19 CBE341 Lecture 3 (9/17) - Dimensional analysis The development of fluid mechanics has depended heavily on experimental results, as most flow problems of practical interest cannot be solved exactly by analytical methods. As experimental work is both time-consuming and expensive, it is natural to examine if one can gather the desired information from as few experiments as possible, while working with a suitably scaled model of the geometry of interest. Dimensional analysis is an important tool to achieve this goal. It also exposes dimensionless groups (or parameters) which can be used to correlate the experimental data, which allows us to use the correlations across geometric scales. Dimensional analysis also allows us to examine the exact equations of motion critically. By recasting these equations in a dimensionless form, one can recognize the relative importance of various terms in the model equations. This may allow rational simplifications of the exact equations. These simplified equations may be solved more easily, and sometimes even analytically. (As we have not discussed the equations of motion, you have to simply take my word for it at the present time.) The dimensional analysis can also suggest when one may be able to do away with the differential equation model for the flow, and “accept” simple macroscopic balance solutions. Nature of Dimensional Analysis Let us first consider the problem of designing suitably scaled experimental systems and correlating the data. Dimensional analysis performed for these specific goals does not require as a starting point the differential equations describing the evolution of various variables. Instead, we simply need to remember that any equation (correlation) we derive must be dimensionally homogeneous. We will now discuss an organized procedure for doing dimensional analysis and selecting dimensionless groups. Buckingham Pi theorem, which is the foundation for dimensional analysis, can be stated as follows.20 Given a physical problem in which the dependent parameter 1qis a function of 1n − independent parameters, 23,,...,,nqq qwe expect ( )123,,...,nqfqq q= (8) or equivalently ( )12,,..., 0ngqq q = (9) If m is the minimum number of independent dimensions (mass. length, time, temperature, electric charge, etc.) required to specify the dimensions of all the parameters, the number of independent dimensionless ratios is, ,i inm≥− If ris the rank of the dimensional matrix, described below, then .inr=− That is, the rank of the dimensional matrix is less than or equal to . .mr m≤ Note that i is the number of independent dimensionless ratios. Denoting the i dimensionless ratios as 12,,...,,iππ π we write (8) & (9) as ( )1123,,...,iGππππ= (10) or ( )12,,..., 0iGππ π= The Buckingham Pi theorem does not predict the functional form of 1 or GG and these must be found experimentally.21 Dimensional Matrix Independent Dimensions List of Parameters 1212111123 Mass (m)Length (L)Time (t) ... TemperatureChar gennnnnnaa abb bccddeeqqq q↓⎡⎤⎢⎢⎢⋅⎢⋅⎢⎢⋅⎢⋅⋅⋅⋅⎢⎢⋅⋅⋅⋅⎣⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ Here , , ,iiiabcetc. denote the exponents, i.e. the dimension of 1q is given by 1111~ ...abcqmLt The matrix 11nnaabb⋅⋅⋅⋅⎡⎤⎢⎥⎢⎥⎢⎥⋅⋅⋅⎢⎥⋅⋅⎢⎥⎢⎥⋅⋅⎢⎥⋅⋅⎣⎦ is known as the dimensional matrix and r is the rank of this matrix. Example: Consider uniform streaming of a fluid past a smooth, stationary sphere. We seek to express the drag force Facting on the sphere in terms of the density ( )ρand viscosity ( )µ of the fluid, diameter of the sphere ( )D and the fluid velocity ( ).V i.e. ( ),, ,Ff VDρµ= (11) { } we have 5 parameters , , , ,nFVDρµ∴= The primary dimensions are ,,.mLt The dimensional matrix can be constructed as follows:22 110011311120 10 1FVDmLtρµ−−−−− As none of the rows can be expressed as a linear combination of the other two (linearly independent), we conclude that the rank of the dimensional matrix is 3. Therefore there are 53 2−=independent dimensionless parameters for this problem. (Note that determination of rank can be done with Gaussian elimination and other methods) It is convenient to select from the list of parameters a number of “core groups” or “repeating parameters” equal to the rank of the dimensional matrix, which requires inclusion of all the primary dimensions. For example, in this problem, we can choose ( ), , VDρ. Note that we have picked three repeating parameters (consistent with the fact that we have three primary dimensions) and that none of the core group can be expressed in terms of other two. Other valid options include ( ), , Vρµ, ( ), , VDµ, and ( ), , Dρµ. Practically, it makes sense to pick as repeating parameters only those parameters which are “inputs” that we set in an experiment. For example, in equation (11), we ask, “For given ,, and ,VDρµ what is the force F?” Thus, it makes sense to exclude F from the list of repeating parameters. Given our core group of ( ), , VDρ we are now left with two other parameters, and .Fµ We seek a dimensionless parameter by expressing Fin terms of ( ), , VDρ.23 { }000100023 1 0 equating exponents of -3 + 1 0 -2- 0 FVD mLtmL m LLmLttL tmαβ γαβγπρααβ γβ=≡∴⎛⎞⎛⎞⋅≡⎜⎟⎜⎟⎝⎠⎝⎠∴+=++==∴122 1,2,2 FVDαβγπρ=− =− =−∴= We seek the other dimensionless parameter by expressing µ in terms of , & .VDρ 22 and find ab cVDVDπµγµπρ== Then, according to the Buckingham Pi Theorem, we expect 1222 FfVD VDµππρρ⎛⎞===⎜⎟⎝⎠ (12) The functional form of fmust be found experimentally. The power of the dimensional analysis is revealed when we compare equations (11) and (12). At first, it appeared that to construct the desired correlation for the force F, we should run experiments where four seemingly independent variables should be changed over suitable ranges – this means