Fundamental Properties of Fluids

J. Gordon Leishman

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

8 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 the air is a type of 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.

In all branches of science and engineering, properties are defined to help describe how things behave around us in the physical world. For example, concepts of mass, weight, energy, work, power, etc., are all essential physical properties relevant to how we describe the world 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 that they apply to bulk matter or a finite group of molecules rather than to each molecule, the group of molecules 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 parameter used to describe the behavior of a fluid, including pressure, density, temperature, viscosity, flow velocity, and the speed of sound.
Be able to use the equation of state to relate gas properties, i.e., pressure, density, and temperature.
Know how to calculate the viscosity of air 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 mostly rigid and have shapes that are difficult to change under the action of external forces.

The difference between solids and fluids is that in a fluid the molecules 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, 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 any local changes that might be associated with individual molecular motion are irrelevant. This latter assumption is easily justified in many practical cases of fluid flows.

As the pressure is reduced, the distance between molecules begins to increase, 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 68 nm or 6.8 $\times 10^{-9}$ m. It is evident that 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 aircraft’s length, such as at the edge of space where the air density is very low. In this case, the air molecules would be spaced sufficiently far apart that interactions with the aircraft would occur only infrequently. So the air loses the characteristics of what might be called a “normal” gas behavior.

The Knudsen number $K\!n$ is often used to quantify such low-density flows, which is defined as the ratio of $\lambda$ to a characteristic length and measures the degree of departure from a continuum. Usually, when $K\!n > 0.01$ , the continuum concept becomes increasingly invalid. As a result, the gas behavior cannot be explained in terms of macroscopically varying quantities and must be described using a rarefied gas model where the behavior of individual molecules must be described, usually using statistical models.

A continuum model is the most common model to describe a fluid. However, spacecraft can encounter rarified or free molecule flows at the upper edges of the atmosphere. Intermediate cases of such flows (neither one nor the other) are often termed low-density flows. 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

The pressure in a fluid is given the symbol $p$ (lower case) and is defined as the compressive force per unit area acting on a surface immersed in the fluid. Consider a general point B in a fluid volume, as shown in the figure below. Let $dA$ be a small elemental area at point B, which could be imagined as being located at the tip of a tiny probe of linear dimension $\delta$ that is introduced into the fluid, with the force on it being $d \vec{F}$ .

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.

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

(1) $\begin{equation*} p = \lim_{dA \rightarrow \delta^2} \left( \frac{dF}{dA} \right) \mbox{~with~$\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.. Higher or lower pressure would be associated with more or fewer molecules impacting a given surface area per unit of time. So to create a large amount of pressure on a given area, a large force must be exerted. 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, which means that the value of pressure can have a different value from one point to another at different points in the fluid. Therefore, the pressure becomes a function of the Cartesian spatial coordinates, i.e., $p = p(x, y, z)$ . In some cases, 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.

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, because a Pa is a small value of pressure, units of kilo-Pascals (kPa) may be used. In the USC system, units of lb in $^{-2}$ (pounds per square inch) are common.

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 both 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

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

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

In regard to the physical interpretation of a pressure force, consider what happens when someone is scuba-diving in the ocean. 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 1,000 times as many 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 used 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 so similar to $\rho$ , then 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 $d\cal{V}$ . 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 density of the flow 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

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

where $\delta$ is a large linear dimension compared with the mean distance between the molecules $\lambda$ . Therefore, density is 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 that $\varrho = \varrho(x, y, z, t)$ .

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 Measures of Fluid Density

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

Specific Volume

The specific volume of a fluid is reciprocal of its density, and is given the symbol $v$ or $SV$ and can be expressed as

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

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 denoted by the symbol $w$ or $SW$ , i.e.,

(5) $\begin{equation*} w \equiv SW = \frac{\mbox{Weight~of~fluid}}{\mbox{Volume~of~fluid}} \end{equation*}$

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, its density is often measured relative to another fluid, which is called the specific gravity, $SG$ . Usually the reference is the density of water, so

(6) $\begin{equation*} SG = \frac{\mbox{Density~of~fluid}}{\mbox{Density~of~water}} = \frac{\mbox{Weight~of~fluid}}{\mbox{Weight~of~water}} \end{equation*}$

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 kinetic 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. 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 $T$ , which is directly proportional to the average kinetic energy ( $KE$ ) of the gas molecules. This relationship is formally written as $KE = (3/2) kT$ , where $k$ is known as Boltzmann’s constant. Therefore, a high-temperature fluid would be one in which the molecules move about at relatively high speeds, as shown in the figure below. In contrast, a low-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., $T = T(x, y, z, t)$ .

