Modeling with Geometric Similarity and Dimensional Analysis

Transcription

1 Modeling with Geometric Similarity and Dimensional Analysis Math 232, Fall 2018 Instructor: Dr Doreen De Leon 1 Introduction In the process of constructing a model, we need to identify and classify the variables that influence the behavior of the physical system We must then determine appropriate relationships between those variables that we retain for incorporation into the model In some situations, the discovery of the nature of the relationship between independent variables and dependent variables comes about by making some reasonable assumption based on a law of nature or on previous experience and construction of a mathematical model In some cases, a series of experiments might be conducted to determine how the variables are related This might result in a figure or a table, and then we may apply an appropriate curve-fitting method to predict the value of the dependent variable Most of our work to this point has focused on modeling observations that were dependent on one variable In reality, of course, most real-world problems are characterized by relations involving several variables Dimensional analysis is a method for helping to determine how the selected variables are related and for significantly reducing the amount of experimental data that must be collected It is based on the premise that physical quantities have dimensions and that physical laws are not altered by changing the units measuring dimension Thus, the phenomenon under investigation can be described by a dimensionally consistent relation among the variables A dimensional analysis provides qualitative information about the model It is especially important when it is necessary to conduct experiments as part of the modeling process, because the method is helpful in testing the validity of including or neglecting a particular factor; in reducing the number of experiments that need to be conducted to make predictions; and in improving the usefulness of the results by providing alternatives for the parameters employed to present them Although dimensional analysis is very useful for validating newly developed mathematical models or for confirming that certain formulas makes sense, applying the tools of dimensional analysis does not always yield desirable results In addition, equations derived using dimensional analysis will still have unknowns for which we must solve 1

2 2 Modeling Using Geometric Similarity Geometric similarity is a concept that you should have encountered in multiple math courses, such as geometry (in middle or high school) and calculus It is another tool that can be used to simplify the mathematical modeling process Definition 21 Two objects are geometrically similar if there is a one-to-one correspondence between points of the objects, such that the ratio of the distances between corresponding points is constant for all pairs of points Examples include: similar triangles and the use of scaling on a map We will look at an example of modeling with geometric similarity before we begin our discussion of dimensional analysis One advantage of two objects being geometrically similar is a simplification in some computations, such as volume and surface area Suppose that we have two similar boxes, a smaller box with dimension l 1, w 1, h 1 and a larger box of dimensions l 2, w 2, h 2 Then, the ratio of their volumes is given by V 1 = l 1w 1 h 1 = k 3, V 2 l 2 w 2 h 2 since l 2 = kl 1, w 2 = kw 1, h 2 = kh 1 Similarly, the ratio of the total surface areas is given by S 1 S 2 = 2l 1h 1 + 2w 1 h 1 + 2w 1 l 1 2l 2 h 2 + 2w 2 h 2 + 2w 2 l 2 = k 2 Note also that the surface area and volume may be expressed as a proportionality in terms of a chosen dimension, the characteristic dimension Suppose we choose the length as the characteristic dimension Then Or, we can say that S 1 l 2 1 S 1 = k 2 = l2 1 S 2 = S 2 l 2 2 l 2 2 = constant holds for any two geometrically similar objects; ie, S l 2 Similarly, we can show that volume is proportional to the cube of the length, or V l 3 Example 21 (Raindrops from a Motionless Cloud, Take 2) Suppose we wish to determine the terminal velocity of a raindrop from a motionless cloud If we draw a free body diagram, we will see that the only forces acting on the raindrop are gravity and air resistance, or drag We will make the following assumptions: atmospheric drag on the raindrop is proportional to the product of the drop s surface area S and the square of its velocity v; 2

3 the weight of the raindrop w is proportional to its mass m (assuming constant gravity); and all raindrops are geometrically similar By Newton s second law, F = F g F d = ma, and under terminal velocity a = 0, so Newton s second law is reduced to F g F d = 0, or F g = F d Assuming that F d Sv 2 and the force of gravity is proportional to the weight w, F g w Since m w, we have F g m Assuming that all raindrops are similar, we can relate surface area and volume Since S l 2 and V l 3, where l is the characteristic dimension, it follows that Therefore, l S 1 2 V 1 3 S V 2 3 Since weight and mass are proportional to volume, applying the transitive rule for proportionality gives S m 2 3 Since F g = F d, we have that or m m 2 3 v 2, v m Dimensional Analysis and Similarity - Section 32 of Heinz 31 Dimensions as Products The study of physics is based on concepts such as mass, length, time, velocity, acceleration, force, energy, work, and pressure To each such concept, a unit of measurement is assigned A physical law, such as Newton s law, is true, provided that the units of measurement are consistent Thus, if mass is measured in kilograms and acceleration in meters per second 3

