PDF Basics: Fluid Mechanics - KNTU

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Objectives

• dimensional homogeneity

• nondimensionalize the equations

• dimensional analysis, in which the combination of dimensional variables, nondimensional variables, and dimensional constants into nondimensional parameters reduces the number of necessary independent parameters in a problem.

زیلانآ • یداعبا

: بیکرت اهریغتم

و یاهرتماراپ هلأسم

( دعب

،راد یب

،دعب تباوث و

...

) ب ه روظنم

شهاک دادعت

یاهریغتم ریگرد

( لقتسم )

رد هلأسم

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Nondimensionalization of Equations

• divide each term in the equation by a collection of variables and constants whose product has those same dimensions, the equation is rendered nondimensional.

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Nondimensionalization of Equations

• As a simple example:

𝑫𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒂𝒍 𝒓𝒆𝒔𝒖𝒍𝒕: 𝒛 = 𝒛_𝟎 + 𝒘_𝟎𝒕 − 𝟏

𝟐 𝒈𝒕^𝟐 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒎𝒐𝒕𝒊𝒐𝒏: 𝒅^𝟐𝒛

𝒅𝒕^𝟐 = −𝒈

𝒘: 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒊𝒏 𝒛 − 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝑎𝑡 𝑡 = 0: 𝑧 = 𝑧₀ & 𝑤 = 𝑤₀

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• To nondimensionalize equation of motion, we need to select scaling parameters, based on the primary dimensions contained in the original equation.

• In a typical fluid flow problem, the scaling parameters usually include:

• Other parameters and fluid properties such as density, viscosity, and gravitational acceleration enter the problem as well.

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Nondimensionalization of Equations

𝑵𝒐𝒏𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒂𝒍𝒊𝒛𝒆𝒅 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔: 𝒛^∗ = 𝒛

𝒛_𝟎 , 𝒕^∗ = 𝒘_𝟎𝒕 𝒛_𝟎

𝒅^𝟐𝒛

𝒅𝒕^𝟐 = 𝒅 𝒅𝒕

𝒅𝒛

𝒅𝒕 = 𝒅 𝒅𝒕

𝒅 𝒛_𝟎𝒛^∗ 𝒅 𝒕^∗𝒛_𝟎 𝒘_𝟎

= 𝒅 𝒅𝒕

𝒘_𝟎𝒅𝒛^∗

𝒅𝒕^∗ = 𝒘_𝟎 𝒅 𝒅 𝒕^∗𝒛_𝟎

𝒘_𝟎

𝒅𝒛^∗ 𝒅𝒕^∗ = 𝒘_𝟎^𝟐

𝒛_𝟎 𝒅 𝒅𝒕^∗

𝒅𝒛^∗

𝒅𝒕^∗ = 𝒘_𝟎^𝟐 𝒛_𝟎

𝒅^𝟐𝒛^∗ 𝒅𝒕^∗𝟐 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒎𝒐𝒕𝒊𝒐𝒏: 𝒅^𝟐𝒛

𝒅𝒕^𝟐 = −𝒈

𝒅^𝟐𝒛

𝒅𝒕^𝟐 = 𝒘_𝟎^𝟐 𝒛_𝟎

𝒅^𝟐𝒛^∗

𝒅𝒕^∗𝟐 = −𝒈 → 𝒘_𝟎^𝟐 𝒈𝒛_𝟎

𝒅^𝟐𝒛^∗

𝒅𝒕^∗𝟐 = −𝟏

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𝑭𝒓𝒐𝒖𝒅 𝒏𝒖𝒎𝒆𝒃𝒓: 𝑭𝒓 = 𝒘_𝟎 𝒈𝒛_𝟎

𝑵𝒐𝒏𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒂𝒍𝒊𝒛𝒆𝒅 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒎𝒐𝒕𝒊𝒐𝒏: 𝒅^𝟐𝒛^∗

𝒅𝒕^∗𝟐 = − 𝟏 𝑭𝒓 ^𝟐

𝑵𝒐𝒏𝒅𝒊𝒆𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒂𝒍 𝒓𝒆𝒔𝒖𝒍𝒕: 𝒛^∗ = 𝟏 + 𝒕^∗ − 𝟏

