Dimensional Analysis 11. 3. 2014 Hyunse Yoon, Ph.D. Assistant Research Scientist IIHR-Hydroscience & Engineering e-mail: [email protected] Dimensions and Units β’ Dimension: A measure of a physical quantity β ππππππ system: Mass (ππ), Length (πΏπΏ), Time (ππ) β πΉπΉπΉπΉπΉπΉ system: Force (πΉπΉ = ππππππ β2 ), Length (L), Time (T) β’ Unit: A way to assign a number to that dimension Dimension Symbol Mass ππ Length Time πΏπΏ ππ SI Unit BG Unit kg (kilogram) lb (pound) m (meter) ft (feet) s (second) s (second) 2 The Principle of Dimensional Homogeneity (PDH) β’ Every additive terms in an equation must have the same dimensions Ex) Displacement of a falling body 1 π§π§ = π§π§0 + ππ0 π‘π‘ β πππ‘π‘ 2 2 πΏπΏ πΏπΏ = πΏπΏ + ππ β π§π§0 : Initial distance at π‘π‘ = 0 β ππ0 : Initial velocity πΏπΏ ππ β 2 ππ ππ 2 3 Nondimensionalization β’ β’ 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: π§π§ πΏπΏ π§π§ β = =Μ π§π§0 πΏπΏ Substitute into the equation, Then, divide by π§π§0 , β1 ππ π‘π‘ πΏπΏππ ππ 0 π‘π‘ β = =Μ πΏπΏ π§π§0 β β 1 π‘π‘ π§π§ π‘π‘ π§π§0 0 β ππ π§π§ β π§π§0 = π§π§0 + ππ0 ππ0 2 ππ0 1 β2 π§π§ = 1 + π‘π‘ + π‘π‘ , 2πΌπΌ β β 2 ππ02 πΌπΌ = πππ§π§0 4 Dimensional vs. Non-dimensional Equation β’ Dimensional equation or 1 2 π§π§ = π§π§0 + ππ0 π‘π‘ β πππ‘π‘ 2 πΉπΉ π§π§, π§π§0 , ππ0 , ππ, π‘π‘ = 0 β 5 variables β’ Non-dimensional equation or 1 β2 π§π§ = 1 + π‘π‘ β π‘π‘ 2πΌπΌ β ππ π§π§ β , π‘π‘ β , πΌπΌ = 0 β β 3 variables 5 Advantages of Nondimensionalization Dimensional: (a) ππ0 fixed at 4 m/s and (b) π§π§0 fixed at 10 m ππ02 πΌπΌ = πππ§π§0 Non-dimensional: (a) and (b) are combined into one plot 6 Dimensional Analysis β’ A process of formulating fluid mechanics problems in terms of non-dimensional variables and parameters 1. Reduction in variables πΉπΉ = π΄π΄1 , π΄π΄2 , β¦ , π΄π΄ππ = 0, π΄π΄ππ = dimensional variables ππ = Ξ 1 , Ξ 2 , β¦ , Ξ ππ<ππ = 0, Ξ ππ = non-dimensional parameters 2. Helps in understanding physics 3. Useful in data analysis and modeling 4. Fundamental to concepts of similarity and model testing 7 Buckingham Pi Theorem β’ β’ IF a physical process satisfies the PDH and involves ππ dimensional variables, it can be reduced to a relation between only ππ dimensionless variables or Ξ βs. The reduction, ππ = ππ β ππ, 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. ππ = Number of dimensional variables ππ = Minimum number of dimensions to describe the variables ππ = ππ β ππ = Number of non-dimensional variables 8 Methods for determining Ξ βs 1. Functional relationship method a. b. c. Inspection (Intuition; Appendix A) Exponent method (Also called as the method of repeating variables) Step-by-step method (Appendix B) 2. Non-dimensionalize