Dimensional Analysis Exercise Solutions

Dimensional Analysis Exercise Solutions 1. Nondimensionalize the differential equation: d2 x −gR2 = , dt2 (x + R)2 x(0) = 0, dx (0) = V0 dt In this example, [x] = L, g is the acceleration due to gravity, R is the radius of the earth, and V0 is initial velocity. Note that this is the same one as in the text- Try to do it without referring back to it! SOLUTION: See the text. 2. We will re-dimensionalize the pendulum. That is, we start with the nondimensional form, d2 θ = − sin(θ) dτ 2 with the substitution: r t g τ=p = t l l/g Compute the differential equation for d2 θ . dt2 SOLUTION: s dθ dθ dt dt l We note that = · . Given our substitution, = so that dτ dt dτ dτ g s dθ l dθ = . Repeat the process and substitute back in to get a final dτ g dt answer: d2 θ g = − sin(θ) 2 dt l 3. Find a proportionality relationship using dimensional analysis of centrifugal force F in terms of mass m, velocity v and radius r. SOLUTION: We set: F = kma v b rc , where k is dimensionless, and compare dimensions: ML Lb = M a · b · Lc 2 T T We see that a = 1, b = 2 so c = −1. This gives: F = 1 kmv 2 r 4. In fluid mechanics, the Reynolds number is a dimensionless number involving fluid velocity v, density ρ, viscosity µ and a characteristic length r. Find this dimensionless product of the variables, given the table of dimensions below (Fill in the missing values first): Variable Dimension v ρ M L−3 µ M L−1 T −1 r SOLUTION: Set up the dimensionless product: La M b M c · · · Ld T a L3b Lc T c from which we get (M, L, T): b+c =0 a − 3b − c + d = 0 −a − c = 0 Since we have one free variable, we’ll set a = 1 to get b = 1, c = −1, d = 1. In this case, we have that the dimensionless product that represents the Reynolds number is given by: vρr µ 5. Certain stars, whose light and radial velocities undergo periodic vibrations, are thought to be pulsing. It is hypothesized that the period t of pulsation depends upon the star’s radius, r, its mass, m, and the gravitational constant, G. (a) Before going into this problem, as a simpler problem, compute the units of force from: F = ma. L SOLUTION: [F ] = [m][a] = M · LT −2 = M T2 (b) Newton’s law of gravitation asserts that the attractive force between two bodies is proportional to the product of their masses divided by the distance between them: F = Gm1 m2 r2 Compute the units of G from this. ML M2 L3 SOLUTION: 2 = [G] 2 ⇒ [G] = T L MT2 (c) Going back to our star, express t as an appropriate product of powers of m, r, and G. 2 3c SOLUTION: t = ma rb Gc ⇒ T = M a Lb MLc T 2c so: 1 3 1 a − c = 0, b + 3c = 0, −2c = 1 ⇒ a = − , b = , c = − 2 2 2 r3/2 Now, t = √ mG 6. Find the volume flow rate of blood flowing in an artery, dV dt , as a function of the pressure drop per unit length, P , the radius r, the density ρ and the viscosity µ. Here is a list of dimensions. Fill in the blanks: Variable dV /dt Dimension P r ρ M/(LT 2 ) L M/L3 µ M/(LT ) Now perform dimensional analysis and apply Buckingham’s π Theorem. SOLUTION: A generic dimensionless variable will have the form: π = (dV )a P b rc ρd µe so that, in dimensions, 1= Mb Md Me L3a · b 2b · Lc · 3d · e e a T L T L L T and by equating exponents (L,M, then T): −b b −a −2b 3a +c −3d +d −e = 0 +e = 0 −e = 0 We have three equations and five unknowns, so we can choose the two free variables. Since we’re going to want to solve for dV /dt, we’ll choose a to be one of the free ones. Arbitrarily, we choose e as the second. There are several ways of writing the solutions. But notice that, given the choice we made for free variables, we can write: −b + c − 3d b+d −2b = −3a + e = −e =a+e Which we could leave as is, or we can go ahead and solve, getting: 1 1 b = − a − e, 2 2 c = −2a − e, d= 1 1 a− e 2 2 Substituting a = 2, e = 0 and a = 0, e = 2, we get: a = 2, b = −1, c = −4, d = 1, e = 0 and a = 0, b = −1, c = −2, d = −1, e = 2 3 Therefore, 2 dV ρ µ2 , π = 2 dt P r4 P r2 ρ By the Buckingham π Theorem, there exists an f so that: !  2 dV ρ µ2 f , =0 dt P r4 P r2 ρ  π1 = By the Implicit Function Theorem, we assume that we can solve this so that there exists a function h: 2  2   ρ µ dV =h 4 dt Pr P r2 ρ so that: dV = r2 · dt s P ·h ρ  µ2 P r2 ρ  7. The power P delivered to a pump depends on the specific weight w of the fluid pumped, the height h to which the fluid is pumped, and the fluid flow rate q. Use dimensional analysis to determine an equation for power. Below are listed the dimensions: Variable P Dimension M L2 T −3 w M L−2 T −2 h L q L3 T −1 Set up the equations from: a  b  3 d  M L2 M L Lc T3 L2 T 2 T From which: a+b −3a − 2b − d 2a − 2b + c + 3d =0 =0 =0 Take a as the free variable, and set it to 1, then b = c = d = −1. This gives one dimensionless variable, π1 = P whq so that the Buckingham π Theorem states that there is a function f so that   P f =0 whq Since f (π) = 0, π is a constant. Thus, P ∝ whq or P = kwhq 4