FUNDAMENTALS OF FLUID MECHANICS Chapter 7 Dimensional … · FL T V LT p FL D L FL T − − −...

127
1 FUNDAMENTALS OF FUNDAMENTALS OF FLUID MECHANICS FLUID MECHANICS Chapter 7 Dimensional Analysis Chapter 7 Dimensional Analysis Modeling, and Similitude Modeling, and Similitude Jyh Jyh - - Cherng Cherng Shieh Shieh Department of Bio Department of Bio - - Industrial Industrial Mechatronics Mechatronics Engineering Engineering National Taiwan University National Taiwan University 12/07/2009 12/07/2009

Transcript of FUNDAMENTALS OF FLUID MECHANICS Chapter 7 Dimensional … · FL T V LT p FL D L FL T − − −...

  • 1

    FUNDAMENTALS OFFUNDAMENTALS OFFLUID MECHANICSFLUID MECHANICS

    Chapter 7 Dimensional AnalysisChapter 7 Dimensional AnalysisModeling, and SimilitudeModeling, and Similitude

    JyhJyh--CherngCherng ShiehShiehDepartment of BioDepartment of Bio--Industrial Industrial MechatronicsMechatronics Engineering Engineering

    National Taiwan UniversityNational Taiwan University12/07/200912/07/2009

  • 2

    MAIN TOPICSMAIN TOPICS

    Dimensional AnalysisDimensional AnalysisBuckingham Pi TheoremBuckingham Pi TheoremDetermination of Pi TermsDetermination of Pi TermsComments about Dimensional AnalysisComments about Dimensional AnalysisCommon Dimensionless Groups in Fluid MechanicsCommon Dimensionless Groups in Fluid MechanicsCorrelation of Experimental DataCorrelation of Experimental DataModeling and SimilitudeModeling and SimilitudeTypical Model StudiesTypical Model StudiesSimilitude Based on Governing Differential EquationSimilitude Based on Governing Differential Equation

  • 3

    Dimensional Analysis Dimensional Analysis 1/41/4

    A typical fluid mechanics problem in which A typical fluid mechanics problem in which experimentation is required, consider the steady flow of an experimentation is required, consider the steady flow of an incompressible Newtonian fluid through a long, smoothincompressible Newtonian fluid through a long, smooth--walled, horizontal, circular pipe. walled, horizontal, circular pipe. An important characteristic of this system, which would An important characteristic of this system, which would be interest to an engineer designing a pipeline, is the be interest to an engineer designing a pipeline, is the pressure drop per unit length that develops along the pipe pressure drop per unit length that develops along the pipe as a result of friction.as a result of friction.

    從不可壓縮牛頓流體流經長、水平、圓管的實驗談起

    探討沿著管流方向單位管長度的壓力降

    看似簡單的問題,分析看似簡單的問題,分析起來,恐怕不是那麼容起來,恐怕不是那麼容易,想要有易,想要有analytical analytical solutionsolution??

  • 4

    Dimensional Analysis Dimensional Analysis 2/42/4

    The first step in the planning of an experiment to study The first step in the planning of an experiment to study this problem would be to decide on the factors, or this problem would be to decide on the factors, or variables, that will have an effect on the pressure drop.variables, that will have an effect on the pressure drop.Pressure drop per unit lengthPressure drop per unit length

    )V,,,D(fp μρ=Δ l

    Pressure drop per unit length depends on FOUR variables:Pressure drop per unit length depends on FOUR variables:sphere size (D); speed (V); fluid density (sphere size (D); speed (V); fluid density (ρρ); fluid viscosity ); fluid viscosity (m)(m)

    在setup實驗過程中,什麼是影響壓力降之因子?

    先想出四個因子……

  • 5

    Dimensional Analysis Dimensional Analysis 3/43/4

    To perform the experiments in a meaningful and To perform the experiments in a meaningful and systematic manner, it would be necessary to change on of systematic manner, it would be necessary to change on of the variable, such as the velocity, which holding all other the variable, such as the velocity, which holding all other constant, and measure the corresponding pressure drop.constant, and measure the corresponding pressure drop.Difficulty to determine the functional relationship between Difficulty to determine the functional relationship between the pressure drop and the various facts that influence it.the pressure drop and the various facts that influence it.

    最「笨」的想法或做法…一個因子變,其他因子holding constant

  • 6

    Series of TestsSeries of Tests出現一拖拉庫的資料與圖表

    如何從中找出影響壓力降的因子與壓力降的關係

    把實驗當作傳家寶,一代傳一代,代代當作使命來執行

    就算已經完成實驗就算已經完成實驗…………

  • 7

    Dimensional Analysis Dimensional Analysis 4/44/4

    Fortunately, there is a much simpler approach to the Fortunately, there is a much simpler approach to the problem that will eliminate the difficulties described problem that will eliminate the difficulties described above.above.Collecting these variables into two Collecting these variables into two nondimensionalnondimensionalcombinations of the variables (called dimensionless combinations of the variables (called dimensionless product or dimensionless groups)product or dimensionless groups)

    Only one dependent and one Only one dependent and one independent variableindependent variableEasy to set up experiments to Easy to set up experiments to determine dependencydetermine dependencyEasy to present results (one graph)Easy to present results (one graph)

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρφ=

    ρ

    Δ VDV

    pD2l

    有沒有簡單的方法

    把四個因子連同壓力降,彙整成兩個無因次之群組或無因次乘積

    自變數或獨立變數

    應變數

    Dependent variable

    應變數y與獨立變數x:y=f(x)

    降低難度而已降低難度而已……

  • 8

    Plot of Pressure Drop Data Using Plot of Pressure Drop Data Using ……

    0002124

    3

    2 TLF)FT)(TFL()L/F(L

    VpD

    ==ρ

    Δ−−

    &l

    0002

    124TLF

    )TFL()L)(LT)(TFL(VD==

    μρ

    −−&

    dimensionless product or dimensionless product or dimensionless groupsdimensionless groups

  • 9

    Buckingham Pi Theorem Buckingham Pi Theorem 1/51/5

    A fundamental question we must answer is how many A fundamental question we must answer is how many dimensionless products are required to replace the original listdimensionless products are required to replace the original list of of variables ?variables ?The answer to this question is supplied by the basic theorem of The answer to this question is supplied by the basic theorem of dimensional analysis that statesdimensional analysis that states

    If an equation involving k variables is dimensionally homogeneous, it can be reduced to a relationship among k-r independent dimensionless products, where r is the minimum number of reference dimensions required to describe the variables.

    Buckingham Pi TheoremBuckingham Pi TheoremPi termsPi terms

    到底無因次群組如何取得?

    答案自來自因次分析的基本原理

    此基本原理稱為 如果一個方程式含有k個變數,

    且該方程式為dimensionally homogeneous

    這個方程式當然是描述「problem」的方程式

    可以簡化成多少個可以簡化成多少個dimensionless dimensionless groupsgroups?長成什麼模樣?長成什麼模樣

  • 10

    Buckingham Pi Theorem Buckingham Pi Theorem 2/52/5

    Given a physical problem in which the dependent variable Given a physical problem in which the dependent variable is a function of kis a function of k--1 independent variables.1 independent variables.

    Mathematically, we can express the functional relationship Mathematically, we can express the functional relationship in the equivalent formin the equivalent form

    )u,.....,u,u(fu k321 =

    0)u,.....,u,u,u(g k321 =

    Where g is an unspecified function, different from f.Where g is an unspecified function, different from f.

    針對每一個問題,我們總能先列出這種關係,只是f?現在還不見得知道

    從數學的角度,我們也能寫成這個樣子。f與g是不同的

  • 11

    Buckingham Pi Theorem Buckingham Pi Theorem 3/53/5

    The Buckingham Pi theorem states that: Given a relation The Buckingham Pi theorem states that: Given a relation among k variables of the formamong k variables of the form

    The k variables may be grouped into kThe k variables may be grouped into k--r independent r independent dimensionless products, or dimensionless products, or ΠΠ terms, expressible in terms, expressible in functional form byfunctional form by

    0)u,.....,u,u,u(g k321 =

    0),,,,,,(or

    ),,,,,(

    rk321

    rk321

    =ΠΠΠΠφ

    ΠΠΠφ=Π

    −r ?? r ?? ΠΠ????

    把把kk個變數併成個變數併成kk--rr個無因次群組或乘積個無因次群組或乘積

  • 12

    Buckingham Pi Theorem Buckingham Pi Theorem 4/54/5

    The number r is usually, but not always, equal to the The number r is usually, but not always, equal to the minimum number of independent dimensions required to minimum number of independent dimensions required to specify the dimensions of all the parameters. Usually the specify the dimensions of all the parameters. Usually the reference dimensions required to describe the variables reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F, L, and T.will be the basic dimensions M, L, and T or F, L, and T.The The theorem does not predict the functional form of theorem does not predict the functional form of φφor or ϕϕ . The functional relation among the independent . The functional relation among the independent dimensionless products dimensionless products ΠΠ must be determined must be determined experimentally.experimentally.The kThe k--r dimensionless products r dimensionless products ΠΠ terms obtained terms obtained from the procedure are independent.from the procedure are independent.Pi theorem只能找出PI terms,讓您說「PI terms間有關係」,卻還不能讓您找出PI term間的真正關係,真正的關係得靠實驗來推測

    最好找到的當然是r,r就是使用來描述每個變數的dimensions個數。因為使用的reference dimensions不是MLT就是FLT,因此r最多也只是3。

    從從variablesvariables PI termsPI terms最多減少最多減少33個個

  • 13

    Buckingham Pi Theorem Buckingham Pi Theorem 5/55/5

    A A ΠΠ term term is not independentis not independent if it can be obtained from a if it can be obtained from a product or quotient of the other dimensionless products of product or quotient of the other dimensionless products of the problem. For example, ifthe problem. For example, if

    then neither then neither ΠΠ55 nor nor ΠΠ66 is independent of the other is independent of the other dimensionless products or dimensionless products or dimensionless groupsdimensionless groups..

