Although both liquids and gases are fluids, there are distinctions between them. At the molecular level, a substance in the liquid phase is similar to one in the solid phase in that the molecules are close to, and interact with, one another. The molecules in a liquid are able to move around a little more freely than those in a solid, where the molecules typically only vibrate around relatively fixed positions. By contrast, the molecules of a gas are not constrained and fly around in a chaotic swarm, with hardly any interaction. On a macroscopic level, there is also a distinction between liquids and gases. If you pour a certain volume of a liquid into a container, the liquid will occupy that same volume, whatever the shape of the container. However, if you introduce a sample of gas into a container, the molecules will fly around and fill the entire container. By definition, the density of a substance is its mass per unit volume, and it’s typically denoted by the letter ρ (the Greek letter rho):
Density = mass/volume
Ρ – M/v
Note that this equation immediately implies that m is equal to ρV. The ratio of the density of any substance to the density of water is known as the specific gravity of the substance:
Specific gravity = ρsubstance / ρwater
Imagine that we have a tank with a lid on top, filled with some liquid. Suspended from this lid is a string, attached to a thin sheet of metal. The weight of the liquid produces a force that pushes down on the metal sheet. If the sheet has length l and width w, and is at depth h below the surface of the liquid, then the weight of the liquid on top of the sheet is
Fg = mg = ρVg = ρ(lwh)g
where ρ is the liquid’s density. If we divide this weight by the area of the sheet (A = lw), we get the pressure due to the liquid:
Pliquid = force/area = Fg liquid /A = ρ(lwh)g/lw = ρgh
Note that the hydrostatic pressure due to the liquid, Pliquid = ρgh, depends only on the density of the liquid and the depth below the surface; in fact, it’s proportional to both of these quantities. One important consequence of this is that the shape of the container doesn’t matter.
If a block is placed in a tank of fluid, because the pressure on each side of the block depends on its average depth, we see that there’s more pressure on the bottom of the block than there is on the top. Therefore, there’s a greater force pushing up on the block than there is pushing down on it. The forces due to the pressure on the other four sides (left and right, front and back) cancel out, so the net force on the block is upward. This net upward force is called the buoyant force (or just buoyancy for short), denoted Fbuoy. We calculate the magnitude of the buoyant force using Archimedes’ principle; in words, Archimedes’ principle says
The strength of the buoyant force is equal to the weight of the fluid displaced by the object.
When an object is partially or completely submerged in a fluid, the volume of the object that’s submerged, which we call Vsub, is the volume of the fluid displaced.1 By multiplying this volume by the density of the fluid, we get the mass of the fluid displaced; then, multiplying this mass, it gives us the weight of the fluid displaced. So, here’s Archimedes’ principle as a mathematical equation:
Buoyant force: Fbuoy = ρfluid Vsub x g
When an object floats, its submerged volume is just enough to make the buoyant force it feels balance its weight. So, if an object’s density is ρobject and its (total) volume is V, its weight will be mg = ρVg. The buoyant force it feels is ρfluidVsub g. Setting these equal to each other, we find that:
Vsub / V = ρobject /ρfluid
So, if ρobject < ρfluid, then the object will float; and the fraction of its volume that’s submerged is the same as the ratio of its density to the fluid’s density. For example, if the object’s density is 2/3 the density of the fluid, then the object will float, and 2/3 of the object will be submerged. If an object is denser than the fluid, it will sink. In this case, even if the entire object is submerged (in an attempt to maximize Vsub and the buoyant force), its weight is still greater than the buoyant force, and down it goes. And if an object just happens to have the same density as the fluid, it will be in static equilibrium anywhere underneath the fluid.
