11 Fundamental Properties of Fluids


Aerodynamics is the underpinning of atmospheric flight, so understanding aerodynamic principles, or, more generally, fluid dynamic principles, is one key to successfully designing all types of flight vehicles. Aeronautical and astronautical engineers must understand the behavior of fluids under a broad range of conditions. Fluids can be liquids or gases, and air is a gas. To understand the action of aerodynamic flows on flight vehicles, it is first necessary to become intimately familiar with the fundamental physical properties used to describe the behavior of fluids and the relationships between them. One must learn about fluid dynamics and aerodynamics before a deeper understanding of the other characteristics of flight vehicles becomes possible.

In all branches of science and engineering, properties are defined to help describe how things behave in the physical world. For example, concepts of mass, weight, energy, work, power, etc., are all essential physical properties relevant to physical behavior and problem-solving. The pertinent properties of fluids include pressure, density, temperature, viscosity, flow velocity, and the speed of sound. These also point properties in that their values can change from one spatial location to another in the fluid; they may also vary with respect to time at a given spatial location. They are called macroscopic properties. In this regard, they apply to bulk matter or a finite group of molecules rather than to each molecule, the matter having net physical dimensions much greater than the mean free path[1] between the molecules; this approach is known as a continuum assumption.

Furthermore, it must be recognized that these fluid properties will not be independent of one another, i.e., they will have some interdependencies, often through the effects of temperature. Changing the value of one property inevitably means that the values of other properties may also vary. For example, increasing the pressure of air, such as by compressing it, is accompanied by an increase in density and temperature. The relationships between pressure, temperature, and density can be established using thermodynamic principles and ideal gas laws that are formally encompassed in an equation of state.

Learning Objectives

  • Understand the continuum model for describing a fluid.
  • Become familiar with the parameters used to describe a fluid’s behavior, including pressure, density, temperature, viscosity, flow velocity, and the speed of sound.
  • Use the equation of state to relate gas properties, i.e., pressure, density, and temperature.
  • Understand the concept of viscosity and how to calculate the viscosity of a gas using Sutherland’s law.
  • Know the differences between diffusion and effusion of a gas.
  • Understand what streamlines are in a flow and how to calculate their locations.
  • Know how to calculate the speed of sound in a fluid.
  • Appreciate the concepts of surface tension and capillary action and how they can affect the behavior of fluids.

What is a Fluid?

Fluids are substances with mass and volume, but they have no predefined shape and can easily flow. They can be liquids (e.g., water) or gases (e.g., air). Unlike solids, fluids have relatively mobile molecules, as illustrated in the figure below. In a solid, the molecules are tightly packed in a lattice and have no mobility other than tiny vibrations around their fixed positions. Solids are primarily rigid and have shapes that are difficult to change under the action of external forces.

The difference between solids and fluids is that the molecules in a fluid are mobile and can be easily deformed by external forces.

However, the molecules are further apart and much more mobile in fluids. This characteristic means that fluids are easily deformed and will flow readily under the action of external forces. It is the tendency for fluids to flow and continuously deform under the action of an applied force that makes them more difficult to understand. Of particular interest to aerospace engineers is the gas called “air.” Like all gases, the air is composed of molecules that are relatively far apart. So, air can be relatively easily squeezed or compressed, which has many consequences for the aerodynamic characteristics of flight vehicles.

Continuum Versus Free Molecule Flow

To describe the behavior of a fluid, a molecular model of the flow is adopted (as in all branches of Newtonian mechanics) based on what is known as a continuum. In a continuum model, it is assumed that the distance between the individual fluid molecules, or more specifically, their mean free path, which can be denoted by the length scale \lambda, is tiny compared to the physical dimensions of the problem, as suggested in the figure below. With a continuum model, the macroscopic fluid properties, such as temperature, density, pressure, and flow velocity, can be considered constant at any point in space and vary continuously from point to point. In a continuum, any local changes associated with individual molecular motion are irrelevant, which is easily justified in most practical cases of fluid flows.

As pressure reduces, the distance between molecules increases, and as a vacuum is approached, the gas becomes rarified, making a continuum model invalid.

So, why does this distinction matter to aerospace engineers? For a gas such as air in the lower atmosphere, \lambda is of the order of 10-8 m. The physical dimensions of a flight vehicle flying in the lower atmosphere will be many orders of magnitude greater than \lambda. Therefore, in this case, the fluid flow about the vehicle would justifiably be considered as a continuum, and all of the standard macroscopic properties used to describe a fluid, such as pressure, temperature, etc., would apply.

However, consider a situation when \lambda becomes the same order as the flight vehicle’s length, such as at the edge of space where the air density is very low. At an altitude of 100 km, the value of \lambda is of the order of 0.1 to 1 m, and at the edge of space at 300 km, \lambda is greater than 100 m. This means that the air molecules would be spaced sufficiently far apart that interactions with the vehicle would occur only infrequently. In this case, the properties will not vary continuously from point to point, i.e., the flow cannot be assumed as a continuum, so it will behave differently to one this is.

The Knudsen number, K\!n, is often used to quantify such low-density flows. The validity of the continuum assumption is inherently tied to the collision rate of the molecules in the gas. Therefore, letting v_f represent the intermolecular collision rate or frequency, i.e., collisions per unit time, and t_f represent the characteristic flow time, then the Knudsen number (which is a non-dimensional similarity parameter) can be defined as

(1)   \begin{equation*} K\!n = \frac{1}{\nu_f \, t_f} \end{equation*}

Alternatively, if \overline{c} is the mean molecular speed, \lambda is the mean free path (the average distance traveled by a molecule before encountering a collision), and L represents a characteristic length scale, then

(2)   \begin{equation*} K\!n = \frac{1}{\nu_f \, t_f} = \frac{1}{ \left( \overline{c} / \lambda \right) \left(L / \overline{c} \right) } = \frac{\lambda}{L} \end{equation*}

Therefore, the ratio of \lambda to a characteristic length, L, becomes a measure of the degree of departure from a continuum.

Usually, when the value of K\!n is more significant than 0.01, it has been found that the continuum concept becomes increasingly invalid. When K\!n is greater than one and approaching 10, the conditions can be described as a free molecular flow. Under such conditions, the mean free path of the molecules becomes of the order of or greater than the characteristic length scale, and inter-molecular interactions are infrequent. As a result, the gas behavior cannot be explained in terms of macroscopically varying quantities; it must be described using rarefied gas model where the behavior of individual molecules must be described, usually using statistical models.

However, the definition of a characteristic length scale may need to be clarified. Choosing a length scale such as the flight vehicle’s length will give a global measure of how well the continuum assumption may apply at a given flight condition. Another choice is to use a length scale related to the local flow characteristics, in which case the continuum concept may become invalid at different flight conditions.

A continuum model is the most common model used to describe a fluid, and this assumption will be used throughout this ebook as it applies to the description of flow properties. However, spacecraft can encounter rarified or free molecule flows at the upper edges of the atmosphere where the Knudsen number is low. Intermediate cases of such flows (neither one nor the other) are often termed low-density flows. However, astronautical engineers are more likely to encounter problems with rarefied flows. For example, the drag on spacecraft in low Earth orbit at the outer edges of the atmosphere cannot be predicted using continuum gas assumptions.

Fluid Pressure

Remember that fluid is full of many relatively mobile molecules. In physical terms, the pressure can be thought of as the magnitude dF of the force d\vec{F} produced in a direction normal to this elemental area from the average reaction force (i.e., the time rate of change of momentum) of the molecules per unit time are impacting upon this surface.

Pressure at a point in a flow is the limiting case of force per unit area. The pressure can be assumed to be constant over a small elemental area, consistent with the continuum assumption.

Therefore, the pressure at point B in a fluid can be defined as

(3)   \begin{equation*} p = \lim_{dA \rightarrow \delta^2} \left( \frac{dF}{dA} \right) \mbox{\small ~with~$\delta \gg \lambda$} \end{equation*}

where \delta is a measurable dimension compared to the mean distance between the fluid molecules, i.e., \lambda.

This latter definition means that the pressure p is the limiting form of the time-averaged force per unit area as the area shrinks to a point, but that “point” is still big enough to be described by a continuum model. With the assumption of a continuum model, the area cannot shrink to zero because the pressure in the fluid will result from the random movement of individual molecules.

Notice that pressure can be interpreted as a normal compressive force per unit area, which can be recognized as equivalent to stress. The physical interpretation of pressure assumes that it is caused by the time rate of change of momentum of the fluid molecules, such as when they might strike the walls of a surface or a container. Higher or lower pressure would be associated with more or fewer molecules impacting a given surface area per unit of time. So, a large force must be exerted to create a large amount of pressure on a given area. Alternatively, this same force must be exerted over a small area (or do both) to get a higher pressure.

Pressure is also a point property, meaning that its value can differ from one point to another throughout the fluid. Therefore, the pressure becomes a function of the Cartesian spatial coordinates, i.e., p = p(x, y, z). Sometimes, the pressure value at a point may also change in time (time is given the symbol t), so more generally, pressure can be written as p = p(x, y, z, t). It is also essential to recognize that pressure is a scalar quantity (it has magnitude but no direction), and at a given point, the pressure has the same value in all directions, which is called Pascal’s Law.

How is the value of \delta quantified?

The term “measurable dimension” regarding pressure or other point properties can be considered equivalent to the diameter of the tip of one’s pen or pencil, so about 0.5 mm or 0.02 (twenty thousandths) of an inch. In the wind tunnel, pressure probes, pressure tubing, and the pressure measuring area of transducers are typically of this dimension. Anything larger means that point properties are not being recorded. Anything smaller means the pressure is acting over an area too small for a sufficiently large integrated effect to be recorded as a valid measurement.

Units of Pressure

Pressure has engineering units of N m^{-2} or Pascals (Pa) in the SI system or lb ft^{-2} (pounds per square foot) in the U.S. customary (USC) system. In practice, kilo-Pascals (kPa) units may be used because a Pascal is a small pressure value. Units of hectopascals are often used in barometric pressure measurements, where one hectoPascal (hPa) is equal to 100 Pascals. One hPa equals one millibar; one bar is 100,000 Pa or 100 kPa. However, a bar is not an SI unit, and its use should be avoided in engineering applications. In the USC system, units of lb in^{-2} (pounds per square inch) are also common; converting from units of lb in^{-2} to units of lb ft^{-2} requires a multiplication factor of 144, i.e. one lb in^{-2} = 144 lb ft^{-2}

Pressure as a Force

Sometimes, the effects of pressure may be interpreted as a force, i.e., as a quantity with magnitude and direction. However, pressure can be viewed as a force only when the area and the orientation of a surface are specified over which the pressure acts. Therefore, a line of action to resolve the effects of the pressure is also needed to find the force on a surface over which the given pressure acts.

For example, if the small elemental surface had an outward-pointing unit normal vector \vec{n}, as shown in the figure below, then the pressure force normal to the surface would be

(4)   \begin{equation*} d\vec{F} = ( p \, dA) \,  (-\vec{n})  = -p \, \vec{n} \, dA \end{equation*}

where the minus sign indicates that the pressure force will act inward in the opposite direction to the outward pointing direction of \vec{n}.

An interpretation of pressure as a force can be specified as the product of pressure and area acting inward in the opposite direction to the surface unit normal vector, \vec{n}.

What is a physical interpretation of a pressure force?

Consider what happens when scuba diving in the ocean or diving deeply into a swimming pool. The pressure exerted by the water above feels like a squeezing force on one’s body. Water is three orders of magnitude denser than air, i.e., it has many times more molecules per unit volume, so only relatively small changes in depth are required for one’s body to feel significant changes in pressure. One will feel this pressure first on the ears, which are very sensitive to pressure changes; they respond over several orders of magnitude. The Eustachian tubes, also called the auditory tubes or pharyngotympanic tubes, are a narrow passage that connects the middle part of the ear to the back of the nose, helping to regulate this external pressure. When the pressure outside the ear changes, one may feel discomfort or even pain as one’s ears adjust to the increased pressure, which can be achieved by yawing or swallowing. The opposite happens when one returns to the surface, which also requires changing the differential pressure in the Eustachian tubes to atmospheric pressure.

Check Your Understanding #1 – Calculation of a pressure force

A piston pushes down on a trapped volume of gas in a cylinder that has a diameter of 3 in. A pressure gauge measures the pressure in the gas to be 11 lb/in2. What is the force applied to the piston? Repeat the problem in SI units if the piston diameter is 76 mm and the pressure is 75.8 kPa.

