PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Stanford University Department of Aeronautics and Astronautics Introduction to Symmetry AnalysisBrian CantwellDepartment of Aeronautics and AstronauticsStanford UniversityChapter 2 - Dimensional AnalysisStanford University Department of Aeronautics and Astronautics What is a dimension?2.1 IntroductionStanford University Department of Aeronautics and Astronautics Figure 2.1 Elliptical orbit of a planet about the SunNewtonian law of gravitation2.2 The Two-Body Problem in a Gravitational Field(2.1)Stanford University Department of Aeronautics and Astronautics Table 2.1 The planets and their orbitsThe mass of the Earth isand the mean diameter isThe eccentricity of a planet’s orbit is(2.2)kgkmStanford University Department of Aeronautics and Astronautics and(2.4)(2.5)(2.6)(2.3)Parameters of the problemThere are six parameters and three fundamental dimensions. So we can expect the solution to depend on three dimensionless numbersThese variables must be related by a dimensionless function of the formStanford University Department of Aeronautics and Astronautics or(2.7)(2.8)The mean radius is defined asTheory tells us thatStanford University Department of Aeronautics and Astronautics Figure 2.2 Kepler’s third law for the Solar System.Stanford University Department of Aeronautics and Astronautics Figure 2.3 Viscous fow past a sphere(2.10)(2.9)Dimensions of the governing parametersThe parameters of the problem are related to one another through a function of the form2.3 The Drag on a SphereStanford University Department of Aeronautics and Astronautics The fact that the parameters have dimensions highly restricts the kind of drag functions that are possible. For example, suppose we guess that the drag law has the form(2.11a)(2.11b)If we introduce the dimensions of each parameter the expression has the form Suppose the units of mass are changed from kilograms to grams. Then the number for the drag will increase by a factor of a thousand. But the expression in parentheses will not increase by this factor and the equality will not be satisfied. In effect the drag of the sphere will seem to depend on the choice of units and this is impossible. The conclusion is that (2.11a) can not possibly describe the drag of a sphere.Stanford University Department of Aeronautics and Astronautics Step1The drag expression must be invariant under a three parameter dilation group.(2.12)Scale the units of mass using the one-parameter group(2.13)(2.14)(2.15)We can derive the required drag expression as follows.The drag expression must be independent of the scaling parameter m and therefore must be of the form.The effect is to transform the parameters as follows.Stanford University Department of Aeronautics and Astronautics (2.16)(2.17)(2.18)The dimensions of the variables remaining areLet the units of length be scaled according toStep 2The effect of this group on the new variables isThe drag relation must be independent of the scaling parameter l.A functional form that accomplishes this is(2.19)Stanford University Department of Aeronautics and Astronautics The dimensions of these variables are(2.20)Finally scale the units of timeStep 3(2.21)(2.22)The effect of this group on the remaining variables isThe drag relation must be independent of the scaling parameter t. Finally(2.23)where(2.24)Stanford University Department of Aeronautics and Astronautics In the limit of vanishing Reynolds number the drag of a sphere is given byIf we insert the expressions for the Drag coefficient and Reynolds number into Equation (2.25) the drag law becomes (2.25)(2.26)Note that at low Reynolds number the drag of a sphere is independent of the density of the surrounding fluid. In this limit there is only one dimensionless parameter in the problem proportional to the product CD x Re.Stanford University Department of Aeronautics and Astronautics One might conjecture that the same kind of law applies to the low Reynolds number flow past a circular cylinder. In this case the drag force is replaced by the drag force per unit span with units(2.27)with drag coefficient(2.28)Assume that in the limit of vanishing Reynolds number the drag coefficient of a circular cylinder follows the same law as for the sphere.(2.29)If we restore the dimensioned variables in (2.29) the result is(2.30)This is a completely incorrect result!Stanford University Department of Aeronautics and Astronautics Measurements of circular cylinder drag versus Reynolds number taken by a variety of investigators.The data shows a huge amount of scatter - why?Figure 2.4 Experimental measurements of the drag of a circular cylinderThe drag of a sphere or a cylinder depends on a wide variety of length and velocity scales that we have ignored!(2.31)Stanford University Department of Aeronautics and Astronautics Figure 2.5 High speed fow past a sphere2.4 The Drag on a Sphere in High Speed FlowThe dimensions of the new variables are(2.32)Stanford University Department of Aeronautics and Astronautics There are now two additional dimensionless variables related to the fact that the sphere motion significantly changes the temperature of the oncoming gas.(2.33)The drag relation is now a function of four dimensionless variables.(2.34)The Mach number is used in (2.34).(2.35)where(2.36)Without loss of generality we can write(2.37)Stanford University Department of Aeronautics and Astronautics Figure 2.6 Bashforth’s drag data for a 7.4 cm-diameter cannonballStanford University Department of Aeronautics and Astronautics Figure 2.7 Compilation of sphere drag as a function of Mach number and Reynoldsnumber from Miller and Bailey .As the Mach number increases the drag coefficient tends to become independent of both Reynolds number and Mach number with(2.38)Stanford University Department of Aeronautics and Astronautics 2.4 The Buckingham Pi TheoremStanford University Department of Aeronautics and AstronauticsStanford University Department of Aeronautics and Astronautics It is important to recognize that the dimensionless parameters generated by the algorithm just described are not unique. For example in the case of the sphere we could have wound up with the following, equally correct, result.(2.41)(2.42)In this form