If the specific gravity is less than 1, this means the substance is less dense than water; if the specific gravity is greater than 1, then the substance is denser than water. The density of liquid water is taken to be a constant, equal to 1000 kg/m3. If we place an object in a fluid, the fluid exerts a contact force on the object. How that force is distributed over any small area of the object’s surface defines the pressure:
Pressure = Force⊥ / Area
The subscript ⊥ (which means perpendicular) is meant to emphasize that the pressure is defined to be the magnitude of the force that acts perpendicular to the surface, divided by the area. Because force is measured in newtons (N) and area is expressed in square meters (m2), the SI unit for pressure is the newton per square meter. This unit is known as the pascal.1
Consider a pipe through which fluid is flowing. The flow rate, f, is the volume of fluid that passes a particular point per unit time; for example, the number of liters of water per minute that are coming out of a faucet. In SI units, flow rate is expressed in m3/s. To find the flow rate, all we need to do is multiply the cross-sectional area of the pipe at any point by the average speed of the flow at that point. Be careful not to confuse flow rate with flow speed; flow rate tells us how much fluid flows per unit time; flow speed tells us how fast it’s moving. If the pipe is carrying a liquid, which we assume is incompressible (that is, its density remains constant), then the flow rate must be the same everywhere along the pipe. Choose any two points, Point 1 and Point 2, in a pipe carrying a liquid. If there aren’t any sources or sinks between these points, all the liquid that flows by Point 1 must also flow by Point 2. In other words, the flow rate at Point 1 must be the same as the flow rate at Point 2: f1 = f2. Rewriting this using f = Av, we get the Continuity Equation:
A1V1 = A2V2
Because the product Av is a constant, the flow speed will increase where the pipe narrows, and decrease where the pipe widens. In fact, we can say that the flow speed is inversely proportional to the cross-sectional area—or to the square of the radius—of the pipe.
The most important equation in fluid mechanics is Bernoulli’s Equation, which is the statement of conservation of energy for ideal fluid flow. First, let’s describe the conditions that make fluid flow ideal.
- The fluid is incompressible.
This works very well for liquids and also applies to gases if the pressure changes are small.
- The fluid’s viscosity is negligible.
Viscosity is the force of cohesion between molecules in a fluid; think of viscosity as internal friction for fluids. For example, maple syrup is sticky and has a greater viscosity than water: there’s more resistance to a flow of maple syrup than to a flow of water. While Bernoulli’s Equation would give good results when applied to a flow of water, it would not give good results if it were applied to a flow of maple syrup.
- The flow is streamline.
In a tube carrying a flowing fluid, a streamline is just what it sounds like: it’s a line in the stream. If we were to inject a drop of dye into a clear glass pipe carrying, say, water, we’d see a streak of dye in the pipe, indicating a streamline. The entire flow is said to be streamline (as an adjective) or laminar if the individual streamlines don’t curl up into vortices but instead remain steady and smooth. When the flow is streamline, the fluid moves smoothly through the tube. If the three conditions described above hold, and the flow rate, f, is steady, Bernoulli’s Equation can be applied to any pair of points along a streamline within the flow. Let ρ be the density of the fluid that’s flowing. Label the points we want to compare as Point 1 and Point 2. Choose a horizontal reference level, and let y1 and y2 be the heights of these points above this level.1 If the pressures at Points 1 and 2 are P1 and P2, and if the flow speeds at these points are v1 and v2, then Bernoulli’s Equation say:
P1 + pgy + ½ pv12 = P1 + pgy + ½ pv22
Consider the two points labeled in the pipe shown below:
Since the heights y1 and y2 are equal in this case, the terms in Bernoulli’s Equation that involve the heights will cancel, leaving us with
P1 + pgy + ½ pv12 = P1 + pgy + ½ pv22
We already know from the Continuity Equation (f = Av) that the speed increases as the cross-sectional area of the pipe decreases; that is, since A2 < A1, we know that v2 > v1, so the equation above tells us that P2 < P1. This shows that the pressure is lower where the flow speed is greater. This is known as the Bernoulli (or Venturi) effect.2
The height of the liquid column above Point 2 is less than the height of the liquid column above Point 1, because the pressure at Point 2 is lower than at Point 1, due to the fact that the flow speed at Point 2 is greater than at Point 1. The Bernoulli Effect also accounts for many everyday phenomena. It’s what allows airplanes to fly, curve balls to curve, and tennis balls hit with top spin to drop quickly.2 The use of pitot tubes is an example of Bernoulli’s equation in action.2 These are specialized measurement devices that determine the speed of fluid flow by determining the difference between the static and dynamic pressure of the fluid at given points along a tube.
1) Bettini, Alessandro (2016). A Course in Classical Physics 2—Fluids and Thermodynamics. Springer. p. 8
2) The Venturi effect”. Wolfram Demonstrations Project. Retrieved 2017-08-06
3) Alberts, B.; Johnson, A.; Lewis, J.; Raff, M.; Roberts, K.; Walters, P. (2002). Molecular Biology of the Cell (4th ed.). New York and London: Garland Science.
4) Moran and Shapiro, Fundamentals of Engineering Thermodynamics, Wiley, 4th Ed, 2000.
5) Kauzmann, W. (1966). Kinetic Theory of Gasses, W.A. Benjamin, New York