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FLUID MECHANICS DIMENSIONAL ANALYSIS 4Th Semester Mechanical Engineering

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FLUID MECHANICS DIMENSIONAL ANALYSIS 4Th Semester Mechanical Engineering FLUID MECHANICS DIMENSIONAL ANALYSIS 4th Semester Mechanical Engineering Prof. M. K. SINHA Mechanical Engineering Department N I T Jamshedpur 831014 Definition Dimensional analysis is a means of simplifying a physical problem by appealing to dimensional homogeneity to reduce the number of relevant variables. It is useful for: • presenting and interpreting experimental data; • attacking problems not amenable to a direct theoretical solution; • checking equations; • establishing the relative importance of particular physical phenomena; • physical modelling. Dimensions • A dimension is the type of measure by which something is quantified. • A unit is a means of fitting a numerical value to that quantity. • SI units are preferred in scientific and engineering works. FUNDAMENTAL DIMENSIONS In fluid mechanics the fundamental or primary dimensions, together with their SI units are: • Mass M (kilogram, kg) • Length L (metre, m) • Time T (second, s) • Temperature Θ (kelvin, K) Additional dimensions In physics additional dimensions, together with their SI units are: • Electric current I (ampere, A) • Luminous intensity C (candela, cd) • Amount of substance n (mole, mol) Derived dimensions Quantity Symbol/s Dimensions Geometry Area A L2 Volume V L3 Second moment of area I L4 Kinematics Velocity U LT-1 Angle θ none Angular velocity ω T-1 Quantity of flow Q L3T-1 Mass flow rate m. MT-1 Derived dimensions Quantity Symbol/s Dimensions Dynamics Force F MLT-2 Moment ,Torque T ML2T-2 Energy, Work, Heat E, W ML2T-2 Power P ML2T-3 Pressure, Stress P, 훕 ML-1T-2 Fluid properties Density ρ ML-3 Viscosity µ ML-1T-1 Kinematic viscosity ν L2T-1 Surface tension σ MT-2 Thermal conductivity k MLT-3 Θ-1 2 -2 -1 Specific heat cp, cv L T Θ Bulk modulus K ML-1T-2 Alternative choices for fundamental dimensions The choice of fundamental dimensions is not unique. It is not uncommon and it may sometimes be more convenient to choose force F as a fundamental dimension rather than Mass M, and have a {FLT} system rather than {MLT}. Principle of Dimensional Homogeneity • It states that all terms in an equation must have the same dimensions, together with the equation holds its physical meaning. • If the terms do not have the same dimension, then the equation is not correct. • Dimensional homogeneity is also a useful tool for checking formulae. • But dimensional analysis cannot determine numerical factors such as it cannot distinguish between ½ gt2 and gt2 Buckingham's Pi Theorem • If a problem involves n physical variables(v1, v2, ....vn) and m independent dimensions(M, L, T or F, L, T or such) then it can be reduced to a relationship between n – m non-dimensional parameters (Π1 , ..., Π n-m). • In other words, the equation expressing the phenomenon as a function of the physical variables f(v1, v2, v3,......, vn) = 0 can be substituted by the following equation expressing it as a function φ of a smaller number of non-dimensional groups: φ(Π1 ,π2 ..., Π n-m ) = 0 This is called Buckingham’s Pi theorem. Construction of non-dimensional πs These non-dimensional πs can be constructed as follows: (i) Choose m dimensionally-distinct variables to act as scales. In fluid mechanics it is common to choose a geometric quantity (e.g. a length), a kinematic quantity (e.g. a velocity) and a property of the fluid (e.g. density) or select independent variables which do not form a π themselves (ii) For each of the n – m remaining variable construct a non- dimensional of the form a b c Π = (variable)(scale1 ) (scale2 ) (scale3 ) ....... where a, b, c, ... are chosen so as to make each π non-dimensional. The power of physical variables(a, b, c,...)in each π group are determined algebraically by the condition that the powers of each fundamental dimension must sum to zero. Application Example. Obtain an expression in non-dimensional form for the pressure gradient(dp/dx) in a horizontal pipe with flow velocity V, pipe diameter D, fluid density ρ, fluid viscosity µ and pipe wall roughness e . Solution: Step 1. Identify the physical variables. dp/dx, ρ, V, D, e, µ Step 2. Write down dimensions [dp/dx] = ML-2T-2 [ρ] = ML-3 [ V] = LT-1 [D] = L Application [e] = L [µ] = ML-1T-1 Step 3. Find number of non-dimensional groups. Number of physical variables n= 6 Number of fundamental dimensions m = 3 (M, L, T) Number of