Dimensional Analysis - University Of Iowa

Dimensional Analysis11. 3. 2014Hyunse Yoon, Ph.D.Assistant Research ScientistIIHR-Hydroscience & Engineeringe-mail: hyun-se-yoon@uiowa.edu

Dimensions and Units Dimension: A measure of a physical quantity– 𝑀𝑀𝑀𝑀𝑀𝑀 system: Mass (𝑀𝑀), Length (𝐿𝐿), Time (𝑇𝑇)– 𝐹𝐹𝐹𝐹𝐹𝐹 system: Force (𝐹𝐹 𝑀𝑀𝑀𝑀𝑇𝑇 2 ), Length (L), Time (T) Unit: A way to assign a number to that 𝐿𝑇𝑇SI UnitBG Unitkg (kilogram)lb (pound)m (meter)ft (feet)s (second)s (second)2

The Principle of DimensionalHomogeneity (PDH) Every additive terms in an equation must have the same dimensionsEx) Displacement of a falling body1𝑧𝑧 𝑧𝑧0 𝑉𝑉0 𝑡𝑡 𝑔𝑔𝑡𝑡 22𝐿𝐿𝐿𝐿 𝐿𝐿 𝑇𝑇– 𝑧𝑧0 : Initial distance at 𝑡𝑡 0– 𝑉𝑉0 : Initial velocity𝐿𝐿𝑇𝑇 2𝑇𝑇𝑇𝑇 23

Nondimensionalization Nondimensionalization: Removal of units from physical quantities by asuitable substitution of variablesNondimensionalized equation: Each term in an equation is dimensionlessE.g.) Displacement of a falling bodyLet:𝑧𝑧𝐿𝐿𝑧𝑧 ̇𝑧𝑧0𝐿𝐿Substitute into the equation,Then, divide by 𝑧𝑧0 , 1𝑉𝑉𝑡𝑡𝐿𝐿𝑇𝑇𝑇𝑇0𝑡𝑡 ̇𝐿𝐿𝑧𝑧0 1𝑡𝑡𝑧𝑧𝑡𝑡𝑧𝑧00 𝑔𝑔𝑧𝑧 𝑧𝑧0 𝑧𝑧0 𝑉𝑉0𝑉𝑉02𝑉𝑉01 2𝑧𝑧 1 𝑡𝑡 𝑡𝑡 ,2𝛼𝛼 2𝑉𝑉02𝛼𝛼 𝑔𝑔𝑧𝑧04

Dimensional vs. Non-dimensional Equation Dimensional equationor1 2𝑧𝑧 𝑧𝑧0 𝑉𝑉0 𝑡𝑡 𝑔𝑔𝑡𝑡2𝐹𝐹 𝑧𝑧, 𝑧𝑧0 , 𝑉𝑉0 , 𝑔𝑔, 𝑡𝑡 0 5 variables Non-dimensional equationor1 2𝑧𝑧 1 𝑡𝑡 𝑡𝑡2𝛼𝛼 𝑓𝑓 𝑧𝑧 , 𝑡𝑡 , 𝛼𝛼 0 3 variables5

Advantages of NondimensionalizationDimensional: (a) 𝑉𝑉0 fixed at 4 m/s and(b) 𝑧𝑧0 fixed at 10 m𝑉𝑉02𝛼𝛼 𝑔𝑔𝑧𝑧0Non-dimensional: (a) and (b) arecombined into one plot6

Dimensional Analysis A process of formulating fluid mechanics problems in terms ofnon-dimensional variables and parameters1. Reduction in variables𝐹𝐹 𝐴𝐴1 , 𝐴𝐴2 , , 𝐴𝐴𝑛𝑛 0,𝐴𝐴𝑖𝑖 dimensional variables𝑓𝑓 Π1 , Π2 , , Π𝑟𝑟 𝑛𝑛 0, Π𝑖𝑖 non-dimensional parameters2. Helps in understanding physics3. Useful in data analysis and modeling4. Fundamental to concepts of similarity and modeltesting7

