Monroe L. Weber-Shirk S chool of Civil and Environmental Engineering When the Steady- State design fails! Hydraulic Transients.

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