Fundamental Properties of Fluids

J. Gordon Leishman

doi:https://doi.org/10.15394/eaglepub.2022.1066.n7

9 Fundamental Properties of Fluids

Introduction

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 deep 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 behaviors and problem-solving. 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 point to point in the fluid; they may also vary with respect to time at a given point. They are further referred to as 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 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 interdependencies. 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 formally embodied in an equation of state.

Learning Objectives

Understand the concept of a continuum model for describing a fluid.
Become familiar with the parameters used to describe the behavior of a fluid, 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.
Know how to calculate the viscosity of a gas using Sutherland’s law.
Understand what streamlines are in a flow and how to calculate their locations.

What is a Fluid?

Fluids are substances with mass and volume but no predefined shape. Fluids can be liquids (e.g., water) or gases (e.g., air). Unlike solids, fluids are substances with 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 for 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 are mobile in a fluid 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, a behavior that has many consequences on 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 properties of the fluid, such as its 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 the pressure is reduced, the distance between molecules increases, i.e., the mean free path increases, and eventually, as a vacuum is approached, the gas becomes rarified to the point a continuum model cannot describe it.

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 an aircraft flying in the lower atmosphere will be many orders of magnitude greater than $\lambda$ . Therefore, the fluid flow about the aircraft, in this case, 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, $\lambda$ 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, and it will behave differently.

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 greater 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 a rarefied gas model where the behavior of individual molecules must be described, usually using statistical models.

However, notice that the definition of a characteristic length scale may still be ambiguous. 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 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) $\begin{equation*} p = \lim_{dA \rightarrow \delta^2} \left( \frac{dF}{dA} \right) \mbox{~with~$\delta \gg \lambda$} \end{equation*}$

where $\delta$ is a large dimension compared with the mean distance between the fluid molecules $\lambda$ .

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 is to assume that it is caused by the time rate of change of momentum of the fluid molecules, such as 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.

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, but 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 $\vec{n}$ .

Leishman image — 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 might be a physical interpretation of a pressure force?

Consider what happens when someone is scuba-diving in the ocean or diving deeply into a swimming pool. There is a pressure exerted by the water above, which feels like a squeezing force effect on your 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 your body to feel significant changes in pressure.

Worked Example #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 110 lb/in². What is the force applied to the piston?

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 will be the pressure times the area, i.e.,

$F = p \, A = 110.0 \times 7.07 = 777.55~\mbox{lb}$

Fluid Density

Another essential property to describe a fluid’s characteristics is its density, which is 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 are still contained within the small elemental volume. Density is defined as the mass of fluid 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 number of molecules in the fluid per unit volume.

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

(5) $\begin{equation*} \varrho = \lim_{d{\cal{V}} \rightarrow \delta^3} \left( \frac{dm}{d\cal{V}} \right) \mbox{~with~$\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. 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 ft $^{3}$ /slug (i.e., ft $^{3}$ slug $^{-1}$ ) the USC system.

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, which is 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. The temperature plays a vital role in high-speed aerodynamics where the air is compressible, and there could be aerodynamic heating from frictional forces.

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⁻¹in 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 energy of its molecules.

Note: 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

To construct a temperature scale, two fixed points are taken. 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. The temperature on one scale is converted to another scale by using

(9) $\begin{equation*} \frac{\mbox{Temperature $T$ on any scale} - T_{\rm \mbox{\tiny LFP}}}{T_{\rm \mbox{\tiny UFP}} - T_{\rm \mbox{\tiny LFP}}} = \mbox{constant} \end{equation*}$

Therefore,

(10) $\begin{equation*} \frac{T (^{\circ}\mbox{C}) - 0}{100} = \frac{T^{\circ}\mbox{F} - 0}{212 - 32} = \frac{T (^{\circ}\mbox{K}) - 273.14}{373.15 - 273.15} = \frac{T (^{\circ}\mbox{R}) - 492.67}{672.67 - 492.67} \end{equation*}$

or

(11) $\begin{equation*} \frac{T (^{\circ}\mbox{C}) }{100} = \frac{T (^{\circ}\mbox{F}) - 0}{180} = \frac{T (^{\circ}\mbox{K}) - 273.14}{100} = \frac{T (^{\circ}\mbox{R}) - 492.67}{80} \end{equation*}$