    23

    4/31

    632

    15 or

    Π=Π

    ΠΠΠ

    PI terms可以拼措成新的PI terms

    由別的由別的pi termspi terms拼措起來的拼措起來的pi termpi term當然就非當然就非independentindependent

  • 14

    Determination of Pi Terms Determination of Pi Terms 1/121/12

    Several methods can be used to form the dimensionless Several methods can be used to form the dimensionless products, or pi term, that arise in a dimensional analysis.products, or pi term, that arise in a dimensional analysis.The method we will describe in detail is called the The method we will describe in detail is called the METHOD of repeating variables.METHOD of repeating variables.Regardless of the method to be used to determine the Regardless of the method to be used to determine the dimensionless products, dimensionless products, one begins by listingone begins by listing all all dimensional variablesdimensional variables that are known (or believed) to that are known (or believed) to affect the given flow phenomenon.affect the given flow phenomenon.Eight steps listedEight steps listed below outline a recommended below outline a recommended procedure for determining the procedure for determining the ΠΠ terms.terms.請follow下列步驟,就可以輕易找出PI terms……

  • 15

    Determination of Pi Terms Determination of Pi Terms 2/122/12

    Step 1 List all the variables.Step 1 List all the variables. 11

    List all the dimensional variables involved.List all the dimensional variables involved.Keep the number of variables to a minimum, so that we Keep the number of variables to a minimum, so that we can minimize the amount of laboratory work.can minimize the amount of laboratory work.All variables must be independent. For example, if the All variables must be independent. For example, if the crosscross--sectional area of a pipe is an important variable, sectional area of a pipe is an important variable, either the area or the pipe diameter could be used, but either the area or the pipe diameter could be used, but not both, since they are obviously not independent.not both, since they are obviously not independent.

    γγ==ρρ××g, that is, g, that is, γγ,,ρρ, and g are not independent., and g are not independent.

    列出「所有」變數

    變數最少最好?但不能把重要的忽略掉 變數要independent

    天龍8步

  • 16

    Determination of Pi Terms Determination of Pi Terms 3/123/12

    Step 1 List all the variables. Step 1 List all the variables. 22

    Let k be the number of variables.Let k be the number of variables.Example: For pressure drop per unit length, k=5. (All Example: For pressure drop per unit length, k=5. (All variables are variables are ΔΔppll,, D,D,μμ,,ρρ, and , and V )V )

    )V,,,D(fp μρ=Δ l

    恰好…為原則

    結論是 k=…

  • 17

    Determination of Pi Terms Determination of Pi Terms 4/124/12

    Step 2 Express each of the variables in terms of Step 2 Express each of the variables in terms of basic dimensions. Find the number of reference basic dimensions. Find the number of reference dimensions.dimensions.

    Select a set of fundamental (primary) dimensions.Select a set of fundamental (primary) dimensions.For example: MLT, or FLT.For example: MLT, or FLT.Example: For pressure drop per unit length , we choose Example: For pressure drop per unit length , we choose FLT.FLT.

    12

    243

    LTVTFL

    TFLLDFLp−−

    −−

    ==μ

    =ρ==Δ

    &&

    &&&lr=3r=3

    MLTMLT或或FLTFLT,不會影響結果,不會影響結果

  • 18

    Determination of Pi Terms Determination of Pi Terms 5/125/12

    Step 3 Determine the required number of pi terms.Step 3 Determine the required number of pi terms.Let k be the number of variables in the problem.Let k be the number of variables in the problem.Let r be the number of reference dimensions (primary Let r be the number of reference dimensions (primary dimensions) required to describe these variables.dimensions) required to describe these variables.The number of pi terms is The number of pi terms is kk--rrExample: For pressure drop per unit length k=5, r = 3, Example: For pressure drop per unit length k=5, r = 3, the number of pi terms is the number of pi terms is kk--rr=5=5--3=2.3=2.

    有了k,有了r,當然就可以得出PI terms個數 k-r

  • 19

    Determination of Pi Terms Determination of Pi Terms 6/126/12

    Step 4 Select a number of repeating variables, Step 4 Select a number of repeating variables, where the number required is equal to the number where the number required is equal to the number of reference dimensions.of reference dimensions.

    Select a set of r dimensional variables that includes all Select a set of r dimensional variables that includes all the primary dimensions ( repeating variables).the primary dimensions ( repeating variables).These repeating variables will all be combined with These repeating variables will all be combined with each of the remaining parameters. No repeating each of the remaining parameters. No repeating variables should have dimensions that are power of the variables should have dimensions that are power of the dimensions of another repeating variable.dimensions of another repeating variable.Example: For pressure drop per unit length ( r = 3) Example: For pressure drop per unit length ( r = 3) select select ρρ , V, D., V, D.

    從k個變數中找出r個重複變數,剩下來k-r個就是非重複變數

    r個重複變數必須包括problem中所涉及的所有primary dimensions

  • 20

    Determination of Pi Terms Determination of Pi Terms 7/127/12

    Step 5 Form a pi term by multiplying one of the Step 5 Form a pi term by multiplying one of the nonrepeatingnonrepeating variables by the product of the variables by the product of the repeating variables, each raised to an exponent that repeating variables, each raised to an exponent that will make the combination dimensionless. will make the combination dimensionless. 11

    Set up dimensional equations, combining the variables Set up dimensional equations, combining the variables selected in Step 4 with each of the other variables selected in Step 4 with each of the other variables ((nonrepeatingnonrepeating variables) in turn, to form dimensionless variables) in turn, to form dimensionless groups or dimensionless product.groups or dimensionless product.There will be k There will be k –– r equations.r equations.Example: For pressure drop per unit lengthExample: For pressure drop per unit length

    組成PI terms

  • 21

    nonrepeating variables

    repeating variables

    配對

  • 22

    Determination of Pi Terms Determination of Pi Terms 8/128/12

    Step 5 (Continued) Step 5 (Continued) 22

    cba1 VDp ρΔ=Π l

    1c,2b,1a0c2b:T

    0c4ba3:L0c1:F

    TLF)TFL()LT()L)(FL( 000c24b1a3

    −=−==⇒=+−

    =−++−=+

    =−−− &

    21 VDp

    ρ

    Δ=Π l

    Repeating variables

  • 23

    Determination of Pi Terms Determination of Pi Terms 9/129/12

    Step 6 Repeat Step 5 for each of the remaining Step 6 Repeat Step 5 for each of the remaining nonrepeatingnonrepeating variables.variables.

    cba2 VD ρμ=Π

    1c,1b,1a0c2b1:T

    0c4ba2:L0c1:F

    TLF)TFL()LT()L)(TFL( 000c24b1a2

    −=−=−=⇒=+−

    =−++−=+

    =−−− &

    ρμ

    =ΠDV2

    處理其他PI terms

  • 24

    Determination of Pi Terms Determination of Pi Terms 10/1210/12

    Step 7 Check all the resulting pi terms to make Step 7 Check all the resulting pi terms to make sure they are dimensionless.sure they are dimensionless.

    Check to see that each group obtained is dimensionless.Check to see that each group obtained is dimensionless.Example: For pressure drop per unit length .Example: For pressure drop per unit length .

    0000002

    00000021

    TLMTLFDV

    TLMTLFV

    Dp

    ==ρ

    μ=Π

    ==ρ

    Δ=Π

    &&

    &&l

    驗證無因次

  • 25

    Determination of Pi Terms Determination of Pi Terms 11/1211/12

    Step 8 Express the final form as a relationship Step 8 Express the final form as a relationship among the pi terms, and think about what is means.among the pi terms, and think about what is means.

    Express the result of the dimensional analysis.Express the result of the dimensional analysis.

    Example: For pressure drop per unit length .Example: For pressure drop per unit length .

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛ρ

    μφ=

    ρ

    ΔDVV

    Dp2l

    ),,,,,( rk321 −ΠΠΠφ=Π

    Dimensional analysis will not provide Dimensional analysis will not provide the form of the function. The function the form of the function. The function can only be obtained from a suitable set can only be obtained from a suitable set of experiments.of experiments.

    列出來,也僅能列出來

    得靠實驗來找出兩者間的真正關係得靠實驗來找出兩者間的真正關係

  • 26

    Determination of Pi Terms Determination of Pi Terms 12/1212/12

    The pi terms can be rearranged. For example, The pi terms can be rearranged. For example, ΠΠ22, could , could be expressed asbe expressed as

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρφ=

    ρ

    Δ VDV

    Dp2l

    μρ

    =ΠVD

    2

    真正關係得靠實驗了

  • 27

    Example 7.1 Method of Repeating Example 7.1 Method of Repeating VariablesVariables

    A thin rectangular plate having a width w and a height h is A thin rectangular plate having a width w and a height h is located so that it is normal to a moving stream of fluid. located so that it is normal to a moving stream of fluid. Assume that the drag, D, that the fluid exerts on the plate Assume that the drag, D, that the fluid exerts on the plate is a function of w and h, the fluid viscosity, is a function of w and h, the fluid viscosity, µµ ,and ,and ρρ, , respectively, and the velocity, V, of the fluid approaching respectively, and the velocity, V, of the fluid approaching the plate. Determine a suitable set of pi terms to study this the plate. Determine a suitable set of pi terms to study this problem experimentally.problem experimentally.

