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# Fenômentos de Transporte

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rapidly) are the velocity, a+, of the fluid at shutoff, the density of the fluid, p, and the bulk modulus of the fluid, p = p(dp/dp)~ Find a complete set of dimensionless groups for this problem. Solution The dimensions of the bulk modulus are (5.2.2-6) Chapter 5: Application of Dimensional Analysis pmu p (v) p 233 1 -1 -2 0 1 -1 1 - 3 0 1 -1 -2 The dimensional matrix in an MLt system is M L t (5.2.2-7) We have four rows that can be combined in various ways to make 3x3 determinants. From the properties of determinants, the only effect on a determinant of switching rows is to change the sign. Hence, only combinations of rows, not permutations of rows, are significant. We can therefore f ~ n n ~ ~ (5.2.2-8) different 3rd-order determinants from our dimensional matrix. We must see if any of these are not equal to zero. From the properties of determinants we know that if any two rows of a determinant are identical, the determinant is zero, so it immediately follows that any determinant that contains both the first and fourth rows of our dimensional matrix will be zero. This eliminates two of our four combinations - the first and fourth rows combined with 1) the second row and 2) the third row. This leaves us with two determinants to investigate further. One contains the first row in combination witb the second and third rows, and the other contains the fourth row in combination with the second and third rows. However, since the first and fourth rows are identical, these last two determinants are identical, leaving only a single determinant to investigate l 9 The notation for combinations is read \u201cn choose r,\u201d and the general formula in terms of permutations of r things from a set of n is = P: = n! r! r! (n - r ) Fraser, D. A. S. (1958). Statistics: An Introduction. New York, NY, John Wiley and Sons, Inc. 234 Chapter 5: Application of Dimensional Analysis and so we have a matrix of rank less than three since we have shown all possible third-order determinants to be zem. We proceed to check second-order determinants. It is easy to find by inspection a 2x2 determinant that is not equal to zero, e.g. [ ;-;I = 1 (5.2.2- 10) so the rank of our dimensional matrix is two, and for this example, the rank of the dimensional matrix, 2, is not equal to the number of fundamental dimensions, 3. We will obtain, therefore, (4 - 2) = 2 dimensionless groups, not ( 4 - 3 ) = 1. An acceptable set can be seen to be (5.2.2-1 1) 5.3 Systematic Analysis of Variables The proof of Buckingham\u2019s theorem is based on a set of algebraic theorems concerned with the class of functions known as dimensional ly homogeneous.2o A systematic approach to finding a set of independent dimensionless products can be based on this algebraic theory. We now develop a systematic way to convert a functional relationship among n independent quantities Q1, Q2, ... Qn to one involving (n-r) dimensionless products lcl, 1c2, ~ 3 , ... z ~ - ~ , where r is the rank of the dimensional matrix of the Q\u2019s *O Brand, L. (1957). The Pi Theorem of Dimensional Analysis. Arch. Rd. Mech. A n d . 1: 35. Chapter 5: Application of Dimensional Analysis 235 In general, the number of dimensionless products in a complete set is equal to the total number of variables minus the maximum number of these variables that will not form a dimensionless product (the rank of the dimensional matrix). The algorithm to find tbe R'S as follows: 1) Select the number of fundamental dimensions. 2) List the Q variables and their dimensions in terms of the fundamental dimensions. 