4 squared, then the force must be measured in Newtons These units of measure belong to the MKS (meter-kilogram-second) system In the English system, mass is in slugs, acceleration is in feet per second squared, and force is in pounds The three primary physical quantities that we will consider in this section are mass, length, and time We associate with these quantities the dimensions M, L, and T, respectively The dimensions are symbols that reveal how the numerical value of a quantity changes when the units of measurement change in certain ways The dimensions of other quantities follow from definitions or from physical laws and are expressed in terms of M, L, and T For example, acceleration is represented as LT 2 There are other, more complicated physical quantities that are not usually defined directly in terms of mass, length, and time, but in terms of other quantities, such as velocity For example, momentum is the product of mass and velocity Thus, its dimension is MLT 1 The basic definition of a quantity may also involve dimensionless constants, which are ignored in finding the dimension Thus, the dimension of kinetic energy, which is one half the product of the mass and the velocity squared, is M ( LT 1) 2 = ML 2 T 2 Quick recap so far: 1 The concept of dimension is based on three physical quantities, mass, length, and time These quantities are measured in some appropriate system of units whose choice does not affect the assignment of dimensions 2 There are other physical quantities, such as area and velocity, that are defined as simple products involving only mass, length, or time 3 There are other, more complex physical properties, such as energy, whose definitions involve quantities other than mass, length, and time, but which we may use definitions and physical laws to write as products only involving mass, length, and/or time 4 To each product, there is assigned a dimension, ie, an expression of the form where n, p, and q are real numbers M n L p T q, (31) When a basic dimension is missing from a product, the corresponding exponent is understood to be 0 When n, p, and q are all zero in an expression of the form (31), so that the dimension reduces to M 0 L 0 T 0, the quantity is said to be dimensionless One must be careful in forming sums of products, because we cannot add products that have unlike dimensions 4

5 Another important concept is that of dimensional homogeneity In general, an equation that is true regardless of the system of units in which the variables are measured is said to be dimensionally homogeneous For example, the equation giving the time a body falls a distance s under gravity (neglecting air ressistance) is 2s t = g This expression is dimensionally homogeneous Example 31 (Simple Pendulum) Consider a simple pendulum of length r, which has a mass m attached to it Let θ denote the initial angle of displacement from the vertical One characteristic of interest is the period (the time required for the pendulum to swing through one complete cycle, returning to its starting point) We represent the period of the pendulum as t We will analyze this problem using dimensional analysis, under the following assumptions: the hinge is frictionless; the mass is concentrated at one end of the pendulum; and the drag force is negligible First, analyze the dimensions of the variables Variable m g t r θ Dimension M LT 2 T L M 0 L 0 T 0 Next, find all of the dimensionless products among the variables Any such product must take the form m a g b t c r d θ e, (32) and therefore, must have dimension (M) a (LT 2 ) b (T ) c (L) d (M 0 L 0 T 0 ) e = M a L b+d T c 2b (33) Therefore, a product of the form (32) is dimensionless if and only if This gives the linear system of equations M a L b+d T c 2b = M 0 L 0 T 0 a = 0 b + d = 0 2b + c = 0 We can see that there are infinitely many solutions The rules for choosing arbitrary variables when doing dimensional analysis are the following 5

6 (1) Choose the dependent variable so it will only appear once (2) Select any variable that will speed up the solving of the system; eg, choose a variable that appears in all (or most) equations With that in mind, we should choose b as the independent variable, and therefore, the solution to the system is a = 0; c = 2b; d = b The other independent variable is e, because it does not appear in the system One dimensionless product is obtained by setting b = 0 and e = 1, giving a = c = d = 0 A second dimensionless product is obtained by setting b = 1 and e = 0, giving a = 0, c = 2, and d = 1 These solutions give the dimensionless products Π 1 = m 0 g 0 t 0 r 0 θ 1 = θ Π 2 = m 0 g 1 t 2 r 1 θ 0 = gt2 r Assuming that t = f(r, m, g, θ), we need to determine more about the function f First, if the units of m were to change by some factor, the measure of the period will not change, because it is measured in units (T ) of time Also note that because m is the only factor whose dimension contains M, it cannot appear in the model If the scale of the units (L) for measuring length were changed, then it also can t change the measure of the period, either In order to ensure that this is correct, then g and r must appear together in the model as some power of g Finally, if we make the units (T ) that measure time smaller by a factor r of, say, 10, then the measure of the period will increase by this same factor Therefore, to r have the dimension of T on the right-hand side of t = f(r, m, g, θ), r must appear as g, because T appears to the power of -2 in the dimension of g Note that we have not found anything to restrict the angle θ Therefore, the equation for the period should take the form r t = g h(θ), where h is a function to be determined or approximated experimentally Question: Is it reasonable to assume that the friction and drag forces are negligible? Answer: The model obtained so far is t = rgh(θ) If θ is kept constant, say θ = θ 0, while r is allowed to vary, we have t 1 r1 /gh(θ 0 ) = t 2 r2 /gh(θ 0 ) = r1 r 2 6

