Hydraulic Fluid Power - Andrea Vacca

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p Subscript n plus 1 Baseline right-parenthesis Over rho EndFraction EndRoot"/>

      By equating the last two equations, it is possible to provide the expression for Ωeq, ser. This can be shown for the case of two orifices below, assuming again the same flow coefficient for all orifices:

      (4.16)StartFraction rho Over 2 EndFraction left-parenthesis StartFraction upper Q Over normal upper Omega Subscript e q comma s e r Baseline upper C Subscript f Baseline EndFraction right-parenthesis squared equals left-parenthesis p 1 minus p 3 right-parenthesis equals left-parenthesis p 1 minus p 2 right-parenthesis plus left-parenthesis p 2 minus p 3 right-parenthesis equals StartFraction rho Over 2 EndFraction left-parenthesis StartFraction upper Q Over upper C Subscript f Baseline EndFraction right-parenthesis squared dot left-parenthesis StartFraction 1 Over normal upper Omega 1 squared EndFraction plus StartFraction 1 Over normal upper Omega 2 squared EndFraction right-parenthesis

      Therefore, the equivalent orifice area is

      (4.17)normal upper Omega Subscript e q comma s e r Baseline equals StartRoot StartStartFraction 1 OverOver left-parenthesis StartFraction 1 Over normal upper Omega 1 squared EndFraction plus StartFraction 1 Over normal upper Omega 2 squared EndFraction right-parenthesis EndEndFraction EndRoot

      This equation can easily be generalized for more orifices:

      (4.18)normal upper Omega Subscript e q comma s e r Baseline equals StartRoot StartStartFraction 1 OverOver sigma-summation Underscript i Endscripts StartFraction 1 Over normal upper Omega Subscript i Superscript 2 Baseline EndFraction EndEndFraction EndRoot semicolon d Subscript e q comma s e r Baseline equals left-parenthesis StartStartFraction 1 OverOver sigma-summation Underscript i Endscripts StartFraction 1 Over d Subscript i Superscript 4 Baseline EndFraction EndEndFraction right-parenthesis Superscript one fourth

      Example 4.2 Orifice Sizing

      An emergency supply system uses an accumulator pressurized at 100 bar to extend a cylinder. This has a piston diameter D = 15 cm and sees a force of 50 kN. The connection between the cylinder and the accumulator is managed through a normally closed solenoid valve. The valve, when energized, implements a restriction equivalent to an orifice (this feature is represented in the valve symbol) with diameter of 4.4 mm and flow coefficient of 0.7. Assuming the accumulator is large enough to maintain a constant supply pressure, calculate the cylinder extension speed. Elaborate a simple system modification (without changing any of the existing components) for reducing the extension velocity to 60% of the previous value. Assume the pressure of the accumulator constant during the cylinder extension.

Schematic illustration of the ISO of the hydraulic emergency system, which includes an accumulator as energy source, an on/off valve, and a linear actuator.

      The ISO schematic of the hydraulic emergency system, which includes an accumulator as energy source, an on/off valve, and a linear actuator. The upstream pressure of the accumulator, constant, pacc = 100 bar; the orifice diameter d0 = 4.4 mm and the flow coefficient Cf = 0.7. The load on the actuator is F = 50 kN and the piston diameter, D = 15 cm.

       Find:

      The extension velocity of the piston, ModifyingAbove x With dot

      Solution:

      The system contains a solenoid valve (SV; which will be further described in Chapter 8), which when energized opens an accumulator to the piston chamber of a linear accumulator. For simplicity, the accumulator can be seen as a constant pressure source. In reality, the pressure inside the accumulator will decrease as the accumulator releases flow, as it will be better described in Chapter 9.

      When the valve is energized, the flow rate across it is defined by the orifice equation:

upper Q equals StartFraction pi dot d Subscript upper O Superscript 2 Baseline Over 4 EndFraction dot upper C Subscript f Baseline dot StartRoot StartFraction 2 left-parenthesis p Subscript a c c Baseline minus p Subscript c y l Baseline right-parenthesis Over rho EndFraction EndRoot

      The pressure inside the cylinder is given by the external force:

p Subscript c y l Baseline equals StartFraction upper F Over upper A Subscript c y l Baseline EndFraction equals StartStartFraction upper F OverOver pi StartFraction upper D squared Over 4 EndFraction EndEndFraction equals StartStartFraction 50 000 left-bracket upper N right-bracket OverOver pi StartFraction left-parenthesis 0.15 left-bracket m right-bracket right-parenthesis squared Over 4 EndFraction EndEndFraction equals 28.3 italic b a r

      Therefore, the actuator speed is

StartLayout 1st Row 1st Column ModifyingAbove x With dot equals StartFraction upper Q Over upper A EndFraction 2nd Column equals StartStartFraction StartFraction pi dot d Subscript upper O Superscript 2 Baseline Over 4 EndFraction dot upper C Subscript f Baseline dot StartRoot StartFraction 2 left-parenthesis p Subscript a c c Baseline minus p Subscript c y l Baseline right-parenthesis Over rho EndFraction EndRoot OverOver pi StartFraction upper D squared Over 4 EndFraction EndEndFraction equals left-parenthesis StartFraction 4.4 left-bracket italic m m right-bracket Over 150 left-bracket italic m m right-bracket EndFraction right-parenthesis squared dot 0.7 StartRoot StartFraction left-parenthesis 100 minus 28.3 right-parenthesis dot 10 Superscript 5 Baseline left-bracket upper N slash m squared right-bracket Over 850 left-bracket italic k g slash m cubed right-bracket EndFraction EndRoot 2nd Row 1st Column Blank 2nd Column equals 0.08 m slash s equals 8 italic c m slash s EndLayout

      In order to reduce the actuator's speed, there are possible alternatives:

      1 Increase the piston diameter.

      2 Decrease the valve size, thus reducing the valve coefficient k.

      3 Reduce the accumulator pressure.

      4 Add an orifice in series with the valve (see figure below).

      All mentioned solutions are reasonable; however, solutions 1, 2, and 3 require modifications to the existing components. Solution 4 can be a simple way to modify an existing system.

Schematic illustration of the system that contains a solenoid valve and when energized opens an accumulator to the piston chamber of a linear accumulator. upper Q equals StartFraction pi dot d Subscript e q Superscript 2 Baseline Over 4 EndFraction dot upper C Subscript f Baseline dot StartRoot StartFraction 2 left-parenthesis p Subscript a c c Baseline minus p Subscript c y l Baseline right-parenthesis Over rho EndFraction EndRoot

      deq is the diameter of the equivalent orifice given by the series connection of SV and O2. The desired speed of the actuator corresponds to the following flow:

upper Q 2 equals ModifyingAbove x With dot Subscript 2 Baseline dot StartFraction pi <p style= Скачать книгу