href="#ulink_e3e5b245-83c4-5e48-9e98-edd3c55f4cb8">Table 2.4).
Table 2.3 Typical values for fluid density.
| Fluid | Density [kg/m3] |
|---|---|
| Mineral oil | 870–900 |
| Water | 1000 |
| Water/glycol | 1060 |
| Water/oil emulsion | 920–940 |
| Vegetable oil | 930 |
| Chlorinated hydrocarbons | 1400 |
| Phosphoric esters | 1150 |
| Silicon‐based fluid | 930–1030 |
Table 2.4 Fluid parameters used for the plot of Figure 2.4.
| Fluid | ρ0 [kg/m3] |
|
γ [K−1] |
|---|---|---|---|
| Mineral oil | 870 | 0.70 · 10−4 | 0.65 · 10−3 |
| HFC | 1050 | 0.30 · 10−4 | 0.70 · 10−3 |
| HFD | 1150 | 0.35 · 10−4 | 0.75 · 10−3 |
Example 2.1 Liquid compressibility
A 10 l rigid container is completely full of hydraulic fluid (density equal to 870 kg/m3; bulk modulus equal to 1700 MPa). Calculate the volume of oil (in liters, l or cubic meters, m3) that must be introduced into the container to achieve a pressure increase from 0 to 200 bar.
Given:
The volume of a container V = 10 l; the density of the fluid inside the container ρ = 870 kg/m3 and the bulk modulus B = 1700 MPa; the pressure difference for the fluid inside the container Δp = 200 bar.
Find:
The volume ΔV of fluid necessary to achieve the given Δp.
Solution:
The volume variation ΔV can be found straight from the definition of the bulk modulus:
which gives
2.6 Fluid Viscosity
Each fluid particle can experience surface forces, due to pressure or friction that are generated by contact with other particles or a solid surface. These surface forces lead to stresses. The concept of stress helps in describing how forces acting on the boundaries of a medium are transmitted throughout the medium itself. The concept of stress is very intuitive for a solid: stresses develop when the material is elastically deformed or strained. For example, when a force is applied to a solid part, such as a cantilever beam, stresses are generated within it. In contrast, if the medium is a fluid, stresses can be generated by motion rather than by deflection. In particular, shear stresses in fluids arise due to viscous flow. For this reason, while solids are elastic, fluids are viscous.
The shear effects can be visualized through the classic example of a flow between parallel plates (Figure 2.6). For a fluid at rest, there will be no shear stresses. This corresponds to the rest condition for a fluid element highlighted in Figure 2.6a. Now suppose that a rightward force dF is applied to the upper plate, dragging it across the fluid at a certain velocity. The relative shearing of the plates produces a shear stress τ, which acts on the fluid element:
(2.13)
where dA is the area of contact of the fluid element with the plate.
Figure 2.6 Shear stress in a fluid: initial rest condition (a) and (b) after a force is applied.
As one can guess from Figure 2.6, under the action of the force applied to the plate, the fluid element will continue deforming, with a rate that can be measured by the angle dα. It is possible to demonstrate that deformation rate can be expressed in terms of the velocity of the upper plate du [15]:
(2.14)
Therefore, the fluid element subjected to a shear stress τ experiences a rate of deformation (or shear rate) given by du/dy. This fact is valid for any fluid. However, the relationship between τ and du/dy varies for different types of fluids. Fluids in which the shear stress is proportional to the shear rate are called Newtonian fluids. Common fluids, such as water and air, and hydraulic fluids behave as Newtonian fluids in most conditions.
The constant of proportionality
