Transcript Exercise 3x

MEC-E5005 Fluid Power Dynamics L (5 cr)
Weekly rehearsals, autumn 2016 (7.10.2016 -16.12.2016)
Location: Maarintalo building (mostly Maari-E classroom)
Time: Fridays 10:15-13:00 (14:00) o’clock
Schedule:
Staff
Asko Ellman, prof. (TTY)
Jyrki Kajaste, university teacher
Contact person:
Heikki Kauranne, university teacher
7.10. Exercise, R-building
14.10. Exercise, Maarintalo
21.10. Exercise, R-building
(28.10. Evaluation week for Period I, no activities)
4.11. Lecture (K1, bulding K202)
11.11. Exercise Milestone: Cylinder model benchmarking/checking
18.11. Exercise
25.11. Exercise Milestone: Valve model benchmarking/checking
2.12. Exercise Milestone: Seal model benchmarking/checking
9.12. Exercise Milestone: Personal simulation work
16.12. Exercise Discussion (Evaluation week for Period II)
Simulation of Fluid Power
•
•
•
•
Modeling of fluid properties
Modeling of valves
Modeling of actuators
Modeling of fluid power systems
Simulation work
• Cylinder system
–
–
–
–
–
Hydraulic cylinder (actuator 1)
Proportional control valve (Regel Ventil)
Load (mass)
Control system (open loop control)
Hydraulic motor (actuator 2)
• Control system
– PID control of systems
• Position control
• Velocity control
Hydraulic circuit to be modeled 1
M
p/U
pA
pB
p/U
x
CONTROL
U
Hydraulic circuit to be modeled 2
M
p/U
pA
pB
p/U
x
CONTROL
U
POSITION
CONTROL
Hydraulic circuit to be modeled 3
, 
p/U
pA
pB
p/U
CONTROL
U
VELOCITY
AND
POSITION
CONTROL
Simulation of dynamics
• Phenomena are time dependent
• Differential equations are solved
• The core of fluid power simulation is solving of
the pressure of a fluid volume (pipe, cylinder
tms.) by integration
– ”Hydraulic capacitance”
Simulation of fluid power - variables
• Essential variables in fluid power technology are
– Flow Rate qv [m3/s]
– Pressure p [Pa], [N/m2]
• The variables in question define the hydraulic
power
• P= pqv (power of a hydraulic component,
pump, valve etc.)
– p pressure difference over a component
– q flow rate through a component
Modeling of a system
pOUT
qv1IN
V
qvOUT
qv2IN
”Fluid volume”: pressure is solved,
flow rates as inputs
p1IN
p2IN
”Valve”: flow rate is solved,
pressures as inputs
• Common way to realize a model of a system is to divide it into
– Fluid volumes (pressure is essential to these volumes)
– Components between fluid volumes (”valves” ja ”pumps”, flow rate is
essential to these volumes)
Building up a system of ”fluid
volumes” and ”valves” (flow sources)
V1
V2
V3
”pump” – ”pipe” – ”valve” – ”pipe” – ”valve” – ”actutor”
Equation for pressure generation combination
• The mechanisms in fluid power which may alter the pressure in a
chamber include a) change in fluid amount b) change in volume.
• ”Equation for pressure generation” may be expressed as follows:
a)
b)
dp K f 
V 


q

v
dt V0 
t 
Negligible changes in total volume• (V0= constant)
Significant changes in total volume (V)
dp
Kf

dt V0  V
Textbook p. 18
Equation 25
Ellman & Linjama: Modeling of Fluid Power Systems
V 

