Transcript Exercise 8x

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: Servo system(s) modeling and tuning
16.12. Exercise Finalizing of the reports
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
CONTINUOUS
library
Hydraulic circuit to be modeled 3
, 
p/U
pA
pB
p/U
CONTROL
U
VELOCITY
AND
POSITION
CONTROL
CONTINUOUS
library
Final report – part 1
FINAL
1. Evaluation of the realized system model.
1. Strengths
1. Application areas
2. Weaknesses
1. Submodels
2. How to improve the model and submodels
3. Parameter values
1.
2.
What parameter values are hard to specify?
What parameters are sensitive?
Maximum length 1 page
Final report – part 2
1.
FINAL
Closed loop control system tuning
1.
2.
3.
Position servo 1 (cylinder as an actuator)
Velocity servo (hydraulic motor as an actuator)
Position servo 2 (hydraulic motor as an actuator)
–
Tune the systems according to Ziegler-Nichols tuning method (P, PI and PID
control).
Specify the parameter values
Test the effect of classical ”thumb rules” for system improving
Analyze the test data and report
–
–
–
1.
2.
3.
4.
5.
6.
Overall performance
Stability
Overshoot
Steady state error
Settling time
Rise time
Only cases 1 and 2
1. Tune P (Ziegler-Nichols)
2. Tune PI (Ziegler-Nichols)
3. Tune PID (Ziegler-Nichols)
1. Change P in PID
2. Change I in PID
3. Change D in PID
Maximum length 1 page for analysis
Plots and tables can be included as attachments
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
Flow rates in a proportional control
valve
35 bar
p1
qA= qPA- qAT
A
35 bar
B
p2
Q
U
P --> A
T
qAT
qPA
P
T
Model for control edge P -> A
70 bar
The glow rates of control edges are combined (addition) to get the net flow rates
for cylinder chambers qA ja qB as well as for pump and tank flow (qP ja qT ). As
the control edges are: PA, PB, AT and BT,
qA= qPA- qAT
qB= qPB- qBT
chamber A
chamber B
qP= qPA+ qPB
qT= qAT+ qBT
pump
tank
Valve leakage parameters, tuning
Leakage parameters for (PA and PB) and (AT and BT) can have
different values!
Previously
%K parameter for leakage
K_0V=(0.45/60000)/sqrt(50e5);
New version
%K parameters ofr leakage
K_0V1=(0.45/60000)/sqrt(60e5);
K_0V2=(0.45/60000)/sqrt(40e5);
In this case the first orifice is ”tighter”
and the second one is ”wider” 
pressure in cylinder chamber becomes
lower.
K_0V1 for
PA and PB blocks
K_0V2 for
AT and BT blocks
%K value, leakage, orifices PA and PB
%K value, leakage, orifices AT and BT
The sum of pressure differences
remain the same (100 bar). Also the
total leakage (0.9 l/min) is the same.
System control
System is controlled with
proportional control valve
Valve voltage range is
-10V … +10V
Voltage U connected ON
at time t_1 (U)
In the test case
Voltage U connected
OFF at time t_2 (-U)
Instructions p. 4
At instant t = 2.923 s voltage –2 V is
connected to the valve and at instant
4.117 s valve control is set to zero.