Show solution/hide solution.

By definition, pressure, p, is the force, F, divided by area, A, so

    \[ p = \frac{F}{A} \]

which can be assumed to act uniformly. The area of the piston is

    \[ A = \frac{\pi D^2}{4} = \frac{\pi \times 3.0^2}{4} = 7.07~\mbox{in$^2$} \]

Therefore, the force in USC units will be the pressure times the area, i.e.,

    \[ F = p \, A = 11.0 \times 7.07 = 77.76~\mbox{lb} \]

In SI units, then A = 76 mm and p = 75.8 kPa, so the force will be

    \[ F = p \, A = (75.8 \times 10^3) \times (76.0 \times 10^{-3}) = 5,761~\mbox{N} = 5.76~\mbox{kN} \]

Fluid Density

Another essential property to describe a fluid’s characteristics is its density, given the symbol \rho or \varrho. Because the symbol p used for pressure looks similar to \rho, it is better to use \varrho for density to preserve clarity, especially when p and \varrho are used in the same equation.

Again, consider some point B in the fluid, as shown in the figure below. Let d\cal{V} be an elemental volume surrounding point B, and dm is the associated mass of fluid inside d\cal{V}. In the continuum assumption, millions of molecules will remain within the small elemental volume. Density is defined as the fluid mass per unit volume, which is also a measure of the number of molecules per unit volume. Mass is usually given the symbol m. Volume is given the symbol \cal{V}, i.e., a curly form of “V.” Notice: Do not confuse volume \cal{V} with velocity V, the latter usually being written in vector form, i.e., \vec{V}.

The flow density at a point depends on the mass of the fluid per unit volume.

The density of the fluid at point B is formally defined as

(5)   \begin{equation*} \varrho = \lim_{d{\cal{V}} \rightarrow \delta^3} \left( \frac{dm}{d\cal{V}} \right) \mbox{\small ~with~$\delta \gg \lambda$} \end{equation*}

where \delta is a large linear dimension compared with the mean distance between the molecules \lambda. Therefore, flow density means the ratio of the mass of a small volume of the fluid relative to the volume that contains it. Flow density is also a scalar quantity and, in general, like the pressure, it can be written, in general, that \varrho = \varrho(x, y, z, t).

Units of Fluid Density

Fluid density has units of kg/m{^{3}} (or more appropriately as kg m{^{-3}}) in the SI system or slugs ft{^{-3}} in the U.S. customary or USC system, where the slug is the base unit of mass.[2][/footnote] In dealing with aerodynamic problems, it is helpful to remember that air has a density of 1.225 kg m{^{-3}} or 0.002378 slugs ft{^{-3}} at sea level standard temperature and pressure. These values are at mean sea level (MSL) as defined in the International Standard Atmosphere (ISA) and are usually designated as the symbol \varrho_0.

Other types of measurement values of fluid density may be used in practice, mainly when dealing with liquids, which may be referenced to the density of water. These values include specific volume, specific weight, and specific gravity.

Specific Volume

The specific volume of a fluid is the reciprocal of its density and is given the symbol {\mathsf{v}} or SV and can be expressed as

(6)   \begin{equation*} {\mathsf{v}}  \equiv SV = \frac{\mbox{Volume~of~fluid}}{\mbox{Mass~of~fluid}} \equiv \frac{1}{\mbox{Density~of~fluid}} = \frac{1}{\varrho} \end{equation*}

The units of specific volume are volume per unit mass, so m{^{3}}/kg (i.e., m{^{3}} kg^{-1}) in the SI system or in the USC system it will have units of ft{^{3}}/slug (i.e., ft{^{3}} slug^{-1}).

Specific Weight

The specific weight is the weight of a unit volume of a fluid. It is often denoted by the symbol {\mathsf{w}} or SW, i.e.,

(7)   \begin{equation*} {\mathsf{w}}  \equiv SW = \frac{\mbox{Weight~of~fluid}}{\mbox{Volume~of~fluid}} = \frac{ \varrho \, {\cal{V}} \, g}{{\cal{V}}} = \varrho \, g \end{equation*}

Specific weight has units of weight per unit volume, so its value depends on acceleration under gravity or “g.” The units of specific weight are N/m{^{3}} (i.e., N m{^{-3}}) in the SI system or lb/ft{^{3}} (i.e., lb ft{^{-3}}) the USC system.

Specific Gravity

When dealing with liquids, their density is often measured relative to another fluid, called the specific gravity, SG. Usually, the reference is the density of water, so

(8)   \begin{equation*} SG = \frac{\mbox{Density~of~fluid}}{\mbox{Density~of~water}} = \frac{\varrho}{\varrho_{\rm water}} = \frac{\mbox{Weight~of~fluid}}{\mbox{Weight~of~water}} = \frac{\varrho \, {\cal{V}} \, g }{\varrho_{\rm water} \, {\cal{V}} \, g } \end{equation*}

Notice that specific gravity is a non-dimensional or unitless quantity. SG is the most common alternative measurement of fluid density.

Fluid Temperature

The temperature in a fluid is given the symbol T and is related to the average internal energy of the molecules at that point in the fluid. Temperature affects fluid properties differently, depending on whether the fluid is a liquid or a gas, as well as the molecular composition of the fluid. Temperature plays a vital role in aerodynamics, where the air is compressible, and aerodynamic heating from frictional effects can occur.

Again, the molecular model can help explain the concept of temperature. This relationship is usually written as E = (3/2) k_B \, T, where k_B is known as Boltzmann’s constant, i.e., this constant is the conversion factor that connects energy to temperature. The Boltzmann constant is defined as 1.380649×10−23 J K−1 in SI units, the Joule (J) being the unit of energy. Therefore, as shown in the figure below, a higher-temperature fluid would be one in which the molecules move about at relatively high speeds. In contrast, a lower-temperature fluid would be one with relatively low molecular speeds. Temperature is also a point, scalar property. In general, the temperature in a fluid will vary from point to point; temperature may also vary with time at a given point, i.e., T = T(x, y, z, t).

A fluid’s temperature depends on the average internal kinetic energy of its molecules.

The relationship that total energy E = (3/2) k_B \, T strictly holds for monoatomic gases, which have three degrees of translational freedom, i.e., kinetic energy only. Diatomic gases, such as nitrogen and oxygen, which comprise 98% of air, also have two degrees of freedom of rotational motion and two degrees of vibrational motion, the latter being important only at higher temperatures. Therefore, their total internal energy will be related using E = (5/2) k_B \, T at lower temperatures and E = (7/2) k_B \, T at higher temperatures.

Units of Temperature

Two fixed points are taken to construct a temperature scale. The first fixed point is the freezing point of water, called the lower fixed point or T_{\rm \mbox{\tiny LFP}}. The second fixed point is the boiling point of water, which is called the upper fixed point or T_{\rm \mbox{\tiny UFP}}. Temperature is measured in units of Centigrade or Celsius ^{\circ}C or Kelvin, K or ^{\circ}K, in the SI system or Fahrenheit ^{\circ}F or Rankine, R or ^{\circ}R, in the USC system.

Celsius or Centigrade Scale

This scale was devised by Anders Celsius in 1710. The interval between T_{\rm \mbox{\tiny LFP}} and T_{\rm \mbox{\tiny UFP}} is 100 units, where each unit is called one degree Celsius (1^{\circ}\mbox{C}). In this scale, then the lower fixed point is T_{\rm \mbox{\tiny LFP}} = 0^{\circ}C (freezing point of water), and the upper fixed point is T_{\rm \mbox{\tiny UFP}} = 100^{\circ}C (boiling point of water).

Fahrenheit Scale

This scale was devised by Gabriel Fahrenheit in 1717. The interval between T_{\rm \mbox{\tiny LFP}} and T_{\rm \mbox{\tiny UFP}} is 180 units, where each unit is called one degree Fahrenheit (1^{\circ}\mbox{F}). In this scale, then the lower fixed point is T_{\rm \mbox{\tiny LFP}} = 32^{\circ}F and the upper fixed point is T_{\rm \mbox{\tiny UFP}} = 212^{\circ}F.

Kelvin Scale

This scale was devised by William Thomson (later Lord Kelvin) in 1848. The zero temperature is absolute zero on this scale. It is the thermodynamic scale for use in SI units. The interval between T_{\rm \mbox{\tiny LFP}} and T_{\rm \mbox{\tiny UFP}} is 100 units, where each unit is called one degree Kelvin (1^{\circ}\mbox{K} or 1\mbox{K}). In this scale, then the lower fixed point is T_{\rm \mbox{\tiny LFP}} = 273.15^{\circ}K, and the upper fixed point is T_{\rm \mbox{\tiny UFP}} = 373.15^{\circ}K.

Rankine Scale

This scale was devised by William John Macquorn Rankine in 1859. In this scale, the zero temperature is also absolute zero. It is the thermodynamic scale for use in USC units. The interval between T_{\rm \mbox{\tiny LFP}} and T_{\rm \mbox{\tiny UFP}} is 80 units, where each unit is called one degree Rankine (1^{\circ}R or 1 R). In this scale, then T_{\rm \mbox{\tiny LFP}} = 492.67^{\circ}R, and T_{\rm \mbox{\tiny UFP}} = 672.67^{\circ}R.

Temperature Scale Conversions

Converting from one temperature scale to another is straightforward because they are all linearly related, as shown in the figure below; this figure should not be used for numerical calculations. Converting to Centigrade or Celsius ^{\circ}C from Fahrenheit ^{\circ}F is performed using

(9)   \begin{equation*} T (^{\circ}\mbox{C}) = \frac{5}{9} \bigg( T (^{\circ}\mbox{F}) - 32 \bigg) \end{equation*}

Converting to Fahrenheit ^{\circ}\mbox{F} from Centigrade or Celsius ^{\circ}\mbox{C} uses

(10)   \begin{equation*} T (^{\circ}\mbox{F}) = \frac{9}{5} \, T (^{\circ}\mbox{C}) + 32 \end{equation*}

Converting to Kelvin K from Centigrade or Celsius ^{\circ}\mbox{C} uses

(11)   \begin{equation*} T (^{\circ}\mbox{K}) = T (^{\circ}\mbox{C}) + 273.15 \end{equation*}

Finally, converting to Rankine R from Fahrenheit ^{\circ}\mbox{F} uses

(12)   \begin{equation*} T (^{\circ}\mbox{R}) = T (^{\circ}\mbox{F}) + 459.67 \end{equation*}


Converting temperature values from one scale to another is relatively easy because the scales are all linearly related.

Notice that it is often suggested that the degree symbol (^{\circ}) not be used when citing temperature units (especially for the Kelvin and Rankine scales). However, many publications can be found with and without the degree symbol. Nevertheless, retaining the degree symbol on the temperature units is entirely acceptable for students and others when working on aerodynamic problems. Finally, it is helpful to remember the standard sea level values of temperature (based on the ISA model), usually given the symbol T_0, which are 15^{\circ}C = 59^{\circ}F = 288.15^{\circ}K = 518.67^{\circ}R.

Using Temperatures in Engineering Problem-Solving

In engineering problem-solving, caution should be applied so that the correct absolute (engineering) units of temperature are used, i.e., units of Kelvin or Rankine, because these scales measure the temperature relative to absolute zero temperature, i.e., the temperature when the average internal energy and motion of the molecules becomes effectively zero. For example, for two temperatures T_1= 20^{\circ}\mbox{C} and T_2 = 40^{\circ}\mbox{C}, then the ratio T_1/T_2 is written correctly as

(13)   \begin{equation*} \frac{T_1}{T_2} = \frac{20 + 273.15}{40 + 273.15} = 0.936~\mbox{~(Correct!)} \end{equation*}

but incorrectly as

(14)   \begin{equation*} \frac{T_1}{T_2} = \frac{20}{40} = 0.5~\mbox{~(\bf Incorrect!!)} \end{equation*}

Why two absolute temperature scales?

Using two absolute temperature scales is a historical result of different conventions and preferences in various fields, such as physics and engineering. The Kelvin (K) absolute temperature scale, which was proposed in 1848 and based on the Celsius (C) unit, is named after Sir William Thomson, a professor of natural philosophy (physics) at the University of Glasgow, who later became Lord Kelvin. Interestingly enough, Kelvin was skeptical of the future of aviation, refusing to join the Royal Aeronautical Society, stating that “I have not the smallest molecule of faith in aerial navigation other than ballooning or of expectation of good results from any of the trials we hear of.” The Rankine (R) scale, also an absolute thermodynamic temperature scale, was proposed in 1859 based on the Fahrenheit (F) unit. This scale is named after the first University of Glasgow engineering professor, William J. M. Rankine.

