Hydraulic Fluid Power. Andrea Vacca

Hydraulic Fluid Power

Hydraulic Fluid Power - Andrea Vacca

Скачать книгу
Over 4 EndFraction equals 0.6 dot 0.08 left-bracket m slash s right-bracket dot StartFraction pi dot left-parenthesis 0.15 right-parenthesis squared left-bracket m squared right-bracket Over 4 EndFraction dot 60 000 equals 50.9 l slash min"/>

      The equivalent series orifice diameter results:

d Subscript e q Baseline equals NestedStartRoot StartFraction 4 dot upper Q 2 Over pi 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 EndFraction NestedEndRoot equals NestedStartRoot StartFraction 4 dot 50.9 slash 60 000 left-bracket m cubed slash s right-bracket Over pi dot 0.7 dot StartRoot StartFraction 2 dot 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 EndFraction NestedEndRoot equals 3.44 italic m m

      Assuming that both orifices have the same flow coefficient (Cf = 0.7), the diameter of orifice O2 results:

normal d Subscript upper O Baseline 2 Baseline equals left-parenthesis StartStartFraction 1 OverOver StartFraction 1 Over normal d Subscript e q Superscript 4 Baseline EndFraction minus StartFraction 1 Over normal d Subscript upper O Superscript 4 Baseline EndFraction EndEndFraction right-parenthesis Superscript 1 slash 4 Baseline equals left-parenthesis StartStartFraction 1 OverOver StartFraction 1 Over left-parenthesis 3.4 left-bracket italic m m right-bracket right-parenthesis Superscript 4 Baseline EndFraction minus StartFraction 1 Over left-parenthesis 4.4 left-bracket italic m m right-bracket right-parenthesis Superscript 4 Baseline EndFraction EndEndFraction right-parenthesis Superscript 1 slash 4 Baseline equals 3.86 italic m m

      Even though the orifice element is described by a single equation (Eq. (4.5)), an orifice can assume different roles in a hydraulic circuit. A first way to classify an orifice function is based on its location in the system: in fact, orifices can be present either in the working and return lines or on the pilot lines of the systems. The working and return lines are represented by the connections to/from the actuators of the system; the power transfer functions achieved by the system occur in these lines. Thus, these lines are usually characterized by significant values of flow rate and pressure. Pilot lines are instead used to transmit pressure information to different locations of the system. These latter lines usually have negligible flow rates and are used for control purposes. According to ISO1219‐1 [1], pilot lines are always indicated with dashed lines, while working and return lines with a solid line.

      4.5.1 Orifices in Pressure and Return Lines

      When an orifice is used in the working or the return line of a system, it can operate as metering or compensating element.

      This classification is important to understand several control strategies used in hydraulic systems, as it will be shown in Part II and Part III of the book. A significant example is now provided to clarify the distinction between these different behaviors of an orifice.

      Example 4.3 Orifice as a Metering Element or a Compensator

      The system in figure consists of a fixed displacement pump and a variable orifice in parallel with a pressure relief valve. The pressure relief valve limits the maximum pressure at the pump outlet to p*. The pump delivers a fixed flow rate QP independently of the pump outlet pressure. Find the flow rate through the orifice, QO, as well as the pump outlet pressure pP, as a function of the orifice area opening, Ω. Describe also the function of the orifice, which can either be metering or compensator.

"Schematic illustration of a fixed displacement pump and a variable orifice in parallel with a pressure relief valve."

       Given:

      The pump flow rate, QP; the setting of the relief valve p*.

       Find:

      1 The flow rate through the orifice, QO

      2 The pressure at pump outlet, pP

      3 The orifice function (metering/compensator)

      Solution:

      Point P is located at the junction of the three main elements of the system (the pump, the relief valve, the variable orifice). Therefore, the operating pressure at point P can be found by intersecting the characteristics curves of these components, while also satisfying the constraints of maximum allowed pressure and available flow.

      To clarify this statement, the characteristic curves of the three components in the (Δp, Q) chart are shown in the figure below. In particular,

       the orifice characteristic curves are plotted according to the orifice equation 4.5 for decreasing values of the area Ω (Ω1 > Ω2 > Ω3…). This trend is as also shown in the plot of Figure 4.4. It is important to observe that, in this case, the pressure drop across the orifice equals pP, since the pressure downstream the orifice is pT = 0 bar.

       the pump curve represented by a horizontal line (constant flow rate). In fact, for this problem the pump provides a constant flow independent on the system pressure.

       the relief valve curve, represented by a vertical line. The relief valve, which will be explained more in detail in Chapter 8, limits the maximum pressure at the junction point P to p*.

 Graph depicts the intersection between the pump characteristic and the orifice curve is at X. When a pressure drop for a given flow rate, the orifice O behaves as a compensator. When a flow rate as a consequence of a given pressure drop, the orifice behaves as metering.

      The behavior of the system can be analyzed for different openings of the variable orifice O.

      In case of a large orifice area (Ω = Ω1), the intersection between the pump characteristic and the orifice curve is at X1. This point is located at a pressure lower than p*: the entire pump flow rates QP goes to the orifice (QP = QO); and the relief valve is closed (QRV = 0). In this case, the orifice Eq. (4.5) can be used to find the pressure at the point P:

p Subscript upper P Baseline equals p 1 equals StartFraction rho Over 2 EndFraction dot left-parenthesis StartFraction upper Q Subscript upper P Baseline Over upper C Subscript f Baseline normal upper Omega 1 EndFraction right-parenthesis squared

      In this condition, the orifice O behaves as a compensator, since it establishes a pressure

Скачать книгу