Buckingham Pi Theorem IF a physical process satisfies the PDH and involves 𝒏𝒏 dimensionalvariables, it can be reduced to a relation between only 𝒓𝒓 dimensionlessvariables or Π’s.The reduction, 𝒎𝒎 𝒏𝒏 𝒓𝒓, equals the maximum number of variables thatdo not form a pi among themselves and is always less than or equal to thenumber of dimensions describing the variables.𝑛𝑛 Number of dimensional variables𝑚𝑚 Minimum number of dimensions to describe the variables𝑟𝑟 𝑛𝑛 𝑚𝑚 Number of non-dimensional variables8

Methods for determining Π’s1. Functional relationship methoda.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 (GDE’s)and initial (IC) and boundary (BC) conditions9

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 numberrequired is equal to the number of reference dimensions.Step 5: Form a pi term by multiplying one of the nonrepeating variables bythe product of the repeating variables, each raised to an exponent thatwill 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 aredimensionless.Step 8: Express the final form as a relationship among the pi terms, andthink about what it means.10

Example 111

Steps 1 through 3 Step 1: List all the variables that are involved in the problem.Δ𝑝𝑝 𝑓𝑓 𝐷𝐷, ℓ, 𝑉𝑉, 𝜇𝜇Step 2: Express each of the variables in terms of basic dimensions.VariableUnitDimension Δ𝑝𝑝𝐷𝐷ℓ𝑉𝑉𝜇𝜇N/m2mmm/sN s/m2𝑀𝑀𝐿𝐿 1 𝑇𝑇 2𝐿𝐿𝐿𝐿𝐿𝐿𝑇𝑇 1𝑀𝑀𝐿𝐿 1 𝑇𝑇 1Step 3: Determine the required number of pi terms.𝑛𝑛 5 for 𝛥𝛥𝛥𝛥, 𝐷𝐷, ℓ, 𝑉𝑉, and 𝜇𝜇𝑚𝑚 3 for 𝑀𝑀, 𝐿𝐿, 𝑇𝑇 𝑟𝑟 𝑛𝑛 𝑚𝑚 5 3 2 (i.e., 2 pi terms)12

Step 4 Select a number of repeating variables, where the number required isequal to the number of reference dimensions (for this example, 𝑚𝑚 3).All of the required reference dimensions must be included within thegroup of repeating variables, and each repeating variable must bedimensionally independent of the others (The repeating variables cannotthemselves 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 piterm. (𝐷𝐷, 𝑉𝑉, 𝜇𝜇) for (𝐿𝐿, 𝑇𝑇, 𝑀𝑀), respectively13

Step 5 Combine 𝐷𝐷, 𝑉𝑉, 𝜇𝜇 with one additional variable (Δ𝑝𝑝 or ℓ), in sequence, to find the two piproductsΠ1 𝐷𝐷 𝑎𝑎 𝑉𝑉 𝑏𝑏 𝜇𝜇𝑐𝑐 Δ𝑝𝑝 𝐿𝐿Equate �𝑇):Solve for,Therefore, 𝑀𝑀𝑐𝑐 1𝑎𝑎𝐿𝐿𝑇𝑇 1𝑏𝑏𝑀𝑀𝐿𝐿 1 𝑇𝑇 1𝑐𝑐𝑀𝑀𝐿𝐿 1 𝑇𝑇 2𝐿𝐿 𝑎𝑎 𝑏𝑏 𝑐𝑐 1 𝑇𝑇 𝑏𝑏 𝑐𝑐 2 𝑀𝑀0 𝐿𝐿0 𝑇𝑇 0𝑐𝑐 1 0𝑎𝑎 𝑏𝑏 𝑐𝑐 1 0 𝑏𝑏 𝑐𝑐 2 0𝑎𝑎 1𝑏𝑏 1𝑐𝑐 1Π1 𝐷𝐷𝑉𝑉 1 𝜇𝜇 1 Δ𝑝𝑝 Δ𝑝𝑝𝑝𝑝𝜇𝜇𝜇𝜇14