Combined Gas Equation & Equation of State

Having introduced the concepts of pressure, density, and temperature, it should also be recognized that these quantities have interdependencies, i.e., changing one value affects the others. There are four gas laws that one learns in physics and chemistry, namely Boyle’s law, Charles’s law, Gay-Lussac’s law, and Avogadro’s law. These are all empirical gas laws because their relationships were derived from laboratory experiments with gasses.

Gas Laws

Boyle’s law states that for a constant mass, the volume of the gas is inversely proportional to its pressure, i.e.,

(15)   \begin{equation*} {\cal{V}} \ \propto \ \dfrac{1}{p} \quad \mbox{or} \quad  p \, {\cal{V}}  = \mbox{constant} \quad \mbox{or} \quad p_1 {\cal{V}}_1 = p_2 {\cal{V}}_2 \end{equation*}

where the conditions 1 and 2 are the initial and final conditions, respectively.

Charles’s law states that for a gas at a constant pressure, its volume is proportional to temperature, i.e.,

(16)   \begin{equation*} {\cal{V}} \ \propto \ T \quad \mbox{or} \quad  T \, {\cal{V}}  = \mbox{constant}  \quad \mbox{or} \quad T_1 {\cal{V}}_1 = T_2 {\cal{V}}_2 \end{equation*}

Gay-Lussac’s law states that at a constant mass and volume, pressure is proportional to temperature, i.e.,

(17)   \begin{equation*} p \ \propto \ T \quad \mbox{or} \quad  p \, T  = \mbox{constant}  \quad \mbox{or} \quad p_1 {\cal{V}}_1 = p_2 {\cal{V}}_2 \end{equation*}

Combined Gas Law

These three laws can be combined, leading to the combined gas law, which is written as

(18)   \begin{equation*} \dfrac{ p \, {\cal{V}}}{T} = \mbox{constant} \quad \mbox{or} \quad \dfrac{ p_1 \, {\cal{V}}_1}{T_1} = \dfrac{ p_2 \, {\cal{V}}_2}{T_2} \end{equation*}

This is a thermodynamic equation. All gases have properties that can be measured, including their pressure, temperature, and volume. Numerous scientific experiments and careful measurements have determined that these variables are quantifiably related to each other in that for a given mass of gas, if any two properties can be determined (i.e., they can be measured or calculated), the combined gas law can be used to determine (calculate) the other.

Equation of State

Avogadro’s law relates the amount of a gas, unusually in terms of moles, is proportional to its volume at a constant pressure and temperature, i.e.,

(19)   \begin{equation*} {\cal{V}}  \ \propto \  n \quad \mbox{or} \quad \dfrac{{\cal{V}}}{n} = \mbox{constant} \quad \mbox{or} \quad \dfrac{{\cal{V}}_1}{n_1} = \dfrac{{\cal{V}}_2}{n_2} \end{equation*}

where n is the mass of the gas expressed in moles. Remember that 1 mole = 6.022\times10^{23} atoms, which is called Avogadro’s number,[3] N_A, so a mole (a base unit) expresses the density of the gas.

These preceding relationships are now formally embodied in an equation of state, which determines the quantitative relationships between a gas’s pressure, density (or mass and volume), and temperature. In physics and chemistry, people inevitably first come across the use of the general or universal equation of state for a gas, i.e.,

(20)   \begin{equation*} p \, {\cal{V}} = n \, {\overline{R}} \, T \end{equation*}

where {\overline{R}} is called the universal gas constant, which is the same for all gases. The universal gas constant, {\overline{R}}, is related to the Boltzmann constant, k_B, by {\overline{R}} = k_B \, N_A.

However, this latter equation of state is not particularly useful for engineering purposes, especially for finding and relating fluid properties at a point. But, if both sides of this general equation are divided by the mass of the gas, then the volume now becomes the specific volume, {\mathsf{v}}, which is the reciprocal of the density of the gas. Recall that the “specific” in the term specific volume means “divided by mass,” i.e.,

(21)   \begin{equation*} {\mathsf{v}} = \frac{{\cal{V}}}{m} = \frac{1}{\varrho} \end{equation*}

Therefore, an alternative but equivalent equation of state for a gas can be written as

(22)   \begin{equation*} p {\mathsf{v}}  = R \, T \end{equation*}

which is more usually written as

(23)   \begin{equation*} p = \varrho \, R \, T \end{equation*}

Equation 23 is the usual form of the equation of state used in engineering, gas dynamics, and aerodynamics. Notice also that this result is independent of the volume of the considered gas. However, in this case, another (different) gas constant, R, is used, which equals the universal gas constant divided by the gas’s mass per mole, i.e.,

(24)   \begin{equation*} R = \frac{k_B \, N_A}{n} \end{equation*}

Therefore, the value of R in Eq. 23 is not universal and depends on the gas type, i.e., the value of the gas constant in this case is gas-specific. Subsequently, caution should be used to ensure that the correct value of R is used in engineering calculations for each gas being considered and also in the correct units.

However, it is useful to remember that all the forms of the equation of state are equivalent, i.e.,

(25)   \begin{equation*} p \, {\cal{V}} = n \,{\overline{R}} \, T \quad \equiv  \quad p \,  \mathsf{v} = R \, T \quad \equiv  \quad p = \varrho \, R \, T \end{equation*}

and which version to use depends, in part, on the problem being solved. Because point properties are often desired in engineering problems with fluids, the most common form of the equation of state is

(26)   \begin{equation*} p = \varrho \, R \, T \end{equation*}

What are the units of the gas constant?

In SI units, the gas-specific constant R, is measured in J kg^{-1} K^{-1}. In USC, the units of R are ft-lb slug^{-1}R^{-1}. A consistent set of units must be used throughout engineering calculations, for which base units of mass, length, and time are preferable. Remember that a Joule (J) is equivalent to a Newton-meter (N m), so it has base units of kg m^{2} s^{-2}.

The main advantage of the equation of state in the engineering form is that if the values of two quantities are known, e.g., pressure and temperature, which are both easily measured, it allows the calculation of the other quantity, i.e., the density is obtained from the equation

(27)   \begin{equation*} \varrho = \frac{p}{R \, T} \end{equation*}

Therefore, using the equation of state reduces the number of independent quantities from three to two. Thus, the equation of state can be interpreted as a two-dimensional surface in the p, \varrho and T state space, i.e., a surface defined by a function f(p, \varrho, T) = 0, as shown in the figure below, so that every point on the surface represents a unique equilibrium thermodynamic state of the gas.

The equation of state is one of the few “handy” equations in fluids engineering.

Strictly speaking, the preceding equation of state in Eq. 23 applies only to an ideal gas, i.e., one where the molecules are sufficiently far apart that intermolecular bonding forces are relatively low and that inter-molecular collisions are perfectly elastic. Under what might be called “normal conditions,” at temperatures and pressures reasonably close to standard atmospheric conditions, air behaves like an ideal gas. Generally, a gas nearly always acts as a perfect gas at normal to moderate temperatures and/or pressures. It behaves less ideally at very low temperatures and/or high pressures.

Check Your Understanding #2 – Calculation of air density

During measurements in a wind tunnel, the pressure and temperature of the air in SI units are found to be 102.3 kPa and 15.7^{\circ}C, respectively. Calculate the density of the air in the tunnel. Repeat the problem if the pressure is measured in USC units as 14.61 pounds per square inch (psi) at a temperature of 71.1^{\circ}F.

Show solution/hide solution.

Because this question involves pressure, temperature, and density, we will use the equation of state, i.e.,

    \[ p = \varrho \, R \, T \]

where p is pressure, \varrho is density, T is absolute temperature, and R is the gas constant, in this case, for air. Rearranging for the density gives

    \[ \varrho = \frac{p}{R \, T} \]

The first part of the problem is in SI units. In this case, the absolute temperature is 15.7 + 273.15 = 288.85 K. The gas constant R for air in SI units is 287.057 J kg^{-1} K^{-1} so the density of the air will be

    \[ \varrho = \frac{p}{R \, T} = \frac{102.3 \times 10^3}{287.057 \times 288.85} = 1.2304~\mbox{kg m${^{-3}}$} \]

Remember that for engineering calculations, we must always use absolute temperature.

The second part of the problem is in USC units. In this case, the absolute temperature is 71.1^{\circ}F + 459.67 = 530.77 R. The pressure is given in terms of common units of pounds per square inch (psi), so to convert to base USC units of pounds per square foot (psf or lb/ft^{2}), it is necessary to multiply by 144. The gas constant for air in USC units is 1716.49 ft-lb slug^{-1}R^{-1} so the density of the air will be

    \[ \varrho = \frac{p}{R \, T} = \frac{14.61 \times 144.0 }{1716.49 \times 530.77} = 0.002309~\mbox{slugs ft${^{-3}}$} \]

Bulk Modulus

The bulk modulus, denoted by the symbol K_b, is a measure of a substance’s resistance to uniform compression, such as under the action of a pressure or some external body force per unit mass. The bulk modulus is equivalent to a compressive “stiffness” and is an essential property in fluid mechanics. In particular, its value is related to the speed of sound propagation in liquids, which plays a fundamental role in various scientific, industrial, and medical imaging[4], and other technological applications[5], where precise measurement and understanding of acoustic properties are crucial.

When a fluid of original volume {\cal{V}} is subjected to a change in external pressure \Delta p, it undergoes a volume change \Delta {\cal{V}}, the relationship being given by

(28)   \begin{equation*} \Delta  p = - K_b \left( \dfrac{\Delta {\cal{V}}}{ {\cal{V}}} \right) \end{equation*}

where the coefficient K_b is called the bulk modulus. Notice that the minus sign means a positive change in applied pressure will decrease the fluid volume, as illustrated in the figure below. In the limit when the change in pressure becomes infinitesimally small, then Eq. 28 becomes

(29)   \begin{equation*} dp = - K_b \left( \dfrac{d {\cal{V}}}{ {\cal{V}}} \right) \end{equation*}

Therefore, the bulk modulus is defined as

(30)   \begin{equation*} K_b = -{\cal{V}} \left( \dfrac{dp}{d{\cal{V}}} \right) \end{equation*}


The value of the bulk modulus is proportional to how difficult it is to compress the fluid under the action of an external pressure.

Conservation of mass of the fluid volume also requires that

(31)   \begin{equation*} \varrho \, {\cal{V}} = \left( \varrho + d \varrho \right) \left( {\cal{V}}  - d {\cal{V}} \right) \end{equation*}

Expanding out and simplifying gives

(32)   \begin{equation*} \varrho\,  d \varrho  - {\cal{V}}  \, d \varrho = 0 \end{equation*}


(33)   \begin{equation*} \dfrac{ {\cal{V}}}{d {\cal{V}}} = \dfrac{\varrho}{d\varrho} \end{equation*}

Therefore, the bulk modulus can also be defined as

(34)   \begin{equation*} K_b = -\varrho \left( \dfrac{dp}{d \varrho} \right) \end{equation*}

In other words, Eqs. 30 and 34 quantify the degree of compressibility of the original volume of fluid under pressure. A high value of the bulk modulus indicates that the substance is difficult to compress, i.e., it is essentially incompressible. It will also be apparent that the bulk modulus has units of pressure.

Liquids have relatively high values of bulk modulus because their molecules are closely packed together and resist pressure and other external forces that can cause compression. Indeed, for all practical purposes, liquids cannot be compressed significantly (their values of bulk moduli are in the GPa or Giga-Pascal range), so, with few exceptions, they can be considered incompressible fluids.