Converting temperature values from one scale to another is relatively easy because the scales are all linearly related. **Figure not to be used for numerical calculations.**

Therefore, when using Eq. {univTconv}, converting to Centigrade or Celsius $^{\circ}$ C from Fahrenheit $^{\circ}$ F gives

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

Converting to Fahrenheit $^{\circ}\mbox{F}$ from Centigrade or Celsius $^{\circ}\mbox{C}$ , gives

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

Converting to Kelvin K from Centigrade or Celsius $^{\circ}\mbox{C}$ , gives

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

Converting to Rankine R from Fahrenheit $^{\circ}\mbox{F}$ , gives

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

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), which are 15 $^{\circ}$ C or 59 $^{\circ}$ F.

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

(16) $\begin{equation*} \frac{T_1}{T_2} = \frac{20 + 273.15}{40 + 273.15} = 0.936~\mbox{~(correct)} \end{equation*}$

but incorrectly as

(17) $\begin{equation*} \frac{T_1}{T_2} = \frac{20}{40} = 0.5~\mbox{~(incorrect)} \end{equation*}$

Why two absolute temperature scales?

Using two absolute temperature scales is a historical result of different conventions and preferences in different fields, in this case, 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 but was based on the Fahrenheit (F) unit. This scale is named after the University of Glasgow engineering professor William J. M. Rankine.

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. These relationships are formally embodied in an equation of state, which determines the quantitative relationships between a gas’s pressure, density (or volume), and temperature.

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

In physics and chemistry, people inevitably first come across the use of the general or universal equation of state for a gas, i.e.,

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

where ${\overline{R}}$ is called the universal gas constant (which is the same for all gases), and $n$ is the mass of the gas expressed in moles as given by the symbol $n$ . Remember that 1 mole = 6.022 $\times$ 10 $^{23}$ atoms, so a mole (a base SI unit) expresses the density of the gas.

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.,

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

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

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

which is more usually written as

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

Equation 21 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. Therefore, the value of $R$ in Eq. 21 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 specific gas.

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

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

and which version to use depends, in part, on the problem being solved.

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.

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

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

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. 21 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 at very high pressures.

Need some help with the equation of state? Here is a short video lesson on the equation of state from Dr. Leishman’s “Math & Physics Hints and Tips” series.

Worked Example #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.

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}$}$

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. 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 consequences of the effects of viscosity come into the behavior of all types of fluids.

Viscometer

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

(24) $\begin{equation*} W_b = \varrho_b \, \frac{4}{3} \pi R^3 \end{equation*}$

where $\varrho_b$ is the mass density of the steel ball. The (upward) buoyancy force on the ball (Archimedes’s principle) will be

(25) $\begin{equation*} B_b = \varrho_{\rm oil} \, \frac{4}{3} \pi R^3 \end{equation*}$

Therefore, the equilibrium equation is

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

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

(27) $\begin{equation*} D_{\mu} = 6 \pi R \mu V \end{equation*}$

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. Viscosity is a point property of the fluid, and, in general, it can be written that $\mu = \mu(x, y, z, t)$ . Therefore, in equilibrium, then

(28) $\begin{equation*} \frac{4}{3} \pi R^3 \left( \varrho_b -\varrho_{\rm oil} \right) - 6\pi R \mu V = 0 \end{equation*}$

and rearranging gives

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

For a ball of the same weight and size, Eq. 29 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 is determined. Stokes’ law (Eq. 27) 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.

Units of Viscosity

Viscosity has units of kg m^-1 s^-1 or Nm^-2s or Pa s in the SI system or slug ft^-1 s^-1 in the USC system. However, the unit typically employed in practice is called the “poise” (P) or gram cm^-1 s^-1. The unit of poise is named after Jean Léonard Marie Poiseuille. 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 is reported in units of micropoise (μP).

