UNIT III DIMENSIONAL ANALYSIS

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1 UNIT III DIMENSIONAL ANALYSIS INTRODUCTION Dimensional analysis is one of the most important mathematical tools in the study of fluid mechanics. It is a mathematical technique, which makes use of the study of dimensions as an aid to the solution of many engineering problems. The main advantage of a dimensional analysis of a problem is that it reduces the number of variables in the problem by combining dimensional variables to form non-dimensional parameters. By far the simplest and most desirable method in the analysis of any fluid problem is that of direct mathematical solution. But, most problems in fluid mechanics such complex Phenomena that direct mathematical solution is limited to a few special cases. Especially for turbulent flow, there are so many variables involved in the differential equation of fluid motion that a direct mathematical solution is simply out of question. In these problems dimensional analysis can be used in obtaining a functional relationship among the various variables involved in terms of non-dimensional parameters. Dimensional analysis has been found useful in both analytical and experimental work in the study of fluid mechanics. Some of the uses are listed: 1) Checking the dimensional homogeneity of any equation of fluid motion. ) Deriving fluid mechanics equations expressed in terms of non-dimensional parameters to show the relative significance of each parameter. ) Planning tests and presenting experimental results in a systematic manner. 4) Analyzing complex flow phenomena by use of scale models (model similitude) DIMENSIONS AND DIMENSIONAL HOMOGENEITY Scientific reasoning in fluid mechanics is based quantitatively on concepts of such physical phenomena as length, time, velocity, acceleration, force, mass, momentum, energy, viscosity, and many other arbitrarily chosen entities, to each of which a unit of measurement has been assigned. These are all known as dimensions. Of course dimensions are of no use without a magnitude being attached. We must know more than that something has a length. It must also have a standardized unit - such as a meter, a foot, a yard etc.

2 Dimensions are properties which can be measured. Units are the standard elements we use to quantify these dimensions. In dimensional analysis we are only concerned with the nature of the dimension i.e. its quality not its quantity. The following common abbreviation is used: Length L Mass M Time T Force F Temperature θ In this module we are only concerned with L, M, T and F (not θ). We can represent all the physical properties we are interested in with L, T and one of M or F (F can be represented by a combination of LTM). These notes will always use the LTM combination. The following table (taken from earlier in the course) lists dimensions of some common physical quantities. For the purpose of obtaining a numerical solution, we adopt for computation the quantities in SI or MKS units. In a more general sense, however, it is desirable to adopt a consistent dimensional system composed of the smallest number of dimensions in terms of which all the physical entities may be expressed. The fundamental dimensions of mechanics are length [L], time [T], mass [M], and force [F], related by Newton s second law of motion, F ma. Base Quantity Length Mass Time Electric Temperature Amount of Substance Luminous Intensity Base Unit meter (m) kilogram (kg) second (s) Current ampere (A) Kelvin (K) mole (mol) candela cd The last two are not used in fluid mechanics and temperature is only used sometimes Quantity SI UNIT Dimension velocity m/s ms LT Acceleration m/s² ms LT Force N kgms MLT kgm/s² Energy or work Joule J kgm²s ML T Nm kgm²/s² Power Watt W Nms ML T

3 Nm/s kgm²s kgm²/s³ Pressure or stress Pascal p Nm ² ML T N/m² Kg/m/s² kgm s ² Density Kg/m³ kgm ML Specific weight Kg/m²/s² kgm s ML T Relative density Viscosity Surface tension A ratio No units Ns/m² Kg/ms N/m Kg/s² Nsm Kgm 1 s 1 kgs No dimension ML 1 T 1 Nm 1 MT Moment Nm Nm ML T Mass moment of inertia Kg m² kgm² ML² Torque Nm Nm ML T Quantity Symbol Dimensions Mass m M Length l L Time t T Temperature T θ Velocity u LT -1 Acceleration a LT - Momentum/Impulse mv MLT -1 Force F MLT - Energy - Work W ML T - Power P ML T - Moment of Force M ML T - Angular momentum - ML T -1 Angle η M 0 L 0 T 0 Angular Velocity ω T -1 Angular acceleration α T - Area A L

