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& 8.0 Dimensional Analysis INTRODUCTION Fluid flow is influenced by several parameters like, the geometry, fluid properties and fluid velocity. In the previous chapters analytical methods used in fluid flow studies were discussed. In the study of flow of real fluids analytical methods alone are found insufficient. Experimental methods and results have contributed heavily for the development of fluid mechanics. The solution of realistic problems usually involves both anlytical and experimental studies. Experiments are used to validate analytical results as well as generalize and extend their applications. Depending either solely on analytical methods or experiments for the design of systems is found to lead to inadequate performance and high cost. Experimental work is rather costly and time consuming, particularly when more than three parameters are involved. Hence it is necessary to plan the experiments so that most information is obtained from fewest experiments. Dimensional analysis is found to be a very useful tool in achieving this objective. The mathematical method of dimensional analysis comes to our help in this situation. The number of parameters can be reduced generally to three by grouping relevant variables to form dimensionless parameters. In addition these groups facilitate the presentation of the results of the experiments effectively and also to generalize the results so that these can be applied to similar situations. Flow through pipes can be considered as an example. Viscosity, density, flow velocity and diameter are found to influence the flow. If the effect of each of these parameters on flow is separately studied the number of experiments will be large. Also these results cannot be generalized and its usefulness will be limited. When the number of these variables are combined to form a dimensionless group like (u D ρ/µ) few experiments will be sufficient to obtain useful information. This parameter can be varied by varying one of the variables which will be the easier one to vary, for example velocity u. The results will be applicable for various combinations of these parameters and so the results can be generalized and extended to new situations. The results will be applicable also for different fluids and different diameters provided the value of the group remains the same. Example 8.1 illustrates the advantage dimensional analysis in experiment planning. The use of the results of dimensional analysis is the basis for similitude and modal studies. The topic is discussed in the next chapter. 263 Fluid Mechanics and Machinery 264 Example 8.1. The drag force F on a stationary sphere in flow is found to depend on diameter D, velocity u, fluid density ρ and viscosity µ. Assuming that to study the influence of a parameter 10 experimental points are necessary, estimate the total experimental points needed to obtain complete information. Indicate how the number of experiments can be reduced. To obtain a curve F vs u, for fixed values of ρ, µ and D, experiments needed = 10. To study the effect of ρ these 10 experiments should be repeated 10 times with 10 values of ρ the total now being 102. The 102 experiments have to repeated 10 times each for different values of µ. Total experiments for u, ρ and µ = 103. To study the effect of variation of diameter all the experiments have to be repeated 10 times each. Hence total experiments required = 104. These parameters can be combined to obtain two dimensionless parameters, F 2 ρu D 2 =f FG ρ uD IJ H µ K (The method to obtain such grouping is the main aim of this chapter) Now only 10 experiments are needed to obtain a comprehensive information about the effect of these five parameters. Experiments can be conducted for obtaining this information by varying the parameter (uDρ/µ) and determining the values for F/ρu2D2. Note : It will be almost impossible to find fluids with 10 different densities and 10 different viscosities. 