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n this chapter, we first review the concepts of dimensions and units. We then review the fundamental principle of dimensional homogeneity, and show how it is applied to equations in order to nondimensionalize them and to identify dimensionless groups. We discuss the concept of similarity between a model and a prototype. We also describe a powerful tool for engi- neers and scientists called dimensional analysis, in which the combination of dimensional variables, nondimensional variables, and dimensional con- stants into nondimensional parameters reduces the number of necessary independent parameters in a problem. We present a step-by-step method for obtaining these nondimensional parameters, called the method of repeating variables, which is based solely on the dimensions of the variables and con- stants. Finally, we apply this technique to several practical problems to illus- trate both its utility and its limitations. DoDooDo OBJECTIVES When you finish reading this chapter, you should be able to Develop a better understanding of dimensions, units, and dimensional homogeneity of equations Understand the numerous benefits of dimensional analysis Know how to use the method of repeating variables to identify nondimensional parameters Understand the concept of dynamic similarity and how to apply it to experimental modeling 270 FLUID MECHANICS Length > 3.2 cm > em 1 2 3 FIGURE 7-1 A dimension is a measure of a physical quantity without numerical values, while a unit is a way to assign a number to the dimension. For example, length is a dimension, but centimeter is a unit. FIGURE 7-2 The water strider is an insect that can walk on water due to surface tension. O Dennis DrennerNisuals Unlimited. 7-1 - DIMENSIONS AND UNITS A dimension is a measure of a physical quantity (without numerical val- ues), while a unit is a way to assign a number to that dimension. For exam- ple, length is a dimension that is measured in units such as microns (um), feet (ft), centimeters (cm), meters (m), kilometers (km), etc. (Fig. 7-1). There are seven primary dimensions (also called fundamental or basic dimensions) —mass, length, time, temperature, electric current, amount of light, and amount of matter. All nonprimary dimensions can be formed by some combination of the seven primary dimensions. For example, force has the same dimensions as mass times acceleration (by Newton's second law). Thus, in terms of primary dimensions, E o 2 Time? / — Emi where the brackets indicate “the dimensions of” and the abbreviations are taken from Table 7-1. You should be aware that some authors prefer force instead of mass as a primary dimension—we do not follow that practice. Dimensions offorce: fForce) = (Mass qn TABLE 7-1 Primary dimensions and their associated primary Sl and English units Dimension Symbol* SI Unit English Unit Mass m kg (kilogram) Ibm (pound-mass) Length L m (meter) ft (foot) Time” t s (second) s (second) Temperature T K (kelvin) R (rankine) Electric current I A (ampere) A (ampere) Amount of light c cd (candela) cd (candela) Amount of matter N mol (mole) mol (mole) * We italicize symbols for variables, but not symbols for dimensions. ! Note that some authors use the symbol T for the time dimension and the symbol O for the temperature dimension. We do not follow this convention to avoid confusion between time and temperature, EXAMPLE 7-1 Primary Dimensions of Surface Tension An engineer is studying how some insects are able to walk on water (Fig. 7-2). A fluid property of importance in this problem is surface tension (o), which has dimensions of force per unit length. Write the dimensions of sur- face tension in terms of primary dimensions. SOLUTION The primary dimensions of surface tension are to be determined. Analysis From Eg. 7-1, force has dimensions of mass times acceleration, or (mL/t2). Thus, Force me Lt Dimensions of surface tension: = = = (mt) q toa = E) = (EU qm 19 Discussion The usefulness of expressing the dimensions of a variable or constant in terms of primary dimensions will become clearer in the discussion of the method of repeating variables in Section 7-4. 272 FLUID MECHANICS Equation of the Day equation 2=C Tre gemoulh 12 +39 p+ q” FIGURE 7-6 The Bernoulli equation is a good example of a dimensionally homogeneous equation. All additive terms, including the constant, have the same dimensions, namely that of pressure. In terms of primary dimensions, each term has dimensions Em/(CD. o o The nondimensionalized Bernoulli equation PY por C BM, *P, P, a | mm FIGURE 7-7 A nondimensionalized form of the Bernoulli equation is formed by dividing each additive term by a pressure (here we use P,.). Each resulting term is dimensionless (dimensions of £13). EXAMPLE 7-2 Dimensional Homogeneity of the Bernoulli Equation Probably the most well-known (and most misused) equation in fluid mechanics is the Bernoulli equation (Fig. 7-6), discussed in Chap. 5. The standard form of the Bernoulli equation for incompressible irrotational fluid flow is Bernoulli equation: P+ davi +pgz=C 4) (a) Verify that