4 Volume V L First Moment of Area Ar L Second Moment of Area I L 4 Density ρ ML - Specific heat- Constant Pressure C p L T - θ -1 Quantity Symbol Dimensions Mass /Unit Area m/a ML - Mass moment ml ML Moment of Inertia I ML Pressure /Stress p /σ ML -1 T - Strain τ M 0 L 0 T 0 Elastic Modulus E ML -1 T - Flexural Rigidity EI ML T - Shear Modulus G ML -1 T - Torsional rigidity GJ ML T - Stiffness k MT - Angular stiffness T/η ML T - Flexibility 1/k M -1 T Vorticity - T -1 Circulation - L T -1 Viscosity µ ML -1 T -1 Kinematic Viscosity τ L T -1 Diffusivity - L T -1 Friction coefficient f /µ M 0 L 0 T 0 Restitution coefficient M 0 L 0 T 0 Specific heat- Constant volume C v L T - θ -1 Determine the dimensions of kinetic energy.

5 E k 1 mv g c m has dimensions of mass or [M] v has dimensions of length per time or [L] / [T] g c is dimensionless Use dimensional homogeneity: E k L [ M ] [ ] T Write down the basic dimensions of pressure p. Solution Pressure is defined as p Force/Area The S.I. unit of pressure is the Pascal which is the name for 1N/m² Since force is MLT - ² and area is L² then the basic dimensions of pressure are [ P ] ML 1 T When solving problems it is useful to use a notation to indicate the MLT dimensions of a quantity and in this case we would write [ P ] ML 1 T Principle of Homogeneity of Dimensions It states that if the dimensions of each term on both the sides of equation are same, then the physical quantity will be correct. Each term that is added or subtracted must have the same dimensions. Terms that are equal to each other must also have the same dimensions. Equations for which both of these statements are true are said to be dimensionally homogeneous. All valid equations must be dimensionally homogeneous, but not all dimensionally homogeneous equations are valid.the principles of dimensional analysis are developed from the principle of dimensional homogeneity which is self-evident. It is characteristic of physical equations that only like quantities, which is those systems having the same dimensions, are added or equated.

6 To check the dimensional correctness of a given physical relation Example, To check the correctness of v u + at, using dimensions Dimensional formula of final velocity v [LT -1 ] Dimensional formula of initial velocity u [LT -1 ] Dimensional formula of acceleration x time, at [LT - x T] [LT -1 ] Dimensions on both sides of each term are the same. Hence, the equation is dimensionally correct...methods OF DIMENSIONAL ANALYSIS..1.Rayleigh's Method This method is based on the fundamental principle of dimensional homogeneity of physical variables involved in a problem. Procedure- 1. The dependent variable is identified and expressed as a product of all the independent variables raised to an unknown integer exponent.. Equating the indices of n fundamental dimensions of the variables involved, n independent equations are obtained.. These n equations are solved to obtain the dimensionless groups. Example Let us illustrate this method by solving the pipe flow problem Step Here, the dependent variable p/l can be written as p l AV a D b c d ρ µ Where, A is a dimensionless constant. Step -----Inserting the dimensions of each variable in the above equation, we obtain,

7 ML T 1 a b ( ) ( ) ( ) c 1 1 L ML ( ML T ) d A LT Equating the indices of M, L, and T on both sides, we get, c + d 1 a + b - c - d - -a - d - step -----There are three equations and four unknowns. Solving these equations in terms of the unknown d, we have a - d b -d - 1 c 1- d Hence, we can be written p d d d AV D 1 1 h ρ l d µ p l AV ρ µ D h VD h ρ d Or pd h ρ l V µ A VDh ρ d The rise of liquid h in a capillary tube depends on tube diameter D, surface tension σ,specific weight of liquid ν and the contact angleθ.using Rayleigh s method,obtain an expression for h. Solution: The dimensions of each variable are as follows: Variables h D σ γ θ Dimensions L L MT 1 ML T M Capillary rise h can be expressed in terms of the independent variables as 0 0 L T 0

8 a b c h KD σ γ θ Where K is a dimensionless constant. The variables in equation can be expressed in terms of their fundamental dimensions as L K a [ ] [ ] b 1 L MT [ ML T ] c Equating the powers of M and L on both sides of the above equation, we get Power of M : b+c 0 b-c Power of L : a-c1 ac+1 Substituting the values of a and b in terms of c,we have h D h h D γ KD σd c c θ γ K θ σd γ Kφ θ σd... Buckingham's π Theorem c Assume a physical phenomenon is described by m number of independent variables like x 1, x, x,..., x n. The phenomenon may be expressed analytically by an implicit functional relationship of the controlling variables as f ( x x, x,..., ) 0 1, x n

