Marine Hydrodynamics (Lecture 5)

2.20 - Marine Hydrodynamics, Spring 2005 Lecture 5 2.20 - Marine Hydrodynamics Lecture 5 Chapter 2 - Similitude (Keyword: EQUAL RATIOS) Similitude: Similarity of behavior for different systems with equal similarity parameters. Prototype (real world) ↔ Model (physical/ analytical/ numerical . . . experiments) Similitude Geometric Similitude Kinematic Similitude Dynamic Similitude .. . Similarity Parameters (SP’s) Length ratios, angles Displacement ratios, velocity ratios Force ratios, stress ratios, pressure ratios ρ, ν Internal Constitution Similitude Boundary Condition Similitude .. . For similitude we require that the similarity parameters SP’s (eg. angles, length ratios, velocity ratios, etc) are equal for the model and the real world. Example 1 Two similar triangles have equal angles or equal length ratios. In this case the two triangles have geometric similitude. Example 2 For the flow around a model ship to be similar to the flow around the prototype ship, both model and prototype need to have equal angles and equal length and force ratios. In this case the model and the prototype have geometric and dynamic similitude. 1 2.1 Dimensional Analysis (DA) to Obtain SP’s 2.1.1 Buckingham’s π theory Reduce number of variables → derive dimensionally homogeneous relationships. 1. Specify (all) the (say N) relevant variables (dependent or independent): x1 , x2 , . . . xN e.g. time, force, fluid density, distance. . . We want to relate the xi ’s to each other I( x1 , x2 , . . . xN ) = 0 2. Identify (all) the (say P) relevant basic physical units (“dimensions”) e.g. M,L,T (P = 3) [temperature, charge, . . . ]. 3. Let π = xα1 1 xα2 2 . . . xαNN be a dimensionless quantity formed from the xi ’s. Suppose xi = Ci M mi Lli T ti , i = 1, 2, . . . , N where the Ci are dimensionless constants. For example, if x1 = KE = 12 M V 2 = 1 M 1 L2 T −2 (kinetic energy), we have that C1 = 12 , m1 = 1, l1 = 2, t1 = −2. Then 2 π = (C1α1 C2α2 . . . CNαN )M α1 m1 +α2 m2 +...+αN mN Lα1 l1 +α2 l2 +...+αN lN T α1 t1 +α2 t2 +...+αN tN For π to be dimensionless, we require ⎧ ⎫ N ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ αi mi = 0 ⎪ ⎬ αi li = 0 P aP × N system of Linear Equations ⎪ ⎪ ⎪ ⎪ αi ti = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩   ⎪ ⎭ Σ notation Since (1) is homogeneous, it always has a trivial solution, (1) αi ≡ 0, i = 1, 2, . . . , N (i.e. π is constant) There are 2 possibilities: (a) (1) has no nontrivial solution (only solution is π = constant, i.e. independent of xi ’s), which implies that the N variable xi , i = 1, 2, . . . , N are Dimensionally Independent (DI), i.e. they are ‘unrelated’ and ‘irrelevant’ to the problem. (b) (1) has J(J > 0) nontrivial solutions, π1 , π2, . . . , πJ . In general, J < N , in fact, J = N − K where K is the rank or ‘dimension’ of the system of equations (1). 