Gases have much lower values of bulk modulus because their molecules are further apart with larger spaces between them, so they can be compressed much more easily. In general, gases cannot be considered as being incompressible fluids, and the value of the bulk modulus of a gas varies depending on its pressure and temperature. The most relevant bulk modulus used in practice is the isothermal (constant temperature) bulk modulus, which can be written as

(35)   \begin{equation*} K_T = -{\cal{V}} \left( \dfrac{dp}{d{\cal{V}}} \right)_T \end{equation*}

The isothermal gas law (Boyle’s law) states that p \, {\cal{V}} = constant, so differentiating with respect to {\cal{V}} using the product rule gives

(36)   \begin{equation*} p \, \dfrac{d{\cal{V}}}{d{\cal{V}}}   + {\cal{V}} \, \dfrac{dp}{d{\cal{V}}}  = p + {\cal{V}} \, \dfrac{dp}{d{\cal{V}}} = 0 \end{equation*}

so that

(37)   \begin{equation*} \dfrac{dp}{d{\cal{V}}} = -\dfrac{p}{{\cal{V}}} \end{equation*}

Substituting this result into Eq. 35 gives

(38)   \begin{equation*} K_T = -{\cal{V}} \left( \dfrac{dp}{d{\cal{V}}} \right)_T = {\cal{V}} \left(  \dfrac{p}{{\cal{V}}} \right) = p \end{equation*}

This means that the isothermal bulk modulus of a gas is its pressure, i.e., the “stiffness” or resistance to compression is directly proportional to its pressure. As the gas’s pressure increases, its resistance to further compression increases proportionally.

Fluid Viscosity

The property of viscosity can be viewed as the fluid’s resistance to shear when different parts of the fluid are in relative motion, i.e., its internal friction or resistance to being deformed. Viscosity can also be viewed as a measure of fluidity, i.e., the higher the viscosity, the lower the fluidity. The symbol \mu (i.e., the Greek symbol “mu”) is a constant known as the coefficient of dynamic viscosity, or more simply, just the fluid’s viscosity. All fluids have viscosity to a lesser or greater degree, so for fluid in relative motion, the property of viscosity causes shear forces to be produced within the fluid. Gases generally have a much lower viscosity than liquids, perhaps a fairly obvious expectation. Some liquids are very viscous, e.g., molasses, corn syrup, oil, etc. However, the effects of viscosity come into play in the behavior of all types of fluids.


One way to begin to understand the concept of viscosity is to consider a demonstration with the three columns of oil, as shown in the figure below. Each oil has a different viscosity, ranging from SAE 20 (the thinnest and less viscous) to SAE 40 (the most viscous); SAE is the Society of Automotive Engineers. Suppose a heavy steel ball is dropped into the oil. In that case, it will descend at a velocity proportional to its viscosity, the oil causing shear stresses on the ball’s surface, giving it a viscous drag as it moves downward under gravity.

An illustration of the viscous nature of a fluid, where the ball will drop slower in a fluid of higher viscosity.

The balance of forces in equilibrium descent is such that the weight of the ball, W_b, less any buoyancy force, B_b, is equal to the viscous drag on the ball, D_{\mu}. The weight will be density times volume times acceleration under gravity, i.e.,

(39)   \begin{equation*} W_b = \varrho_b \left( \frac{4}{3} \pi R^3 \right) \,g \end{equation*}

where \varrho_b is the density of the steel ball. The (upward) buoyancy force on the ball (Archimedes’s principle) will be equal to the weight of oil displaced by the ball, i.e.,

(40)   \begin{equation*} B_b = \varrho_{\rm oil} \left( \frac{4}{3} \pi R^3 \right) \,g \end{equation*}

Therefore, the equilibrium equation is

(41)   \begin{equation*} W_b - B_b - D_{\mu} = 0 = \frac{4}{3} \pi R^3 \, g \left( \varrho_b -\varrho_{\rm oil} \right) - D_{\mu} \end{equation*}

The drag force D_{\mu} on a sphere of radius R dropping through a fluid of high viscosity \mu at low speed V_d (this is called a creeping flow) is given by Stokes’s law, i.e.,

(42)   \begin{equation*} D_{\mu} = 6 \pi R \mu V_d \end{equation*}

where \mu is the viscosity. Therefore, in equilibrium, then

(43)   \begin{equation*} \frac{4}{3} \pi R^3 \, g \left( \varrho_b -\varrho_{\rm oil} \right) - 6 \pi R \mu V_d = 0 \end{equation*}

and rearranging to solve for V_d gives

(44)   \begin{equation*} V_d = \frac{2 ( \varrho_b - \varrho_{\rm oil}) }{9 \, \mu} \, g \, R^2 \end{equation*}

For a ball of the same weight and size, Eq. 44 shows it will drop in the oil at a velocity that is inversely proportional to the oil’s viscosity, \mu, i.e., the higher the viscosity, the slower the ball drops. The forgoing is the principle used in the falling-sphere viscometer. The time it takes for a steel sphere of known size and weight can be measured using two lines on the tube, from which the ball’s velocity, V_d, is determined. Stokes’ law (Eq. 42) can be used to determine the viscosity, \mu, of the oil (or other liquid) from the resulting velocity by knowing the size and weight of the sphere as well as the density of the liquid, i.e.,

(45)   \begin{equation*} \mu = \frac{2 ( \varrho_b - \varrho_{\rm oil}) }{9 \, V_d} \, g \, R^2 \end{equation*}

Units of Viscosity

Viscosity has units of kg m-1 s-1 or Nm-2s or Pa s (i.e., a Pascal-second often called the Poiseuille, named after Jean Léonard Marie Poiseuille) in the SI system, or slug ft-1 s-1 in the USC system. However, the unit of viscosity typically employed in practice is called the “Poise” (P) or gram cm-1 s-1. The unit of poise is 1 Poise = 0.1 Pa s = 1 Poiseuille, i.e., 1 Pa s = 10 Poise. The viscosity of liquids is usually a low numerical value, so it is often reported in units of centipoise (cP). In contrast, the viscosity of gases, which are much less viscous than liquids, is usually reported in units of micropoise (μP). In base units, at MSL ISA then \mu_0 = 1.789 x 10-5 kg m-1 s-1 = 1.789 x 10-5 Pa s = 3.737 x 10-7 slug ft-1 s-1.

Shear in a Fluid

In Newton’s Philosophiæ Naturalis Principia Mathematica of 1687, he uses the Latin word tenacitas to mean viscosity. He then defines the concept of viscosity as “Resistentia, quae oritur ex inopia lubricitatis partium fluidi, caeteris paribus, proportionalis est velocitati, qua partes fluidi separantur ab invicem,” which can be translated as “The resistance which arises from the lack of slipperiness originating in a fluid, all other things being equal, is proportional to the velocity by which the parts of the fluid are being separated from each other.”

Newton performed experiments like that shown in the figure below with a fluid of depth h between a moving upper plate and a stationary lower plate, i.e., a flow containing a velocity gradient in one direction with the velocity in the fluid increasing moving up from one point to another. The upper (faster) layer draws the lower (slower) layer along through a force on the lower layer, so there must be a shear force in the fluid between the layers. Simultaneously, the lower layer tends to retard the upper layer by an equal and opposite force (i.e., Newton’s Third Law).


The viscosity of the fluid is related to its resistance to shearing deformation.

The fluid next to the bottom plate wants to stay at rest, and the fluid touching the top plate is dragged along (because of viscosity) with the velocity U, i.e., a velocity gradient forms the fluid between the two plates. Maintaining this gradient requires the application of a force, F, where

(46)   \begin{equation*} F \ \propto \ \frac{A \, U}{h} \end{equation*}

where A is the area of the plate. Notice that the ratio U/h is the slope of the velocity profile or the velocity gradient. In terms of force per unit area, which is a stress, \tau, then

(47)   \begin{equation*} \tau = \frac{F}{A} = \mu \left( \frac{U}{h} \right) \end{equation*}

where the constant of proportionality, \mu, is the “stiffness” or viscosity of the fluid.

In general, for the straight and parallel motion of a given fluid, the tangential stress produced between two adjacent fluid layers is proportional to the velocity gradient in a direction perpendicular to the layers, i.e.,

(48)   \begin{equation*} \tau =\mu \left( \frac{du}{dy} \right) \end{equation*}

where {u} is the velocity at some distance y. The quantity du/dy is the {u} velocity gradient in the y direction. The velocity gradient du/dy is equivalent to a strain rate, so this preceding equation is just a statement of a fluid’s linear stress/strain rate relationship. Equation 48 is called Newton’s law of viscosity. Remember that maintaining a velocity gradient and the shear stresses in a fluid requires continuous application of a force; if the force were to stop, the shear stresses would become zero.

To explain this latter point further, consider a fluid element as it flows in a fluid with a velocity gradient, as shown in the right-side figure above. If du/dy is positive, then the upper surface of the element will move faster than the lower surface, so over some time dt, the upper AC surface will travel further than the lower surface DE by a distance

(49)   \begin{equation*} dx = \left( \frac{du}{dy} \right)\, dy \, dt \end{equation*}

Consequently, a shear deformation or strain is produced in the fluid. This strain can be calculated from the geometry of the deformation shown in the figure above. The shear, \gamma, which is the angle between the lines EB and EC is

(50)   \begin{equation*} \gamma = \frac{ \left( \displaystyle{\frac{du}{dy}} \right)\, dy \, dt}{dy} = \left( \frac{du}{dy} \right)\, dt \end{equation*}

to a small angle approximation. Rearranging the equation gives

(51)   \begin{equation*} \frac{d \gamma}{dt} = \frac{du}{dy} \end{equation*}

and so the shear stress in the fluid is

(52)   \begin{equation*} \tau=\mu \left( \frac{d \gamma}{dt} \right) \end{equation*}

This latter result is another way of writing Newton’s law of viscosity. Notice that the shear stress depends on the strain rate, i.e., d \gamma/ dt. Remember that the shear stress in a solid is proportional to strain, so a constantly applied strain will create constant deformation and stress. In a fluid, however, shear stress is only produced by a strain rate because the fluid flows and deforms continuously.

Velocity Gradients in a Fluid

Newton’s law of viscosity should be written more precisely using the partial derivative on the velocity gradient, i.e., it should be written as

(53)   \begin{equation*} \tau=\mu \left( \frac{\partial u}{\partial y} \right) \end{equation*}

because the {u} velocity in a fluid may vary in other directions as well, e.g., the flow is three-dimensional, so there could be {u} velocity gradients in the x and z directions, i.e., \partial u/\partial x, and \partial u/\partial z. In general, these gradients can be written in the matrix (or tensor) form as

(54)   \begin{equation*} \frac{\partial u^i}{\partial x^j} = \left[ \begin{array}{ccc} \displaystyle{  \frac{\partial u}{\partial x}} & \displaystyle{ \frac{\partial u}{\partial y} }  & \displaystyle{ \frac{\partial u}{\partial z} } \\[12pt] \displaystyle{  \frac{\partial v}{\partial x} } & \displaystyle{ \frac{\partial v}{\partial y} } & \displaystyle{  \frac{\partial v}{\partial z} } \\[12pt] \displaystyle{  \frac{\partial w}{\partial x} } & \displaystyle{ \frac{\partial w}{\partial y} } & \displaystyle{  \frac{\partial w}{\partial z} } \end{array} \right] \end{equation*}

The velocity gradient tensor is symmetric, meaning that the cross-derivatives are related, i.e., \partial u /\partial y = \partial v /\partial x, \partial u /\partial z =\partial w / \partial x, and \partial v / \partial z = \partial w / \partial y. This symmetry reduces the number of unique velocity gradients from nine to six.

What is the kinematic viscosity?

Dynamic viscosity is a measure of shear resistance, i.e., viscosity only matters with motion or dynamics. In many fluid problems involving viscosity, the magnitude of the viscous forces compared to the magnitude of inertia forces is critically important, that is, the forces causing an acceleration of the fluid. Because the viscous forces are proportional to \mu and the inertia forces are proportional to \varrho, the ratio of \mu/ \varrho is often involved in solving the problem. This ratio of \mu to density \varrho is called the kinematic viscosity and given the symbol \nu (Greek symbol “nu”), i.e.,

    \[ \nu = \frac{\mu}{\varrho} \]

Therefore, the kinematic viscosity is a derived parameter. Values of kinematic viscosity have units of m^2s^{-1} in the SI system or ft^2s^{-1} in the USC system.

Mechanisms of Viscosity

The viscous properties of a fluid arise from two sources: 1. Intermolecular momentum transfer between the molecules; 2. Bonding between the molecules. Therefore, the viscosity of a fluid depends on whether it is a gas or a liquid, i.e., its characteristics depend primarily on the physics associated with the mean relative spacing between the molecules.

The molecules are relatively close together in a liquid but not as mobile as in a gas. In this case, viscosity results more from molecular bonding and less from intermolecular momentum transfer. As shown in the schematic below, stronger bonding results in higher resistance to deformation, i.e., higher viscosity. In general, intermolecular bonding can be influenced by factors such as the molecules’ size, molecular weight, strength of their bonds, and the temperature of the liquid.