8.1 METHODS OF DETERMINATION OF DIMENSIONLESS GROUPS 1. Intuitive method: This method relies on basic understanding of the phenomenon and then identifying competing quantities like types of forces or lengths etc. and obtaining ratios of similar quantities. Some examples are: Viscous force vs inertia force, viscous force vs gravity force or roughness dimension vs diameter. This is a difficult exercise and considerable experience is required in this case. 2. Rayleigh method: A functional power relation is assumed between the parameters and then the values of indices are solved for to obtain the grouping. For example in the problem in example 1 one can write (π1, π2) = F a ρbDcµdUe The values of a, b, c, d, and e are obtained by comparing the dimensions on both sides the dimensions on the L.H.S. being zero as π terms are dimensionless. This is also tedious and considerable expertise is needed to form these groups as the number of unknowns will be more than the number of available equations. This method is also called ‘‘indicial” method. 3. Buckingham Pi theorem method: The application of this theorem provides a fairly easy method to identify dimensionless parameters (numbers). However identification of the influencing parameters is the job of an expert rather than that of a novice. This method is illustrated extensively throughout this chapter. VED P-2\D:\N-fluid\Fld8-1.pm5 Dimensional Analysis 8.2 265 THE PRINCIPLE OF DIMENSIONAL HOMOGENEITY The principle is basic for the correctness of any equation. It states ‘‘If an equation truly expresses a proper relationship between variables in a physical phenomenon, then each of the additive terms will have the same dimensions or these should be dimensionally homogeneous.’’ For example, if an equation of the following form expresses a relationship between variables in a process, then each of the additive term should have the same dimensions. In the expression, A + B = C/D, A, B and (C/D) each should have the same dimension. This principle is used in dimensional analysis to form dimensionless groups. Equations which are dimensionally homogeneous can be used without restrictions about the units adopted. Another application of this principle is the checking of the equations derived. Note : Some empirical equations used in fluid mechanics may appear to be non homogeneous. In such cases, the numeric constants are dimensional. The value of the constants in such equations will vary with the system of units used. 8.3 BUCKINGHAM PI THEOREM The statement of the theorem is as follows : If a relation among n parameters exists in the form f(q1, q2, ........ qn) = 0 then the n parameters can be grouped into n – m independent dimensionless ratios or π parameters, expressed in the form g(π1, π2 ........ πn–m) = 0 or (8.3.1) π1 = g1 (π2, π3 ...... πn–m) where m is the number of dimensions required to specify the dimensions of all the parameters, q1, q2, .... qn. It is also possible to form new dimensionless π parameters as a discrete function of the (n – m) parameters. For example if there are four dimensionless parameters π1, π2, π3 and π4 it is possible to obtain π5, π6 etc. as π5 = π1 π3π4 or π6 = π 10.5 π 22/ 3 The limitation of this exercise is that the exact functional relationship in equation 8.3.1 cannot be obtained from the analysis. The functional relationship is generally arrived at through the use of experimental results. Irrespective of the method used the following steps will systematise the procedure. Step 1. List all the parameters that influence the phenomenon concerned. This has to be very carefully done. If some parameters are left out, π terms may be formed but experiments then will indicate these as inadequate to describe the phenomenon. If unsure the parameter can be added. Later experiments will show that the π term with the doubtful VED P-2\D:\N-fluid\Fld8-1.pm5 Chapter 8 8.3.1 Determination of π Groups Fluid Mechanics and Machinery 266 parameters as useful or otherwise. Hence a careful choice of the parameters will help in solving the problem with least effort. Usually three type of parameters may be identified in fluid flow namely fluid properties, geometry and flow parameters like velocity and pressure. Step 2. Select a set of primary dimensions, (mass, length and time), (force, length and time), (mass, length, time and temperature) are