Lecture 1 | Pressure at a Depth, Pascal's Principle

Fluid Statics

Lecture 1 | Pressure at a Depth, Pascal's Principle

Fluid Statics

Many of the empirical observation within the field of fluid mechanics lead to equations or laws that are dependent on the mass of the fluid or object being observed. Often times we cannot directly measure the mass but know information about the volume of the fluid or object. Thus we can use the material property called mass density to determine the mass of a given volume of material. The mathematical definition of mass density is...

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

In other words, mass density is mass per unit volume, and it is a material property (i.e. it is a constant for a given material).

*Note: There are other types of density (e.g. energy density, charge density, number density, area density, etc…). Since we will be working with mass density in fluid mechanics, we often drop the mass and just call it density.

Check out this video by Mark Drollinger highlighting 5 facts about density. Hopefully, this short video will demonstrate the importance of understanding mass density.

When a body of fluid is near a massive object such as the Earth, a pressure gradient within the fluid is formed. In this section we will quantify these pressure gradients by finding an expression for pressure at some depth below the surface of an incompressible fluid. We will also explore applications of Pascal’s law which tells us how a change in pressure within a an incompressible fluid is transmitted to other locations within the fluid.

The Science Nuts explain hydrostatics with a few quick everday examples, enjoy!

Pre-lecture Study Resources

Watch the pre-lecture videos and read through the OpenStax text before doing the pre-lecture homework or attending class.

Fluid Statics | Pressure

Consider comparing the differences between gold and Styrofoam (polystyrene foam). Perhaps you wish to compare the weight of two materials. Well, you can obtain one pound of gold and one pound of Styrofoam, so perhaps weight is not a good quantity when trying to compare the two materials. But now that you have a pound of each material, you notice that pound of gold takes up much less volume than the pound of Styrofoam. Cleverly, you define a new quantity called mass density, which is the mass per unit volume of each material. Mathematically this is written as…

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

It turns out, that the density is a material property. One pound of gold has the same density 2 pounds, 1,000 pounds, 1/100^th of a pound, etc..

We must be a little bit more careful though. Consider water, if we place a container of water under enough pressure, we can slightly change the density. Likewise, if we change the temperature of a container of water, the density also slightly changes. Thus, the density of any material is dependent on temperature and pressure. However, for solids and liquids, the density is very nearly constant for a wide range of temperatures and pressure, thus throughout our fluid mechanics studies, we will ignore these small density variations due to temperature and pressure changes. Another way to state this approximation is to say that we will assume fluids are incompressible (i.e. their densities are constant for any temperature or pressure).

While standing on a horizontal patch of ice, the normal force that you apply on the ice is equal to the force of gravity from the earth on you ($m \, g$). The ice supports this normal force no matter how you stand on the ice (e.g. lay down or stand with high heels on). However, you might already known that you would feel much more comfortable lying down on thin ice than standing with high heels on. Even though the ice is supporting the same weight ($m \, g$), the ice is more likely to break if you wear the high heels, thus we must introduce a new quantity to differentiate between these two scenarios. The new quantity introduced is called pressure. Pressure is defined as the perpendicular component of force divided by the area that the force is applied to. Mathematically this is written as…

$P = \frac{F_{\perp}}{A}$

Notice how it is only the perpendicular component of force. Since area is a scalar and the perpendicular component of force is a scalar, then pressure must also be a scalar.

The SI unit for pressure is $\frac{N}{m^2}$, which is given a special name called a Pascal (Pa).

$1 Pa = 1 \frac{N}{m^2}$

*NOTE: Pressure is often written as $P = \frac{F}{A}$, where it is implied that the force is only the perpendicular component.

Pressure plays an important role in fluid mechanics because pressure differences cause an associated net force.

$pressure \, difference \Longrightarrow \sum{\vec{F}}$

As we will see in the hydrostatics section, there exists a pressure gradient within a fluid near a massive object like the Earth - the pressure increases the deeper you go in the fluid. So pressure and the associated net force due to pressure differences are the foundations upon which we will build our fluid mechanics models.

