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Dimensional Analysis Dimensional Analysis 11. 3. 2014 Hyunse Yoon, Ph.D. Assistant Research Scientist IIHR-Hydroscience & Engineering e-mail: [email protected] Dimensions and Units • Dimension: A measure of a physical quantity – system: Mass ( ), Length ( ), Time ( ) – system: Force ( = ), Length (L), Time (T) −2 • Unit: A way to assign a number to that dimension Dimension Symbol SI Unit BG Unit Mass kg (kilogram) lb (pound) Length m (meter) ft (feet) Time s (second) s (second) 2 The Principle of Dimensional Homogeneity (PDH) • Every additive terms in an equation must have the same dimensions Ex) Displacement of a falling body 1 = + 2 2 0 0 − = + 2 − 2 – : Initial distance at = 0 – : Initial velocity 0 0 3 Nondimensionalization • Nondimensionalization: Removal of units from physical quantities by a suitable substitution of variables • Nondimensionalized equation: Each term in an equation is dimensionless E.g.) Displacement of a falling body Let: = = = = −1 ∗ ∗ 0 Substitute into the equation, 0 ̇ 0 ̇ 1 = + ∗ 2 ∗ 2 ∗ 0 0 Then, divide by , 0 0 0 0 − 0 1 0 = 1 + + , = 2 ∗ ∗ ∗2 0 2 0 4 Dimensional vs. Non-dimensional Equation • Dimensional equation 1 = + 2 2 or 0 0 − , , , , = 0 ⇒ 5 variables 0 0 • Non-dimensional equation 1 = 1 + ∗ ∗ ∗2 or − 2 , , = 0 ⇒ 3 variables ∗ ∗ 5 Advantages of Nondimensionalization = 2 0 0 Dimensional: (a) fixed at 4 m/s and Non-dimensional: (a) and (b) are (b) fixed at 10 m combined into one plot 0 6 0 Dimensional Analysis • A process of formulating fluid mechanics problems in terms of non-dimensional variables and parameters 1. Reduction in variables = , , … , = 0, = dimensional variables = , , … , = 0, = non-dimensional parameters 1 2 2. Helps Π 1inΠ understanding2 Π< physicsΠ 3. Useful in data analysis and modeling 4. Fundamental to concepts of similarity and model testing 7 Buckingham Pi Theorem • IF a physical process satisfies the PDH and involves dimensional variables, it can be reduced to a relation between only dimensionless variables or ’s. Π • The reduction, = , equals the maximum number of variables that do not form a pi among themselves and is always less than or equal to the number of dimensions − describing the variables. = Number of dimensional variables = Minimum number of dimensions to describe the variables = = Number of non-dimensional variables − 8 Methods for determining Π’s 1. Functional relationship method a. Inspection (Intuition; Appendix A) b. Exponent method (Also called as the method of repeating variables) c. Step-by-step method (Appendix B) 2. Non-dimensionalize governing differential equations (GDE’s) and initial (IC) and boundary (BC) conditions 9 Exponent Method (or Method of Repeating Variables) • Step 1: List all the variables that are involved in the problem. • Step 2: Express each of the variables in terms of basic dimensions. • Step 3: Determine the required number of pi terms. • Step 4: Select a number of repeating variables, where the number required is equal to the number of reference dimensions. • Step 5: Form a pi term by multiplying one of the nonrepeating variables by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless. • Step 6: Repeat Step 5 for each of the remaining nonrepeating variables. • Step 7: Check all the resulting pi terms to make sure they are dimensionless. • Step 8: Express the final form as a relationship among the pi terms, and think about what it means. 10 Example 1 11 Steps 1 through 3 • Step 1: List all the variables that are involved in the problem. = , , , • Step 2: Express each of theΔ variables inℓ terms of basic dimensions. Variable Unit N/m2 m m m/s N⋅s/m2 Δ ℓ Dimension −1 −2 −1 −1 −1 • Step 3: Determine the required number of pi terms. = 5 for , , , , and = 3 for , , ℓ = = 5 3 = 2 (i.e., 2 pi terms) ∴ − − 12 Step 4 • Select a number of repeating variables, where the number required is equal to the number of reference dimensions (for this example, = 3). • All of the required reference dimensions must be included within the group of repeating variables, and each repeating variable must be dimensionally independent of the others (The repeating variables cannot themselves be combined to form a dimensionless product). • Do NOT choose the dependent variable as one of the repeating variables, since the repeating variables will generally appear in more than one pi term. ⇒ ( , , ) for ( , , ), respectively 13 Step 5 • Combine , , with one additional variable ( or ), in sequence, to find the two pi products Δ ℓ = = −1 −1 −1 −1 −2 1 Π = Δ = +1 +−−1 −−−2 0 0 0 Equate exponents: Mass( ): + 1 = 0 Length( ): + 1 = 0 Time( ): 2 = 0 − − Solve for, − − − = 1 = 1 = 1 