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) 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀 𝐿𝐿 𝑇𝑇 −2 𝐹𝐹𝐹𝐹𝐹𝐹 𝐹𝐹 𝑀𝑀𝑀𝑀𝑇𝑇 • 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) 𝑀𝑀 Time s (second) s (second) 𝐿𝐿 𝑇𝑇

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 𝑧𝑧 𝑧𝑧0 𝑉𝑉0𝑡𝑡 − 𝑔𝑔𝑡𝑡 = + 𝐿𝐿 𝐿𝐿 2 𝐿𝐿 𝐿𝐿 𝑇𝑇 − 2 𝑇𝑇 – : Initial distance at = 0 𝑇𝑇 𝑇𝑇 – : Initial velocity 𝑧𝑧0 𝑡𝑡 𝑉𝑉0

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,𝑧𝑧 𝐿𝐿 𝑉𝑉 𝑡𝑡 𝐿𝐿𝑇𝑇 𝑇𝑇 𝑧𝑧 0 ̇ 𝑡𝑡 0 ̇ 𝑧𝑧 𝐿𝐿 𝑧𝑧1 𝐿𝐿 = + ∗ 2 ∗ 2 ∗ 𝑡𝑡 𝑧𝑧0 𝑡𝑡 𝑧𝑧0 Then, divide by , 0 0 0 𝑧𝑧 𝑧𝑧 𝑧𝑧 𝑉𝑉 0 − 𝑔𝑔 0 1𝑉𝑉 𝑉𝑉 0 = 1 + + , = 𝑧𝑧 2 ∗ ∗ ∗2 𝑉𝑉0 𝑧𝑧 𝑡𝑡 𝑡𝑡 𝛼𝛼 2𝛼𝛼 𝑔𝑔𝑧𝑧0 4 Dimensional vs. Non-dimensional Equation

• Dimensional equation 1 = + 2 2 or 𝑧𝑧 𝑧𝑧0 𝑉𝑉0𝑡𝑡 − 𝑔𝑔𝑡𝑡 , , , , = 0 ⇒ 5 variables

𝐹𝐹 𝑧𝑧 𝑧𝑧0 𝑉𝑉0 𝑔𝑔 𝑡𝑡

• Non-dimensional equation 1 = 1 + ∗ ∗ ∗2 or 𝑧𝑧 𝑡𝑡 − 𝑡𝑡 , , = 0 ⇒2𝛼𝛼 3 variables ∗ ∗ 𝑓𝑓 𝑧𝑧 𝑡𝑡 𝛼𝛼 5 Advantages of Nondimensionalization

= 2 𝑉𝑉0 𝛼𝛼 𝑔𝑔𝑧𝑧0 Dimensional: (a) fixed at 4 m/s and Non-dimensional: (a) and (b) are (b) fixed at 10 m combined into one plot 𝑉𝑉0 6 𝑧𝑧0 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 𝐹𝐹 𝐴𝐴1 𝐴𝐴2 𝐴𝐴𝑛𝑛 𝐴𝐴𝑖𝑖 2. Helps𝑓𝑓 Π 1inΠ understanding2 Π𝑟𝑟<𝑛𝑛 physicsΠ𝑖𝑖 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. 𝒏𝒏 𝒓𝒓 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 dimensions𝒎𝒎 𝒏𝒏 − describing𝒓𝒓 the variables.

= Number of dimensional variables = Minimum number of dimensions to describe the variables 𝑛𝑛 = = Number of non-dimensional variables 𝑚𝑚

𝑟𝑟 𝑛𝑛 − 𝑚𝑚

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 variables𝑝𝑝 𝑓𝑓 𝐷𝐷 inℓ terms𝑉𝑉 𝜇𝜇 of basic dimensions.