𝟐 𝑭𝒓^𝟐 𝒕^∗𝟐 𝑎𝑡 𝑡 = 0: 𝑧 = 𝑧₀ → 𝑧^∗ = 1

& 𝑤 = 𝑤₀ ∴ 𝒘 = 𝒅𝒛

𝒅𝒕 = 𝒅 𝒛_𝟎𝒛^∗ 𝒅 𝒕^∗𝒛_𝟎 𝒘_𝟎

= 𝒘_𝟎𝒅𝒛^∗

𝒅𝒕^∗ = 𝒘_𝟎 → 𝒂𝒕 𝒕 = 𝟎:𝒅𝒛^∗

𝒅𝒕^∗ = 𝟏

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Advantages of Nondimensionalization

Advantages of

Nondimensionalization of an Equation

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Trajectories of a steel ball falling in a vacuum: a) z₀ variations, w₀ fixed; b) w₀ variations, z₀ fixed

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Trajectories of a steel ball falling in a vacuum. Data a & b are nondimensionalized and combined onto one plot.

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Nondimensionalization of Equations

Mass conservation equation:

Navier-Stokes equations:

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Nondimensionalization of Equations

To nondimensionalize these equations, divide all lengths by a reference length, L,

and all velocities by a reference speed, 𝑽_∞, which usually is taken as the freestream velocity.

and the pressure by 𝝆𝑽_∞^𝟐 (twice the freestream dynamic pressure).

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Nondimensionalization of Equations

Mass conservation equation:

Navier-Stokes equations (x-dir):

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Nondimensionalization of Equations

Reynolds number

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Nondimensionalization of Equations

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Dimensional Analysis & Similarity

• There are 3 necessary conditions for complete similarity between a model and a prototype.

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• In a general flow field, complete similarity between a model and prototype is achieved only when there is geometric, kinematic, and dynamic similarity.

• You are already familiar with one 𝚷 “Froude number, Fr”.

𝚷: 𝒅𝒆𝒏𝒐𝒕𝒆 𝒂 𝒏𝒐𝒏𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒂𝒍 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒆𝒓

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Dimensional Analysis & Similarity

• In a general dimensional analysis problem,

• Where k is the total number of 𝚷^′𝒔.

• To ensure complete similarity, the model and prototype must be geometrically similar, and all independent pi groups must match between model and prototype.

1 dependent 𝜫: 𝜫_𝟏

Other 𝜫^′𝒔: 𝜫_𝟐, … 𝒆𝒕𝒄. : 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔

𝜫_𝟐,𝒎 = 𝜫_𝟐,𝒑; 𝜫_𝟑,𝒎 = 𝜫_𝟑,𝒑; … ; 𝜫_𝒌,𝒎 = 𝜫_𝒌,𝒑

𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏𝒔𝒉𝒊𝒑 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝜫^′𝒔: 𝜫_𝟏 = 𝒇(𝜫_𝟐, 𝜫_𝟑, … , 𝜫_𝒌)

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• Under these conditions the dependent pi of the model is guaranteed to also equal the dependent pi of the prototype.

𝒊𝒇 𝜫_𝟐,𝒎 = 𝜫_𝟐,𝒑; 𝜫_𝟑,𝒎 = 𝜫_𝟑,𝒑; … ; 𝜫_𝒌,𝒎 = 𝜫_𝒌,𝒑 𝒕𝒉𝒆𝒏 𝜫_𝟏,𝒎 = 𝜫_𝟏,𝒑

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Dimensional Analysis & Similarity

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• The Reynolds number is the most well known and useful dimensionless parameter in all of fluid mechanics.

• In this problem there is only one independent pi; & it was said if the independent pi match (the Reynolds numbers match); then the independent pi also match.

magnitude of the aerodynamic drag on the car Reynolds number

𝒊𝒇 𝜫_𝟐,𝒎 = 𝜫_𝟐,𝒑 (𝑹𝒆_𝒑 = 𝑹𝒆_𝒎) 𝒕𝒉𝒆𝒏 𝜫_𝟏,𝒎 = 𝜫_𝟏,𝒑

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Dimensional Analysis & Similarity

• This enables engineers to measure the aerodynamic drag on the model car and then use this value to predict the aerodynamic drag on the prototype car.

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Dimensional Analysis & Similarity

• The power of using dimensional analysis and similarity to supplement experimental analysis is: actual values of the dimensional parameters (density, velocity, etc.) are irrelevant. As long as the corresponding independent Pi’s are set equal to each other, similarity is achieved, even if different fluids are used.

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Buckingham Pi Theorem

هیرظن • یاپ

ماهگنیکاب :

کی • هدیدپ یکیزیف

ار رد رظن دیریگب هک

رد نآ رتماراپ هتسباو

یعبات n-1 زا

رتماراپ لقتسم

دشاب .