each additive term in the Bernoulli equation has the same dimensions. (b) What are the dimensions of the constant C? SOLUTION We are to verify that the primary dimensions of each additive term in Eq. 1 are the same, and we are to determine the dimensions of constant C. Analysis (a) Each term is written in terms of primary dimensions, o — Force À Length 1 ) = fm Ui = tintas = (e o (tras Time? Length?) tr, k v o ( Mass (Ea o (Es x a) o (m) 2º?" J” Avolume Time / / ” (Length? x Time?) let tpoz -f Mass Length, ny o (es x ego o (8) P9% = Yyolume Time? SS = Length? x Time?) At, Indeed, all three additive terms have the same dimensions. (b) From the law of dimensional homogeneity, the constant must have the same dimensions as the other additive terms in the equation. Thus, ca-(5) Discussion If the dimensions of any of the terms were different from the others, it would indicate that an error was made somewhere in the analysis. Primary dimensions of the Bernoulli constant: Nondimensionalization of Equations The law of dimensional homogeneity guarantees that every additive term in an equation has the same dimensions. It follows that if we divide each term in the equation by a collection of variables and constants whose product has those same dimensions, the equation is rendered nondimensional (Fig. 7-7). If, in addition, the nondimensional terms in the equation are of order unity, the equation is called normalized. Normalization is thus more restric- tive than nondimensionalization, even though the two terms are sometimes (incorrectly) used interchangeably. Each term in a nondimensional equation is dimensionless. In the process of nondimensionalizing an equation of motion, nondimen- sional parameters often appear—most of which are named after a notable scientist or engineer (e.g., the Reynolds number and the Froude number). This process is referred to by some authors as inspectional analysis. As a simple example, consider the equation of motion describing the ele- vation z of an object falling by gravity through a vacuum (no air drag), as in Fig. 7-8. The initial location of the object is z, and its initial velocity is vo in the z-direction. From high school physics, dz qo Dimensional variables are defined as dimensional quantities that change or vary in the problem. For the simple differential equation given in Eg. 7-4, there are two dimensional variables: z (dimension of length) and t (dimen- sion of time). Nondimensional (or dimensionless) variables are defined as quantities that change or vary in the problem, but have no dimensions; an example is angle of rotation, measured in degrees or radians which are dimensionless units. Gravitational constant g, while dimensional, remains constant and is called a dimensional constant. Two additional dimensional constants are relevant to this particular problem, initial location z, and initial vertical speed wo. While dimensional constants may change from problem to problem, they are fixed for a particular problem and are thus distin- guished from dimensional variables. We use the term parameters for the combined set of dimensional variables, nondimensional variables, and dimensional constants in the problem. Equation 7-4 is easily solved by integrating twice and applying the initial conditions. The result is an expression for elevation z at any time É Equation of motion: (7-4) Dimensional result: Zz=2Z, + Wmt— i gt? (7-5) The constant 3 and the exponent 2 in Eg. 7-5 are dimensionless results of the integration. Such constants are called pure constants. Other common examples of pure constants are q and e. To nondimensionalize Eg. 7-4, we need to select scaling parameters, based on the primary dimensions contained in the original equation. In fluid flow problems there are typically at least three scaling parameters, e.g., L, V, and P, — P, (Fig. 7-9), since there are at least three primary dimensions in the general problem (e.g., mass, length, and time). In the case of the falling object being discussed here, there are only two primary dimensions, length and time, and thus we are limited to selecting only two scaling parameters. We have some options in the selection of the scaling parameters since we have three available dimensional constants g, zo, and wo. We choose zy and Wos You are invited to repeat the analysis with g and z; and/or with g and wo. With these two chosen scaling parameters we nondimensionalize the dimen- sional variables z and t. The first step is to list the primary dimensions of al/ dimensional variables and dimensional constants in the problem, Primary dimensions of all parameters: BU tB=18 0d =tL5 O Quo = (LO to) = IL The second step is to use our two scaling parameters to nondimensionalize z and t (by inspection) into nondimensional variables z* and t*, : al; , z Wot Nondimensionalized variables: 7 = Z tt = E (1-8) 0 0 273 CHAPTER 7 w = component of velocity in the z-direction ) z = vertical distance q = gravitational acceleration in the negative z-direction FIGURE 7-8 Object falling in a vacuum. Vertical velocity is drawn positively, so w < O for a falling object. VP, Po ——— FIGURE 7-9 In a typical fluid flow problem, the scaling parameters usually include a characteristic length L, a characteristic velocity V, and a reference pressure difference Py — P... Other parameters and fluid properties such as density, viscosity, and gravitational acceleration enter the problem as well. 2715 CHAPTER 7 EXAMPLE 7-3 Illustration of the Advantages of Nondimensionalization Your little brother's high school physics class conducts experiments in a large vertical pipe whose inside is kept under vacuum conditions. The stu- dents are able to remotely release a steel ball at initial height z; between O and 15 m (measured from the bottom of the pipe), and with initial vertical speed wo between O and 10 m/s. A computer coupled to a network of photo- sensors along the pipe enables students to plot the trajectory of the steel ball (height z plotted as a function of time t) for each test. The students are unfamiliar with dimensional analysis or nondimensionalization techniques, and therefore conduct several “brute force” experiments to determine how the trajectory is affected by initial conditions z; and mp. First they hold wo fixed at 4 m/s and conduct experiments at five different values of x: 3, 6, 9, 12, and 15 m. The experimental results are shown in Fig. 7-12a. Next, they hold z fixed at 10 m and conduct experiments at five different values of mo: 2, 4, 6, 8, and 10 m/s. These results are shown in Fig. 7-12b. Later that evening, your brother shows you the data and the trajectory plots and tells you that they plan to conduct more experiments at different values of Zy and wo. You explain to him that by first nondimensionalizing the data, the problem can be reduced to just one parameter, and no further experiments are required. Prepare a nondimensional plot to prove your point and discuss. SOLUTION A nondimensional plot is to be generated from all the available trajectory data. Specifically, we are to plot Z* as a function of t*. Assumptions The inside of the pipe is subjected to strong enough vacuum o o — ay Va pressure that aerodynamic drag on the ball is negligible. 0 os | 15 ; 25 3 Properties The gravitational constant is 9.81 m/s2. (b) ú Es ú Analysis Equation 7-4 is valid for this problem, as is the nondimensional- ization that resulted in Eg. 7-9. As previously discussed, this problem com- FIGURE 7-12 bines three of the original dimensional parameters (g, zo, and mo) into one nondimensional parameter, the Froude number. After converting to the dimensionless variables of Eq. 7-6, the 10 trajectories of Fig. 7-12a and b are replotted in dimensionless format in Fig. 7-13. It is clear that all the trajectories are of the same family, with the Froude number as the only remaining parameter. Fr? varies from about 0.041 to about 1.0 in these exper- iments. If any more experiments are to be conducted, they should include combinations of zy and wo that produce Froude numbers outside of this range. A large number of additional experiments would be unnecessary, since all the trajectories would be of the same family as those plotted in Fig. 7-13. Discussion At low Froude numbers, gravitational forces are much larger than inertial forces, and the ball falls to the floor in a relatively short time. At large values of Fr on the other hand, inertial forces dominate initially, and the ball rises a significant distance before falling; it takes much longer for the ball to hit the ground. The students are obviously not able to adjust the gravitational constant, but if they could, the brute force method would require many more experiments to document the effect of g. If they nondimensionalize first, how- ever, the dimensionless trajectory plots already obtained and shown in Fig. 7-13 would be valid for any value of g; no further experiments would be required unless Fr were outside the range of tested values. Trajectories of a steel ball falling in a vacuum: (a) w, fixed at 4 m/s, and (b) zo fixed at 10 m (Example 7-3). 165 144) 4 FP=10 FIGURE 7-13 Trajectories of a steel ball falling If you are still not convinced that nondimensionalizing the equations and the parameters has many advantages, consider this: In order to reasonably docu- ment the trajectories of Example 7-3 for a range of all three of the dimensional in a vacuum. Data of Fig. 7-12a and b are nondimensionalized and combined onto one plot. 216 FLUID MECHANICS FIGURE 7-14 Throwing a baseball on the moon (Example 7-4). parameters g, zo, and w, the brute force method would require several (say a minimum of four) additional plots like Fig. 7-12a at various values (levels) of Was plus several additional sets of such plots for a range of g. A complete data set for three parameters with five levels of each parameter would require 5? = 125 experiments! Nondimensionalization reduces the number of parameters from three to one—a total of only 5! = 5 experiments are required for the same resolution. (For five levels, only five dimensionless trajectories like those of Fig. 7-13 are required, at carefully chosen values of Fr.) Another advantage of nondimensionalization is that