  • 28

    Example 7.1 Example 7.1 SolutionSolution1/51/5

    Drag force on a PLATEDrag force on a PLATE

    Step 1:List all the dimensional variables involved. Step 1:List all the dimensional variables involved. D,w,h, D,w,h, ρρ,,μμ,V,V k=6 dimensional parameters.k=6 dimensional parameters.Step 2:Select primary dimensions M,L, and T. Express Step 2:Select primary dimensions M,L, and T. Express each of the variables in terms of basic dimensionseach of the variables in terms of basic dimensions

    )V,,,h,w(fD μρ=

    1311

    2

    LTVMLTML

    LhLwMLTD−−−−

    ==ρ=μ

    ===

    &&&

    &&&

    探討平板的阻力

  • 29

    Example 7.1 Example 7.1 SolutionSolution2/52/5

    Step 3: Determine the required number of pi terms. Step 3: Determine the required number of pi terms. kk--rr=6=6--3=33=3

    Step 4:Select repeating variables w,V,Step 4:Select repeating variables w,V,ρρ..Step 5~6:combining the repeating variables with each Step 5~6:combining the repeating variables with each of the other variables in turn, to form dimensionless of the other variables in turn, to form dimensionless groups or dimensionless products.groups or dimensionless products.

  • 30

    Example 7.1 Example 7.1 SolutionSolution3/53/5

    1c,2b,2a0b2:T

    0c3ba1:L0c1:M

    TLM)ML()LT()L)(MLT(VDw 000c3b1a2cba1

    −=−=−=>>=−−

    =−++=+

    ==ρ=Π −−−

    ρ=Π 221 Vw

    D

    0c,0b,1a0b:T

    0c3ba1:L0c:M

    TLM)ML()LT()L(LVhw 000c3b1acba2

    ==−=>>=

    =−++=

    ==ρ=Π −−

    wh

    2 =Π

  • 31

    Example 7.1 Example 7.1 SolutionSolution4/54/5

    1c,1b,1a0b1:T

    0c3ba1:L0c1:M

    TLM)ML()LT()L)(TML(Vw 000c3b1a11cba3

    −=−=−=>>=−−

    =−++−=+

    ==ρμ=Π −−−−

    ρμ

    =ΠwV3

  • 32

    Example 7.1 Example 7.1 SolutionSolution5/55/5

    Step 7: Check all the resulting pi terms to make sure Step 7: Check all the resulting pi terms to make sure they are dimensionless. they are dimensionless. Step 8: Express the final form as a relationship among Step 8: Express the final form as a relationship among the pi terms.the pi terms.The functional relationship isThe functional relationship is

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛ρ

    μφ=

    ρ

    ΠΠφ=Π

    wV,

    wh

    VwD

    or),,(

    22

    321

  • 33

    Selection of Variables Selection of Variables 1/31/3

    One of the most important, and difficult, steps in applying One of the most important, and difficult, steps in applying dimensional analysis to any given problem is the selection dimensional analysis to any given problem is the selection of the variables that are involved.of the variables that are involved.There is no simple procedure whereby the variable can be There is no simple procedure whereby the variable can be easily identified. Generally, one must rely on a good easily identified. Generally, one must rely on a good understanding of the phenomenon involved and the understanding of the phenomenon involved and the governing physical laws. governing physical laws. If extraneous variables are included, then too many pi If extraneous variables are included, then too many pi terms appear in the final solution, and it may be difficult, terms appear in the final solution, and it may be difficult, time consuming, and expensive to eliminate these time consuming, and expensive to eliminate these experimentally. experimentally.

    因次分析中最重要,也是最困難的步驟

    沒有所謂簡單的程序可以決定那些變數要?那些不要?取決於對物理現象與Physical laws的了解

    太多變數浪費分析問題的時間

  • 34

    Selection of Variables Selection of Variables 2/32/3

    If important variables are omitted, then an incorrect result If important variables are omitted, then an incorrect result will be obtained; and again, this may prove to be costly will be obtained; and again, this may prove to be costly and difficult to ascertain. and difficult to ascertain. Most engineering problems involve certain simplifying Most engineering problems involve certain simplifying assumptions that have an influence on the variables to be assumptions that have an influence on the variables to be considered.considered.Usually we wish to keep the problems as simple as Usually we wish to keep the problems as simple as possible, perhaps even if some accuracy is sacrificedpossible, perhaps even if some accuracy is sacrificed

    遺漏變數導致錯誤或誤差過大

    在工程分析中透過假設,忽略某些變數,讓問題變得比較簡單

    假設也要合理,不能為簡化而簡化

  • 35

    Selection of Variables Selection of Variables 3/33/3

    A suitable balance between simplicity and accuracy is an A suitable balance between simplicity and accuracy is an desirable goal.~~~~~desirable goal.~~~~~Variables can be classified into three general group:Variables can be classified into three general group:

    Geometry: lengths and angles.Geometry: lengths and angles.Material properties: relate the external effects and the Material properties: relate the external effects and the responses.responses.External Effects: produce, or tend to produce, a change External Effects: produce, or tend to produce, a change in the system. Such as force, pressure, velocity, or in the system. Such as force, pressure, velocity, or gravity.gravity.

    在簡化與正確間取得balance

    流體本身(如黏度、密度)、固體本身(如物體長、寬、高等幾何尺寸)、固流體間(如速度、壓力、重力、外力)

  • 36

    Determination of Reference Dimension Determination of Reference Dimension 1/31/3

    When to determine the number of pi terms, it is important When to determine the number of pi terms, it is important to know how many reference dimensions are required to to know how many reference dimensions are required to describe the variables.describe the variables.In fluid mechanics, the required number of reference In fluid mechanics, the required number of reference dimensions is three, but in some problems only one or two dimensions is three, but in some problems only one or two are required.are required.In some problems, we occasionally find the number of In some problems, we occasionally find the number of reference dimensionsreference dimensions needed to describe all variables is needed to describe all variables is smaller than the number of basic dimensionssmaller than the number of basic dimensions. Illustrated in . Illustrated in Example 7.2Example 7.2.. 特別注意使用到的reference dimensions個數低於3

    使用到的reference dimensions個數大多為3,但也可能低於3

  • 37

    Example 7.2 Determination of Pi TermsExample 7.2 Determination of Pi Terms

    An open, cylindrical tank having a diameter D is An open, cylindrical tank having a diameter D is supported around its bottom circumference and is filled to supported around its bottom circumference and is filled to a depth h with a liquid having a specific weight a depth h with a liquid having a specific weight γγ. The . The vertical deflection, vertical deflection, δδ , of the center of the bottom is a , of the center of the bottom is a function of D, h, d, function of D, h, d, γγ, and E, where d is the thickness of , and E, where d is the thickness of the bottom and E is the modulus of elasticity of the bottom the bottom and E is the modulus of elasticity of the bottom material. Perform a dimensional analysis of this problem.material. Perform a dimensional analysis of this problem.

    圓筒內裝liquid放在一個平臺上,平臺下凹深度δδ

  • 38

    Example 7.2 Example 7.2 SolutionSolution1/31/3

    The vertical deflectionThe vertical deflection

    )E,,d,h,D(f γ=δ

    For F,L,T. Pi terms=6For F,L,T. Pi terms=6--2=42=4For M,L,T Pi terms=6For M,L,T Pi terms=6--3=33=3

    212

    223

    TMLE,FLETML,FL

    Ld,Ld

    Lh,Lh

    LD,LD

    L,L

    −−−

    −−−

    ==

    =γ=γ

    ==

    ==

    ==

    =δ=δ

    &&&&

    &&&&

    &&&&

    &&&&

    &&&&

    &&&&

    平臺下凹深度δ與那些變數有關?

    使用不同的dimensional system

  • 39

    Example 7.2 Example 7.2 SolutionSolution2/32/3

    For F,L,T system, Pi terms=6For F,L,T system, Pi terms=6--2=42=4

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛γ

    Φ=δ

    ⇒γ

    =Π=Π=Πδ

    =ΠDE,

    Dd,

    Dh

    DDE,

    Dd,

    Dh,

    D 4321

    D and D and γγ are selected as repeating variablesare selected as repeating variables

    4433

    2211

    ba4

    ba3

    ba2

    ba1

    EDdD

    hDD

    γ=Πγ=Π

    γ=Πγδ=Π

  • 40

    Example 7.2 Example 7.2 SolutionSolution3/33/3

    For M,L,T system, Pi terms=6For M,L,T system, Pi terms=6--3=3 ?3=3 ?

    A closer look at the dimensions of the variables listed A closer look at the dimensions of the variables listed reveal that only two reference dimensions, reveal that only two reference dimensions, L and MTL and MT--22are required.are required.

    直接目測法,看似3個,其實只有2個

  • 41

    Determination of Reference Dimension Determination of Reference Dimension 2/32/3

    ( )σγ=Δ ,,Dfh

    222 TM

    TLMLL

    Dh σγΔ

    LF

    LFLL

    Dh

    3

    σγΔ

    EXAMPLEEXAMPLE

    MLT SYSTEMMLT SYSTEM FLT SYSTEMFLT SYSTEM

    Pi term=4-3=1 Pi term=4-2=2

    另一個例子

  • 42

    Determination of Reference Dimension Determination of Reference Dimension 3/33/3

    2200T0211L1100M

    Dh

    −−−

    σγΔ

    0000T1311L

    1100FDh

    −−

    σγΔ

    Dh

    ⎟⎟⎠

    ⎞⎜⎜⎝

    γ

    σΦ=

    Δ⇒

    γ

    σ=Π 222 DD

    hD

    Set Dimensional MatrixSet Dimensional MatrixMLT SYSTEMMLT SYSTEM FLT SYSTEMFLT SYSTEM

    Rank=2 Pi term=4Rank=2 Pi term=4--2=22=2

    利用dimensional matrix的rank來判斷

  • 43

    Uniqueness of Pi Terms Uniqueness of Pi Terms 1/41/4

    The Pi terms obtained depend on the somewhat arbitrary The Pi terms obtained depend on the somewhat arbitrary selection of repeating variables. For example, in the selection of repeating variables. For example, in the problem of studying the pressure drop in a pipe.problem of studying the pressure drop in a pipe.