Step 6 Repeat Step 5 for each of the remaining nonrepeating variables.Π2 𝐷𝐷𝑎𝑎 𝑉𝑉 𝑏𝑏 𝜇𝜇𝑐𝑐 ℓ 𝐿𝐿Equate �𝑇):Solve for,Therefore,𝑎𝑎𝐿𝐿𝑇𝑇 1𝑏𝑏𝑀𝑀𝐿𝐿 1 𝑇𝑇 1 𝑀𝑀𝑐𝑐 𝐿𝐿𝑎𝑎 𝑏𝑏 𝑐𝑐 1 𝑇𝑇 𝑏𝑏 𝑐𝑐 𝑀𝑀0 𝐿𝐿0 𝑇𝑇 0𝑐𝑐𝐿𝐿𝑐𝑐 0𝑎𝑎 𝑏𝑏 𝑐𝑐 1 0 𝑏𝑏 𝑐𝑐 0𝑎𝑎 1𝑏𝑏 0𝑐𝑐 0Π2 𝐷𝐷 1 𝑉𝑉 0 𝜇𝜇0 ℓ ℓ𝐷𝐷15

Step 7 Check all the resulting pi terms.One good way to do this is to express the variables in terms of 𝐹𝐹, 𝐿𝐿, 𝑇𝑇 ifthe basic dimensions 𝑀𝑀, 𝐿𝐿, 𝑇𝑇 were used initially, or vice �UnitN/m2mmm/sN s/m2Dimension𝐹𝐹𝐿𝐿 2𝐿𝐿𝐿𝐿𝐿𝐿𝑇𝑇 1𝐹𝐹𝐹𝐹𝐿𝐿 2Δ𝑝𝑝𝑝𝑝𝐹𝐹𝐿𝐿 2 𝐿𝐿0 0 0Π1 ̇ ̇𝐹𝐹𝐿𝐿 𝑇𝑇𝜇𝜇𝜇𝜇𝐹𝐹𝐹𝐹𝐿𝐿 2 𝐿𝐿𝑇𝑇 1ℓ𝐿𝐿 ̇ 𝐹𝐹 0 𝐿𝐿0 𝑇𝑇 0Π2 ̇𝐷𝐷𝐿𝐿16

Step 8 Express the final form as a relationship among the pi terms.Π1 𝑓𝑓 Π2or Think about what it means.Since Δ𝑝𝑝 ℓ,where 𝐶𝐶 is a constant. Thus,Δ𝑝𝑝𝑝𝑝ℓ ℓ 𝐶𝐶 𝜇𝜇𝜇𝜇𝐷𝐷Δ𝑝𝑝 1𝐷𝐷217

Problems with One Pi Term The functional relationship that must exist for one pi term iswhere 𝐶𝐶 is a constant.Π 𝐶𝐶 In other words, if only one pi term is involved in a problem, itmust be equal to a constant.18

Example 2𝜔𝜔 𝑓𝑓 𝐷𝐷, 𝑚𝑚, 𝛾𝛾19

Steps 1 through gN/m3Dimension𝑇𝑇 1𝐿𝐿𝑀𝑀𝑀𝑀𝐿𝐿 2 𝑇𝑇 2𝑛𝑛 4 for 𝜔𝜔, 𝐷𝐷, 𝑚𝑚, and 𝛾𝛾𝑚𝑚 3 for 𝑀𝑀, 𝐿𝐿, 𝑇𝑇 𝑟𝑟 𝑛𝑛 𝑚𝑚 4 3 1 (i.e., 1 pi term)𝑚𝑚 repeating variables 𝐷𝐷, 𝑚𝑚, 𝛾𝛾20

