1. Chapter 4 DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE

2. Dimensionless parameters significantly deepen our understanding of fluid-flow phenomena in a way which is analogous to the case of a hydraulic jack, where the ratio of piston diameters determines the mechanical advantage, a dimensionless number which is independent or the overall size of the jack. They permit limited experimental results to be applied to situations involving different physical dimensions and often different fluid properties. The concepts of dimensional analysis introduced in this chapter plus an understanding of the mechanics of the type of flow under study make possible this generalization of experimental data. The consequence of such generalization is manifold, since one is now able to describe the phenomenon in its entirety and is not restricted to discussing the specialized experiment that was performed. Thus, it is possible to conduct fewer (but highly selective) experiments to uncover the hidden facets of the problem and thereby achieve important savings in time and money.

3. Equally important is the fact that, researchers are able to discover new features and missing areas of knowledge of the problem at hand. This directed advancement of our understanding of a phenomenon would be impaired if the tools of dimensional analysis were not available. Many of the dimensionless parameters may be viewed as a ratio of a pair of fluid forces, the relative magnitude indicating the relative importance of one of the forces with respect to the other. If some forces in a particular flow situation are very much larger than a few others, it is often possible to neglect the effect of the smaller forces and treat the phenomenon as though it were completely determined by the major forces. This means that simpler (but not necessarily easy) mathematical and experimental procedures can be used to solve the problem. For situations with several forces of the same magnitude (inertial, viscous, and gravitational forces) special techniques are required.

4. 4.1 DIMENSIONAL HOMOGENEITY AND DIMENSIONLESS RATIOS Solving practical design problems in fluid mechanics requires both theoretical developments and experimental results. By grouping significant quantities into dimensionless parameters, it is possible to reduce the number of variables appealing and to make this compact result (equations or data plots) applicable to all similar situations. If one were to write the equation of motion ∑F = ma for a fluid particle, including all types of force terms that could act (gravity, pressure, viscous, elastic, and surface-tension forces), an equation of the sum of these forces equated to ma (the inertial force) would result.

5. Each term must have the same dimensions - force. The division of each term of the equation by any one of the terms would make the equation dimensionless. For example, dividing through by the inertial force term would yield a sum of dimensionless parameters equated to unity. The relative size of any one parameter, compared with unity, would indicate its importance. If divide the force equation through by a different term, say the viscous force term, another set of dimensionless parameters would result. Without experience in the flow case it is difficult to determine which parameters will be most useful.

6. An example of the use of dimensional analysis and its advantages is given by considering the hydraulic jump. The momentum equation for this case (4.1.1) The right-hand side - the inertial forces; left-hand side - the pressure forces due to gravity. These two forces are of equal magnitude, since one determines the other in this equation. The term γy 1 2 /2 has the dimensions of force per unit width, and it multiplies a dimensionless number which is specified by the geometry or the hydraulic jump.

7. If one divides this equation by the geometric term 1 - y 2 /y 1 and a number representative of the gravity forces, one has (4.1.2) The left-hand side - the ratio of the inertia and gravity forces, even though the explicit representation of the forces has been obscured through the cancellation of terms that are common in both the numerator and denominator. This ratio is equivalent to a dimensionless parameter, actually the square of the Froude number. This ratio of forces is known once the ratio y 2 /y 1 is given, regardless or what the values y 2 and y 1 are. From this observation one can obtain an appreciation or the increased scope that Eq. (4.1.2) affords over Eq. (4.1.1) even though one is only a rearrangement of the other.

8. In writing the momentum equation which led to Eq. (4.1.2) only inertia and gravity forces were included in the original problem statement. But other forces, such as surface tension and viscosity, are present (were neglected as being small in comparison with gravity and inertia forces). However, only experience with the phenomenon, or with phenomena similar to it, would justify such an initial simplification. For example, if viscosity had been included because one was not sure of the magnitude of its effect, the momentum equation would become This statement is more complete than that given by Eq. (4.1.2). However, experiments would show that the second term on the left- hand side is usually a small fraction of the first term and could be neglected in making initial tests on a hydraulic jump.