7 In other words, the model predicts that t will vary as r for constant θ A plot of t versus r does yield a somewhat linear curve We could do a similar analysis, holding r constant and varying θ, in which case we obtain t 1 t 2 = h(θ 1) h(θ 2 ) A plot of t versus θ for several observations would gives us an idea of the nature of h, which we may be able to approximate by regression To use the model to predict t, it might be convenient to use the equation g t r = h(θ), g and plot t r by g r g versus θ Then, for a given value of θ, we could determine t, multiply it r for a given r, and then determine t The application of dimensional analysis to a real-world problem is based on the assumption that the solution to the problem is given by a dimensionally homogeneous equation in terms of the appropriate variables Given this, we must determine the form of the desired equation by finding a dimensionless equation and then solving for the dependent variable Example 32 (Wind Force on a Van) Suppose you are driving a van down a highway with gusty winds How does the speed of your vehicle affect the wind force you are experiencing? The force F of wind on the van is certainly affected by the speed v of the van and the surface area A of the van directly exposed to the wind s direction Thus, we might hypothesize that the force is proportional to some power of the speed times some power of the surface area; ie, F = kv a A b, (34) for some dimensionless constant k following: Analyzing the dimensions of the variables gives the Variable F k v A Dimension MLT 2 M 0 L 0 T 0 LT 1 L 2 Thus, dimensionally, equation (34) becomes MLT 2 = ( M 0 L 0 T 0) ( LT 1) a ( L 2 ) b 7

8 Obviously there is something wrong here, because the dimension M for mass does not appear on the right-hand side So, we must revise equation (34) What is missing from our original assumptions regarding the factors that affect the wind force on the van? After some reflection, we might consider that density ρ has an effect Including density as a factor makes the model F = kv a A b ρ c (35) Since density is mass per unit volume, the dimension of density is ML 3 Thus, dimensionally, equation (35) is MLT 2 = ( M 0 L 0 T 0) ( LT 1) a ( L 2 ) b ( ML 3 ) c Equating exponents on both sides of the equation gives the system of equations c = 1 a + 2b 3c = 1, a = 2 whose solution is a = 2, b = 1, and c = 1 Thus, the model is F = kv 2 Aρ Table 31 gives a summary of the dimensions of some common physical quantities Table 31: Dimensions of physical quantities Mass M Momentum MLT 1 Length L Work ML 2 T 2 Time T Density ML 3 Velocity LT 1 Viscosity ML 1 T 1 Acceleration LT 2 Pressure ML 1 T 2 Specific weight ML 2 T 2 Surface tension MT 2 Force MLT 2 Power ML 2 T 3 Frequency T 1 Rotational inertia ML 2 Angular velocity T 1 Torque ML 2 T 2 Angular acceleration T 2 Entropy ML 2 T 2 Angular momentum ML 2 T 1 Heat ML 2 T 2 Energy ML 2 T 2 Dynamic viscosity ML 1 T 1 8

9 32 The Process of Dimensional Analysis Our goal now is to investigate how to use dimensionless products to find all possible dimensionally homogeneous equations among the variables The key result is Buckingham s Pi Theorem, which summarizes the theory of dimensional analysis 1 The theorem loosely states that if we have a physically meaningful equation involving a certain number, say n, of physical variables, and those variables are expressed in terms of m fundamental (or primary) physical quantities, then the original expression is equivalent to an equation involving n m dimensionless variables constructed from the original variables This provides a method for computing sets of dimensionless parameters from the given variables, even if the form of the equation is still unknown However, the choice of dimensionless parameters is not unique; Buckingham s theorem only provides a way of generating sets of dimensionless parameters, and will not choose the most physically meaningful In mathematical terms, if we have a physically meaningful equation, such as f(q 1, q 2,, q n ) = 0, where the q i are n physical variables, and they are expressed in terms of m independent physical units, then the above equation can be restated as φ(π i, Π 2,, Π n m ) = 0, where the Π i are dimensionless parameters constructed from the q i by n m equations of the form Π i = q m 1 1 q m 2 2 qn mn, where the exponents m i are rational numbers The use of Π i to represent the dimensionless parameters was introduced by Edgar Buckingham in his original 1914 paper On physically similar systems; illustrations of the use of dimensional equations, in Physical Review, 4, pp Formal statement of the theorem: Theorem 31 (Buckingham s Pi Theorem) Any physically meaningful relation F (Q 1, Q 2,, Q n ) = 0, where Q i are n physical variables such that Q i 0 and the Q i are expressed in terms of m independent physical units, is equivalent to a relation of the form ψ(π 1, Π 2,, Π n m ) = 0, 1 A portion of this discussion is taken from Principles of Mathematical Modeling, second edition, by Clive L Dym and a Wikipedia entry on Buckingham s Pi Ttheorem 9