9 Now if m be the number of fundamental dimensions like mass, length, time, temperature etc., involved in these m variables, then according to Buckingham's π theorem The phenomenon can be described in terms of (n - m) independent dimensionless groups like π 1,π,..., π n-m, where π terms, represent the dimensionless parameters and consist of different combinations of a number of dimensional variables out of the m independent variables defining the problem. Therefore the analytical version of the phenomenon given by above Equation can be reduced to ( π π, π,... ) 0 f π n m 1, According to Buckingham's π theorem This physically implies that the phenomenon which is basically described by m independent dimensional variables is ultimately controlled by (n-m) independent dimensionless parameters known as π terms. Steps for determining the π terms: List all the variables both independent and dependent involved in the phenomenon. Let m be the total number of variables involved in the phenomenon. List the fundamental dimensions of each of the variables. Let n be the total number of fundamental dimensions (mass, length, time, etc.) involved to represent all the variables. Express the problem in terms of the dimensionless parameters. Here the number of dimensionless parameters will be (n-m). Identify the repeating variables. The number of repeating variables should be same as the number of fundamental dimensions involved to represent all the variables. Form the π terms by combining the repeating variables with each of the remaining variables. Solve the dimensional equations to obtain the dimensionless parameters. Wrong choice of physical properties: If, when defining the problem, extra - unimportant - variables are introduced then extra л groups will be formed. They will play very little role influencing the physical behavior of the problem concerned and should be identified during experimental work. If an important / influential variable was missed then a л group would be missing. Experimental

10 analysis based on these results may miss significant behavioral changes. It is therefore, very important that the initial choice of variables is carried out with great care. Guideline for selecting repeating variables: The dependent variables should not be selected as repeating variables. They must include all of the fundamental dimensions. They must not form a π term by themselves. The efficiency η of a fan depends on the diameter D of the rotor, kinematic viscosity γ of the fluid, the angular velocity ω and the discharge Q. Using Buckingham s π theorem, show that the discharge Q through a centrifugal pump can be expressed as Q ω η ϕ ω, D D γ The dimensions of each variable are as follows: η D ω Q γ M L T L 1 T The number of variables describing the phenomena is n5 1 L T 1 L T The number of fundamental dimensions required to describe the variables is m According to Buckingham's π theorem, the number of dimensionless л terms is n-m 5- π terms Now by using Buckingham's π theorem the problem can be expressed as f ( π π, π ) 0 1, Using D and ω as repeating variables, π terms can be written as

11 π η π π 1 D D a a b ω Q 1 b ω γ First л term: Second л term: π 1 η D a 1 b π ω Q The variables in the equation can be expressed in terms of their fundamental dimensions as a 1 b 1 M L T [ L] [ T ] [ L T ] Equating the powers of M,L and T on both sides,we have Power of T: b 1 0 Power of L: b a a Substituting these values in D a1 b 1 Q π ω Q we have π D ω Q π ωd Third л term: a b π D ω γ The variables in the equation fundamental dimensions as M L T a 1 b 1 [ L] [ T ] [ L ] T π a b D ω γ can be expressed in terms of their Equating the powers of M,L and T on both sides, we have Power of T:

12 b b Power of L: a a Substituting these values into Substituting the values of 1 π a b 1 γ D ω γ, we have π D ω γ π ωd π π and in ( π π, π ) 0 Q f η, ωd 1, f we have Q η φ ωd γ, 0 ωd ωd, γ The discharge Q through an orifice is a function of the diameter d, the pressure difference p, the density ρ, and the viscosity µ, show that, where φ is some unknown function. Let s write out the dimensions of the variables ρ d p:(force/area) µ Q ML - L ML -1 T - ML -1 T -1 1 L T 5 variables involved in the problem: d, p, µ, ρ, and Q. Choose the three recurring (governing) variables; d, ρ, and Q. From Buckingham's π theorem we have m-n 5 - non-dimensional groups. From Buckingham's π theorem we have m-n 5 - non-dimensional groups.

