1. MEC-E5005 Fluid Power Dynamics L (5 cr)
Weekly rehearsals, autumn 2016 ( )
Location: Maarintalo building (mostly Maari-E classroom)
Time: Fridays 10:15-13:00 (14:00) o’clock
Schedule:
Exercise, R-building
Exercise, Maarintalo
Exercise, R-building
( Evaluation week for Period I, no activities)
Lecture (K1, bulding K202)
Exercise Milestone: Cylinder model benchmarking/checking
Exercise
Exercise Milestone: Valve model benchmarking/checking
Exercise Milestone: Seal model benchmarking/checking
Exercise Milestone: Personal simulation work
Exercise Discussion (Evaluation week for Period II)
Staff
Asko Ellman, prof. (TTY)
Jyrki Kajaste, university teacher
Contact person:
Heikki Kauranne, university teacher

2. Simulation of Fluid Power
Modeling of fluid properties
Modeling of valves
Modeling of actuators
Modeling of fluid power systems

3. Simulation work Cylinder system Control 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

4. Hydraulic circuit to be modeled 1pA pB
p/U
p/U x
CONTROL U

5. Hydraulic circuit to be modeled 2pA pB
p/U
p/U x
CONTROL U
POSITION
CONTROL

6. Hydraulic circuit to be modeled 3
,  pA pB
p/U
p/U
CONTROL U
VELOCITY
AND
POSITION
CONTROL

7. 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”

8. 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= p qv (power of a hydraulic component, pump, valve etc.)
p pressure difference over a component
q flow rate through a component

9. Modeling of a system
pOUT
qvOUT V
qv1IN
qv2IN
p1IN
p2IN
”Fluid volume”: pressure is solved,
flow rates as inputs
”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)

10. Building up a system of ”fluid volumes” and ”valves” (flow sources)
”pump” – ”pipe” – ”valve” – ”pipe” – ”valve” – ”actutor”

11. 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:
Ellman & Linjama: Modeling of Fluid Power Systems a) b)
Textbook p. 18
Equation 25
Negligible changes in total volume (V0= constant)
Significant changes in total volume (V)
This equation can be applied in hydraulic cylinder calculations.

12. Cylinder – variables Variables LEAKAGE LEAKAGE xmax pB pA x dx/dt, x
Chamber pressures (time derivatives),
textbook p 75
LEAKAGE
LEAKAGE
xmax pB pA x
dx/dt, x
Variables
Input
Flow rates
Piston speed
Absolute position of piston
Output
Chamber pressures
Piston force (net pressure force) F
qvA
qvB

13. Cylinder – parameters Parameters – constants(?) LEAKAGE LEAKAGE
AB(xmax-x) volume in chamber B
xmax-x length of liquid column
xmax
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
B chamber
V0B ”dead volume” of chamber + liquid volume in pipes
AB difference of piston and piston rod areas (annulus) x
AAx volume in chamber A
x length of liquid column

14. Cylinder – liquid volumes
LEAKAGE
LEAKAGE
AB(xmax-x) volume in chamber B
xmax-x length of liquid column
xmax
Constant and changing volumes
A chamber
V0 ”dead volume” of chamber + liquid volume in pipes
AAx piston position dependent extra volume
x ”absolute position of piston”
B chamber
ABxmax B chamber maximum volume (piston at end position)
ABx liquid volume displaced by annular piston x
pipes
AAx volume in chamber A
x length of liquid column

15. Chamber A realization, example3 4 2 1
To get absolute position x for piston
integrate piston velocity to get get change in position related to start point
add start position value. 2 3 5 4 1 5
Piston leakage is not included.

16. Chamber B realization, example3 4 2 1 2 3 5 4 1
(xmax-x) length of liquid piston
AB(xmax-x) volume of liquid piston 5
Piston leakage is not included.

17. Hydraulic circuit modeling
Phase 3
Proportional valve
pA, pB, pP, pT, U
qA, qB, qP, qT in
out M
Calculate
qA, qB, qP, qT by
using control edge models pA pB qB x qA U
CONTROL U qP qT

18. Cylinder model test Benchmarking 1 Attention!
Do not connect seal model!
Test values fo parameters
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= 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
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

19. Testing of cylinder model
Use for example Display module to check the end values of signals
Attention!
Do not connect seal model!
Test 1a Test 2
p_A= ________ bar p_A= ________ bar
p_B= ________ bar p_B= ________ bar
F= __________ N F= __________ N
Test 1b Test 3
F= __________ N F= __________ N

20. 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!

21. How to include simulation plots in your document?
Hard copy
Alt + Print Screen (for example Simulink plot)
Curve plotting using plot command
Scope Parameters  History  Save data to workspace
Variable name: Signal_1
Format: Array
Simulate
In Matlab workspace
>> figure
>> plot(Signal_1(:,1),Signal_1(:,2))
Comma (,)

22. Liquid spring dx/dt F= ? ktot= ? TEST 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.
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?
dx/dt
F= ?
ktot= ? kB kA

23. 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

24. 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
The spring constant ... (V= Ax), liquid in pipes must also be included.
liquid in cylinder chambers + pipes liquid only in cylinder chambers
The volume change of a piston
A piston area
Kf bulk modulus
V liquid total volume (chamber + pipe)
x liquid column length

25. Oscillation M kB kA kA 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 M kA kB kA

26. 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
Natural angular velocity … and natural frequency
In cylinder case the sum of the cylinder chamber spring constants defines the total spring constant.

27. 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
m= %inertia mass
%Natural frequency
f=1/(2*pi)*sqrt(k_tot/m)

28. 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
Position of spool defines flow areas of orifices.

29. Turbulent orifice 𝑞 v = 𝐶 q 𝐴 0 2∆𝑝 𝜌 Textbook Starting form page 24
Flow rate
𝑞 v = 𝐶 q 𝐴 ∆𝑝 𝜌 V1
q12IN
p1OUT V2
-q12IN
p2OUT
Flow coefficient Cq
Orifice area A0
Pressure difference p
Liquid density 
Flow coefficient
Reynolds number

30. Simulink realization 𝑞 v = 𝐶 q 𝐴 0 2∆𝑝 𝜌
Useful moduls for Simulink realization OR

31. Control edge model 1 Problem!
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)
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: [ ]

32. 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: [ ]
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).
Where qv orifice= 40/60000 m3/s and p= 35105 Pa.
 How to calculate K10V

33. 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.
The initial assumption is that the pressure losses are identical pPA = pAT= 50 bar.
With zero control signal
In this leakage case
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.

34. K values for orifices from valve data sheet
K_10V=(40/60000)/sqrt(35e5); %full opening (10 V)
K_0V=(0.45/60000)/sqrt(50e5); %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.
Actuator channels A and B blocked, no flow.
All orifices (K_0V):
bar
50 bar
50 bar
0.45 l/min
0.45 l/min
0 bar
0 bar
Spool at center position: from the data sheet
Pump pressure 100 bar
Total flow below 0.9 l/min
0.9 l/min
100 bar