Theory

Fluid mechanics

How density and pressure describe a fluid, how Stevin's law, Pascal's principle and Archimedes' principle explain liquids at rest and flotation, and how the continuity equation and Bernoulli's theorem describe fluids in motion.

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Density and pressure
Stevin’s law

Density and pressure

1. Fluid

A fluid is a system that has no shape of its own and that, at rest, cannot withstand shear stresses: a force tangent to one of its surfaces, however small, sets it in motion. Liquids and gases are fluids. A liquid has a volume of its own and a free surface; a gas occupies the whole volume available. A fluid is called incompressible if its density is the same at every point and does not change with pressure. Liquids almost always behave as incompressible fluids, because reducing their volume by a small fraction requires very high pressures, and throughout this topic liquids are treated as incompressible.

2. Density

The density of a homogeneous body of mass $m$ and volume $V$ is

ρ = \frac{m}{V},

and it is measured in $kg/ m^{3}$ . Typical values are $1.00 \cdot 1 0^{3} kg/ m^{3}$ for fresh water, about $1.03 \cdot 1 0^{3} kg/ m^{3}$ for seawater, $13.6 \cdot 1 0^{3} kg/ m^{3}$ for mercury and about $1.2 kg/ m^{3}$ for air at room temperature. Density makes it possible to go from volume to mass and hence to weight: a block of aluminum ( $ρ = 2.70 \cdot 1 0^{3} kg/ m^{3}$ ) of volume $0.20 d m^{3} = 2.0 \cdot 1 0^{- 4} m^{3}$ has mass $0.54 kg$ .

3. Pressure

Let $F$ be the force that a fluid exerts on a flat surface of area $A$ . The pressure is the ratio between the component of $F$ perpendicular to the surface and the area:

p = \frac{F _{⊥}}{A} .

If the force is not uniform, the pressure at a point is the limit of $\frac{Δ F _{⊥}}{Δ A}$ over smaller and smaller surfaces around the point. Pressure is a scalar and is measured in pascals, $1 Pa = 1 N/ m^{2}$ . A fluid at rest exerts only perpendicular forces on every surface, because a tangential component would set the fluid itself in motion; moreover, at a point the pressure has the same value whatever the orientation of the surface on which it is measured. Therefore the pressure force on a small surface of area $Δ A$ has magnitude $p Δ A$ and is directed toward the surface, perpendicular to it. The same force produces very different pressures depending on the area: a person of weight $600 N$ standing on $0.040 m^{2}$ exerts $1.5 \cdot 1 0^{4} Pa$ on the floor, and resting on a heel of $1.0 c m^{2}$ exerts $6.0 \cdot 1 0^{6} Pa$ .

Other units of pressure

Besides the pascal, the following units are used: the bar, $1 bar = 1 0^{5} Pa$ , the atmosphere, $1 atm = 1.013 \cdot 1 0^{5} Pa$ , which is the average air pressure at sea level, and the millimeter of mercury, $1 mmHg \approx 133 Pa$ , so that $1 atm = 760 mmHg$ . In the exercises $p_{0} = 1.013 \cdot 1 0^{5} Pa$ denotes atmospheric pressure, and every value must be converted to pascals before being used in the formulas.

Stevin’s law

Consider an incompressible fluid of density $ρ$ at rest in the uniform gravitational field $g$ , and let $y$ be the height measured along the vertical, upward. For two points of the fluid at heights $y_{1}$ and $y_{2}$ we have

p_{2} = p_{1} + ρ g (y_{1} - y_{2}) .

In particular, all points at the same height have the same pressure and, if the free surface is in contact with air at pressure $p_{0}$ , at depth $h$ below the surface the pressure is

p = p_{0} + ρ g h .

The law states that the pressure in a liquid depends only on depth, and grows linearly with it, independently of the shape of the container. In water $ρ g = 9.81 \cdot 1 0^{3} Pa/m$ , so at a depth of $10 m$ the pressure is $1.013 \cdot 1 0^{5} + 0.981 \cdot 1 0^{5} \approx 1.99 \cdot 1 0^{5} Pa$ , about twice atmospheric pressure.

Absolute pressure

p

(in

1 0^{5} Pa

) at depth

h

(in meters) in water: the line

p = p_{0} + ρ g h

rises by about

0.98 \cdot 1 0^{5} Pa

every

10 m

Proof

Inside the fluid, consider an imaginary cylinder with vertical axis, bases of area $A$ at heights $y_{2} < y_{1}$ and height $y_{1} - y_{2}$ . The liquid contained in the cylinder is at rest, so the resultant of the forces acting on it is zero. The pressure forces on the lateral surface are horizontal and, since the fluid is in equilibrium in the horizontal direction as well, their resultant is zero. Along the vertical three forces act: the force $p_{1} A$ downward on the upper base, the force $p_{2} A$ upward on the lower base and the weight $m g = ρ A (y_{1} - y_{2}) g$ , where we have used the fact that the density is uniform. Equilibrium along the vertical gives

p_{2} A - p_{1} A - ρ g A (y_{1} - y_{2}) = 0,

and dividing by $A$ we obtain the claim. With a horizontal cylinder the weight has no component along the axis, therefore the pressures on the two bases are equal, so points at the same height have the same pressure. Finally, taking point $1$ on the free surface, where $p_{1} = p_{0}$ , and setting $h = y_{1} - y_{2}$ , we obtain $p = p_{0} + ρ g h$ .

