Hydraulic Fluid Power. Andrea Vacca

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Hydraulic Fluid Power - Andrea Vacca

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expression and the direction of the flow force for Case A and Case B.

      Solution:

      This example shows two possible cases of poppet valve design. With poppet valves, the flow exit area can be almost perfectly sealed when the valve is closed (poppet against the seat). This feature is very difficult to achieve with the spool valve geometries in Figure 3.17, due to the necessary clearance between the spool and the valve body.

       Flow force derivation – Case A

upper F Subscript f l comma steady comma upper A Baseline equals minus rho StartFraction upper Q squared Over normal upper Omega EndFraction cosine theta

      The flow force, in this case, tends to close the valve poppet.

       Flow force derivation – Case B

      Additionally, in this case, Eq. (3.43) can be applied to the CV shown in figure; however, it has to be noted that the Eq. (3.47) includes the overall contribution of the solid walls:

upper F Subscript x comma steady Baseline equals rho StartFraction upper Q squared Over normal upper Omega EndFraction cosine theta

      For this case, the pressure acting on CV at the left side (next to the entrance) is not acting on the sliding element, but on the valve body. It can be assumed that this pressure is equal to the entrance pressure p1 and uniformly distributed at the surface (i.e. small opening area Ω). Next, assuming that with the whole poppet downstream, the flow area Ω is at the pressure p2:

upper F Subscript f l comma steady comma upper B Baseline equals normal upper Delta p StartFraction pi left-parenthesis upper D squared minus d squared right-parenthesis Over 4 EndFraction minus rho StartFraction upper Q squared Over normal upper Omega EndFraction cosine theta

      For typical valve geometries, it is easy to show that the first term is generally higher than the second one so that in this case the flow force tends to open the poppet.

      This can be proven by writing the orifice equation that relates Q and Δp:

upper Q equals c Subscript f Baseline normal upper Omega StartRoot StartFraction 2 normal upper Delta p Over rho EndFraction EndRoot

      and considering that generally

StartFraction pi left-parenthesis upper D squared minus d squared right-parenthesis Over 4 EndFraction much-greater-than normal upper Omega

      1 3.1 The spool valve shown in the figure below has a maximum displacement of xv = 5 mm. The area ratio for the valve, in terms of opening area per unit travel of the spool, is w = 25 mm. Determine the maximum steady state flow force, considering the max working pressure conditions of 200 bar. Assume the jet angle to be 69° and a valve orifice coefficient of 0.7.

      2 3.2 In an agricultural spraying system, the nozzles are supplied with water through 500 ft of aluminum (roughness, e = 5 10−6 ft) tubing from an engine driven pump. In its most efficient operating range, the pump output is 1500 gpm at a discharge pressure not exceeding 65 psi. For satisfactory operation, the sprinklers must operate at 30 psi or higher pressure.Minor losses and elevation changes may be neglected. Which is the smallest pipe size that can be used between: a 4‐in., a 5‐in., or a 6‐in. internal diameter pipe?

      3 3.3 Water is pumped from a reservoir on a construction job site using the pipe system of the figure below. The flow rate must be 600 gm and water must leave the spay nozzle at 120 ft/s. The tubing piping system consists of a re‐entrant entrance, a regular 90°threaded elbow, two regular 45°threaded elbows, and a fully open gate valve (GV). The total length of the pipe is 700 ft, and the pipe has a diameter of 4 in.Calculate the total head loss of the system, including major and minor losses.Calculate the pump head required for the system.The following data can be used:Drawn tubing roughness: 5·10−6 ftRe‐entrant entrance loss coefficient: 0.890° elbow loss coefficient: 1.545° elbow loss coefficient: 0.4GV loss coefficient: 0.15

      4 3.4 A pumping system is used to fill an elevated hydraulic tank as shown in the figure below. The hose used has a 1.91‐cm (0.75 in.) diameter and 7.62‐m (25 ft) length. The internal roughness is 0.5 mm (1.97 10−2 in) and the gage pressure of the oil at the pump outlet is 379 kPa (55 psi). The oil density is 900 kg/m3 (62.4 lbm/ft3) and the viscosity of the oil is 4 10−6 m2/s (1.20 10−5 ft2/s).The minor loss coefficient due to the bends in the hose are much smaller than the minor loss coefficient due to the GV, positioned at the pump outlet. GV has a loss coefficient of 2. Determine the volumetric flow rate of the oil into the elevated tank.

      5 3.5 Oil flows through two pipes with the same diameter, length, and friction factor. The flow rate through the second pipe is twice that through the first pipe. Both flows are turbulent and fully developed. Which statement is correct about the pressure drop over the pipe length, Δp, for the two pipes?Δp2 = 0.25Δp1Δp2 = 0.5Δp1Δp2 = Δp1Δp2 = 2Δp1Δp2 =4 Δp1

      6 3.6 Oil of density 850 kg/m3 is flowing through the 180° elbow shown below. At the inlet of the elbow, the gage pressure is 1 bar. The oil discharges to the atmosphere. Assume properties are uniform over the inlet and outlet areas. Additionally, A1 = 25 cm2 and A2 = 2.5 cm2. The inlet velocity is 3.5 m/s. Find the horizontal component of the force to hold the elbow in place.

      7 3.7 Water flows from a tank that sits on frictionless wheels. The pipe has a diameter of 0.5 m. The tank is open to atmospheric (location (1)) and it is connected with a 50‐m‐long pipe to a second pipe that is 75 m long, bolted at location (3). A filter with a loss coefficient K = 8 is placed along the second pipe and the flow exits to the atmosphere at location (2). All other minor losses are negligible.Use the generalized Bernoulli's equation to derive an expression for the velocity in the pipe as a function of the friction coefficient.Determine the pressure at the location (3), assuming a friction coefficient of 0.015.Use the CV approach and the momentum equation to determine the tension of the bolts of the flange at location (3).

      8 3.8 A pump is used for a fountain jet as in the schematic shown in the figure below. The pump is connected to an electric motor that rotates at 1000 rpm. Following data for the system are given: D = 0.100 m; L = 1 m (L is the length of each branch, as shown in the figure); and relative roughness, /D = 0.02, for all pipe sections. The inlet of the pump is at the same elevation as the free surface of reservoir, as shown in the figure.Factors for minor losses are Kin = 0.5; Kbend = 1, Kconv= negligibleFor the water, assume ρ = 1000 kg/m3; μ = 1 × 10−3N s/m2Knowing that the pump characteristic curve is given by following equation:[Hp] = m, [Q] = m3/s, the units of the coefficients 10 and 625 are not written for convenience.Find the flow rate through the pump, in m3/s for Case A.Indicate the operating point of the system, in a (Q, H) plot (qualitative

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