Hydraulic Transients
When the SteadyState design fails!
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Hydraulic Transients: Overview
In all of our flow analysis we have assumed
gradually
either _____
steady_____
state operation or ________
varied 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?
Hydraulic Transients
Unsteady Pipe Flow: time varying flow and pressure
 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
Catastrophic transients
unstable pump or turbine
operation
pipe 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
ODE
Gradually varied (“Lumped”) _________
conduit walls are assumed rigid
fluid assumed incompressible
flow is function of _____
time only
PDE
Rapidly varied (“Distributed”) _________
fluid assumed slightly compressible
conduit walls may also be assumed to be elastic
flow is a function of time and ________
location
Establishment of Flow:
Final Velocity
How long will it take?
1
H
EGL
2
2
V
2g
HGL
V2
L
p1 V12
p2 V22

 z1 

 z 2  h f   hL
 2g
 2g
g = 9.8 m/s2
H = 100 m
1.5
K = ____
f = 0.02
L = 1000 m
V 2 D=1m
0.5
Ken= ____
1.0
Kexit= ____
Final Velocity
H  z1  z2  h f 
2
2
LV
hf  f
D 2g
 hL
h
L
 K
V2  L

H
f


K


2g  D

2 gH
Vf 
fL
K 
D
V2
2g
g = 9.8 m/s2
H = 100 m
 K = 1.5
f = 0.02
L = 1000 m
D=1m
9.55 m/s
What would V be without losses? _____
44 m/s
Establishment of Flow:
Navier Stokes? Initial Velocity
velocity (m/s)
before head loss becomes significant
mdV
10
F  ma
F=
9
dt
8
F = pA = g HA
7
m = r AL
HA  AL
dV
dt
t
V 
H
V
gt
L
0
HAt  ALV
0
5
10
2 gH
fL
K 
D
g = 9.8 m/s2
H = 100 m
 K = 1.5
f = 0.02
L = 1000 m
D=1m
0
HA dt  AL  dV
0
6
5
4
3
2
1
Vf 
15
20
25
time (s)
V 
HAt
AL
V 
H
L
gt
30
Flow Establishment:
Full Solution
F 
d
å
(mV )
dt
F =________,
gravity ________
drag
t 0 =-
g hl D
4L
F = t 0 Lp D

L  V 2  d   ALV 

 A  H   K  f    

D
2
g
dt
g

 



t
V
 dt  
0
0
L
2

V
L

 
g  H   K  f  
D  2g 


t
dV
 dt  
V
0
0
dV
gH 1  K f  2
 
 V
L 2 L D
Flow Establishment:
tanh!
dV
1
- 1 bV
ò0 a 2 - b2V 2 = ab tanh a
if V 
V
 bV 

t 
tanh 

 a 
ab
1
V 
a
b
1
tanh abt 
b
a 
gH
L
Vf 
a
a
b
 gH  K f  
2 gH
V
tanh  t
 

 2L  L D  
fL


K 
D
V < Vf
1  K f 
b
 

2 L D
Time to reach final velocity
1
1  bV
t
tanh 
ab
 a
t0.9V f
 
 1
1 V
tanh  

V 
 ab
 f 
 0.9V f
1
1

tanh 
 V
ab
 f

 

tanh 1 (0.9)
gH  K f 
 

2L  L D 
Vf 
a
b
tanh 1 (0.9)  1.47
Time to reach 0.9Vf increases as:
tanh 1 (0.9)
L increases
t0.9V 
gH 
L

K

f
H decreases


2 L2 
D
Head loss decreases
f
Flow Establishment
t0.9V  14.34 s
f
12
10
velocity (m/s)
g = 9.8 m/s2
H = 100 m
K = 1.5
f = 0.02
L = 1000 m
D=1m
8
6
4
2
0
0
10
20
30
time (s)
 gH  K f  
2 gH
V
tanh  t
 



fL
2
L
L
D




K 
D
Was f constant?
Re =
VD
n
107
40
Household plumbing example
 Have you observed the gradual increase in flow
when you turn on the faucet at a sink? No? Good!
 50 psi - 350 kPa - 35 m of head
 K = 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
V > Vf?
dV
1
1
a + bV
- 1 bV
t =ò 2
= ctnh
=
ln
2
2
¥ a - b V
ab
a 2ab a - bV
a
if V >
b
V
20
If V0=
velocity (m/s)
a
V = ctnh (abt )
b
a + bV0
1
tV0 =
ln
2ab a - bV0
15
10
5
0
0
5
10
15
time (s)
a
V = ctnh é
ab t + tVo ù
ë
û
b
(
)
ctnh ( x ) =
sinh(2 x)
cosh (2 x ) - 1
Why does velocity approach final velocity so rapidly?
20
Lake Source Cooling Intake
Schematic
Motor
Lake Water Surface
Steel Pipe
?
100 m
Intake Pipe, with
flow Q and cross
sectional area Apipe
1m
Pump inlet
length of intake pipeline is 3200 m
What happens during startup?
What happens if pump is turned off?
Wet Pit,
with plan
view area
Atank
Transient with varying driving
force
F