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρΦ=

    ρ

    Δ VDV

    Dp12

    l

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρΦ=

    μΔ VD

    VDp

    2

    2l

    )V,,,D(fp μρ=Δ lSelecting D,V, and ρ as repeating variables:

    Selecting D,V, and μ as repeating variables:

    選擇不同的repeating variables產生不同的PI terms

  • 44

    Uniqueness of Pi Terms Uniqueness of Pi Terms 2/42/4

    Both are correct, and both would Both are correct, and both would lead to the same final equation for lead to the same final equation for the pressure drop. the pressure drop. There is not a There is not a unique set of pi termsunique set of pi terms which which arises from a dimensional analysisarises from a dimensional analysis. . The functions The functions ΦΦ11 and and ΦΦ22 are will are will be different because the dependent be different because the dependent pi terms are different for the two pi terms are different for the two relationships.relationships.

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρΦ=

    ρ

    Δ VDV

    Dp12

    l

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρΦ=

    μΔ VD

    VDp

    2

    2l

    兩者都對,換言之沒有唯一的

    PI terms集合。請注意Φ1與Φ2是不相同的。

  • 45

    Uniqueness of Pi Terms Uniqueness of Pi Terms 3/43/4

    ( )321 ,ΠΠΦ=Π

    ( ) ( )'2223'211 ,, ΠΠΦ=ΠΠΦ=Π

    EXAMPLEEXAMPLE

    b3

    a2

    '2 ΠΠ=ΠForm a new pi termForm a new pi term

    All are correct

    前面已經談過,PI terms可以組合成新的PI term

  • 46

    Uniqueness of Pi Terms Uniqueness of Pi Terms 4/44/4

    μΔ

    ρ×

    ρ

    ΔV

    DpVDV

    Dp 22

    ll

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρΦ=

    μΔ VD

    VDp

    2

    2l

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρΦ=

    ρ

    Δ VDV

    Dp12

    l Selecting D,V, and ρ as repeating variables:

    Repeating variablesRepeating variables的最妥當選擇?的最妥當選擇?

  • 47

    Common Dimensionless GroupsCommon Dimensionless Groups 1/21/2

    A list of variables that commonly arise in fluid mechanical problems.Possible to provide a physical interpretation to the dimensionless groups which can be helpful in assessing their influence in a particular application.

    流體力學領域常見的無因次群組

    每一群組都以紀念當事人命名,每一群組都有其物理意義。非常重要!

  • 48

    Froude Number Froude Number 1/21/2

    In honor of William Froude (1810~1879), a British civil engineerIn honor of William Froude (1810~1879), a British civil engineer, , mathematician, and naval architect who pioneered the use of towimathematician, and naval architect who pioneered the use of towing ng tanks for the study of ship design.tanks for the study of ship design.Froude number is the ratio of the forces due to the acceleratioFroude number is the ratio of the forces due to the acceleration of a n of a fluid particles (inertial force) to the force due to gravity (grfluid particles (inertial force) to the force due to gravity (gravity avity forces).forces).Froude number is significant for flows with free surface effectsFroude number is significant for flows with free surface effects..Froude number less than unity indicate subcritical flow and valuFroude number less than unity indicate subcritical flow and values es greater than unity indicate supercritical flow.greater than unity indicate supercritical flow.

    3

    2222

    gLLV

    gLVFr

    gLVFr

    ρ

    ρ==>>=

  • 49

    Froude Number Froude Number 2/22/2

    *

    *s*

    s

    2s

    ss

    s dsdVVV

    dsdVV

    dtdVa

    l===

    l

    ssVVV *s*s ==

    mdsdVVVF *

    *s*

    s

    2

    Il

    =

    lll gV

    gV

    dsdVV

    gV

    FFrF

    2

    *

    *s*

    s

    2

    G

    I ≡≡==

  • 50

    Reynolds Number Reynolds Number 1/21/2

    In honor of Osborne Reynolds (1842~1912), the British engineer In honor of Osborne Reynolds (1842~1912), the British engineer who first demonstrated that this combination of variables could who first demonstrated that this combination of variables could be be used as a criterion to distinguish between laminar and turbulentused as a criterion to distinguish between laminar and turbulent flow.flow.The Reynolds number is a measure of the ration of the inertia foThe Reynolds number is a measure of the ration of the inertia forces rces to viscous forces.to viscous forces.If the Reynolds number is small (Re

  • 51

    Reynolds Number Reynolds Number 2/22/2

    Flows with very small Reynolds numbers are commonly referred to Flows with very small Reynolds numbers are commonly referred to as as ““creeping flowscreeping flows””..For large Reynolds number flow, the viscous effects are small For large Reynolds number flow, the viscous effects are small relative to inertial effects and for these cases it may be possirelative to inertial effects and for these cases it may be possible to ble to neglect the effect of viscosity and consider the problem as one neglect the effect of viscosity and consider the problem as one involving a involving a ““nonviscousnonviscous”” fluid.fluid.Flows with Flows with ““largelarge”” Reynolds number generally are turbulent. Flows Reynolds number generally are turbulent. Flows in which the inertia forces are in which the inertia forces are ““smallsmall”” compared with the viscous compared with the viscous forces are characteristically laminar flowsforces are characteristically laminar flows..

  • 52

    Euler numberEuler number

    In honor of Leonhard Euler (1707~1783), a famous Swiss In honor of Leonhard Euler (1707~1783), a famous Swiss mathematician who pioneered work on the relationship between mathematician who pioneered work on the relationship between pressure and flow.pressure and flow.EulerEuler’’s number is the ratio of pressure force to inertia forces. It iss number is the ratio of pressure force to inertia forces. It isoften called theoften called the pressure coefficient, Cp.pressure coefficient, Cp.

    22 Vp

    VpEu

    ρΔ

    ≡ρ

    =

  • 53

    CavitationCavitation NumberNumber

    For problems in which For problems in which cavitationcavitation is of concern, the dimensionless is of concern, the dimensionless group is commonly used, where group is commonly used, where ppvv is the vapor is the vapor pressure and ppressure and prr is some reference pressure.is some reference pressure.The The cavitationcavitation number is used in the study of number is used in the study of cavitationcavitation phenomena.phenomena.The smaller the The smaller the cavitationcavitation number, the more likely number, the more likely cavitationcavitation is to is to occur.occur.

    2

    vr

    V21

    ppCaρ

    −=

    221

    vr V/)pp( ρ−

    Ca越小,表示cavitation越易發生

  • 54

    Cauchy Number & Mach Number Cauchy Number & Mach Number 1/21/2

    The Cauchy number is named in honor of The Cauchy number is named in honor of AugustinAugustin Louis de Louis de Cauchy (1789~1857), a French engineer, mathematician, and Cauchy (1789~1857), a French engineer, mathematician, and hydrodynamicisthydrodynamicist..The Mach number is named in honor of Ernst Mach (1838~1916), The Mach number is named in honor of Ernst Mach (1838~1916), an Austrian physicist and philosopher.an Austrian physicist and philosopher.Either number may be used in problems in which fluid Either number may be used in problems in which fluid compressibility is important.compressibility is important.

    CaEVMa

    EV

    EV

    cVMa

    EVaC

    22

    2

    =

    ρ

    ==ρ

    =υυυυ

  • 55

    Cauchy Number & Mach Number Cauchy Number & Mach Number 2/22/2

    Both numbers can be interpreted as representing an index of the Both numbers can be interpreted as representing an index of the ration of inertial force to compressibility force, where V is thration of inertial force to compressibility force, where V is the flow e flow speed and c is the local sonic speed.speed and c is the local sonic speed.Mach number is a key parameter that characterizes compressibilitMach number is a key parameter that characterizes compressibility y effects in a flow.effects in a flow.When the Mach number is relatively small (say, less than 0.3), tWhen the Mach number is relatively small (say, less than 0.3), the he inertial forces induced by the fluid motion are not sufficientlyinertial forces induced by the fluid motion are not sufficiently large large to cause a significant change in the fluid density, and in this to cause a significant change in the fluid density, and in this case the case the compressibility of the fluid can be neglected.compressibility of the fluid can be neglected.For truly incompressible flow, c=For truly incompressible flow, c=∞∞ so that M=0.so that M=0.

  • 56

    StrouhalStrouhal Number Number 1/21/2

    In honor of In honor of VincenzVincenz StrouhalStrouhal (1850~1922), who used this parameter (1850~1922), who used this parameter in his study of in his study of ““singing wires.singing wires.”” The most dramatic evidence of this The most dramatic evidence of this phenomenon occurred in 1940 with the collapse of the Tacoma phenomenon occurred in 1940 with the collapse of the Tacoma Narrow bridges. The shedding frequency of the vortices coincidedNarrow bridges. The shedding frequency of the vortices coincidedwith the natural frequency of the bridge, thereby setting up a with the natural frequency of the bridge, thereby setting up a resonant condition that eventually led to the collapse of the brresonant condition that eventually led to the collapse of the bridge.idge.This parameter is important in unsteady, oscillating flow probleThis parameter is important in unsteady, oscillating flow problems ms in which the frequency of the oscillation is in which the frequency of the oscillation is ωω ..

    VtS lω=

  • 57

    StrouhalStrouhal Number Number 2/22/2

    This parameter represents a measure of This parameter represents a measure of the ration of inertial force due to the the ration of inertial force due to the unsteadiness of the flow (local unsteadiness of the flow (local acceleration) to the inertial forces due to acceleration) to the inertial forces due to change in velocity from point to point in change in velocity from point to point in the flow field (convective acceleration). the flow field (convective acceleration). This type of unsteady flow may develop This type of unsteady flow may develop when a fluid flows past a solid body (such when a fluid flows past a solid body (such as a wire or cable) placed in the moving as a wire or cable) placed in the moving stream.stream.