13 f f ( Q, d, ρ, µ, p) ( π, π ) π Q a1 π Q d a d b1 ρ b c1 ρ First term π 1 : M L T c 0 µ p 1 a1 b1 c1 1 1 ( L T ) ( L) ( ML ) ML T Power of M: 0 c c 1-1 Power of L: 0 a 1 + b 1 - c a 1 + b 1 Power of T : 0 -a 1-1 a 1-1 b 1 1 Substituting the values of a1, b1, c1 in 1 1 c1 π Q a d b ρ µ we get 1 π Q 1 d ρ dµ µ ρq Second л term π : Note p is a pressure (force/area) with dimensions ML -1 T - ) Power of M] 0 c + 1 c -1 power of L] 0 a + b - c a + b Power of T] 0 -a - a -

14 b 4 So the physical situation is described by this function of non-dimensional numbers, The question wants us to show: Take the reciprocal of square root of π : Convert π 1 by multiplying by this new group, π a then we can say

15 The thrust force, F generated by a propeller is found to depend on the following parameters: diameter D, forward velocity u, density ρ, viscosity µ and rotational speed N. determine the dimensionless parameters to correlate the phenomenon. The influencing parameters with dimensions are listed below using MLT set. S.No. Parameters Unit Dimensions 1 Thrust force, F N ML/T Diameter, D m L Forward velocity m/s L/T 4 Density ρ kg/m M/L 5 Viscosity µ kg/ms M/LT 6 Rotational speed, N 1/s 1/T There are 6 variables and three dimensions. So three π terms can be obtained. Choosing D, u and ρ as repeating variables, a a b c ML L b M Let π 1 Fu D ρ orm L T L a T T L Comparing indices of M,L and T 1+c0,1+a+b-c0,--a0 a -, b -, c -1 Substituting the values of a,b,c c c π 1 u F D ρ Thrustforce Inertiaforce Letπ a b c µ u D ρ orm 0 0 L T 0 M LT L T a a L b M L c c

16 Comparing the indices M,L and T 1+c0,-1+a+b-c0,-1-a0 a 1, b 1, c 1 Substituting the values of a, b and c π µ ρud or ρud µ Inertiaforcwe Viscousforce Letπ 1 L T T M L a c a b c b Nu D orm L T L a c ρ Comparing the indices of M,L and T C0, a+b-c0, -1-a0 a -1,b1 Substituting the values of a,b and c π ND u rotationalspeed Forwardspeed udρ ND Flu D ρ f, µ u... Limitations of dimensional analysis: 1. Dimensional analysis does not give any due regarding the selection of variables.. The complete information is not provided by dimensional analysis.. The values of coefficient and the nature of function can be obtained only by experiments or from mathematical analysis... DISCUSSION ON DIMENSIONLESS PARAMETERS Forces encountered in flowing fluids include those due to inertia, viscosity, pressure, gravity, surface tension and compressibility. These forces can be written as dv dv Inertia force m. a ρ V V V V L dt ρ ds ρ

17 du Viscous force τ A µ A µ V L dy Pressure force ( p) A ( p) L Gravity force m g g ρ L Surface tension force σ L E A E L E Compressibility force v v ; where v is the Bulk modulus The ratio of any two forces will be dimensionless. Inertia forces are very important in fluid mechanics problems. So, the ratio of the inertia force to each of the other forces listed above leads to fundamental dimensionless groups. These are, 1. Reynolds number. Froude s number. Euler s number 4. Weber s number 5. Mach number..1. Reynolds number: It is the ratio of the inertia force of a flowing fluid and the viscous force of the fluid. Mathematically, R e ρvl VL (1) µ ν where V is the velocity of the flow, L is the characteristics length, ρ, µ and ν are the density, dynamic viscosity and kinematic viscosity of the fluid respectively. If R e is very small, there is an indication that the viscous forces are dominant compared to inertia forces. Such types of flows are commonly referred to as creeping/viscous flows. Conversely, for large R e, viscous forces are small compared to inertial effects and flow problems are characterized as inviscid analysis. This number is also used to study the transition between the laminar and turbulent flow regimes.... Froude s number: It is the square root of the ratio of inertia force of a flowing fluid to the gravity.

18 Fr F F r r FI F ρl V ρl g V G gl Mathematically, it is written as, V Fr () g. L where V is the velocity of the flow, L is the characteristics length descriptive of the flow field and g is the acceleration due to gravity. This number is very much significant for flows with free surface effects such as in case of open-channel flow. In such types of flows, the characteristics length is the depth of water. F r less than unity indicates subcritical flow and values greater than unity indicate super-critical flow. It is also used to study the flow of water around ships with resulting wave motion. It is important for flows that are influenced by gravity force such as flow over notches and weirs, flow over spillway of dams, flow through open channels; etc.it is useful in many engineering applications such as design of hydraulic structure, design of ship, etc.... Euler s number: It is defined as the square root of the ratio of the inertia force of a flowing fluid to the pressure force. Mathematically written as, where E u p () 1 ρv p is the difference in local pressure and free stream pressure, V is the velocity of the flow, ρ is the density of the fluid. The denominator in Eq. () is called dynamic pressure. E u is the ratio of pressure force to inertia force and it is also called as the pressure coefficient C p. In the study of cavitations phenomena, similar expressions are

