FUNDAMENTALS OF FLUID MECHANICS Chapter 7 Dimensional Analysis Modeling, and Similitude 1 MAIN TOPICS Dimens ional A nal ysi s Buckingham Pi Theorem Determination of Pi Terms Comments about Dimensional Analysis Common Dimensionless Groups in Fluid Mechanics Correlation of Experimental Data Modeling and Similitude Typical Model Studies Similitude Based on Governing Differential Equation 2 Dimensional Analysis 1/4 A typical fluid mechanics problem in which experimentation is required consider the steady flow of an incompressible Newtonian fluid through a long, smoothsmooth-- walled, horizontal, circular pipe. An important characteristic of this system, which would be interest to an engineer designing a pipeline, is the pressure drop per unit length that develops along the pipe as a result of friction. 3 Dimensional Analysis 2/4 The first step in the planning of an experiment to study this problem would be to decide the factors, or variables, that will have an effect on the pressure drop.. Pressure droppp per unit leng th p f (D, , ,V ) Pressure drop per unit length depends on FOUR variables: sphere size (D); speed (V); fluid density ( ρ); fluid viscosity (m) 4 Dimensional Analysis 3/4 To perform the experiments in a meaningful and systematic manner, it would be necessary to change the variable, such as the velocity, which holding all other constant, and measure the corresponding pressure drop. Difficulty to determine the functional relationship between the pressure drop and the various facts that influence it. 5 Series of Tests 6 Dimensional Analysis 4/4 Fortunately, there is a much simpler approach to the problem that will eliminate the difficulties described above. Collecting these variables into two nonnon--dimensionaldimensional combinations of the variables (called dimensionless product or dimensionless groups) Only one dependent and one Dp VD independent variable 2 Easy to set up experiments to V determine dependency Easy to present results (one graph) 7 Plot of Pressure Drop Data Using … 3 Dp L(F/ L ) 0 0 0 F L T V2 (FL4T2 )(FT1)2 4 2 1 VD (FL T )(LT )(L) 0 0 0 F L T (FL2T) dimensionless product or dimensionless groups 8 Buckingham Pi Theorem 1/5 A fundamental question we must answer is how many dimensionless products are required to replace the original list of variables ? The answer to this question is supplied by the basic theorem of dimensional analysis that states 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. Pi terms Buckingham Pi Theorem 9 Buckingham Pi Theorem 2/5 Given a physical problem in which the dependent variable is a function of kk--11 independent variables.. u1 f (u 2 , u 3 ,....., u k ) Mathematically, we can express the functional relationship in the equivalent form g(u1, u 2 , u 3 ,,,....., u k ) 0 wwherehere g is an unspecified function, different from f. 10 Buckingham Pi Theorem 3/5 The Buckingham Pi theorem states that: Given a relation among k variables of the form g(u1, u 2 , u 3 ,....., u k ) 0 The kiblk variables may bditbe grouped into k--rr idindepend ent dimensionless products, or Π terms, expressible in ftilfbfunctional form by 1 ( 2 , 3 ,,,, kr ) r ?? Π?? or (1, 2 , 3 ,,,, kr ) 0 11 Buckingham Pi Theorem 4/5 The numbfber of r illblis usually, but not always, equal lh to the minimum number of independent dimensions required to specify the dimensions of all the parameters . Usually the reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F , L, and T .. ThThisis theorem does not predict the particular functional form of or . The functional relation among the independent dimensionless products Π must be determined experimentally.. The k--rr dimensionless Π terms obtained from the procedure are independent. 12 Buckingham Pi Theorem 5/5 A Π term is not independent if it can be obtained from a product or quotient of the other dimensionless products of the problem. For example, if 2 3 / 4 1 or 1 5 6 2 2 3 3 then neither Π5 nor Π6 is independent of the other dimensionless products or dimensionless groups.. 13 Determination of Pi Terms 1/12 Several methods can be used to form the dimensionless products, or pi term, that arise in a dimensional analysis. Regardless of the method to be used to determine the dimensionless products, one beggyins by listin gall ((dimensionaldimensional)) variables that are known (or believed) to affect the given flow phenomenon.. Eight steps listed below outline a recommended procedure for determining the Π terms. 