Simple Hydraulic System2
20 Sep 2012
(Updated 21 Sep 2012)
A simple model that highlights the effects of changing various parameters in a hydraulic system.
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1) The pressure relief valve is set to maintain the pressure of the entire hydraulic circuit at 100000 Pascals.
2) The pump is set to pump at 10 rpm. The pressure created in response to the fluid flow from the pump moves the double acting cylinder. The position of the piston can be seen in the ‘position’ scope, and the pressure in the ‘pressure’ scope.
Increasing motor speed
1) Double click on pump subsystem and increase the angular velocity of the pump driving shaft from 10 to 100 rpm.
2) Notice the change in hydraulic pressure and rod position in the scopes.
3) The pressure is limited at 1e5 pascals due to the pressure relief valve.
4) The max position of the rod is 0.05m because Force = Pressure*Area of piston = 100000*5e-4 = 50 N. The displacement of the rod is then determined by x = F/k = 50/1000 = 0.05m. k is the spring constant of the spring attached to the mass.
5) Notice also how the pressure changes when the rod is moving (workdone by the rod)
Increasing pressure relief valve setting
1) Change the Valve Pressure setting to 2e5 Pa.
2) Notice now the rod is able to move to 0.1 m. Force = Pressure*PistonArea = 2e5 * 5e-4 = 100N. Displacement of rod = F/K = 100/1000 = 0.1m.
Changing Piston Area
1) Change the piston Area from 5e-4 m^2 to 5e-5 m^2 (reducing the area).
2) Notice the max rod displacement is 0.01m. Force = Pressure*PistonArea = 2e5 * 5e-5 = 10N. Displacement of rod = F/K = 10/1000 = 0.01m.
Changing spring constant
1) Change the spring constant k (double click mass subsystem) from 1000 to 500 N/m.
2) Notice the rod position increases to from 0.01 to 0.02m. Force = Pressure*PistonArea = 2e5 * 5e-5 = 10N. Displacement of rod = F/K = 10/500 = 0.02m.
3) Change Damping constant from 100 to 10.
4) Notice the transient portion of the rod position vibrates more because of lower damping value.
| Required Products
MATLAB 8.0 (R2012b)
|21 Sep 2012
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