• 𝒒_𝟏 = 𝒇 𝒒_𝟐, 𝒒_𝟑, … , 𝒒_𝒏 𝒐𝒓 𝒈 𝒒_𝟏, 𝒒_𝟐, 𝒒_𝟑, … , 𝒒_𝒏

یارب • هطبار

یم g n ناوت

رتماراپ هداد

هدش ار

تروصب تبسن n-m

یب دعب لقتسم هورگ

یدنب درک

. نیا تبسن اه

ار یاهرتماراپ یاپ

یم دنیوگ و

تروصب ریز

هتشون یم

نوش د .

• 𝑮 𝚷_𝟏, 𝚷_𝟐, … , 𝚷_𝒏−𝒎 = 𝟎 or

• 𝚷_𝟏 = 𝑮_𝟏 𝚷_𝟐, 𝚷_𝟑, … , 𝚷_𝒏−𝒎 = 𝟎

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• Now: how to generate the nondimensional parameters, i.e., the pi’s

• several methods; the most popular (and simplest) method is the method of repeating variables, popularized by Edgar Buckingham (1867–1940)

6 steps

• As a simple first example, consider a ball falling in a vacuum as discussed before.

Pi’s Variables?

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• List the parameters (dimensional variables, nondimensional variables, and dimensional constants) and count them.

• Let n be the total number of parameters in the problem, including the dependent variable.

Step 1

𝑬𝒙𝒂𝒎𝒑𝒍𝒆: 𝑳𝒊𝒔𝒕 𝒐𝒇 𝒓𝒆𝒍𝒆𝒗𝒂𝒏𝒕 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒆𝒓𝒔:

𝒛 = 𝒇 𝒕, 𝒘_𝟎, 𝒛_𝟎, 𝒈 𝒏 = 𝟓

𝒅𝒕^𝟐 = −𝒈

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Buckingham Pi Theorem

• List the primary dimensions for each of the n parameters.

Step 2

𝒛 𝒕 𝒘_𝟎 𝒛_𝟎 𝒈

𝑳^𝟏 𝒕^𝟏 𝑳^𝟏𝒕^−𝟏 𝑳^𝟏 𝑳^𝟏𝒕^−𝟐

𝑬𝒙𝒂𝒎𝒑𝒍𝒆:

𝒅𝒕^𝟐 = −𝒈

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• Guess the reduction j.

• As a first guess, set j equal to the number of primary dimensions represented in the problem.

• The expected number of Pi’s (k) is equal to n minus j, according to the Buckingham Pi theorem.

• Example: As a first guess, j is set equal to 2, the number of primary dimensions represented in the problem (L and t).

– Reduction: 𝒋 = 𝟐

• If this value of j is correct, the number of Pi’s predicted by the Buckingham Pi theorem is:

Step 3

𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝜫’𝒔: 𝒌 = 𝒏 − 𝒋 = 𝟓 − 𝟐 = 𝟑

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Buckingham Pi Theorem

• Choose j repeating parameters that will be used to construct each Pi.

• Example:

• We need to choose two repeating parameters since j=2.

Step 4

𝑹𝒆𝒑𝒆𝒂𝒕𝒊𝒏𝒈 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒆𝒓𝒔: 𝒘_𝟎 & 𝒛_𝟎

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• Generate the Pi’s one at a time by grouping the j repeating parameters with one of the remaining parameters, forcing the product to be dimensionless.

Step 5

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Buckingham Pi Theorem

𝜫_𝟏 = 𝒛𝒘_𝟎^𝒂^𝟏𝒛_𝟎^𝒃^𝟏

𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒔 𝒐𝒇 𝜫_𝟏: 𝜫_𝟏 = 𝑳^𝟎𝑻^𝟎 = 𝒛𝒘_𝟎^𝒂^𝟏𝒛_𝟎^𝒃^𝟏 = 𝑳^𝟏 𝑳^𝟏𝑻^{−𝟏 𝒂}^𝟏𝑳^𝒃^𝟏 𝑻𝒊𝒎𝒆: 𝒕^𝟎 = 𝒕^−𝒂^𝟏 𝟎 = −𝒂_𝟏 𝒂_𝟏 = 𝟎