d
(mv)
dt
Q
d  Apipe LV 
Apipe H  hl  

dt 
g

Apipe g
H  hl dt  dQ
where
L  Q2

hl   K  f  2
D  Apipe 2 g

Lake elevation - wet pit water level
H = ______________________________
L
What is z=f(Q)?
Q 
Apipe g
L
H  hl t
dzwetpit
dt
=
Q
f(Q)
Awetpit
Finite Difference Solution!
Is f constant?
Wet Pit Water Level and Flow
Oscillations
4
Q
3
1
2
0.5
1
0
0
3
Q (m /s)
1.5
z
-0.5
-1
-1
-2
-1.5
-3
-2
-4
0
200
400
600
time (s)
800
What is happening on the vertical lines?
1000
z (m)
2
1200
constants
Wet Pit with Area Equal to Pipe
Area
2
20
Q
1.5
z
10
0.5
5
0
0
-0.5
-5
-1
-10
Pipe collapse
Water Column Separation
-1.5
-2
0
200
Why is this unrealistic?
400
600
time (s)
800
1000
-15
-20
1200
z (m)
Q (m3/s)
1
15
Overflow Weir at 1 m
2
2
Q
1.5
z
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
0
200
400
600
800
time (s)
1000
-2
1200
z (m)
Q (m3/s)
1
1.5
Period of Oscillation:
Frictionless Case
Apipe g
H  hl dt  dQ
z = -H
z = 0 at lake surface
L
Wet pit mass balance
dQ

 Apipe g
dt

Apipe g
dt 2
d z
dt 2
gApipe
LAwetpit
dt 2

dQ
dt
z
L

Q
dt
L
Awetpit d 2 z
2
Awetpit
z
dz
Awetpit d 2 z
z0
 gA

pipe
z  C1 cos t
 LA
wetpit


 gA


pipe

C
sin
t
2



 LAwetpit




Period of Oscillations
g Ap
T  2
3170m 24m 2
2
9.81m / s 1.7m
T = 424 s
Pendulum Period?
L
T  2
g
2
2
4
Q
1.5
Q (m3/s)
T  2
plan view area of wet pit (m2)
24
pipeline length (m)
3170
inner diameter of pipe (m)
1.47
gravity (m/s2)
9.81
z
3
1
2
0.5
1
0
0
-0.5
-1
-1
-2
-1.5
-3
-2
-4
1200
0
200
400
600
time (s)
800
1000
z (m)
L Awet pit
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
Valve Closure in Pipeline
V2
V
Sudden valve closure at t = 0 causes change
in discharge at the valve
↑p at valve
What will make the fluid slow down?____
Instantaneous change would require
infinite force
__________
Impossible to stop all the fluid
instantaneously What do you think happens?
Transients: Distributed System
Tools
Conservation of mass
Conservation of momentum
Conservation of energy
We’d like to know
pressure change
rigid walls
elastic walls
propagation speed of pressure wave
time history of transient
Pressure change due to velocity
change
HGL
V0  V
V0
a
steady flow
unsteady flow
V0  V V0  a
velocity
V0
density
0
0  
0
0  
pressure
P0
P0  P
P0
P0  P
V0  V  a
Momentum Equation
HGL
V0  V
V0
1
a
2
M1  M 2  W  Fp1  Fp2  Fss
M 1x  M 2 x  Fp1 x  Fp2
M 1x   1V12 A1
M 2 x   2V22 A2
Neglect head loss!
Mass conservation
1V1 A1   2V2 A2
1V1 A1 V2  V1   p1 A1  p2 A2
A1  A2
1V1V  p
 p = p2 - p 1
Magnitude of Pressure Wave
V0  V
V0
1
a
2
1V1V  p
V1  V0  a
a  V0
p   aV
- a DV
DH =
g
Dp = gDH
increase in HGL.
Decrease in V causes a(n) _______
Propagation Speed:
Rigid Walls
V0  V
V0
0
a
0  
A(V0  a) 0  A(V0  a  V )( 0   ) Conservation of mass
 0

V  (V0  a) 
 1
  0  

  

V  (V0  a) 
  0   
Solve for V
Propagation Speed:
Rigid Walls
V0  V
V0
0
p    0 (V0  a)V
 


V  (V0  a) 
  0   
a
0  
momentum
mass
  
   0
V0  a

p   0 (V0  a) 
  0   
p  a 2  Need a relationship between pressure and density!
2
Propagation Speed:
Rigid Walls
K 
p
 
a 
2
p

definition of bulk modulus of elasticity
a
K

Example:
Find the speed of a pressure wave in a water pipeline
assuming rigid walls.
2.2 x 109
a 
 1480 m/s
K  2.2 GPa (for water)
1000
3
  1000 Kg/m
speed of sound in water
Propagation Speed:
Elastic Walls
V0  V
V0
0  
0
a 
K
0
D
a
Additional parameters
D = diameter of pipe
t = thickness of thin walled pipe
E = bulk modulus of elasticity for pipe
a 
K 0
effect of water compressibility
K D
1
effect of pipe elasticity
E t
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 GPa
What is the increase in pressure?
solution
Time History of Hydraulic
Transients: Function of ...
 Time history of valve operation (or other control
device)
 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
Time History of Hydraulic
Transients
1
H
V=Vo
3
V=0
H
V= -Vo
V=0
a
L
a
t
2
t
H
V=0
L