- Voltage -2 V at t1=2.923 s
- Voltage 0 V at t2=4.117 s
Plot (measured and simulated
pressure data)
figure
plot(p_a(:,1),p_a(:,2)*1e-5,'b')
hold
plot(t_mit,pA_mit,'r')
title(’A volume’)
figure
plot(p_b(:,1),p_b(:,2)*1e-5,'b')
hold
plot(t_mit,pB_mit,'r')
title(’B volume’)
Valve Model Benchmarking
figure
plot(U(:,2),qv(:,2))
hold on
plot(U(:,2),qv(:,3),'r')
plot(U(:,2),qv(:,4),'k')
plot(U(:,2),qv(:,5),'m')
Simulation
• As a fucnction of control voltage U
• flow rates into cylinder chambers A and B
• flow rate a) from pump and b) into tank
• 20 second simulation
• Valve input -10 V -> +10 V (ramp)
Friction force [N]
Seal model for hydraulic cylinder
Velocity [m/s]
Stribeck curve
Friction in lubricated contact
I Lubricant and contacts between surface roughness peaks, very thin lubricant film
II Both boundary and hydradynamic lubrication prevail, thin lubricant film
III Lubricant film separates surfaces from each other, thick lubricant film
Friction factor
v/Fn
Relative velocity*viscosity/load
Static friction model
Pressure dependence
Substitute for Sgn function
Realization of static friction model
Parameters
Piston velocity as input
dx/dt - 𝑥
Divide by constant vs
Square
Change the sign
Exponential function
Subtraction, two constants (FS and FC )
Multiplication
Addition (FC and expression)
Signum function
Multiplication
Multiplication with a constant (b)
Addition
vs
FS
FC
b
describes the velocity realated to minimum force
static friction force
Coulomb friction force
viscous friction coefficient
Gain block -> 1/ vs
Math function -> square
Gain block -> -1
Math function -> exp
Subtract
Product
Add
Testing:
Sgn
vs
n. 0.02 m/s
Product
FS
1000 N
FC
500 N
Gain block -> b
b
1000 N/(m/s)
Add
Dynamic seal model
LuGre model
How to build the simulation model
Seal state of deformation
Integrate
Input
1
Time derivative of seal
deformation (bending rate) is
integrated and used as an input
(bend)
Stribeck
2
Input from Load model
Seal force equation
Inputs from the model above
z
0
1
b
bend of the cylinder seal [m]
spring constant of the cylinder seal [N/m]
damping constant of the cylinder seal [Ns/m]
viscous friction coefficient [Ns/m]
*
* Pressure dependent, can be be omitted, = 1
Seal parameters
Include these in your parameter file
z_max=0.1e-3;
maximum bend
sigma_0= F_s/z_max;
spring constant
sigma_1=1*sqrt(sigma_0*m);
damping constant
Including pump model
qB
qA
out
qOUT
p1IN
qS
p2IN
qT
V
qPRV
in qP
q=f(p)
T
𝑉rad
η

1
hm
𝑉rad ηvol
pOUT
Dp
P=T
q
q1IN
𝑃=
V
d𝑊
d𝑡
Tehty työ saadaan integroimalla teho
q2IN
Simulink
1
𝑊= 𝑃
𝑠
Pressure controlled pump
The dynamics of a pressure controlled pump can be described by
using a) an integrator or b) first order system (Viersma).
Laplace domain
𝐾p
𝑄=
𝑃
𝜏s + 1
p_set
Pressure setting value
Input
P value
Actual
pressure
Time constant  describes the dynamics. Pump has typically
meaningful inertia masses (related to a PRV).
Look http://www.eaton.com/ecm/groups/public/@pub/@eaton/@hyd/documents/content/pll_1439.pdf p. 13
Cylinder end forces 1
Implementation by
using Switch blocks
x
Cylinder ends can be modeled as stiff springs and dampers
Inputs
”left” end
free motion area
”right” end
x
Recommendations for parameter values
Kend – maximum force (pressure) of cylinder causes an ”appropriate” deformation
(e.g. 0.2 mm)
Damping coefficient
𝑏 = 0.5 𝐾end 𝑚
m
is the effective inertia load
xmax
Cylinder end forces 2
Constants (3)
Simulation model
- Multiplications by constants
- Subrtactions
Cylinder end forces 3
Activation of end forces
For example Switch block can be utilized.
- Top connection: threshold value is surpassed
- Middle connection: the signal to be controlled
- Lowest connection: threshold value is passed underneath
Piston in free motion zone – no force
Left cylinder end force
Piston position from Load
Right cylinder end force
End force
Parameter values for cylinder end
model
m= 234;
%load mass
%Ends
K_end= A_a*p_P/0.2e-3;
b_end= 0.5*sqrt(K_end*m);
%spring constant
%damping coefficient
Performance of pumps and motors 1
PUMP
Wilson’s model
Flow rate (output)
Pump torque (input)
 Pump angle set value (0 - 1)
Vi displacement (per revolution)
n rotational speed (1/s)
Cs laminar flow loss coefficient
Dp pressure difference over pump
 fluid kinematic viscosity
 fluid density
Cf Coulomb friction coefficient
Cv viscous friection coefficient
Tc constant torque loss
Ideal pump
Reference:
and motor Ellmann, A., Kauranne, H. Kajaste, J. & Pietola, M.
