The hydrostatic equilibrium


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1 Chapter 10 The hydrostatic equilibrium 10.1 The force on the infinitesimal parcel Now we will compute the total force acting on an infinitesimal parcel of fluid at rest. Consider a rectangular parallelepiped parcel with sides δx, δy and δz (Fig. 10.1). Let us denote by ρf p δxδyδz the component of the force due to the pressure along the faces of the parcel. Consider what occurs along the x direction. The x component of this force ρf p,x δxδyδz is given by the sum of the forces on the two surfaces normal to the x direction ρf p,x δxδyδz = p (x δx2 ), y, z, t δyδz p (x + δx2 ), y, z, t δyδz = = 1 [ p (x δx2 ) δx, y, z, t p (x + δx2 )], y, z, t δxδyδz. But the first two factors in the last member define the opposite of the derivative of the pressure with respect to x, so that this expression can be rewritten as ρf p,x δxδyδz = x δxδyδz, 63
2 64 Franco Mattioli (University of Bologna)..... (x, y, z) i p(x δx/2, y, z). δx δy δz i p(x + δx/2, y, z) Fig. 10.1: The forces in the x and y directions depend only on the difference in pressure between the two opposite faces of the parcel in the corresponding direction. The vertical component also depends on the weight of the parcel. that is, ρf p,x = x. The negative sign means that if the pressure increases with x, then the parcel is subject to a force directed toward the negative xaxis. Analogously, one obtains and ρf p,y = y ρf p,z = z. However, as far as the zdirection is concerned, also the weight (9.2) of the parcel due to the gravity field must be taken into account. The addition of this contribution to the pressure term yields the total vertical force ρ(f p,z + F g,z ) = z ρg. The total force F can thus be expressed in vector form as F = p ρ + g. (10.1)
3 Principles of Fluid Dynamics ( 65 Hence, the force on the parcel does not depend on the absolute value of the pressure, but only on its spatial variations. This means that the motion of a fluid does not change if it is uniformly compressed or decompressed everywhere. The use of the geopotential Φ in place of the gravity acceleration is often preferred, so that F = p ρ Φ. On the other hand, once the specific volume α = 1 ρ is introduced, (10.1) can also be written as F = α p Φ The hydrostatic equilibrium If the fluid is at rest, then the forces acting on it must balance and (10.1) yields p + ρg = 0. (10.2) If gravity does not change its direction, then the first two components of this equation are x = 0, y = 0. These expressions state that in a motionless fluid the pressure does not vary in the horizontal directions. In other words, the surfaces at constant pressure are horizontal planes. The vertical component of (10.2) takes the form z = ρg, (10.3) and is called hydrostatic equation. This equation permits us to evaluate the vertical behavior of the pressure, once the vertical behavior of the density is known.
4 66 Franco Mattioli (University of Bologna) Both fields, in fact, depend only on the vertical coordinate. By differentiating this equation with respect to the horizontal variables and applying the previous identities, it results that the horizontal derivatives of the density also vanish. Hence, the surfaces of constant density are also horizontal planes. Problem 10.1 Evaluate the vertical force exerted by the pressure stresses on an infinitesimal parallelepiped particle made of a material different from the fluid in which it is immersed. Solution. Since the lateral sides of the particle are subjected to stresses cancelling each other, the force is related to the difference between the pressure at the bottom and the pressure at the top of the parcel δxδyδz = ρgδ = g δm, z where δm is the mass of the parcel of fluid moved by the particle. Thus, the particle is subjected to a total force equal to the weight of the displaced fluid. If the density of the particle is different from the density of the fluid, a static equilibrium is possible neither for the particle, nor for the fluid. Comment. This is the enunciation of the Archimedes law for an infinitesimal particle. Problem 10.2 Two vessels containing water are placed in such a way that the height of their free surface is different. If we connect the bottoms of the two vessels by means of a Ushaped tube, is the static equilibrium of the two vessel preserved? Solution. The two free surfaces are in contact with air, and therefore they are a the same atmospheric pressure, but they are not on the same geopotential surface. A motion will develop until the two free surfaces will be at the same level (one of the two surfaces might be situated inside the tube). Problem 10.3 Two vessels containing water have their free surfaces at a different level. If we connect the free surfaces by means of a tube filled with water with the shape of an inverted U, is the static equilibrium of the two vessel preserved? Solution. As in the previous problem, the equilibrium is not possible, because the free surfaces do not lie on the same geopotential surface. A motion will develop until the two free surfaces are at the same level, or when the upper vessel is completely emptied. Comment. This is the principle of the siphon, used to decant a liquid from a vessel to another. The situation just described can be expressed by saying that a fluid stratifies along the vertical coordinate. Each level thus is characterized by the same magnitude of the various quantities.