Therefore, − − = = −1 −1 Δ Π1 Δ 14 Step 6 • Repeat Step 5 for each of the remaining nonrepeating variables. = = −1 −1 −1 2 Π = ℓ = +−+1 −− 0 0 0 Equate exponents: Mass( ): = 0 Length( ): + + 1 = 0 Time(): = 0 − Solve for, − − = 1 = 0 = 0 Therefore, − = = −1 0 0 ℓ Π2 ℓ 15 Step 7 • Check all the resulting pi terms. • One good way to do this is to express the variables in terms of , , if the basic dimensions , , were used initially, or vice versa. Variable Unit N/m2 m m m/s N⋅s/m2 Δ ℓ Dimension −2 −1 −2 = = = −2 Δ 0 0 0 Π1 ̇ −2 −1 ̇ = = = ℓ 0 0 0 Π2 ̇ ̇ 16 Step 8 • Express the final form as a relationship among the pi terms. = or Π1 Π2 = Δ ℓ • Think about what it means. Since , Δ ∝ ℓ = Δ ℓ where is a constant. Thus, ⋅ 1 Δ ∝ 2 17 Problems with One Pi Term • The functional relationship that must exist for one pi term is = Π where is a constant. • In other words, if only one pi term is involved in a problem, it must be equal to a constant. 18 Example 2 = , , 19 Steps 1 through 4 Variable Unit 1/s m kg N/m3 Dimension −1 −2 −2 = 4 for , , , and = 3 for , , = = 4 3 = 1 (i.e., 1 pi term) ∴ − − repeating variables = , , 20 Step 5 (and 6) = = −2 −2 −1 Π = = + −2 −2−1 0 0 0 Equate exponents: Mass( ): + = 0 Length( ): = 0 Time( ): 1 = 0 − 2 or, −2 − 1 1 = 1 = = 2 2 − − = = 1 1 − −1 2 2 ∴ Π 21 Steps 7 through 8 = = −1 −1 2 = 0 0 0 ̇ −3 ̇ Π • The dimensionless function is = where is a constant. Thus, = ⋅ Therefore, if is increased will decrease. 22 Example 3 = , Δ 23 Example 3 –Contd. Variable Unit N/m2 m N/m Δ Dimension −1 −2 −2 = 3 for , , = = 3 3 = 0 ⇒ No pi term? ∴ − − 24 Example 3 –Contd. • Since the repeating variables form a pi among them: = = −1 −2 Δ 0 0 0 ̇ −2 ̇ should be reduced to 2 for and . −2 25 Example 3 –Contd. Select and as the repeating variables: = = = Thus, −2 −1 −2 0 0 0 Π Δ : + 1 = 0 : 1 = 0 : 2 = 0 − or, −2 − = 1 and = 1 Hence, − = Δ Π 26 Example 3 –Contd. • Alternatively, by using the system Variable Unit N/m2 m N/m Δ Dimension −2 −1 = 3 for , , and = 2 for and Δ = 3 2 = 1 ⇒ 1 pi term ∴ − 27 Example 3 –Contd. With the system: = = = Thus, −1 −2 0 0 0 Π1 Δ : 1 + = 0 : 2 + = 0 or, − − = 1 and = 1 Hence, − = Δ Π 28 Common Dimensionless Parameters for Fluid Flow Problems • Most common physical quantities of importance in fluid flow problems are (without heat transfer): Length Velocity Density Viscosity Gravity Surface Compre Pressure tension ssibility change −1 −3 −1 −1 −2 −2 −1 −2 Δ−2 −2 = 8 variables = 3 dimension = - = 5 pi terms ( , , , , ) ∴ 29 1) Reynolds number = • Generally of importance in all types of fluid dynamics problems • A measure of the ratio of the inertia force to the viscous force Inertia force = = = Viscous force 3 ⋅ 2 – If 1 (referred to as “creeping flow”), fluid density is less important – If is large, may neglect the effect of viscosity ≪ • distinguishes among flow regions: laminar or turbulent value varies depending upon flow situation 30 2) Froude number = • Important in problems involving flows with free surfaces • A measure of the ratio of the inertia force to the gravity force (i.e., the weight of fluid) Inertia force = = = Gravity force 3 2 ⋅ 3 31 3) Weber number = 2 • Problems in which there is an interface between two fluids where surface tension is important • An index of the inertial force to the surface tension force Inertia force = = = Surface tension force 3 2 ⋅ • Important parameter at gas-liquid or liquid-liquid interfaces and when these surfaces are in contact with a boundary 32 4) Mach number = = ⁄ • : speed of sound in a fluid (a symbol is also used) • Problems in which the compressibility of the fluid is important • An index of the ratio of inertial forces to compressibility forces Inertia force = = = Compressibility force 3 2 ⋅ 2 2 2 2 2 (Note: Cauchy number, = = ) • Paramount importance in high speed 2flow2 ( 2) ⁄ • If < 0.3, flow can be considered as incompressible ≥ 33 5) Pressure Coefficient = Δ 2 • Problems in which pressure differences, or pressure, are of interest • A measure of the ratio of pressure forces to inertial forces Pressure force = = = Inertia force 2 2 Δ Δ Δ 2 3 • Euler number: ⋅ = • Cavitation number: 2 = − 1 2 2 34 Appendix A: Inspection Method • Steps 1 through 3 of the exponent method are the same: = , , , Δℓ Δℓ −3 −4 2 −2 −1 = = 5 3 = 2 − − 35 Appendix A: Inspection Method – Contd. • Let contain the dependent variable ( in this example) • Then, combine it with other variables so that a non-dimensional product will result: Π1 Δℓ To cancel , = = −3 Δℓ To cancel , ̇ −4 2 ̇ 2 1 1 1 = = Δℓ Then, to cancel , 2 ̇ 2 −1 2 ̇ 1 = = Δℓ 0 2 ̇ ̇ = Δℓ ∴ Π1 2 36 Appendix A: Inspection Method – Contd.
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