1. L. M. LyeDOE Course1 Design and Analysis of Multi-Factored Experiments Dimensional Analysis

2. L. M. LyeDOE Course2 Dimensional Analysis Dimensional analysis is a mathematical method which is of considerable value in problems which occur in engineering. Dimensional analysis is essentially a means of utilizing partial knowledge of a problem when the details are too obscure to permit an exact analysis. Dimensional Analysis is also known as Partial Analysis – problem is partially solve and still require experimentation to obtain a functional relationship.

3. L. M. LyeDOE Course3 Basic Dimensions Variables used in engineering are usually expressed in terms of a limited number of basic dimensions namely: Mass (M), Length (L), Time (T), and sometimes Temperature (  ). E.g. Velocity, V= length/time or LT -1 Force, F = mass x accel (N) = MLT -2 Viscosity,  = N.s/m 2 = ML -1 T -1 Flow rate, Q = m 3 /s = L 3 T -1 Hence most physical quantities we deal with have only 3 basic dimensions: M, L, T.

4. L. M. LyeDOE Course4 Uses of Dimensional Analysis Four main uses of DA: –Checking the consistency of units, making sure that the LHS has the same units as the RHS of an equation. i.e dimensional homogeneity. E.g. E (ML 2 T -2 )= mc 2 (ML 2 T -2 )  apples + apples = apples. –Determining the units of empirical coefficients – e.g. V = C R 1/2 S 1/2 What’s the units of Chezy’s C? –Understanding complex phenomenon by expressing functional relationships in terms of dimensionless parameters to reduce the dimensionality of the problem and to simplify the analysis. This helps to establish the form of an equation relating a number of variables. –To assist in the analysis of experimental results.

5. L. M. LyeDOE Course5 General Ideas If any physical quantity, J, is considered, it will be possible to reduce it to some function of the three fundamental dimensions, M, L, and T. i.e. J = f [M, L, T] If the magnitude of J is compared for two similar systems, then: J”/J’ = f [M, L, T]”/ f [M, L, T]’= J Evidently, this ratio must be dimensionless. This is true if the function is in the form of a product: J = K [M a L b T c ]

6. L. M. LyeDOE Course6 General ideas (continued) Where K is a numeric value, and a, b, and c are powers or indices who magnitudes have to be determined. E.g. J = velocity, then K=1, a=0, b=1, c= -1, since velocity has dimensions LT -1. It can be argued that dimensional relationships are arbitrary, since the magnitudes depend on the choice of units (feet, metres, pounds, kilograms, etc.) For this reason, an equation which is a statement of a physical law is often used in a dimensionless form. Dimensionless equations are completely general, and are therefore frequently the basis for representation of experimental data.

7. L. M. LyeDOE Course7 Methods of Dimensional Analysis There are many methods of dimensional analysis. The techniques have been progressively refined over the years. Rayleigh’s Indicial method – oldest method Buckingham  theorem – most famous method Hunsaker and Rightmire’s method – quick method Matrix Method – modern method using a computer for matrix inversion. All methods are absolutely dependent on the correct identification of all the factors which govern the physical events being analyzed.

8. L. M. LyeDOE Course8 The omission of a single factor may give quite misleading results. Hence, dimensional analysis is not unlike DOE. Choice of factors that influence the response must be carefully chosen. DOE can be used for factors that cannot be expressed in physical quantities e.g. categorical variables (colours, gender, material types, etc.) However, DOE can be combined with DA to obtain functional relationships in an efficient way. More on this later.

9. L. M. LyeDOE Course9 Rayleigh’s method Consider a problem involving a scale model test of a hydraulic machine. The thrust force F, velocity v, viscosity  and density  of the fluid is given including a typical size of the system, L, is also given. Two questions must be posed, namely: –How to analyze or plot the data in the most informative way, and –How to relate the performance of the model to that of the working prototype.