In a liquid, the source of viscosity is forces produced by intermolecular bonding, which become stronger for a more viscous fluid.

For example, liquids with large, heavy molecules tend to have a higher viscosity than liquids with small, light molecules because the larger molecules have more intermolecular bonds and are more resistant to flow. Liquids like benzene, diethyl ether, gasoline, ethanol, and water flow very readily and have a low viscosity. Others, such as honey, heavy oils, glycerin, motor oil, molasses, and maple syrup, flow very slowly and have a high viscosity. There is also a correlation between viscosity and molecular shape. Liquids of long, flexible molecules tend to have higher viscosities than those of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, increasing the viscosity of the liquid.

The molecules are relatively further apart in a gas, so the bonding is much lower. The mechanism of viscosity results from intermolecular momentum transfer as the fairly mobile molecules diffuse throughout the gas. Nevertheless, gases are still viscous and exhibit the characteristics of viscosity. This effect becomes more apparent when the gas has an initial velocity gradient, as shown in the schematic below.

In a gas, the source of viscosity is intermolecular momentum transfer from the diffusion of molecules between gas layers, which manifests as resistance to deformation.

The random motion of the gas molecules between fluid layers means that collisions inevitably occur, and a momentum interchange occurs such that slower molecules gain some momentum from the faster molecules. The consequence of intermolecular momentum transfer is a shear force between the gas layers in regions of velocity gradient and resistance to further deformation, which manifests viscosity. Remember that fluids must be continuously deformed to produce stresses, so in the absence of any additional shear rates, the velocity gradients will diminish as the momentum interchange balances throughout the layers of the gas.

Newtonian Versus Non-Newtonian Fluids

For a fluid in which the value of \mu is constant and independent of the strain rate (i.e., \mu is independent of the magnitude of the velocity gradient), it is called a Newtonian fluid. Many fluids, including air and water, behave as Newtonian fluids, i.e., the fluid behaves so that it obeys a linear stress/strain relationship as shown in the figure below, and, therefore, the value of \mu can be assumed constant. Remember again that in a fluid, the shear stresses produced by viscosity are related to the strain rate produced in the fluid by its deformation.

The stress/strain rate for various fluids. Air behaves as a Newtonian fluid in that its viscosity remains independent of its shear rate.

However, not all fluids behave in this linear way. For example, fluids such as certain oils, blood, inks, and most paints behave nonlinearly or in a non-Newtonian manner in that their viscosity changes as a function of the strain rate. The viscosity may increase or decrease with an increasing strain rate, depending on the nature of the fluid. For example, a dilatant fluid experiences shear thickening, and a pseudoplastic fluid exhibits shear thinning. The behavior of such non-Newtonian fluids is less well understood, so their behavior is less predictable, but they have many practical uses. For example, paints behave as pseudoplastic fluids that experience thinning when mixed or agitated during brushing or spraying. After application, the paint thickens, so it does not run off the surface. The study of non-Newtonian fluids is called rheology.

Why is ketchup so hard to get out of the bottle?

Ketchup is infamous for being hard to get out of a bottle unless you know the secret. Shake it first! In the bottle, ketchup is a reasonably viscous fluid and does not pour easily because of the relatively strong cohesive bonding between the ketchup molecules, which includes various polymer chains. To liquefy ketchup, vigorously shake the bottle, which agitates and deforms the fluid and creates a strain rate. This process also stretches the polymer molecules to experience less cohesive bonding, reducing viscosity. The process takes a few seconds, but you can easily pour the ketchup onto your burger. Today, ketchup often comes in squeeze bottles, in which the ketchup is forced through an orifice by creating pressure in the bottle. This process also creates a strain rate, which reduces the viscosity of the ketchup so it can be poured. After this point, the ketchup thickens again, so it does not run off your burger!

Effects of Temperature on Viscosity

Temperature significantly affects the magnitude of the viscosity of both gases and liquids. Consequently, the viscous characteristics of gases change differently from liquids when subjected to changes in temperature, as shown in the figure below. For example, the viscosity of a liquid generally decreases with increasing temperature; this effect occurs because of the reduction in bonding forces as the motion of the molecules causes them to move further away from each other, which dominates over any increase in inter-molecular momentum transfer. Consequently, liquids become less viscous and easier to flow with increasing temperature. However, gases, including air, typically show increased viscosity with increasing temperature because of the increase in intermolecular momentum transfer; this effect increases the resistance to the fluid’s deformation.

The effects of temperature on the viscosity of a fluid depend on whether it is a gas or a liquid. The viscosity of gases increases with increasing temperature.

Sutherland’s Law of Viscosity for Gases

The coefficient of dynamic viscosity \mu (or often just the coefficient of viscosity (or just the viscosity) for a gas can be calculated using Sutherland’s formula or Sutherland’s law. This empirical (i.e., experimentally derived) law was first published in 1893 and can be written as a function of absolute temperature, T, as

(55)   \begin{equation*} \mu (T) = \mu_{\rm ref} \left( \frac{T}{T_{\rm ref}} \right)^{1.5} \left( \frac{T_{\rm ref} + S}{T+S} \right) \end{equation*}

where reference values (subscript “ref”) are in appropriate SI or USC units. The parameter S is known as Sutherland’s constant. A graphical interpretation of Sutherland’s law is shown in the figure below.

Sutherland’s law is a semi-empirical equation representing the variation in the viscosity of gases with respect to the gas’s temperature.

For air in SI units, then T_{\rm ref} = 323^{\circ}K are \mu_{\rm ref} = 1.716 \times 10^{-5} kg m^{-1} s^{-1} (also known as units of Pa~s), with a Sutherland constant of S = 110.0^{\circ} K. In USC units at T_{\rm ref} = 518.67^{\circ}R, then S = 198.72^{\circ}R and \mu_{\rm ref} = 3.737 \times 10^{-7} slugs s^{-1} ft^{-1}. Coefficients for other gases are widely available in reference books and online data sources.

Sutherland’s law is widely used in various engineering fields to predict the viscosity of gases at different temperatures. It applies over various temperatures for air and gases such as oxygen, nitrogen, and helium. However, it is essential to note that it is not universally applicable to all gases. For example, Sutherland’s law does not apply to gases that exhibit significant deviations from ideal behavior, such as rarefied gases or gases containing large molecules. Additionally, Sutherland’s law does not apply to liquids, which show a much more complex relationship between viscosity and temperature because of the effects of intermolecular bonding. In general, Sutherland’s law should be used cautiously, and its applicability should be verified for each specific gas and temperature range of interest.

Check Your Understanding #3 – Calculating the value of viscosity

If a measurement in air gives a temperature of 52^{\circ}F, calculate the dynamic viscosity coefficient. Hint: Use Sutherland’s Law. What happens to the viscosity of air as its temperature increases, and why?

Show solution/hide solution.

Sutherland’s Law can be expressed as

    \[ \mu (T) = \mu_{\rm ref} \left( \frac{T}{T_{\rm ref}} \right)^{1.5} \left( \frac{T_{\rm ref} + S}{T+S} \right) \]

where T_{\rm ref} = 518.67^{\circ}R, S = 198.72^{\circ}R and \mu_{\rm ref} = 3.737 \times 10^{-7} slugs s^{-1} ft^{-1}. In this case, the absolute temperature is

    \[ 52 \ ^{\circ}\mbox{F} \equiv  511.67 \ ^{\circ}\mbox{R} \]

Inserting the values gives

    \[ \mu = 3.737 \times 10^{-7}  \left( \frac{511.67}{518.67} \right)^{1.5} \left( \frac{518.67 + 198.72}{511.67 + 198.72} \right) \]


    \[ \mu = 3.698\times 10^{-7}~\mbox{ slugs s$^{-1}$ ft$^{-1}$} \]

Bonding between the molecules in a gas is relatively low compared to a liquid. Therefore, the intermolecular momentum transfer between the molecules increases more with increasing temperature, which manifests as an increase in viscosity.

Temperature Effects on the Viscosity of Liquids

The Andrade equation, named after British physicist Edward Andrade, is a commonly used semi-empirical model to predict the effects of temperature on the viscosity of liquids. This equation relates the viscosity of a liquid to temperature based on an exponential relationship, i.e.,

(56)   \begin{equation*} \mu (T) = A \exp \left( \frac{B}{T} \right) \end{equation*}

where \mu is the viscosity of the liquid and T is the absolute temperature, and the coefficients A and B depend on the liquid. Values of A and B for this two parameter model are widely published for many liquids. This equation represents the behavior that viscosity decreases with increasing temperature for a liquid.

There are also other semi-empirical models for the effects of temperature on the viscosity of liquids. The three-parameter model is

(57)   \begin{equation*} \mu = A \exp \left( \frac{B}{T - C} \right) \end{equation*}

and the four-parameter model is

(58)   \begin{equation*} \mu = A \exp \left( \frac{B}{T} + C \, T + D \, T^2 \right) \end{equation*}

Again, the coefficients, A, B, C, and D for most liquids can be found in published sources.

What does it mean when my car needs to use 20W50 engine oil? 

Engine oils are classified by their viscosity grades, designated by numbers such as SAE 5 to SAE 50 or higher; the higher the number, the higher the viscosity (thicker oil). Using a single-grade oil, such as SAE 50, can lead to problems in extreme temperature conditions. The oil must maintain adequate viscosity at high temperatures to provide sufficient lubrication and prevent engine wear. However, a highly viscous oil like SAE 50 can become too thick at low temperatures and may not properly lubricate the engine components. To address these temperature-related issues, most modern oils are formulated as multigrade oils denoted by a combination of two numbers, such as SAE 20W50. The “W” represents winter and indicates the oil’s viscosity at low temperatures. The oil has an SAE 20 oil viscosity at lower temperatures, providing better initial lubrication of the engine parts. At higher temperatures, it maintains the viscosity of an SAE 50 oil to offer adequate lubrication. This characteristic involves blending additives into the base oil that modify the intermolecular interactions and control its temperature-dependent viscosity.

Flow Velocity

In fluid dynamics, the focus is on fluids in motion, so the velocity of the fluid is a significant quantity that must be defined carefully. By definition, a velocity is a vector quantity, so the velocity of any given fluid packet will have both a magnitude (a speed) and a direction. When the concept of the velocity of a fluid is considered, which will have relative motion between the fluid packets, then its velocity becomes more subtle to describe than for a solid body where all the parts will move in unison.

For example, for a solid body in translational motion, it is evident that all points of the body will be traveling at the same velocity, i.e., with the same speed and direction. However, different parts of a fluid in motion will most likely travel at varying velocities so that they will be in relative motion. This is just one reason why the motion of fluids is somewhat more challenging to describe, both physically and mathematically.

Consider the flow about an airfoil and try to follow the path of a small group of fluid particles initially upstream of the airfoil at point 1, as shown in the figure below. This group is called a fluid element because it is a small elemental flow volume. The speed and direction of this fluid element will change as it moves downstream from point 1 to point 2, point 2 being nearer to the nose of the airfoil. Point 2 at the nose is called a stagnation point because the air is brought to rest and stagnates. Therefore, the flow velocity at 1 or 2 or 3, or any other point, is just the velocity of an infinitesimally small fluid element as it passes through that point.

A fluid element’s velocity depends on its speed and direction; therefore, flow velocity is a vector quantity.

Velocity has magnitude and direction, but it is still a point property in that its value will change from point to point in the flow, and it can also change with respect to time, i.e., \vec{V} = \vec{V} (x, y, z, t). Flow velocities are measured in units of m s^{-1} in the SI system or ft s^{-1} in the USC system.


Tracking where the fluid moves in space and time is essential to understanding flow problems. A pathline is the trajectory or path a fluid element traces in time. By definition, a streamline is a line drawn tangential to the local velocity vector field, i.e., there is no flow perpendicular to a streamline. In steady flow problems, when the flow properties do not change with respect to time, called a steady flow, then the pathlines in the fluid are identical to the streamlines.

The equation of a streamline is straightforward to determine. For a two-dimensional flow in the xy plane, then \vec{V} = (u, v) so the slope of a streamline is just dy/dx = v/u, which is an ordinary differential equation. Therefore, this differential equation could be solved with the known velocity field to trace a streamline in a given plane. In three-dimensions, i.e., in x, y and z space then \vec{V} = (u, v, w). In this case, a direction vector, say d\vec{l} = (dx, dy, dz), can be defined that points along the streamline, i.e., in a direction parallel to the streamline, as shown in the figure below.

