Monroe L. Weber-Shirk S chool of Civil and Environmental Engineering When the Steady- State design fails! Hydraulic Transients.
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Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
When the Steady-State design fails!
Hydraulic TransientsHydraulic Transients
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, long
pipeline?What happens if we suddenly stop the flow of
water through a tunnel leading to a turbine?
steady state graduallyvaried
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 transientsunstable pump or turbine operationpipe breaks
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.
Analysis of Transients
Gradually varied (“Lumped”) _________conduit walls are assumed rigidfluid assumed incompressibleflow is function of _____ only
Rapidly varied (“Distributed”) _________fluid assumed slightly compressibleconduit walls may also be assumed to be elasticflow is a function of time and ________
ODE
PDE
time
location
Establishment of Flow:Final Velocity
2V
EGL
HGL
1
H
g
V
2
22
V2
L
Lf hhzg
Vpz
g
Vp2
222
1
211
22
Ken= ____Kexit= ____
g = 9.8 m/s2
H = 100 mK = ____f = 0.02L = 1000 mD = 1 m
1.5
0.51.0
major
minor
How long will it take?
Final Velocity
Lf hhzzH 21
22
f f2
VLh
D g
g
VKhL
2
2
2
f2
V LH K
g D
9.55 m/s2
ff
gHV
LK
D
g = 9.8 m/s2
H = 100 m K = 1.5f = 0.02L = 1000 mD = 1 m
What would V be without losses? _____44 m/s
Establishment of Flow:Initial Velocity
dt
dVALHA
before head loss becomes significant
maF mdVF
dt=
gtL
HV
Vt
dVALdtHA00
ALVHAt AL
HAtV
g = 9.8 m/s2
H = 100 m K = 1.5f = 0.02L = 1000 mD = 1 m
01
2
3
4
5
6
7
8
9
10
0 5 10 15 20 25 30
time (s)
velo
city
(m
/s)
gtL
HV
2
ff
gHV
LK
D
F =
m =
pA HAg=
ALr
Navier Stokes?
________, ________
Flow Establishment:Full Solution
)(mVdt
dF
020
1 f2
tV dV
dtgH K
VL L D
2
f2
VL d ALVA H K
D g dt g
20 0 f
2
t V Ldt dV
VLg H K
D g
F =å gravity drag 0 4lh D
L
gt =-
0F L Dt p=
a
bV
abt 1tanh
1
abtb
aV tanh
2 f tanh
f 2
gH gH KV t
L L L DKD
L
gHa
1 f
2
Kb
L D
12 2 20
1tanh
V dV bVa b V ab a
-=-ò
b
aV if
b
aV f
Flow Establishment:tanh!
V < Vf
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
f2
f
fV
f
Vt
ab V gH KL L D
47.1)9.0(tanh 1
b
aV f
Time to reach 0.9Vf increases as:
L increases
H decreases
1
0.9
2
tanh (0.9)
f2
fVtgH L
KL D
Head loss decreases
Flow Establishment
g = 9.8 m/s2
H = 100 mK = 1.5f = 0.02L = 1000 mD = 1 m
s 34.149.0 fVt
2 f tanh
f 2
gH gH KV t
L L L DKD
0
2
4
6
8
10
12
0 10 20 30 40
time (s)
velo
city
(m/s)
Was f constant?
ReVDn
= 107
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 = 0.02L = 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!
V > Vf?
if a
Vb
>
0
0
0
1ln
2V
a bVt
ab a bV+
=-
( ) oV
aV ctnh ab t t
bé ù= +ë û
12 2 2
1 1ln
2
V dV bV a bVt ctnh
a b V ab a ab a bV-
¥
+= = =
- -ò
( )( )
sinh(2 )
cosh 2 1x
ctnh xx
=-
0
5
10
15
20
0 5 10 15 20
time (s)ve
loci
ty (m
/s)
If V0=( ) a
V ctnh abtb
=
Why does velocity approach final velocity so rapidly?
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
Plastic Pipe
3100 m
Pump inletlength of intake pipeline is 3200 m
1 m
Motor
What happens during startup?