extrapolation to untested values of one or more of the dimensional parameters is possible. For example, the data of Example 7-3 were taken at only one value of grav- itational acceleration. Suppose you wanted to extrapolate these data to a dif- ferent value of g. Example 7-4 shows how this is easily accomplished via the dimensionless data. EXAMPLE 7-4 Extrapolation of Nondimensionalized Data The gravitational constant at the surface of the moon is only about one-sixth of that on earth. An astronaut on the moon throws a baseball at an initial speed of 21.0 m/s at a 5º angle above the horizon and at 2.0 m above the moon's surface (Fig. 7-14). (a) Using the dimensionless data of Example 7-3 shown in Fig. 7-13, predict how long it takes for the baseball to fall to the ground. (b) Do an exact calculation and compare the result to that of part (a). SOLUTION Experimental data obtained on earth are to be used to predict the time required for a baseball to fall to the ground on the moon. Assumptions 1 The horizontal velocity of the baseball is irrelevant. 2 The surface of the moon is perfectly flat near the astronaut. 3 There is no aero- dynamic drag on the ball since there is no atmosphere on the moon. 4 Moon gravity is one-sixth that of earth. Properties The gravitational constant on the moon is &roon = 9.81/6 = 1.63 m/s2. Analysis (a) The Froude number is calculated based on the value of &roon and the vertical component of initial speed, Wo = (21.0 m/s) sin(5º) = 1.830 m/s from which w (1.830 m/s? 20 Wo . Fr = qmemão (1.63 m/s%20m) — 18 This value of Fr? is nearly the same as the largest value plotted in Fig. 7-13. Thus, in terms of dimensionless variables, the baseball strikes the ground at t* = 2.75, as determined from Fig. 7-13. Converting back to dimensional variables using Eq. 7-6, Ê 2.75(2.0 Estimated time to strike the ground: t= A = aa =3.01s (b) An exact calculation is obtained by setting z equal to zero in Eg. 7-5 and solving for time t (using the quadratic formula), 278 FLUID MECHANICS Prototype: Model: Vm al E —— Fom FIGURE 7-16 Kinematic similarity is achieved when, at all locations, the velocity in the model flow is proportional to that at corresponding locations in the prototype flow, and points in the same direction. (by a constant scale factor) to the velocity at the corresponding point in the prototype flow (Fig. 7-16). Specifically, for kinematic similarity the velocity at corresponding points must scale in magnitude and must point in the same relative direction. You may think of geometric similarity as /ength-scale equivalence and kinematic similarity as time-scale equivalence. Geometric similarity is a prerequisite for kinematic similarity. Just as the geometric scale factor can be less than, equal to, or greater than one, so can the velocity scale factor. In Fig. 7-16, for example, the geometric scale factor is less than one (model smaller than prototype), but the velocity scale is greater than one (velocities around the model are greater than those around the prototype). You may recall from Chap. 4 that streamlines are kinematic phenomena; hence, the streamline pattern in the model flow is a geometrically scaled copy of that in the prototype flow when kinematic similarity is achieved. The third and most restrictive similarity condition is that of dynamic sim- ilarity. Dynamic similarity is achieved when all forces in the model flow scale by a constant factor to corresponding forces in the prototype flow (force-scale equivalence). As with geometric and kinematic similarity, the scale factor for forces can be less than, equal to, or greater than one. In Fig. 7-16 for example, the force-scale factor is less than one since the force on the model building is less than that on the prototype. Kinematic similarity is a necessary but insufficient condition for dynamic similarity. 1t is thus pos- sible for a model flow and a prototype flow to achieve both geometric and kinematic similarity, yet not dynamic similarity. All three similarity condi- tions must exist for complete similarity to be ensured. In a general flow field, complete similarity between a model and prototype is achieved only when there is geometric, kinematic, and dynamic similarity. We let uppercase Greek letter Pi (TI) denote a nondimensional parameter. You are already familiar with one TI, namely the Froude number, Fr. In a general dimensional analysis problem, there is one II that we call the dependent II, giving it the notation II,. The parameter TI, is in general a function of several other Is, which we call independent IT's. The func- tional relationship is Functional relationship between IP's: DM = (IL, 6, ..., 1) RD) where k is the total number of II's. Consider an experiment in which a scale model is tested to simulate a prototype flow. To ensure complete similarity between the model and the prototype, each independent TI of the model (subscript m) must be identical to the corresponding independent TI of the prototype (subscript p), i.e., Ip, m = 1, o Mô, m > 13, po eos Mim — Mo pr To ensure complete similarity, the model and prototype must