10. L. M. LyeDOE Course10 Solution Let’s postulate that the force F is related to the other given quantities: [1] F = f [ , v, , L] The form of the function is completely unknown, but it has been proposed earlier that: –The function must be in the form of a power product –There must be a dimensional balance between both sides of the equation

11. L. M. LyeDOE Course11 From [1], the equation maybe rewritten as: [2] F = K [  a v b  c, L d ] So obviously, for dimensional homogeneity, the dimensions on the LHS must equal those on the RHS. Expressing each quantity in [2] in terms of its dimensions, MLT -2 = K[ (ML -3 ) a (LT -1 ) b (ML -1 T -1 ) c L d ] Equating the indices for M, L, and T, M: 1 = a + c L: 1 = -3a + b - c + d T: -2 = -b - c or 2 = b + c

12. L. M. LyeDOE Course12 Thus we have 3 equations but 4 unknowns, so a complete solution is not attainable. We can only obtain a partial solution. Let’s express all indices in terms of say c. a = 1 – c b = 2 – c d = 2 – c Substituting for a, b, and d in [2]: F = K [  1-c v 2-c  c L 2-c ]

13. L. M. LyeDOE Course13 Or, Since the function represents a product, it may be restated as:

14. L. M. LyeDOE Course14 Or as: Where: K and c are unknown and must be obtained by experimentation. Key points to note from the above equation: –Two groups have emerged from the analysis. If you check, it will be found that both groups are dimensionless. –Dimensionless groups are independent of units and of scale and are therefore equally applicable to the model or to the prototype

15. L. M. LyeDOE Course15 –Both groups represent ratios of forces: thrust force/inertial force, and inertial force/viscous force. –All three fundamental dimensions are present. Therefore, if the model is to truly represent the prototype, then both model and prototype must conform to the law of dynamic similarity, i.e. the magnitude of each dimensionless group must be the same for the model as for the prototype. –The dimensionless groupings of variables are not unique. Different dimensionless groups can emerge. –The Raleigh indicial method is okay as long as the number of variables is small.

16. L. M. LyeDOE Course16 Another example The velocity of propagation of a pressure wave through a liquid can be expected to depend on the elasticity of the liquid represented by the bulk modulus K, and its mass density r. Establish by D. A. the form of the possible relationship. Assume: u = C K a  b U = velocity = L T -1,  = M L -3, K = M L -1 T -2 L T -1 = M a L -a T -2a x M b L -3b M: 0 = a + b L: 1 = -a – 3b T: -1 = - 2a

17. L. M. LyeDOE Course17 Therefore: a = ½, b = -½, and a possible equation is: Rayleigh’s method is not always so straightforward. Consider the situation of flow over a U-notched weir. Q = f( , , H, g) [Q] = [C  a  b H c g d ] [ ] => dimensions of Using the M, L, T system, [L 3 T -1 ] = [ML -3 ] a [M L -1 T -1 ] b [L] c [L T -2 ] d M: 0 = a + b(1) L: 3 = -3a –b +c + d (2) T: -1 = - b – 2d (3)

18. L. M. LyeDOE Course18 We have only 3 equations, but there are 4 unknowns. Need to express a, b, c, in terms of d. b = 1 – 2d a = -b = 2d -1 c = 3 + 3a + b – d = 1 + 3d Q = C  (2d-1)  (1-2d) H (1+3d) g (d) or

19. L. M. LyeDOE Course19 Buckingham  theorem (1915) This is perhaps the mother of all dimensional analysis methods. Many other methods were built upon this method. Buckingham proposed that: –If a physical phenomenon was a function of m quantities and n fundamental dimensions, dimensional analysis would produce (m-n)  groups; –Each  should be a function of n governing variables plus one more quantity; –The governing quantities must include all fundamental dimensions; –The governing quantities must not combine among themselves to form a dimensionless group; –As each  is dimensionless, the final function must be dimensionless, and therefore dimensionally: – f [  1,  2, …,  m-n ] = M 0 L 0 T 0

20. L. M. LyeDOE Course20 Example 1 Using the previous problem, there were 5 quantities (F, , v, , L) and 3 dimensions (M, L, T), from which we can derive 2 groups. First choose 3 governing variables that together must contain M, L, and T. Let’s choose: , v, L. Combine them with one other variable to get  1 :  1 =  a v b L c F = M 0 L 0 T 0 Therefore: (ML -3 ) a (LT -1 ) b L c (MLT -2 ) = M 0 L 0 T 0

21. L. M. LyeDOE Course21 Example 1 (continued) Equating indices: M: a+1 = 0, therefore a = -1 T: -b – 2 = 0, therefore b = -2 L: -3a + b + c + 1 = 0 Substituting for a and b, c = -2. So,  1 = F/  v 2 L 2 Repeat the process with the other variable:  2 =  a v b L c  = M 0 L 0 T 0 =  /  vL, as before

