Numerical Hydraulics

W. Kinzelbach withMarc Wolf andCornel Beffa

Lecture 3: Computation of pressure surges

The phenomenon

Till time t=0: Steady state flow QAt time t = 0: Instantaneous closing of valveObservation: Sudden pressure rise at valve

→ pressure surge (water hammer)

ValveReservoir

And some typical damages

Sayano-Shushenskaya plant in southern Siberia

Water pipe damage due to pressure surge

San Bruno: PG&E Power Outage and Pressure Surge Preceded Blast

Where pressure surges occur…

• water distribution systems• waste water transfers• storm water rising mains• power station cooling systems• oil pipelines• RAS (activated sludge) onsite pipelines• hydropower stations• and any fluid system in which the inertia (mass

and velocity) of the fluid is significant.

The phenomenon

Pressure wave propagates with wave velocity c

If valve closing time is smaller than the run time of the wave to the reflection point and back the surge is called

Joukowski surge

Pressure vs. time at valve

50 100 150 200 250 300-1.5

-1

-0.5

0

0.5

1

1.5x 10

7 Druck am letzten Knoten gegen Zeit

Damping of amplitude throughfriction

Negative pressure wave

The negative pressure wave cannot become lower than thevapour pressure of the fluid.If the pressure falls below the vapour pressure, a vapour bubble is formed. The water columnseparates from the valve.When the pressure increasesagain the bubble collapses. This phenomenon is calledcavitation.

Pressure surge with vapour bubble formation

Computed and observed pressure surge at twoplaces along a pipe

Today the reliable computation of pressure surges is possible

Measures against pressure surges

Surge shaft

Measures against pressure surges

Surge vessels (Windkessel)

Special valves

The equations of unsteady pipe flow

• Continuity

• Momentum equation (per unit volume) ( inclination angle of

pipe)

( ) ( )0 0

( ) ( )

A Av v A Av v

t x t x x A t x

with p and A A p

sin( ) 0R

v v pv g g I

t x x

Further transformations (1)

• Continuity: As density and cross-sectional area depend on x and t only via the pressure p, the chain rule can be applied.

0

1 10

d p p dA p p vv v

dp t x A dp t x x

d dA p p vv

dp A dp t x x

Using the moduli of elasticity of water EW und of the pipe Epipe

1 1 1 / 1

'W pipe

d dA D e

dp A dp E E E

e is the pipe wall thickness, E‘ is the combined modulus of elasticity of the system

Further transformations (2)

• Compressibility: Definition

• Using pressure tank formula

/ / 1W

W

dV V d

dp dp E

1 1 /

pipe

dA dA dD D e

A dp A dD dp E

2

2 /

pipe

pipe

p xD xe

D e p

Further transformations(3)

• Momentum equation

21( )

2

2

R

R

h vI Hydraulics I

x D g

v vTaking into account the flow direction I

Dg

The equations of unsteady pipe flow

• Continuity

• Momentum equation

10

'

p p vv

E t x x

1sin( ) 0

2

v vv v pv g

t x x D

2 PDE with 2 unknown functions p(x,t) und v(x,t)plus initial and boundary conditions

(1)

(2)

Boundary conditions

• Pressure boundary condition: p given– e.g. water level in reservoir, controlled pump

• Velocity/Flux boundary condition: v given– e.g. flow controlled (v from Q/A)

• Combination: Relation between pressure and flux given– Z. B. function of pressure reduction valve, characteristic

curve of pump

• Closing of a valve at the end of a pipe– Initially flow Q, then according to closing function reduction

to zero withing closing time of the valve.

Linearised equations

• Delete all terms in (1) and (2) which are non-linear (for convenience: = 0):

10

'

p v

E t x

1

0v p

t x

General solution by elimination: - Take partial derivative of first equation

with respect to t- Take partial derivative of second

equation with respect to xSubtraction yields:

Linearised equations

• Wave equation (for p, analogously for v)

which has general solution

• Wave with wave velocity

• Example: Modulus of elasticity of steel = 200‘000 MN/m2, Modulus of elasticity of water = 2‘000 MN/m2, wall thickness e = 0.02 m, D = 1 m, = 1000 kg/m3 yields c = 1333 m/s

2 2

2 2

'0

p E p

t x

'Ec

( , ) ( )f x t f x ct

Joukowski surge

• Estimate of surge pressure after instantaneous closing of valve (neglecting friction, linearized equations): „Worst case“

• General solution:

0 ( ) ( )p p F x ct f x ct

0

1( ) ( )v v F x ct f x ct

c

Proof by insertion into linearised equations!!