be geometrically similar, and all independent II groups must match between model and prototype. Under these conditions the dependent II of the model (II, ) is guaranteed to also equal the dependent TI of the prototype (IT, ,). Mathematically, we write a conditional statement for achieving similarity, (ii Hm = ando Mm= Mp. and em = pr then Th,m= Mp (712) Consider, for example, the design of a new sports car, the aerodynamics of which is to be tested in a wind tunnel. To save money, it is desirable to test a small, geometrically scaled model of the car rather than a full-scale prototype of the car (Fig. 7-17). In the case of aerodynamic drag on an automobile, it turns out that if the flow is approximated as incompressible, there are only two IP's in the problem, Fo pVvêr? The procedure used to generate these IIºs is discussed in Section 7-4. In Eg. 7-13, F, is the magnitude of the aerodynamic drag on the car, p is the air density, Vis the car's speed (or the speed of the air in the wind tunnel), L is the length of the car, and q is the viscosity of the air. II, is a nonstandard form of the drag coefficient, and TI, is the Reynolds number, Re. You will find that many problems in fluid mechanics involve a Reynolds number (Fig. 7-18). The Reynolds number is the most well known and useful dimensionless parameter in all of fluid mechanics. VL mM =f(l) where H, and m=25 (-13) In the problem at hand there is only one independent TI, and Eg. 7-12 ensures that if the independent Is match (the Reynolds numbers match: Hm = Hop), then the dependent TI's also match (Il, m = T1,5). This enables engineers to measure the aerodynamic drag on the model car and then use this value to predict the aerodynamic drag on the prototype car. EXAMPLE 7-5 Similarity between Model and Prototype Cars The aerodynamic drag of a new sports car is to be predicted at a speed of 50.0 mi/h at an air temperature of 25ºC. Automotive engineers build a one- fifth scale model of the car to test in a wind tunnel. It is winter and the wind tunnel is located in an unheated building; the temperature of the wind tunnel air is only about 5ºC. Determine how fast the engineers should run the wind tunnel in order to achieve similarity between the model and the prototype. SOLUTION We are to utilize the concept of similarity to determine the speed of the wind tunnel. Assumptions 1 Compressibility of the air is negligible (the validity of this approximation is discussed later). 2 The wind tunnel walls are far enough away so as to not interfere with the aerodynamic drag on the model car. 3 The model is geometrically similar to the prototype. 4 The wind tunnel has a moving belt to simulate the ground under the car, as in Fig. 7-19. (The moving belt is necessary in order to achieve kinematic similarity everywhere in the flow, in particular underneath the car.) Properties For air at atmospheric pressure and at T = 25ºC, p = 1.184 kg/m? and yu = 1.849 x 10º kg/m - s. Similarly, at T = 5ºC, p = 1.269 kg/m? and = 1.754 x 10ºkgm.s. Analysis Since there is only one independent TI in this problem, the simi- larity equation (Eq. 7-12) holds if IL m = Ip, where II, is given by Eq. 7-13, and we call it the Reynolds number. Thus, we write Val. poViL =fnmn q,,—Re,— "222 Hom = Re Amo um ' Hp 219 CHAPTER 7 Prototype car Model car Vm ERR FIGURE 7-17 Geometric similarity between a prototype car of length L, and a model car of length L,. pu e FIGURE 7-18 The Reynolds number Re is formed by the ratio of density, characteristic speed, and characteristic length to viscosity. Alternatively, it is the ratio of characteristic speed and length to kinematic viscosity, defined as v=ulp. Discussion By arranging the dimensional parameters as nondimensional ratios, the units cancel nicely even though they are a mixture of Sl and Eng- lish units. Because both velocity and length are squared in the equation for Il,, the higher speed in the wind tunnel nearly compensates for the model's smaller size, and the drag force on the model is nearly the same as that on the prototype. In fact, if the density and viscosity of the air in the wind tun- nel were identical to those of the air flowing over the prototype, the two drag forces would be identical as well (Fig. 7-20). The power of using dimensional analysis and similarity to supplement experimental analysis is further illustrated by the fact that the actual values of the dimensional parameters (density, velocity, etc.) are irrelevant. As long as the corresponding independent IT"s are set equal to each other, similarity is achieved—even if different fluids are used. This explains why automobile or aircraft performance can be simulated in a water tunnel, and the perfor- mance of a submarine can be simulated in a wind tunnel (Fig. 7-21). Sup- pose, for example, that the engineers in Examples 7-5 and 7-6 use a water tunnel instead of a wind tunnel to test their one-fifth scale model. Using the properties of water at room temperature (20' C is assumed), the water tunnel speed required to achieve similarity is easily calculated as ovo (PA (Lo POR fr) o [1.002 x 103 kg/m -s) a) o . = (600 mi eo x 105 kgjm - 580 kg/m 16) = 16:1 milh As can be seen, one advantage of a water tunnel is that the required water tunnel speed is much lower than that required for a wind tunnel using the same size model. 