10 where Π i are dimensionless parameters constructed from the Q i by n m equations of the form Π i = Q m 1 1 Q m 2 2 Q mn n, where m i Q Alternately, the relation can be written in the form Π 1 = φ(π 2, Π 3,, Π n m ) 33 Similarity Suppose that we are interested in the effects of wave action on a large ship at sea, heat loss of a submarine and the drag force it experiences in its underwater environment, or the wind effects on an aircraft wing Frequently, we study a scaled-down model in a simulated environment to predict accurately the performance of the physical system The actual physical system for which the predictions are to be made is called the prototype The question is: How do we scale experiments in the laboratory to ensure that the effects observed for the model will be the same as those experienced by the prototype? The dimensional products resulting from dimensional analysis of the problem can provide insight into how the scaling for a model should be done The idea comes from Buckingham s Pi Theorem If the physical system can be described by a dimensionally homogeneous equation in n variables, then it can be put into the form ψ (Π 1, Π 2,, Π p ) = 0 for a set of p dimensionless products (where p = n m, m being the number of independent physical units) Assume that the dependent variable of the problem appears only in the product Π p and that Π p = φ (Π 1, Π 2,, Π p 1 ) For the solution to the model and the prototype to be the same, it is sufficient that the value of all independent dimensionless products Π 1, Π 2,, Π p 1 be the same for both the model and the prototype For example, suppose the Reynold s number vrρ appears as one of the dimensionless products in a fluid mechanics problem, where v represents the fluid velocity, r is a characteristic µ dimension, such as the length of a ship, ρ represents the fluid density, and µ represents the fluid viscosity Then, let v m, r m, ρ m, and µ m be the values for the scaled model For the effects on both the prototype and the model to be the same, we need vrρ µ = v mr m ρ m µ m 10

11 This equation is known as a design condition to be satisfied by the model For example, if we need to scale the length of the prototype by a factor of 10 (so, r = 10r m ), then the same Reynold s number can be obtained by using the same fluid (so ρ m = ρ and µ m = µ) and changing the velocity (so v m = 10v) If this is impractical, then a different fluid can be used so that the equation is satisfied 4 Examples of Low Complexity 41 Kepler s Third Law Our goal is to calculate the orbital period of a planet, T P On what variables does the orbital period depend? (mean) distance from the sun, r; gravitational force F G ; and the mass m of the planet Analyzing the dimensions of the variables gives Variable r F G m T P Dimension D MLT 2 M T Next, we must find all dimensionless products among the variables Any such product must be of the form r a F b Gm c T d P, and so, must have dimension L a (MLT 2 ) b M c T d Therefore, the product is dimensionless if and only if a + b = 0 b + c = 0 2b + d = 0 This system has infinitely many solutions of the form a = b c = b d = 2b 11

12 We find Π 1 by letting b = 1, giving a = 1, c = 1, d = 2, or Π 1 = r 1 F g m 1 T 2 P Since there is only one nondimensional product, it must be constant, so we have where c is a constant, or r 1 F g m 1 T 2 P = c, T P = c 1 mr F G, where c 1 is a nonnegative constant Since this formula does not rely on the fact that the paths of the planets are elliptical, we may assume that the planets revolve around the sun in a circular orbit Applying Newton s second law, we have ( ) 4π 2 r F G = ma = m T 2 P = 4π 2 rm TP 2 Solving for T P gives and so c 1 = 2π mr T P = 2π, F G 42 Drag Force of Object in a Fluid Our goal is to analyze the drag force of an object moving through a fluid We make the assumption that the objects are spherical The motion of the spherical objects is damped by the surrounding fluid The relevant variables are damping force F d ; radius of the sphere r; the relative velocity between the sphere and the fluid, v; and the fluid viscosity µ Analyzing the dimensions of the variables gives Next, we must find all dimensionless products among the variables Any such product must be of the form F m 1 d r m 2 v m 3 µ m 4 = (MLT 2 ) m 1 L m 2 (LT 1 ) m 3 (ML 1 T 1 ) m 4, 12