Step 5 (and 6)Π 𝐷𝐷 𝑎𝑎 𝑚𝑚𝑏𝑏 𝛾𝛾 𝑐𝑐 𝜔𝜔 𝐿𝐿Equate �𝑇): 𝑀𝑀𝑏𝑏 𝑐𝑐𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝐿𝐿 2 𝑇𝑇 2𝑐𝑐𝐿𝐿 𝑎𝑎 2𝑐𝑐 𝑇𝑇 2𝑐𝑐 1 𝑀𝑀0 𝐿𝐿0 𝑇𝑇 0𝑇𝑇 1𝑏𝑏 𝑐𝑐 0𝑎𝑎 2𝑐𝑐 0 2𝑐𝑐 1 0or,𝑎𝑎 1𝑏𝑏 1121𝑐𝑐 Π 𝐷𝐷 1 𝑚𝑚2 𝛾𝛾 2 𝜔𝜔 12𝜔𝜔 𝑚𝑚𝐷𝐷 𝛾𝛾21

Steps 7 through 8Π 𝜔𝜔 𝑚𝑚𝐷𝐷𝛾𝛾 ̇The dimensionless function iswhere 𝐶𝐶 is a constant. Thus,𝑇𝑇 1𝐿𝐿𝐹𝐹𝐿𝐿 1 𝑇𝑇 2𝐹𝐹𝐿𝐿 3 ̇ 𝐹𝐹 0 𝐿𝐿0 𝑇𝑇 0𝜔𝜔 𝑚𝑚 𝐶𝐶𝐷𝐷 𝛾𝛾𝜔𝜔 𝐶𝐶 𝐷𝐷Therefore, if 𝑚𝑚 is increased 𝜔𝜔 will decrease.𝛾𝛾𝑚𝑚22

Example 3Δ𝑝𝑝 𝑓𝑓 𝑅𝑅, 𝜎𝜎23

Example 3 ��𝜎N/m2mN/m𝑀𝑀𝐿𝐿 1 𝑇𝑇 2𝐿𝐿𝑀𝑀𝑇𝑇 2𝑚𝑚 3 for 𝑀𝑀, 𝐿𝐿, 𝑇𝑇 𝑟𝑟 𝑛𝑛 𝑚𝑚 3 3 0 No pi term?24

Example 3 –Contd. Since the repeating variables form a pi among them:Δ𝑝𝑝𝑝𝑝𝑀𝑀𝐿𝐿 1 𝑇𝑇 2 𝐿𝐿0 0 0 ̇ ̇𝑀𝑀𝐿𝐿 𝑇𝑇𝜎𝜎𝑀𝑀𝑇𝑇 2𝑚𝑚 should be reduced to 2 for 𝑀𝑀𝑇𝑇 2 and 𝐿𝐿.25

Example 3 –Contd.Select 𝑅𝑅 and 𝜎𝜎 as the repeating variables:Thus,or,Hence,Π 𝑅𝑅𝑎𝑎 𝜎𝜎 𝑏𝑏 Δ𝑝𝑝 �𝑇𝑇 2𝑏𝑏𝑏𝑏 1 0𝑎𝑎 1 0 2𝑏𝑏 2 0𝑀𝑀𝐿𝐿 1 𝑇𝑇 2 𝑀𝑀0 𝐿𝐿0 𝑇𝑇 0𝑎𝑎 1 and 𝑏𝑏 1Π Δ𝑝𝑝𝑝𝑝𝜎𝜎26

Example 3 –Contd. Alternatively, by using the 𝐹𝐹𝐹𝐹𝐹𝐹 /mDimension𝐹𝐹𝐿𝐿 2𝐿𝐿𝐹𝐹𝐿𝐿 1𝑛𝑛 3 for Δ𝑝𝑝, 𝑅𝑅, and 𝜎𝜎𝑚𝑚 2 for 𝐹𝐹 and 𝐿𝐿 𝑟𝑟 3 2 1 1 pi term27