7-4 -» THE METHOD OF REPEATING VARIABLES AND THE BUCKINGHAM PI THEOREM We have seen several examples of the usefulness and power of dimensional analysis. Now we are ready to learn how to generate the nondimensional parameters, i.e., the II's. There are several methods that have been developed for this purpose, but the most popular (and simplest) method is the method of repeating variables, popularized by Edgar Buckingham (1867-1940). The method was first published by the Russian scientist Dimitri Riabou- chinsky (1882-1962) in 1911. We can think of this method as a step-by-step procedure or “recipe” for obtaining nondimensional parameters. There are six steps, listed concisely in Fig. 7-22, and in more detail in Table 7-2. These steps are explained in further detail as we work through a number of example problems. As with most new procedures, the best way to learn is by example and practice. As a simple first example, consider a ball falling in a vacuum as discussed in Section 7-2. Let us pretend that we do not know that Eg. 7-4 is appropriate for this problem, nor do we know much physics concerning falling objects. In fact, suppose that all we know is that the instantaneous 281 CHAPTER 7 Prototype V, o ——» F Hop cm “E Dip Model Vm= Vo to m pa=ta Foms fia E FIGURE 7-20 For the special case in which the wind tunnel air and the air flowing over the prototype have the same properties (om = Ppr Hm = Hp), and under similarity conditions (Vm = Vol ol) the aerodynamic drag force on the prototype is equal to that on the scale model. If the two fluids do not have the same properties, the two drag forces are not necessarily the same, even under dynamically similar conditions. FIGURE 7-21 Similarity can be achieved even when the model fluid is different than the prototype fluid. Here a submarine model is tested in a wind tunnel. Courtesy NASA Langley Research Center. 282 FLUID MECHANICS The Method of Repeating Variables Step |: List the parameters in the problem and count their total number n. Step 2: List the primary dimensions of each of the n parameters. Step 3: Set the reduction j as the number of primary dimensions. Calculate k, the expected number of I['s, k=n-j Step 4: Choose j repeating parameters. Step 5: Construct the k |l's, and manipulate as necessary. Step 6: Write the final functional relationship and check your algebra. FIGURE 7-22 A concise summary of the six steps that comprise the method of repeating variables. wi = initial vertical speed 9 = gravitational acceleration in the negative z-direction n= | elevation ) Es elevation of ball (ts Woy Zo, 9) (datum plane) FIGURE 7-23 Setup for dimensional analysis of a ball falling in a vacuum. Elevation z is a function of time £, initial vertical speed w,, initial elevation zy, and gravitational constant g. TABLE 7-2 Detailed description of the six steps that comprise the method of repeating variables* Step 1 List the parameters (dimensional variables, nondimensional variables, and dimensional constants) and count them. Let n be the total number of parameters in the problem, including the dependent variable. Make sure that any listed independent parameter is indeed independent of the others, i.e., it cannot be expressed in terms of them. (E.g., don't include radius r and area A = qr?, since rand À are not independent.) List the primary dimensions for each of the n parameters. Guess the reduction j. As a first guess, set j equal to the number of primary dimensions represented in the problem. The expected number of II's (k) is equal to n minus j, according to the Buckingham Pi theorem, Step 2 Step 3 The Buckingham Pi theorem: k=n-j (14) If at this step or during any subsequent step, the analysis does not work out, verify that you have included enough parameters in step 1. Otherwise, go back and reduce j by one and try again. Choose j repeating parameters that will be used to construct each II. Since the repeating parameters have the potential to appear in each TI, be sure to choose them wisely (Table 7-3). Generate the II's one at a time by grouping the j repeating parameters with one of the remaining parameters, forcing the product to be dimensionless. In this way, construct all k Il's. By convention the first TI, designated as Il,, is the dependent II (the one on the left side of the list). Manipulate the II's as necessary to achieve established dimensionless groups (Table 7-5). Check that all the II's are indeed dimensionless. Write the final functional relationship in the form of Eq. 7-11. Step 4 Step5 Step 6 * This is a step-by-step method for finding the dimensiontess II groups when performing a dimensional analysis. elevation z of the ball must be a function of time t, initial vertical speed wo, initial elevation zo, and gravitational constant g (Fig. 7-23). The beauty of dimensional analysis is that the only other thing we need to know is the pri- mary