Fluid Mechanics

Fluid mechanics is the physics of flowing matter, which includes, but is not limited to, cars moving through the traffic grid, waste flowing through the sewer system, gases moving through an engine, or sap moving sucrose from the leaves to the distal parts of a tree. Other examples of fluid mechanics include buoyancy (why you'll float in the Dead Sea), surface tension, wound healing, pattern formation in boiling liquids (the so-called Rayleigh-Bènard convection), and the motion of ants or flocks of birds moving in unison.

Lava, flowing molten rock, is an example of a fluid

Fluids have a bad reputation because their detailed behavior is governed by the Navier-Stokes equations, which pose great mathematical difficulties. Happily, it is often unnecessary to solve these equations to obtain great insight to the behavior of fluids. One can often make progress with the right set of approximations. To start, we'll analyze the mechanics of a very simple problem, that of floating.

If you jump into water, chances are you'll float, whether it be salt or freshwater, but if you throw an anvil, it goes straight to the bottom. The reason why one floats and the other sinks is very simple and can be obtained by a straightforward consideration of a pool of water, free of people, anvils, or other debris. Consider some selection of water, $\Gamma_S$ within the pool, of volume $V_{\textrm{H}_2\textrm{O}}$ of arbitrary shape, indicated by the black closed circle in the diagram below.

Because the pool is in a gravitational field, $\Gamma_S$ is pulled down by the force

$$F=m_{\textrm{H}_2\textrm{O}}g = V_{\textrm{H}_2\textrm{O}}\rho_{\textrm{H}_2\textrm{O}}g$$

Now, the pool maintains a stable shape within the container, therefore the forces on any such arbitrary volume are balanced. Concretely, any volume $V_{\textrm{H}_2\textrm{O}}$ in the pool experiences an upward force $V_{\textrm{H}_2\textrm{O}}\rho_{\textrm{H}_2\textrm{O}}g$, called the buoyant force.

The pool doesn't know that $\Gamma_S$ is a volume of water, if we replaced $\Gamma_S$ with the same volume of some other substance with the same weight as $\Gamma_S$, it would feel the same buoyant force and would float in place. In fact, if we replaced $\Gamma_S$ with any object of the same volume, it would feel the same buoyant force, regardless of its weight. Hence, we have arrived at Archimede's principle.

Archimede's principle

The buoyant force on a body is vertical and is equal and opposite to the weight of the fluid displaced by the body.

$$F_\text{buoyant}=\rho_\text{fluid}V_\text{object}g$$

This idea holds whether the fluid is water, molasses, air, or some other simple fluid.

Ice cubes

Archimedes' principle can explain why frozen ice cubes rise to the top of your drink. Upon freezing, water undergoes a volume expansion, so that a volume $V$ of liquid water, ends up as a volume $V^\prime > V$ of frozen ice. Because the force it feels is proportional to its large volume, while the pull of gravity is proportional to its unchanged weight, the ice feels a net force upward $F = \rho g\left(V^\prime - V\right)$.

Physical therapy

Even for objects that ultimately sink, Archimede's principle suggests an apparent weight reduction. When walking in water, a human of mass $m$ who usually feels like they weigh $mg$ will feel a reduced weight of $mg-V_\text{human}\rho_{\textrm{H}_2\textrm{O}}g$. Because the density of flesh is approximately 0.97 g/mL, the weight of a typical human will be

$$\begin{aligned} &mg - \frac{m}{\rho_\text{flesh}}\rho_{\textrm{H}_2\textrm{O}}g \\ &= mg\left(1-\frac{\rho_{\textrm{H}_2\textrm{O}}}{\rho_\text{flesh}}\right)\\ &\approx 0.03\times mg \end{aligned}$$

just 3% of their normal weight. This makes swimming pools a convenient place for physical therapists who need to start teaching people to walk before their legs are strong enough to walk normally.

For fluids in active flow, we need something better than the balance of forces. Fortunately, we can get a long way with one simple assumption, the conservation of matter. When fluid moves from one position to the next, it must do so in such a way that no fluid matter is destroyed.

For instance, if we inject fluid into the mouth of a tube at a rate of $J_\text{in}=\SI{1}{\liter\per\second}$ , we should find that $\SI{1}{\liter}$ of the fluid comes out the other side every second, i.e. that $J_\text{in}=J_\text{out}$.

For a toy model, consider a the device below that consists of one level section of tube $\Gamma_\text{A}$ with cross section of area $A_A$, connected by a linker section to another level tube $\Gamma_\text{B}$ with cross section $A_B$.

We want to find a relationship that connects the velocity and pressure of the fluid in either section of tube.

