Chapter 15

Fluid Mechanics

reminder : density $\rho$ of a given substance defined as the mass of a unit volume

$\rho~=~m/V~ kg/m^3$

Measuring $\rho$ in kg/m³, is not very convenient, since it yields fairly large numbers for ordinary materials (1 m³ is a relatively large volume), more commonly one expresses densities in g/cm³ or in terms of specific gravity, i.e. the ratio between the density of the substance and the density of water. Given that $\rho_{water}~=~1~g/cm^3$, specific gravity is numerically equal to density measured in g/cm³.

Question: isn't it strange that the density of water (at 4⁰ to be precise) is exactly 1 g/cm³? Any idea why ?

Remember that densities vary with pressure and temperature. The variation is not very big for solids and liquids, but is quite sizeable for gases.

Pressure

In the following we wil make ample use of a new quantity, the pressure, defined as the force per unit surface. If I have a solid object (e.g. a cylinder) of mass m resting on a table, and A is the area of contact, then the pressure it exerts on the table is, if h is the height of the cylinder,

P = F/A = mg/A = $\rho$ V g/A = $\rho$ghA/A = $\rho$gh

Similarly, if I have a certain amount of liquid in a container, it follows that the pressure exerted on the bottom of the container is $\rho$gh, with $\rho$ the liquid density and h its level above the bottom.

More precisely, if the pressure at a certain depth h₁ is P₁, then the pressure at a depth 2 will be

$P_{2} = P_{1}+\rho g(h_{2}-h_{1})$

So far so good, but things become less intuitive if I put some object within the volume of the liquid. Let's start by putting a flat fish like a flounder in the water, and let's assume that it is at rest; from what we just said, its upper surface must feel a pressure $\rho gh$, where h is the depth of the flounder. But, given that it is at rest, there must be an equal and opposite pressure acting on its lower surface. And the result would have been the same if I had taken a round fish: if it is at rest, the same force per unit surface must be felt by every section of its surface.

These simple arguments could be generalized and verified by accurate measurements, leading to the conclusion that a fluid exerts on all the surfaces of a submerged object a pressure which is perpendicular to the surface itself and numerically equal to $\boldmath{P_{1}+\rho gh}$, where P₁ is the pressure at the surface and h the depth of the submerged object.

A legitimate question would be: if the pressure at a given depth of a fluid is due to the weight of the column of fluid above, how can a force be exerted in any direction and not just in the vertical direction?

The answer is that, in a fluid, pressure is due to the motion of the molecules and their collisions against each other or against any submersed object or even against the walls of the container. Think of a container full of liquid : even though it is clear that the walls feel a pressure (if they weren't strong enough the container would burst), this cannot be directly due to the weight of the liquid, since the weight force is parallel to the side walls. The idea of fluid pressure being due to collisions by fluid molecules is even more clear if we think of a gas. The walls of a gas filled balloon undoubtedly are feeling a pressure, but this is obviously not due to the weight of the gas...

But sometimes the weight of a gas cannot be ignored : the column of air above us is high enough, that its weight, and its pressure, are not negligible. Can we estimate the magnitude of atmospheric pressure? The expression $P = \rho gh$, with h the height of the air column (some tens of kilometers), would not give the right answer, since the density is not constant (you all know that there is "less air" at higher altitudes).

But atmospheric pressure can easily be measured, and this was done for the first time by Evangelista Torricelli, a student of Galileo, who was asked to figure out why one could pump water out of mines only up to a height of about 10 meters. The belief of the time was that water pumps worked since they created a vacuum above the water surface, and water then rushed in to fill the vacuum (Natura abhorret vacuum). Torricelli instead realized that the agent was the atmospheric pressure, and in doing so he produced the first (mercury based) barometer. Atmospheric pressure is quite useful, without it we could neither sip a drink through a straw, nor use shaving cream....

For historical reasons, many different units are used for pressure: the correct SI unit is the pascal 1 P = 1 N/m², but other units are mm (or inches) of mercury, atmospheres, bars, pounds per square inch (psi)... 1 atmosphere = 760 mm Hg ; how much is that in pascals or psi's?

Absolute pressures can be measured with a Torricelli-like barometer, but often it is more convenient just to measure the pressure excess (or defect) with respect to the atmospheric pressure: this quantity is called gauge pressure. To measure gauge pressure one can use an open ended, mercury filled, U-shaped tube, one end of which is connected to the pressure under measurement,the other end to the open air (i.e. to atmospheric pressure). Differences in the height of the mercury column in the two branches will indicate how much the pressure to be measured differs from atmospheric.

We have seen that the pressure in a fluid at a depth h, is given by

$P = P_{0}+\rho gh$

where P₀ is the pressure at the surface. Suppose that we exert a force on a liquid (e.g. by means of a piston), this force is going to be transmitted throughout the body of the liquid, and every section of the fluid will be experiencing the same pressure increase. Formally, this effect is described by Pascal's principle:

the pressure applied at one point in one enclosed fluid is transmitted undiminished to every part of the fluid and to the walls of the container

This property of fluids is exploited extensively in hydraulic lifts, etc.

Food for thought:

even though a hydraulic press allows to lift a heavy weight with little effort, there is no violation of energy (and work) conservation. You can convince yourself of this by figuring out the total displacement of the two sides of the hydraulic lift (and you can figure this out by thinking in terms of the volume of fluid being displaced).