What happens if pump is turned off?
Transient with varying driving force
)( vmdt
dF
g
LVA
dt
dhHA pipe
lpipe
2
2f
2lpipe
L Qh K
D A g
thHL
gAQ l
pipe
dQdthHL
gAl
pipe H = ______________________________Lake elevation - wet pit water level
f(Q)
Finite Difference Solution!
Q
where
wetpit
wetpit
dz Qdt A
=What is z=f(Q)?
Is f constant?
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 1200time (s)
Q (
m3 /s
)
-4
-3
-2
-1
0
1
2
3
4
z (m
)
Q z
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 1200time (s)
Q (
m3 /s
)
-20
-15
-10
-5
0
5
10
15
20
z (m
)
Q z
Why is this unrealistic?
Overflow Weir at 1 m
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 200 400 600 800 1000 1200time (s)
Q (
m3 /s
)
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
z (m
)
Q z
Period of Oscillation: Frictionless Case
dQdthHL
gAl
pipe
zL
gA
dt
dQ pipe Q
dt
dzAwetpit
zL
gA
dt
zdA pipewetpit 2
2
dt
dQ
dt
zdAwetpit 2
2
z = -H
02
2
zLA
gA
dt
zd
wetpit
pipe
wetpit
pipe
wetpit
pipe
LA
gAtC
LA
gAtCz sincos 21
Wet pit mass balance
z = 0 at lake surface
Period of Oscillations
pA
A
g
LT pitwet 2
2
2
2 7.1
24
/81.9
31702
m
m
sm
mT
plan view area of wet pit (m2) 24pipeline length (m) 3170inner diameter of pipe (m) 1.47gravity (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
Tg
Transients
In previous example we assumed that the velocity was the same everywhere in the pipe
We did not consider compressibility of water or elasticity of the pipe
In the next example water compressibility and pipe elasticity will be central
VV2
Valve Closure in Pipeline
Sudden valve closure at t = 0 causes change in 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
Transients: Distributed SystemTools
Conservation of massConservation of momentumConservation of energy
We’d like to knowpressure change
rigid wallselastic walls
propagation speed of pressure wavetime history of transient
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
aV0
V0 V
HGL
V0 a V0 V a
Momentum Equation
2121 ppxx FFMMx
12
111 AVM x 22
222 AVM x
221112111 ApApVVAV
aV0
V0 V
HGL
222111 AVAV
1 2
Mass conservation
A1 A2
p = p2 - p1pVV 11
sspp FFFWMM 2121
Neglect head loss!
Magnitude of Pressure Wave
pVV 11
aV0
V0 V
1 2
1V aV 0
Vap
a VH
g- D
D =
0Va
p HgD = D
Decrease in V causes a(n) _______ in HGL.increase
Propagation Speed:Rigid Walls
Conservation of mass
a
V0 V0 V0 0
1)(
0
00
aVV
0
0 )( aVV
Solve for V
))(()( 0000 VaVAaVA
Propagation Speed:Rigid Walls
a
V0 V0 V0 0
momentumVaVp )( 00
aV 0 0
2ap
0
0 )( aVV mass
0
200 )( aVp
Need a relationship between pressure and density!
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.
GPa 2.2K
3Kg/m 1000
m/s 14801000
10 x 2.2 9
a
speed of sound in water
(for water)
Propagation Speed:Elastic Walls
a
V0 V0 V0 0
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
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 wall thickness, steel pipeline? The initial flow velocity was 5 m/s.
E for steel is 200 GPaWhat is the increase in pressure?
Time History of Hydraulic Transients: Function of ...
Time history of valve operation (or other control device)
Pipeline characteristicsdiameter, thickness, and modulus of elasticity length of pipeline frictional characteristics
tend to decrease magnitude of pressure wave
Presence and location of other control devicespressure relief valves surge tanks reservoirs
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
t L
a
V= -Vo
L
t 2L
a
1
2
3
4
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
t 3L
a
t 4L
a
5
6
7
8
Pressure variation over time
reservoir level
Pressure variation at valve: velocity head and friction losses neglected
4L
a
8L
a
12L
a
H
time
Pre
ssur
e he
ad
Neglecting head loss!
