel 500 m after a rapid
valve closure in a 1 m diameter, 1 cm wallthickness, steel pipeline? The initial flow
velocity was 5 m/s.
E for steel is 200 GPaWhat is the increase in pressure?
Ti Hi f H d li
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Time History of Hydraulic
Transients: Function of ...
Time history of valve operation (or other controldevice)
Pipeline characteristics
diameter, thickness, and modulus of elasticity
length of pipeline
frictional characteristics tend to decrease magnitude of pressure wave
Presence and location of other control devices
pressure relief valves
surge tanks
reservoirs
Ti Hi f H d li
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Time History of Hydraulic
Transients
V=Vo V=0
a
H
L
V=0
H
L
t
L
a
t
V= -Vo V=0
a
H
L
tL
a
V= -Vo
L
t2L
a
1
2
3
4
Ti Hi t f H d li
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Time History of Hydraulic
Transients
V= -Vo V=0
a
H
L
V=0H
L
V=Vo V=0
a
H
L
V= Vo
L
a
Lt
2
t
3L
a
t3L
a
t
4L
a
5
6
7
8
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Pressure variation over time
reservoir
level
Pressure variation at valve: velocity head and friction
losses neglected
4L
a
8L
a
12L
a
H
time
Pressurehead
Neglecting head loss!
Real traces
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Lumped vs. Distributed
For LSC wet pit
T = 424 s
= 4*3170 m/1400 m/s = ____
4LT
a>>
pressure fluctuation period
lumped
pA
A
g
LT
pitwet2
9.1 s
For _______ system
4La
= __________________________
What would it take to get a transient with a period of
9 s in Lake Source Cooling? ____________Fast valve
M th d f C t lli
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Methods of Controlling
Transients
Valve operation
limit operation to slow changes
if rapid shutoff is necessary consider diverting the flowand then shutting it off slowly
Surge tank
acts like a reservoir closer to the flow control point
Pressure relief valve automatically opens and diverts some of the flow when
a set pressure is exceeded
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Surge Tanks
Reservoir
Tail water
T
Penstock Reduces amplitude of pressure
fluctuations in ________ by reflecting
incoming pressure waves
Decreases cycle time of pressurewave in the penstock
Start-up/shut-down time for turbine
can be reduced (better response to
load changes)
Surge tank
tunnel
Surge tanks
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Use of Hydraulic Transients
There is an old technology thatused hydraulic transients to liftwater from a stream to a higher
elevation. The device was called aRam Pumpand it made arhythmic clacking noise.
How did it work? High pressure pipe
Stream
Ram Pump
Source pipe
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Minimum valve closure time
How would you stop a pipeline full of water
in the minimum time possible without
bursting the pipe?
pipe lA g
H h dt dQL
pipe
l
A g pz h dt dQ
L g
pH z
g
V
EGL
HGL
H
L
( ) 2pa g Vr r m= - + +
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Simplify: no head loss and hold
pressure constant
pipe
l
A g pz h dt dQ
L g
pipeA g p z dt dQL g
0
pipeA g p
z t Q
L g
0
pipe
Q Lt
pA g z
g
V
EGL
HGL
H
L
Integrate from 0 to t and from Q
to 0 (changes sign)
0V Ltp
g zg
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Back to Ram Pump:
Pump Phase
Coordinate system?
P1 = _____
P2
= _____
z2-z1 = ___
High pressure pipe
StreamSource pipe
00
2
4
6
8
10
12
0 10 20 30 40
time (s)
velocity(m/s)
z3
z1
3z g
z
-z1p
zg
3 1z z
l
dV g pz h
dt L g
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Reflections
What is the initial head loss term if the pump
stage begins after steady state flow has been
reached? _____What is ?_____
What is when V approaches zero?
______Where is most efficient pumping? ___________
How do you pump the most water? ______
l
dV g pz h
dt L g
z1
l
pz h
g
z3
l
pz h
g
3 1z zLow V (low hl)
Maintain high V
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Ram: Optimal Operation
What is the theoretical maximum ratio of
pumped water to wasted water?
Rate of decrease in PE of wasted waterequals rate of increase in PE of pumped
water
1 3 1w pumped Q z Q z z
1
3 1
pumped
w
Q z
Q z z
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High Q and Low loses?
0
2
4
6
8
10
12
0 10 20 30 40
time (s)
velocity(m/s)
l
dV g pz h
dt L g
l
dV g pz h
dt L g
3 1dV g
dz
Lz
t
1zdV g
dt L
Acceleration
Deceleration (pumping)Insignificant head loss
Keep V high for max Q
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Cycle times
1acc
acc
gtdVt z
dt L
3 1deceldecel gtdV t z zdt L
1 3 1acc decelgt gt z z z
L L
1
3 1
acc
decel
t z
t z z
Change in velocities must match
decel accdV dV t tdt dt
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Summary (exercise)
When designing systems, pay attention to
startup/shutdown
Design systems so that high pressure waves
never occur
High pressure waves are reflected at reservoirs
or surge tanks
B i f P k
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Burst section of Penstock:
Oigawa Power Station, Japan
Chaudhry page 17
C ll d i f P k
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Collapsed section of Penstock:
Oigawa Power Station, Japan
Chaudhry page 18
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Values for Wet Pit Analysis
Flow rate before pump failure (m3/s) 2
plan view area of wet pit (m2) 24
pipeline length (m) 3170inner diameter of pipe (m) 1.47
elevation of outflow weir (m) 10
time interval to plot (s) 1000
pipe roughness (m) 0.001
density (kg/m3) 1000dynamic viscosity (Ns/m2) 1.00E-03
gravity (m/s2) 9.81
P l it El ti
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Pressure wave velocity: Elastic
Pipeline
E = 200 GPa
D = 1 m
t = 1 cm
t
D
E
K
Ka
1
0 m/s1020
01.0
1
10200
102.21
1000102.2
9
9
9
x
x
xa
0.5 s to travel 500 m
Hgp
m5209.8m/s
m/s)m/s)(-5(1020
2
g
VaH
psi740=MPa5.1=m))(520m/s)(9.8kg/m(100023
p
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Ram Pump
Water inlet
Air Chamber
Rapid valve
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Ram pump
High pressure pipe
Stream
Ram Pump
Source pipeH1
H2
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Ram animation
http://schou.dk/animation/hydraram.swf8/4/2019 06 Hydraulic Transients
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Ram Pump
D
f
L
K
L
gHV
V
ab
t
f
f
Vf
2
)9.0(tanh9.0tanh
1
1
11
9.0Time to establish flow
0
2
4
6
8
10
12
0 10 20 30 40
time (s)
velocity(m/s)
2
dV gH
dt L
dt
dVALHA
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Surge Tanks
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Real pressure traces
At valve At midpoint
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Presentacin tomada de la web:
Profesor
Monroe L. Weber - Shrik
At valve At midpoint