Joukowski surge

• After t = 0 only the backward running wave F(x+ct) is found in the upstream

• v at the valve is 0• Maximum p is given by:

• Solution:

0 ( )p p p F x ct

0

1( )v F x ct

c

0p cv

Example continued: c=1333 m/s, Q0=1 m3/s, L=100 m yields: p=1.7E6 N/m2

Numerical solution of the complete equations

• Difference method– Discretisation of space and time– x and t

• Difference equations for time step t, t+t• Problem: Discretisation „softens“ pressure

front numerically• Way out: Method of characteristics

– Follows the pressure signal in moving coordinate system

• Normal difference method– Softening of pressure front

• Method of characteristics– Grid is adapted to frontal velocity (feasible, as

v<<c, c+v ≈ c-v ≈ c)

Method of characteristics

Front ofpressure wave

Front ofpressure wave ct = x

ct < x

x

x

Method of characteristics

• Replacing equations (1) and (2) by 2 linear combinations

yields:

1( ) ( ) sin( ) 0

2

v vv v p pv c v c g

t x c t x D

1( ) ( ) sin( ) 0

2

v vv v p pv c v c g

t x c t x D

(1) (2) (1) (2)c and c

Method of characteristics

• With total derivative along x(t)

the equations have the form:

1sin( ) 0

2

v vDv Dp dxg with v c c

Dt c Dt D dt

1sin( ) 0

2

v vDv Dp dxg with v c c

Dt c Dt D dt

Forward characteristic

Backward characteristic

D dx

Dt t dt x

(c is actually relative wave velocity with respect to average water movement.)

Difference scheme

Chose time step such that x = c t

In every time step there are two unknowns at each of the N+1 nodes:

Divide pipe of length L in N sections, length of one section x = L/N

Node 1

section 1

Node N+1

section Nx

1 1,j ji iv p

To determine these unknowns 2N+2 equations are required. From quantities at time j quantities at time j+1 are computed.The new times j+1 become the old times j of the next time step.

Upper index time step, lower index node

Difference scheme

x

t

x = c t

ii-1 i+1

j

j +13

1

2

1 2

3

1j j

i i

j

p pp

x x

1j ji i

i

p pp

t t

1 1

1 1

, ,

.j j j j

i i i i

i forward i backward

p p p pDp Dpviz

Dt t Dt t

Total derivative or derivative along characteristic line

space

time

Using c + v ≈ c - v ≈ cnode i communicates within time intervalt with node i-1 via the forward characteristic and with node i+1via thebackward characteristic

Difference form of equations

• Equations for nodes 2 to N: 2N-2 equations

1 11 11 11

sin( ) 02

j jj j j ji ii i i i

v vv v p pg

t c t D

forward characteristic

backward characteristic

1 11 11 11

sin( ) 02

j jj j j ji ii i i i

v vv v p pg

t c t D

The pressure loss term is linearised by evaluating it at the old time jEquations can be solved for 1 1,j j

i iv p

The further equations are determined by the boundary conditions and the one characteristic which can be used at the respective boundary

Method of characteristics• Example: Reservoir with pipe which is closed instantaneously at t=0• 2 further equations from boundary conditions

In the example:

• 2 further equations from characteristic equationsIn the example:

111

sin( ) 02

j jj j jN NN N N

v vv p pg

t c t D

From forward characteristic for i=N+1

From backward characteristic for i=1

12 21 2 2 11

sin( ) 02

j jj j jB

v vv v p pg

t c t D

1 11 10 ,j j

N Bv after closing of valve p reservoir pressure p

Simplified case for basic Matlab-Program

=0, friction neglected, equations nodes 2 to N

1 11 11 11

sin( ) 02

j jj j j ji ii i i i

v vv v p pg

t c t D

1 11 11 11

sin( ) 02

j jj j j ji ii i i i

v vv v p pg

t c t D

forward characteristic

backward characteristic

= 0

= 0

Solution by subtracting resp. adding the two equations

)(5.0)(5.0

)(5.0

)(5.0

11111

11111

ji

ji

ji

ji

ji

ji

ji

ji

ji

ji

ppvvcp

ppc

vvv

Simplified case for basic Matlab-Program

2 equations from boundary conditions1 11 10 ,j j

N Bv after closing of valve p reservoir pressure p

2 equations from characteristics for i = 1 and i = N+1

111

sin( ) 02

j jj j jN NN N N

v vv p pg

t c t D

From forward characteristic for i=N+1

From backward characteristic for i=1

12 21 2 2 11

sin( ) 02

j jj j jB

v vv v p pg

t c t D

= 0

= 0

jN

jN

jN cvpp

11

)(1

221

1j

Bjj pp

cvv

Additions

• Formation of vapour bubble

• Branching pipes

• Closing functions

• Pumps and pressure reduction valves

• ….

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

Example (1)

Tank 1Tank 2

connecting pipe

valve

L=500 m

D = 0.2 m, e = 0.01 m, roughness k = 0.0005 m = 1000 kg/m3, Ew = 2000 MN/m2, Epiper = 210000 MN/m2

pressure downstream reservoir 80 mWS, pressure upstream reservoir 90 mWS closing time of valve1 s, Q before closing: 0.2 m3/sloss coefficient valve 2, time of calculation 60 s,number of pipe sections n = 10

Use Program„Hydraulic System“

Example (2)

Tank 1Tank 2

ValveL=500 m

D = 0.2 m, e = 0.01 m, roughness k = 0.0005 m = 1000 kg/m3, Ew = 2000 MN/m2, Epipe = 210000 MN/m2

pressure of both downstream reservoirs 80 mWS, pressure upstream reservoir 90 mWS closing time 1 s, Q before closing of valve: 0.2 m3/sloss coefficient of valve 2, computation time 60 s,number of pipe sections n = 10

Tank 3

L=500 m