TFL01 Factor Selection

TFL01 Factor Selection - Factor Selection Basic Concepts 1...

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Unformatted text preview: Factor Selection Basic Concepts 1 Unit Objectives Define basic terms Identify how to select factors Determine non dimensional groupings Use non dimensional grouping to design experimental models Use non dimensional groupings to calculate scaling factors 2 1 Basic Concepts I Factors – Parameters to be studied, to see if they have an influence on a specific outcome Levels – Parameter magnitudes to be tested Nuisance Variables – Parameters that could have an affect, but which are not being studied Noise – Unknown influential parameters 3 Basic Concepts II Factor Interactions Level – Effects where one factor “magnifies” the effect of a second factor Confidence – The higher the confidence level the worse the discrimination – To over come this increase the number of experiments (n), or reduce variance (2) by doing accurate experiments 4 2 Factor Selection Factors can be variables or attributes Identify the salient factors and interactions to be examined The number of factors should be 3 or 4, and the number of levels should be 2 or 3 Factor selection can be facilitated by using a Cause-and-Effect or Fishbone Diagram 5 Cause-and-Effect Diagrams Materials Material Type Weight Roughness Contact Area Machine Method Velocity Lubrication Sliding Friction Humidity Temperature Man Measurement Milieu 6 3 Non-Dimensional Analysis Non Dimensional Groupings – Buckingham Theorem Common Non Dimensional Groups Similitude – Geometric, Kinematic, Dynamic, and Transport Scaling 7 Buckingham Theorem I If a process is deemed to be controlled by n parameters (P1, P2, P3, ..., Pn) such as velocity (v), pressure (p), heat transferred (q), etc. containing in total k dimensions (e.g. M, L, T and/or in total), then there will be n-k dimensionless groupings (i.e. groupings); where M is mass, L is length, T is time, and is temperature. 8 4 Buckingham Theorem II Each grouping has to be composed of k salient parameters (P1, P2, P3, ...., Pk), which have to appear in each grouping, plus one of the other remaining n-k parameters (Pk+1, Pk+2, ...., Pn). The salient parameters should then be raised to their own powers (i.e. a, b, c, ...., k), and as a group the salient parameters should contain all of the k dimensions, but should not form a nondimensional grouping in themselves. 9 Buckingham Theorem III Finally, each grouping is then taken to be non-dimensional thereby allowing the power values (a, b, c, ...., k) to be determined from a system of equations. 1 P1a1 P2b1 P3c1 Pkk1 Pk 1 2 P1a2 P2b2 P3c2 Pkk2 Pk 2 10 n k P1an k P2bn k P3cn k Pkkn k Pn 5 Example 1 It is necessary to develop a nondimensional grouping to allow an experimental model of an object falling on to a cantilever to be established. K v m d 11 Example 1 (Solution) I The relevant parameter are identified as being the mass of the falling object, m, the velocity of the object, v, the cantilever spring constant, K, and the cantilever thickness, d. The dimensions involved are M,L, and T, therefore there will be three salient parameters, which will be taken to be m, v, and d. 12 6 Example 1 (Solution) II Consequently, in this case n = 4 (m, v, K, and d) and k = 3 (M, L, and T), thus No of groupings = = = n–k 4–3 1 The powers associated with each parameter can then be identified. 13 Example 1 (Solution) III mavbd c K Lb ML M b Lc 2 M 0 L0 T 0 T TL ( a 1) ( b c ) ( 2 b ) M L T M 0 L0 T 0 a 1 0 a 1 b20 b 2 a bc0 c2 14 m 1v 2 d 2 K 7 Example 1 (Solution) IV A physical interpretation of this grouping can then be noted by an appropriate manipulation of the terms. Kd 2 mv 2 2 1 2 Kd 1 2 2 mv Spring Energy KE of Object 15 Example 2 It is necessary to identify a dimensionless grouping to model the flow in a pipe. 16 8 Example 2 (Solution) I Relevant parameter are pipe diameter, d (L), flow velocity, v (L/T), flow density, (M/L3), and dynamic viscosity, (M/LT). Dimensions involved are mass, M, length, L, and time, T, therefore there will only be three salient parameters, which are taken to be pipe diameter, d, flow velocity, v, and flow density, (they do not form a dimensionless group on their own and they encompass three dimensions, M, L, and T), therefore n = 4 (m, v, , and ) and k = 3 (M, L, and T). 