Variable

Unit N/m2 m m m/s N⋅s/m2 Δ𝑝𝑝 𝐷𝐷 ℓ 𝑉𝑉 𝜇𝜇 Dimension −1 −2 −1 −1 −1 • Step 3: Determine𝑀𝑀𝐿𝐿 the𝑇𝑇 required𝐿𝐿 number of 𝐿𝐿pi terms. 𝐿𝐿𝑇𝑇 𝑀𝑀𝐿𝐿 𝑇𝑇 = 5 for , , , , and = 3 for , , 𝑛𝑛 𝛥𝛥𝛥𝛥 𝐷𝐷 ℓ 𝑉𝑉 𝜇𝜇 = = 5 3 = 2 (i.e., 2 pi terms) 𝑚𝑚 𝑀𝑀 𝐿𝐿 𝑇𝑇 ∴ 𝑟𝑟 𝑛𝑛 − 𝑚𝑚 − 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).

𝑚𝑚 • 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

𝐷𝐷 𝑉𝑉 𝜇𝜇 𝐿𝐿 𝑇𝑇 𝑀𝑀 13 Step 5

• Combine , , with one additional variable ( or ), in sequence, to find the two pi products 𝐷𝐷 𝑉𝑉 𝜇𝜇 Δ𝑝𝑝 ℓ = = 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑎𝑎 −1 𝑏𝑏 −1 −1 𝑐𝑐 −1 −2 1 Π 𝐷𝐷 =𝑉𝑉 𝜇𝜇 Δ𝑝𝑝 𝐿𝐿 𝐿𝐿𝑇𝑇 𝑀𝑀𝐿𝐿 =𝑇𝑇 𝑀𝑀 𝐿𝐿 𝑇𝑇 𝑐𝑐+1 𝑎𝑎+𝑏𝑏−𝑐𝑐−1 −𝑏𝑏−𝑐𝑐−2 0 0 0 𝑀𝑀 𝐿𝐿 𝑇𝑇 𝑀𝑀 𝐿𝐿 𝑇𝑇 Equate exponents: Mass( ): + 1 = 0 Length( ): + 1 = 0 𝑀𝑀 𝑐𝑐 Time( ): 2 = 0 𝐿𝐿 𝑎𝑎 𝑏𝑏 − 𝑐𝑐 −

Solve for, 𝑇𝑇 −𝑏𝑏 − 𝑐𝑐 − = 1 = 1 = 1 Therefore, 𝑎𝑎 𝑏𝑏 − 𝑐𝑐 − = = −1 −1 Δ𝑝𝑝𝑝𝑝 Π1 𝐷𝐷𝑉𝑉 𝜇𝜇 Δ𝑝𝑝 𝜇𝜇𝜇𝜇 14 Step 6

• Repeat Step 5 for each of the remaining nonrepeating variables.

= = 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑎𝑎 −1 𝑏𝑏 −1 −1 𝑐𝑐 2 Π 𝐷𝐷=𝑉𝑉 𝜇𝜇 ℓ 𝐿𝐿 𝐿𝐿𝑇𝑇 = 𝑀𝑀𝐿𝐿 𝑇𝑇 𝐿𝐿 𝑐𝑐 𝑎𝑎+𝑏𝑏−𝑐𝑐+1 −𝑏𝑏−𝑐𝑐 0 0 0 𝑀𝑀 𝐿𝐿 𝑇𝑇 𝑀𝑀 𝐿𝐿 𝑇𝑇 Equate exponents: Mass( ): = 0 Length( ): + + 1 = 0 Time(𝑀𝑀): 𝑐𝑐 = 0 𝐿𝐿 𝑎𝑎 𝑏𝑏 − 𝑐𝑐 Solve for, 𝑇𝑇 −𝑏𝑏 − 𝑐𝑐 = 1 = 0 = 0 Therefore, 𝑎𝑎 − 𝑏𝑏 𝑐𝑐 = = −1 0 0 ℓ Π2 𝐷𝐷 𝑉𝑉 𝜇𝜇 ℓ 𝐷𝐷 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. 𝐹𝐹 𝐿𝐿 𝑇𝑇 Variable 𝑀𝑀 𝐿𝐿 𝑇𝑇 Unit N/m2 m m m/s N⋅s/m2 Δ𝑝𝑝 𝐷𝐷 ℓ 𝑉𝑉 𝜇𝜇 Dimension −2 −1 −2 𝐹𝐹𝐿𝐿 𝐿𝐿 𝐿𝐿 𝐿𝐿𝑇𝑇 𝐹𝐹𝐹𝐹𝐿𝐿 = = = −2 Δ𝑝𝑝𝑝𝑝 𝐹𝐹𝐿𝐿 𝐿𝐿 0 0 0 Π1 ̇ −2 −1 ̇ 𝐹𝐹 𝐿𝐿 𝑇𝑇 𝜇𝜇𝜇𝜇 𝐹𝐹𝐹𝐹𝐿𝐿 𝐿𝐿𝑇𝑇 = = = ℓ 𝐿𝐿 0 0 0 Π2 ̇ ̇ 𝐹𝐹 𝐿𝐿 𝑇𝑇 𝐷𝐷 𝐿𝐿 16