    For example, in a certain Reynolds number range, a periodic flowFor example, in a certain Reynolds number range, a periodic flow will will develop downstream from a cylinder placed in a moving stream duedevelop downstream from a cylinder placed in a moving stream due to to a regular patterns of vortices that are shed from the body.a regular patterns of vortices that are shed from the body.

  • 58

    Weber Number Weber Number 1/21/2

    Named after Moritz Weber (1871~1951), a German professor of Named after Moritz Weber (1871~1951), a German professor of naval mechanics who was instrumental in formalizing the general naval mechanics who was instrumental in formalizing the general use of common dimensionless groups as a basis for similitude use of common dimensionless groups as a basis for similitude studies.studies.Weber number is important in problem in which there is an interfWeber number is important in problem in which there is an interface ace between two fluids. In this situation the surface tension may plbetween two fluids. In this situation the surface tension may play an ay an important role in the phenomenon of interest.important role in the phenomenon of interest.

    σρ

    =l2VWe

  • 59

    Weber Number Weber Number 2/22/2

    Weber number is the ratio of inertia forces to surface tension fWeber number is the ratio of inertia forces to surface tension forces.orces.Common examples of problems in which Weber number may be Common examples of problems in which Weber number may be important include the flow of thin film of liquid, or the formatimportant include the flow of thin film of liquid, or the formation of ion of droplets or bubbles.droplets or bubbles.The flow of water in a river is not affected significantly by The flow of water in a river is not affected significantly by surefacesurefacetension, since inertial and gravitational effects are dominant tension, since inertial and gravitational effects are dominant (We>>1).(We>>1).

  • 60

    Correlation of Experimental DataCorrelation of Experimental Data

    Dimensional analysis only provides the dimensionless groups Dimensional analysis only provides the dimensionless groups describing the phenomenon, and not the specific relationship describing the phenomenon, and not the specific relationship between the groups.between the groups.To determine this relationship, suitable experimental data must To determine this relationship, suitable experimental data must be be obtained.obtained.The degree of difficulty depends on the number of pi terms.The degree of difficulty depends on the number of pi terms.

    因次分析:由變數間的關聯 無因次群組間的關聯

    因次分析僅告訴我們,這些無因次群組是有關係的。關係為何?得靠實驗Data來找它們間的關聯性。找尋關聯性的複雜度隨PI terms個數增加而增加,且絕非等比級數

  • 61

    Problems with One Pi TermProblems with One Pi Term

    The functional relationship for one Pi term.The functional relationship for one Pi term.

    C1 =Πwhere C is a constant. The value of the constant must still be determined by experiment.

    只有一個PI term

  • 62

    Example 7.3 Flow with Only One Pi TermExample 7.3 Flow with Only One Pi Term

    Assume that the drag, D, acting on a spherical particle that falAssume that the drag, D, acting on a spherical particle that falls very ls very slowly through a viscous fluid is a function of the particle diaslowly through a viscous fluid is a function of the particle diameter, meter, d, the particle velocity, V, and the fluid viscosity, d, the particle velocity, V, and the fluid viscosity, μμ. Determine, . Determine, with the aid the dimensional analysis, how the drag depends on twith the aid the dimensional analysis, how the drag depends on the he particle velocity. particle velocity.

  • 63

    Example 7.3 Example 7.3 SolutionSolution

    The drag The drag

    ),V,d(fD μ= 132

    LTVMLTFLLdFD

    −−

    ==ρ

    =μ==

    &&

    &&&

    VD

    dVCDCdV

    D1

    μ==μ

    For a given particle and fluids, the drag varies For a given particle and fluids, the drag varies directly with the velocitydirectly with the velocity

  • 64

    Problems with Two or More Pi Term Problems with Two or More Pi Term 1/21/2

    Problems with two pi termsProblems with two pi terms)( 21 ΠΦ=Π

    the functional relationship the functional relationship among the variables can the among the variables can the be determined by varying be determined by varying ΠΠ22 and measuring the and measuring the corresponding value of corresponding value of ΠΠ11..

    The empirical equation The empirical equation relating relating ΠΠ22 and and ΠΠ11 by using by using a standard curvea standard curve--fitting fitting technique.technique.

    An empirical relationship is An empirical relationship is valid over the range of valid over the range of ΠΠ22..

    Dangerous to Dangerous to extrapolate beyond extrapolate beyond valid rangevalid range

    有兩個PI terms

    若有經驗公式也要講清楚適用範圍

    利用curve fitting技術

  • 65

    Problems with Two or More Pi Term Problems with Two or More Pi Term 2/22/2

    Problems with three pi terms.Problems with three pi terms.

    ( )321 ,ΠΠΦ=Π

    Families curve of curvesFamilies curve of curves

    To determine a suitable empirical equation To determine a suitable empirical equation relating the three pi terms.relating the three pi terms.

    To show data correlations on simple graphs.To show data correlations on simple graphs.

    三個PI terms:固定其中一個,找出其他兩個的關係,製作出Families curve of curves

  • 66

    Modeling and SimilitudeModeling and SimilitudeTo develop the procedures for designing To develop the procedures for designing models so that the model and prototype models so that the model and prototype will behave in a similar fashionwill behave in a similar fashion…………..

    製造模型&比擬

    探討設計模型的程序,使得設計出來的模型可用來預測雛型的行為,即模型可以用來比擬雛型。換言之,模型與雛型間必須達到所謂的「相似性(Similarity)」

  • 67

    Model vs. Prototype Model vs. Prototype 1/21/2

    Model ? A model is a representation of a physical system that maModel ? A model is a representation of a physical system that may y be used to predict the behavior of the system in some desired rebe used to predict the behavior of the system in some desired respect. spect. Mathematical or computer models may also conform to this Mathematical or computer models may also conform to this definition, our interest will be in physical model.definition, our interest will be in physical model.Prototype? The physical system for which the prediction are to bPrototype? The physical system for which the prediction are to be e made. made. Models that resemble the prototype but are generally of a differModels that resemble the prototype but are generally of a different ent size, may involve different fluid, and often operate under diffesize, may involve different fluid, and often operate under different rent conditions. conditions. Usually a model is smaller than the prototype. Usually a model is smaller than the prototype. Occasionally, if the prototype is very small, it may be advantagOccasionally, if the prototype is very small, it may be advantageous eous to have a model that is larger than the prototype so that it canto have a model that is larger than the prototype so that it can vevemore easily studied. For example, large models have been used tomore easily studied. For example, large models have been used tostudy the motion of red blood cells.study the motion of red blood cells.

    模型,物理系統的代表,可用來預測系統的行為。有物理、數學與電腦模型。

    雛型,第一版的物理系統,成品的前身,檢驗預測結果的依據

    模型「類似」雛型,不見得是雛型,可以較大或較小,操作環境不盡然相似,有可能出現distortion

  • 68

    Model vs. Prototype Model vs. Prototype 2/22/2

    With the successful development of a valid model, it is possibleWith the successful development of a valid model, it is possible to to predict the behavior of the prototype under a certain set of predict the behavior of the prototype under a certain set of conditions.conditions.There is an inherent danger in the use of models in that predictThere is an inherent danger in the use of models in that predictions ions can be made that are in error and the error not detected until tcan be made that are in error and the error not detected until the he prototype is found not to perform as predicted.prototype is found not to perform as predicted.It is imperative that the model be properly designed and tested It is imperative that the model be properly designed and tested and and that the results be interpreted correctly.that the results be interpreted correctly.

    所有在模型試驗的結果不能保證移到prototype就會ok,因為有太多無法預測的變數造成風險的存在。即便如此,好好地設計模型,好好的進行模型試驗,正確地解讀試驗資料,仍然是有必要的。

  • 69

    Similarity of Model and PrototypeSimilarity of Model and Prototype

    What conditions must be met to ensure the similarity of model anWhat conditions must be met to ensure the similarity of model and d prototype?prototype?Geometric SimilarityGeometric Similarity

    Model and prototype have same shape.Model and prototype have same shape.Linear dimensions on model and prototype correspond within Linear dimensions on model and prototype correspond within constant scale factor.constant scale factor.

    Kinematic SimilarityKinematic SimilarityVelocities at corresponding points on model and prototype differVelocities at corresponding points on model and prototype differonly by a constant scale factor.only by a constant scale factor.

    Dynamic SimilarityDynamic SimilarityForces on model and prototype differ only by a constant scale Forces on model and prototype differ only by a constant scale factor.factor.

    如何確保模型與雛型的相似性

    在談「相似性」時,相似性包括「幾何上的相似性」、「運動上的相似性」與「動力上的相似性」

    即Model與prototype間的…與…維持某一比例關係

  • 70

    Theory of Models Theory of Models 1/51/5

    The theory of models can be readily developed by using The theory of models can be readily developed by using the principles of dimensional analysis.the principles of dimensional analysis.For given problem which can be described in terms of a For given problem which can be described in terms of a set of pi terms asset of pi terms as

    ),,,,( n321 ΠΠΠφ=ΠThis relationship can be formulated This relationship can be formulated with a knowledge of the general with a knowledge of the general nature of the physical phenomenon nature of the physical phenomenon and the variables involved. and the variables involved.

    This equation applies to any system that is governed by the same variables.

    利用因次分析的原理所發展出來的理論

    要能如此,表示model與prototype所處環境的變數相同

  • 71

    Theory of Models Theory of Models 2/52/5

    A similar relationship can be written for a model of this A similar relationship can be written for a model of this prototype; that is,prototype; that is,

    where the form of the function will be the same as long where the form of the function will be the same as long as the as the same phenomenonsame phenomenon is involved in both the is involved in both the prototype and the model.prototype and the model.