13 Variable F d r v µ Dimension MLT 2 L LT 1 ML 1 T 1 and so, must have dimension M m 1 m 4 L m 1+m 2 +m 3 m 4 T 2m 1 m 3 m 4 Therefore, the product is dimensionless if and only if m 1 + m 4 = 0 m 1 + m 2 + m 3 m 4 = 0 2m 1 m 3 m 4 = R 2 R 1 R R 2+R 3 R R R 1 R Therefore, m 4 is a free variable, and we may solve for all other variables in terms of m 4 Alternately, we may choose m 1 to be the free variable (since it appears in each equation, as well), giving m 4 = m 1 m 2 = m 4 = m 1 m 3 = m 4 = m 1 Let m 1 = 1 Then m 2 = m 3 = m 4 = 1, and we have one dimensionless product, Therefore, where c is a constant, and Π 1 = F d rvµ F d rvµ = c, F d = crvµ Other methods may be used to show that c = 6π, so F d = 6πrvµ 13

14 5 Applications of Medium-Complexity 51 Terminal Velocity of a Raindrop Revisited Consider the problem of determining the terminal velocity v of a raindrop falling from a motionless cloud We have looked at this problem before, taking a simpler view of it Now, we will look at the problem using dimensional analysis First, what are the variables influencing the behavior of the raindrop? Obviously, the terminal velocity will depend on the size of the raindrop given by, say, its radius r The density ρ of the air and the viscosity µ of the air will also affect the behavior (since viscosity measures the resistance to motion, which, in gases, is caused by collisions between fast-moving molecules) Another important variable is the acceleration due to gravity g Although the surface tension of the raindrop is a factor that influences the behavior of the raindrop s descent, we will ignore it here If necessary, surface tension can be taken into account in a later, refined model Analyzing the dimensions of the variables gives Variable v r g ρ µ Dimension LT 1 L LT 2 ML 3 ML 1 T 1 Next, we must find all dimensionless products among the variables Any such product must be of the form v a r b g c ρ d µ e, (51) and, so, must have dimension ( LT 1 ) a L b ( LT 2) c ( ML 3 ) d ( ML 1 T 1) e Therefore, a product of the form (51) is dimensionless if and only if the following system of equations is satisfied The system has infinitely many solutions d + e = 0 a + b + c 3d e = 0 a 2c e = 0 b = 3 2 d 1 2 a c = 1 2 d 1 2 a e = d, 14

15 where a and d are arbitrary One dimensionless product Π 1 is obtained by setting a = 1, d = 0, and another, Π 2 is obtained by setting a = 0, d = 1 So, Π 1 = vr 1 2 g 1 2, and Π2 = r 3 2 g 1 2 ρµ 1 You should check the results to verify that the products are dimensionless Then, according to Buckingham s Pi Theorem, there is a function ψ such that ( ) ψ vr 1 2 g , r 2 g 2 ρµ 1 = 0 Assuming that we can solve this equation for vr 1 2 g 1 2 as a function of Π 2, it follows that ( ) v = r 3 2 g 1 2 ρ rgφ, µ where φ is some function of Π 2 52 A Damped Pendulum We have already looked at a simple pendulum, in which we ignored the effects of a drag force and of friction Now, we will analyze the spring, this time incorporating the effects of drag on the motion of the pendulum We continue to assume that the hinge is frictionless and that the mass is concentrated at the end of the pendulum Let F d represent the total drag force Our goal is to determine the period of a pendulum of length r with a mass m attached to it; denote θ as the initial angle of displacement from the vertical First, we need to determine a model for the drag force For our pendulum, it is reasonable to assume that either the drag force is proportional to the velocity, v: F d v; ie, F d = kv; or the drag force is proportional to the square of the velocity: F d v v ; ie, F d = kv v (sometimes simply written as F d = kv 2, when we are considering v as speed) For simplicity, we will assume that F d = kv To simplify our analysis, we will use the dimensional constant k = F d, which has dimension v MLT 2 LT 1 = MT 1 15