Step 8

• Express the final form as a relationship among the pi terms.

= or Π1 𝑓𝑓 Π2 = Δ𝑝𝑝𝑝𝑝 ℓ 𝑓𝑓 𝜇𝜇𝜇𝜇 𝐷𝐷 • Think about what it means. Since ,

Δ𝑝𝑝 ∝ ℓ = Δ𝑝𝑝𝑝𝑝 ℓ where is a constant. Thus, 𝐶𝐶 ⋅ 𝜇𝜇𝜇𝜇 𝐷𝐷

𝐶𝐶 1

Δ𝑝𝑝 ∝ 2 𝐷𝐷 17 Problems with One Pi Term

• The functional relationship that must exist for one pi term is

=

Π 𝐶𝐶 where is a constant.

𝐶𝐶 • In other words, if only one pi term is involved in a problem, it must be equal to a constant.

18 Example 2

= , ,

𝜔𝜔 𝑓𝑓 𝐷𝐷 𝑚𝑚 𝛾𝛾

19 Steps 1 through 4

Variable

Unit 1/s m kg N/m3 𝜔𝜔 𝐷𝐷 𝑚𝑚 𝛾𝛾 Dimension

−1 −2 −2 𝑇𝑇 𝐿𝐿 𝑀𝑀 𝑀𝑀𝐿𝐿 𝑇𝑇 = 4 for , , , and = 3 for , , 𝑛𝑛 𝜔𝜔 𝐷𝐷 𝑚𝑚 𝛾𝛾

𝑚𝑚 𝑀𝑀 𝐿𝐿 𝑇𝑇 = = 4 3 = 1 (i.e., 1 pi term)

∴ 𝑟𝑟 𝑛𝑛 − 𝑚𝑚 − repeating variables = , ,

𝑚𝑚 𝐷𝐷 𝑚𝑚 𝛾𝛾 20 Step 5 (and 6)

= = 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑏𝑏 −2 −2 𝑐𝑐 −1 Π 𝐷𝐷=𝑚𝑚 𝛾𝛾 𝜔𝜔 𝐿𝐿 𝑀𝑀 𝑀𝑀=𝐿𝐿 𝑇𝑇 𝑇𝑇 𝑏𝑏+𝑐𝑐 𝑎𝑎−2𝑐𝑐 −2𝑐𝑐−1 0 0 0 𝑀𝑀 𝐿𝐿 𝑇𝑇 𝑀𝑀 𝐿𝐿 𝑇𝑇 Equate exponents: Mass( ): + = 0 Length( ): = 0 𝑀𝑀 𝑏𝑏 𝑐𝑐 Time( ): 1 = 0 𝐿𝐿 𝑎𝑎 − 2𝑐𝑐 or, 𝑇𝑇 −2𝑐𝑐 − 1 1 = 1 = = 2 2 𝑎𝑎 − 𝑏𝑏 𝑐𝑐 −

= = 1 1 − −1 2 2 𝜔𝜔 𝑚𝑚 ∴ Π 𝐷𝐷 𝑚𝑚 𝛾𝛾 𝜔𝜔 𝐷𝐷 𝛾𝛾

21 Steps 7 through 8

= = −1 −1 2 = 𝜔𝜔 𝑚𝑚 𝑇𝑇 𝐹𝐹𝐿𝐿 𝑇𝑇 0 0 0 ̇ −3 ̇ 𝐹𝐹 𝐿𝐿 𝑇𝑇 Π 𝐷𝐷 𝛾𝛾 𝐿𝐿 𝐹𝐹𝐿𝐿 • The dimensionless function is