To start, let us take our system to be the fluid that's between the discs $\partial_\text{A}$ and $\partial_\text{B}$ at time zero, which we call $\Sigma$. In order for $\Sigma$ to flow to the right, there must be a net force to push it along. If the fluid pressure in $\Gamma_\text{A}$, $P_A$ is greater than the fluid pressure in $\Gamma_{B}$, $P_B$, then the fluid to the left of $\Sigma$, $\Sigma_L$ will push with greater force than the fluid to the right, $\Sigma_R$, and hence $\Sigma$ will flow. These two forces will be $P_A A_A$ and $P_B A_B$ respectively.

Applying the work energy principle, we have $W = \Delta T + \Delta U$. For our parcel of fluid, work is performed by the two pressures in moving $\Sigma$ along the tube. Consider the flow undertaken in some span of time $\Delta t$. In $\Gamma_A$, $\Sigma$ will move the distance $v_A\Delta t$, and in $\Gamma_B$, $\Sigma$ will move the distance $v_B\Delta t$. Hence, the net work $W$ on the fluid is given by the work done on our fluid by the fluid to the left $\Sigma_L$: $F dx = P_A A_A v_A \Delta t$, minus the work done by our fluid to the fluid on the right $\Sigma_R$: $P_B A_B v_B \Delta t$.

$$W = P_A A_A v_A \Delta t - P_B A_B v_B \Delta t$$

However, we have the conservation condition $J_\text{in} = J_\text{out}$ which applies for the discs $\partial_A$ and $\partial_B$. Hence, it must be true that $A_A v_A \Delta t = A_B v_A \Delta t$.

In other words, the volume of fluid $\Delta m$ that flows through $\partial_A$ is equal to the mass of fluid that flows out from $\partial_B$. Hence,

$$W = \Delta V \left(P_A-P_B\right)$$

Now, the work done on $\Sigma$ is equal to the change in energy of the fluid. The kinetic energy of our fluid $\Sigma$ should be the same as before with the exception that the quantity $\Delta m$ of liquid which initially moved with velocity $v_A$ in $\Gamma_A$ is now traveling with velocity $v_B$ in $\Gamma_B$, and therefore $\Delta T = \frac12 \rho\Delta V v_B^2 - \frac12 \rho\Delta V v_A^2$. So, we have $\Delta V \left(P_A - P_B\right) = \frac12 \rho \Delta V \left( v_B^2 - v_A^2 \right)$, or $P_A + \frac12\rho v_A^2 = P_B + \frac12\rho v_B^2$, which is Bernoulli's relation for fluid flow in an arbitrary tube of level height.

In this derivation, the tubes were kept at equal level for simplicity's sake. It is trivial to recalculate our relation for the case when the two tube sections are of differing heights in a gravitational field, as occurs for the plumbing system in an apartment building. In this case, the work-energy principle is given by $W = \Delta T + \Delta U$ and we have thus have the full Bernoulli relation

$$P_A + \frac12\rho v_A^2 + \rho gh_A = P_B + \frac12\rho v_B^2 + \rho gh_B$$

Note that our calculation did not depend in any way on the particular setup that we used (the two tubes and linker section). The same calculation applies to a tube of arbitrary shape which carries out an arbitrary trajectory through a gravitational field. Thus, the relation can be used to connect any two cross sections of a fluid's flow.

Bernoulli's relation has a number of applications, particularly in the use of hydraulics.

Fork-lift operator

Suppose a sleepy forklift operator parked clumsily on a hill without applying the emergency brake. Suddenly, the forklift slips and starts rolling backward down the hill. In order to stop before slamming into a schoolbus full of teddy bears, it will be necessary to apply a normal force of 150 N to the disk brake. Suppose the force is delivered by a piston of surface area 2 m$^2$ which is connected by a hydraulic system to a foot lever of surface area 0.02 m$^2$. With what force must the operator step on the lever in order to apply the required force? Assume there is no significant height difference from the brake to the lever.

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In this problem, the fluid is transmitting the force from the operators foot to the brake pad. Brake pads hover very close to the disc they press, so we can assume that the fluid velocity is negligible. Indeed, once the foot is clamped down on the lever, the arrangement will be static. In this case, Bernoulli's relation simplifies to

$$P_\text{foot} = P_\text{brake}$$

As pressure is given by the force per unit area, we have $$\frac{F_\text{foot}}{A_\text{foot}} = \frac{F_\text{brake}}{A_\text{brake}}$$

yielding

$$F_\text{foot} = F_\text{brake} \times \frac{A_\text{foot}}{A_\text{brake}} 1.5 \text{N}$$

The asymmetry in the piston areas gives the operator a significant mechanical advantage over the braking system, making it very easy for the operator to apply the huge force required.