Atmospheric pressure ($P_{atm}$)

We live in a mixture of gasses (the atmosphere) that surrounds us in all directions. Consider a column of atmosphere directly above us. This column of atmosphere has an associated weight, and just like our ice example, this atmospheric weight is supported by the surface of the Earth, or our heads as we stand on the Earth’s surface. Thus if we consider the force that this column of atmosphere applies over the area of the column, we can define a pressure due to the atmosphere near the Earth’s surface. The actual atmosphereic pressure is complicated, depending on temperature, height, humidity, etc… However, at sea level in standard conditions according to the International Standard Atmosphere model (ISA), the standard atmosphereic pressure is $101.325 \, kPa$. This value is also defined as 1 atmosphere ($atm$).

$1 \, atm = 101.325 \, kPa$

Gauge pressure ($P_{g}$)

The devices we use to measure pressure within a closed container often measure the pressure difference between the inside of a container and the outside. For example, if you use a tire pressure gauge to measure the pressure in your car tires, you are really measuring the difference between the outside atmosphereic pressure and the total pressure inside the tire (as seen in the figure below).

Absolute pressure ($P_{abs}$)

The absolute pressure is then the sum of the atmospheric pressure and gauge pressure.

$P_{abs} = P_{atm} + P_{g}$

This is a representation of a tire with different pressures but only one pressure reading. There is the environmental pressure and the pressure inside the tire. The total pressure is the net sum of all of the pressure so the P total is equal to the P atmospheric plus P gauge.

Pressure of a tire

Pressure at a depth

Have you ever gone swimming in a really deep body of water? If so, you are probably familiar with the sensation you get when diving down to very deep depths below the surface of the water. This sensation is a result of your inner ears letting you know the pressure of the water around you is increasing. How can we mathematically show this pressure increase as we go deeper into the water, also referred to as “pressure at a depth”?

Consider a container of water open to the atmosphere at the top as shown in the figure below.

Pressure acting on a box in an open container of water

Let the density of the water be $\rho_f$. Now imagine a smaller rectangular volume element of water in the container as indicated by slightly darker blue. The downward force on this volume element of water at the top of the surface is due to the atmospheric pressure $P_0$ multiplied by the area $A$ of the top of the volume element. Since this volume of water is stationary, (i.e. it is not accelerating), we know that there must be an upward force at the bottom of the volume element which is greater than the top force because it must also support the weight of the volume element. This statement is analgous to the following situation: 5 books are stacked on top of each other, the top book has a normal force acting on it upwards from the second from the top equal to the weight of the top book only, while the book on the bottom of the stack has a normal force acting on it upwards from a table equal to the combined weight of all 5 books. There are also forces from the pressure on either side (left, and right) of the volume element which must be equal to each other because the volume element is not accelerating in either direction. The descriptive representation of this scenario is nicely visualized with an accompanying FBD for the volume element of water as shown below. (The horizontal forces are not shown since they all cancel out.)

Diagram of vertical forces acting on volume element of water

Using Newton's second law we can translate our physical representation (FBD) into a mathematical representation as shown below.

Mathematical representations of pressure

Notice the end result for the pressure at the bottom of the volume element is not dependent on the area or volume, it only depends on the depth below the surface $d$, the density of the fluid $\rho_f$, the acceleration due to gravity caused by the planet the body of fluid is near $g$, and the pressure above the surface of the fluid $P_0$.

*NOTE: Our analysis of pressure at a depth used the assumption that the fluid is incompressible (i.e. the fluid has a constant density). This is a good assumption for liquids like water that do not compress much. In contrast, the atmosphere is still considered a fluid, but it is a gas which is easily compressible, thus our pressure at a depth analysis is only valid for very small depths.

Pascal's law

Pascal's law is formally stated as follows, "for an incompressible and enclosed fluid, a change in pressure at one location results in the same change in pressure, without a diminish in magnitude, at all locations within the enclosed fluid and walls".