2 2.1.2 Model Law Instead of relating the N xi ’s by I(x1 , x2 , . . . xN ) = 0, relate the J π’s by F (π1 , π2 , . . . πJ ) = 0, where J = N − K < N For similitude, we require (πmodel )j = (πprototype )j where j = 1, 2, . . . , J. If 2 problems have all the same πj ’s, they have similitude (in the πj senses), so π’s serve as similarity parameters. Note: • If π is dimensionless, so is π × const, π const , 1/π , etc. . . • If π1 , π2 are dimensionless, so is π1 × π2 , π1 , π1const1 π2 × π2const2 , etc. . . In general, we want the set (not unique) of independent πj ’s, for e.g., π 1 , π 2 , π 3 or π 1 , π 1 × π 2 , π 3 , but not π 1 , π 2 , π 1 × π 2 . Example: Force on a smooth circular cylinder in steady, incompressible flow Application of Buckingham’s π Theory. F U ρ,ν D Figure 1: Force on a smooth circular cylinder in steady incompressible fluid (no gravity) A Fluid Mechanician found that the relevant dimensional quantities required to evaluate the force F on the cylinder from the fluid are: the diameter of the cylinder D, the fluid velocity U , the fluid density ρ and the kinematic viscosity of the fluid ν. Evaluate the non-dimensional independent parameters that describe this problem. 3 xi : F, U, D, ρ, ν → N = 5 xi = ci M mi Lli T ti → P = 3 P = 3 mi li ti  F 1 1 -2 U 0 1 -1 N =5   D ρ ν 0 1 0 1 -3 2 0 0 -1 π = F α1 U α2 Dα3 ρα4 ν α5 For π to be non-dimensional, the set of equations αi mi =0 αi li =0 αi ti =0 has to be satisfied. The system of equations above after we substitute the values for the mi ’s, li ’s and ti ’s assume the form: ⎛ ⎞ α 1 ⎛ ⎞ ⎛ ⎞ ⎜ α2 ⎟ 1 0 0 1 0 0 ⎜ ⎟ ⎜ ⎟ ⎝ 1 ⎠ ⎝ 1 1 −3 2 0 ⎠ ⎜ α3 ⎟ = ⎝ ⎠ −2 −1 0 0 −1 α4 0 α5 The rank of this system is K = 3, so we have j = 2 nontrivial solutions. Two families of solutions for αi for each fixed pair of (α4 , α5 ), exists a unique solution for (α1 , α2 , α3 ). We consider the pairs (α4 = 1, α5 = 0) and (α4 = 0, α5 = 1), all other cases are linear combinations of these two. 4 1. Pair α4 = 1 and α5 = 0. ⎛ ⎞⎛ ⎞ ⎛ ⎞ 1 0 0 α1 −1 ⎝ 0 1 0 ⎠ ⎝ α2 ⎠ = ⎝ 4 ⎠ 2 0 0 1 α3 which has solution ⎛ ⎞ ⎛ ⎞ α1 −1 ⎝ α2 ⎠ = ⎝ 2 ⎠ 2 α3 ρU 2 D2 F ≡ Cd , which is the Drag coefficient. ∴ π1 = F α1 U α2 Dα3 ρα4 ν α5 = Conventionally, π1 → 2π1−1 and ∴ π1 = F 1 ρU 2 D2 2 2. Pair α4 = 0 and α5 = 1. ⎛ ⎞⎛ ⎞ ⎛ ⎞ 1 0 0 α1 0 ⎝ 0 1 0 ⎠ ⎝ α2 ⎠ = ⎝ −2 ⎠ −1 0 0 1 α3 which has solution ⎛ ⎞ ⎛ ⎞ α1 0 ⎝ α2 ⎠ = ⎝ −1 ⎠ α3 −1 ν UD ≡ Re , which is the Reynolds number. ∴ π2 = F α1 U α2 Dα3 ρα4 υ α5 = Conventionally, π2 → π2−1 , ∴ π2 = UD ν Therefore, we can write the following equivalent expressions for the non-dimensional inde­ pendent parameters that describe this problem: F (π1 , π2 ) = 0 F (Cd , Re ) = 0 F UD F( , )=0 2 2 1/2 ρU D ν or or or 5 π1 = f (π2 ) Cd = f (Re ) F UD = f( ) 2 2 1/2 ρU D ν Appendix A Dimensions of some fluid properties Quantities Dimensions (M LT ) Angle θ none (M 0 L0 T 0 ) Length L L Area A L2 Volume ∀ L3 Time t T Velocity V LT −1 Acceleration V̇ LT −2 Angular velocity ω T −1 Density ρ M L−3 Momentum L M LT −1 Volume flow rate Q L3 T −1 Mass flow rate Q M T −1 Pressure p M L−1 T −2 Stress τ M L−1 T −2 Surface tension Σ M T −2 Force F M LT −2 Moment M M L2 T −2 Energy E M L2 T −2 Power P M L2 T −3 Dynamic viscosity μ M L−1 T −1 Kinematic viscosity ν 6 L2 T −1