a
L
4
V= -Vo
L
L
t
L
a
t
2L
a
Time History of Hydraulic
Transients
H
5
V= -Vo
H
7
V=Vo
V=0
a
L
2L
t

a
V=0
a
t
6
3L

a
L
8
H
V=0
V= Vo
L
t
L
3L
a
t
4L
a
Pressure head
Pressure variation over time
H
reservoir
level
Neglecting head loss!
4L
a
8L
a
12L
a
time
Pressure variation at valve: velocity head and friction
losses neglected
Real traces
Lumped vs. Distributed
4L
T >>
a
lumped system
For _______
pressure fluctuation period
T = __________________________
For LSC wet pit
T  2
T = 424 s
4L
9.1 s

= 4*3170 m/1400 m/s = ____
L Awet pit
g Ap
a
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
Reservoir
 Reduces amplitude of pressure
tunnel by reflecting
fluctuations in ________
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
Penstock
T
Tail water
Surge tanks
Use of Hydraulic Transients
 There 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
Source pipe
Stream
Ram Pump
Minimum valve closure time
How would you stop a pipeline full of water
in the minimum time possible without
bursting the pipe?
r a = - (Ñp + r g ) + mÑ2 V
Apipe g
L
 H  hl  dt  dQ
Apipe g  p


 z   hl  dt  dQ

L   g


 p

H  
 z 
 g

H
EGL
HGL
V
L
Simplify: no head loss and hold
pressure constant
Apipe g  p


 z   hl  dt  dQ

L   g


H
EGL
HGL
V
Apipe g  p

 z  dt  dQ

L g

Apipe g  p

 z  t  Q0

L g

Q0 L
t
 p

Apipe g 
 z 
g

L
Integrate from 0 to t and from Q
to 0 (changes sign)
V0 L
t
 p

g
 z 
g


dV  g  p

 z  hl 

dt
L g

Back to Ram Pump:
Pump Phase
10
velocity (m/s)
 Coordinate system?
 P1 = _____
0
z3  g
 P2 = _____
 z2-z1 = ___
-z1
12
z1
6
4
2
0
0
10
20
40
High pressure pipe
Source pipe
Stream
30
time (s)
p
 z  z3  z1
g
z
8
z3
Reflections

dV  g  p

 z  hl 

dt
L g

What is the initial head loss term if the pump
stage begins after steady state flow has been
z1
reached? _____
p
z3
What is  g  z  hl ?_____
What is p  z  hl when V approaches zero?
g
z3  z1
______
Low V (low hl)
Where is most efficient pumping? ___________
How do you pump the most water? ______
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
Qw z1  Qpumped  z3  z1 
Q pumped
Qw
z1

z3  z1
High Q and Low loses?
Acceleration
12
velocity (m/s)
10
8
6
Insignificant head loss

dV  g  p

 z  hl 

dt
L g

dV  g

  z1 
dt
L
Deceleration (pumping)
4
Keep V high for max Q
2
0
0
10
20
time (s)
30
40

dV  g  p

 z  hl 

dt
L g

dV  g

 z3  z1 
dt
L
Cycle times
 gtacc
dV
tacc 
  z1 
dt
L
 gtdecel
dV
tdecel 
 z3  z1 
dt
L
gtacc
gtdecel
 z1  
 z3  z1 
L
L
tacc
z1

tdecel z3  z1
Change in velocities must match
dV
dV
tdecel 
tacc
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)
2
plan view area of wet pit (m2)
24
pipeline length (m)
3170
inner 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)
1000
dynamic viscosity (Ns/m2)
1.00E-03
gravity (m/s2)
9.81
Pressure wave velocity: Elastic
Pipeline
E = 200 GPa
D=1m
t = 1 cm
K 0
a 
K D
1
E t
a 
2.2 x109 1000
1
2.2 x10
9
1
 1020 m/s
200 x109 0.01
0.5 s to travel 500 m
 aV
 (1020 m/s)(-5 m/s)
H 

 520 m
g
9.8m/s 2
p  gH
p  (1000 kg/m 3 )(9.8 m/s 2 )(520 m) = 5.1 MPa = 740 psi
Ram Pump
Air Chamber
Rapid valve
Water inlet
Ram pump
H2
High pressure pipe
Source pipe
H1
Stream
Ram Pump
Ram animation
Ram Pump
Time to establish flow
t0.9V
12
 0.9V
f
 tanh 1 

ab
 Vf
1
f




tanh 1 (0.9)
gH1  K f 






2L L D 
velocity (m/s)
10
HA  AL
8
dV
dt
6
4
dV g
 H2
dt L
2
0
0
10
20
time (s)
30
40
Surge Tanks
Real pressure traces
At valve
At midpoint