16 Next, analyze the dimensions of the variables Variable m g t r θ k Dimension M LT 2 T L M 0 L 0 T 0 MT 1 Then, find all of the dimensionless products among the variables Any such product must take the form m a g b t c r d θ e k f, (52) and therefore, must have dimension (M) a (LT 2 ) b (T ) c (L) d (M 0 L 0 T 0 ) e (MT 1 ) f = M a+f L b+d T c 2b f (53) Therefore, the product (52) is dimensionless if and only if a + f = 0 b + d = 0 2b + c f = 0 We have three equations in six unknowns, and we would like to choose solutions so that t appears in only one of the dimensionless products So, we choose c, e, and f as the arbitrary variables, giving the solutions a = f b = 1 (c f) 2 d = b Setting c = 1, e = f = 0, we obtain a = 0, b = 1 2, d = 1 Setting e = 1, c = f = 0, we 2 obtain a = b = d = 0 And, setting c = e = 0, f = 1, we obtain a = 1, b = 1 2, d = 1 2 Therefore, we have the following dimensionless products Π 1 = g 1 2 tr 1 2 Π 2 = θ, and Π 3 = m 1 g 1 2 r 1 2 k From Buckingham s Pi Theorem, there is a function ψ such that ( g ψ t r, θ, k ) r = 0 m g 16

17 g Assuming that we can solve this equation for t, we obtain r t = where φ is some function of Π 2 and Π 3 ( r g φ θ, k m To test the model, we could do something similar to what we discussed in Example 31, but this would be somewhat problematic, since our goal is to hold parameters of the function φ constant, while varying other parameters In this case, however, we cannot vary r while keeping Π 3 constant We can, however, try to vary r so that r/m stays constant r g g To use the model in a predictive sense, one could plot t r versus k r m g of θ To be effective, this would require several trials with various values of the data is obtained, a model can be determined using regression ), for various values k r m g Once Choosing Among Competing Models Since dimensional analysis only involves algebra, as opposed to application of physical principles, it might be a good idea to develop multiple models under different assumptions before performing experiments, which could potentially be very costly There are three models we can look at for the pendulum problem: (1) no drag forces: t = r g h(θ); (2) drag force proportional to v: t = (3) drag force proportional to v 2 : t = ( r g φ θ, k m r g φ r g ( θ, k 1r m ) ) ; and Experimentation would indicate which model is best under different circumstances 17

18 6 Applications of High-Complexity 61 Draining Cylinder Consider a right cylinder of cross-sectional area A filled to depth H with a perfect liquid of density ρ A hole of area a is made in the bottom of the container, and the liquid drains under the influence of gravity with acceleration g 2 Determine T, the time for the container to empty, in terms of the parameters A, H, ρ, a, and g Analyzing the dimensions of the variables gives Variable T A H ρ a g Dimension T L 2 L ML 3 L 2 LT 2 Next, we must find all dimensionless products among the variables Any such product must be of the form T m 1 A m 2 H m 3 ρ m 4 a m 5 g m 6, (61) and, so, must have dimension T m 1 ( L 2) m 2 L m 3 ( ML 3) m 4 ( L 2 ) m 5 ( LT 2 ) m 6 Therefore, a product of the form (51) is dimensionless if and only if the following system of equations is satisfied m 1 2m 6 = 0 m 4 = 0 2m 2 + m 3 3m 4 + 2m 5 + m 6 = 0 Clearly, this system has infinitely many solutions, with m 4 = 0 m 1 = 2m 6 m 3 = 2m 2 2m 5 m 6 Find Π 1 by letting m 6 = 1, m 2 = 0, m 5 = 0 to obtain m 1 = 2 and m 3 = 1 Find Π 2 by letting m 6 = 0, m 2 = 1, and m 5 = 0 to obtain m 1 = 0 and m 3 = 2 Finally, find Π 3 by letting m 6 = 0, m 2 = 1, and m 5 = 1 to obtain m 1 = 0 and m 3 = 0 Our dimensionless constants are then Π 1 = T 2 H 1 g Π 2 = AH 2 Π 3 = A 1 a 2 This example is adapted from James Graham-Eagle s paper, The Draining Cylinder in the The College Mathematics Journal, Volume 40, Number 5, November 2009, pp

