Transcript Exercise 2x
MEC-E5005 Fluid Power Dynamics L (3 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 (28.10. Evaluation week for Period I ?) 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. Modeling of a system pOUT q1IN V qOUT q2IN ”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) Phase 2 Hydraulic circuit modeling Cylinder - qA, qB, dx/dt in - pA, pB, F out Inertia mass - F - dx/dt, x F M (dV/dt) dx/dt in out qA pA pB dx/dt, x qB CONTROL U Building up a system of ”fluid volumes” and ”valves” (flow sources) Mechanics V1 V2 V3 ”pump” – ”pipe” – ”valve” – ”pipe” – ”valve” – ”actutor” Net force of a cylinder Hydraulic force F= pAAA-pBAB Net force is affected also by seal forces: - Spring force related to bending of seals - Friction force related to sliding - Seal model will be constructed later! Load model • Input: force – connected to: cylinder net force • Output: state of motion (piston velocity dx/dt and Multiplication/division by a constant position x) – Connected to: state of motion cylinder chambers’ Initial condition can be volume change included In our simulation work the load is plainly inertia mass. The state of motion can be calculated by applying Newton’s second law. F ma OR Initial value Integrating input signal twice where m = mass of an object ja a = acceleration” In our model we are interested in the state of motion of the load and piston ( piston velocity dx/dt and position x). Thus … - At first solve the acceleration of load a - Integrate the acceleration -> velocity (change in velocity) - Velocity= 0 at the start of the simulation (unless you give it another initial value) - Inegrate velocity or integrate acceleration twice to get the change in position - At the start of the simulation position = x0 -> take that into account to get the absolute position value Benchmarking 1 Cylinder model testAttention! Test values fo parameters Do not connect seal model! • Cylinder size 32/201000 AA ja AB (check piston area values in the Matlab workspace) • B= 1.6109 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 = 110-3 m/s and b) dx/dt = -110-3 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= 110-6 m3/s 2.2 connect to chamber B flow rate input qB= 110-6 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 = 110-3 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