PROPERTIES OF FLUIDS AND THEIR UNITS
2.2 COMPRESSIBILITY OF LIQUIDS
2.2.1 Example: Bulk Modulus of Multiple Containers
Consider a system consisting of a pump, a valve, a flexible hose, and a cylin-the system bulk modulus accounting for entrained air and cylin-the compliance
Calculate the volumes of the hose and the cylinder:
Hose volume =π0.52
4 × 60 = 11.78 in.3 Cylinder volume =π22
4 × 36 = 113.1 in.3 Total fluid volume = 10 + 11.38 + 113.1 = 134.9 in.3 of the containing volumes.
der. The salient properties of the system are shown inTable 2.2. Calculate
18 PROPERTIES OF FLUIDS
Table 2.2
: Properties of a system using different container characteristicsCharacteristic Size Units
Pump and valve volume 10.0 in.3
Pump wall material linear modulus
30E+6 lbf/in.2
Flexible hose length 60.0 in.
Flexible hose inner diameter 0.5 in.
Flexible hose bulk modulus [4]
7000 lbf/in.2
Cylinder length 36.0 in.
Cylinder inner diameter 2.0 in.
Cylinder wall thickness 0.075 in.
Cylinder wall material linear modulus
30E+6 lbf/in.2 Uncontaminated oil bulk
modulus
0.26E+6 lbf/in.2 Undissolved air volume in
reservoir at atmospheric pressure
5.0 %
System operating pressure 1000 lbf/in.2gauge Atmospheric pressure 14.7 lbf/in.2abs.
Calculate bulk modulus for each of the compliant container volumes:
Pump bulk modulus, Equation 2.13 = 30E+6
2.5 = 12E+6 lbf/in.2
Cylinder bulk modulus, Equation 2.14 = 0.075 × 30E+6 2
= 1.125E+6 lbf/in.2
The oil with the undissolved air must also have its bulk modulus eval-uated at a working pressure of 1000 lbf/in.2 The following calculation as-sumes that the temperature of the oil in the reservoir is equal to the tem-perature in the system. Although this is unlikely to be exactly true, the
approximation will be adequate in most instances:
Total volume = 0.9463 + 0.0007244
= 0.947
Use Equation 2.9 to calculate the bulk modulus of air when compressed to 1000 lbf/in.2:
βA= γAp = 1.4 (14.7 + 1000)
= 1421 lbf/in.2
Noting that the pressure in this expression is an absolute pressure.
Now use Equation 2.6 to calculate the effective bulk modulus of the oil contaminated with air:
Although we shall only need to use the reciprocal of the bulk modulus, we will calculate the bulk modulus of the liquid alone under working conditions:
βOIL= 0.2282E+6 lbf/in.2
We shall now use Equation 2.12 to calculate the effective bulk modulus of the oil in the system accounting for the compliancies of the various
20 PROPERTIES OF FLUIDS containing volumes and the air in the oil:
1
So the effective bulk modulus of the oil in the system is:
βe= 56.8E+3 lbf/in.2
Before leaving this example, it should be noted that calculations of this nature can seldom be performed with such apparent accuracy because the input information will not be known very accurately. The example should be used to gain some appreciation of the relative magnitude of compliance effects. Inspect the terms in the specific application of Equation 2.12 to find the terms with the largest values. Because the equation has the bulk moduli in reciprocal form, it is these relatively large terms that will dominate the final inversion. In this example, the flexible hose term at 12.47E−6 is easily the largest term and the air contaminated oil term follows at 4.382E−6.
Although a thin walled cylinder was chosen, its effect is less than one tenth of the hose term. In summary, using flexible hoses has been the main factor in reducing the bulk modulus of the uncontaminated oil from 260E+3 to a system value of 56.8E+3 lbf/in.2 For many purposes, such a reduction may be of no significance. On the other hand, any system where precise positional or velocity control is desired will need attention to maintaining high values of effective bulk modulus for the system.
In summary, bulk modulus of a typical hydrocarbon based fluid with no entrapped air is about 270, 000 lbf/in.2 (or 1860 MPa). Entrapped air, however, has a significant influence upon its value and typically the effective value in a working system may be 200, 000 lbf/in.2 (or 1380 MPa) at best and often much less if flexible hoses are present.
The oil spring:
Another effect of bulk modulus is that any constrained oil volume behaves like a spring. Figure 2.6shows a double acting cylinder.There is oil on both sides of the piston. For generality, the two sides of the cylinder are not matched for piston area. Consider the situation where the oil flow to or from the cylinder is blocked so volumes V1and V2are trapped.
The effective bulk modulus of the trapped oil accounting for entrapped air and hose and cylinder compliance is βe. Suppose a incremental force ∆F is applied to the piston and the rod moves ∆x relative to the cylinder. The pressure in V1 will fall by ∆p1 and the pressure in V2 will rise by ∆p2. If this were a conventional metallic spring, we would write the expression for stiffness as:
k = ∆F
∆x (2.15)
Using the characteristics of the cylinder, this expression can be written:
k = −∆p1A1
∆V1/A1
+ ∆p2A2
−∆V2/A2
The sign convention implies that an increase in pressure is positive and that an increase in fluid volume is likewise positive. Now recast Equation 2.5 in the form:
−∆p
∆V = β V
Figure 2.6
: Trapped oil in a cylinder treated as a spring.22 PROPERTIES OF FLUIDS Hence the spring rate expression can be written:
k = βeA21
We now need to investigate the variation of stiffness as the relative values of V1and V2change. This is most easily done by introducing a total oil volume VT that does not change as the piston moves. The stiffness expression will then have the form:
This expression may be plotted or the rules of calculus applied to show that the stiffness has a minimum value when:
A21
V1 = A22 VT − V1
=A22 V2
Now most cylinders that will be used for precise positioning of loads will use a double, symmetrical rod configuration. In the usual situation where the pipe connections from the valve have equal volumes, the minimum stiffness condition occurs when the piston is in the center of its travel.
The significance of the oil spring stiffness will be seen later when dy-namic models of systems are discussed.