some of the sets used popularly. Step 3. List the dimensions of all parameters in terms of the chosen set of primary dimensions. Table 8.3.1. Lists the dimensions of various parameters involved. Table 8.3.1. Units and Dimensions of Variables Variable Unit (SI) Dimension MLT θ system FLT θ system Mass kg M FT2/L Length m L L Time s T T N ML/T2 F deg C or K θ θ Area m2 L2 L2 Volume m3 L3 L3 Volume flow rate m3/s L3/T L3/T Mass flow rate kg/s M/T FT/L Velocity m/s L/T L/T Rad/s 1/T 1/T N ML/T2 F N/m2 M/LT2 F/L2 Nm ML2/T2 FL Work, Energy J, Nm ML2/T2 FL Power W, J/s ML2/T3 FL/T Density kg/m3 M/L3 FT2/L4 M/LT FT/L2 Force Temperature Angular velocity Force Pressure, stress, Bulk modulus Moment Dynamic viscosity kg/ms, Ns/m2 Kinematic viscosity m2/s L2/T L2/T Surface tension N/m M/T2 F/L Specific heat J/kg K L2/T2 θ L2/T 2θ Thermal conductivity W/mK ML/T3 θ F/Tθ transfer coefficient W/m2 K M/T3 θ F/LTθ Expansion coefficient (m/m)/K 1/T 1/T Convective heat VED P-2\D:\N-fluid\Fld8-1.pm5 Dimensional Analysis 267 Step 4. Select from the list of parameters a set of repeating parameters equal to the number of primary dimensions. Some guidelines are necessary for the choice. (i) the chosen set should contain all the dimensions (ii) two parameters with same dimensions should not be chosen. say L, L2, L3, (iii) the dependent parameter to be determined should not be chosen. Step 5. Set up a dimensional equation with the repeating set and one of the remaining parameters, in turn to obtain n – m such equations, to determine π terms numbering n – m. The form of the equation is, π1 = qm+1 . q1a . q2b . q3c ..... qmd As the LHS term is dimensionless, an equation for each dimension in terms of a, b, c, d can be obtained. The solution of these set of equations will give the values of a, b, c and d. Thus the π term will be defined. Step 6. Check whether π terms obtained are dimensionless. This step is essential before proceeding with experiments to determine the functional relationship between the π terms. Example 8.2. The pressure drop ∆P per unit length in flow through a smooth circular pipe is found to depend on (i) the flow velocity, u (ii) diameter of the pipe, D (iii) density of the fluid ρ, and (iv) the dynamic viscosity µ. (a) Using π theorem method, evaluate the dimensionless parameters for the flow. (b) Using Rayleigh method (power index) evaluate the dimensionless parameters. Choosing the set mass, time and length as primary dimensions, the dimensions of the parameters are tabulated. S.No. Parameter Unit used Dimension (N/m2/m (N = kgm/s2) M/L2T 2 m L 1 Pressure drop/m, ∆P 2 Diameter, D 3 Velocity, u m/s L/T 4 Density, ρ kg/m3 M/L3 5 Dynamic viscosity, µ kg/ms M/LT There are five parameters and three dimensions. Hence two π terms can be obtained. As ∆P is the dependent variable D, ρ and µ are chosen as repeating variables. Let π1 = ∆P Daρbuc, Substituting dimensions, M L2 T 2 La M b Lc L3b T c Using the principle of dimensional homogeneity, and in turn comparing indices of mass, length and time. 1+b=0 –2–c=0 ∴ b = – 1, – 2 + a – 3b + c = 0 ∴ a + c = – 1 ∴ c = – 2, Substituting the value of indices we obtain VED P-2\D:\N-fluid\Fld8-1.pm5 Hence a = 1. Chapter 8 M0L0T0 = Fluid Mechanics and Machinery 268 π1 = ∆PD/ρu2; ∴ This represents the ratio of pressure force and inertia force. Check the dimension : M L2 T 2 L L3 T 2 = M0L0T 0 M L2 Let π2 = µ Da ρbuc, substituting dimensions and considering the indices of M, L and T, M0L0T0 = 1 + b = 0 or M a M b Lc L LT L3b T c b = – 1, – 1 + a – 3b + c = 0, a + c = – 2, – 1 – c = 0, c=–1 a=–1 Substituting the value of indices, π2 = µ/uρD ∴ M T L3 1 = M0L0T 0 LT L M L check, This term may be recognised as inverse of Reynolds number. So π2 can be modified as π2 = ρuD/µ also π2 = (uD/v). The significance of this π term is that it is the ratio of inertia force to viscous force. In case D, u and µ had been choosen as the repeating, variables, π1 = ∆PD2/u µ and π2 = ρDu/µ. The parameter π1/π2 will give the dimensionless term. ∆P D/ρu2. In this case π1 represents the ratio pressure force/viscous force. This flow phenomenon is influenced by the three forces namely pressure force, viscous