- Solids - Will maintains a fixed shape and will not flow.
- Liquids - Does not maintain a fixed shape, will flow, but is not readily compressible.
- Gases - Does not maintain a fixed shape, will flow, and will expand or contract depending on the container it is in.
- Plasma - Ionized atoms in which electrons are stripped from the nucleus.

Typically the progression amongst these phases is associated with increasing temperature or internal energy.

Here we are concerned with fluids, in other words matter which flows, ie. either gasses or liquids.

The density of a substance is defined as

$\large \rho=\frac{m}{V}$

The specific gravity of a substance is it's density relative to the density of water at $4.0^{o}\mathrm{C}$ which is $1.00\mathrm{g/cm^{3}}=1\times10^{3}\mathrm{kg/m^{3}}$

So we can say that the specific gravity of a substance is equal to it's density in $\mathrm{g/cm^{3}}$.

Pressure is defined as the force per unit area in a direction perpendicular to the surface of an object

$\large P=\frac{F}{A}$

There are **lots** of units for pressure.

The SI unit is $\mathrm{N/m^{2}}$ or pascal ($\mathrm{Pa}$).

You will also be familiar with $\mathrm{psi}$ ($\mathrm{lb/in^{2}}$)

$1 \mathrm{psi}=6.895×10^{3}\,\mathrm{Pa}$

Some other common units are:

$1\,\mathrm{atm}=101.3\,\mathrm{kPa}$

$1\,\mathrm{Torr}=1\mathrm{mm_{Hg}}=133\,\mathrm{Pa}$

$1\,\mathrm{bar}=1.0\times10^{5}\,\mathrm{Pa}$

Click here for a conversion table.

At a given depth in a static fluid the pressure must be the same in every direction, otherwise the fluid would not be static. The force due to the fluid on any solid surface it touches must be perpendicular to the surface, otherwise there would be some flow of the fluid over the surface.

To quantify the force due to pressure on a surface of area $A$ at a depth $h$ we should consider it's origin, which is gravitational force acting on liquid above the point at which we want to know the pressure.

We need to also take in to account the external pressure, so we should come up with a general formula which will let us take that in to account.

We can consider a slab of water of thickness $dy$ at height $y$ above the bottom. The pressure acting on the bottom is $P$, the pressure on the top at height $y+dy$ differs by $dP$ (making it $P+dP$).

Gravity acts on the slab and the force due to gravity on the slab is $dF_{G}=\rho g A\,dy$

The equilibrium condition tells us that all the forces must balance so

$PA-(P+dP)A-\rho gA\,dy=0$

giving us a differential form of the relationship between pressure and depth

$\frac{dP}{dy}=-\rho g$

We will now apply our equation for pressure

$\frac{dP}{dy}=-\rho g$

to an open container of fluid of uniform density

$\int_{P_{1}}^{P_{2}}\,dP=-\int_{y_{1}}^{y_{2}}\rho g\,dy$

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

The depth $h=y_{2}-y_{1}$ and the pressure at the top $P_{2}$ is atmospheric pressure $P_{0}$

We can therefore say that the pressure at depth $h$, $P=P_{1}$ is

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

We can see that we need to take in to account the pressure of the atmosphere above the container!

If we now consider the same situation but with a hole in the bottom, to which a hose is attached the pressure change is again

$\int_{P_{1}}^{P_{2}}\,dP=-\int_{y_{1}}^{y_{2}}\rho g\,dy$

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

so the pressure **difference** is again $\rho gh$, but we should remember that the atmospheric pressure is the same at both ends, so if you want to find the effective pressure on the water at the end of the hose you should not add on the atmospheric pressure.

Note that while we consider atmospheric pressure to be the same at both ends here and this is very good approximation for small height differences, over large distances air pressure and density are very much not constant!

If you measure the pressure in your tires you are actually measuring the pressure difference between the atmospheric pressure. The absolute pressure is therefore the sum of the gauge pressure $P_{G}$, which is what you measure, and the atmospheric pressure $P_{0}$

$P=P_{0}+P_{G}$

There are many different methods for measuring pressure.

Pascal's principle, due to Blaise Pascal is an important principle that is the basis of the hydraulic press

**Pascal's Principle:** If an external pressure is applied to a confined fluid the pressure at every point within the fluid increases by that amount.

$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}$

Earlier we looked at the forces on an element within a fluid. Now we will look at an object that has displaced some of the fluid. In this case there is a force we call the buoyant force which reflects the difference between the force on the top and bottom of the object due to the fluid . In a fluid of uniform density $\rho_{f}$

$F_{B}=F_{2}-F_{1}=\rho_{f}gA(h_{2}-h_{1})=\rho_{f}gA\Delta h=\rho_{f}Vg=m_{f}g$

We can note from this equation that the buoyant force does not depend on the depth of the object or on the density of the object. The buoyant force is determined by the weight of the fluid which is displaced by the object.

This is a statement of Archimedes' principle discovered by Archimedes in the 3rd century BC.

Legend has it that Archimedes arrive at his principle while working on a task for King Hiero II who wished to know if a crown he had asked to be made from pure gold actually was…without damaging it!

