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: 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) Staff Asko Ellman, prof. (TTY) Jyrki Kajaste, university teacher Contact person: Heikki Kauranne, university teacher

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

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

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

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

Hydraulic circuit to be modeled 3 ,  pA pB p/U 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= p qv (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 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)

Building up a system of ”fluid volumes” and ”valves” (flow sources) ”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: 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.

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

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

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

Chamber A realization, example 3 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.

Chamber B realization, example 3 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.

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

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= 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 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

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

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? 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 (,)

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

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

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

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

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.

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=234 %inertia mass %Natural frequency f=1/(2*pi)*sqrt(k_tot/m)

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.

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

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

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

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.

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): 0.45 l/min @ 50 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