SGChapt05 - V. Modeling, Similarity, and Dimensional...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: V. Modeling, Similarity, and Dimensional Analysis To this point, we have concentrated on analytical methods of solution for fluids problems. However, analytical methods are not always satisfactory due to: (1) limitations due to simplifications required in the analysis, (2) complexity and/or expense of a detailed analysis. The most common alternative is to: Use experimental test & verification procedures. However, without planning and organization, experimental procedures can : (a) be time consuming, (b) lack direction, (c) be expensive. This is particularly true when the test program necessitates testing at one set of conditions, geometry, and fluid with the objective to represent a different but similar set of conditions, geometry, and fluid. Dimensional analysis provides a procedure that will typically reduce both the time and expense of experimental work necessary to experimentally represent a desired set of conditions and geometry. It also provides a means of "normalizing" the final results for a range of test conditions. A normalized (non-dimensional) set of results for one test condition can be used to predict the performance at different but fluid dynamically similar conditions ( including even a different fluid). The basic procedure for dimensional analysis can be summarized as follows: | v v V-1 | e-Text Main Menu | Textbook Table of Contents | Study Guide 1. Compile a list of relevant variables (dependent & independent) for the problem being considered, 2. Use an appropriate procedure to identify both the number and form of the resulting non-dimensional parameters. This procedure is outlined as follows for the Buckingham Pi Theorem Definitions: n = the number of independent variables relevant to the problem j’ = the number of independent dimensions found in the n variables j = the reduction possible in the number of variables necessary to be considered simultaneously k = the number of independent pi terms that can be identified to describe the problem, k = n - j Summary of Steps: 1. List and count the n variables involved in the problem. 2. List the dimensions of each variable using {MLTΘ} or {FLTΘ}. Count the number of basic dimensions ( j’) for the list of variables being considered. 3. Find j by initially assuming j = j’ and look for j repeating variables which do not form a pi product. If not successful, reduce j by 1 and repeat the process. 4. Select j scaling, repeating variables which do not form a pi product. 5. Form a pi term by adding one additional variable and form a power product. Algebraically find the values of the exponents which make the product dimensionless. Repeat the process with each of the remaining variables. 6. Write the combination of dimensionless pi terms in functional form: Πk = f( Π1, Π2, …Πi) Consider the following example for viscous pipe flow. The relevant variables for this problem are summarized as follows: ∆P = pressure drop µ = viscosity ρ = density ε = roughness V = velocity L = length {∆ P Seven pipe flow variables: dependent D = diameter ρ, V, D, µ, ε, L } independent | v v V-2 | e-Text Main Menu | Textbook Table of Contents | Study Guide Use of the Buckingham Pi Theorem proceeds as follows: 1. Number of independent variables: n = 7 2. List the dimensions of each variable ( use m L t Θ ): ∆P variables ρ -1 -2 V -3 dimensions mL t mL Lt -1 D µ L ε mL t -1 -1 L L L The number of basic dimensions is j’ = 3. 3. Choose j = 3 with the repeating variables being ρ, V, and D. They do not form a dimensionless pi term. No combination of the 3 variables will eliminate the mass dimension in density or the time dimension in velocity. 4. This step described in the above step. The repeating variables again are ρ, V, and D and j = 3. Therefore, k = n – j = 7 – 3 = 4 independent Π terms. 