𝑳𝒆𝒏𝒈𝒕𝒉: 𝑳^𝟎 = 𝑳^𝟏𝑳^𝒂^𝟏𝑳^𝒃^𝟏 𝟎 = 𝟏 + 𝒂_𝟏 + 𝒃_𝟏 𝒃_𝟏 = −𝟏 𝜫_𝟏 = 𝒛

𝒛_𝟎

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𝜫_𝟐 = 𝒕𝒘_𝟎^𝒂^𝟐𝒛_𝟎^𝒃^𝟐

𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒔 𝒐𝒇 𝜫_𝟐: 𝜫_𝟐 = 𝑳^𝟎𝑻^𝟎 = 𝒕𝒘_𝟎^𝒂^𝟐𝒛_𝟎^𝒃^𝟐 = 𝑻^𝟏 𝑳^𝟏𝑻^{−𝟏 𝒂}^𝟐𝑳^𝒃^𝟐 𝑻𝒊𝒎𝒆: 𝑻^𝟎 = 𝑻^𝟏𝑻^−𝒂^𝟐 𝟎 = 𝟏 − 𝒂_𝟐 𝒂_𝟐 = 𝟏

𝜫_𝟐 = 𝒘_𝟎𝒕 𝒛_𝟎

𝑳𝒆𝒏𝒈𝒕𝒉: 𝑳^𝟎 = 𝑳^𝒂^𝟐𝑳^𝒃^𝟐 𝟎 = 𝒂_𝟐 + 𝒃_𝟐 𝒃_𝟐 = −𝟏

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𝜫_𝟑 = 𝒈𝒘_𝟎^𝒂^𝟑𝒛_𝟎^𝒃^𝟑

𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒔 𝒐𝒇 𝜫_𝟐: 𝜫_𝟑 = 𝑳^𝟎𝑻^𝟎 = 𝒈𝒘_𝟎^𝒂^𝟑𝒛_𝟎^𝒃^𝟑 = 𝑳^𝟏𝑻^−𝟐 𝑳^𝟏𝑻^{−𝟏 𝒂}^𝟑𝑳^𝒃^𝟑 𝑻𝒊𝒎𝒆: 𝑻^𝟎 = 𝑻^−𝟐𝑻^−𝒂^𝟑 𝟎 = −𝟐 − 𝒂_𝟑 𝒂_𝟑 = −𝟐

𝜫_𝟑 = 𝒈𝒛_𝟎 𝒘_𝟎^𝟐

𝑳𝒆𝒏𝒈𝒕𝒉: 𝑳^𝟎 = 𝑳^𝟏𝑳^𝒂^𝟑𝑳^𝒃^𝟑 𝟎 = 𝟏 + 𝒂_𝟑 + 𝒃_𝟑 𝒃_𝟑 = 𝟏

𝜫_{𝟑,𝒎𝒐𝒅𝒊𝒇𝒊𝒆𝒅} = 𝒈𝒛_𝟎 𝒘_𝟎^𝟐

− 𝟏𝟐

= 𝒘_𝟎

𝒈𝒛_𝟎 = 𝑭𝒓

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• Check that all the Pi’s are indeed dimensionless. Write the final functional relationship in the form.

• Example:

Step 6

𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏𝒔𝒉𝒊𝒑 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝜫^′𝒔: 𝜫_𝟏 = 𝒇(𝜫_𝟐, 𝜫_𝟑, … , 𝜫_𝒌) 𝑹𝒆𝒍𝒂𝒕𝒊𝒐𝒏𝒔𝒉𝒊𝒑 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝜫^′𝒔: 𝜫_𝟏 = 𝒇(𝜫_𝟐, 𝜫_𝟑)

𝒛

𝒛_𝟎 = 𝒇(𝒘_𝟎𝒕

𝒛_𝟎 , 𝒘_𝟎 𝒈𝒛_𝟎)

𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒓𝒆𝒔𝒖𝒍𝒕 𝒐𝒇 𝒅𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒂𝒍 𝒂𝒏𝒂𝒍𝒚𝒔𝒊𝒔: 𝒛^∗ = 𝒇(𝒕^∗, 𝑭𝒓)

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• fundamental limitation of dimensional analysis and the method of repeating variables: This method cannot predict the exact mathematical form of the equation.

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Buckingham Pi Theorem

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Guidelines for Repeating parameter

• 1- Never pick the dependent variable. Otherwise, it may appear in all the 𝚷^′𝒔, which is undesirable.

In the present problem we cannot choose z

𝒅𝒕^𝟐 = −𝒈

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Guidelines for Repeating parameter

• 2- The chosen repeating parameters must not by themselves be able to form a dimensionless group. It would be impossible to generate the rest of 𝚷^′𝒔.

We could not, for example, choose 𝒕, 𝒘_𝟎, and 𝒛_𝟎, because (𝒕𝒘_𝟎/𝒛_𝟎) is a 𝚷.