19 Therefore, according to Buckingham s Pi Theorem, there is a function ψ such that ψ ( T 2 H 1 g, AH 2, A 1 a ) = 0 If we assume that we can solve for T 2 H 1 g, we will obtain a function of the form ( H A T = g φ H, a ) 2 A Let us see if we can work towards determining an exact expression Suppose that both A and a are scaled by the same value β It seems reasonable that T is not affected by this change This implies that ( H βa T = g φ H, βa ) ( H βa = 2 βa g φ H, a ) 2 A This demonstrates that φ(x, y) is independent of x, so we now have H ( a ) T = g φ A Can we do better? Consider the evolution of the system as a function of time Since the volume in the cylinder is Ah, where h represents the height of the fluid at any time, the rate at which the volume decreases is d(ah) This equals the rate at which the liquid drains dt from the hole in the bottom of the container, av, where v is the speed at which the liquid flows from the hole Since A is constant, equating the two quantities yields the differential equation A dh dt = av If we assume that v depends only on h and g, which seems reasonable since the pressure at the bottom of the container is what drives the liquid through the hole, and pressure depends on h and g, then the differential equation is separable and may be solved by integrating both sides over the duration of the experiment = 0 H H 0 1 T v dh = a 0 A dt 1 v dh = a A T = a A H ( a ) g φ A 19

20 Differentiating the above equation with respect to H gives ) 1 v(h) = 1 2 gh a A φ ( a A Since v does not depend on a or A, it follows that φ(x) = C x now have that H T = g C A a for some constant C So, we Now, if a = A, then the entire base of the container is gone and the liquid falls as a rigid body under the influence of gravity Since, in evacuating the container, the liquid falls a distance H from rest, it follows that 2H T = when a = A g Thus, C = 2, and we obtain or Torricelli s law T = A a 2H g, Note: Typically, Torricelli s law appears as a problem in Calculus and differential equations textbooks The standard solution approach involves equating the rate of loss of water in the container with the rate at which it passes through the hole, applying Torricelli s law to determine the latter expression For our problem involving a right circular cylinder, this reduces to solving the separable first order equation A dh dt = a 2gh 62 Explosion Analysis In excavating and mining operations, it is obviously very important to be able to predict the size of an explosion resulting from the use of TNT (or some other explosive agent) in a given type of soil Since direct experimentation is expensive (and destructive), we would like to be able to use small laboratory or field tests and then use scaling to determine effects of explosions of greater magnitude Step 1: Identify the Problem Predict the volume of the crater formed by a spherical explosive placed at a depth d below the surface of the soil Step 2: Make Simplifying Assumptions Initially, our assumptions will be as follows 20

21 The explosive is spherical The craters are geometrically similar The crater size depends on the following variables: the radius r of the crater; the density ρ of the soil; and the mass m of the explosive Step 3: Construct the Model We apply dimensional analysis, using the above assumptions We search for all dimensionless products among the three variables Any such product has the form r m 1 ρ m 2 m m 3, which has dimension M m 2+m 3 L m 1 3m 2 A product of this form is dimensionless if and only if m 1 3m 2 = 0 m 2 + m 3 = 0 Since we are interested in solving for the radius, we will let m 1 be the free variable, and set m 1 = 1 Then m 2 = 1 3 and m 3 = 1 3 Therefore, we obtain one dimensionless product, ( ρ 3 Π 1 = r m)1 Since there is only one dimensionless product, it must be a constant, giving ( ρ 3 r m)1 = c, or r = c ( ) 1 m 3 ρ Step 4: Solve and Interpret the Model Therefore, the crater dimension of the radius varies with the cube root of the mass of the explosive Since the crater volume is proportional to r 3, the volume of the crater will thus be proportional to m/ρ, or V m ρ 21

22 Step 5: Validate the Model It has been shown experimentally that this is an adequate model for small explosions (less than 300 lb of TNT) at zero depth in the soils that have good cohesion For larger explosions, though, this model is unsatisfactory Since the model is not good for larger explosions, we must re-evaluate the model, incorporating more (or different) variables Assumptions: Suppose that we now take into account gravity and the charge energy E of the explosive The charge energy is defined as the product of the mass of the explosive W and its specific energy We apply dimensional analysis again, searching for all dimensionless products among the variables r, ρ, g, and E, ie, for dimensionless products of the form r m 1 ρ m 2 g m 3 E m 4 Analyzing the dimensions of the variables, we see that So, any dimensionless product must have dimension Variable r ρ g E Dimension L ML 3 LT 2 ML 2 T 2 L m 1 (ML 3 ) m 2 (LT 2 ) m 3 (ML 2 T 2 ) m 4 = M m 2+m 4 L m 1 3m 2 +m 3 +2m 4 T 2m 3 2m 4 For the product to be dimensionless, the m 1,, m 4 must solve the linear system m 1 3m 2 + m 3 + 2m 4 = 0 m 2 + m 4 = 0 2m 3 2m 4 = 0 Solve this system R 1+3R 2 R R 2 R R 1 R 3 R Again, we may let m 1 be the free variable since we wish to solve for r So, let m 1 = 1 Then m 4 = 1 4, m 2 = m 3 = 1 4 Thus, we obtain one dimensionless constant, which will be constant, or r ( ρg ) 1 4 E 22 = c,