Shear in a Fluid

Another way to understand the concept of viscosity is to consider a flow containing a velocity gradient in one direction, e.g., the velocity in the fluid increases, moving up from one point to another. Suppose two adjacent layers of this fluid are considered, which in this case is between two plates, one stationary and the other moving, as illustrated in the figure below. 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.

In Newton’s Philosophiæ Naturalis Principia Mathematica of 1687, he defines the concept of viscosity 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 above with a fluid of depth $h$ between a moving upper plate and a stationary lower plate. 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

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

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

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

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.,

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

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 32 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

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

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 in the above figure, is

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

to a small angle approximation. Rearranging the equation gives

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

and so the shear stress in the fluid is

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

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

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

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

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

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. The values of kinematic viscosity have units of m $^2$ s $^{-1}$ in the SI system or ft $^2$ s $^{-1}$ in the USC system.

Mechanisms of Viscosity

The viscous properties of a fluid arise from two sources: 1. Inter-molecular 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. In this case, viscosity results more from molecular bonding and less from inter-molecular 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 liquid.

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. In gases, 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, which is related to the diffusion of molecules with layers of the gas and manifests as a resistance to deformation and a source of viscosity.

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 a variety of fluids. Air behaves as 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 in a nonlinear or 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. Generally, 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

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

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 measured viscosity of gases.

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 $\mu_{\rm ref} = 3.63 \times 10^{-7}$ 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 a wide range of 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.

Worked Example #3 – Calculation 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?

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 $\mu_{\rm ref} = 3.63 \times 10^{-7}$ 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

$\mu = 3.63 \times 10^{-7} \left( \frac{511.67}{518.67} \right)^{1.5} \left( \frac{518.67 + 198.72}{511.67 + 198.72} \right)$

Therefore,

$\mu = 3.5917\times 10^{-7}~\mbox{ slugs s$^{-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.,

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

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 $B$ 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

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

and the four-parameter model is

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

Again, the coefficients, $A$ , $B$ , $C$ , and $D$ for most liquids can be found in many 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 in the same direction. However, different parts of a fluid in motion will most likely travel at different 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 that is 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 hence 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.

The velocity of a fluid element depends on both its speed and direction, i.e., the 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 ms $^{-1}$ in the SI system or ft s $^{-1}$ in the USC system.

Streamlines

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 $x$ – $y$ 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 out 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

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

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.,

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

Therefore, in this case, to trace the positions of the streamlines in a flow $\vec{V} = (u, v, w)$ , it is necessary to solve simultaneously three ordinary differential equations (ODEs), i.e.,

(45) $\begin{eqnarray*} \frac{dz}{dy} & = & \frac{w}{v} \\[8pt] \frac{dz}{dx} & = & \frac{w}{u} \\[8pt] \frac{dy}{dx} & = & \frac{v}{u} \end{eqnarray*}$

Worked Example #4 – Calculating the equation of a streamline

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

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, to find streamlines means that all three ODEs must be solved simultaneously. Numerical integration methods available include Euler’s method, Runge-Kutta methods, etc. Second-order Runge-Kutta methods are commonly used to calculate streamlines. In two-dimensional flow, the streamlines are found by solving simultaneously the two differential equations

$\frac{d x}{dt} = u$

and

$\frac{d y}{dt} = v$

for $x$ and $y$ , starting from some initial points at $t=0$ . If the time step is small enough, an explicit Euler method will be sufficient for calculating the streamlines. 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 for. 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 using the numerical tracing of streamlines and flow vectors from measurements of a velocity field using the Particle Image Velocimetry (PIV) method.

Speed of Sound

The speed of sound in any medium depends on how quickly vibrational energy can be transferred through the medium from molecule to molecule. 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.

Calculation of the Speed of Sound

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.,

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

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.,

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

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.