22. L. M. LyeDOE Course22 Example 2 The head loss per unit length (  h/L) of pipe in turbulent flow through a smooth pipe depends on v, D, , g, and . Determine the general form of the equation. F(  h/L, v, D, , , g) = 0; 6 variables, 3 dimensions, 3  terms. That is:  1 = f(  2,  3 ). Choose 3 repeating variables: v, D, and 

23. L. M. LyeDOE Course23 M: z 1 + 1 = 0 L: x 1 + y 1 – 3z 1 – 1 = 0 T: -x 1 – 1 = 0 Therefore: x 1 = -1, z 1 = -1, and y 1 = -1. or

24. L. M. LyeDOE Course24 Final solution: Where Re = Reynold’s number, and Fr = Froude number =

25. L. M. LyeDOE Course25 Rightmire and Hunsaker’s method (1947) This method is easier to use and quicker than Buckingham’s method. The method is similar the Buckingham’s method by use of repeating variables, but express them in term of the variables themselves. e.g. D to represent Length, [L] = D v to represent Time, [T] = L/v = D/v  to represent Mass, [M] =  L 3 =  D 3

26. L. M. LyeDOE Course26 Example  1 : g = L T -2 = v 2 D D -2 = v 2 D -1, therefore  1 =  2 :  = M L -1 T -1 = Therefore:  2 =  3 =, as before.

27. L. M. LyeDOE Course27 Another example A spherical drop of liquid of diameter D oscillates under the influence of its surface tension. Investigate the frequency of oscillation f. F(f, , D,  ) = 0 Answer:

28. L. M. LyeDOE Course28 Matrix method Best for problems with many variables. Can be solved using a matrix inversion routine. Consider a problem with 7 variables in 4 dimensions (M, L, T,  ) First form the dimensional matrix: a b c de f g MLTMLT AB

29. L. M. LyeDOE Course29 We need to transform the above matrix to: I = A -1 A D = A -1 B A -1 = adj A/ |A| Whatever operation was done to get the unit matrix on the left must be also done on the right to get D. a b c d e f g abcdabcd 10 0 0 0 1 0 0 0 0 1 0 0 0 0 1

30. L. M. LyeDOE Course30 Consider the earlier problem of the head loss per unit length. F(  h/L, v, g, D, ,  ) = 0  v D g   h/L MLTMLT 1 0 0 0 1 0 -3 1 1 1 -1 0 0 -1 0 -2 -1 0 AB Inverse of A can be obtained by Matlab, Minitab or by hand

31. L. M. LyeDOE Course31 D = The three  terms are: The matrix method can obtain the  terms all in one go instead of one term at a time.

32. L. M. LyeDOE Course32 Comments about Dimensional Analysis Most important but most difficult problem in applying DA to any problem is the selection of the variables involved. There is no easy way of identifying the correct variables without specialized knowledge about the phenomenon being investigated. If you select too many variables, you get too many  terms and may require much additional experimentation to eliminate them. If important variables are omitted, then an incorrect result will be obtained, which may prove to be costly and difficult to ascertain.

33. L. M. LyeDOE Course33 Type of variables: Geometry – length, angles, diameter, or area. Material properties – , , elasticity, etc. External effects – any variable that tends to produce a change in the system e.g. forces, pressures, velocities, gravity, etc. You need to keep the number of variables to a minimum, and that they are independent. E.g. D and Area need not be included together because one is derived from the other. Therefore, heavy thinking is required in variable selection  similar to DOE.

34. L. M. LyeDOE Course34 Points to consider in the selection of variables: Clearly define the problem. What is the main response variable of interest? That is, what is Y? Consider the basic laws that govern the phenomenon. Even a crude theory may be useful. Start the selection process by grouping the variables in the 3 broad classes: geometry, material properties, and external effects. Consider other variables that may not fall into one of the three categories, e.g. time, temperature, colour, equipment, etc.

35. L. M. LyeDOE Course35 Be sure to include all quantities that enter the problem even though some of them may be held constant e.g. g. For D.A. it is the dimensions of the quantities that are important – not specific values. Make sure that all variables are independent – look for relationships among subsets of the variables (same as DOE). Remember that after a dimensional analysis, you still need to carry out the experiment to relate the dimensionless groups. Hence DOE may be needed unless you have only 1 or 2  terms. Remember the first lecture?