5 Principles of Fluid Dynamics ( Pressure in a stratified fluid In general, solving (10.3), the pressure can be calculated as p = η z gρ(z)dz + p 0, (10.4) where p 0 is the pressure at a given level. The main difference between liquids and gases is that as the pressure tends to zero in a liquid the density tends to a finite value, while in a gas it tends to zero. Furthermore, over a liquid we always have the vapor of the gas, and possibly still other gases. In a column of gas of infinite height it is possible to evaluate the pressure at a certain level z p = gρ(z) dz. (10.5) z The only realistic application of this equation is provided by a motionless atmosphere. In this case for z = 0 the atmospheric pressure at sea level can be recovered as p a = 0 gρ(z) dz. (10.6) When the atmospheric pressure p a is constant, the surface of a homogeneous liquid at rest must be horizontal. The integration of (10.3) is immediate p = ρgz + p a, (10.7) where the origin of the vertical coordinate has been placed over the free surface of the liquid. In the case in which the free surface of a liquid is not horizontal, but is described by the equation z = η(x, y) we have p = ρg(η z) + p a. (10.8) It follows that the horizontal gradient of the pressure within the fluid is H p = ρg H η, where the operator H is the horizontal component of the operator. Thus, a static equilibrium is not possible, unless η = 0.
6 68 Franco Mattioli (University of Bologna) It is not difficult to extend these results to a fluid given by the superposition of a certain number of homogeneous fluids. In particular, all the interfaces between two adjacent layers, where a discontinuity of the density is present, must be horizontal in order to have a static solution. If instead an incompressible liquid is continuously stratified, then the pressure can be calculated as η p = gρ(z)dz + p a, (10.9) provided the vertical stratification of the density is known. z 10.4 Archimedes law Consider a closed surface S drawn inside the fluid. The total force P on the volume enclosed by it, according to (9.3), is equal to P = S pn ds. Let us multiply this expression by the unit vector i, which can therefore pass under the integral symbol, and apply the Gauss Theorem (G.1) i P = pi n ds = (pi) d = S i d = i x x d Repeating the same steps with the other two unit vectors j and k we obtain the three components of the following equation P = p d (see also problem [G.5]). An application of (10.2) leads to P = ρg d = Mg k, where M is the mass of the fluid contained in the volume. Hence, along the considered surface a force develops equal to the opposite of the weight of the fluid contained in it in static conditions. If the same volume is occupied by a body of different weight, it will either move upward if the pressure force is greater, or downward if it is lower. This is the proof of the wellknown Archimedes law fore bodies of finite extension.
7 Principles of Fluid Dynamics ( 69 This law, in the form S pn ds = ρg d, can be seen as an integral version of the hydrostatic equation, valid for masses of finite volume. It can be assumed as the starting point to derive the hydrostatic equation (10.2) by applying the various steps seen in this section in the reverse order for an infinitesimal volume. Thus, the starting point for the idrostatic balance, can be that the pressure force must equal the weight of the included fluid Historical notes and essential bibliography Since antiquity men surmised that air had a mass. In the fourth century BC Aristotle carried out an experiment aimed at an evaluation of its weight. In fact, he weighed a leather bag before full of air and then empty. He did not detect any difference, so that he concluded that air had no weight. This experiment not only failed for the inaccuracy of the used balances, but especially for the absolute unawareness of Archimedes law. Only about two thousand years later Galileo was able to show in a scientific way that air has indeed weight, by means of an experiment in which air was compressed [20]. The study of liquids, whose mass can be easily measured, was strongly developed in the third century BC with Archimedes, who established his famous law (named principle, because at that time it was not possible to derive it from other more basic principles). Even in this case Galileo returned to the subject in 1586 with one of his first works, which led to the construction in 1608 of an efficient hydrostatic balance (la bilancetta, i.e., the little balance), with which it was possible to easily measure the density of a body. In 1586 Stevin [50] published a work in which it was shown that the pressure at the bottom of a vessel depends on the height of the liquid contained in it, and not on the shape of the vessel. At that time this result was known as the hydrodynamic paradox [51]. In 1664, Pascal [39] realized that the pressure in a fluid did not depend on the horizontal coordinates. Later, he formulated the law that took his name.
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