Hydraulic Fluid Power. Andrea Vacca

Hydraulic Fluid Power

Hydraulic Fluid Power - Andrea Vacca

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alt="upper V Subscript a i r comma r comma 1 Baseline equals upper V Subscript s a t Baseline left-parenthesis 1 minus StartFraction p 1 Over p Subscript s a t Baseline EndFraction right-parenthesis left-parenthesis StartFraction p Subscript s a t Baseline Over p 1 EndFraction right-parenthesis Superscript StartFraction 1 Over gamma EndFraction"/>

Graph depicts the dissolved and undissolved air in a hydraulic oil. The volume in the x axis is referred to the reference pressure.

      Source: Adapted from Nervegna [14].

      This expression can then be used to evaluate the volume fraction αg in the above expressions.

      As previously discussed, in the majority of cases, hydraulic components do not see instances of vapor cavitation; therefore it can be usually assumed that αv = 0. However, in certain conditions, vapor cavitation can occur. If the reader has interest in these cases, for the evaluation of the parameter αv, it is recommended to refer to more specialized literature, such as [25, 26].

      Example 2.2 Volumetric flow rate of a hydraulic pump

      In the field of hydraulics, it is very common to express flow rates as volumetric flow rates. As it will be discussed in the next chapter, there are two main reasons for that. First, it is straightforward to evaluate the motion velocity of a hydraulic actuator if the volumetric flow through it is known. Second, it is quite common to use volumetric flow meters for the measurement of the flow rate.

      This example describes what is the effect of fluid compressibility on the volumetric flow rate.

      A hydraulic pump supplies flow to a hydraulic cylinder that lifts a certain load. Due to this load, the pump outlet is pressurized at 100 bar. The inlet flow, from a tank open to atmosphere, is 100 l/min. Determine the volumetric flow exiting the pump. The density of the fluid is 870 kg/m3; bulk modulus of the fluid is 18 000 bar.

      Consider the following cases:

      1 The pump inlet pressure is the same as the tank pressure.

      2 The pump inlet pressure is −0.3 bar gauge pressure.

 Schematic illustration of the hydraulic pump with the pump outlet pressure p1, p2 and the pump inlet flow Q1, Q2.

       Given:

      The pump inlet flow, Q1 = 100 l/min; the pump outlet pressure p2 = 100 bar (gauge pressure); the Bunsen coefficient of the oil, αa = 9 % ; the density of the fluid, ρ = 870 [kg/m3]; the bulk modulus of the fluid B = 18 000 bar.

      The inlet pump pressure in two cases: (a) p1 = pT = 1 bar (absolute pressure); (b) p1 = 0.7 bar (absolute pressure).

       Find:

      The outlet pump flow rate, Q2, for the two cases (a) and (b).

      Solution

       Case (a)

      There is no pressure loss from the tank to the pump inlet. Therefore, assuming that the fluid is at saturation condition in the tank, there is no undissolved air at the pump inlet. This means that αg = 0 and the fluid is entirely liquid.

      Assuming steady‐state flow conditions, the same mass flows at the same rate at the inlet and outlet of the pump, so that

rho 1 upper Q 1 equals rho 2 upper Q 2

      Considering that

rho 2 equals StartStartFraction rho 1 OverOver 1 minus StartFraction normal upper Delta p Over upper B Subscript l Baseline EndFraction EndEndFraction

      we have

upper Q 2 equals upper Q 1 left-parenthesis 1 minus StartFraction normal upper Delta p Over upper B EndFraction right-parenthesis equals 100 left-bracket l slash italic min right-bracket dot left-parenthesis 1 minus StartFraction 100 left-bracket italic b a r right-bracket Over 18 000 left-bracket italic b a r right-bracket EndFraction right-parenthesis equals 99.44 l slash italic min

      Due to the effect of the fluid compressibility, the volumetric flow rate at the pump outlet is lower than the volumetric flow rate at the inlet by 0.56%.

      In this case, a certain amount of dissolved air is present at the pump inlet, being p1 < pT = pSAT. The amount of air at the inlet port 1 can be evaluated as

upper V Subscript a i r comma r comma 1 Baseline equals upper V Subscript s a t Baseline left-parenthesis 1 minus StartFraction p 1 Over p Subscript s a t Baseline EndFraction right-parenthesis left-parenthesis StartFraction p Subscript s a t Baseline Over p 1 EndFraction right-parenthesis Superscript StartFraction 1 Over gamma EndFraction

      where psat = pT, considering the process as isothermal (γ = 1):

upper V Subscript a i r comma r comma 1 Baseline left-parenthesis percent-sign right-parenthesis equals 0.09 left-parenthesis 1 minus StartFraction 0.7 Over 1 EndFraction right-parenthesis left-parenthesis StartFraction 1 Over 0.7 EndFraction right-parenthesis equals 0.09 dot 0.429 equals 0.039

      Therefore, the density of the fluid at the inlet section is

StartLayout 1st Row 1st Column rho 1 2nd Column equals 3rd Column alpha Subscript g Baseline rho Subscript g Baseline plus left-parenthesis 1 minus alpha Subscript g Baseline right-parenthesis rho Subscript l Baseline equals StartFraction upper V Subscript g Baseline Over left-parenthesis upper V Subscript l Baseline plus upper V Subscript g Baseline right-parenthesis EndFraction dot rho Subscript g Baseline plus left-parenthesis 1 minus StartFraction upper V Subscript g Baseline Over left-parenthesis upper V Subscript l Baseline plus upper V Subscript g Baseline right-parenthesis EndFraction right-parenthesis rho Subscript l Baseline 2nd Row 1st Column Blank 2nd Column equals 3rd Column 0.0375 dot 1.225 left-bracket italic k g slash m cubed right-bracket plus left-parenthesis 1 minus 0.0375 right-parenthesis dot 870 left-bracket italic k g slash m cubed right-bracket equals 837.4 italic k g slash m cubed EndLayout

      In the above expression, it is considered that the density of the air at standard conditions is ρg = 1.225 kg/m3.

      The density at the pump outlet can be calculated considering that at high pressure (p2 = 100 bar) all the fluid is liquid:

rho 2 equals StartStartFraction rho Subscript upper T Baseline OverOver 1 minus StartFraction normal upper Delta p Over upper B EndFraction EndEndFraction equals StartStartFraction 870 left-bracket italic k g slash m cubed right-bracket OverOver 1 minus StartFraction 100 left-bracket italic b a r right-bracket Over 18 000 left-bracket italic b a r right-bracket EndFraction <hr><noindex><a href=Скачать книгу