Nondimensionalization, Scaling, and Units

Topic 1: Nondimensionalization, Scaling, and Units Course Notes, Math 558 Spring 2012 Barenblatt.book Holmes.new.book This section is a selection of material related to include Chapters 0–3 of [1], Chapters 1–3 of [2]. 1 Dimensional Analysis 1.1 Example 1. Throw a ball Consider the following example: We throw a ball directly upwards from the surface of the Earth. What is its maximum height? We need two physical laws to describe this system: 1. Newton’s Second Law. The acceleration of a body is directly proportional to its force and inversely proportional to its mass; moreover, the acceleration is parallel to the form. This is typically said as “Force equals mass times acceleration” or F = ma. 2. Newton’s law of universal gravitation. Every body in the universe attracts every other body with a gravitational force, which force is proportional to each of the masses of the objects and inversely proportional to the square of the distance between the objects. In equations, if x is the vector from mass 1 to mass 2, then the force on mass 1 due to mass 2 is Gm1m2x F12 = ; kxk3 and the force on mass 2 due to mass 1 is Gm1m2x F21 = −F12 = − : kxk3 Since this problem is one-dimensional, we can think of the force, acceleration, and position as scalars (distance from the ground). So if we define mB as the mass of the ball, mE as the mass of the earth, then we have the differential equation for x(t), the height above the ground, as 2 d x GmBmE m = − ; (1) eq:d B dt2 d2 where d = R + x is the distanceeq:d between the center of the ball and the center of the earth (R is the radius of the earth). Simplifying (1) we obtain d2x gR2 = − ; (2) eq:x dt2 (R + x)2 2 where we write g = GME=R . Clearly this problem, being second-order, requires two integration constants, but these can be specified by x(0) and x0(0), the initial position and velocity. Let us denote the maximal height by xmax. 1 This equation can actually be solved exactly, but it is tedious and requires special functions (inverse sinh, for example) and in general is a big mess. Let us delay this solution for now. Idea 1. The first idea we might have is to say that x(t) R for all t, and thus x(t) + R ≈ R: (3) eq:approx eq:x If we make this approximation, then (2) becomes d2x = −g; (4) eq:constcoeff dt2 which we can easily solve as g x(t) = − t2 + x0(0)t + x(0): 2 0 ∗ Writing x(0) = 0 and x (0) = v0, it is easy to see that the time of maximal height is t = v0=g, and therefore 2 the maximal height is xmax = v0=2g. Of course, what we have done here is what is known as an uncontrolledeq:approx approximation! We have no idea, a priorieq:x, how the error we have introduced in the approximation in (3) percolates into errors in the solution of (2). Studying the effects of this approximation is exactly what we will study in Topic 2 of this course in a few weeks. eq:constcoeff Idea 2. Let us imagine that we didn’t know how to solve (4), and look at the units of the problem. There are three physical dimensions that appear in this problem, length, time, and mass; we denote these as L, T , M. Clearly, the units of velocity are L=T , acceleration is L=T 2, force is ML=T 2, etc. (If we didn’t know these, we could deduce the first two from the definitions and the third from Newton’s Second Law!) Now, let us make the Ansatz that the maximal height depends only on g, mB, and v0, so that xmax = f(g; mB; v0): If this is true, then these quantities must have the same units, i.e. [xmax] = [f(g; mB; v0)]; and let us make the further Ansatz that this function can be written as a monomial, so that a b c [xmax] = [m v0g ]: (We will justify this second Ansatz later.) This then becomes L b L c L = M a = M aLb+cT −b−2c: T T 2 Equating powers, we obtain the system a = 0; b + c = 1; −b − 2c = 0; or, a = 0; b = 2; c = −1: Therefore we have v2 x = α 0 : max g This is consistent with the exact answer derived above, although is a weaker statement (here we only know that there is a constant out front and not that this constant is α = 