17 Example 2 (Solution) II No of groupings d avb c Lb M c M L b 3c M 0 L0 T 0 TL LT ( c 1) ( a b 3c 1) ( b 1) M L M 0 L0 T 0 T a =n–k =4–3 =1 c 1 0 b 1 0 a b 3c 1 0 d 1v 1 1 c 1 b 1 a 1 18 9 Example 2 (Solution) III vd vd vd 4 4 vd Qv 1 4 (v / l ) dl mv A Inertia Force Viscosity Force 19 Similitude Dimensionless groups provide a measure of a physical system’s behavior that is independent of a unit system. A system’s fundamental behavior can then often be expressed as functions of these non-dimensional groupings. Many groupings exist for different areas of engineering, however, most are associated with fluid flow and heat transfer. 20 10 Common Groupings Biot Number (Bi) Froude Number (Fr) Grashof Number (Gr) Mach Number (M) Nusselt Number (Nu) Prandtl Number (Pr), and Reynolds Number (Re) 21 System Modeling Since these groupings are in actuality either ratios of different force or energy terms within a system, they become constants that must be maintained when attempting to model a large physical system on a small scale. Each characteristic ratio between a model and its associated actual system must remain constant; this is called similitude. 22 11 Types of Similitude Geometric – Length ratios remain constant Kinematic – Velocity, acceleration, and flow rate ratios remain constant Dynamic – Force ratios remain constant Transport – Transport phenomena ratios remain constant (e.g. heat transfer ratios) 23 Similitude Issues It is almost impossible to meet all dynamic and transport similitude requirements simultaneously; only a few forces ratios within a model can be equal to their equivalent ratios in the actual system at the same time. In most cases it is necessary to identify the dominant phenomena, and only match the force and/or transport ratios associated with these particular phenomena. 24 12 Example 3 A 1:5 scale model has to be used to study the behavior of a new low speed aerofoil traveling at 70 m/s at sea level (i.e. 1 atm) and at 20 oC. If the air in the wind tunnel is also at the same pressure and temperature, what should be the air speed in the tunnel? 25 Example 3 (Solution) I Reynolds number (Re) has to be the same in both the model and aerofoil cases. vl vl Re model foil v l model v l foil vmodel vfoil lfoil l model 70 5 350 m/s 26 26 13 Example 3 (Part 2) What are the Mach (M) numbers associated with both of these cases (i.e. for the model and the aerofoil)? 27 Example 3 (Solution) II M model vmodel a where a RT 1.4 287 293 343 m/s 350 343 1.02 (Supersonic shock waves) M foil vfoil a 70 343 0.2 (very subsonic) 28 28 14 Example 3 (Part 3) If a cryogenic wind tunnel run on liquid nitrogen is used, does this help address this problem? Note: the boiling point for nitrogen is –196 oC ) 29 Example 3 (Solution) III model pmodel R Tmodel foil pfoil R Tfoil 110 5 287 83 4.2 kg/m 3 1.458 10 6 Tmodel 1 110.4 / Tmodel 1.458 10 6 83 1 110.4 / 83 110 5 287 293 1.2 kg/m 3 1.458 10 6 Tfoil 1 110.4 / Tfoil 1.458 10 6 293 1 110.4 / 293 model foil 5.7 10 6 Ns/m 2 18.110 6 Ns/m 2 30 15 Example 3 (Solution) IV (Re) model (Re) foil 4.2 vmodel 1 1.2 70 5 5.7 10 6 18.110 6 vmodel 31.6 m/s a model 1.4 287 83 182.6 m/s M model 31.6 182.6 0.17 By using a cryogenic tunnel it is possible to match two non-dimensional groupings. 31 Scaling Non Dimensional groupings can also be used to identify how different phenomena scale (change) with respect to physical size It is then possible to identify how, say, an elastic stress would increase as the size of a system increased physically 32 16 Example 4 If the dominant forces associated with a particular system are inertia and Coulomb friction, how would the velocity ratio be affected by the length scale ratio? 33 Example 4 (Solution) I The velocity ratio can be expressed in terms of the length and time ratios. vm l m t a va t m l a A time ratio can also be established knowing that the inertia and friction forces are the dominant dynamic characteristics. F F Friction ( F ) Constant m a Inertia ( I ) Im Ia m m m g a ma g mm a m ma a a 34 17 Example 4 (Solution) II If both friction coefficients are assumed to be equal (i.e. m=a), the previous ratio can be used to relate a length to time. m mm g a ma g mm am ma aa am 1 aa lm ta2 1 2 t m la ta l a tm lm 35 Example 4 (Solution) III If this relationship is then combined with the initial velocity ratio an equation can be established that relates the velocity ratio to the geometry scale ratio. vm l m t a va t m la l lm a la lm lm la 36 18 Summary Factors are the conditions that are deliberately changed in an experiment Nuisance variables and noise are other experimental conditions Only a few factors should be chosen at one time and only at a few levels Factors often form non dimensional groupings These groupings are used in modeling 37 19 ...
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