    ),,,,,( nmm3m2m1 ΠΠΠφ=Π

    The prototype and the model must have The prototype and the model must have the the same phenomenonsame phenomenon..

    因為model與prototype的變數相同,透過PI原理所推導出來的PI terms當然就都一樣

    稱model與prototype(所處環境)具有相同的現象

    區隔model或非model

  • 72

    Theory of Models Theory of Models 3/53/5

    Model design (the model is designed and operated) Model design (the model is designed and operated) conditions, also called similarity requirements or modeling conditions, also called similarity requirements or modeling laws.laws.

    The The form of form of ΦΦ is the same for model and prototype, it is the same for model and prototype, it follows thatfollows that

    nmnm33m22 ..... Π=ΠΠ=ΠΠ=Π

    m11 Π=Π

    This is the This is the desired prediction equationdesired prediction equation and and indicates that the measured of indicates that the measured of ΠΠ1m1m obtained obtained with the model will be equal to the with the model will be equal to the corresponding corresponding ΠΠ11 for the prototype as long as for the prototype as long as the other the other ΠΠ parameters are equal. parameters are equal.

    要能如此

    才能

    稱為Design conditions

    稱為Prediction equation

  • 73

    Theory of Models Theory of Models 4/54/5 –– SummarySummary11

    The prototype and the model must have the same The prototype and the model must have the same phenomenon. phenomenon.

    ),,,,( n321 ΠΠΠφ=ΠFor prototypeFor prototype

    For modelFor model ),,,,,( nmm3m2m1 ΠΠΠφ=Π

    Model與prototype所處環境具有same phenomenon

  • 74

    Theory of Models Theory of Models 5/55/5 –– SummarySummary22

    The model is designed and operated under the following The model is designed and operated under the following conditions conditions (called design conditions, also called similarity (called design conditions, also called similarity requirements or modeling laws)requirements or modeling laws)

    有前提有前提+有條件+有條件 才能用才能用ModelModel預測預測PrototypePrototype

    The measured of The measured of ΠΠ1m1m obtained with the model will be obtained with the model will be equal to the corresponding equal to the corresponding ΠΠ11 for the prototype. for the prototype.

    nmnm33m22 ..... Π=ΠΠ=ΠΠ=Π

    m11 Π=Π Called prediction equationCalled prediction equation

  • 75

    Theory of Models Theory of Models EXAMPLE 1EXAMPLE 1

    Example: Considering the drag force on a sphereExample: Considering the drag force on a sphere..

    The prototype and the model must have the same phenomenon. The prototype and the model must have the same phenomenon.

    Design conditions.Design conditions.

    Then Then ……

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρ=

    ρ mmmm

    12m

    2mm

    m DVfDV

    F

    ),,V,D(fF μρ=

    prototype122

    VDfDV

    F⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρ=

    ρ

    prototypeelmod

    VDVD⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρ=⎟⎟

    ⎞⎜⎜⎝

    ⎛μ

    ρ

    prototype22

    elmod22 DV

    FDV

    F⎟⎟⎠

    ⎞⎜⎜⎝

    ρ=⎟

    ⎟⎠

    ⎞⎜⎜⎝

    ρ

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρ=

    ρ

    VDfDV

    F122

  • 76

    Theory of Models Theory of Models EXAMPLE 2EXAMPLE 2

    Example: Determining the drag force on a thin rectangular plate Example: Determining the drag force on a thin rectangular plate ((ww×× hh in size)in size)

    The prototype and the model must have the same phenomenon. The prototype and the model must have the same phenomenon.

    Design conditions.Design conditions.

    Then Then ……

    ( )V,,,h,wfD ρμ=

    prototype22

    Vw,hw

    VwD

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρΦ=

    ρ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρΦ=

    ρ mmmm

    m

    m2

    mm2

    m

    m wV,hw

    VwD

    μρ

    ρ=

    VwwV,hw

    hw

    m

    mmm

    m

    m

    m

    2

    mm

    2

    m2

    mm2

    m

    m22 DV

    VwwD

    VwD

    VwD

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛ρρ

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛=>>

    ρ=

    ρ

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρΦ=

    ρ

    Vw,hw

    VwD

    22

  • 77

    Example 7.5 Prediction of Prototype Example 7.5 Prediction of Prototype Performance from Model Data Performance from Model Data 1/21/2

    A long structural component of a bridge has the cross section A long structural component of a bridge has the cross section shown in Figure E7.5. It is known that when a steady wind blows shown in Figure E7.5. It is known that when a steady wind blows past this type of bluff body, vortices may develop on the downwipast this type of bluff body, vortices may develop on the downwind nd side that are shed in a regular fashion at some definite frequenside that are shed in a regular fashion at some definite frequency. cy. Since these vortices can create harmful periodic forces acting oSince these vortices can create harmful periodic forces acting on the n the structure, it is important to determine the shedding frequency. structure, it is important to determine the shedding frequency. For For the specific structure of interest, D=0.1m, H=0.3m, and a the specific structure of interest, D=0.1m, H=0.3m, and a representative wind velocity 50km/hr. Standard air can be assumerepresentative wind velocity 50km/hr. Standard air can be assumed. d. The shedding frequency is to be determined through the use of a The shedding frequency is to be determined through the use of a smallsmall--scale model that is to be tested in a water tunnel. For the scale model that is to be tested in a water tunnel. For the model Dm=20mm and the water temperature is 20model Dm=20mm and the water temperature is 20℃℃..

  • 78

    Example 7.5 Prediction of Prototype Example 7.5 Prediction of Prototype Performance from Model Data Performance from Model Data 2/22/2

    Determine the model dimension, Determine the model dimension, HmHm, and the velocity at which the , and the velocity at which the test should be performed. If the shedding frequency test should be performed. If the shedding frequency ωω for the for the model is found to be 49.9Hz, what is the corresponding frequencymodel is found to be 49.9Hz, what is the corresponding frequencyfor the prototype? for the prototype? For air at standard condition For air at standard condition For water at 20For water at 20℃℃,,

    35 m/kg23.1,sm/kg1079.1 =ρ−×=μ −

    3water

    3water m/kg998,sm/kg101 =ρ−×=μ

  • 79

    Example 7.5 Example 7.5 SolutionSolution1/41/4

    Step 1:List all the dimensional variables involved. Step 1:List all the dimensional variables involved. ωωD,H,V,D,H,V,ρρ,,μμ k=6 dimensional variables.k=6 dimensional variables.Step 2:Select primary dimensions F,L and T. List the Step 2:Select primary dimensions F,L and T. List the dimensions of all variables in terms of primary dimensions. dimensions of all variables in terms of primary dimensions. r=3 primary dimensionsr=3 primary dimensions

    TMLTFLLTV

    LHLDT2241

    1

    −−−

    =μ=ρ=

    ===ω

    &&&

    &&&

  • 80

    Example 7.5 Example 7.5 SolutionSolution2/42/4

    Step 3: Determine the required number of pi terms.Step 3: Determine the required number of pi terms.kk--rr=6=6--3=33=3

    Step 4:Select repeating variables D,V, Step 4:Select repeating variables D,V, μμ..Step Step 5~6:combining the repeating variables with each of 5~6:combining the repeating variables with each of the other variables in turn, to form dimensionless groups.the other variables in turn, to form dimensionless groups.

    μρ

    =μρ=Π

    =μ=Πω

    =μω=Π

    VDVD

    DHVHD

    VDVD

    333

    222111

    cba3

    cba2

    cba1

  • 81

    Example 7.5 Example 7.5 SolutionSolution3/43/4

    The functional relationship isThe functional relationship is

    The prototype and the model must have the same The prototype and the model must have the same phenomenon.phenomenon.

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρφ=

    ω VD,DH

    VD

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρφ=

    ω

    m

    mmm

    m

    m

    m

    mm DV,DH

    VD

    StrouhalStrouhal numbernumber

  • 82

    Example 7.5 Example 7.5 SolutionSolution4/44/4

    The Design conditions.The Design conditions.

    ThenThen……..m

    m

    DH

    DH

    =m

    mmm DVVDμ

    ρ=

    μρ

    mm60...DDHH mm === s/m9.13...VD

    DVmm

    mm ==ρ

    ρμμ

    =

    Hz0.29...D

    DVV

    mm

    m==ω=ω

  • 83

    Model ScalesModel Scales

    The ratio of a model variable to the corresponding The ratio of a model variable to the corresponding prototype variable is called the scale for that variable.prototype variable is called the scale for that variable.

    2

    m2

    1

    m1

    m2

    m1

    2

    1l

    l

    l

    l

    l

    l

    l

    ll ==λ⇒=

    ρρ

    =λρm

    VVm

    V =λ

    μμ

    =λμm

    Length ScaleLength Scale

    Velocity ScaleVelocity Scale

    Density ScaleDensity Scale

    Viscosity ScaleViscosity Scale

  • 84

    Validation of Models DesignValidation of Models Design

    The purpose of model design is to predict the effects of The purpose of model design is to predict the effects of certain proposed changes in a given prototype, and in this certain proposed changes in a given prototype, and in this instance some actual prototype data may be available.instance some actual prototype data may be available.Validation of model design ?Validation of model design ?The model can be designed, constructed, and tested, and The model can be designed, constructed, and tested, and the model prediction can be compared with these data. the model prediction can be compared with these data. If If the agreement is satisfactorythe agreement is satisfactory, then the model can be , then the model can be changed in the desired manner, and the corresponding changed in the desired manner, and the corresponding effect on the prototype can be predicted with increased effect on the prototype can be predicted with increased confidence.confidence.