= 𝜔𝜔 𝑚𝑚 𝐷𝐷 𝛾𝛾 𝐶𝐶 where is a constant. Thus,

𝐶𝐶 = 𝛾𝛾 𝜔𝜔 𝐶𝐶 ⋅ 𝐷𝐷 Therefore, if is increased will decrease. 𝑚𝑚

𝑚𝑚 𝜔𝜔 22 Example 3

= ,

Δ𝑝𝑝 𝑓𝑓 𝑅𝑅 𝜎𝜎

23 Example 3 –Contd.

Variable Unit N/m2 m N/m Δ𝑝𝑝 𝑅𝑅 𝜎𝜎 Dimension −1 −2 −2 𝑀𝑀𝐿𝐿 𝑇𝑇 𝐿𝐿 𝑀𝑀𝑇𝑇

= 3 for , ,

𝑚𝑚 𝑀𝑀 𝐿𝐿 𝑇𝑇 = = 3 3 = 0 ⇒ No pi term?

∴ 𝑟𝑟 𝑛𝑛 − 𝑚𝑚 −

24 Example 3 –Contd.

• Since the repeating variables form a pi among them:

= = −1 −2 Δ𝑝𝑝𝑝𝑝 𝑀𝑀𝐿𝐿 𝑇𝑇 𝐿𝐿 0 0 0 ̇ −2 ̇ 𝑀𝑀 𝐿𝐿 𝑇𝑇 𝜎𝜎 𝑀𝑀𝑇𝑇 should be reduced to 2 for and . −2 𝑚𝑚 𝑀𝑀𝑇𝑇 𝐿𝐿

25 Example 3 –Contd.

Select and as the repeating variables:

𝑅𝑅 𝜎𝜎 = = = Thus, 𝑎𝑎 𝑏𝑏 𝑎𝑎 −2 𝑏𝑏 −1 −2 0 0 0 Π 𝑅𝑅 𝜎𝜎 Δ𝑝𝑝 𝐿𝐿 𝑀𝑀𝑇𝑇 𝑀𝑀𝐿𝐿 𝑇𝑇 𝑀𝑀 𝐿𝐿 𝑇𝑇 : + 1 = 0 : 1 = 0 𝑀𝑀 𝑏𝑏 : 2 = 0 𝐿𝐿 𝑎𝑎 − or, 𝑇𝑇 −2𝑏𝑏 − = 1 and = 1 Hence, 𝑎𝑎 𝑏𝑏 − = Δ𝑝𝑝𝑝𝑝 Π 𝜎𝜎 26 Example 3 –Contd.

• Alternatively, by using the system

𝐹𝐹𝐹𝐹𝐹𝐹

Variable

Unit N/m2 m N/m Δ𝑝𝑝 𝑅𝑅 𝜎𝜎 Dimension −2 −1 𝐹𝐹𝐿𝐿 𝐿𝐿 𝐹𝐹𝐿𝐿 = 3 for , , and = 2 for and 𝑛𝑛 Δ𝑝𝑝 𝑅𝑅 𝜎𝜎

𝑚𝑚 𝐹𝐹 𝐿𝐿 = 3 2 = 1 ⇒ 1 pi term

∴ 𝑟𝑟 −

27 Example 3 –Contd.

With the system:

𝐹𝐹𝐹𝐹𝐹𝐹 = = = Thus, 𝑎𝑎 𝑏𝑏 𝑎𝑎 −1 𝑏𝑏 −2 0 0 0 Π1 𝑅𝑅 𝜎𝜎 Δ𝑝𝑝 𝐿𝐿 𝐹𝐹𝐿𝐿 𝐹𝐹𝐿𝐿 𝐹𝐹 𝐿𝐿 𝑇𝑇 : 1 + = 0 : 2 + = 0 𝐹𝐹 𝑏𝑏 or, 𝐿𝐿 − 𝑎𝑎 − 𝑏𝑏 = 1 and = 1 Hence, 𝑎𝑎 𝑏𝑏 − = Δ𝑝𝑝𝑝𝑝 Π 𝜎𝜎