A streamline is a curve drawn parallel to the local velocity field of the fluid flow. Hence, by definition, there is no flow perpendicular to streamline.

By definition, there is no flow across a streamline, so the equation of a streamline in three-dimensional space is just

(59)   \begin{equation*} d\vec{l} \times \vec{V} = \vec{0} \end{equation*}

noting that \vec{0} is the zero vector.

The meaning of this latter equation becomes clearer by expanding out the vector equation in terms of its scalar components, i.e.,

(60)   \begin{equation*} d\vec{l} \times \vec{V} = (w dy - v dz) \, \vec{i} + (u dz - w dx) \, \vec{j} + (v dx - u dy) \, \vec{k}  = \vec{0} \end{equation*}

In terms of the components, then

(61)   \begin{eqnarray*} \frac{dz}{dy} = \frac{w}{v} \quad & \mbox{or} & \quad   \frac{dz}{w} = \frac{dy}{v}  \\[8pt] \frac{dz}{dx} = \frac{w}{u} \quad & \mbox{or} & \quad   \frac{dz}{w} = \frac{dx}{u}  \\[8pt] \frac{dy}{dx} = \frac{v}{u} \quad & \mbox{or} & \quad   \frac{dy}{v} = \frac{dx}{u} \end{eqnarray*}

so it will be apparent that the equation of a streamline is

(62)   \begin{equation*} \frac{dx}{u} = \frac{dy}{v} = \frac{dz}{w} \end{equation*}

Check Your Understanding #4 – Calculating the equation of a streamline

If a two-dimensional velocity field in the xy plane is defined as (u,v) = (y, -x), then what are the mathematical equations of the streamlines?

Show solution/hide solution.

In this case, the governing equation for the streamline is

    \[ \frac{dy}{dx} = \frac{v}{u} = \frac{-x}{y} \]

Separating the variables and integrating them gives

    \[ \int y dy = -\int x dx + C_1 \]

where C_1 is a constant, so then

    \[ \frac{y^2}{2} = -\frac{x^2}{2} + C_1 \]

or just

    \[ x^2 + y^2 = \mbox{~constant} = C_2 \]

which for different values of C_2 are concentric circular streamlines centered around the origin.

In a more general sense, the streamline equations are often solved using a parameter t, which can be thought of as time or another parameter that varies along the streamline, i.e.,

(63)   \begin{equation*} \frac{dx}{dt} = u(x, y, z), \quad \frac{dy}{dt} = v(x, y, z), \quad \frac{dz}{dt} = w(x, y, z) \end{equation*}

To find the streamline, integrate each equation with respect to t, i.e.,

(64)   \begin{equation*} x(t) = x_0 + \int_{t_0}^{t} u\bigg( x(t), y(t), z(t) \bigg) \, dt \end{equation*}

(65)   \begin{equation*} y(t) = y_0 + \int_{t_0}^{t} v \bigg( x(t), y(t), z(t) \bigg) \, dt \end{equation*}

(66)   \begin{equation*} z(t) = z_0 + \int_{t_0}^{t}  w \bigg( x(t), y(t), z(t) \bigg) \, dt \end{equation*}

where (x, y, z) = (x_0, y_0, z_0) is the starting point of the streamline at t = t_0.

Numerical integration methods available include Euler’s method, Runge-Kutta methods, etc. Second-order Runge-Kutta methods are commonly used to calculate streamlines. If the time step is small enough, an explicit Euler method will be sufficient for calculating the streamlines in two dimensions. For example, for the x component, a one-step explicit method for a given time step \Delta t is of the form

    \[ x^{(t+\Delta t)} = x^t + u(x^t,y^t) \, \Delta t \]

and for the y component

    \[ y^{(t+\Delta t)} = y^t + v(x^t,y^t) \, \Delta t \]

where t represents the current time step and t +\Delta t represents the next time step. This algorithm is easily programmed, and if {u} and {v} are given as some simple function of x and y, then the streamlines can be solved. However, the value of \Delta t must be small enough to prevent errors from accumulating in the values of x and y, and some trial and error may be involved. Some trial and error may also be needed to find suitable initial points to integrate from so that the nature of the flow field becomes apparent.

MATLAB has various solvers for initial value problems for ODEs and can be used to trace streamlines in simple flows; an example is shown below. However, the accuracy of the streamline calculations depends strongly on the spatial and quality of the velocity field that can be calculated or measured and not just the accuracy of the numerical method.


An example of the numerical tracing of streamlines from measurements of a velocity field made using the Particle Image Velocimetry (PIV) method.


A streamtube is a concept used in fluid dynamics to represent a bundle of streamlines. These streamlines delineate the boundaries of a particular flow or control volume as a tubular structure, as shown in the figure below. Within a streamtube, the mass flow rate of fluid entering one end must equal the mass flow rate exiting at the other end by the principle of conservation of mass. This means that the mass flow rate within the boundaries of the streamtube is constant; there is no mass flow over the boundaries of a streamtube. Streamtubes are often considered axisymmetric flows in that the fluid properties are radially symmetric or uniform at any one section. A uniform axisymmetric flow through a streamtube is a common assumption when learning fluid dynamics.

A streamtube is considered to be a tubular collection of individual streamlines.

Speed of Sound

The speed of sound in any medium depends on how quickly vibrational energy can be transferred from molecule to molecule through the medium. All gases are compressible, so pressure disturbances produced at one point will quickly propagate to another point but at a finite speed. This propagation speed is called the speed of sound, given the symbol a, and its value differs from gas to gas.

Speed of Sound in a Gas

It can be shown, in general, that the speed of sound is related to changes in pressure and density of the fluid medium, i.e.,

(67)   \begin{equation*} a = \sqrt{\frac{dp}{d\varrho}} \end{equation*}

which applies at constant entropy, i.e., the vibrational energy transfer is frictionless and reversible. Because liquids and solids are very difficult to compress and change their density, the speed of sound in such media is generally greater than in gases, e.g., sound travels about four times faster in water than in air.

The speed of sound in a gas depends on the temperature of the gas. Indeed, it can be shown that the speed of sound is proportional to the square root of its absolute temperature, i.e.,

(68)   \begin{equation*} a = \sqrt{\gamma R T } \end{equation*}

where \gamma is the ratio of specific heats, and R is the gas constant. Notice that the values of \gamma and R differ for different gases; a useful table is given below.

Thermodynamic properties of some gases.
Gas \gamma R ( J kg ^{-1} ^{\circ}K^{-1}) R (ft lb slug^{-1} ^{\circ}R^{-1})
Air 1.4 287.05 1717.0
Nitrogen 1.4 296.8 1775.0
Hydrogen 1.41 4124.2 24663.0
Helium 1.66 2077.1 12421.0
Oxygen 1.395 259.84 1554.0
Carbon Dioxide 1.289 188.92 1130.0
Carbon Monoxide 1.4 296.84 1775.0

Caution should be exercised to ensure that in equations involving the gas constant R, the value of R is not only for the correct gas but also in the appropriate engineering units. For air the gas constant, R, is 286.9 J kg^{-1} ^{\circ}K^{-1} in the SI system and 1716.49 ft lb slug^{-1} ^{\circ}R^{-1} in the USC system. Also, \gamma = 1.4 for air, which is non-dimensional.

Check Your Understanding #5 – Calculating the speed of sound in a gas

At 300^{\circ}C, estimate the speed of sound in: (a) nitrogen, (b) hydrogen, and (c) helium. Hint: The ratio of specific heats and the gas constants for these gases are listed in the table above.

Show solution/hide solution

(a) For nitrogen, \gamma = 1.40, R = 296.8 J kg ^{-1} ^{\circ}K^{-1}, and T= 300^{\circ}C+ 273.15 = 573.15^{\circ}K.

    \[ a = \sqrt{\gamma \, R \, T} = \sqrt{ 1.40 \times 296.8 \times 573.15} = 488.0~\mbox{m/s} \]

(b) For hydrogen, \gamma = 1.41, R = 4124.2 J kg^{-1} ^{\circ}K^{-1}, and T= 573.15^{\circ}K.

    \[ a = \sqrt{\gamma \, R \, T} = \sqrt{ 1.41 \times 4124.2 \times 573.15} = 1,825.63~\mbox{m/s} \]

(c) For helium, \gamma = 1.66, R = 2077.1 J kg ^{-1} ^{\circ}K^{-1}, and T= 573.15^{\circ}K.

    \[ a = \sqrt{\gamma \, R \, T} = \sqrt{ 1.66 \times 2077.1\times 573.15} = 1,405.8~\mbox{m/s} \]

Doppler Effect

Sound is a pressure disturbance, and the speed of sound propagation in any gas at a given temperature will be constant. Let the frequency of the sound source be f_s, which will be the frequency of the sound heard by any listener if the sound source is stationary. However, the perceived frequency of sound propagation will change if the location of the sound source S and the listener locations (comparing locations L1 and L2) are in relative movement to each other, which is known as the Doppler effect, as illustrated in the figure below.

Illustration of the Doppler effect.
The perceived frequency of sound will depend on the relative velocity between the location of the sound source and a listener’s location, i.e., a Doppler effect.

The frequency heard by the listener, f_L, is given by

(69)   \begin{equation*} f_L = \left( \dfrac{a + V_L}{a + V_S} \right) f_S = \left( \dfrac{1 + M_L}{1 + M_S} \right) f_S \end{equation*}

where V_S and M_s are the relative velocity and Mach number of the source, respectively, and V_L and M_L are the relative velocity and Mach number, respectively, of the listener, and f_S is the frequency of the sound source. It is important that the correct signs are used on V_S and V_L. If V_L = 0 (listeners are stationary), then the value of V_S will be positive if the sound source moves away from the listener, and V_S will be negative if the source moves toward the listener. Therefore, as shown in the figure above, the listener at L_1 will hear a higher frequency than the listener at L_2.

Notice that when the sound source starts to move supersonically, i.e., when V_S = a, the frequency becomes infinite, which means that the sound becomes concentrated over a wavefront, known as a Mach wave. In three dimensions, this wavefront is called a Mach cone. In this case, a listener upstream of the sound source would not hear it coming until the wavefront reaches the listener, in which case there would be a loud “bang” as the wavefront passes by, as shown in the figure below.

If the sound source is moving supersonically, then the sound energy will coalesce along a wavefront called a Mach wave. In three-dimensions, this is a conical wavefront.

In this case, the lower frequency heard by a stationary listener after the passage of the sound source traveling at a Mach number of M_s will be

(70)   \begin{equation*} f_L = \left( \dfrac{1}{1 + M_S} \right) f_S \end{equation*}

The speed of sound is critical for all flight vehicles, which create pressure disturbances as they fly. The airspeed of an aircraft relative to the speed of sound affects the flow physics and the forces produced on the aircraft, this ratio being called the flight Mach number, M_{\infty}. If the aircraft flies much slower than the speed of sound, the conditions are said to be subsonic, and compressibility effects are minor. However, if the aircraft moves faster and approaches or exceeds the speed of sound, called supersonic, compressibility effects become essential, and the flow physics will change. In this case, the issue of the “sonic boom” generated by the aircraft also becomes a consideration for people on the ground.

Speed of Sound in a Liquid

The speed of sound in a liquid varies depending on the type of liquid and its properties, including its density, \varrho, and its bulk modulus, K_b. The speed of sound in a liquid can be calculated using

(71)   \begin{equation*} a = \sqrt{\frac{K_b}{\varrho}} \end{equation*}

where the bulk modulus of the liquid, which, as previously explained, is a measure of its incompressibility. Equation 71 is often referred to as the Newton-Laplace equation.

Sound travels much faster in media with higher values of bulk modulus because it more readily transmits pressure changes. For example, the speed of sound in freshwater at average room temperature is approximately 1,482 m/s (~4,850 ft/s). The speed of sound in seawater is generally higher because of the presence of salts and minerals, averaging around 1,533 m/s (~5,029 ft/s) at similar temperatures. In general, the speed of sound can vary widely in different liquids; e.g., in ethanol, it is around 1,160 m/s (~3,807 ft/s), and in mercury, it is about 1,450 m/s (~4,760 ft/s). For reference, some numerical values of the bulk modulus for liquids are given in the table below, and others are available online or at other authoritative sources. Notice that bulk modulus has units of pressure. However, it should be appreciated that the temperature and impurities in the liquid can also affect its speed of sound.

