γ is the polytropic constant2. By differentiating the expression of Eq. (2.28),
(2.29)
from which, by applying the general definition for the bulk modulus (Eq. (2.4)),
(2.30)
In order to evaluate the equivalent density, viscosity, and bulk modulus for the fluid mixture, by using the expressions presented in the previous paragraphs, it is necessary to first know the amount on undissolved gases. The simpler method for doing such estimation is based on the equilibrium assumption, according to which the gas and the vapor are released or dissolved instantaneously. In reality, these processes are not instantaneous, as they are characterized by time dynamics on the order of tens of milliseconds [25, 26]. It is documented that the air release or vaporization processes happen more rapidly than the opposite dissolving processes [23].
According to the equilibrium assumption, the amount of undissolved air or vapor can be evaluated according to the Henry–Dalton law (Eq. (2.22)), with the graphical representation of Figure 2.13. The saturation line represents the equilibrium states according to Eq. (2.22). For example, considering a hydraulic fluid with 9% of dissolved air at atmospheric pressure (po ≅ 1 bar, state 0), if the pressure is brought to a lower pressure (state 1), the amount of air that can be dissolved is reduced (Vair, d, o < Vair, d, 1), meaning that a certain amount of air, Vair, r, 1, is released:
(2.31)
In the process that brings the Vair, r, 1 from the saturation conditions (psat = p0) to the actual pressure p1, the calculated volume is subject to an expansion that can be approximated as polytropic; therefore the final expression that can be used to estimate the volume of undissolved air is
(2.32)
Figure 2.13 Graphical representation of 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.
The Bunsen coefficient for the oil is 9%.
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
Considering
