Transcript
  • 8/12/2019 Me 3560 Presentation Ch Vii

    1/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    1

    ME 3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,

    Similitude and Modeling

  • 8/12/2019 Me 3560 Presentation Ch Vii

    2/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    2

    7.1 Dimensional AnalysisMany problems of interest in fluid mechanics cannot be solved usingthe integral and/or differential equations.

    Examples of problems that are studied in the laboratory with the use of models are:

    Wind motions around a football stadium.

    The flow of water through a large hydro-turbine.

    The airflow around the deflector on a semi-truck.

    The wave motion around a pier or a ship.

    Airflow around aircraft.

  • 8/12/2019 Me 3560 Presentation Ch Vii

    3/16

  • 8/12/2019 Me 3560 Presentation Ch Vii

    4/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    4

    Any equation can be expressed in terms of dimensionless parameters

    simply by dividing each term by one of the other terms.

    Consider 22

    221

    211

    22 zg

    V p zg

    V p++=++

    Divide both sides by gz2 to obtain:

    2

    1

    1

    21

    1

    1

    2

    22

    2

    2 12

    12 z

    zgz

    V gz

    pgz

    V gz

    p

    ++=++

    The dimensionless parameters, V 2 / gz and p / gz can be used to predictthe performance of a prototype with a model tested in the laboratory.

    Similitude is the study that allows the prediction of the quantities to beexpected on a prototype from the measurements on a model.

  • 8/12/2019 Me 3560 Presentation Ch Vii

    5/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    5

    Thus, Dimensional Analysis and Similitude seek to approach the

    following problems:

    Express a given dimensional , functional relation in a Nondimensionalform.

    Use Nondimensional parameters in similitude testing of models touse the results obtained to predict performance of prototypes.

    Nondimensionalize equations (algebraic or differential) to determinerelevant nondimensional parameters.

  • 8/12/2019 Me 3560 Presentation Ch Vii

    6/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    6

    7.2 Buckingham Pi TheoremThe number of dimensionless products required to replace the originallist of variables describing a physical phenomenon is established byBuckingham Pi Theorem, which is the basic theorem of dimensional

    analysis and states that: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.

    The dimensionless products are frequently referred to as pi terms

  • 8/12/2019 Me 3560 Presentation Ch Vii

    7/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    7

    The Pi theorem is based on the idea of dimensional homogeneity.

    It is assumed that for any physically meaningful equation involving k variables, such as: ),...,,( 321 k uuu f u =

    The dimensions of the variable on the L.H.S. of the equal sign must beequal to the dimensions of any term that stands by itself on the R.H.S. of the equation.

    Thus, it is possible to rearrange the equation into a set of dimensionlessproducts (pi terms) so that

    ),...,,( 321 r k = The required number of pi terms is fewer than the number of originalvariables by r . r is the minimum number of reference dimensions required to describethe original list of variables.

  • 8/12/2019 Me 3560 Presentation Ch Vii

    8/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    8

    Usually the reference dimensions required to describe the variables will

    be the basic dimensions M , L, and T or F , L, and T .

    In some instances perhaps only two dimensions, such as L and T , arerequired, or maybe just one, such as L.

    In a few rare cases the variables may be described by somecombination of basic dimensions, such as M / T 2 and L, and in this case r would be equal to two rather than three.

  • 8/12/2019 Me 3560 Presentation Ch Vii

    9/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    9

    7.3 Determination of the Pi Terms1. List all the variables involved in the problem.2. Express each of the variables in terms of basic dimensions.3. Determine the required number of Pi terms.

    4. Select a number of repeating variable, where the number required isequal to the number of reference dimensions.5. Form a Pi term by multiplying one of the non repeating variables by

    the product of the repeating variables each raised to an exponent that

    will make the combination dimensionless.6. Repeat the previous step for the remaining non repeating variables.7. Check all resulting Pi terms to make sure they are dimensionless and

    independent.8. Express the final form as a relationship between Pi terms.

  • 8/12/2019 Me 3560 Presentation Ch Vii

    10/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    10

    Geometry . The geometric characteristics can usually be described by aseries of lengths and angles. In most problems the geometry of thesystem plays an important role, and a sufficient number of geometricvariables must be included to describe the system.

