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Numerical Hydraulics . Lecture 3: Computation of pressure surges. W. Kinzelbach with Marc Wolf and Cornel Beffa. The phenomenon. Valve. Reservoir. Till time t=0: Steady state flow Q At time t = 0: Instantaneous closing of valve Observation: Sudden pressure rise at valve

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numerical hydraulics

Numerical Hydraulics

Lecture 3: Computation of pressure surges

W. Kinzelbach with

Marc Wolf and

Cornel Beffa

the phenomenon
The phenomenon

Valve

Reservoir

Till time t=0: Steady state flow Q

At time t = 0: Instantaneous closing of valve

Observation: Sudden pressure rise at valve

→pressure surge (water hammer)

and some typical damages
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
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 phenomenon6
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

Damping of

amplitude through

friction

negative pressure wave
Negative pressure wave

The negative pressure wave can

not become lower than the

vapour pressure of the fluid.

If the pressure falls below the

vapour pressure, a vapour bubble

is formed. The water column

separates from the valve.

When the pressure increases

again the bubble collapses.

This phenomenon is called

cavitation.

slide9

Computed and observed

pressure surge at two

places along a pipe

Today the reliable computation of

pressure surges is possible

measures against pressure surges11
Measures against pressure surges

Surge vessels (Windkessel)

Special valves

the equations of unsteady pipe flow
The equations of unsteady pipe flow
  • Continuity
  • Momentum equation (per unit volume) (a inclination angle of pipe)
further transformations 1
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.

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

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

further transformations 2
Further transformations (2)
  • Compressibility: Definition
  • Using pressure tank formula
the equations of unsteady pipe flow16
The equations of unsteady pipe flow
  • Continuity
  • Momentum equation

(1)

(2)

2 PDE with 2 unknown functions p(x,t) und v(x,t)

plus initial and boundary conditions

boundary conditions
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
Linearised equations
  • Delete all terms in (1) and (2) which are non-linear (for convenience: a = 0):
  • General solution by elimination:
  • Take partial derivative of first equation with respect to t
  • Take partial derivative of second equation with respect to x
  • Subtraction yields:
linearised equations19
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, r = 1000 kg/m3 yields c = 1333 m/s

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

Proof by insertion into linearised equations!!

joukowski surge21
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 Dp is given by:
  • Solution:

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

numerical solution of the complete equations
Numerical solution of the complete equations
  • Difference method
    • Discretisation of space and time
    • Dx and Dt
  • Difference equations for time step t, t+Dt
  • Problem: Discretisation „softens“ pressure front numerically
  • Way out: Method of characteristics
    • Follows the pressure signal in moving coordinate system
method of characteristics
Method of characteristics
  • 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)

Front of

pressure wave

cDt < Dx

Dx

Front of

pressure wave

cDt = Dx

Dx

method of characteristics24
Method of characteristics
  • Replacing equations (1) and (2) by 2 linear combinations

yields:

method of characteristics25
Method of characteristics
  • With total derivative along x(t)

the equations have the form:

Forward characteristic

Backward characteristic

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

difference scheme
Difference scheme

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

Node N+1

Node 1

section N

section 1

x

Upper index time step, lower index node

Chose time step such that Dx = c Dt

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

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.

difference scheme27

3

3

2

1

2

1

Difference scheme

time

Using c + v ≈ c - v ≈ c

node i communicates within time interval

Dt with node i-1 via the forward

characteristic and with node i+1via the

backward characteristic

j +1

Dt

j

Dx = c Dt

space

i-1

i

i+1

Dx

Total derivative or derivative along characteristic line

difference form of equations
Difference form of equations
  • Equations for nodes 2 to N: 2N-2 equations

forward characteristic

backward characteristic

The pressure loss term is linearised by evaluating it at the old time j

Equations can be solved for

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

method of characteristics29
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 equations

In the example:

From forward characteristic for i=N+1

From backward characteristic for i=1

simplified case for basic matlab program

forward characteristic

= 0

backward characteristic

= 0

Simplified case for basic Matlab-Program

a=0, friction neglected, equations nodes 2 to N

Solution by subtracting resp. adding the two equations

simplified case for basic matlab program31
Simplified case for basic Matlab-Program

2 equations from boundary conditions

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

From forward characteristic for i=N+1

= 0

From backward characteristic for i=1

= 0

additions
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
Example (1)

Use Program

„Hydraulic System“

Tank 2

Tank 1

connecting pipe

valve

L=500 m

D = 0.2 m, e = 0.01 m, roughness k = 0.0005 m

r = 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/s

loss coefficient valve 2, time of calculation 60 s,

number of pipe sections n = 10

example 2
Example (2)

Tank 3

L=500 m

Tank 2

Tank 1

Valve

L=500 m

D = 0.2 m, e = 0.01 m, roughness k = 0.0005 m

r = 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/s

loss coefficient of valve 2, computation time 60 s,

number of pipe sections n = 10