Numerical values of the bulk modulus of some common liquids.
Liquid K_b (Pa) K_b (lb/ft^2) \varrho (kg/m^3) \varrho (lb/ft^3) a (m/s) a (ft/s)
Freshwater 2.2 \times 10^9 4.59 \times 10^7 1,000 62.4 ~1,482 ~4,850
Seawater 2.4 \times 10^9 5.02 \times 10^7 1,025 64.0 ~1,533 ~5,029
Ethanol 1.06 \times 10^9 2.22 \times 10^7 780 49.2 ~1,160 ~3,807
Mercury 2.85 \times 10^{10} 5.96 \times 10^8 13,600 849 ~1,450 ~4,760
Glycerol 4.5 \times 10^9 9.41 \times 10^7 1,260 78.7 ~1,920 ~6,299
Benzene 1.07 \times 10^9 2.23 \times 10^7 876 54.7 ~1,280 ~4,200
Methanol 0.81 \times 10^9 1.69 \times 10^7 792 49.4 ~1,100 ~3,609
Olive oil 1.3 \times 10^9 2.72 \times 10^7 920 57.4 ~1,370 ~4,485
Acetone 1.16 \times 10^9 2.42 \times 10^7 790 49/3 ~1,210 ~2,964
Carbon tetrachloride 1.3 \times 10^9 2.72 \times 10^7 1,590 99.3 ~930 ~3,051

Diffusion & Effusion

One interesting characteristic of fluids, is their rapid diffusion or mixing ability at the molecular level, even without turbulence or other agitation. Gaseous molecules travel at relatively high speeds, so they collide frequently with other molecules as they travel in many directions. Understanding and predicting the diffusion of gases and other fluids, as well as aerosols[6] or particulates suspended in fluids (colloids),[7] is fundamental to many engineering applications.


Diffusion is a process in which fluid molecules move through a concentration gradient from an area of higher concentration to a lower one until they are more evenly mixed; the idea is shown in the figure below. This movement is driven by the kinetic energy and natural mixing of the molecules and continues until a uniform equilibrium is reached. Diffusion is generally faster in gases because the molecules are further apart and move more freely, especially in gases with low molecular weight. Diffusion at the molecular level is relatively slow, but its rate is considerably enhanced by turbulence or other forced processes such as convection.

Diffusion is a process by which fluid molecules spread out evenly in a space because of their random motion.

The diffusion characteristics of fluids can be predicted using Fick’s laws, after the work of Adolf Fick. In the case where the concentration gradient is time-invariant, the diffusion flux, J, which is the amount of substance per unit area per unit time, is proportional to the negative gradient of the concentration, C, i.e.,

(72)   \begin{equation*} J = -D \, \frac{dC}{dx} \end{equation*}

where D is called the diffusion coefficient, i.e., a measure of the diffusivity. Equation 72 is called Fick’s first law of diffusion. This law is purely empirical in origin, and was developed by following the work of Thomas Graham. The diffusion coefficient, D, depends on the properties of the fluid, including its viscosity and temperature. Fick’s second law applies to non-steady-state diffusion, where the concentration within the diffusion medium changes with time, which is expressed by

(73)   \begin{equation*} \frac{\partial C}{\partial t} = D \, \frac{\partial^2 C}{\partial x^2} \end{equation*}

Check Your Understanding #6 – Diffusion of a gas in air

An initial volume of ammonia gas diffuses along a 1 m length of duct that has a cross-sectional area of 0.35 m^2. The opposite end of the duct initially has a 10% ammonia concentration compared to the other end. The temperature and pressure are constant. The diffusivity of ammonia in air under these conditions is 2.2 \times 10^{-5} m^2/s. Estimate the rate of diffusion of the ammonia through the duct. Approximately how long will it take before the ammonia becomes uniformly diffused?

Show solution/hide solution.

The concentration gradient is given by

    \[ \frac{dC}{dx} = \frac{C_1 - C_2}{L} = \frac{1 - 0.1}{1.0} = 0.9 \ \text{mol}~\text{m}^{-4} \]

The flux J is

    \[ J = -D \left( \frac{dC}{dx} \right) = -2.2 \times 10^{-5} \times 0.9 = -1.98 \times 10^{-5} \ \text{mol} ~\text{m}^{-2}~\text{s}^{-1} \]

where the diffusivity is 2.2 x 10-5 m2 s-1; this means that ammonia diffuses through air at a rate of 2.2 x 10-5 square meters per second, which is quite low. Therefore, in this case the rate of diffusion will be

    \[ \overbigdot{n} = J \times A = -1.98 \times 10^{-5} \times 0.35 \, \text{m}^2 = -6.93 \times 10^{-6} \, \text{mol} ~\text{s}^{-1} \]

The diffusion time is given by

    \[ t \approx \frac{L^2}{6D} \]

which comes from Fick’s second law. Substituting values gives

    \[ t \approx \frac{(1 .0)^2}{6 \times 2.2 \times 10^{-5}} \approx 7,575 \, \text{s} \]

Therefore, the time it will take for ammonia to uniformly diffuse throughout the 1 m long section is approximately 2.1 hours.


Effusion refers to the escape of fluid molecules through a porous surface or a tiny hole, the underlying process being similar to diffusion, as shown in the figure below. In this case, molecules move through the holes in a metered fashion at a rate related to their kinetic energy and molecular size, i.e., the faster the molecules move and the smaller they are, the more frequently they will pass through a given size of hole in a given time, and so move from one side of the porous surface to the other.

Effusion is a process by which fluid molecules diffuse through a porous media.

The behavior of effusion, as well as diffusion, can be predicted using Graham’s law. Graham’s law states that the rate of effusion or diffusion of a given gas is inversely proportional to the square root of its relative molecular mass, i.e.,

(74)   \begin{equation*} \frac{r_1}{r_2} = \sqrt{ \frac{M_{r_{2}}}{M_{r_{1}} } } \end{equation*}

where {r_1} and {r_2} are the rates of effusion or diffusion of gases 1 and 2, respectively, and M_{r_{1}} and M_{r_{2}} are the relative molecular masses of gases 1 and 2, respectively.

The proof of this result can be developed from the kinetic theory of gases, which states that the average kinetic energy of gas molecules is proportional to their absolute temperature, T, i.e.,

(75)   \begin{equation*} KE_{\text{av}} = \frac{1}{2} m \,  V^2 \end{equation*}

where m is the mass of a gas molecule and V is the average root-mean-square speed of the molecules. The average kinetic energy of one mole of gas (which contains N_A molecules, where N_A is Avogadro’s number) is also proportional to the absolute temperature, i.e.,

(76)   \begin{equation*} KE_{\text{av}} = \frac{3}{2} \, k_B \, T \end{equation*}

where k_B is Boltzmann’s constant. Because the kinetic energy for one molecule or for one mole are proportional to the temperature, then

(77)   \begin{equation*} KE_{\text{av}} = \frac{1}{2} \,  m \,  V^2 = \frac{3}{2} \,  k_B \, T \end{equation*}

A gas molecule’s mass m can be related to its relative molecular mass M_r by m = M_r/N_A. Substituting gives

(78)   \begin{equation*} \frac{1}{2} \left( \frac{M_r}{N_A} \right) V^2 = \frac{3}{2} \,  k_B \,  T \end{equation*}

and solving for V^2 gives

(79)   \begin{equation*} V^2 = \frac{3 \,  k_B \,  T \,  N_A}{M_r} \end{equation*}

Because k_B \,  N_A = R, where R is the universal gas constant, then

(80)   \begin{equation*} V = \sqrt{\frac{3 \,  R \,  T}{M_r}} \end{equation*}

showing that the root-mean-square speed, V, of the molecules is inversely proportional to the square root of their relative molecular mass, M_r.

It can be deduced, therefore, that the rate of effusion, r, of a gas is proportional to the root-mean-square speed V of its molecules, i.e.,

(81)   \begin{equation*} r \ \propto \ V \end{equation*}

So, for two different gases (1 and 2) at the same temperature, then

(82)   \begin{equation*} \frac{r_1}{r_2} = \sqrt{ \frac{M_{r_{2}}}{M_{r_{1}}} } \end{equation*}

which is known as Graham’s law. Stated in words, Eq. 82 shows that the effusion rate is inversely proportional to the square root of its relative molecular mass.

Graham’s law also holds for diffusion, albeit only approximately, but has been proven to work well in practice. Likewise, the effusion or diffusion of one gas mixed with another can be predicted only approximately with Graham’s law, because the interactions between the different gas molecules affect the combined rate. Notice that Fick’s laws focus on the diffusion process through a medium, driven by concentration gradients. Graham’s law focuses on the rate of diffusion in terms of their molar masses.

Thomas Graham’s original work, which pertained to the effusion and diffusion of gases, can also be adapted for liquids. Indeed, in practice, Graham’s law represents the diffusion or effusion of a wide range of liquids, albeit approximately. However, it is essential to note that the processes of effusion and diffusion in liquids are somewhat more complicated than gaseous diffusion, because of the more frequent intermolecular interactions and viscosity effects, i.e., cohesive bonding. In practical applications, the precise prediction of diffusion rates in liquids often requires some empirical data.

Graham’s law of effusion is also known to work for gases that exit through somewhat larger holes and orifices, in which case the law is often modified to

(83)   \begin{equation*} r = \dfrac{k_h}{ V} \end{equation*}

where k_h is an empirical constant related to the type of gas and the size and shape of the hole or opening. If different gases exit through the same hole size, then the ratio of their effusion rates (but not their rates per se) can be predicted using Eq. 82.

Surface Tension & Capillary Action

Surface tension is the tendency of a liquid-gas interface, or a free-surface between two immiscible fluids, to behave like a stretched elastic membrane or “skin.” Capillary action is directly related to surface tension, which is usually understood as describing the behavior of a liquid-gas interface in narrow spaces, usually with the formation of a surface meniscus. Capillary action and surface tension are both phenomena that arise from the relative effects of the behavior of intermolecular forces in fluids. The fluid property of surface tension is widely used in various applications, including lubrication, painting or coating technologies, printing with inks, the production of emulsions and foams, bubble formation, microfluidics, and various manufacturing processes.

Surface Tension

There is a natural tendency for all liquids to minimize their net energy state by minimizing their surface area. For example, the often-observed behavior of a liquid droplet on a horizontal surface wetted by a liquid, such as water “beading” on a surface covered with a thin film of oil or wax, is because of surface tension effects.

Consider the figure below, which shows beads of water sitting on a solid surface. The physics of surface tension can be explained in terms of the relative effects of cohesive forces between the liquid molecules and the adhesive forces between the liquid and the gas at the interface. Away from the surface, the liquid molecules are pulled uniformly in every direction by the cohesive forces from neighboring molecules, giving zero resultant force. As the interface or free surface is approached, however, the forces on the molecules are no longer uniform. At all the liquid–gas interfaces, there is a much greater attraction between the liquid molecules (cohesion) than to the molecules in the gas (adhesion). Therefore, in this case, the free surface of the water tends to be pulled inward by the water molecules below the surface, giving it a characteristic surface curvature.

The larger droplets are “flatter” because the effects of gravity increase more quickly, and so overcome the effects of surface tension.

For small water droplets sitting on a surface, as shown in the figure above, it will be noticed that the droplet is almost perfectly spherical because, in this configuration, there is the least surface area for a given volume. For larger droplets, the shape becomes somewhat flatter and bulges because of the increasingly important effects of gravity, which is roughly proportional to the weight of the droplet, i.e., proportional to a^3 where a is the approximate radius of the droplet.

This latter situation is very similar to that of a water-filled balloon; the water’s weight accounts for the gravitational effect, and the balloon’s skin stretches as the water volume increases, miming the effects of surface tension. Because the surface area is proportional to a^2, the gravitational to surface tension ratio depends on the ratio a^3/a^2 = a; this latter effect becomes increasingly important for larger droplets.

Surface tension, which is usually represented by the symbol \sigma (although sometimes \gamma is used – so symbol conflict arises again), has units of force per unit length. The base units of surface tension are Newtons per meter (N/m) in SI units, or pounds per foot (lb/ft) in USC units. The Young-Laplace equation describes the pressure difference across a curved liquid-gas interface resulting from surface tension.