The most likely way that he achieved this was by the application of his principle to a balance with the crown on one side and a pure gold brick on the other.

The net force of an object is determined by the sum of the buoyant force and the gravitational force. If an object is more dense than the water then it will sink to a point where either the fluid density is the same as the object density, or to the bottom where it can be supported by normal force.

The net force on an object will be $\vec{F}_{B}+\vec{F}_{G}=\rho_{F}V_{O}g-\rho_{O}V_{O}g$.

For the scale to balance in air

$\rho_{c}V_{c}gl_{1}=\rho_{b}V_{b}gl_{2}$

For it to balance in water

$(\rho_{c}-\rho_{w})V_{c}gl_{1}=(\rho_{b}-\rho_{w})V_{b}gl_{2}$

Which would mean that

$V_{c}gl_{1}=V_{b}gl_{2}$

which can only be true if $\rho_{c}=\rho_{b}$. If the scale tips the densities are not equal and the fraud is exposed!

For objects less dense than water Archimedes' principle also applies.

An object floats when the gravitational force balances the buoyant force

$F_{B}=mg$

$\rho_{F}V_{displ}g=\rho_{O}V_{O}g$

$\frac{V_{displ}}{V_{0}}=\frac{\rho_{O}}{\rho_{f}}$

The fraction of the volume which is displaced, or equivalently, the fraction of the object which is submerged, is determined by ratio of the object's density to that of the fluid.

I want to inflate a ball so that it will float in water with exactly half of the ball above the surface and the other half below when the temperature is 20$\mathrm{^{o}C}$. The ball will be 20 cm in diameter and made from rubber which is 1 cm thick and has density 1800 $\mathrm{kg\, m^{-3}}$. The density of air at 20$\mathrm{^{o}C}$ under normal conditions, such as those that exist above the surface of the water, is 1.2 $\mathrm{kg\, m^{-3}}$. The density of water is 1000 $\mathrm{kg\, m^{-3}}$. The inside of the ball is to be filled with pressurized air, which will have a higher mass density than that outside the ball. What should the mass density of air inside the ball be so that ball floats as desired?

Weight of displaced fluid

$(\frac{1}{2}1000+\frac{1}{2}1.2)\frac{4}{3}\pi0.1^{3}$

Weight of ball including dense air

$1800\frac{4}{3}\pi(0.1^{3}-0.09^{3})+\rho_{air}\frac{4}{3}\pi0.09^{3}$

At equilibrium these two are equal, so

$\rho_{air}=\frac{(\frac{1}{2}1000+\frac{1}{2}1.2)\frac{4}{3}\pi0.1^{3}-1800\frac{4}{3}\pi(0.1^{3}-0.09^{3})}{\frac{4}{3}\pi0.09^{3}}$

$\rho_{air}=17.55\mathrm{kg\,m^{-3}}$

As long a balloon is free to expand the pressure will be the same on the outside and inside of the balloon, and so the density of the helium balloon will remain less than the surrounding atmosphere as long as the composition doesn't change to lighter elements. Of course at some point the balloon will stop expanding and so the while the density of the gas inside will remain the same the density outside will be decreasing as the balloon continues to ascend, until the density inside and outside is the same and the balloon reaches float altitude. So if you want to go to the edge of space you need a really big balloon. This would be a good time to jump!

A standard spherical latex balloon has mass 3 g. Helium has a density of $0.18\,\mathrm{kg\,m^{-3}}$ and air has a density of $1.2\,\mathrm{kg\,m^{-3}}$. What is the minimum diameter to which you should inflate the balloon with helium so that it will rise? You may neglect the volume of air which the latex displaces (but not the mass of the latex).

The condition for the balloon to rise is that the bouyant force must equal the weight

$m_{b}g+\rho_{He}Vg=\rho_{Air}Vg$

$m_{b}=(\rho_{Air}-\rho_{He})\frac{4\pi}{3}r^{3}=(\rho_{Air}-\rho_{He})\frac{\pi}{6}d^{3}$

$d^{3}=\frac{6m_{b}}{(\rho_{Air}-\rho_{He})\pi}$

$d=(\frac{6\times0.003}{(1.2-0.18)\pi})^{1/3}=0.178\mathrm{m}$

A smooth flow of a fluid is called laminar flow. Flow at higher velocities frequently becomes turbulent.

We will mostly focus on laminar flow.

If we consider a steady flow of fluid in an enclosed tube the amount of mass which flows past a point in a time interval $\Delta t$

$\frac{\Delta m}{\Delta t}$

is constant. If we express the mass in terms of volume and density

$\frac{\Delta m_{1}}{\Delta t}=\frac{\rho_{1}\Delta V_{1}}{\Delta t}=\frac{\rho_{1}A_{1}\Delta l_{1}}{\Delta t}=\rho_{1}A_{1}v_{1}$

and use the fact that

$\frac{\Delta m_{1}}{\Delta t}=\frac{\Delta m_{2}}{\Delta t}$

We can say that

$\rho_{1}A_{1}v_{1}=\rho_{2}A_{2}v_{2}$

which for an incompressible fluid gives

$A_{1}v_{1}=A_{2}v_{2}$

Bernoulli's principle due to Daniel Bernoulli connects the velocity of a flowing fluid to the pressure.