 qv  t 
This equation can
be applied in
hydraulic cylinder
calculations.
Cylinder – variables
LEAKAGE
Chamber pressures (time derivatives),
textbook p 75
LEAKAGE
xmax
pA
pB
x
dx/dt, x
Variables
F
qvA
qvB
Input
- Flow rates
- Piston speed
- Absolute position of piston
Output
- Chamber pressures
- Piston force (net pressure force)
Cylinder – parameters
LEAKAGE
LEAKAGE
AB(xmax-x) volume in chamber B
• xmax-x length of liquid column
xmax
x
Parameters – constants(?)
A chamber
- Beff effective bulk modulus
- pressure, temperature, free air(!) and elasticity of walls
- V0A ”dead volume” of chamber + liquid volume in pipes
- AA piston area
AAx volume in chamber A
• x length of liquid column
B chamber
- Beff effective bulk modulus
- V0B ”dead volume” of chamber + liquid volume in pipes
- AB difference of piston and piston rod areas (annulus)
Cylinder – liquid volumes
LEAKAGE
LEAKAGE
AB(xmax-x) volume in chamber B
• xmax-x length of liquid column
xmax
Constant and changing volumes
x
pipes
A chamber
- V0 ”dead volume” of chamber + liquid volume in pipes
- AAx piston position dependent extra volume
- x ”absolute position of piston”
B chamber
V0 ”dead volume” of chamber + liquid volume in pipes
- ABxmax B chamber maximum volume (piston at end position)
- ABx liquid volume displaced by annular piston
- x ”absolute position of piston”
AAx volume in chamber A
• x length of liquid column
Chamber A realization, example
3
4
2
1
To get absolute position x
for piston
1. integrate piston
velocity to get get
change in position
related to start point
2. add start position
value.
2
3
5
4
1
5 Piston leakage is not included.
Chamber B realization, example
3
4
2
1
2
3
5
4
1
(xmax-x)
AB(xmax-x)
length of liquid piston
volume of liquid piston
5 Piston leakage is not included.
Phase 3
Hydraulic circuit modeling
Proportional valve
- pA, pB, pP, pT, U in
- qA, qB, qP, qT
out
Calculate
qA, qB, qP, qT
by
using control edge
models
M
pB
pA
qA
qB
x
U
CONTROL
U
qP
qT
Benchmarking 1
Cylinder model testAttention!
Test values fo parameters
Do not connect seal model!
•
Cylinder size 32/201000  AA ja AB (check piston area values in the Matlab workspace)
•
B= 1.6109 Pa
•
x0= 0.5 m
•
Extra liquid volumes at cylinder ends: 3.2 cm3 (piston side), 2.0 cm3 (rod side)
•
Pipes d_pipe= 0.012 m and L_pipe= 0.75 m
1. ”plug cylinder ports” and ”push/pull” piston with rod, use velocities
a) dx/dt = 110-3 m/s and b) dx/dt = -110-3 m/s
-> test pressure changes using 10 second simulations
-> test force changes using 10 second simulations
2. ”lock” the piston rod (dx/dt= 0) and
2.1 connect to chamber A flow rate input qA= 110-6 m3/s
2.2 connect to chamber B flow rate input qB= 110-6 m3/s
-> test pressure changes using 10 second simulations
-> test force changes using 10 second simulations
3. connect the following signal values to piston velocity and flow rates
dx/dt = 110-3 m/s
qA= +dx/dt  AA
qB= -dx/dt  AB
-> test pressure changes using 10 second simulations
-> test force changes using 10 second simulations
Testing of cylinder model
Use for example Display module to check the end values of signals
p_A_lopussa
q_A_testi
q_A
q_B_testi
q_A
p_A
q_B
q_B
p_B
v_testi
dx/dt
v
p_A ja p_B
F
Sylinteri
p_B_lopussa
F_paine
Testi 1a
v= 1e-3
Testi 1b
v= -1e-3
Testi 2
v= 0
Testi 3
v= 1e-3
Attention!
Do not connect seal model!
1
s
Integrator
delta_x
Test 1a Testi 1a
p_A= xxx bar
p_A= ________
p_B= yyy bar bar
p_B= ________
F= zzz kN bar
F= __________ N
F_lopussa
Testi 2
p_A= xxx bar
p_B= yyy bar
F= zzz kN
Testi 1b
Testi 3
p_A= xxx bar
p_A= xxx bar
p_B= yyy bar
Test 1b p_B= yyy bar
F=
zzz
kN
p_A= ________ bar F= zzz kN
p_B= ________ bar
F= __________ N
Test 2
p_A= ________ bar
p_B= ________ bar
F= __________ N
Test 3
p_A= ________ bar
p_B= ________ bar
F= __________ N
Cylinder model benchmarking/checking
Documenting
• Use blank Word document (or another word processor)
• Models to be documented (copying: Edit  Copy
Model To Clipboard)
– Chamber A
– Chamber B
– Cylinder model ( Force)
• Test results:
– End values after 10 second simulation time (pA, pB, F)
• This document material is usable in your personal
simulation model report!
How to include simulation plots in
your document?
1. Hard copy
1. Alt + Print Screen (for example Simulink plot)
2. Curve plotting using plot command
1. Scope Parameters  History  Save data to workspace
1. Variable name: Signal_1
2. Format: Array
3. Simulate
4. In Matlab workspace
5. >> figure
6. >> plot(Signal_1(:,1),Signal_1(:,2))
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
Comma (,)
-1
0
1
2
3
4
5
6
7
8
9
10
Liquid spring
• Liquids are incompressible
• Cylinder filled with (slightly compressible)
liquid acts as a spring (spring constant?)
• The cylinder and mass attached to the
piston rod form a spring –mass system
which has a tendency to vibrate with
nominal frequency f.
dx/dt
F= ?
ktot= ?
• TEST
– Set cylinder chamber flow rates to 0
– Small piston velocity input
– Study change in cylinder force as
function of displacement
– Evaluate cylinder spring constant (F=
kx)
– Two springs (A and B) … in series or in
parallel?
kB
kA
Liquid spring and mass
• Liquids are incompressible
• Cylinder filled with (slightly
compressible) liquid acts as
a spring (spring constant?)
• The cylinder and mass
attached to the piston rod
form a spring –mass system
which has a tendency to
vibrate with nominal
frequency f.
M
k