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

𝐿𝐿 𝑉𝑉−1 𝜌𝜌−3 −𝜇𝜇1 −1 𝑔𝑔−2 𝜎𝜎−2 −𝐾𝐾1 −2 Δ−2𝑝𝑝 −2 𝐿𝐿 𝐿𝐿𝑇𝑇 𝑀𝑀𝐿𝐿 𝑀𝑀𝐿𝐿 𝑇𝑇 𝐿𝐿𝑇𝑇 𝑀𝑀𝑇𝑇 𝑀𝑀𝐿𝐿 𝑇𝑇 𝑀𝑀𝐿𝐿 𝑇𝑇 = 8 variables = 3 dimension 𝑛𝑛 = - = 5 pi terms ( , , , , ) 𝑚𝑚 ∴ 𝑟𝑟 𝑛𝑛 𝑚𝑚 𝑅𝑅𝑅𝑅 𝐹𝐹𝐹𝐹 𝑊𝑊𝑊𝑊 𝑀𝑀𝑀𝑀 𝐶𝐶𝑝𝑝 29 1) Reynolds number

= 𝜌𝜌𝜌𝜌𝜌𝜌 • Generally of importance in all types𝑅𝑅𝑅𝑅 of fluid dynamics problems 𝜇𝜇 • A measure of the ratio of the inertia force to the viscous force

Inertia force = = = Viscous force 3 𝑉𝑉 𝜌𝜌𝐿𝐿 𝑉𝑉 ⋅ 𝑚𝑚𝑚𝑚 𝐿𝐿 𝜌𝜌𝜌𝜌𝜌𝜌 𝑉𝑉 2 𝜏𝜏𝜏𝜏 𝜇𝜇 𝐿𝐿 𝜇𝜇 – If 1 (referred to as “creeping flow”), fluid𝐿𝐿 density is less important – If is large, may neglect the effect of viscosity 𝑅𝑅𝑅𝑅 ≪ • 𝑅𝑅𝑅𝑅 distinguishes among flow regions: laminar or turbulent value varies depending upon flow situation 𝑅𝑅𝑒𝑒𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

30 2) Froude number

= 𝑉𝑉 𝐹𝐹𝐹𝐹 𝑔𝑔𝑔𝑔 • 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 𝑉𝑉 2 𝑚𝑚𝑚𝑚 𝜌𝜌𝐿𝐿 𝑉𝑉 ⋅ 𝑉𝑉 3 𝐿𝐿 𝛾𝛾𝑉𝑉 𝜌𝜌𝜌𝜌 𝐿𝐿 𝑔𝑔𝑔𝑔

31 3) Weber number

= 2 𝜌𝜌𝑉𝑉 𝐿𝐿 𝑊𝑊𝑊𝑊 • Problems in which there is an interface𝜎𝜎 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 𝑉𝑉 2 𝑚𝑚𝑚𝑚 𝜌𝜌𝐿𝐿 𝑉𝑉 ⋅ 𝜌𝜌𝑉𝑉 𝐿𝐿 𝐿𝐿 𝜎𝜎𝜎𝜎 𝜎𝜎𝜎𝜎 𝜎𝜎 • Important parameter at gas-liquid or liquid-liquid interfaces and when these surfaces are in contact with a boundary

32 4) Mach number

= = 𝑉𝑉 𝑉𝑉 𝑀𝑀𝑀𝑀 𝑘𝑘⁄𝜌𝜌 𝑎𝑎 • : speed of sound in a fluid (a symbol is also used) • Problems in which the compressibility of the fluid is important • An𝑎𝑎 index of the ratio of inertial forces to𝑐𝑐 compressibility forces

Inertia force = = = Compressibility force 3 𝑉𝑉 2 𝑚𝑚𝑚𝑚 𝜌𝜌𝐿𝐿 𝑉𝑉 ⋅ 𝑉𝑉 2 2 2 2 𝐿𝐿 2 𝜌𝜌𝑐𝑐 𝐿𝐿 𝜌𝜌𝑐𝑐 𝐿𝐿 𝑐𝑐 (Note: Cauchy number, = = ) • Paramount importance in high speed 2flow2 ( 2) ⁄ • If < 0.3, flow can be considered𝐶𝐶𝐶𝐶 𝑉𝑉as incompressible𝑐𝑐 𝑀𝑀𝑎𝑎 𝑉𝑉 ≥ 𝑐𝑐 𝑀𝑀𝑀𝑀 33 5) Pressure Coefficient