    Material Properties. Since the response of a system to applied external

    effects such as forces, pressures, and changes in temperature isdependent on the nature of the materials involved in the system, thematerial properties that relate the external effects and the responses mustbe included as variables.

    External Effects . This terminology is used to denote any variable thatproduces, or tends to produce, a change in the system. For example,forces applied to a system, pressures, velocities, or gravity.

    Factors to Consider when Selecting the Variables Involved in a

    Phenomenon:

  • 8/12/2019 Me 3560 Presentation Ch Vii

    11/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    11

    When selecting the variables involved in a problem it is also needed to:

    Keep the number of variables to a minimum, Make sure that all variables selected are independent.Clearly define the problem. What is the main variable of interest (thedependent variable)?Consider the basic laws that govern the phenomenon.Start the variable selection process by grouping the variables into threebroad classes: geometry, material properties, and external effects.Consider other variables that may not fall into one of the abovecategories. For example, time will be an important variable if any of thevariables are time dependent.

    Be sure to include all quantities that enter the problem even thoughsome of them may be held constant (e.g., the acceleration of gravity, g).

  • 8/12/2019 Me 3560 Presentation Ch Vii

    12/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    12

    Problem 7.12: At a sudden contraction in a pipe the diameter changes

    from D 1 to D 2. The pressure drop, p , which develops across thecontraction is a function of D 1 and D 2, as well as the velocity, V , in thelarger pipe, and the fluid density, , and viscosity, . Use D 1, V , and asrepeating variables to determine a suitable set of dimensionlessparameters. Why would it be incorrect to include the velocity in thesmaller pipe as an additional variable?

    Problem 7.17: A thin elastic wire is placed between rigid supports. Afluid flows past the wire, and it is desired to study the static deflection, , at the center of the wire due to the fluid drag. Assume that:

    ),,,,,( V E d l f =

    where l is the wire length, d the wire diameter, the fluid density, thefluid viscosity, V the fluid velocity, and E the modulus of elasticity of the wire material. Develop a suitable set of pi terms for this problem.

  • 8/12/2019 Me 3560 Presentation Ch Vii

    13/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    13

    7.6 Common Dimensionless Groups in FluidMechanics

  • 8/12/2019 Me 3560 Presentation Ch Vii

    14/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    14

    7.8 Modeling and Similitude A model is a representation of a physical system that may be used to

    predict the behavior of the system in some desired respect .

    The physical system for which the predictions are to be made is calledthe prototype.

    Mathematical or computer models may also conform to this definition,

    but our interest will be in physical models: models that resemble theprototype but are generally of a different size, may involve differentfluids, and often operate under different conditions (pressures, velocities,etc.).Usually a model is smaller than the prototype.Occasionally, if the prototype is very small, itmay be advantageous to have a model that is

    larger than the prototype.

  • 8/12/2019 Me 3560 Presentation Ch Vii

    15/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    15

    7.8.1 Theory of ModelsThe process to determine the performance of a prototype based on thestudy of a model can be summarized as follows:1. Establish the dependent variable and the variables that affect it.

    ),...,,(321 k

    uuu f u =

    ),...,,( 321 r k = 2. Determine the Pi terms for the problem:

    3. Prepare a model and experimental conditions such that the Pi terms inthe model match those in the prototype:

    pr k mr k pm pm )()(3322 ;...; ===

    Therefore:m p 11 =

  • 8/12/2019 Me 3560 Presentation Ch Vii

    16/16

    ME3560 Fluid Mechanics

    Chapter VII. Dimensional Analysis,Similitude and Modeling

    Summer I 2014

    16

    By matching the Pi terms between model and prototype, geometric

    similitude, kinematic similitude and dynamic similitude are matched.1TypicallyratioScale >==

    m

    p

    L

    LGeometric Similitude:

    Kinematic Similitude: It is related to velocity, angular velocity,

    acceleration, etc. between the model and the prototypeDynamic Similitude: It is related to fluid properties, forces, moments,pressures, etc.


Recommended