Example 3 –Contd.With the 𝐹𝐹𝐹𝐹𝐹𝐹 system:Thus,or,Hence,Π1 𝑅𝑅𝑎𝑎 𝜎𝜎 𝑏𝑏 Δ𝑝𝑝 𝐿𝐿𝐹𝐹:𝐿𝐿:𝑎𝑎𝐹𝐹𝐿𝐿 11 𝑏𝑏 0 2 𝑎𝑎 𝑏𝑏 0𝑏𝑏𝐹𝐹𝐿𝐿 2 𝐹𝐹 0 𝐿𝐿0 𝑇𝑇 0𝑎𝑎 1 and 𝑏𝑏 1Π Δ𝑝𝑝𝑝𝑝𝜎𝜎28

Common Dimensionless Parametersfor Fluid Flow Problems Most common physical quantities of importance in fluid flowproblems are (without heat �𝑉𝜌𝜌𝐿𝐿𝑇𝑇 1𝑀𝑀𝐿𝐿 3Viscosity Gravity𝜇𝜇𝑀𝑀𝐿𝐿 1 𝑇𝑇 1𝑔𝑔𝐿𝐿𝑇𝑇 ��𝐾𝐾Δ𝑝𝑝𝑀𝑀𝑇𝑇 2𝑀𝑀𝐿𝐿 1 𝑇𝑇 2𝑀𝑀𝐿𝐿 2 𝑇𝑇 2𝑛𝑛 8 variables𝑚𝑚 3 dimension 𝑟𝑟 𝑛𝑛 - 𝑚𝑚 5 pi terms (𝑅𝑅𝑅𝑅, 𝐹𝐹𝐹𝐹, 𝑊𝑊𝑊𝑊, 𝑀𝑀𝑀𝑀, 𝐶𝐶𝑝𝑝 )29

1) Reynolds number 𝜌𝜌𝜌𝜌𝜌𝜌𝑅𝑅𝑅𝑅 𝜇𝜇Generally of importance in all types of fluid dynamics problemsA measure of the ratio of the inertia force to the viscous forceInertia force 𝑚𝑚𝑚𝑚 Viscous force 𝜏𝜏𝜏𝜏 ���𝐿 𝜌𝜌𝜌𝜌𝜌𝜌𝜇𝜇𝐿𝐿2𝑉𝑉 – If 𝑅𝑅𝑅𝑅 1 (referred to as “creeping flow”), fluid density is less important– If 𝑅𝑅𝑅𝑅 is large, may neglect the effect of ���𝑐 distinguishes among flow regions: laminar or turbulent value variesdepending upon flow situation30

2) Froude number𝐹𝐹𝐹𝐹 𝑉𝑉𝑔𝑔𝑔𝑔Important in problems involving flows with free surfacesA measure of the ratio of the inertia force to the gravity force (i.e., theweight of fluid)Inertia force 𝑚𝑚𝑚𝑚 Gravity force 𝛾𝛾𝑉𝑉𝑉𝑉2𝐿𝐿 𝑉𝑉𝜌𝜌𝜌𝜌 𝐿𝐿3𝑔𝑔𝑔𝑔𝜌𝜌𝐿𝐿3𝑉𝑉 31

3) Weber number 𝜌𝜌𝑉𝑉 2 𝐿𝐿𝑊𝑊𝑊𝑊 𝜎𝜎Problems in which there is an interface between two fluids where surfacetension is importantAn index of the inertial force to the surface tension force𝑚𝑚𝑚𝑚Inertia force Surface tension force 𝜎𝜎𝜎𝜎𝜌𝜌𝐿𝐿3𝑉𝑉 𝜎𝜎𝜎𝜎𝑉𝑉2𝐿𝐿 𝜌𝜌𝑉𝑉 𝐿𝐿𝜎𝜎Important parameter at gas-liquid or liquid-liquid interfaces and whenthese surfaces are in contact with a boundary32