Dimensional Analysis - Part 01

Transcription

1 Dimensional Analysis - Part 01 Bridgman April 20, 2017 MATH-310 Introduction to Mathematical Modeling

2 Units? We don t need no stinkin units!

3 Consider the following, typical Calculus I problem: Example A ball falls from a height of h feet. Disregarding any effects due to air resistance, estimate the speed of the ball when it hits the ground, give that its acceleration due to gravity is g feet per second squared. 1/11

4 Consider the following, typical Calculus I problem: Example A ball falls from a height of h feet. Disregarding any effects due to air resistance, estimate the speed of the ball when it hits the ground, give that its acceleration due to gravity is g feet per second squared. To solve this we must formulate the corresponding differential equation, giving d 2 y dt 2 = g, with y(0) = h, and y (0) = 0, y (t) = gt y(t) = 1 2 gt2 + h y (t) = gt 0 = 2gh. 1/11

5 Now consider a different perspective on the same problem: Example A ball falls from a height h. Disregarding any effects due to air resistance, estimate its speed when it hits the ground, given that the acceleration due to gravity is g. 2/11

6 Now consider a different perspective on the same problem: Example A ball falls from a height h. Disregarding any effects due to air resistance, estimate its speed when it hits the ground, given that the acceleration due to gravity is g. If we assume that the dimensions of h and g belong to the quantities themselves, then we can guess at the final speed without solving the differential equation. We know that the dimensions of h are length (L), that the dimensions g are length per time squared (LT 2 ) and that the dimensions of speed are length per time (LT 1 ). The only combination of g and h with the dimension of speed is gh a dimensionless constant. Thus, an estimate for the speed would be v gh. 2/11

7 Introduction

8 Dimensional analysis is the analysis of a relationship by considering its units of measure. It is a method for helping to determine how the selected variables are related and for 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 changed by changing the units measuring these dimensions. Definition We will denote the measure of a quantity using square brackets, [...]. The measures that are commonly used are [length] = L; [mass] = M; and [time] = T. 3/11

9 Example What are the units of measure for Force in the following? F = ma We see that on the right side of the equation, [ma] = [m][a] = M LT 2. So, Force must have units of measure of [F ] = MLT 2. 4/11

10 Example (Physical Quantities in MLT System) Mass M Momentum MLT 1 Length L Work ML 2 T 2 Time T Density ML 3 Velocity LT 1 Pressure ML 1 T 2 Acceleration LT 2 Torque ML 2 T 2 Force MLT 2 Energy ML 2 T 2 However, there are also some quantities that do not have dimensions, such as an angle measure. These are referred to as dimensionless quantities. Other dimensionless quantities include trigonometric functions, exponential functions, and/or logarithms. 5/11

11 Some guidelines thus far, We have based the concept of dimension on 3 physical quantities: mass m, length l and time t. The measure of a product is the product of the measures. i.e. [ma] = [m][a] We can exponentiate dimensions i.e. [ m] = [m] Terms must have the same units of measure to add them. Physical relationships must be dimensionally consistent. More complex physical entities can be expressed as products involving mass, length and time by algebraic simplification. 6/11

12 1. What is the dimension of ω in sin (ωt) where t is time?. 2. Show that the following equation is dimensionally inconsistent, if v is velocity, x is position and t is time. v 2 = t 2 + x t 3. The following are 2 versions of Poiseuille s equation which describes the flow of fluid through a cylindrical tube. flow rate = πpr 4 8ηl or flow rate = πρpr 4 The constants are Pressure [P] = ML 1 T 2 Length [l] = L Radius [r] = L Fluid Density [ρ] = ML 3 Viscosity [η] = ML 1 T 1 Determine the dimension(s) of the flow rate. Are these equations the same? 8ηl 7/11

13 4. The various constants of physics often have physical dimensions (dimensional constants) because their values depend on the system in which they are expressed. Consider Newton s Law of Gravitation, F = Gm 1m 2 r 2 which states the the attractive force between 2 bodies is proportional to the product of their masses divided by the square of the distance between the bodies. In this case G represents the gravitational constant. Find the dimension of G so that Newton s Law is dimensionally compatible. 5. Find a dimensionless product relating the torque [τ] = ML 2 T 2 produced by a car engine, the engine s rotation rate [ψ] = T 1, the volume of air displaced V and the air density ρ. 8/11

14 Unit-Free

15 In the discussion of dimensional analysis, we are careful to distinguish between the word unit and the word dimension. By units we mean specific physical quantities like seconds, hours, days, etc. all of which have dimensions of time. Any fundamental dimension D has the property that its units may be scaled by an appropriate conversion factor λ to D in a new system of units. Thus, D = λd. Extending this concept to derived quantities, if then gives q in a new system of units. [q] = D a 1 1 Da 2 2 Dan n q = λ a 1 1 λa 2 2 λan n q 9/11

16 Definition A physical law, f (q 1, q 2,..., q n) = 0, is unit-free if f ( q 1, q 2,..., q n ) = 0 if and only if f (q 1, q 2,..., q n ) = 0 for all derived quantities, q i = λ i q i. Consider the physical law, f (x, t, g) x 1 2 gt2 = 0 which relates the distance x (in centimeters) a body falls in a constant gravitational field g (in cms/sec 2 ) to the time t (in seconds). 10/11

17 We now convert x to inches and t to minutes with respective conversion factors, x = λ 1 x = x (in/cm), t = λ 2 t = 1 60 t (min/sex) 11/11

18 We now convert x to inches and t to minutes with respective conversion factors, x = λ 1 x = x (in/cm), t = λ 2 t = 1 60 t (min/sex) Then, as [g] = LT 2, ḡ = λ 1 λ 2 2g 11/11

19 We now convert x to inches and t to minutes with respective conversion factors, x = λ 1 x = x (in/cm), t = λ 2 t = 1 60 t (min/sex) Then, as [g] = LT 2, ḡ = λ 1 λ 2 2g Thus, f ( x, t, ḡ) = x 1 2ḡ t 2 = λ 1 x 1 2 (λ 1λ 2 2g)(λ 2 t) = λ 1 (x 1 2 gt2 ) = 0 Therefore our physical law is unit-free. 11/11