14 Determination of Pi Terms 2/12 Step 1 List all the variables.. 1 List all the dimensional variables involved. Keep the number of variables to a minimum, so that we can minimize the amount of laboratory work . All variables must be independent. For example, if the cross-sectional area of a pipe is an important variable , either the area or the pipe diameter could be used, but not both, since they are obviously not independent.. Ex. γ=ρ×g, that is, γ,, ρ, and g are not independent. 15 Determination of Pi Terms 3/12 Step 1 List all the variables. 2 Let k be the number of variablesvariables.. Example: For pressure drop per unit length, k=5. (All variables are p ,, D,,, , and V)V ) p f (D,,,V) 16 Determination of Pi Terms 4/12 Step 2 Express each of the variables in terms of basic dimensionsand find the number of reference dimensions.. Select a set of fundamental (primary) dimensions. For example: MLT, or FLT. Example: For pressure drop per unit length , we choose FLT. p FL3 D L FL4T2 2 1 r=3 FL T V LT 17 Determination of Pi Terms 5/12 Step 3 Determine the required number of pi terms. Let k be the number of variables in the problem. Let r be the number of reference dimensions (primary dimensions) required to describe these variables . The number of pi terms is kk--rr Exampldilhk3le: For pressure drop per unit length k=5, r = 3, the number of pi terms is kk--r=5r=5--3=23=2.. 18 Determination of Pi Terms 6/12 Step 4 Selbfiibllect a number of repeating variables, where the number required is equal to the number of reference dimensions. Select a set of r dimensional variables that includes all the primary dimensions (repeating variables). These repeating variables will all be combined with each of the remaining parameters. Example: For pressure drop per unit length ( r = 3) select ρ , V, D. p FL3 D L FL4T2 FL2T V LT1 19 Determination of Pi Terms 7/12 Step 5 Form a pibi term by mulililtiplying one of fh the nonrepeating variables by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless. 1 Set up dimensional equations, combining the variables selected in Step 4 with each of the other variables (nonrepeating variables) in turn, to form dimensionless groups or dimensionless product. There will be k – r equations. Example: For pressure drop per unit length 20 Determination of Pi Terms 8/12 Step 5 (Continued) 2 p DaVbc 1 3 a 1 b 4 2 c 0 0 0 (FL )(L) (LT ) (FL T ) F L T F :1 c 0 L : 3 a b 4c 0 p D T : b 2c 0 1 a 1,b 2,c 1 V2 21 Determination of Pi Terms 9/12 Step 6 Repeat Step 5 for each of the remaining nonnon--rereppgeating variables. a b c 2 D V 2 a 1 b 4 2 c 0 0 0 (FL T)(L) (LT ) (FL T ) F L T F :1 c 0 L : 2 a b 4c 0 T :1 b 2c 0 2 a 1,b 1,c 1 DV 22 Determination of Pi Terms 10/12 Step 7 Check all the resulting pi terms to make sure they are dimensionless. Check to see that each group obtained is dimensionless. Example: For pressure drop per unit length . p D 0 0 0 0 0 0 1 F L T M L T V2 0 0 0 0 0 0 2 F L T M L T DV 23 Determination of Pi Terms 11/12 Step 8 Express the final form as a relationship amonggp the pi terms, and think about what is means. Express the result of the dimensional analysis. 1 ( 2 , 3 ,,,, kr ) Examppple: For pressure drop ppg per unit length . p D Dimensional analysis will not provide the form of the function . The function 2 V DV can only be obtained from a suitable set of experimentseriments.. 24 Determination of Pi Terms 12/12 The pi terms can be rearranged. For example, Π2, could be expressed as VD 2 p D VD V2 25 Example 7.1 Method of Repeating Variables AhiA thin rectangu lar pl ate h avi ng a wid idhth w and a h hihhieight h is located so that it is normal to a moving stream of fluid. Assume that the drag, D, that the fluid exerts on the plate is a function of w and h, the fluid viscosity, µ ,and ρ, respectively, and the velocity , V , of the fluid approaching the plate. Determine a suitable set of pi terms to study this pppyroblem experimentally. 26 Example 7. 1 Solution1/5 Drag force on a PLATE D f (w, h,,, V) Step 1:List all the dimensional variables involved. DhD,w,h, ρ,μ,V k6k=6 dimensi onal paramet ers.