TAMU CHEN 304 - L3 (15 pages)

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Texas A&M University
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Chen 304 - Chem Engr Fluid Ops
Chem Engr Fluid Ops Documents
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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 19 Given a physical problem in which the dependent parameter q1 is a function of n 1 independent parameters q2 q3 qn we expect 8 q1 f q2 q3 qn or equivalently 9 g q1 q2 qn 0 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 i n m If r is the rank of the dimensional matrix described below then i n r That is the rank of the dimensional matrix is less than or equal to m r m Note that i is the number of independent dimensionless ratios Denoting the i dimensionless ratios as 1 2 i we write 8 9 as 10 1 G1 2 3 i or G 1 2 i 0 The Buckingham Pi theorem does not predict the functional form of G or G1 and these must be found experimentally 20 Dimensional Matrix Independent Dimensions List of Parameters Mass m Length L Time t Temperature Charge q1 q2 a1 b 1 c1 d1 e1 a2 b2 q3 qn an bn cn dn en Here ai bi ci etc denote the exponents i e the dimension of q1 is given by q1 m a1 Lb1 t c1 The matrix a1 b 1 an bn 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 F acting on the sphere in terms of the density and viscosity of the fluid diameter of the sphere D and the fluid velocity V i e F f V D 11 we have n 5 parameters F V D The primary dimensions are m L t The dimensional matrix can be constructed as follows 21 F V m 1 1 0 L 1 3 1 t 2 0 1 D 0 1 1 1 0 1 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 5 3 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 V D 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 V V D and D 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 V D and what is the force F Thus it makes sense to exclude F from the list of repeating parameters Given our core group of V D we are now left with two other parameters F and We seek a dimensionless parameter by expressing F in terms of V D 22 1 F V D m0 L0t 0 mL m L 3 L m 0 L0t 0 2 t L t 1 0 equating exponents of m 3 1 0 2 0 1 2 2 F 1 V 2 D 2 We seek the other dimensionless parameter by expressing in terms of V D 2 aV b D c and find 2 VD Then according to the Buckingham Pi Theorem we expect 1 F f 2 2 2 V D VD 12 The functional form of f must 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 experiments with different fluids spheres of different sizes and flow velocities In contrast equation 12 reveals that when viewed in terms of dimensionless variables there is only one independent dimensionless group that we choose in an experiment and find the other Thus one can simply perform all the experiments with a single sphere and any single Newtonian fluid and vary flow velocity and generate the data needed to construct the correlation in equation 12 A correlation developed by running experiments with water as the fluid can be used to estimate what 23 would happen for any other fluid Experiments can be run at room temperature even though the actual application may be at an elevated temperature etc Explanation via Linear Algebra Reconsider our dimensional matrix for drag force on a sphere F V m 1 1 0 L 1 3 1 t 2 0 1 D 0 1 1 1 0 1 The rank of this matrix is 3 and thus there are 2 independent dimensionless ratios To identify the independent dimensionless ratios we need to solve for the null space of the dimensional matrix a 1 1 0 0 1 b 0 1 3 1 1 1 c 0 2 0 1 0 1 …


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