governing differential equations (GDEβs) and initial (IC) and boundary (BC) conditions 9 Exponent Method (or Method of Repeating Variables) β’ β’ β’ β’ β’ β’ β’ β’ 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 β’ β’ Step 1: List all the variables that are involved in the problem. Ξππ = ππ π·π·, β, ππ, ππ Step 2: Express each of the variables in terms of basic dimensions. Variable Unit Dimension β’ Ξππ π·π· β ππ ππ N/m2 m m m/s Nβ s/m2 πππΏπΏβ1 ππ β2 πΏπΏ πΏπΏ πΏπΏππ β1 πππΏπΏβ1 ππ β1 Step 3: Determine the required number of pi terms. ππ = 5 for π₯π₯π₯π₯, π·π·, β, ππ, and ππ ππ = 3 for ππ, πΏπΏ, ππ β΄ ππ = ππ β ππ = 5 β 3 = 2 (i.e., 2 pi terms) 12 Step 4 β’ β’ β’ Select a number of repeating variables, where the number required is equal to the number of reference dimensions (for this example, ππ = 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. β (π·π·, ππ, ππ) for (πΏπΏ, ππ, ππ), respectively 13 Step 5 β’ Combine π·π·, ππ, ππ with one additional variable (Ξππ or β), in sequence, to find the two pi products Ξ 1 = π·π· ππ ππ ππ ππππ Ξππ = πΏπΏ Equate exponents: Mass(ππ): Length(πΏπΏ): Time(ππ): Solve for, Therefore, = ππ ππ+1 ππ πΏπΏππ β1 ππ πππΏπΏβ1 ππ β1 ππ πππΏπΏβ1 ππ β2 πΏπΏ ππ+ππβππβ1 ππ βππβππβ2 = ππ0 πΏπΏ0 ππ 0 ππ + 1 = 0 ππ + ππ β ππ β 1 = 0 βππ β ππ β 2 = 0 ππ = 1 ππ = β1 ππ = β1 Ξ 1 = π·π·ππ β1 ππβ1 Ξππ = Ξππππ ππππ 14 Step 6 β’ Repeat Step 5 for each of the remaining nonrepeating variables. Ξ 2 = π·π·ππ ππ ππ ππππ β = πΏπΏ Equate exponents: Mass(ππ): Length(πΏπΏ): Time(ππ): Solve for, Therefore, ππ πΏπΏππ β1 ππ πππΏπΏβ1 ππ β1 = ππππ πΏπΏππ+ππβππ+1 ππ βππβππ = ππ0 πΏπΏ0 ππ 0 ππ πΏπΏ ππ = 0 ππ + ππ β ππ + 1 = 0 βππ β ππ = 0 ππ = β1 ππ = 0 ππ = 0 Ξ 2 = π·π·β1 ππ 0 ππ0 β = β π·π· 15 Step 7 β’ β’ Check all the resulting pi terms. One good way to do this is to express the variables in terms of πΉπΉ, πΏπΏ, ππ if the basic dimensions ππ, πΏπΏ, ππ were used initially, or vice versa. Variable Ξππ π·π· β ππ ππ Unit N/m2 m m m/s Nβ s/m2 Dimension πΉπΉπΏπΏβ2 πΏπΏ πΏπΏ πΏπΏππ β1 πΉπΉπΉπΉπΏπΏβ2 Ξππππ πΉπΉπΏπΏβ2 πΏπΏ 0 0 0 Ξ 1 = =Μ = Μ πΉπΉ πΏπΏ ππ ππππ πΉπΉπΉπΉπΏπΏβ2 πΏπΏππ β1 β πΏπΏ =Μ πΉπΉ 0 πΏπΏ0 ππ 0 Ξ 2 = =Μ π·π· πΏπΏ 16 Step 8 β’ Express the final form as a relationship among the pi terms. Ξ 1 = ππ Ξ 2 or β’ Think about what it means. Since Ξππ β β, where πΆπΆ is a constant. Thus, Ξππππ β = ππ ππππ π·π· Ξππππ β = πΆπΆ β ππππ π·π· Ξππ β 1 π·π·2 17 Problems with One Pi Term β’ The functional relationship that must exist for one pi term is where πΆπΆ is a constant. Ξ = πΆπΆ β’ In other words, if only one pi term is involved in a problem, it must be equal to a constant. 