1=2). However, notice that we had to insert much less information into the problem to obtain this solution and did not have to know how to solve an ODE. 2 1.2 Example 2. Drag on a sphere Imagine a sphere moving through a fluid, we want to compute the force due to drag on the sphere. We postulate that it should depend on the dimensional quantities R; v; ρ, µ: R is the radius of the sphere, v is the velocity of the sphere, ρ is the density of the fluid, and µ is the (dynamic) viscosity of the fluid. The units of these quantities are L M M [R] = L; [v] = ; [ρ] = ; [µ] = : T L3 LT By the previous logic, we should write a b c d [DF ] = [R v ρ µ ]; or ML L b M c M d = La T 2 T L3 LT Expanding these out and equating powers, we obtain the system a + b − 3c + d = 1; c + d = 1; b + d = 2: There are only three equations, but four unknowns! So there will not be a unique solution and there is (at least) one free variable. For now, let us make the choice of d as the free variable, and then we obtain a = 2 − d; b = 2 − d; c = 1 − d; giving us µ d D = αR2−dv2−dρ1−dµd = αR2v2ρ ; F Rvρ α where is a scalar. Let us denote µ Π = (5) eq:defofPi Rvρ and we have 2 2 d DF = αR v ρΠ : (6) eq:ad We might want to know the dimensions of Π, so we compute M=LT [Π] = = 1: L(L=T )(M=L3) eq:ad We say that Π is dimensionless. Now, since Π is dimensionless, (6) is dimensionally correct no matter what the choice of α, or d. Therefore we could write 2 2 d1 DF = R v ρ(α1Π ); or 2 2 d2 DF = R v ρ(α2Π ); or in fact any linear combination of such terms, namely n ! 2 2 X dk DF = R v ρ αkΠ : k=1 In general, given any function of Π, this expression is dimensionally correct, so we can write 2 2 DF = R v ρf(Π) 3 for some unknown function f. Since Π is nondimensional, f(Π) is nondimensional for any function f, and therefore we cannot proscribe the solution further. Now, notice that we made a choice of d as a free parameter in the derivation above. Would anything have changed had we made another choice there? For example, let’s now say that c is free. Solving, we would obtain a = 1 + c; b = 1 + c; d = 1 − c; giving Rvρc D = αR1+cv1+cρcµ1−c = αRvρ : F µ Thus defining Rvρ Πe = ; (7) eq:defofPitilde µ we have c DF = αRvµΠe ; and by the same argument of arbitrary powers, we have DF = Rvµg(Π)e : Are these two different expressions really different? Note, first of all, that Πe = 1=Π. Moreover, if we set them equal, we obtain ρR2v2f(Π) = Rvµg(1=Π); Rvρ f(Π) = g(1=Π); µ f(Π) = Πg(1=Π): Thus there is a functional relationship between f and g; if we know one, we know the other. So these expressions are equivalent even though they do not seem so at first glance. Now, one might ask how one determines theHolmes.new.book unknown function f (or g), and this is something that should be done by experiment. See Figure 1.3 of [2]. The nondimensional quantities Π and Πe show up so often in fluid dynamics that they are given names: Πe is known as the Reynolds number, and Π is the Pécletnumber. 1.3 Using dimensional analysis for scale models Let us imagine that we have specific values for the quantities R; v; ρ, µ in mind, but we want to know the drag a sphere would experience without building the object itself and making a measurement. Can we figure out how to build a scale model of the sphere, and then embed it in another physical experiment to get the measurement. We know that µ D = ρR2v2f(Π); Π = : F Rvρ As long as we know f(Π), then we are done. So let us build a model, with model parameters Rm; vm; ρm; µm so that Πm = Π. Then f(Πm) = f(Π), and we can measure the former to get the value for the latter. The restriction that Π = Π means that m µ µ = m : Rvρ Rmvmρm Assuming that we’re using the same fluid, so that ρm = ρ, µm = µ, we then have Rv Rv = Rmvm; or, vm = : Rm 4 As an example, let us say that we wanted to measure the drag force on a sphere of radius 1000m at a given velocity v. This is much to large to build, but we could, for example, build a sphere with radius Rm = 2m and then choose velocity vm = 500v.

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