dimensions of each of these quantities. As we go through each step of the method of repeating variables, we explain some of the subtleties of the technique in more detail using the falling ball as an example. Step 1 There are five parameters (dimensional variables, nondimensional variables, and dimensional constants) in this problem; n = 5. They are listed in func- tional form, with the dependent variable listed as a function of the indepen- dent variables and constants: List of relevant parameters: z=Hbwoyzw9) n=5 284 FLUID MECHANICS TABLE 7-3 Guidelines for choosing repeating parameters in step 4 of the method of repeating variables* Guideline Comments and Application to Present Problem 1. Never pick the dependent variable. Otherwise, it may appear in all the T's, which is undesirable. 2. The chosen repeating parameters must not by themselves be able to form a dimensionless group. Otherwise, it would be impossible to generate the rest of the II's. 3. The chosen repeating parameters must represent al/ the primary dimensions in the problem. 4. Never pick parameters that are already dimensionless. These are TI's already, all by themselves. 5. Never pick two parameters with the same dimensions or with dimensions that differ by only an exponent. 6. Whenever possible, choose dimensional constants over dimensional variables so that only one II contains the dimensional variable. 7. Pick common parameters since they may appear in each of the II's. 8. Pick simple parameters over complex parameters whenever possible. In the present problem we cannot choose z, but we must choose from among the remaining four parameters. Therefore, we must choose two of the following parameters: t, Wo, Zo, and g. In the present problem, any two of the independent parameters would be valid according to this guideline. For illustrative purposes, however, suppose we have to pick three instead of two repeating parameters. We could not, for example, choose t, wo, and Zy, because these can form a II all by themselves (two/Z9). Suppose for example that there were three primary dimensions (m, L, and t) and two repeating parameters were to be chosen. You could not choose, say, a length and a time, since primary dimension mass would not be represented in the dimensions of the repeating parameters. An appropriate choice would be a density and a time, which together represent all three primary dimensions in the problem. Suppose an angle 9 were one of the independent parameters. We could not choose 0 as a repeating parameter since angles have no dimensions (radian and degree are dimensionless units). In such a case, one of the II's is already known, namely 0. In the present problem, two of the parameters, z and Z,, have the same dimensions (length). We cannot choose both of these parameters. (Note that dependent variable z has already been eliminated by guideline 1.) Suppose one parameter has dimensions of length and another parameter has dimensions of volume. In dimensional analysis, volume contains only one primary dimension (length) and is not dimensionally distinct from length—we cannot choose both of these parameters. If we choose time t as a repeating parameter in the present problem, it would appear in all three II's. While this would not be wrong, it would not be wise since we know that ultimately we want some nondimensional height as a function of some nondimensional time and other nondimensional parameter(s). From the original four independent parameters, this restricts us to wo, Zo, and g. In fluid flow problems we generally pick a length, a velocity, and a mass or density (Fig. 7-25). It is unwise to pick less common parameters like viscosity wu or surface tension o, since we would in general not want w or o, to appear in each of the II's. In the present problem, m and Z are wiser choices than g. It is better to pick parameters with only one or two basic dimensions (e.g., a length, a time, a mass, or a velocity) instead of parameters that are composed of several basic dimensions (e.g., an energy or a pressure). * These guidelines, while not infallible, help you to pick repeating parameters that usually lead to established nondimensional TI groups with minimal effort. In similar fashion we create the first independent TI (II,) by combining the repeating parameters with independent variable t. First independent II: = twgz: Dimensions of ll: EI) = EL = Etwgzb) = CUL Lt) CHAPTER 7 Equating exponents, Time: th=(Ure 0=1-a a=] Length: ALY=(Lob] O=a+b b=-a b=-1 TG is thus wot h=— (71-17) 2% Finally we create the second independent II (II;) by combining the repeat- ing parameters with g and forcing the TI to be dimensionless (Fig. 7-26). ii . — 237D; Second independent II: 1 = gwozo FIGURE 7-25 Dimensions of l;; EG) = (LN = fgwgzy = (LT ALE atos Itis wise to choose common parameters as repeating parameters Equating exponents, since they may appear in each of your dimensionless TI groups. Time: =) 0=-2-a; a=-2 Length: 4LO) = (LILSLi 0=l+a+b b=-1-a b=1 IE is thus º —- Mo o = 42 (-18) o lh) = mL OTOOCONO = 41) All three T's have been found, but at this point it