3) Select a subset of the Q variables equal in number to the number of fundamental dimensions such that none of the selected quantities is dimensionless, the set includes all the fundamental dimensions and no two variables of the subset have the same dimensions. 4) The dimensionless products (p's) are found one at a time by forming the product of the subset variables each raised to an unknown power times one of the remaining Q variables to a known power. (This process is repeated n-k times, using all the remaining Q variables.) 5 ) Apply dimensional homogeneity to each of the products obtained to determine the unknown exponents. Algebraic manipulation of the dimensionless products that does not change their number will not destroy the completeness of the set. Thus, a II may be replaced by any power of itself, by its product of any other ~t raised to any power, or by its product with a numerical constant. d e 5.3-1 Drqg force on a S D ~ The drag force, F, on a smooth sphere of diameter D suspended in a stream of flowing fluid depends on the freestream velocity, v,, the fluid density, p, and the fluid viscosity, p. (5.3-2) 236 Chapter 5: Application of Dimensional Analysis Using the FLt system, (a) Write the dimensional matrix. (b) Show that the rank of the dimensional matrix is 3. (c) Using (I?, D, p) as the fundamental set, determine the dimensionless group incorporating v,. Solution a) The dimensional matrix in the indicated system is b) We can easily fmd a non-zero third-order determinant, e.g., 1 0 : :-:I 1 1 = (l)[(l)(-1) -(1)(0)] = - 1 (5.3 -3) (5.3-4) c) Choosing Q1= F, Q2 = D, and Q3 = p as the subset, (n - k) = (6 - 3) = 3 F a + c = O 3 a = - 1 L: l + b - 2 ~ = 0 3 b = l t: - l + c = O 3 c = l (5.3-5) Example 5.3-2 Dimensionless prouus for -flow over a -flat ulate For flow over a flat plate, the pertinent variables in predicting the velocity in the xdirection are Chapter 5: Application of Dimensional Analysis v, = f (x, y, v,, p, p) (5.3-6) Using the MLt system (a) Write tbe dimensional matrix. (b) Show that the rank of the dimensional matrix is 3. (c) Using (x, p, p) as the fundamental set, determine the dimensionless group incorporating v,. Solution "x x Y V - P P M I O 0 0 0 1 1 L 1 1 1 1 - 3 - 1 4-1 0 0-1 0 - 1 0 1 1 1 - 3 - 1 - 1 0 -1 (5.3-7) = 0 - 1 - 3 = - 4 + 0 * r a n k = 3 (5.3-8) c) Choosing Q1= x, Q2 = p, and Q3 = p as the subset 7c = v, xa p b p c [4] [LIa [3Ib [B]' M: b + c = O =+ a = l L: l + a - 3 b - c = 0 * b = l t: - 1 - c = o * c = - 1 237 (5.3-9) 238 Chapter 5: Application of Dimensional Analysis W(Lt2) L L L/t M/(Lt) M/L3 L U D l e 5.3-3 Consistencv 0 f d imension less pro U D s across pvstem of di- F/L2 F/L2 F/(Ft2/M)2 F/L2 = M2/(Ft4) L L Ft2/M L I L L Ft2/M L L J 0 ° * 5 L/t Ft2/(Mt) L/t - - (LF/M)O.~ = Fth4 M / [ L 0 ° * 5 J (Ft2/L)/(Lt) M/[(Ft2/M)t M/(Lt) M/L3 (Ft2/L)/L3 M/(Ft2/M)3 M/L3 1 = M2/(Ft3) - - ( ~ ~ ) 0 . 5 ~ - 1 . 5 = F a 2 = Ft2/L4 = M4/(F3t6) L L Ft2/M L ML/(Ft2) The following table shows a list of possible variables for flow in a pipe. Apply the algorithm for obtaining dimensionless groups in each of the five fundamental systems of units MLt, FML, FLt, FMt, and FMLt, and show that the results obtained are equivalent. Details of the application of the algorithm for the FML, FLt, and FMt systems are omitted and left as an exercise in Problem 5.20. The rank of the dimensional matrix may be shown in each case to be equal to the number of fundamental dimensions. The algorithm presented will work for systems of four fundamental dimensions as well as three if the dimensional conversion factor & is included in the set of Q\u2019s purely as a formal procedure. For the MLt system (5.3- 10) 21 From Newton\u2019s law, F = M U 2 . 22 From Newton\u2019s law, t = (ML/F)0*5. 23 From Newton\u2019s law, M = Ft2/L. 24 From Newton\u2019s law, L = Ft2/M. Chapter 5: Application of Dimensional Analysis 239