EFFECT OF PARAMETER UNCERTAINTY ON RELIABILITY OF HYDRAULIC TRANSMISSION SYSTEM SIMULATION
Proceedingsof IMECE2005 2005 ASME International Mechanical Engineering Congress and Exposition November 5-11, 2005, Orlando, Florida USA
Performance of pumps and motors 2
MOTOR
Wilson’s model
Flow rate (input)
Vi Dp
qV 2  Vi n  Cs
2
Motor torque (output)
 Motor angle set value (0 - 1)
Vi displacement (per revolution)
n rotational speed (1/s)
Cs laminar flow loss coefficient
Dp pressure difference over motor
 fluid kinematic viscosity
 fluid density
Cf Coulomb friction coefficient
Cv viscous friection coefficient
Tc constant torque loss
Vi Dp
Vi Dp
T 
 Cf
 CvVi n  Tc
2
2
Ideal
pump and
motor
Reference:
Ellmann, A., Kauranne, H. Kajaste, J. & Pietola, M.
EFFECT OF PARAMETER UNCERTAINTY ON RELIABILITY OF HYDRAULIC TRANSMISSION SYSTEM SIMULATION
Proceedingsof IMECE2005 2005 ASME International Mechanical Engineering Congress and Exposition November 5-11, 2005, Orlando, Florida USA
Motor torque
Motor flow rate submodel
Pump and load inertia
Pipe pressure
If the flows are ”IN” and
”OUT” (as in this case)
the signs must be (+) and
().
Attention!
We used another kind of
structure in ”servo
motor model” (about
two pages later)
Motor, pipes and load(s)
Very preliminary parameter values
%Pump/motor
V_rad=13.3e-6/(2*pi);
I_pump=3.3e-3+10e-3;
C_s=150e-9;
C_f=0.16;
C_v=100e3;
T_c=0.1;
nu= 32e-6;
ro=870;
V_pipe=pi/4*0.012^2*0.75;
Torque load
(constant)
Inertia load
Servo system (motor) model
This can be zero OR you could compensate the constant torque load (T_load)
Feedback
For example
2 [s]
0 [RPMs]
1000 [RPMs]
0 [s]
PID-block
(continuous library)
The
variable to
be
controlled
is
rotational
velocity in
RPMs.
Includes (2) pipes, motor
torque and motor flow rate
submodels
Motor model 2
Flows (A and B) from
valve to pipes are (+)
and flows from pipes
to valve are ().
In this case the
direction of flow
(+/) is determined
outside of the pipe
models.
See next page for pipe models!
This is flow from A to B.
Flow inputs in the
pipe models are
both plus signed ->
the flow direction
must be
determined
outside of the the
pipe models.
Pipe models 2
Both models are identical (inputs are + / +). The flow
directions must be decided outside of the pipe
models.
SAUER - DANFOSS
http://www.bibus.hu/fileadmin/editors/countries/bihun/product_data/sauer-danfoss/documents/sauerdanfoss_orbital_motors_catalogue_en.pdf
Tuning of cylinder system
• Oscillations during and after transients (start, stop)
– Oscillation frequency  depends on mass and effective bulk modulus
• Pressure levels during stationary phases
– Valve (individual control edge, PA – AT and PB – BT) dependent
• Different leakage constants for K_0V (PA – AT)
• No slipping of piston during stationary phases
– Seal friction is not adequate to keep the piston stationary with zero
valve imput
– Blocking of piston movement can be realized by zero position
adjustment (additional constant input)
• Too long stroke of piston
– The capacity of valve may not be 40 l/min @ 35 bar during test case
because of high viscosity value (higher than the nominal value)
– Test for example value 32 l/min
Zero position adjustment of
proportional control valve
Zero position adjustment of valve to keep the cylinder piston steady.
Zero position
adjustment
included
Viscosity and valve capacity
If the fluid viscosity changes also the valve capacity may alter. If the
viscosity changes from 10 cSt to 800 cSt, the pressure difference will be
almost triple. If the viscosity is doubled the pressure loss will grow with a
factor 1.189.
The valve data sheet and measurement data documents include no
information about fluid viscosity. The 40 l/min valve capacity may easily
be altered by 20% due to changes in fluid viscosity. Also minor losses in
valve manifolds and also pipe friction affect the flow.
Typically the change in pressure difference due to viscosity alteration can
be estimated by using equation
p2   2 
  
p1   1 
0.25
PLOTS
figure
plot(t_mit,x_mit)
hold on
plot(x_sim(:,1),x_sim(:,2),’r’)
figure
plot(t_mit,pA_mit,’b’)
hold on
plot(pA_sim(:,1),pA_sim(:,2)*1e-5,’r’)
figure
plot(t_mit,pB_mit,’b’)
hold on
plot(pB_sim(:,1),pB_sim(:,2) *1e-5,’r’)