  • 85

    Distorted ModelsDistorted Models

    In many model studies, to achieve dynamic similarity In many model studies, to achieve dynamic similarity requires duplication of several dimensionless groups. requires duplication of several dimensionless groups. In some cases, complete dynamic similarity between In some cases, complete dynamic similarity between model and prototype may not be attainable. If one or more model and prototype may not be attainable. If one or more of the similarity requirements are not met, for example, of the similarity requirements are not met, for example, if , then it follows that the prediction equatif , then it follows that the prediction equation ion

    is not true; that is, is not true; that is, MODELS for which one or more of the similarity MODELS for which one or more of the similarity requirements are not satisfied are called requirements are not satisfied are called DISTORTED DISTORTED MODELSMODELS. .

    m22 Π≠Π

    m11 Π=Π m11 Π≠Π

  • 86

    Distorted Models Distorted Models EXAMPLEEXAMPLE--1 1/31 1/3

    Determine the drag force on a surface ship, complete Determine the drag force on a surface ship, complete dynamic similarity requires that both Reynolds and Froude dynamic similarity requires that both Reynolds and Froude numbers be duplicated between model and prototype.numbers be duplicated between model and prototype.

    To match Froude numbers To match Froude numbers between model and prototypebetween model and prototype

    2/1p

    pp2/1

    m

    mm

    )g(

    VFr

    )g(VFr

    ll===

    p

    ppp

    m

    mmm

    VReVRe

    υ==

    υ=

    ll

    Froude numbersFroude numbers

    Reynolds numbersReynolds numbers2/1

    p

    m

    p

    mVV

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛=

    l

    l

  • 87

    Distorted Models Distorted Models EXAMPLEEXAMPLE--11 2/32/3

    To match Reynolds numbers between model and To match Reynolds numbers between model and prototypeprototype

    p

    m

    p

    m

    p

    mVV

    l

    l=

    υυ

    2/3

    p

    m

    p

    m2/1

    p

    m

    p

    m⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛=⎟

    ⎟⎠

    ⎞⎜⎜⎝

    ⎛=

    υυ

    l

    l

    l

    l

    l

    l

    If If llmm/ / llp p equals 1/100(a typical length scale for ship equals 1/100(a typical length scale for ship model tests) , then model tests) , then υυmm//υυp p must be 1/1000.must be 1/1000.>>> The kinematic viscosity ratio required to >>> The kinematic viscosity ratio required to duplicate Reynolds numbers cannot be attained.duplicate Reynolds numbers cannot be attained.

  • 88

    Distorted Models Distorted Models EXAMPLEEXAMPLE--11 3/33/3

    It is impossible in practice for this model/prototype scale It is impossible in practice for this model/prototype scale of 1/100 to satisfy both the Froude number and Reynolds of 1/100 to satisfy both the Froude number and Reynolds number criterianumber criteria; ; at best we will be able to satisfy only at best we will be able to satisfy only one of them.one of them.If water is the only practical liquid for most model test of If water is the only practical liquid for most model test of freefree--surface flows, a surface flows, a fullfull--scale test is required to obtain scale test is required to obtain complete dynamic similaritycomplete dynamic similarity..

  • 89

    Distorted Models Distorted Models EXAMPLEEXAMPLE--22 1/21/2

    In the study of open channel or freeIn the study of open channel or free--surface flows. surface flows. Typically in these problems both the Reynolds number Typically in these problems both the Reynolds number and Froude number are involved. and Froude number are involved.

    To match Froude numbers To match Froude numbers between model and prototypebetween model and prototype

    2/1pp

    pp2/1

    mm

    mm

    )g(

    VFr

    )g(VFr

    ll=== Froude numbersFroude numbers

    p

    pppp

    m

    mmmm

    VReVRe

    μ

    ρ==

    μρ

    =ll Reynolds numbersReynolds numbers

    ll

    lλ=⎟

    ⎟⎠

    ⎞⎜⎜⎝

    ⎛=

    2/1

    p

    m

    p

    mVV

  • 90

    Distorted Models Distorted Models EXAMPLEEXAMPLE--22 2/22/2

    To match Reynolds numbers between model and To match Reynolds numbers between model and prototypeprototype

    m

    p

    m

    p

    p

    m

    p

    mVV

    l

    l

    ρ

    ρ

    μμ

    =

    If If llmm/ / ll p p equals 1/100(a typical length scale for ship equals 1/100(a typical length scale for ship model tests) , then model tests) , then υυmm//υυp p must be 1/1000.must be 1/1000.>>>The kinematic viscosity ratio required to >>>The kinematic viscosity ratio required to duplicate Reynolds numbers cannot be attained.duplicate Reynolds numbers cannot be attained.

    p

    m

    mp

    pm2/3

    m

    p

    mp

    pm

    p

    m

    p

    m

    //

    //

    VV

    υυ

    =ρρ

    μμ=λ

    ρρ

    μμ==

    l

    l

    l

    l

    l

  • 91

    Typical Model StudiesTypical Model Studies

    Flow through closed conduits.Flow through closed conduits.Flow around immersed bodies.Flow around immersed bodies.Flow with a free surface.Flow with a free surface.

  • 92

    Flow Through Closed Conduits Flow Through Closed Conduits 1/51/5

    This type of flow includes pipe flow and flow through This type of flow includes pipe flow and flow through valves, fittings, and metering devices. valves, fittings, and metering devices. The conduits are often circular, they could have other The conduits are often circular, they could have other shapes as well and may contain expansions or contractions.shapes as well and may contain expansions or contractions.Since there are no fluid interfaces or free surface, the Since there are no fluid interfaces or free surface, the dominant forces are inertial and viscous forces so that the dominant forces are inertial and viscous forces so that the Reynolds number is an important similarity parameter.Reynolds number is an important similarity parameter.

  • 93

    Flow Through Closed Conduits Flow Through Closed Conduits 2/52/5

    For low Mach numbers (Ma

  • 94

    Flow Through Closed Conduits Flow Through Closed Conduits 3/53/5

    To meet the requirement of geometric similarityTo meet the requirement of geometric similarity

    To meet the requirement of Reynolds numberTo meet the requirement of Reynolds number

    ll

    l

    l

    l

    lll

    l

    l

    lλ=

    εε

    ==⇒ε

    = mmi

    im

    m

    mi

    m

    im

    mm

    mm

    m

    mmmV

    VVVl

    lll

    ρρ

    μμ

    =⇒μ

    ρ=

    μρ

    If the same fluid is used, thenIf the same fluid is used, then ll

    lλ=⇒= /VV

    VV

    mm

    m

    1m

    m ==μμ

    l

    l

  • 95

    Flow Through Closed Conduits Flow Through Closed Conduits 4/54/5

    The fluid velocity in the model will be larger than that in The fluid velocity in the model will be larger than that in the prototype for any length scale less than 1. Since length the prototype for any length scale less than 1. Since length scales are typically much less than unity.scales are typically much less than unity.Reynolds number similarity may be difficult to achieve Reynolds number similarity may be difficult to achieve because of the large model velocities required.because of the large model velocities required.

    VVm >lll

    λ=⇒= /VVV

    Vm

    m

    m

    1

  • 96

    Flow Through Closed Conduits Flow Through Closed Conduits 5/55/5

    With these similarity requirements satisfied, it follows that With these similarity requirements satisfied, it follows that the dependent pi term will be equal in model and the dependent pi term will be equal in model and prototype. For example,prototype. For example,

    The prototype pressure dropThe prototype pressure drop21 V

    Δ=ΠDependent pi termDependent pi term

    m

    2

    mmp

    VVp Δ⎟⎟

    ⎞⎜⎜⎝

    ⎛ρρ

    =Δ mpp Δ≠ΔIn general

  • 97

    Example 7.6 Reynolds Number SimilarityExample 7.6 Reynolds Number Similarity

    Model test are to be performed to study the flow through a largeModel test are to be performed to study the flow through a largevalve having a 2valve having a 2--ftft--diameter inlet and carrying water at a diameter inlet and carrying water at a flowrateflowrate of of 30cfs. The working fluid in the model is water at the same 30cfs. The working fluid in the model is water at the same temperature as that in the prototype. Complete geometric similartemperature as that in the prototype. Complete geometric similarity ity exits between model and prototype, and the model inlet diameter exits between model and prototype, and the model inlet diameter is is 30 in. Determine the required 30 in. Determine the required flowrateflowrate in the model. in the model.

  • 98

    Example 7.6 Example 7.6 SolutionSolution

    DD...

    VAAV

    QQ mmmm ===

    To ensure dynamic similarity, the model tests should be run so that

    m

    m

    m

    mmm D

    DVVVDDVReeR =⇒

    ν=

    ν=

    Same fluid used in the model and prototypeSame fluid used in the model and prototype

    cfs75.3)s/ft30((2ft)

    (3/12ft)Q 3m ==

  • 99

    Flow Around Immersed Bodies Flow Around Immersed Bodies 1/71/7

    This type of flow includes flow around aircraft, This type of flow includes flow around aircraft, automobiles, golf balls, and building.automobiles, golf balls, and building.For these problems, geometric and Reynolds number For these problems, geometric and Reynolds number similarity is required.similarity is required.Since there are no fluid interfaces, surface tension is not Since there are no fluid interfaces, surface tension is not important. Also, gravity will not affect the flow pattern, so important. Also, gravity will not affect the flow pattern, so the Froude number need not to be considered.the Froude number need not to be considered.For incompressible flow, the Mach number can be omitted.For incompressible flow, the Mach number can be omitted.

  • 100

    Flow Around Immersed Bodies Flow Around Immersed Bodies 2/72/7

    A general formulation for these problems isA general formulation for these problems is

    ⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛μ

    ρεφ

    l

    ll

    l V,,i

    Where Where ll is some characteristic length of the system and is some characteristic length of the system and lliirepresents other pertinent lengths, represents other pertinent lengths, εε/ / ll is the relative is the relative roughness of the surface, and roughness of the surface, and ρρVVll//μμ is the Reynolds number.is the Reynolds number.