is prudent to examine them to see if any manipulation is required. We see immediately that TI, and II, are the same as the nondimensionalized variables z* and t* defined by flo) = mL OOTOOCONO = 41) Eg. 7-6—no manipulation is necessary for these. However, we recognize |O that the third II must be raised to the power of —3 to be of the same form : as an established dimensionless parameter, namely the Froude number of Et = EmOLOOTOOCONO = (1) Eq. 7-8: Modified TI; n (3) * Mo Er q-19) 3 3, modified vê NV 9% o Such manipulation is often necessary to put the TI's into proper estab- lished form. The TI of Eq. 7-18 is not wrong, and there is certainly no mathematical advantage of Eq. 7-19 over Eq. 7-18. Instead, we like to FIGURE 7-26 say that Eg. 7-19 is more “socially acceptable” than Eq. 7-18, since it is a The II groups that result from the named, established nondimensional parameter that is commonly used in method of repeating variables are the literature. In Table 7-4 are listed some guidelines for manipulation of guaranteed to be dimensionless your nondimensional TI groups into established nondimensional parameters. because we force the overall Table 7-5 lists some established nondimensional parameters, most of exponent of all seven primary which are named after a notable scientist or engineer (see Fig. 7-27 and the dimensions to be zero. Historical Spotlight on p. 289). This list is by no means exhaustive. When- ever possible, you should manipulate your TIºs as necessary in order to con- vert them into established nondimensional parameters. TABLE 7-5 281 CHAPTER 7 Some common established nondimensional parameters or IT's encountered in fluid mechanics and heat transfer* Name Definition Ratio of Significance ut Archimedes number Ar= Po, -—p) n Gravitational force Viscous force Length Diameter Length Width Surface thermal resistance Internal thermal resistance Gravitational force Surface tension force Pressure — Vapor pressure L L Aspect ratio AR= w or D Biot number Bi= E = dl Bond number Bo= or pol! Os . P-P, Cavitation number Ca (sometimes o,) = Vê TA ( ti NP-P) sometimes mn? . 8, Darcy friction factor f=—a pv F Drag coefficient Cp= 2pVºA v2 Eckert number Ec= cpT Euler number Eu = Ar (sometimes [E pv 20V' my Fanning friction factor Cj=—— pV' , : t Fourier number Fo (sometimes 7) = 2 v (sometimes o AT|Lp? cr = 88! | p Froude number Grashof number u CT — Ta) Jakob number Ja= ST Ta) ha A Knudsen number Kn = L Lewis number Le=-+ a POoDas Dus A Lift coefficient C=5 A 2p VA Inertial pressure Wall friction force Inertial force Drag force Dynamic force Kinetic energy Enthalpy ARE YOUR PI'S DIMENSIONLESS? Pressure difference Dynamic pressure Wall friction force Inertial force Physical time Thermal diffusion time Inertial force FIGURE 7-28 Gravitational force A quick check of your algebra Buoyancy force is always wise. Viscous force Sensible energy Latent energy Mean free path length Characteristic length Thermal diffusion Species diffusion Lift force Dynamic force (Continued) 288 FLUID MECHANICS TABLE 7-5 (Continued) Name Definition Ratio of Significance Mach number Nusselt number Peclet number Power number Prandtl number Pressure coefficient Rayleigh number Reynolds number Richardson number Schmidt number Sherwood number Specific heat ratio Stanton number Stokes number Strouhal number Weber number Ma (sometimes M) = r Flow speed Speed of sound Nu = Lh Convection heat transfer k Conduction heat transfer - PLVc, LV Bulk heat transfer k a Conduction heat transfer w Power Np="53 Bontional imo pDw Rotational inertia p=" = UC Viscous diffusion a k Thermal diffusion P Static pressure difference Dynamic pressure g9BIATID pc, Buoyancy force Ra = - ku Viscous force pVL VL Inertial force Re="— => EE u v Viscous force Ri= Lg Ap Buoyancy force q” Inertial force : Cp k (sometimes y) = — CG PLA pcoV , = oDGV Stk (sometimes St) = 18uL é fL St (sometimes S or Sr) = v “pViL os We Viscous diffusion Overall mass diffusion Species diffusion Enthalpy Internal energy Meattranster Thermal capacity Particle relaxation time Characteristic flow time Characteristic flow time Period of oscillation Inertial force Surface tension force * A is a characteristic area, D is a characteristic diameter, fis a characteristic frequency (Hz), L is a characteristic length, tis a characteristic time, T is a characteristic (absolute) temperature, Vis a characteristic velocity, Wis a characteristic width, Wis a characteristic power, w is a characteristic angular velocity (rad/s). Other parameters and fluid properties in these I's include; c = speed of sound, c,, c, = specific heats, D, = particle diameter, Dag — species diffusion coefficient, h = convective heat transfer coefficient, hg = temperature, V volume flow rate, free path length, yu = viscosity, » = Kkinematic viscosity, p density, ps = solid density, p, = vapor density, o, = surface tension, and rw = thermal diffusivi latent heat of evaporation, k = thermal conductivity, P = pressure, Ti = saturation coefficient of thermal expansion, À = mean luid density, p, liquid density, p, = particle hear stress along a wall 
