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. Scaling laws: 2 2 For example, ( , , , ) : 5 variables (Re) : 2 variables F F f L V F VL g V L C g ρ µ ρ µ ρ = = = 2 2 If Re Re then where and mean model and prototype, respectively. m p F m F p p p p p m m m m C C m p F V L F V L ρ ρ = =   =     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. ( ) 1 2 2 2 For example, ( , , , ); 5 Number of dimensions: 3 3, 5 3 2 ; Re (Re) F F F f L V n MLT j k n j F VL C C g V L ρ µ ρ µ ρ = = = = Π = = 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 ρ cannot form a pi group. Therefore, j = 3 and k = n j = 2 (two independent dimensionless groups).
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