non-dimensional groups(πs): n – m = 3 Step 4. Select dimensionally independent scaling variables(m = 3) Such as geometric (D), kinematics (V), fluid property(ρ). Application Step 5. Create first pi group by combining(D,V, ρ) with dp/dx. a b c π1 = dp/dx D V ρ Taking the dimensions of both sides: M0L0T0 =( ML-2T-2 ) (L)a (LT-1)b (ML-3)c Equate powers of fundamental dimensions: M: 0 = 1+ c L: 0 = -2 + a + b – 3c T: 0 = -2 – b Solving algebraically to obtain a = 1, b = -2, c = -1 -2 -1 Hence, π1 = dp/dx D V ρ = Application a b c Similarly, π2 = e D V ρ In terms of dimensions: M0L0T0 = L (L)a (LT-1)b (ML-3)c Equating exponents: M: 0 = c L: 0 = 1 + a +b + c T: 0 = -b Solving algebraically to obtain a = - 1, b = o, c = 0 -1 π2 = eD = e/D Application a b c Π3 = µ D V ρ In terms of dimensions: M0L0T0 = (ML-1T-1 ) (L)a (LT-1)b (ML-3)c Equating exponents: M:0 = 1 + c L: 0 = -1 + a +b – 3c T: 0 = -1 –b Solving algebraically to obtain a = -1, b = -1, c = -1 Hence, Π3 = µ/(ρVD) Application Therefore, from pi theorem, the following functional relationship is obtain: π1 = f(π2 π3 ) and With some modifications the non-dimensional relationship then becomes Similarity and Model Testing Flow conditions for a model test are completely similar if all relevant dimensionless parameters have the same corresponding values for model and prototype . ( π1)m = (π1)p , ( π2)m = (π2)p .......... • Enables extrapolation from model to full scale • However, complete similarity usually not possible • Therefore, often it is necessary to use Re, or Fr, or Ma scaling, i.e., select most important Π and accommodate others as best possible . Types of Similarity • Geometric Similarity (similar length scales): A model and prototype are geometrically similar if and only if all body dimensions in all three coordinates have the same linear-scale ratios. • Kinematic Similarity (similar length and time scales): The motions of two systems are kinematically similar if homologous (same relative position) particles lie at homologous points at homologous times . • Dynamic Similarity (similar length, time and force (or mass) scales): in addition to the requirements for kinematic similarity the model and prototype forces must be in a constant ratio . Example Example. A prototype gate valve which will control the flow in a pipe system conveying paraffin is to be studied in a model. Perform dimensional analysis to obtain the relevant nondimensional groups. A 1/5 scale model is built to determine the pressure drop across the valve with water as the working fluid. (a) For a particular opening, when the velocity of paraffin in the prototype is 3.0 m s–1 what should be the velocity of water in the model for dynamic similarity? (b) What is the scale ratio of the quantity of flow? (c) Find the pressure drop in the prototype if it is 60 kPa in the model. (The density and viscosity of paraffin are 800 kg m–3 and 0.002 kg m–1 s–1 respectively. Take the kinematic viscosity of water as 1.12x10-6 m2s-1 ). Solution • The pressure drop is expected to depend upon the gate opening h, the overall depth H, the velocity V, density ρ and viscosity µ . • Required variables with their respective dimensions [∆ p] = ML–1T–2 [h] = L [H] = L [V] = LT–1 [ρ ] = ML–3 [µ ] = ML–1T–1 Number of variables: n = 6. Number of independent dimensions: m = 3 (M, L, T) Number of non-dimensional groups: n – m = 3 Solution Select scaling variables: Geometric (H), kinematic (V), fluid property (ρ). Form dimensionless groups by non-dimensionalising the remaining variables ∆p, h and µ . a b c 0 0 0 -1 -2 a -1 b -3 c π1 = ∆p H V ρ , M L T =( ML T ) (L) (LT ) (ML ) Equating exponents and solving algebraically yields a = 0, b = -2, c = -1, Similarly, π2 = h/H and Solution After some modifications in π3 we can write dimensional analysis in functional form (a) Dynamic similarity requires that all non-dimensional groups be the same in model and prototype; i.e. Solution From last, we have a velocity scale ratio Hence, (b)The scale ratio of the quantity of flow is Solution (c) For pressure drop Non-dimensional Groups in Fluid Mechanics Name Non-dimensional Interpretation Applications groups Reyonlds Number Viscous flows Re Froude number Free surface flows Fr Euler number Pressure or pressure Eu difference flows Cauchy number Compressible flows Ca Mach number Compressible flows Ma Strouhal number Oscillating flows St Weber number Free surface flows We Thank you • Please visit the following site for books https://drive.google.com/open?id=0BwUR7uFnfZiuandwYWpETTR1aFk .
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