The temperature in a fluid depends on the average kinetic energy of the fluid molecules.

The temperature has units of Centigrade or Celsius $^{\circ}$ C or Kelvin, K, in the SI system or Fahrenheit $^{\circ}$ F or Rankine, R, in the USC system. 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 temperature relative to absolute zero temperature, i.e., the temperature when the average kinetic energy of the molecules becomes effectively zero.

Converting from one temperature scale to another is easy because they are all linearly related, i.e., Centigrade or Celsius $^{\circ}$ C from Fahrenheit $^{\circ}$ F:

(7) $\begin{equation*} ^{\circ}\mbox{C} = \frac{5}{9} \left( ^{\circ}\mbox{F} - 32 \right) \end{equation*}$

Fahrenheit $^{\circ}\mbox{F}$ from Centigrade or Celsius $^{\circ}\mbox{C}$ :

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

Kelvin K from Centigrade or Celsius $^{\circ}\mbox{C}$ :

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

Rankine R from Fahrenheit $^{\circ}\mbox{F}$ :

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

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 both with and without the degree symbol. Nevertheless, retaining the degree symbol on the temperature units is entirely acceptable when working on aerodynamic problems. Finally, it is helpful to remember that standard sea level values of temperature (based on the ISA model), which are 15 $^{\circ}$ C or 59 $^{\circ}$ F.

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 the pressure in a gas, its density (or volume), and its temperature.

An equation of state is a thermodynamic equation. All gases have properties that can be measured, including the pressure, temperature, and volume containing the gas. 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), then the equation of state can be used to determine (calculate) the other.

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

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

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.

However, this latter equation is not particularly useful for engineering purposes. But, if both sides of this general equation are divided by the mass of the gas, then the volume now becomes the specific volume, which is the reciprocal of the density of the gas. Therefore, an alternative equation of state for a gas can now be written as

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

which is the usual form used in all of engineering, gas dynamics, and aerodynamics. However, in this case, another (different) gas constant $R$ is obtained, equal to the universal gas constant divided by the gas’s mass per mole. Therefore, the value of $R$ is not universal and depends on the gas type, i.e., the value of the gas constant 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.

The main advantage of the equation of state 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

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

Therefore, the use of 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 fluid dynamics, and can be interpreted as a two-dimensional surface in terms of $p$ , $\varrho$ and $T$ . Each and every point on the surface represents a state of thermodynamic equilibrium.

Strictly speaking, the preceding equation of state in Eq. 12 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 very much like an ideal gas. Generally, a gas nearly always acts as a perfect gas at normal to moderate temperatures and/or pressures and 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 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 then we will use of 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 $^{2}$ ) then it is necessary to multiply by 144. The gas constant for air in USC units is 1716.59 ft lb slug $^{-1}$ R $^{-1}$ so the density of the air will be

$\varrho = \frac{p}{R T} = \frac{14.61 \times 144.0 }{1716.59 \times 530.77} = 0.002309~\mbox{slugs ft$^{-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.

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). 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 and hence 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

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

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

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

Therefore, the equilibrium equation is

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

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.,

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

Therefore, in equilibrium then

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

and rearranging gives

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

For a ball of the same weight and size, Eq. 19 shows it will drop in the oil at a velocity that is inversely proportional to the oil’s viscosity, 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. 17) can be used to determine the viscosity 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.

