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Hydraulic Fluid Power. Andrea Vacca
Читать онлайн.Название Hydraulic Fluid Power
Год выпуска 0
isbn 9781119569107
Автор произведения Andrea Vacca
Жанр Физика
Издательство John Wiley & Sons Limited
Assuming uniform flow at both inlet and outlet sections,
(3.45)
where
Therefore, from Eq. (3.44),
The objective of the flow force analysis usually corresponds to the force acting along the direction of motion of the moving element (in this case, the horizontal axis). This is the force that the actuation mechanism has to counter‐react to balance the spool. The vertical force is usually compensated through symmetric design of the spool and housing. For example, in this case, a second exit area would be typically present in the lower end of the spool. Therefore, for the case in Figure 3.17, the attention is focused on the horizontal component of Eq. (3.46):
The term Fx corresponds to the forces acting on the CV through the CS. These include the pressure forces on the inlet and outlet sections, the pressure forces at the walls, and the frictional forces due to fluid shear. The pressure forces at both sections 1 and 2 can be considered as vertical3, and typically the effects of the fluid shear inside the valve is negligible when compared to the flow forces. For this reason, the above expression (3.47) summarizes all the forces that the walls exert on the fluid inside the CV.
The flow force Ffl can be seen as the reaction force of the above Fx, which is the force that the fluid exerts to the bounding surfaces of the CV.
The flow force presents both a stationary component and a transient component. The stationary component corresponds to a given position of the spool or the poppet of the valve and a constant flow rate. The transient component relates to variations of the spool (or poppet) position, as well as flow variations. In typical problems, the transient component is neglected.
The first term refers to transient conditions, and it can be neglected when studying a stationary condition. Therefore, when studying hydraulic control valves, the second term, is normally the most important one.
(3.49)
The negative sign in Eq. (3.48) implies that for the spool valve geometry of Figure 3.17, the flow force tends to reduce the opening area of the spool toward the outlet section 2. This is also qualitatively visualized in the details of Figure 3.17, which shows the two different pressure distributions at the two opposite spool lands: a uniform distribution at the left side, which comes from low fluid velocities, and a gradually decreasing pressure distribution at near the exit land, which results from an accelerating flow.
According to Eq. (3.48), the amount of the flow force is linked to the flow rate with a quadratic relation, meaning that flow forces can become severe at high flow rates. Additionally, the flow force depends on the exit angle θ of the flow, often referred as flow jet angle, which is usually a quantity difficult to estimate. Luckily, multiple studies on the flow jet angle for different valve geometries are available in the literature, with Merritt's textbook being a meaningful example [32]. According to this source, the angle to be used for geometries such as the one in Figure 3.17 is 69°. This value, however, can be slightly affected by the amount of opening and by the clearance between the spool and the valve body.
The following example shows how the evaluation of the direction of the flow force might not be straightforward, even for very simple geometric cases.
Example 3.3 Flow force evaluation for two different valve designs
Two different poppet valve designs implement the same flow path. The two valve designs are shown in the figures below. The Case A uses a large diameter to guide the sliding element when compared to Case B.
Determine an expression for the flow force acting on the valve poppet respectively for Case A and Case B. Assume that the valve poppet angle θ perfectly guides the flow at the valve exist section Ω.
Given:
Poppet valve geometry for two cases (figures above): flow rate Q; exit area Ω; jet force angle θ; and valve pressure drop Δp = p1 − p2