Archimedes' Principle

The story goes that, while floating in his bathtub, Archimedes saw the solution to a problem that his "boss", the king of Syracuse, has posed to him, i.e. how to determine whether a crown was made of pure gold without taking it apart. It is said that the thought occurred to him from noticing that the more he immersed his body into the water, the more the water level rose. We can recognize two steps into Archimedes' reasoning:

1.

how to measure the volume (and therefore the density) of an object of irregular shape? Answer: fill a container with liquid up to a certain level, fully immerse the body in it, and measure the new liquid level. The height increase times the container's cross section gives the volume of the object. Density of the material can then be found from $\rho$ = m/V.

2.

under what conditions will a body float or sink? Here the reasoning is the following : we start from the previous result, i.e. an immersed body displaces an amount of liquid equal to its volume. When the body is not in the liquid, the liquid is in equilibrium, therefore each volume of water feels a force equal and opposite to its own weight. When the body is immersed, nothing in principle changes, therefore the body will feel an upward force equal to.... the weight of the amount of water it displaces.

Applying this result, we can immediately find the correct condition for floating or sinking :

a body immersed in a fluid (which could equally well be a liquid or a gas) is subjected to its own weight, $w_{b}~=~mg~=\rho_{b}V_{b}g$, and to the Archimedean push, $w_{f}~=~\rho_{f}V_{b}g$.The condition of equilibrium is then

w_b = w_f, i.e. $\rho_{b}~=~\rho_{f}$

If the body's (average) density is less than that of the fluid the body will float, if it is more it will sink. It is also straightforward to find what portion of a floating body is submersed : if we call V_s the submersed portion for a body of total volume V, one has (can you prove it?):

$V_{s}/V~=~ \rho_{b}/\rho_{f}$ which is <1 when $\rho_{b}~<~\rho_{f}$

Fluidodynamics In general, the motion of a fluid is a rather complicated phenomenon (think of white-water rapids, or a boiling pot or the air and sewaves in a heavy storm, etc.). But in many situations we can make simplifying assumptions and obtain some general results concerning the motion of fluids. In the following we will assume that the fluid is

• incompressible, i.e. it has constant density (this is a very good assumption for liquids, but it has to be handled carefully when dealing with gases)
• non-viscous, i.e. there is no internal friction among layers
• irrotational, i.e. the flow is such that the paths of different fluid particles do not intersect each other. This type of flow is usually called laminar flow, and the non-laminar case is called turbulent.

The first result we btain when these conditions are (at least approximately) satisfied, is how the velocity of a fluid changes as it goes through a path of variable cross-section. Let's consider a fluid flowing through a conduit of cross-section A. If V is the volume of fluid that goes through a section of the conduit in the time $\Delta t$, one has

V = A x = A v $\Delta$t

where x is the distance covered by a fluid element in the time $\Delta t$.But given that the total amount of fluid is conserved, and that the fluid is not compressible, this quantity has to be the same everywhere along the fluid's path, i.e., if we take two positions of cross-section A₁ and A₂ we have

$A_{1}v_{1}\Delta t~=~A_{2}v_{2}\Delta t \longrightarrow A_{1}v_{1}~=~A_{2}v_{2}$

that shows how the speed of the flow increases as the fluid encounters a region of narrower cross section, or it decreases in wider regions. This result goes under the name of continuity equation.

The next result, i.e. an expressions that relates pressure to velocity, is somewhat less intuitive, but it has extremely far-reaching consequences and applications. To derive the result (known as Bernoulli's equation) one has to apply to the fluid's motion the work-energy theorem, stating that the change in mechanical energy of a system is given by the net work performed by non-conservative forces. If we consider a small volume of fluid at positions 1 and 2, the change in mechanical energy is given by

(1/2mv₁²+mgy₁) - (1/2mv₂²+mgy₂)

The non conservative forces doing the work are effectively represented by the pressure in the fluid, and net work will be done if there is a net pressure differential $\Delta P$ between the two faces of our fluid element. If our small volume of fluid moves through a distance s, the net work done is

$W~=~F\times s~=~\Delta P\times A\times s~=~\Delta P\times V$

After a few simple step, one gets

$P_{1}+1/2\rho v_{1}^{2}+\rho gy_{1}~=~P_{2}+1/2\rho v_{2}^{2}+\rho gy_{2}$= constant

In the general case, each of the three terms has to be taken into account. But it is useful to consider a couple of special cases :

1.

flow through a pipe of constant cross section. In this case the continuity equation ensures that v₁=v₂, therefore we get

$\Delta P~=~-\rho g\Delta y$

i.e. the same result we had already found for the static case

2.

horizontal flow, y=const. In this case we get

$P_{1}+1/2\rho v_{1}^{2}~=~P_{2}+1/2\rho v_{2}^{2}$= constant

which tells us that in a fluid moving horizontally, an increase in velocity is accompanied by a decrease in pressure.

Consequences and Applications:

• airplane lift
• atomizer
• curved balls
• roof blowing off in strong winds (Quick Quiz 15.10)
• Venturi meters (to measure fluid speed, e.g. speed of airplane w.r.t. surrounding air, or speed of fluid in a pipe)
• prairie dog tunnel ventilation (Question 15.22)
• etc.