Lecture_Note_Ch_5

Lecture_Note_Ch_5 - ME 342 Fluid Mechanics Lecture Note on...

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Lecture Note on Ch. 5: Dimensional Analysis and Similarity Prof. Chang-Hwan Choi Stevens Institute of Technology Department of Mechanical Engineering ME 342 Fluid Mechanics Spring 2008
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Motivation
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Introduction Dimensional analysis: – A method for reducing the number and complexity of experimental variables that affect a given physical phenomenon, by using a sort of compacting technique. •S c a l i n g l a w s : 22 For example, ( , , ): 5 variables (Re) : 2 variables F Ff L V FV L g VL Cg ρµ ρ µ =  ⇒=   If Re Re then where and mean model and prototype, respectively. mp F m F p pp p p mm m m CC L L == A full-scale, 10-meter- diameter windmill
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The Principle of Dimensional Homogeneity If an equation truly expresses a proper relationship between variables in a physical process, it will be dimensionally homogeneous ; that is, each of its additive terms will have the same dimensions. In fluid mechanics, the four basic dimensions are usually taken to be mass M , length L , time T , and temperature Θ , or an MLT Θ system for short. Dimensions of fluid- mechanics properties
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The Pi Theorem One of the methods of reducing a number of dimensional variables into a smaller number of dimensionless groups. 1. If a physical process satisfies the principles of dimensional homogeneity and involves n dimensional variables, it can be reduced to a relation between only k dimensionless variables or Π s (pi or Π meaning a product of variables). The reduction j = n k equals the maximum number of variables that do not form a pi among themselves and is always less than or equal to the number of dimensions describing the variables. 2. Find the reduction j , then select j scaling (or repeating) variables that do not form a pi among themselves. Each desired pi group will be a power product of these j variables plus one additional variable, which is assigned any convenient nonzero exponent. Each pi group thus found is independent. () 12 22 For example, ( , , ); 5 Number of dimensions: 3 3, 5 3 2 ; Re (Re) FF Ff L V n MLT j k n j FV L CC g VL ρµ ρ µ = = ⇒≤ =−≥ = Π= = = =
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Example Problem (Example 5.2 at page 298-300) Number of variables: n =5 Dimensions of each variable: F ={ MLT -2 }, L ={ L }, V ={ LT -1 }, ρ ={ ML -3 }, µ ={ ML -1 T -1 } Find j and select repeating j variables: j is less than or equal to 3 ( MLT ). By careful inspection of the dimensions of each variable, we find that L , V , and cannot form a pi group. Therefore, j = 3 and k = n j = 2 (two independent dimensionless groups).
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Lecture_Note_Ch_5 - ME 342 Fluid Mechanics Lecture Note on...

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