9. In the last equation one can consider the ratio y 2 /y 1 to be a dependent variable which is determined for each of the various values of the force ratios, V 1 2 /gy 1 and F viscous /γy 1 2, which are the independent variables. From the previous discussion it appears that the latter variable plays only a minor role in determining the values of y 2 /y 1. Nevertheless, if one observed that the ratios of the forces, V 1 2 /gy 1 and F viscous /γy 1 2, had the same values in two different tests, one would expect, on the basis of the last equation, that the values of y 2 /y 1 would be the same in the two situations. If the ratio of V 1 2 /gy 1 was the same in the two tests but the ratio F viscous /γy 1 2, which has only a minor influence for this case, was not, one would conclude that the values of y 2 /y 1 for the two cases would be almost the same.

10. This is the key to much of what follows. For if one can create in a model and force ratios that occur on the full-scale unit, then the dimensionless solution for the model is valid for the prototype also. Often it is not possible to have all the ratios equal in the model and prototype. Then one attempts to plan the experimentation in such a way that the dominant force ratios are as nearly equal as possible. The results obtained with such incomplete modeling are often sufficient to describe the phenomenon in the detail that is desired. Writing a force equation for a complex situation may not be feasible, and another process, dimensional analysis, is then used if one knows the pertinent quantities that enter into the problem. In a given situation several of the forces may be of little significance, leaving perhaps two or three forces of the same order or magnitude. With three forces of the same order or magnitude, two dimensionless parameters are obtained; one set of experimental data on a geometrically similar model provides the relations between parameters holding for all other similar flow cases.

11. 4.2 DIMENSIONS AND UNITS The dimensions of mechanics are force, mass, length, and time; they are related by Newton's second law of motion, F = ma (4.2.1) For all physical systems, it would probably be necessary to introduce two more dimensions, one dealing with electromagnetics and the other with thermal effects. For the compressible work in this text, it is unnecessary to include a thermal unit, because the equations or state link pressure, density, and temperature. Newton's second law of motion in dimensional form is F = MLT -2 (4.2.2) which shows that only three of the dimensions are independent. F is the force dimension, M the mass dimension, L the length dimension, and T the time dimension. One common system employed in dimensional analysis is the MLT system.

12. Table 4.1 Table 4.1 Dimensions of physical quantities used in fluid mechanics

13. 4.3 THE П THEOREM The Buckingham Π theorem proves that, in a physical problem including n quantities in which there are m dimensions, the quantities can be arranged into n - m independent dimensionless parameters. Let A 1, A 2, A 3.... A n be the qualities involved, such as pressure, viscosity, velocity, etc. All the quantities are known to be essential to the solution, and hence some functional relation must exist (4.3.1) If Π 1, Π 2,..., represent dimensionless groupings of the quantities A 1, A 2, A 3,..., then with m dimensions involved, an equation of the following form exists (4.3.2)

14. The method of determining the Π parameters is to select m of the A quantities, with different dimensions, that contain among them the m dimensions, and to use them as repeating variables together with one of the other A quantities for each Π. For example, let A 1, A 2, A 3 contain M, L and T, not necessarily in each one, but collectively. Then the Π parameters are made up as (4.3.3) - the exponents are to be determined  each Π is dimensionless. The dimensions of the A quantities are substituted, and the exponents of M, L, and T are set equal to zero respectively. These produce three equations in three unknowns for each Π parameter, so that the x, y, z exponents can be determined, and hence the Π parameter. If only two dimensions are involved, then two of the A quantities are selected as repeating variables, and two equations in the two unknown exponents are obtained for each Π term. In many cases the grouping of A terms is such that the dimensionless arrangement is evident by inspection. The simplest case is that when two quantities have the same dimensions, e.g., length, the ratio or these two terms is the Π parameter.