5. Form the Π terms: Π1 = ρa Vb Dc µ−1 = (mL-3)a ( Lt-1)b Lc ( mL-1t-1 )−1 In order for the Π term to have no net dimensions, the sum of the exponents for each dimension must be zero. Therefore, we have: mass: time: length: a - 1 =0, a= 1 - b + 1 = 0, b = 1 -3a + b + c + 1 = 0, c = 3 – 1 – 1 = 1 We therefore have Π1 = ρ V D /µ = Re = Reynolds number Repeating the process by adding the roughness ε | v v V-3 | e-Text Main Menu | Textbook Table of Contents | Study Guide Π2 = ρa Vb Dc ε1 = (mL-3)a ( Lt-1)b Lc ( L )1 Solving: mass: time: Length: Π2 = ε / D a =0, a= 0 - b = 0, b = 0 -3a + b + c + 1 = 0, c = – 1 Roughness ratio Repeat the process by adding the length L. Π3 = ρa Vb Dc L1 = (mL-3)a ( Lt-1)b Lc ( L )1 Solving: mass: time: length: a =0, a= 0 - b = 0, b = 0 -3a + b + c + 1 = 0, c = – 1 Π3 = L / D length-to-diameter ratio These three are the independent Π terms. Now obtain the dependent Π term by adding ∆P Π4 = ρa Vb Dc ∆P1 = (mL-3)a ( Lt-1)b Lc ( mL-1t-2 )1 Solving: mass: time: length: Π4 = ∆P a + 1 = 0 , a = -1 - b - 2 = 0, b = -2 -3a + b + c - 1 = 0, c = 0 / ρ V2 Pressure coefficient | v v V-4 | e-Text Main Menu | Textbook Table of Contents | Study Guide Application of the Buckingham Pi Theorem to the previous list of variables yields the following non-dimensional combinations: ∆P ρ VD L ε = f ,, ρ V2 µ D D or { Cp = f Re,L, ε } Thus, a non-dimensional pressure loss coefficient for viscous pipe flow would be expected to be a function of (1) the Reynolds number, (2) a nondimensional pipe length, and (3) a non-dimensional pipe roughness. This will be shown to be exactly the case in Ch. VI, Viscous Internal Flow. A list of typical dimensionless groups important in fluid mechanics is given in the accompanying table. From these results, we would now use a planned experiment with data analysis techniques to get the exact form of the relationship among these non dimensional parameters. The next major step is concerned with the design and organization of the experimental test program Two key elements in the test program are: * design of the model * specification of the test conditions, particularly when the test must be performed at conditions similar, but not the same as the conditions of interest. Similarity and non-dimensional scaling The basic requirement is in this process to achieve 'similarity' between the 'experimental model and its test conditions' and the 'prototype and its test conditions' in the experiment. | v v V-5 | e-Text Main Menu | Textbook Table of Contents | Study Guide Table 5.2 Dimensional Analysis and Similarity Parameter Definition RE = Reynolds number Qualitative ratio of effects Inertia Viscosity Flow speed Sound speed Inertia Gravity Inertia Surface tension Pressure Inertia ρ UL µ U A U2 Fr = gL MA = Mach number Froude number Weber number We = ρ U2 L γ Cavitation number (Euler number) Ca = p - pv ρU 2 C pµ k U2 Ec = c p To c γ= p cv ωL St = U Eckert number Specific-heat ratio Strouhal number L Gr = Grashof number Temperature ratio Pressure coefficient Compressible flow Free-surface flow Free-surface flow Cavitation Heat convection Dissipation Enthalpy Internal energy Oscillation Mean speed Wall roughness Body length ε Roughness ratio Always Dissipation Conduction Kinetic energy Enthalpy Pr = Prandtl number Importance β ∆TgL3 ρ2 µ2 Tw To p − p∞ Cp = 1/ 2 ρ U2 Compressible flow Oscillating flow Turbulent,rough walls Buoyancy Viscosity Wall temperature Stream temperature Static pressure Dynamic pressure Lift coefficient CL = L 1/ 2 ρ U2 A CD = D 1/ 2ρ U2 A Heat transfer Aerodynamics, hydrodynamics Lift force Dynamic force Drag coefficient Natural convection Lift force Dynamic force Aerodynamics hydrodynamics Aerodynamics, hydrodynamics | v v V-6 | e-Text Main Menu | Textbook Table of Contents | Study Guide In this context, “similarity” is defined as Similarity: All relevant dimensionless parameters have the same values for the model & the prototype. Similarity generally includes three basic classifications in fluid mechanics: (1) Geometric similarity (2) Kinematic similarity (3) Dynamic similarity Geometric similarity In fluid mechanics, geometric similarity is defined as follows: Geometric Similarity All linear dimensions of the model are related to the corresponding dimensions of the prototype by a constant scale factor SFG Consider the following airfoil section (Fig. 5.4): Fig. 5.4 Geometric Similarity in Model Testing For this case, geometric similarity requires the following: | v v V-7 | e-Text Main Menu | Textbook Table of Contents | Study Guide rm L m Wm = = =⋅ ⋅ ⋅ rp L p Wp SFG = In addition, in geometric similarity, All angles are preserved. All flow directions are preserved. Orientation with respect to the surroundings must be same for the model and the prototype, ie., Angle of attack )m = angle of attack )p Kinematic Similarity In fluid mechanics, kinematic similarity is defined as follows: Kinematic Similarity The velocities at 'corresponding' points on the model & prototype are in the same direction and differ by a constant scale factor SFk. Therefore, the flows must have similar streamline pattterns Flow regimes must be the same. These conditions are demonstrated for two flow conditions, as shown in the following kinematically similar flows (Fig. 5.6). Fig. 5.6a Kinematically Similar Low Speed Flows | v v V-8 | e-Text Main Menu | Textbook Table of Contents | Study Guide Fig. 5.6b Kinematically Similar Free Surface Flows The conditions of kinematic similarity are generally met automatically when geometric and dynamic similarity conditions are satisfied. Dynamic Similarity In fluid mechanics, dynamic similarity is typically defined as follows: Dynamic Similarity This is basically met if model and prototype forces differ by a constant scale factor at similar points. This is illustrated in the following figure for flow through a sluice gate (Fig. 5.7). Fig. 5.7 Dynamic Similarity for Flow through a Sluice Gate | v v V-9 | e-Text Main Menu | Textbook Table of Contents | Study Guide This is generally met for the following conditions: 1. Compressible flows: model & prototype Re, Ma, are equal Rem = Rep, Mam= Map , γm = γp 2. Incompressible flows a. No free surface Rem = Rep b. Flow with a free surface Rem = Rep , Frm = Frp Note: The parameters being considered, e.g., velocity, density, viscosity, diameter, length, etc., generally relate to the flow, geometry, and fluid characteristics of the problem and are considered to be independent variables for the subject problem. The result of achieving similarity by the above means is that relevant non dimensional dependent variables, e.g., CD, Cp, Cf, or Nu, etc., are then equal for both the model and prototype. This result would then indicate how the relevant dependent results, e.g. drag force, pressure forces, viscous forces, are to be scaled for the model to the prototype. Equality of the relevant non-dimensional independent variables Re, Ma, x/L, etc., indicates how the various independent variables of importance should be scaled. | v v V-10 | e-Text Main Menu | Textbook Table of Contents | Study Guide An example of this scaling is shown as follows: Example: The drag on a sonar transducer prototype is to be predicted based on the following wind tunnel model data and prototype data requirements. Determine the model test velocity Vm necessary to achieve similarity and the expected prototype force Fp based on the model wind tunnel test results. Given: Prototype Model sphere sphere D 1 ft 6 in V 5 knots unknown? F ? 5.58 lbf ρ 1.98 ν slugs ft 2 -5 ft 1.4 *10 0.00238 3 slugs ft 1.56 * 10-4 ft s 3 2 s From dimensional analysis: C D = f{ R e } F D 2 = f VD υ ρ V2 For the prototype, the actual operating velocity and Reynolds number are: Prototype: na ⋅ mi ft 1hr ft na ⋅ mi ft 1hr ft 6080 6080 = 8.44 Vp = 5 = 8.44 hr na ⋅ mi 3600 s s hr na ⋅ mi 3600 s s V-11 | v v Vp = 5 | e-Text Main Menu | Textbook Table of Contents | Study Guide Re = p 8. 44 ft * 1ft 5 s = = 6. 03 * 10 2 1. 4 *10 −5 ft s VD υ p Equality of Reynolds number then yields the required model test velocity of Re = Re = m VD p υ m = > Vm = 188 ft s Based on actual test results for the model, i.e. measured Fm, equality of model and prototype drag coefficients yields ∴ C D = C D = > Fp = Fm p m ρp 2 Vp ρm 2 Vm D 2 m D2 p = 37. 4 lb f Note: All fluid dynamic flows and resulting flow characteristics are not Re dependent. Example: The drag coefficient for bluff bodies with a fixed point of separation; e.g., radar antennae, generally have a constant, fixed number for CD which is not a function of Re. CD = const ≠ f (R ) . e | v v V-12 | e-Text Main Menu | Textbook Table of Contents | Study Guide ...
View Full Document

This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

Ask a homework question - tutors are online