Consider a small, curved element of the liquid surface, i.e., part of a spherical or cylindrical cap defined by two principal radii of curvature, r_{1} and r_{2}, as shown in the figure below. Although signs for these values can vary, sign convention usually dictates positive curvature when it is convex and negative when it is concave. The curvature of the surface affects how the surface tension forces act. The surface tension acts tangentially along the boundary of the patch. For an infinitesimal boundary length dl, the force from surface tension is

(84)   \begin{equation*} F_{\rm st}= \sigma \, dl \end{equation*}

A model of the curved liquid-gas interface resulting from the effects of surface tension.

The pressure difference across the surface, \Delta p, which because it is a liquid/gas interface, is usually referred to as the capillary pressure or the Laplace pressure, creates a normal force on the patch, \Delta p \, dA. The surface tension force must balance the capillary pressure force to maintain equilibrium. The net inward force from surface tension on the patch can be approximated by considering the contributions from both radii of curvature, r_{1} and r_{2}.

It can be shown that the force from surface tension in one direction is proportional to \sigma and inversely proportional to r_{1}. This outcome is because a smaller radius of curvature (sharper curve) results in a larger force. Similarly, the other radius of curvature, r_{2}, also contributes an inward force proportional to \sigma but is inversely proportional to r_{2}.

Therefore, the total force from the surface curvature must balance the capillary pressure, i.e.,

(85)   \begin{equation*} \sigma \left( \frac{dA}{r_1} + \frac{dA}{r_2} \right) = \Delta p \, dA \end{equation*}

and so leading to

(86)   \begin{equation*} \Delta p = p_{\rm liq} - p_{\rm gas} = \sigma \left( \frac{1}{r_1} + \frac{1}{r_2} \right) \end{equation*}

This latter relationship is called the Young-Laplace equation. Notice that in the case of spherical shapes such as bubbles or droplets of radius R, then r_1 = r_2 = R and so

(87)   \begin{equation*} \Delta p = \dfrac{2 \sigma}{R} \end{equation*}

For cylindrical cases, then r_1 = R and r_2 = 0, so

(88)   \begin{equation*} \Delta p = \dfrac{\sigma}{R} \end{equation*}

In the general case, for a free surface and where there is an applied “over-pressure,” \Delta p, at the gas/liquid interface in equilibrium, there is a balance between the applied pressure, the hydrostatic pressure, and the effects of surface tension. In this case, the Young-Laplace equation is modified to give

(89)   \begin{equation*} \Delta p = \varrho \, g \ z - \sigma \left( \frac{1}{r_1} + \frac{1}{r_2} \right) \end{equation*}

It should be appreciated that the magnitude of the surface tension depends on the type of liquid and gas and their temperatures. Generally, liquids with stronger intermolecular forces exhibit higher values of surface tension. Water, for example, has a relatively high surface tension because of the significant cohesive bonding between its hydrogen molecules. Surface tension is an important physical property of most liquids. It plays a crucial role in a variety of engineering and scientific problems, including droplet formation, capillarity and meniscus formation, bubble dynamics, surface coatings, and biological and medical applications.

Values of surface tension for a liquid-air interface at 25oC.
Fluid \sigma (N/m) \sigma (lb/ft)
Water 0.0728 0.00498
Ethanol 0.0223 0.00153
Methanol 0.0226 0.00154
Glycerol 0.0631 0.00432
Mercury 0.485 0.0332
Benzene 0.0289 0.00198
Acetone 0.0237 0.00162
Olive Oil 0.032 0.00219

Check Your Understanding #7 – Calculating surface tension

A student does experiments in the lab to study the properties of surface tension. The student finds that cylindrical steel needles of different diameters, d, and lengths, L, will “float” on the water’s surface. A close inspection shows the needle is not under the water, so the floatation effect is not because of buoyancy but entirely of surface tension effects. The student eventually finds that a needle with a diameter of more than 1.6 mm will “break” the surface tension and sink in the water; this outcome is also independent of the length of the needle, L. Use this information to estimate the surface tension of water, \sigma_w. The density of steel, \varrho_s, is 7,830 kg/m{^3}.

Show solution/hide solution

The needle deforms the water surface, and the surface tension forces, F_{\rm st}, act vertically upward, as shown in the figure. If these tensions are assumed to give a resultant force that is nearly vertical, then force equilibrium between the weight of the steel needle and the surface tension, F_{\rm st} gives

    \[ 2 F_{\rm st} = \varrho_s \, g \, {\cal{V}} = \varrho_s \, g \left( \frac{\pi}{4} d^2 L \right) \]

Introducing the surface tension value, \sigma_w, gives

    \[ 2 \sigma_w \, L = \varrho_s \, g \left( \frac{\pi}{4} d^2 L \right) \]

for which \sigma_w is to be determined for d = 1.6 mm = 0.0016 m. Rearranging the equation gives

    \[ \sigma_w = \varrho_s \, g \left( \frac{\pi}{8} d^2 \right) \]

thereby confirming that the surface tension effects will not depend on the length of the needle. Inserting the known numerical values gives

    \[ \sigma_w = 7,830 \times 9.81 \times \left( \frac{\pi}{8} 0.0016^2 \right) = 0.077 ~\mbox{N/m} \]

This value of \sigma_w compares favorably with the accepted value of 0.0728 N/m for water-air at a temperature of 25^{\circ}C.

Capillary Action

Narrow diameter cylindrical tubes are called capillary tubes. If one end of these tubes is dipped in a liquid, as shown in the figure below, it is observed that the liquid in the capillary “wets” the inside of the tube and either rises or falls relative to the surrounding liquid level. This phenomenon is called capillary action. Capillary action occurs at a molecular level because of the intermolecular cohesive attraction between the molecules of the liquid and the adhesive attraction between the capillary walls. Leonardo da Vinci first observed capillary action with water in the late 15th century; he was obsessed with the behavior of water and made hundreds of observations although made no determination of any physical laws. Today, the physics of capillary action arises in many diffident engineering and scientific fields, including microfluidics.

Capillary action results from surface tension and the ratio between cohesion and adhesion, the net balance of which will cause a liquid to rise of descend in a tube.

In capillary action, there is a balance of three main forces at play, namely adhesion, cohesion, and surface tension. Adhesion happens when different molecules are attracted to each other. Cohesion is when molecules that are the same cling to each other. If any of the two phenomena, i.e., surface tension and the ratio between cohesion and adhesion, increase, the liquid will rise in the capillary tube. Furthermore, the narrower the tube, the higher the liquid will rise.

Another interesting feature of capillary tubes is the meniscus formation at the liquid-gas interface. A concave meniscus occurs when the attraction between the particles of the liquid and the container (adhesion) is more than the attraction of the particles of the liquid to each other (cohesion), causing the liquid to climb the walls of the container. This behavior occurs between water and glass; any water-based fluids will have a concave meniscus in glass or other wettable containers. A convex meniscus occurs when the adhesion energy is less than the cohesion energy. Convex menisci arise, for example, between mercury and glass, which was apparent in the old mercury thermometers.

Jurin’s law, named after the 18th-century physicist James Jurin, mathematically describes the capillary action observed when a liquid rises or falls in a narrow tube. The physics involve the balance between adhesive forces (attraction between the liquid and the walls) and cohesive forces (the attraction between the liquid molecules). When a narrow tube is placed vertically into a liquid, it is found that the liquid either rises or falls. If the liquid wets the tube (such as water in a glass tube), then the rise will rise; if it does not wet the tube (such as mercury in a glass tube), then it will fall. According to Jurin’s law, the height h to which the liquid rises or falls is given by

(90)   \begin{equation*} h = \frac{2 \sigma \cos \theta}{\rho g r} \end{equation*}

where \sigma is the surface tension of the liquid, \theta is the contact angle between the liquid and the tube, \rho is the density of the liquid, g is the acceleration under gravity, and r = d/2 is the radius of the tube. Surface tension, \sigma, is the force per unit length at the surface of a liquid because of molecular attractions, causing the liquid surface to behave like a stretched elastic membrane. The contact angle, \theta, is the angle at which the liquid interface meets the solid surface, being less than 90o for a liquid that wets the surface and greater than 90o for one that does not. The smaller the diameter d or radius of the tube, r, the higher the liquid will rise or fall, as capillary forces have a stronger effect in narrower tubes.

Summary of Properties of Air at MSL Standard

Values of the properties of air at standard mean sea-level (MSL) conditions are helpful to have on hand, and these are given in the table below. They are often called MSL ISA conditions, where ISA means International Standard Atmosphere. Air properties at MSL ISA conditions are often used as a standard reference.

Table of ISA Properties at MSL.
Property Symbol SI units USC units
Pressure p_0 1.01325\times10^{5} Nm^{-2} (Pa) 2116.4 lb/ft^2 (29.92 inches of Hg)
Density \varrho_0 1.225 kg m{^{-3}} 0.002378 slugs ft{^{-3}}
Temperature T_0 288.15 K 518.67 R
Dynamic viscosity \mu_0  1.789\times10^{-5} kg m^{-1} s^{-1} 3.737\times10^{-7} slugs ft^{-1} s^{-1}
Speed of sound a_0 340.3 m s^{-1} 1116.47 ft s^{-1}
Gas constant R 287.057 J kg^{-1} K^{-1} 1716.49 ft-lb slug^{-1}R^{-1}

Summary & Closure

The study of fluids and their behavior underpins a wide range of engineering disciplines and is necessary for solving many practical engineering problems, especially in the aerospace field. The relationships between pressure, density, temperature, viscosity, flow velocity, and the speed of sound help engineers predict and understand fluid behavior in different conditions, such as the flow around airfoils, wings, and complete aircraft. The thermodynamic equation of state helps to calculate fluid properties, considering the interrelated effects of pressure, temperature, and density. The concept of viscosity is important to understand, most physical behaviors with fluids being related to viscous effects. Using the correct units in SI or USC for fluid properties is also essential in ensuring accurate results and maintaining effective communications between engineers. It should be remembered that many fluid parameters and associated physical quantities are not generally expressed in base units, and caution is required in applications.

5-Questions Self-Assessment Quickquiz

For Further Thought or Discussion

  • Sometimes, people may feel their ears experience a “popping” sensation when the surrounding pressure changes suddenly, such as going up in an elevator in a tall building. Why?
  • Consider some other engineering applications with important pressure effects.
  • Use Sutherland’s law and write a short piece of MATLAB code to calculate the coefficient of viscosity of air as a function of temperature.
  • Explain the physical mechanism(s) as to why the viscosity of a gas increases with increasing temperature.
  • Why does the speed of sound decrease at higher altitudes in the atmosphere?
  • Does the air’s viscosity in the atmosphere increase or decrease with altitude, and why?
  • Why does a hurricane “spin down” as it crosses over land? What are the fluid mechanisms at work here?
  • Two party balloons are filled, one with air and the other with helium. They are left to sit for several days. What will happen and why?
  • An astronaut spills some water on the ISS. What happens to the water and why?

Other Useful Online Resources

To learn more about fluids and their properties, try some of these online resources:

  • Great early film on the differences between normal gases and rarefied gases.
  • Understanding viscosity. YouTube video.
  • Viscosity demo: Water and oil. YouTube Video.
  • What happens when liquids of different viscosity are poured into a container?
  • Pressure Demo: Water column. YouTube video.
  • A really nice video on understanding viscosity.
  • A good video showing some of the properties of non-Newtonian fluids.


  1. Which is the distance the molecules travel between collisions.
  2. [footnote]The name "slug" as a unit of mass originates from the concept of mass as inertia, i.e., "sluggish," and has been referred to as the "engineer's mass unit."
  3. Experiments by Jean Baptiste Perrin led to the definition of the Avogadro number as the number of molecules in one gram-molecule of oxygen. Perrin named it the Avagardro's number in honor of Avogadro. Standardization of the SI system then led to the formal definition of the Avogadro constant.
  4. In medical ultrasound imaging, the speed of sound in tissues and fluids (like blood) is critical for accurate imaging and diagnosis. Variations in sound speed can indicate abnormalities in tissue density or fluid composition, aiding in detecting tumors, lesions, or other medical conditions.
  5. sound is the primary means of communication and detection in underwater environments. The speed of sound in water determines how quickly signals can propagate, impacting communication, navigation, and detection systems used in underwater applications such as sonar.
  6. Aerosols can have a dispersed phase of liquids (e.g., fog) or solids (e.g., smoke or dust).
  7. A colloid is a mixture in which one substance consisting of microscopically dispersed insoluble particles is suspended throughout another substance.


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Introduction to Aerospace Flight Vehicles Copyright © 2022 – 2024 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

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