15. Example 4.1 The discharge through a horizontal capillary tube is thought to depend upon the pressure drop per unit length, the diameter, and the viscosity. Find the form of the equation. Solution The quantities are listed with their dimensions:

16. Then Three dimensions are used, and with four quantities there will be one Π parameter: Substituting in the dimensions gives The exponents of each dimension must be the same on both sides of the equation. With L first,

17. And similarly for M and T From which x 1 = 1, y 1 = -1, z 1 = -4, and After solving for Q, From which dimensional analysis yields no information about the numerical value of the dimensionless constant C; experiment (or analysis) shows that it is π/128 [Eq. (5.4.10a)].

18. Example 4.2 A V-notch weir is a vertical plate with a notch of angle φ cut into the top of it and placed across an open channel. The liquid in the channel is backed up and forced to flow through the notch. The discharge Q is some function of the elevation H of upstream liquid surface above the bottom of the notch. In addition, the discharge depends upon gravity and upon the velocity of approach V 0 to the weir. Determine the form of discharge equation. Solution A functional relation Is to be grouped into dimensionless parameters. φ is dimensionless; hence, it is one Π parameter. Only two dimensions are used, L and T. If and H are the repeating variables.

19. Then From which, and This can be written In which both f and f 1 are unknown functions. After solving for Q, Either experiment or analysis is required to yield additional information about the function f 1.

20. If H and V 0 were selected as repeating variables in place of g and H, From which, and Since any of the Π parameters can be inverted or raised to any power without affecting their dimensionless status, The unknown function f 2 has the same parameters as f 1, but it could not be the same function. The last form is not very useful, in general, because frequently V 0 may be neglected with V-notch weirs. This shows that a term of minor importance should not be selected as a repeating variable.

21. Example 4.3 The thrust due to any one of a family of geometrically similar airplane propellers is to be determined experimentally from a wind-tunnel test on a model. Use dimensional analysis to find suitable parameters for plotting test results. Solution The thrust F T depends upon speed of rotation ω, speed of advance V 0, diameter D, air viscosity μ, density ρ, and speed of sound c. The function is to be arranged into four dimensionless parameters, since there are seven quantities and three dimensions. Starting first by selecting ρ, ω, and D as repeating variables.

22. By writing the simultaneous equations in x l, y l, z l, etc., as before and solving them gives, Solving for the thrust parameter leads to.. Since the parameters can be recombined to obtain other forms, the second term is replaced by the product of the first and second terms, VDρ/μ, and the third term is replaced by the first term divided by the third term, V 0 /c; thus Of the dimensionless parameters, the first is probably of the most importance since it relates speed of advance to speed of rotation. The second parameter is a Reynolds number and accounts for viscous effects. The last parameter, speed of advance divided by speed of sound, is a Mach number, which would be important for speeds near or higher than the speed of sound. Reynolds effects are usually small, so that a plot of F T /ρω 2 D 4 against V 0 /ωD should be most informative.

23. The steps in a dimensional analysis may be summarized as follows: 1. 1.Select the pertinent variables (requires some knowledge of the process). 2. 2.Write the functional relations, e.g., 3. 3.Select the repeating variables. (Do not make the dependent quantity a repeating variable.) These variables should contain all the m dimensions or the problem. Often one variable is chosen because it specifies the scale, another the kinematic conditions; and in the cases of major interest in this chapter one variable which is related to the forces or mass of the system, for example, D, V, ρ, is chosen. 4. 4.Write the Π parameters in terms of unknown exponents, e.g.,

24. 5. 5.For each of the Π expressions write the equations of the exponents, so that the sum of the exponents of each dimension will be zero. 6. 6.Solve the equations simultaneously. 7. 7.Substitute back into the Π expressions of step 4 the exponents to obtain the dimensionless Π parameters. 8. 8.Establish the functional relation or solve for one of the Π's explicitly: 9. 9.Recombine, if desired, to alter the forms of the Π parameters, keeping the same number or independent parameters.