= Δ𝑝𝑝 𝐶𝐶𝑝𝑝 2 • Problems in which pressure differences,𝜌𝜌𝑉𝑉 or pressure, are of interest • A measure of the ratio of pressure forces to inertial forces

Pressure force = = = Inertia force 2 2 Δ𝑝𝑝𝐿𝐿 Δ𝑝𝑝𝐿𝐿 Δ𝑝𝑝 2 3 𝑉𝑉 • Euler number: 𝑚𝑚𝑚𝑚 𝜌𝜌𝐿𝐿 𝑉𝑉 ⋅ 𝜌𝜌𝑉𝑉 𝐿𝐿 = 𝑝𝑝 • Cavitation number: 𝐸𝐸𝐸𝐸 2 𝜌𝜌𝑉𝑉 = 𝑝𝑝 − 𝑝𝑝𝑣𝑣 𝐶𝐶𝐶𝐶 1 2 2𝜌𝜌𝑉𝑉 34 Appendix A: Inspection Method

• Steps 1 through 3 of the exponent method are the same:

= , , ,

Δ𝑝𝑝ℓ 𝑓𝑓 𝐷𝐷 𝜌𝜌 𝜇𝜇 𝑉𝑉

Δ𝑝𝑝ℓ 𝐷𝐷 𝜌𝜌 𝜇𝜇 𝑉𝑉 −3 −4 2 −2 −1 𝐹𝐹𝐿𝐿 𝐿𝐿 𝐹𝐹𝐿𝐿 𝑇𝑇 𝐹𝐹𝐿𝐿 𝑇𝑇 𝐿𝐿𝑇𝑇 = = 5 3 = 2

𝑟𝑟 𝑛𝑛 − 𝑚𝑚 −

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: • 1 ℓ Π Δ𝑝𝑝 To cancel , = = 𝐹𝐹 −3 Δ𝑝𝑝ℓ 𝐹𝐹𝐿𝐿 𝐿𝐿 To cancel , ̇ −4 2 ̇ 2 𝜌𝜌 1 𝐹𝐹𝐿𝐿 𝑇𝑇 1 𝑇𝑇 1 𝑇𝑇 = = Δ𝑝𝑝ℓ 𝐿𝐿 Then, to cancel , 2 ̇ 2 −1 2 ̇ 𝜌𝜌 𝑉𝑉 𝑇𝑇 1 𝐿𝐿𝑇𝑇 𝐿𝐿 𝐿𝐿 = = Δ𝑝𝑝ℓ 0 2 𝐷𝐷 ̇ 𝐿𝐿 ̇ 𝐿𝐿 𝜌𝜌𝑉𝑉 𝐿𝐿 = Δ𝑝𝑝ℓ𝐷𝐷 ∴ Π1 2 𝜌𝜌𝑉𝑉 36 Appendix A: Inspection Method – Contd.

• Select the variable that was not used in , which in this case , and repeat the process: 1 Π 𝜇𝜇 To cancel , = = 𝐹𝐹 −2 2 𝜇𝜇 𝐹𝐹𝐿𝐿 𝑇𝑇 𝐿𝐿 To cancel , ̇ −4 2 ̇ 𝜌𝜌1 𝐹𝐹𝐿𝐿 𝑇𝑇 1 𝑇𝑇 = = 𝑇𝑇 2 𝜇𝜇 𝐿𝐿 Then, to cancel , ̇ −1 ̇ 𝐿𝐿 𝜌𝜌 𝑉𝑉 1 𝑇𝑇 𝐿𝐿𝑇𝑇1 𝐿𝐿 = = 𝜇𝜇 0 ̇ 𝐿𝐿 ̇ 𝐿𝐿 𝜌𝜌𝑉𝑉 𝐷𝐷 𝐿𝐿 = 𝜇𝜇 ∴ Π2 𝜌𝜌𝜌𝜌𝜌𝜌 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 – Contd.

39 Appendix B: Step-by-step Method – Contd.

40