1

Numerical Hydraulics

W. Kinzelbach withMarc Wolf andCornel Beffa

Lecture 4: Computation of pressure surges continued

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Additions

• Formation of vapour bubble

• Branching pipes

• Different closing functions

• Pumps and pressure reduction valves

• ….

• Consistent initial conditions through steady state computation of flow/pressure distribution

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Closing function• Expressed as Q=Q(t) or by degree of closure , depending from

position of valve, = f(t)• Valve closed: = 0• Valve completely open: = 1• In between: function corresponding to ratio of loss coefficients

or Q (%)

time t0

100

tclose0

open closed

0

2

00

v

v

p

p

Index 0 refers to open valve

p

p

v

v

0

0

0

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Valve as boundary condition• Valve directly in front of a downstream reservoir with pressure pB2

• Valve at Node N+1• Linear closing function = 1-t/tclose

• Determine new pressure and velocity at valve

1 11 11 0

1 2 2 20

2

2

j jN Nj

N B

v v p gp p with

g v

jN

jN

jN

jN

jN

jN vv

D

tcvvcpp

211

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From boundary condition

From forward characteristic

(1)

(2)

Inserting (1) into (2) yields quadratic equation for 11

jNv

For t < tclose:

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Valve as boundary condition

tvv

Dcpvp

p

v

p

cv

p

cvv j

NjN

jN

jNB

jN 222 2

0

20

22

0

20

2

0

20

211

jN

jN

jN

jN

jN

jN vv

D

tcvvcpp

211

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Only one of the two solutions is physically meaningful

For t > tclose:

011

jNv

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Pump

)4/(2

211

11 Dvpvv

D

tcvvcpp j

NjN

jN

jN

jN

jN

jN

Given characteristic function of pump:

)(Qfp

Pump at node i:Simply insert into characteristic equation. e.g. forward characteristic:

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Formation of vapour bubble

jN

jN

jj vvtAVolVol 111

1 5.0

If the pressure falls below the vapour pressure of the fluid (at temperature T) a vapour bubble forms, which fixes the pressure at the vapour pressure of the fluid. The bubble grows as long as the pressure in the fluid does not rise. It collapses again when the pressure increases above the vapour pressure.

Additional equation: Forward characteristic in N:

1 10 : 0j jIf Vol Vol

Volume balance of vapour bubble: volume Vol at valve

As long as the vapour bubble exists, the boundary condition v = 0 at the valve must be replaced by the pressure boundary condition p = pvapour .

If Vol becomes 0 the vapour bubble has collapsed.The velocity in the volume equation is negative, as bubble grows as long aswave moves away from valve.

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1 1( ) ( )j j j

N N N vapourv v p p Tc c

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Branching of pipe

i-1 i i+1

kk+1

Note that continuity requires that Ai-1vi-1=Akvk+Aivi

Characteristics along i -1 … k+1 and along i -1 … i+1

With different lengths of pipes the reflected waves return at different times.At the branching, partial reflection takes place. The pressure surge signalin a pipe grid therefore becomes much more complicated, but at the same time less extreme, as the interferences weaken the maximum.

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Consistent initial conditions by steady state computation of flow/pressure

In the example:

Tank 1 Tank 2connecting pipe

2 201 2

02 2 i

P PL v vv v

g D g g g

In a grid with branchings a steady state computation of the whole grid is required

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Measures against pressure surges

• Slowing down of closing process• Surge vessel (Windkessel)• Surge shaft• Special valves

air

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Surge shaft oscillations

Task: Write a program in Matlab for the calculation of the surge shaft oscillations

Simplified theory: see next page

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Surge shaft oscillations

• The following formulae can be used (approximation of rigid water column) :

Solve for Z(t), Estimate the frequency under neglection of friction.Data: l = 200 m, d1 = 1.25 m, d2 = 4 m, Q = 2 m3/s at time t = 0 local losses negligible, = 0.04, computation time from t = 0 to t = 120 s, instantaneous closing of valve at time t = 0.

1 2f

E E f

v vhdv Z lg gI with I and hldt l d g

dt

dZAvA 21

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Surge shaft oscillationsThe surge in the following surge shaft is to be calculated using the

program Hydraulic System. Vary parameters and compare!

250 m ü. M.

Further data: Closing time 1 s, area surge shaft 95 m2, roughness pressure duct k=0.00161 m, modulus of elasticity pressure duct = 30 GN/m2, modulus of elasticity pressure duct = 30 GN/m2, Loss coefficient valve 2.1 (am ->av) and 2.0 (am<- av) resp., linear closing law, cross-section valve 1.5 m

w stands for wall thickness, in the pressureduct in the rock it is assumed as 2 m effectively.