36. L. M. LyeDOE Course36 More on  terms Specific  terms obtained depend on the somewhat arbitrary selection of repeating variables. For example, if we choose:  , D, g instead of , D, v, we would end up with a different set of  terms. Both results would be correct, and both would lead to the same final equation for  h/L, however, the function relating the different  terms would be different.

37. L. M. LyeDOE Course37 Hence, there is not a unique set of  terms which arises from a dimensional analysis. However the required number of  terms is fixed, and once a correct set is determined all other possible sets can be developed from this set by combination of products of the original set. e.g. if we have problem involving 3  terms,  1 = f(  2,  3 ) we can combine the  terms to give a new  term:

38. L. M. LyeDOE Course38 Then the relationship could be expressed as: All these would be correct; however, the required number of  terms cannot be reduced by this manipulation; only the form of the  terms is altered. or even as

39. L. M. LyeDOE Course39 Which form of  terms is best? There is no simple answer. Best to keep it simple. Some  terms are well-known dimensionless numbers like Reynolds, Froude, Mach, Weber, Cauchy, Euler, etc. In pipe flow problems, the Reynold’s number is prominent while in open channel and ocean engineering problems, the Froude number is more relevant. So it all depends on the field of investigation.

40. L. M. LyeDOE Course40 Correlation of Experimental Data One of the most important uses of dimensional analysis is an aid in the efficient handling, interpretation, and correlation of experimental data. As noted earlier, DA cannot provide a complete answer and a specific relationship among the  terms cannot be determined without experimentation. The degree of difficulty obviously depends on the number of  terms.

41. L. M. LyeDOE Course41 Problems with: 1  term   = C where C = a constant 2  terms   1 =   )  simple regression problem > 2  terms   1 =  (  2,  3 )  multiple regression problem With more and more  terms, a DOE approach may be needed and may require the use of RSM if relationship is nonlinear.

42. L. M. LyeDOE Course42 Example The pressure drop per unit length,  p/L for the flow of blood through a horizontal small diameter tube is a function of flow rate Q, diameter D, and the blood viscosity . For a series of test with D = 2 mm and  = 0.004 Ns/m 2, the following data were obtained for  p measured over a length of 300 mm. Q (m 3 /s): 3.6 4.9 6.3 7.9 9.8 (x 10 -6 )  p (N/m 2 ): 1.1 1.5 1.9 2.4 3.0 (x 10 4 ) Perform a DA for this problem and make use of the data to determine a general relationship between  p and Q, one that is valid for other values of D, L and .

43. L. M. LyeDOE Course43 Solution 4 variables, F (  p/L, D, Q,  ) = 0, i.e. 4 – 3 = 1  term From D.A. (try this yourself), we get: Substituting the values used in the experiment, = constant. Try this yourself =

44. L. M. LyeDOE Course44 Using the data obtained from the experiment, Average for constant C = 40.5, hence: = [ 40.6, 40.7, 40.1 40.4 40.7]

45. L. M. LyeDOE Course45 Example 2 A liquid flows with a velocity v through a hole in the side of a tank. Assume that v = f(h, g, ,  ). Where h is the height of water above the hole,  is the density of the fluid, and  the surface tension. The density  is 1000 kg/m 3, and  = 0.074 N/m. The data obtained by changing h and measuring v are: V (m/s) 3.13 4.43 5.42 6.25 7.00 h (m) 0.50 1.00 1.50 2.00 2.50 Plot the data using appropriate dimensional variables. Could any of the original variables have been omitted?

46. L. M. LyeDOE Course46 Solution 5 variables, F(v, h, g, ,  ) = 0, 2  terms. From dimensional analysis,  gh 2 /  3.31 13.3 28.8 53.0 82.9  v/(gh) 1/2 : 1.41 1.41 1.41 1.41 1.41  Plotting the data will show that is  Independent of which means that  and  can be omitted. Of course this is well-known if one were to apply the Bernoulli equation to solve the problem.

47. L. M. LyeDOE Course47 References Chapter 2, 3, 4 notes from course website. Thomas Szirtes (1998): Applied Dimensional Analysis and Modeling, McGraw Hill, 790 pages. Most books in physics and fluid mechanics. See by Islam and Lye (2007) on combined use of DA and DOE.