force and inertia force. Rayleigh method: (Also called method of Indices). The following functional relationship is formed first. There can be two p terms as there are five variables and three dimensions. DPaDbrcmdue = (p1 p2), Substituting dimensions, Ma L2 a T 2 a Lb Mc Md Le L3 c Ld T d T e = L0 M0 T0 Considering indices of M, L and T, three equations are obtained as below a + c + d = 0, – 2a + b – 3c – d + e = 0, – 2a + d – e = 0 There are five unknowns and three equations. Hence some assumptions are necessary based on the nature of the phenomenon. As DP, the dependent variable can be considered to appear only once. We can assume a = 1. Similarly, studying the forces, m appears only in the viscous force. So we can assume d = 1. Solving a = 1, d = 1, b = 0, c = – 2, e = – 3, (p1 p2) = DPm/r2 u3. Multiply and divide by D, then p1 = DPD/ru2 and p2 = m/ruD. Same as was obtained by p theorem method. This method requires more expertise and understanding of the basics of the phenomenon. Example 8.3. The pressure drop ∆P in flow of incompressible fluid through rough pipes is found to depend on the length l, average velocity u, fluid density, ρ, dynamic viscosity µ, diameter D and average roughness height e. Determine the dimensionless groups to correlate the flow parameters. VED P-2\D:\N-fluid\Fld8-1.pm5 Dimensional Analysis 269 The variables with units and dimensions are listed below. S.No. Variable Unit Dimension 1 ∆P N/m2 M/LT2 2 l L L 3 u m/s L/T 4 ρ kg/m3 M/L3 5 µ kg/ms M/LT 6 D L L 7 e L L There are seven parameters and three dimensions. So four π terms can be identified. Selecting u, D and ρ as repeating variables, (as these sets are separate equations, no problem will arise in using indices a, b and c in all cases). Let Consider π1, π1 = ∆P uaDbρc, π2 = L uaDb ρc, π3 = µ ua Db ρc, π4 = e uaDbρc M0L0T0 = La M LT 2 T a Lb Mc L3c Equating the indices of M, L and T, 1 + c = 0, c = – 1, – 1 + a + b – 3c = 0, – 2 – a = 0, a = – 2, b = 0. Substituting the value of indices we get ∴ Consider π2, ρu2 π1 = ∆P/ρ M0L0T0 = L La T Lb a Mc L3c Equating indices of M, L and T, c = 0, 1 + a + b – 3c = 0, a = 0, ∴ b = – 1, Consider π3 M0L0T0 = ∴ π2 = L/D M La b M c L 3c LT T a L Comparing the indices of µm, L and T, ∴ Consider π4, c = – 1, – 1 + a + b – 3c = 0, – 1 – a = 0 or a = – 1, ρDu π3 = µ/ρ M0L0T0 = L La Ta µ or ρuD/µ Lb Mc L3c This gives, c = 0, 1 + a + b – 3c = 0, – a = 0, b = – 1 ∴ π4 = e/D These π terms may be checked for dimensionless nature. The relationship can be expressed as VED P-2\D:\N-fluid\Fld8-1.pm5 ∆P ρu ∴ b=–1 2 =f LM L , e , ρuD OP ND D µ Q Chapter 8 gives 1 + c = 0 or Fluid Mechanics and Machinery 270 8.4 IMPORTANT DIMENSIONLESS PARAMETERS Some of the important dimensionless groups used in fluid mechanics are listed in Table 8.4.1. indicating significance and area of application of each. Table 8.4.1 Important Dimensionless Parameters Name Description Significance Applications ρuD/µ or uD/v Inertia force/ All types of fluid Number, Re Viscous force dynamics problems Froude Number Inertia force/ Flow with free Gravity force surface (open Reynolds u/(gl)0.5 or Fr u2/gl Euler Number P/ρu2 Eu Cauchy Number channel and ships) Pressure force/ Flow driven by Inertia force pressure Inertia force/ compressible flow ρu2/Ev (Ev– Compressibility bulk modulous) force Mach Number u/c, c–Velocity Inertia force/ M of sound Compressibility Ca Compressible flow force Strouhal ωl/u, Local inertia Unsteady flow with Number ω–Frequency of Force/ frequency of St oscillation Convective oscillation inertia force Weber Number Inertia force/ Problems influenced ρu2l/σ, σ = Surface tension by surface tension Surface tension force free surface flow Lift coefficient L/(1/2 ρAu2) Lift force/ Aerodynamics CL L = lift force Dynamic force We 8.5 CORRELATION OF EXPERIMENTAL DATA Dimensional analysis can only lead to the identification of relevant dimensionless groups. The exact functional relations between them can be established only by experiments. The degree of difficulty involved in experimentation will depend on the number of π terms. VED P-2\D:\N-fluid\Fld8-1.pm5 Dimensional Analysis 271 8.5.1 Problems with One Pi Term In this