06 Hydraulic Transients

8/4/2019 06 Hydraulic Transients

1/59

Monroe L. Weber-ShirkSchool ofCivil and

Environmental Engineering

When the Steady-

State design fails!

Hydraulic Transients
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2/59

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


3/59

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


4/59

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.


5/59

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


6/59

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?


7/59

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


8/59

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?


9/59

________, ________

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=


10/59

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


11/59

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


12/59

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


13/59

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!


14/59

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?


15/59

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?


16/59

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?


17/59

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


18/59

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?


19/59

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


20/59

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


21/59

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


22/59

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


23/59

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


24/59

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


25/59

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


26/59

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!


27/59

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


28/59

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


29/59

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


30/59

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


31/59

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


32/59

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


33/59

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


34/59


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


35/59


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


36/59

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


37/59

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


38/59

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


39/59

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


40/59

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


41/59

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= - + +


42/59

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


43/59

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


44/59

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


45/59

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


46/59

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


47/59

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


48/59

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


49/59

Burst section of Penstock:

Oigawa Power Station, Japan

Chaudhry page 17

C ll d i f P k


50/59

Collapsed section of Penstock:

Oigawa Power Station, Japan

Chaudhry page 18


51/59

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


52/59

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


53/59

Ram Pump

Water inlet

Air Chamber

Rapid valve


54/59

Ram pump

High pressure pipe

Stream

Ram Pump

Source pipeH1

H2


55/59

Ram animation
http://schou.dk/animation/hydraram.swf


56/59

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


57/59

Surge Tanks


58/59

Real pressure traces

At valve At midpoint


59/59

Presentacin tomada de la web:

Profesor

Monroe L. Weber - Shrik

At valve At midpoint

06 Hydraulic Transients

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