Useful thermodynamic properties of some gases.
Gas	$\gamma$	$R$ (SI)	$R$ (USC)
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 $R$ then the value of $R$ is not only for the correct gas but is also in the appropriate engineering units. Remember that 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, remember that $\gamma$ = 1.4 for air, which is non-dimensional.

Worked Example #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.

(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. However, the perceived frequency of sound propagation will change if the location of the sound source S and the listener locations (say, L₁ and L₂) are in relative movement to each other, which is known as the Doppler effect, as illustrated in the figure below.

While the speed of sound is constant in a given media, the perceived frequency of the 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

(48) $\begin{equation*} f_L = \left( \frac{a + V_L}{a + V_S} \right) f_S \end{equation*}$

where $V_S$ is the relative velocity of the source, $V_L$ is the relative velocity of the listener, and $f_S$ is the frequency of the sound source. It is important the correct signs are used on $V_S$ and $V_L$ ; the value of $V_S$ will be positive if the sound source is moving away from the listener and $V_L$ will be positive if the listener is moving toward the source. In the figure above, the listener at $L_1$ will hear a higher frequency than the listener at $L_2$ .

The speed of sound is important 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 as well as the forces produced on that aircraft, this ratio being called the flight Mach number, $M_{\infty}$ . If the aircraft flies much slower than the speed of sound then the conditions are said to be subsonic and compressibility effects are small. However, if the aircraft moves faster and approaches or exceeds the speed of sound, which is called supersonic, then compressibility effects become important and the flow physics will change.

Surface Tension

Surface tension is the tendency of a liquid-gas interface (or free-surface) to behave like a stretched elastic membrane or “skin.” 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. Surface tension, which is usually represented by the symbol $\sigma$ or $\gamma$ , has units of force per unit length. The base units of surface tension is Newtons per meter (N/m) in SI, or pounds per foot (lb/ft) in USC.

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. Consider the figure below, which shows beads of water sitting on a solid surface. 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 concept of surface tension. Notice that the larger droplets are “flatter” because the effects of gravity increase more quickly and so overcome the effects of surface tension.

For tiny droplets of water sitting on a surface, as shown in the figure above, it is often 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 droplet’s shape becomes somewhat flatter and bulges because of the increasingly important effects of gravity, which is roughly proportional to weight, i.e., proportional to $a^3$ where $a$ is the approximate droplet radius. This 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 volume of water increases, which mimics the effects of surface tension. Because the surface area is proportional to $a^2$ , the ratio of gravitational to surface tension depends on the ratio $a^3/a^2 = a$ ; this latter effect becomes increasingly important for larger droplets.

The magnitude of the surface tension depends on the type of liquid and on its temperature. Generally, liquids with stronger intermolecular forces exhibit higher surface tension. Water, for example, has a higher surface tension because of the relatively large cohesive bonding between its hydrogen molecules. This property of surface tension is widely used in various applications, including painting or coating technologies, printing with inks, the production of emulsions and foams, and in various other manufacturing processes.

Worked Example #6 – Calculating surface tension

A student does experiments in the lab to study the property of surface tension. The student finds that cylindrical steel needles of different diameters, $d$ , and different lengths, $L$ , will “float” on the surface of water. A close inspection shows that no part of the needle is under the water, so the floatation effect is not because of buoyancy and must be entirely because 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, and that 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 ${^3}$ .

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

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

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 0.07326 N/m for water at a temperature of 15 $^{\circ}$ C.

Summary & Closure

Understanding the fundamental properties of fluids and their behavior is crucial in the field of aerospace engineering. The relationships between pressure, density, temperature, viscosity, flow velocity, and the speed of sound help engineers to predict and understand fluid behavior in different conditions, such as in air or fluid flow around an aircraft or spacecraft. The thermodynamic equation of state helps engineers to calculate the state of the fluid, taking into account the effects of pressure, temperature, and density. Using the correct units, either in SI or USC, is also important in ensuring accurate results and maintaining effective communications between engineers.

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?
Think about some other engineering applications where pressure effects are important.
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?

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.

License

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Introduction to Aerospace Flight Vehicles Copyright © 2022, 2023 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n7