1.6 Example 5: Position control for a hydraulic actuator
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1. This means a positive flow rate indicates flow into the component Figure 1 37 shows typical results Note how different the magnitudes of the flow rates are due to the unequal areas Figure 1 37 Hydraulic actuator flow rates 1 HJO00 1 flaw rate at port 1 L min 15 2 HJQU0 1 flow rate at port 2 Lmin 1 0 2 4 B g 10 Time 2 8 Plot the valve spool fractional displacement This gives an idea of how close to saturation the valve is during the duty cycle If a value of or is reached the valve is saturated 312. the duty cycle submodel is also 10 By this means the duty cycle will directly represent the actuator displacement in m 4 When you set the parameters for H 000 click on the External variables button to call up the dialog box shown in Figure 1 34 Figure 1 34 External variables of HJ000 External Variables 71x HJOO0 Double hydraulic chamber single rod jack with no orifices at flow ports Aa ARARO shit Agee of oie ois son ATL GOS ee cesta on as ae This indicates that a positive velocity means the rod is moving to the right The greater the displacement the further it is to the right In the current case a zero displacement and velocity means that the rod and piston are stationary and the piston is at the extreme left end of the jack The meaning of the sign of the acceleration and external force should be clear A positive external force opposes the other variables i e makes a negative con tribution to the acceleration Hence it is trying to reduce the velocity and dis placement Remove the dialog box by clicking on Close Step 2 Run simulation and plot results 1 Runa simulation setting a final time of 12 s and a communication interval of 0 05 s 29 Chapter 1 Tutorial examples 2 Plot the following graphs e Actuator displacement and duty cycle output on the same graph e Flow rate at the two actuator ports on the same graph e Flow rate at pump outlet and flow rate at relief valve outlet on the same
3. Chapter 1 Tutorial examples Figure 1 32 The relief valve flow rate pressure drop characteristics 1 RVOO 1 flow rate af relief valve port 1 L min 160 140 120 100 S0 BO 40 20 E 20 40 60 60 100 120 140 160 differential pressure bar 1 6 Example 5 Position control for a hydraulic actuator Objectives e Use a simple proportional control system to achieve a prescribed cycle in a hydraulic system e Show the consequences of using an unequal area actuator e Show saturation in a servo valve e Study stability and instability in the control system 26 Hydraulic Library 4 2 User Manual Figure 1 33 The position control system Hydraulic actuator or jack a sensor EH 5B a 1 3 4 Saar F d ete a 9 k pap Lk p AHT a ne cycle in conversion block iz F Hydraulic Servo valve accumulator m OO f The system sketch for this exercise is shown in Figure 1 33 The hydraulic actuator or jack moves a load and there is control using position feedback The position sensor is used to convert the actuator displacement to a signal A position duty cycle is specified by a duty cycle submodel The duty cycle position is compared with the position indicated by the sensor to produce an error The error is subjected to a gain and the signal transferred to the servo valve A further duty cycle supplies an external force to the actuator via the position transducer Step 1 Build the system and set par
4. ameters 1 Build the new system and save it as actuator The position sensor is found in the category labeled Mechanical A signal port is used to pass the displacement into the feedback loop 2 Use the Premier submodel button to select the simplest possible submodels combinations 27 Chapter 1 Tutorial examples 3 Set parameters for the submodels using the suggested values in the following table Submodel Number on Title Value sketch if any HJ000 piston diameter mm diameter of rod mm PUOO I length of stroke m UDO0 l duration of stage 1 s output at end of stage 2 null duration of stage 2 s output at start of stage 3 null UDOO 2 output at the end of stage 1 null 1000 output at end of stage 1 null 1000 valve natural frequency Hz valve damping ratio null SVOO valve rated current mA 2 DTO000 gain for signal output 1 m GA00 3 value of gain null GA00 4 value of gain null 2 Note 3 2 3 3 0 3 0 0 x 0 l l 0 0 l 5 l l 2 2 2 0 l 0 0 0 50 28 Hydraulic Library 4 2 User Manual e The parameters of H J000 give a very unequal area actuator and the plots can demonstrate the consequences of this e The external force to the right of the actuator is a constant value of 1000 N e The gain in the displacement sensor converts the jack position that is in the range 0 to 1 m to a signal in the range 0 to 10 The gain for the submodel GA00 attached to
5. graph e Fractional spool position Figure 1 35 The required and the actual displacement 1 UDOO 2 user defined duty cycle output rull 2 HJOO0 1 rod displacement m Time z The first plot Figure 1 35 gives an idea of how closely the actual performance matches the required duty cycle 3 Plot the output from the summing junction strictly speaking a differencing junction that gives you the position error in m 4 Try changing the gain attached to the servo valve the servo valve natural frequency and damping ratio 5 Include a high gain value that makes the system unstable 6 Try introducing a dead band up to about 10 30 Hydraulic Library 4 2 User Manual Figure 1 36 Pump and relief valve flow rates 1 AYOO 1 flow rate at relief valve port 1 L min 2 PUOOT 1 flow rate at port 2 Lmin 0 2 4 E 5 Time s 10 A typical plot for the flow rates from the pump and relief valve outlets is shown in Figure 1 36 If you had chosen the pump inlet flow rate instead of the pump outlet flow rate negative values would have appeared on the graph This is easily explained if you click on the External variables button of the Variable List dialog box For both ports of the pump a positive flow rate indicates flow out of the pump It follows that the flow rate at the pump inlet must be negative 7 Plot the two flow rates in the actuator HJOOO For the this submodel flow rate is an input on both flow ports
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