k
M
Cylinder – liquid spring
• Spring constant of cylinder chamber  two springs in parallel
• The cylinder chambers are assumed to be closed vessels. The flow rates
are zero. The changes in volume are also assumed small compared with
total volumes of the chambers. (V remains constant).
• The change in pressure
…
and force
Kf
p  
V
V
Kf
p  
Ax
V
F  Ap
The volume
change of a piston
A2 K f
F  
x
V
• The spring constant ... (V= Ax), liquid in pipes must also be included.
liquid in cylinder chambers + pipes
2
A Kf
k
V
A
Kf
V
x
liquid only in cylinder chambers
piston area
bulk modulus
liquid total volume (chamber + pipe)
liquid column length
AK f
k
x
•
•
•
•
•
•
•
•
•
•
•
•
Oscillation
Connect load mass m to cylinder (F= ma)
Calculate aceleration a
Calculate (integrate) acceleration to get mass velocity dx/dt
Calculate (integrate) velocity dx/dt to get mass displacement (x)
Use mass velocity as input for cylinder piston velocity
Use zero values for flow rate (cylinder inputs are plugged)
Include scope modules for signals
Connect an external force input for cylinder. Reset the force value to zero
at a certain instant.
• t= 0
• Ftot= Fhydr + Fext
• t= T
• Ftot= Fhydr
Study displacements and other signals
What is the a) period b) frequency of oscillation ?
Change sine or chirp input instead of constant force, test how the input
frequency affects the oscillation amplitude
• What is the displacement amplitude at different frequencies?
The system can also be controlled with flow rate (”valve”), reset the
external force to zero and connect sine or chirp input to one of the flow
rate inputs
• What is the displacement amplitude at different frequencies?
M
kB kA
kA
Second order dynamical system –
nominal frequency
• Hydraulic spring and mass (at small displacement amplitudes) form a
harmonic oscillator (second order dynamic system)
• In a simple harmonic motion the returning spring force is directly
proportional to distance from the equilibrium point
• In an oscillating system the potential and kinetic energy change
1 2
1 2
kx  mv
2
2
• Natural angular velocity …
k