The cylinder of liquid between the heights

y_{2}

and

y_{1}

is in equilibrium under its weight and the pressure forces on the two bases

1. Absolute pressure and gauge pressure

The pressure $p$ that appears in Stevin’s law is the absolute pressure. The difference $p - p_{0}$ between the absolute pressure and atmospheric pressure is called gauge pressure (or relative pressure): it is the quantity measured by manometers and by tire pressure gauges, which read zero in open air. In a liquid with its free surface in contact with the atmosphere, the gauge pressure at depth $h$ is $ρ g h$ . The distinction matters in the calculation of forces: on a wall that has the liquid on one side and air on the other, atmospheric pressure acts on both faces, so the resultant force is computed with the gauge pressure alone.

Communicating vessels and U-tube

In communicating vessels containing the same liquid at rest, the free surfaces, all at pressure $p_{0}$ , lie at the same height, whatever the shape of the vessels. This is a direct consequence of Stevin’s law, because points of the same liquid at the same pressure have the same height. If a U-tube contains two immiscible liquids, of densities $ρ_{1}$ and $ρ_{2}$ , we choose the height of the interface: the points $A$ and $B$ at that height in the two arms lie in the same connected liquid, so they have the same pressure. Denoting by $h_{1}$ and $h_{2}$ the heights of the liquid columns above that level, we have $p_{0} + ρ_{1} g h_{1} = p_{0} + ρ_{2} g h_{2}$ , that is,

ρ_{1} h_{1} = ρ_{2} h_{2} .

The heights are inversely proportional to the densities, and comparing the columns makes it possible to measure the density of a liquid relative to that of water.

U-tube with oil and water: the points

A

and

B

lie at the same height in the same liquid, so

ρ_{1} g h_{1} = ρ_{2} g h_{2}

Torricelli's barometer

A tube about one meter long, closed at one end and filled with mercury, is turned upside down into a dish of mercury. The mercury falls to a height $h$ above the surface of the dish and above it there remains a volume in which the pressure is negligible. The point of the tube at the level of the surface of the dish has the same pressure $p_{0}$ as the air, and by Stevin’s law $p_{0} = 0 + ρ_{Hg} g h$ . The barometer therefore measures atmospheric pressure through a height: with $h = 760 mm$ we obtain $p_{0} = 13.6 \cdot 1 0^{3} \cdot 9.81 \cdot 0.760 \approx 1.01 \cdot 1 0^{5} Pa$ . With water, which is $13.6$ times less dense, the equivalent column is $\frac{p _{0}}{ρ g} \approx 10.3 m$ high.

2. Force on a vertical wall

A liquid of density $ρ$ at rest, with its free surface in contact with air, wets a rectangular vertical wall of width $w$ up to the height $H$ , and on the other side of the wall there is air at pressure $p_{0}$ . Then the resultant force that the liquid exerts on the wall is perpendicular to the wall and has magnitude

F = \frac{1}{2} ρ g w H^{2} .

The force grows with the square of the depth, which is why dams have a base thicker than their top. For a gate $4.0 m$ wide with water $3.0 m$ deep we have $F = \frac{1}{2} \cdot 1000 \cdot 9.81 \cdot 4.0 \cdot 9.0 \approx 1.77 \cdot 1 0^{5} N$ .

Proof

The pressure is not uniform on the wall, so we divide it into horizontal strips of height $d z$ , at depth $z$ , on each of which the pressure can be considered constant. On the strip, of area $w d z$ , the gauge pressure is $ρ g z$ , because atmospheric pressure acts on both faces of the wall and its contribution cancels. The forces on the strips are all perpendicular to the wall, so they add as numbers and the total force is the integral

F = \int_{0}^{H} ρ g z w d z = ρ g w [\frac{z ^{2}}{2}]_{0}^{H} = \frac{1}{2} ρ g w H^{2} .

The same result can be written $F = ρ g \frac{H}{2} (w H)$ : it is the wetted area multiplied by the gauge pressure at half height, which is the average pressure because the pressure grows linearly with depth.

The theory continues with 7 more sections (pascals principle and the hydraulic press, archimedes principle, flotation…), plus all worked examples, the solutions and the PDF.

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