18 Example 2 ππ = ππ π·π·, ππ, πΎπΎ 19 Steps 1 through 4 Variable ππ π·π· ππ πΎπΎ Unit 1/s m kg N/m3 Dimension ππ β1 πΏπΏ ππ πππΏπΏβ2 ππ β2 ππ = 4 for ππ, π·π·, ππ, and πΎπΎ ππ = 3 for ππ, πΏπΏ, ππ β΄ ππ = ππ β ππ = 4 β 3 = 1 (i.e., 1 pi term) ππ repeating variables = π·π·, ππ, πΎπΎ 20 Step 5 (and 6) Ξ = π·π· ππ ππππ πΎπΎ ππ ππ = πΏπΏ Equate exponents: Mass(ππ): Length(πΏπΏ): Time(ππ): = ππ ππ+ππ ππ ππ ππ πππΏπΏβ2 ππ β2 ππ πΏπΏ ππβ2ππ ππ β2ππβ1 = ππ0 πΏπΏ0 ππ 0 ππ β1 ππ + ππ = 0 ππ β 2ππ = 0 β2ππ β 1 = 0 or, ππ = β1 ππ = 1 1 2 1 ππ = β β΄ Ξ = π·π· β1 ππ2 πΎπΎ β2 ππ = 1 2 ππ ππ π·π· πΎπΎ 21 Steps 7 through 8 Ξ = β’ ππ ππ π·π· πΎπΎ =Μ The dimensionless function is where πΆπΆ is a constant. Thus, ππ β1 πΏπΏ πΉπΉπΏπΏβ1 ππ 2 πΉπΉπΏπΏβ3 =Μ πΉπΉ 0 πΏπΏ0 ππ 0 ππ ππ = πΆπΆ π·π· πΎπΎ ππ = πΆπΆ β π·π· Therefore, if ππ is increased ππ will decrease. πΎπΎ ππ 22 Example 3 Ξππ = ππ π π , ππ 23 Example 3 βContd. Variable Unit Dimension Ξππ π π ππ N/m2 m N/m πππΏπΏβ1 ππ β2 πΏπΏ ππππ β2 ππ = 3 for ππ, πΏπΏ, ππ β΄ ππ = ππ β ππ = 3 β 3 = 0 β No pi term? 24 Example 3 βContd. β’ Since the repeating variables form a pi among them: Ξππππ πππΏπΏβ1 ππ β2 πΏπΏ 0 0 0 =Μ = Μ ππ πΏπΏ ππ ππ ππππ β2 ππ should be reduced to 2 for ππππ β2 and πΏπΏ. 25 Example 3 βContd. Select π π and ππ as the repeating variables: Thus, or, Hence, Ξ = π π ππ ππ ππ Ξππ = πΏπΏ ππ: πΏπΏ: ππ: ππ ππππ β2 ππ ππ + 1 = 0 ππ β 1 = 0 β2ππ β 2 = 0 πππΏπΏβ1 ππ β2 = ππ0 πΏπΏ0 ππ 0 ππ = 1 and ππ = β1 Ξ = Ξππππ ππ 26 Example 3 βContd. β’ Alternatively, by using the πΉπΉπΉπΉπΉπΉ system Variable Ξππ π π ππ Unit N/m2 m N/m Dimension πΉπΉπΏπΏβ2 πΏπΏ πΉπΉπΏπΏβ1 ππ = 3 for Ξππ, π π , and ππ ππ = 2 for πΉπΉ and πΏπΏ β΄ ππ = 3 β 2 = 1 β 1 pi term 27 Example 3 βContd. With the πΉπΉπΉπΉπΉπΉ system: Thus, or, Hence, Ξ 1 = π π ππ ππ ππ Ξππ = πΏπΏ πΉπΉ: πΏπΏ: ππ πΉπΉπΏπΏβ1 1 + ππ = 0 β2 + ππ β ππ = 0 ππ πΉπΉπΏπΏβ2 = πΉπΉ 0 πΏπΏ0 ππ 0 ππ = 1 and ππ = β1 Ξ = Ξππππ ππ 28 Common Dimensionless Parameters for Fluid Flow Problems β’ Most common physical quantities of importance in fluid flow problems are (without heat transfer): Length πΏπΏ πΏπΏ Velocity Density ππ ππ πΏπΏππ β1 πππΏπΏβ3 Viscosity Gravity ππ πππΏπΏβ1 ππ β1 ππ πΏπΏππ β2 Surface tension Compre ssibility Pressure change ππ πΎπΎ Ξππ ππππ β2 πππΏπΏβ1 ππ β2 πππΏπΏβ2 ππ β2 ππ = 8 variables ππ = 3 dimension β΄ ππ = ππ - ππ = 5 pi terms (π π π π , πΉπΉπΉπΉ, ππππ, ππππ, πΆπΆππ ) 29 1) Reynolds number β’ β’ ππππππ π π π π = ππ 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 ππππ = = Viscous force ππππ β’ πππΏπΏ3 ππ ππ πΏπΏ ππ πΏπΏ = ππππππ ππ πΏπΏ2 ππ β β If π π π π βͺ 1 (referred to as βcreeping flowβ), fluid density is less important β If π π π π is large, may neglect the effect of viscosity π π ππππππππππ distinguishes among flow regions: laminar or turbulent value varies depending upon flow situation 30 2) Froude number πΉπΉπΉπΉ = β’ β’ ππ ππππ 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 ππππ = = Gravity force πΎπΎππ ππ 2 πΏπΏ = ππ ππππ πΏπΏ3 ππππ πππΏπΏ3 ππ β 31 3) Weber number β’ β’ β’ ππππ 2 πΏπΏ ππππ = ππ 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 ππππ Inertia force = = Surface tension force ππππ πππΏπΏ3 ππ β ππππ ππ 2 πΏπΏ = ππππ πΏπΏ ππ Important parameter at gas-liquid or liquid-liquid interfaces and when these surfaces are in contact with a boundary 32 4) Mach number ππππ = β’ β’ β’ ππ ππβππ = ππ ππ ππ: speed of sound in a fluid (a symbol ππ is also used) Problems in which the compressibility of the fluid is important An index of the ratio of inertial forces to compressibility forces ππ πππΏπΏ3 ππ β Inertia force ππππ ππ 2 πΏπΏ = = = 2 Compressibility force ππππ 2 πΏπΏ2 ππππ 2 πΏπΏ2 ππ β’ β’ (Note: Cauchy number, πΆπΆπΆπΆ = ππ 2 βππ 2 = ππππ2 ) Paramount importance in high speed flow (ππ β₯ ππ) If ππππ < 0.3, flow can be considered as incompressible 33 5) Pressure Coefficient πΆπΆππ = β’ β’ Ξππ ππππ 2 Problems in which pressure differences, or pressure, are of interest A measure of the ratio of pressure forces to inertial forces Pressure force ΞπππΏπΏ2 ΞπππΏπΏ2 Ξππ = = = ππ Inertia force ππππ ππππ 2 πππΏπΏ3 ππ β πΏπΏ β’ Euler number: β’ Cavitation number: ππ πΈπΈπΈπΈ = ππππ 2 πΆπΆπΆπΆ = ππ β πππ£π£ 1 ππππ 2 2 34 Appendix A: Inspection Method β’ Steps 1 through 3 of the exponent method are the same: Ξππβ = ππ π·π·, ππ, ππ, ππ Ξππβ πΉπΉπΏπΏβ3 π·π· πΏπΏ ππ πΉπΉπΏπΏβ4 ππ 2 ππ πΉπΉπΏπΏβ2 ππ ππ πΏπΏππ β1 ππ = ππ β ππ = 5 β 3 = 2 35 Appendix A: Inspection Method β Contd. β’ β’ Let Ξ 1 contain the dependent variable (Ξππβ in this example) Then, combine it with other variables so that a non-dimensional product will result: To cancel πΉπΉ, To cancel ππ, Then, to cancel πΏπΏ, πΉπΉπΏπΏβ3 πΏπΏ Ξππβ =Μ =Μ 2 ππ ππ πΉπΉπΏπΏβ4 ππ 2 Ξππβ 1 πΏπΏ = Μ ππ 2 ππ ππ 2 Ξππβ 1 π·π· = Μ πΏπΏ ππππ 2 β΄ Ξ 1 = 1 πΏπΏππ β1 2 =Μ πΏπΏ =Μ πΏπΏ0 1 πΏπΏ Ξππβ π·π· ππππ 2 36 Appendix A: Inspection Method β Contd. β’ Select the variable that was not used in Ξ 1 , which in this case ππ, and repeat the process: To cancel πΉπΉ, To cancel ππ, Then, to cancel πΏπΏ, ππ πΉπΉπΏπΏβ2 ππ πΏπΏ2 =Μ =Μ ππ πΉπΉπΏπΏβ4 ππ 2 ππ ππ 1 πΏπΏ2 =Μ ππ ππ ππ ππ 1 =Μ πΏπΏ ππππ π·π· β΄ Ξ 2 = 1 =Μ πΏπΏ πΏπΏππ β1 1 =Μ πΏπΏ0 πΏπΏ ππ ππππππ 37 Appendix B: Step-by-step Method* *by Ipsen (1960). The pi theorem and Ipsen method are quite different. Both are useful and interesting. 38 Appendix B: Step-by-step Method β Contd. 39 Appendix B: Step-by-step Method β Contd. 40
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