23 giving Therefore, we have that ( ) 1 E 4 r = c ρg V ( ) 3 E 4 (62) ρg Experimental evidence indicates that scaling based on gravity holds for large explosion (more than 100 tons of TNT) where the stresses caused by the explosion are much larger than the material strength of the soil The model (62) predicts that the volume of the crater will decrease as gravity increases This obviously has relevance in the context of craters formed on other planets (or moons) 621 More Complex Explosive Models We can arrive at more complicated models by considering the question of whether or not the material properties of the soil play a less significant role as the size of the charge increases and as gravity increases To simplify this question, we assume that the only soil property of interest is the density ρ (Note that this is what we have already done, so we are not adding anything in this area) We will also fine-tune the model, using more detail to describe it than we previously did Three variables are required to characterize an explosive: the size of the explosive, which can be described in terms of the charge mass W, the charge energy E, or the radius α of the spherical explosive; the energy yield, which can be measured by the specific energy, Q e density per unit volume, Q V ; and or the energy the explosive density δ The variables can be related by the following equations: W = E Q e Q V = δq e ( ) ( 3 W α e = 4π δ ) 23

24 We may, therefore, select a subset of the variables for our model formulation Using V, W, Q e, δ, ρ, g, and d, we may search for all dimensionless products Such products have the form V m 1 W m 2 Q m 3 e δ m 4 ρ m 5 g m 6 d m 7 Analyzing the dimensions of the variables, we see that Variable V W Q e δ ρ g d Dimension L 3 M L 2 T 2 ML 3 ML 3 LT 2 L So, any dimensionless product must have dimension or Such a product is dimensionless if (L 3 ) m 1 M m 2 (L 2 T 2 ) m 3 (ML 3 ) m 4 (ML 3 ) m 5 (LT 2 ) m 6 L m 7, L 3m 1+2m 3 3m 4 3m 5 +m 6 +m 7 M m 2+m 4 +m 5 T 2m 3 2m 6 3m 1 + 2m 3 3m 4 3m 5 + m 6 + m 7 = 0 m 2 + m 4 + m 5 = 0 2m 3 2m 6 = 0 Applying Gaussian elimination gives R 1+R 2 R Since we had three equations in seven unknowns, we expected that we would have four free variables We may choose m 1 (since we wish to solve for the volume) and three other variables as the free variables If we choose m 5, m 6, and m 7 (as we would do in a linear algebra class), we obtain the dimensionless products by finding a basis for the solution space Since solving gives m 2 = 1 3 (m 7 m 6 ) m 1 m 3 = m 6 m 4 = m 1 m (m 6 m 7 ), setting each free variable equal to 1 and the other three equal to 0 gives the following 24

25 dimensionless products Π 1 = V δ W ( ) ( g W Π 2 = Q e δ ( ) 1 δ 3 Π 3 = d W Π 4 = ρ δ We really want to determine relationships in terms of the density ρ of the soil, but since ρ and δ have exactly the same dimensions, we may rewrite our dimensionless products as Π 1 = V ρ W ( ) ( g W Π 2 = Q e δ ( ρ ) 1 3 Π 3 = d W Π 4 = ρ δ We do this so that Π 1 is consistent with the dimensionless product implied by the relation V W/ρ we obtained previously ) 1 3 ) 1 3 Applying Buckingham s Pi Theorem, we obtain the model or Solving for V, we obtain ψ (Π 1, Π 2, Π 3, Π 4 ) = 0, ( ( ) ( ) ) 1 V ρ g W W = φ 3 ( ρ ) 1 3, d, ρ Q e δ W δ V = W ρ φ ( ( g Q e ) ( W δ ) ) 1 3 ( ρ ) 1 3, d, ρ W δ Experimental data is necessary to validate this model Some experimental data is available in [4] Experiments have shown that the logic in our simpler model is sound, ie, that the result of an increase in gravity is a decrease in volume for a fixed charge yield Obviously, this implies that increasing the size of the charge can compensate for an increase in gravity 25

26 References [1] Clive L Dym, Principles of Mathematical Modeling, Second Edition, Elsevier, 2004 [2] Frank Giordano, et al, Mathematical Modeling, Fourth Edition, Brooks/Cole, 2009 [3] Stefan Heinz, Mathematical Modeling, Springer (2011) [4] RM Schmidt, A Centrifuge Cratering Experiment: Development of a Gravity-Scaled Yield Parameter In Impact and Explosion Cratering, edited by DJ Roddy et al, (Pergamon, New York, 1977), pp