58 160 Intermediate Mechanics of Fluids Professor Fred Stern Typed by Stephanie Schrader Fall 2006 Chapter 5 1 Chapter 5 Dimensional Analysis and Modeling The Need for Dimensional Analysis Dimensional analysis is a process of formulating fluid mechanics problems in terms of nondimensional variables and parameters 1 Reduction in Variables If F A1 A2 An 0 Then f 1 2 r n 0 Thereby reduces number of experiments and or simulations required to determine f vs F F functional form Ai dimensional variables j nondimensional parameters j Ai i e j consists of nondimensional groupings of Ai s 2 Helps in understanding physics 3 Useful in data analysis and modeling 4 Fundamental to concept of similarity and model testing Enables scaling for different physical dimensions and fluid properties 58 160 Intermediate Mechanics of Fluids Professor Fred Stern Typed by Stephanie Schrader Fall 2006 Chapter 5 2 Dimensions and Equations Basic dimensions F L and t or M L and t F and M related by F Ma MLT 2 Buckingham Theorem In a physical problem including n dimensional variables in which there are m dimensions the variables can be independent nondimensional arranged into r n m parameters r where usually m m F A1 A2 An 0 f 1 2 r 0 Ai s dimensional variables required to formulate problem i 1 n j s nondimensional parameters consisting of groupings of Ai s j 1 r F f represents functional relationships between An s and r s respectively m rank of dimensional matrix m i e number of dimensions usually 58 160 Intermediate Mechanics of Fluids Professor Fred Stern Typed by Stephanie Schrader Fall 2006 Chapter 5 3 Dimensional Analysis Methods for determining i s 1 Functional Relationship Method Identify functional relationships F Ai and f j by first determining Ai s and then evaluating j s a Inspection b Step by step Method c Exponent Method intuition text class 2 Nondimensionalize governing differential equations and initial and boundary conditions Select appropriate quantities for nondimensionalizing the GDE IC and BC e g for M L and t Put GDE IC and BC in nondimensional form Identify j s Exponent Method for Determining j s 1 determine the n essential quantities 2 select m of the A quantities with different dimensions that contain among them the m dimensions and use them as repeating variables together with one of the other A quantities to determine each 58 160 Intermediate Mechanics of Fluids Professor Fred Stern Typed by Stephanie Schrader Fall 2006 Chapter 5 4 For example let A1 A2 and A3 contain M L and t not necessarily in each one but collectively then the j parameters are formed as follows 1 A1x1 A 2y1 A 3z1 A 4 2 A1x 2 A 2y 2 A 3z 2 A 5 n m A1x n m A 2yn m A 3z n m A n Determine exponents such that i s are dimensionless 3 equations and 3 unknowns for each i In these equations the exponents are determined so that each is dimensionless This is accomplished by substituting the dimensions for each of the Ai in the equations and equating the sum of the exponents of M L and t each to zero This produces three equations in three unknowns x y t for each parameter In using the above method the designation of m m as the number of basic dimensions needed to express the n variables dimensionally is not always correct The correct is the rank of the dimensional matrix i e the value for m next smaller square subgroup with a nonzero determinant 58 160 Intermediate Mechanics of Fluids Professor Fred Stern Typed by Stephanie Schrader Fall 2006 Dimensional matrix M L t Chapter 5 5 A1 An a11 a1n a31 o o a3n o o aij exponent of M L or t in Ai n x n matrix Rank of dimensional matrix equals size of next smaller sub group with nonzero determinant Example Hydraulic jump 58 160 Intermediate Mechanics of Fluids Professor Fred Stern Typed by Stephanie Schrader Fall 2006 Chapter 5 6 Say we assume that V1 V1 g y1 y2 or V2 V1y1 y2 Dimensional analysis is a procedure whereby the functional relationship can be expressed in terms of r nondimensional parameters in which r n number of variables Such a reduction is significant since in an experimental or numerical investigation a reduced number of experiments or calculations is extremely beneficial 58 160 Intermediate Mechanics of Fluids Professor Fred Stern Typed by Stephanie Schrader Fall 2006 Chapter 5 7 1 g fixed vary 2 fixed vary g 3 g fixed vary In general or Represents many many experiments F A1 A2 An 0 dimensional form f 1 2 r 0 nondimensional form with reduced of variables 1 1 2 r It can be shown that Fr y V1 Fr 2 gy1 y1 neglect drops out as will be shown thus only need one experiment to determine the functional relationship 1 X Fr x x2 0 0 2 61 1 Fr x 1 x 2 1 2 1 2 5 1 1 7 3 9 For this particular application we can determine the functional relationship through the use of a control volume analysis neglecting and bottom friction 58 160 Intermediate Mechanics of Fluids Professor Fred Stern Typed by Stephanie Schrader Fall 2006 Chapter 5 8 x momentum equation Fx Vx V A y12 y 22 V1 V1 y1 V2 V2 y 2 2 2 2 y1 y 22 V22 y 2 V12 y1 2 g continuity equation V2 V1y1 V2y2 Note each term in equation must have some units principle of dimensional homogeneity i e in this case force per unit width N m V1 y1 y2 y 2 y 1 2 V12 y1 1 1 2 y1 g y2 y12 pressure forces due to gravity inertial forces y2 3 1 y1 y1 now divide equation by gy 2 V12 1 y 2 y 2 1 gy1 2 y1 y1 dimensionless equation ratio of inertia forces gravity forces Froude number 2 note Fr Fr y2 y1 do not need to know both y2 and y1 only ratio to get Fr 58 160 Intermediate Mechanics of Fluids Professor Fred Stern Typed by Stephanie Schrader Fall 2006 Chapter 5 9 Also shows in an experiment it is not necessary to vary y1 y2 V1 and V2 but only Fr and y2 y1 Next can get an estimate of hL from the energy equation along free surface from 1 2 V12 V22 y1 y2 h L 2g 2g hL y 2 y1 3 4 y1 y 2 f due to assumptions made in deriving 1 D steady flow energy equations Exponent method to determine j s for Hydraulic jump use V V1 y1 as repeating variables F g V1 y1 y2 0 M M L L LL 3 2 L LT T T m 3 r n m 3 n 6 Assume m m to avoid evaluating rank of 6 x 6 dimensional matrix 1 Vx1 y1y1 z1 LT 1 x1 L y1 ML 3 z1 ML 1T 1 L x1 y1 3z1 1 0 y1 3z1 1 x1 1 1 …
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