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. In that case, the upper (faster) layer draws the lower (slower) one along through a force on the lower layer, as illustrated in the figure below, so there is a shear focus 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 “Principia” 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 similar to that shown in the figure above to show that 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.,

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

This latter equation is called Newton’s law of viscosity. The symbol $\mu$ (i.e., the Greek symbol “mu”) is a constant known as the coefficient of dynamic viscosity, or more simply just as the viscosity of the fluid. 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 foregoing equation is just a statement of a linear stress/strain rate relationship for a fluid.

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 a time period $dt$ , the upper AC surface will travel further than the lower surface DE by a distance

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

The consequence is that 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

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

to a small angle approximation. Rearranging the equation gives

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

and so the shear stress in the fluid is

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

This latter result is just 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.

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

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

because the $u$ velocity in a fluid may vary in other directions as well, e.g., there could be $u$ velocity gradients in the $x$ and $z$ directions, i.e., $\partial u/\partial x$ and $\partial u/\partial z$ . In fact, up to nine velocity gradients can exist in a fluid. i.e., three velocity components and three directions.

Viscosity has units of kg m $^{-1}$ s $^{-1}$ or Nm $^{-2}$ s or Pa s in the SI system or slug ft $^{-1}$ s $^{-1}$ in the USC system. Viscosity is a point property of the fluid, and, in general, it can be written that $\mu = \mu(x, y, z, t)$ .

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 the solution to the problem. This ratio of $\mu$ to density $\varrho$ is called the kinematic viscosity $\nu$ (Greek symbol “nu”), i.e.,

(26) $\begin{equation*} \nu = \frac{\mu}{\varrho} \end{equation*}$

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 so 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 size of the molecules, their molecular weight, the strength of their bonds, and the temperature of the 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.

In a liquid, the source of viscosity is forces produced by intermolecular bonding, which become stronger for a more viscous 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 relatively mobile molecules diffuse throughout the gas. Nevertheless, gases are still viscous and exhibit the characteristics of viscosity. This effect becomes more apparent when there is an initial velocity gradient in the gas, as shown in the schematic below. 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.

In a gas, the source of viscosity is intermolecular momentum transfer. This process is related to the diffusion of molecules with layers of the gas and manifests as a resistance to deformation and so a source of viscosity.

The consequence of this process 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 further 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, you need to shake the bottle vigorously, which agitates and deforms the fluid and creates a strain rate. This process also stretches the polymer molecules to experience less cohesive bonding, so reduced viscosity is the result. 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

The temperature has important effects on 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 the reduction in bonding forces dominates over any increase in inter-molecular momentum transfer. The consequence is that liquids usually show a reduction in viscosity with increasing temperature. However, gases, including air, typically show increased viscosity with increasing temperature because of the increase in intermolecular momentum transfer.

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

Sutherland’s Law 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 can be written as a function of absolute temperature, $T$ , as

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

where reference values (subscript “ref”) are in appropriate SI or USC units. The parameter $S$ is known as Sutherland’s constant.

For air in SI units, then the reference values at $T_{\rm ref} = 323^{\circ}$ 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 $^{-1}$ .

Sutherland’s law is widely used in various engineering fields to predict the viscosity of gases at different temperatures. It applies the viscosity of the gas is dominated by intermolecular momentum transfer. For air, as well as gases such as oxygen, nitrogen, and helium, Sutherland’s law has been shown to be applicable over a wide range of temperatures. It is important to note, however, 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 exhibit a much more complex relationship between viscosity and temperature because of the effects intermolecular bonding. In general, Sutherland’s law should be used with caution, 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 coefficient of dynamic viscosity. 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

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

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, then 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.

Flow Velocity

In the field of 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, for a fluid in motion, different parts of the fluid will most likely be traveling at different velocities and so in relative motion to each other, which is 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 referred to as a fluid element because it is just 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 for that matter, 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 its direction, i.e., 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 an important part of understanding flow problems. A pathline is called the trajectory or path that a fluid element traces out 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

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

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.,

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

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.,

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

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 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. 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 that 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.,

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

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 as fast in water compared to 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.,

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

where $\gamma$ is the ratio of specific heats, and $R$ is the gas constant. Notice that the values of $\gamma$ and $R$ are different for different gases, a useful table being 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

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

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 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 a gas’s viscosity 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.

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Introduction to Aerospace Flight Vehicles 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