The pressure is a force per unit area, the work done by the pressure at the left end of the tube in moving the the fluid at the end of the tube through $\Delta l$ does work equal to

$W_{1}=P_{1}A_{1}\Delta l_{1}$

At the other end the force exerted on the fluid is in the opposite direction to the motion and the work done is

$W_{2}=-P_{2}A_{2}\Delta l_{2}$

The mass $m$ of fluid moved changes it height by $y_{2}-y_{1}$ and the work done by gravity on the fluid in is

$W_{g}=-mg(y_{2}-y_{1})$

The net work done is the sum of these three works

$W=P_{1}A_{1}\Delta l_{1}-P_{2}A_{2}\Delta l_{2}-mgy_{2}+mgy_{1}$

The net work is also equal to the change in kinetic energy

$W=\frac{1}{2}mv_{2}^2-\frac{1}{2}mv_{1}^2$

The mass of fluid is given by $m=\rho A_{1}\Delta l_{1}=\rho A_{2}\Delta l_{2}$ and substituting this and rearranging gives us Bernoulli's equation

$P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gy_{1}=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gy_{2}$

A special case of Bernoulli's equation is Toricellis' theorem, which was actually discovered a century before Bernoulli.

If we apply Bernoulli's equation

$P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gy_{1}=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gy_{2}$

to the above situation we can see that $P_{1}=P_{2}$, and $v_{2}$ can be approximated as zero.

So then

$\frac{1}{2}\rho v_{1}^{2}+\rho gy_{1}=\rho gy_{2}$

$v_{1}=\sqrt{2g(y_{2}-y_{1})}$

The magnitude of the velocity of the water exiting the tap is the same as if it had fallen from the top of the water.

$x=v_{x}\sqrt{\frac{2y_{1}}{g}}=\sqrt{2g(y_{2}-y_{1})}\sqrt{\frac{2y_{1}}{g}}$

$x=2\sqrt{y_{1}(y_{2}-y_{1})}$

Max range when

$\frac{dx}{dy_{1}}=0$

$x=2z^{\frac{1}{2}}$

$z=y_{1}(y_{2}-y_{1})$

$\frac{dx}{dy_{1}}=\frac{dx}{dz}\frac{dz}{dy_{1}}$

$\frac{dx}{dy_{1}}=2\times\frac{1}{2}\frac{y_{2}-2y_{1}}{\sqrt{y_{1}(y_{2}-y_{1})}}=0$

$y_{2}=2y_{1}$

$y_{1}=\frac{1}{2}y_{2}$

$P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gy_{1}=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gy_{2}$

At a constant height

$P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2}$

The pressure in a fast moving column of air will be less than that outside it, so an object in the column will experience a force that will tend to keep it in the column.

The airfoil produces dynamic lift by compressing the streamlines above the wing and expanding them below.

From the continuity equation

$A_{1}v_{1}=A_{2}v_{2}$

We can see that the velocity on top of the wing is greater than it is below. Note that the time taken for the air to pass over both surfaces is not the same.

We can also apply the equation

$P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2}$

to understand a Venturi tube

The reason the pressure is not the same in the third part of the venturi tube as in the first is because the air is flowing slower there. This is due to viscosity. The viscosity is a property of the fluid and can greatly affect the rate of flow. If you take viscosity to an extreme it can be the basis for the world's longest running laboratory experiment.

Viscosity can be defined from looking at a fluid trapped between two plates in which one plate moves with velocity $v$ with respect to the other. The fluid in contact with each plate must have the same velocity as the plate it is in contact with and the fluid in between is subject to a velocity gradient. The viscosity $\eta$ $(\mathrm{N\,s\,m^{-2}})$ is defined in terms of the force require to maintain that velocity gradient (which is required to move the plate). In the following equation $A$ is the area of the plate.

$F=\eta A\frac{v}{l}$

In a cylindrical tube of radius $R$ and length $l$ the **volume** rate of flow $Q$ $(\mathrm{m^{3}s^{-1}})$ is related to the change in pressure from the beginning of the tube $P_{1}$ to the end of the tube $P_{2}$ by Poiseulle's equation

$Q=\frac{\pi R^{4}(P_{1}-P_{2})}{8\eta l}$

Surface tension, $\gamma$, is due to attractive forces between the molecules in a liquid, and is the force $F$ per unit length $l$

$\gamma=\frac{F}{l}$

At at an interface between a liquid and another medium the liquid forms a contact angle which is determined by the balance of forces, including surface tension. Surface tension in thin tubes leads to capillarity where the movement up or down the walls of the tube is due the balance of surface tension with gravitational and viscous forces.

If you eliminate viscosity, as occurs in a superfluid, you can see some very interesting effects.

The aerodynamic role of the dimples on a golf ball is an interesting case.

Armed with Bernoulli's equation we can come to an understanding of why the dimples, which convert the laminar flow to a turbulent one actually make the ball go further!