𝒅𝒕^𝟐 = −𝒈

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• 3- The chosen repeating parameters must represent all primary dimensions.

In the present problem we cannot choose L and 𝒘_𝟎

𝒅𝒕^𝟐 = −𝒈

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Guidelines for Repeating parameter

• 4- Never pick parameters that are already dimensionless. These are 𝚷^′𝒔.

For example 𝜽

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• 5- Never pick two parameters with same dimensions or dimensions that differ by only an exponent.

In the present problem we cannot choose z and 𝒛_𝟎. Or in another problem: z[m] and V[m³] (volume)

𝒅𝒕^𝟐 = −𝒈

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Guidelines for Repeating parameter

• 6- If possible: choose dimensional constants over dimensional variables.

In the present problem choose 𝒕_𝟎 instead of 𝒕

𝒅𝒕^𝟐 = −𝒈

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• 7- Pick common parameters

Do not choose 𝝁 or surface tension 𝝈_𝒔

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• 8- Pick simple parameters over complex parameters

It is better to pick parameters with only one or two basic dimensions (e.g., a length, a time, a mass, or a velocity) instead of parameters that are composed of several basic dimensions (e.g., an energy or a pressure).

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Buckingham Pi Theorem

DRAG FORCE ON A SMOOTH SPHERE:

The drag force, F, on a smooth sphere depends on the relative speed, V, the sphere diameter, D, the ﬂuid density, 𝝆, and the ﬂuid viscosity, 𝝁. Obtain a set of dimensionless groups that can be used to correlate experimental data.

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Buckingham Pi Theorem

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Significant dimensionless groups in fluid mechanics

یاهورین • لوادتم

رد نایرج لایس

دنترابع زا

:

یاهورین – یسرنیا

یاهورین – یشان

زا هتیزوکسیو (

تجزل )

یاهورین – یراشف

یورین – هبذاج

یورین – ششک

یحطس

یورین – یشان

زا مکارت یریذپ

• Inertia force )یسرنیا یورین(

– 𝝆𝑽^𝟐𝑳^𝟐: _𝐦^𝐤𝐠_𝟑 ^𝐦_𝐬 ^𝟐 𝐦^𝟐 = ^{𝐤𝐠.𝐦}_𝐬_𝟐 ≈ 𝝆𝑽^𝟐𝑨 = 𝝆𝑽𝑨𝑽 = ሶ𝒎𝑽

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• compare the relative magnitudes of various ﬂuid forces to the inertia force

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• زدلونیر ددع

• 1880s by Osborne Reynolds, the British engineer, who studied the transition between laminar and turbulent ﬂow regimes in a tube.

• رلوا ددع

• In aerodynamic, Leonhard Euler, the Swiss mathematician. Euler is credited with being the ﬁrst to recognize the role of pressure in ﬂuid motion;

≈ 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

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Significant dimensionless groups in fluid mechanics

ددع • نویساتیواک

نویساتیواک • :

لیکشت (bubble/void) بابح

رد لایس و

داجیا یحاون

liquid-free .

رد

رثا تارییغت یناهگان

راشف رد

یحاون مک

راشف خر

یم دهد . بابح اه

هب یحاون اب

راشف رتلااب

هک یم

،دنسر یم

دنکرت و

(shock wave) کوش داجیا

یم دننک .

لماع • یلصا

شیاسرف تسا

.

ندیکرت • بوانتم

بابح اه

هب حطس cyclic stress یزلف

دراو یم دنک .

Inertial cavitation • :

هسورپ یا

تسا هک

رد نآ بابح اه

هب تعرس یم

دنکرت .

Non-inertial cavitation • :

بابح یاراد

رییغت لکش

یناسون اب

لامعا کی

یژرنا

یجراخ (

نادیم کیتسوکآ

)

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Significant dimensionless groups in fluid mechanics

• دورف ددع

• Froud number: William Froude was a British architect. Together with his son, Robert Edmund Froude, he discovered that the parameter.

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• ربو ددع

• Weber number: ratio of inertia to surface tension forces. The value of the Weber number is indicative of the existence of, and frequency of, capillary waves at a free surface.

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Significant dimensionless groups in fluid mechanics

• خام ددع

• Mach number: Austrian physicist Ernst Mach introduced the parameter. Mach number is a key parameter that characterizes compressibility effects in a ﬂow.

• ratio of inertia forces to forces due to compressibility. For truly incompressible ﬂow (and note that under some conditions even liquids are quite compressible), 𝒄 = ∞ so that 𝑴 = 𝟎.

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