    Dependent pi term=Dependent pi term=

    Model of the National Bank of Commerce, San Antonio, Texas, for measurement of peak, rms, and mean pressure distributions. The model is located in a long-test-section, meteorological wind tunnel.

  • 101

    Flow Around Immersed Bodies Flow Around Immersed Bodies 3/73/7

    Frequently, the dependent variable of interest for this type Frequently, the dependent variable of interest for this type of problem is the drag, D, developed on the body.of problem is the drag, D, developed on the body.

    To meet the requirement of geometric similarityTo meet the requirement of geometric similarity

    22D

    V21

    DClρ

    =

    The dependent pi terms would usually be expressed in the form of a drag coefficient

    ll

    l

    l

    l

    lll

    l

    l

    lλ=

    εε

    ==⇒ε

    = mmi

    im

    m

    mi

    m

    im

  • 102

    Flow Around Immersed Bodies Flow Around Immersed Bodies 4/74/7

    To meet the requirement of Reynolds number similarityTo meet the requirement of Reynolds number similarity

    m

    m

    mm

    mm

    m

    mmmV

    VVVl

    l

    l

    lll

    υυ

    =ρρ

    μμ

    =⇒μ

    ρ=

    μρ

    m

    2

    m

    2

    mm2m2

    mm

    m22

    DVVD

    V21

    D

    V21

    D⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛⎟⎟⎠

    ⎞⎜⎜⎝

    ⎛ρρ

    =⇒ρ

    =ρ l

    l

    ll

    The same fluid is used, thenThe same fluid is used, then ll

    lλ=⇒= /VV

    VV

    mm

    m

  • 103

    Flow Around Immersed Bodies Flow Around Immersed Bodies 5/75/7

    The fluid velocity in the model will be larger than that in The fluid velocity in the model will be larger than that in the prototype for any length scale less than 1. Since length the prototype for any length scale less than 1. Since length scales are typically much less than unity.scales are typically much less than unity.Reynolds number similarity may be difficult to achieve Reynolds number similarity may be difficult to achieve because of the large model velocities required.because of the large model velocities required.

  • 104

    Flow Around Immersed Bodies Flow Around Immersed Bodies 6/76/7

    How to reduce the fluid velocity in the model ?How to reduce the fluid velocity in the model ?

    A different fluid is used in the model such thatA different fluid is used in the model such that 1/m

  • 105

    Flow Around Immersed Bodies Flow Around Immersed Bodies 7/77/7

    How to reduce the fluid velocity in the model ?How to reduce the fluid velocity in the model ?

    Same fluid with different density.. Same fluid with different density.. ρρmm>>ρρ

    An alternative way to reduce An alternative way to reduce VVmm is to is to increase the air pressure in the tunnel so that increase the air pressure in the tunnel so that ρρmm>>ρρ. The pressurized tunnels are obviously . The pressurized tunnels are obviously complicated and expensive.complicated and expensive.

  • 106

    Example 7.7 Model Design Conditions and Example 7.7 Model Design Conditions and Predicted Prototype PerformancePredicted Prototype Performance

    The drag on an airplane cruising at 240 mph in standard air is tThe drag on an airplane cruising at 240 mph in standard air is to be o be determined from tests on a 1:10 scale model placed determined from tests on a 1:10 scale model placed inaina pressurized pressurized wind tunnel. To minimize compressibility effects, the airspeed iwind tunnel. To minimize compressibility effects, the airspeed in the n the wind tunnel is also to be 240 mph. Determine the required air wind tunnel is also to be 240 mph. Determine the required air pressure in the tunnel (assuming the same air temperature for mopressure in the tunnel (assuming the same air temperature for model del and prototype), and the drag on the prototype corresponding to aand prototype), and the drag on the prototype corresponding to ameasured force of 1 lb on the model. measured force of 1 lb on the model.

  • 107

    Example 7.7 Example 7.7 SolutionSolution1/21/2

    The Reynolds numbers in model and prototype are the same. Thus,

    μρ

    ρ ll VV

    m

    mmm

    VVmm=V, =V, llmm// ll=1/10=1/10

    μμ

    =μμ

    =ρρ m

    mm

    mm 10VV

    l

    l

    The same fluid with The same fluid with ρρmm==ρρ and and μμmm==μμ cannot be cannot be used if Reynolds number similarity is to be used if Reynolds number similarity is to be maintained.maintained.

  • 108

    Example 7.7 Example 7.7 SolutionSolution2/22/2

    10m =ρρ

    We assume that an increase in pressure does not We assume that an increase in pressure does not significantly change the viscosity so that the required significantly change the viscosity so that the required increase in density is given by the relationshipincrease in density is given by the relationship

    10p

    p mm =ρρ

    =For ideal gas

    ...DVD

    VD

    2m

    2mm2

    1m

    2221

    =ρ ll

  • 109

    Flow Around Immersed Bodies at High Flow Around Immersed Bodies at High Reynolds Number Reynolds Number 1/31/3

    Unfortunately, in many situations the flow characteristics Unfortunately, in many situations the flow characteristics are not strong influenced by the Reynolds number over the are not strong influenced by the Reynolds number over the operating range of interest.operating range of interest.Consider the variation in the drag coefficient with the Consider the variation in the drag coefficient with the Reynolds number for a smooth sphere of diameter d Reynolds number for a smooth sphere of diameter d placed in a uniform stream with approach velocity, V.placed in a uniform stream with approach velocity, V.At high Reynolds numbers the drag is often essentially At high Reynolds numbers the drag is often essentially independent of the Reynolds number.independent of the Reynolds number.

  • 110

    Flow Around Immersed Bodies at High Flow Around Immersed Bodies at High Reynolds Number Reynolds Number 2/32/3

    The effect of Reynolds number on the drag coefficient, CD for a smooth sphere with CD = D/ ½ AρV2, where A is the projected area of sphere, πd2/4.

  • 111

    Flow Around Immersed Bodies at High Flow Around Immersed Bodies at High Reynolds Number Reynolds Number 3/33/3

    For problems involving high velocities in which the Mach For problems involving high velocities in which the Mach number is greater than about 0.3, the influence of number is greater than about 0.3, the influence of compressibility, and therefore the Mach number (or compressibility, and therefore the Mach number (or Cauchy number), becomes significant.Cauchy number), becomes significant.In this case complete similarity requires not only In this case complete similarity requires not only geometric and Reynolds number similarity but also Mach geometric and Reynolds number similarity but also Mach number similarity so thatnumber similarity so that

    l

    lmmmm

    m

    cc

    cV

    cV

    υυ

    =⇒=

  • 112

    Flow with a Free Surface Flow with a Free Surface 1/81/8

    This type of flow includes flow in canals, rivers, spillways, This type of flow includes flow in canals, rivers, spillways, and stilling basics, as well as flow around ship.and stilling basics, as well as flow around ship.For this class of problems, gravitational, inertial forces, For this class of problems, gravitational, inertial forces, and surface tension are important and, therefore, the and surface tension are important and, therefore, the Froude number and Weber number become important Froude number and Weber number become important similarity parameters.similarity parameters.Since there is a free surface with a liquidSince there is a free surface with a liquid--air interface, air interface, forces due to surface tension may be significant, and the forces due to surface tension may be significant, and the Weber number becomes another similarity parameter that Weber number becomes another similarity parameter that needs to be considered along with the Reynolds number.needs to be considered along with the Reynolds number.

  • 113

    Flow with a Free Surface Flow with a Free Surface 2/82/8

    A general formulation for these problems isA general formulation for these problems is

    To meet the requirement of Froude number similarityTo meet the requirement of Froude number similarity

    ⎟⎟

    ⎜⎜

    σρ

    μρε

    φl

    l

    l

    ll

    l 2i V,gV,V,,Dependent pi term=Dependent pi term=

    Where Where ll is some characteristic length of the system and is some characteristic length of the system and lliirepresents other pertinent lengths, represents other pertinent lengths, εε/ / ll is the relative is the relative roughness of the surface, and roughness of the surface, and ρρVVll//μμ is the Reynolds number.is the Reynolds number.

    ll gV

    gV

    mm

    m =ggm =

    ll

    lλ== mm

    VV

  • 114

    Flow with a Free Surface Flow with a Free Surface 3/83/8

    To meet the requirement of To meet the requirement of Reynolds number and Froude Reynolds number and Froude numbernumber similaritysimilarity

    m

    m

    mm

    mm

    m

    mmmV

    VVVl

    l

    l

    lll

    υυ

    =ρρ

    μμ

    =⇒μ

    ρ=

    μρ

    The working fluid for the prototype is normally either freshwateThe working fluid for the prototype is normally either freshwater r or seawater and the length scale is small. or seawater and the length scale is small. It is virtually impossible to satisfy It is virtually impossible to satisfy , so models , so models involving freeinvolving free--surface flows are usually distorted.surface flows are usually distorted.

    ( ) 2/3m / lλ=υυ

    2/3m )( lλ=υυ

  • 115

    Flow with a Free Surface Flow with a Free Surface 4/84/8

    The problem is further complicated if an attempt is made The problem is further complicated if an attempt is made to model surface tension effects, since this requires the to model surface tension effects, since this requires the equality of Weber numbers, which leads to the conditionequality of Weber numbers, which leads to the condition

    ( )22m

    2mmm

    2

    m

    m2

    mm

    VV

    //VV

    ll

    lllλ==

    ρσρσ

    ⇒σ

    ρ=

    σρ

    For the kinematic surface tension For the kinematic surface tension σσ//ρρ. It is evident that the s. It is evident that the same ame fluid cannot be used in model a