case a direct functional relationship will be obtained but a constant c has to be determined by experiments. The relationship will be of the form π1 = c. This is illustrated in example 8.4. Example 8.4. The drag force acting on a spherical particle of diameter D falling slowly through a viscous fluid at velocity u is found to be influenced by the diameter D, velocity of fall u, and the viscosity µ. Using the method of dimensional analysis obtain a relationship between the variables. The parameters are listed below using M, L, T dimension set. S.No. Parameter Symbol Unit Dimension 1 Drag Force F N or kgm/s2 ML/T2 2 Diameter D m L 3 Velocity u m/s L/T 4 Viscosity µ kg/ms M/LT There are four parameters and three dimensions. Hence only one π term will result. π1 = F Da ub µc, Substituting dimensions, M0L0T0 = ML T 2 La Lb T b Mc Lc T c , Equating indices of M, L and T 0 = 1 + c, c = – 1, 1 + a + b – c = 0, 2 + b + c = 0, b = – 1, c = – 1 π1 = F/uDµ µ ∴ F/uDµ µ = constant = c or F = cuDµ or drag force varies directly with velocity, diameter and viscosity. A single test will provide the value of the constant. However, to obtain a reliable value for c, the experiments may have to be repeated changing the values of the parameters. In this case an approximate solution was obtained theoretically for c as 3π. Hence drag force F in free fall is given by F = 3πµuD. This can be established by experiments. This relation is known as Stokes law valid for small values of Reynolds Number (Re << 1). This can be used to study the settling of dust in still air. Inclusion of additional variable, namely density will lead to another π term. In example 8.2 two π terms were identified. If the dimensional analysis is valid then a single universal relationship can be obtained. Experiments should be conducted by varying one of the group say π1 and from the measurement the values of the other group π2 is calculated. A suitable graph (or a computer program) can lead to the functional relationship between the π terms. Linear semilog or log/log plots may have to be used to obtain such a relationship. The valid range should be between the two extreme values used in the experiment. Extrapolation may lead to erroneous conclusions. This is illusration by example 8.5. VED P-2\D:\N-fluid\Fld8-1.pm5 Chapter 8 8.5.2 Problems with Two Pi Terms Fluid Mechanics and Machinery 272 Example 8.5. In order to determine the pressure drop in pipe flow per m length an experiment was conducted using flow of water at 20°C through a 20 mm smooth pipe of length 5 m. The variation of pressure drop observed with variation of velocity is tabulated below.The density of water = 1000 kg/ m3. Viscosity = 1.006 × 10–3 kg/ms. Velocity, m/s 0.3 0.6 0.9 1.5 2.0 3 5 Pressure drop, N 404 1361 2766 6763 11189 22748 55614 Determine the functional relationship between the dimensionless parameters (D ∆P/ρu2) and (ρuD/ µ). Using the data the two π parameters together with log values are calculated and tabulated below. u 0.3 0.6 0.9 1.5 2.0 3 5 0.01798 0.01512 0.01366 0.01202 0.01119 0.01011 0.00890 ρuD/µ 5964 11928 17894 29821 39761 59642 99400 logRe 3.78 4.08 4.25 4.48 4.6 4.78 4.997 – 1.745 – 1.821 – 1.865 – 1.92 – 1.951 – 1.995 – 2.051 D∆P/ρu 2 log(D∆P/ρu) A plot of the data is shown in Fig. Ex. 8.5 (a). The correlation appears to be good. Scatter may indicate either experimental error or omission of an influencing parameter. As the direct plot is a curve., fitting an equation can not be done from the graph. A log log plot results in a straight line, as shown in the Fig. 8.5 (b). To fit an equation the following procedure is used. The slope is obtained by taking the last values: = {– 2.051 – (– 1.745)}/(4.997 – 3.78) = – 0.2508 When extrapolating we can write, the slope using the same – 2.051 – (x)/(5 – 0) = – 0.2508 This gives x = – 0.797. This corresponds to the value of 0.16. Hence we can write, D∆P ρu2 = 0.16 FG PuD IJ H µ K −0.2508 = 0.16 × Re–0.2508 0.022 0.020 4 0.016 ru 0.012 2 2 0.014 D DP 2 ru D DP 0.018 –2 10 8 6 0.010 0.008 –3 0 20.000 40.000 60.000 80.000 10.0000 4 × 10 ruD m (a) P-2\D:\N-fluid\Fld8-1.pm5 2 4 4 6 8 10 ruD m (b) Figure Ex. 8.5 VED 3 10 2 4 5 6 8 10