m
In cylinder case the sum of the cylinder chamber
spring constants defines the total spring
constant.
and natural frequency
1
f 
2
k
m
Natural frequency
%Liquid volume of chamber A
V_tot_a=V_0_a+A_a*x_0 %(cylinder end +pipe) + chamber (piston position!)
%Stiffness of chamber A
k_A= Kf*A_a^2/V_tot_a
%Liquid volume of chamber B
V_tot_b=V_0_b+A_b*(L-x_0) %(cylinder end +pipe) + chamber (piston position!)
%Stiffness of chamber B
k_B= Kf*A_b^2/V_tot_b
%Total stiffness
k_tot=k_A+k_B
%Inertia mass
2
m=234
%inertia mass
f
%Natural frequency
f=1/(2*pi)*sqrt(k_tot/m)
AK
k
V
Directional control valve
Spool valve
Proportional control valve
Tesxtbook
p. 58, Fig. 56
The model for proportional control
valve can be constructed based on
an assumption : the valve consists
of four (4) orificices (control edges).
PA - PB - AT - BT
A
Position of spool defines
flow areas of orifices.
T
B
P
T
Turbulent orifice
Textbook
Starting form page 24
Flow rate
𝑞v = 𝐶q 𝐴0
2∆𝑝
𝜌
V1
q12IN
p1OUT
p2OUT
-
Flow coefficient Cq
Orifice area A0
Pressure difference p
Liquid density 
-q12IN
V2
Flow coefficient
Reynolds number
Simulink realization
𝑞v = 𝐶q 𝐴0
2∆𝑝
𝜌
Useful moduls for Simulink
realization
OR
Control edge model 1
Problem!
1
p1
2
p2
Paine-ero
|u|
sqrt
Abs
Math
Function
paine-ero
1
Q
Sign
ohjausjännite
Product1
3
U
K
PA ja BT
We really do not know the exact
sizes of flow paths in the valve
(unless we measure them).
Model for an individual control edge and an
alternative Lookup Table for PB and AT control
edges (K curve)
K
PB ja AT
With Abs and Sign modules you take care that
the input value for Sqare Root function (Sqrt) is
always positive.
Try for example
Ramp
simulation 20 s, slope 1, initial value -10
Table data:
[0 1 10]
Breakpoints 1:
[-10 0 10]
Control edge model 2 – parameter K10V
Inputs for control edge model are pressures (p1 ja p1) and control voltage (U). Flow rate (q) is the
system output. The valve capacity (”K characteristics”) as a function of voltage can be modeled
for instance with 1-D Lookup Table. In a symmetrical valve the characteristics of PA and PB as
well as AT and BT are similar but mirros images (see figure below).
The parameters for 1-D Lookup Table can be obtained from valve’s datasheet. To form the
characteristics only three points are needed.
Input-vector: [-10 0 10]
Output-vector:
[0 K_0V K_10V] (control edges PA and BT).
[K_10V K_0V 0] (control edges PB and AT)
With full opening (U= 10 V or U= -10 V) the flow rate is 40 l/min
as the pressure difference is 35 bar (for valve in the modeling case).
1
p1
2
p2
Paine-ero
|u|
sqrt
Abs
Math
Function
1
ohjausjännite
Product1
U
K10V 
qv orifice
p
p1  p2  K10V p
 How to calculate K10V
Where qv orifice= 40/60000 m3/s and p= 35105 Pa.
Q
Sign
3
qv orifice  K10V
paine-ero
K
PA ja BT
K
PB ja AT
Control edge model 3 - leakage
Valve lekage at voltage 0 V and parameter K0V
In our case study the leakage (max.) through valve is 0.9 l/min at pressure difference 100 bar. Then the
flows P -> A -> T and P -> B -> T are 0.45 l/min which is in SI system 0.45/60000 m3/s. The leakage can
be assumed to divide into two similar flow paths (PAT and PBT).
We examine flow path P -> A -> T. The orifices are in series so orifices PA and AT share the same flow
rate q and total pressure difference (100 bar) is the sum of individual pressure difference.
pkok  pPA  p AT
The initial assumption is that the pressure losses are identical pPA = pAT= 50 bar.
With zero control signal
In this leakage case
qv orifice  K 0V
K 0V 
p1  p2  K 0V p
qv orifice
p
Qv orifice= 0.45/60000 m3/s and p= 50105 Pa. We can start with these parameters and tune the values
based on measurement results. This means for example that PA ja AT leakage orifices can have different
parameter values.
K values for orifices from valve data
sheet
K_10V=(40/60000)/sqrt(35e5);
K_0V=(0.45/60000)/sqrt(50e5);
%full opening (10 V)
%leakage at ”closed”position (0 V)
About the leakage
It can be assumed that all of the four orifices are similar as the valve is in central
(”closed”) position. However the valve is of spool type and there is some leakage.
All orifices (K_0V):
0.45 l/min @ 50 bar
50 bar
50 bar B
A
0.45 l/min 0.45 l/min
0 bar
0 bar
T
Actuator channels A
and B blocked, no flow.
0.9 l/min
P
T
100 bar
Spool at center position: from
the data sheet
• Pump pressure 100 bar
• Total flow below 0.9 l/min