1 numerical hydraulics w. kinzelbach with marc wolf and cornel beffa lecture 4: computation of...
Post on 18-Dec-2015
215 Views
Preview:
TRANSCRIPT
1
Numerical Hydraulics
W. Kinzelbach withMarc Wolf andCornel Beffa
Lecture 4: Computation of pressure surges continued
2
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
3
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
4
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
11
From boundary condition
From forward characteristic
(1)
(2)
Inserting (1) into (2) yields quadratic equation for 11
jNv
For t < tclose:
5
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
11
Only one of the two solutions is physically meaningful
For t > tclose:
011
jNv
6
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:
7
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.
11
1 1( ) ( )j j j
N N N vapourv v p p Tc c
8
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.
9
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
10
Measures against pressure surges
• Slowing down of closing process• Surge vessel (Windkessel)• Surge shaft• Special valves
air
11
Surge shaft oscillations
Task: Write a program in Matlab for the calculation of the surge shaft oscillations
Simplified theory: see next page
12
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
13
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.
top related