Real traces
Lumped vs. Distributed
For LSC wet pit = 424 s = 4*3170 m/1400 m/s = ____
4LT
a>>
pressure fluctuation period
lumped
pA
A
g
LT pitwet 2
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
Methods of Controlling Transients
Valve operation limit operation to slow changes if rapid shutoff is necessary consider diverting the flow
and 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
Surge Tanks
ReservoirTunnel/Pipeline
Tail water
T
Penstock Reduces amplitude of pressure fluctuations in ________ by reflecting incoming pressure waves
Decreases cycle time of pressure wave in the penstock
Start-up/shut-down time for turbine can be reduced (better response to load changes)
Surge tank
tunnel
Surge tanks
Use of Hydraulic TransientsThere is an old technology that
used hydraulic transients to lift water from a stream to a higher elevation. The device was called a “Ram Pump”and it made a rhythmic clacking noise.
How did it work? High pressure pipe
Stream
Ram Pump
Source pipe
Minimum valve closure time
How would you stop a pipeline full of water in the minimum time possible without bursting the pipe?
pipel
A gH h dt dQ
L
pipel
A g pz h dt dQ
L g
pH z
g
V
EGL
HGL
H
L
( ) 2pa g Vr r m=- Ñ + + Ñ
Simplify: no head loss and hold pressure constant
pipel
A g pz h dt dQ
L g
pipeA g pz dt dQ
L g
0pipeA g p
z t QL 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 Lt
pg z
g
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)
velo
city
(m/s)
z3
z1
3z g
z
-z1
pz
g
3 1z z
l
dV g pz h
dt L g
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 z
Low V (low hl)
Maintain high V
Ram: Optimal Operation
What is the theoretical maximum ratio of pumped water to wasted water?
Rate of decrease in PE of wasted water equals rate of increase in PE of pumped water
1 3 1w pumpedQ z Q z z
1
3 1
pumped
w
Q z
Q z z
High Q and Low loses?
0
2
4
6
8
10
12
0 10 20 30 40
time (s)
velo
city
(m/s)
l
dV g pz h
dt L g
l
dV g pz h
dt L g
3 1
dV g
dz
Lz
t
1zdV g
dt L
Acceleration
Deceleration (pumping)Insignificant head loss
Keep V high for max Q
Cycle times
1acc
acc
gtdVt z
dt L
3 1decel
decel
gtdVt z z
dt L
1 3 1acc decelgt gt
z z zL L
1
3 1
acc
decel
t z
t z z
Change in velocities must match
decel acc
dV dVt t
dt dt
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
Burst section of Penstock:Oigawa Power Station, Japan
Chaudhry page 17
Collapsed section of Penstock:Oigawa Power Station, Japan
Chaudhry page 18
Values for Wet Pit Analysis
Flow rate before pump failure (m3/s) 2plan view area of wet pit (m2) 24pipeline length (m) 3170inner diameter of pipe (m) 1.47elevation of outflow weir (m) 10time interval to plot (s) 1000pipe roughness (m) 0.001density (kg/m3) 1000dynamic viscosity (Ns/m2) 1.00E-03gravity (m/s2) 9.81
Pressure wave velocity: Elastic Pipeline
E = 200 GPaD = 1 mt = 1 cm
t
D
E
K
Ka
1
0 m/s 1020
01.0
1
10200
102.21
1000102.2
9
9
9
x
x
xa
0.5 s to travel 500 m
Hgp
m 5209.8m/s
m/s) m/s)(-5 (1020
2
g
VaH
psi 740 = MPa 5.1 = m) )(520m/s )(9.8kg/m (1000 23p
Ram Pump
Water inlet
Air Chamber
Rapid valve
Ram pump
High pressure pipe
Stream
Ram Pump
Source pipeH1
H2
Ram animationRam animation
Ram Pump
D
f
L
K
L
gHV
V
abt
f
fV f
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)
velo
city
(m/s)
2
dV gH
dt L
dt
dVALHA
Surge Tanks
Real pressure traces
At valve At midpoint
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