19 used where p the difference in liquid stream pressure and liquid-vapour pressure is. The dimensional parameter is called cavitation number. Euler number is important for flows in which the pressure drop is important such as flow through pipes, ducts, pressure rise due to sudden closures of valves, etc...4. Weber s number It is defined as the square root of the ratio of the inertia force of the flowing fluid to the surface tension force. We We We FI F s ρl V σl V σ ρl Weber number is important for flows in which the surface tension is important such as flows with an interface, flow through microchannels, flow of blood in arteries and veins...5. Mach number It is defined as the square root of the ratio of the inertia force of the following fluid to the elastic force. Bulk modulus elasticity of fluid Ma Ma Ma F F wherec ρl V EL V E I E ρ V If the flow is isentropic then, Ma C where, E Bulk modulus elasticity of fluid E ρ Mach number is important for flows in which the compressibility is important such as flow of air past high-speed aircraft,projectiles,missiles,etc.if the flow velocity is greater than the 0. times the sonic velocity in that medium then mach number plays an important role to decide the types of flow that is whether the flow is subsonic or supersonic.

20 In addition, there are few other dimensionless numbers that are of importance in fluid mechanics. They are listed table.4. Parameter Prandtl number Eckert number Specific heat ratio Roughness ratio P r Mathematical expression µ c p k Qualitative definition Dissipation Conduction V Ec Kinetic energy cp T0 Enthalpy c γ c ε L p v Enthalpy Internal energy Wall roughness Body length Importance Heat convection Dissipation Compressible flow Turbulent rough walls Grashof number Temperature ratio Pressure coefficient Lift coefficient Drag coefficient G r T w T C C C 0 p L D β ( T ) g L ρ µ p p ( 1 ) ρ V L ( 1 ) A ρv D ( 1 ) A ρ V Buoyancy Viscosity Wall temperature Stream temperature Static pressure Dynamic pressure Lift force Dynamic force Drag force Dynamic force Natural convection Heat transfer Hydrodynamics, Aerodynamics Hydrodynamics, Aerodynamics Hydrodynamics, Aerodynamics.4. Models & Hydraulic Similitude Similitude is defined as similarities between the model and prototype in all respects. Prototype: A prototype is the first full size or 'pilot' model of a device or process. Sometimes the term is used to mean the first complete item of what subsequently becomes a production series. 1. Is fully functional, but not fault-proof.. Is an actual version of the intended product?. Used for performance evaluation and further improvement of product. 4. Contains complete interior and exterior. 5. Is relatively expensive to produce.

21 Model: A model is any arrangement of parts that demonstrates the way they work together. Scale is arbitrary. 1. Not necessarily functional (don't need to work).. Can be to any scale (usually smaller but can also be of the original size or bigger).. Used for Display or/and [Visual] Demonstration of product. 4. May consist of only the exterior of the object/product it replicates. 5. Relatively cheap to manufacture. Type of Hydraulic Similarities: These similarities are of three types. In order that the relationships determined for a model can be applied to a real life application (prototype) there has to be a physical similarity between the parameters involved in each one. The two systems are said to be physically similar in respect to specified physical quantities when the ratio of the corresponding magnitudes of these quantities between the two systems is everywhere the same. Within the general term physical similarity there are a number of types of similarity some of which are listed below. Geometric similarity: This is basically the similarity of shape. Any length of one system is related to that of another system by a ratio which is normally called the scale. All parts of the scale model of a car should be in direct scale to the full scale item if it is truly geometrically similar. This should ideally include such features as the surface roughness. This does not include non dimensional features e.g. weight... For length scale ratio L L L mod el m Lr prototype L, where L r is the scale factor for length. p For area scale ratio Amod el L m L r prototype L p A Kinematic Similarity: This is basically the similarity of motion between model and prototype, which implies that the geometric similarity and similarity of time intervals. i.e ratios of length are fixed (L r ) and ratios of time intervals (t r ) are fixed. The velocities (ds/dt) of corresponding parts should also be in fixed ratios and the ratios of acceleration (dv/dt) are in ratio. For kinematic similarity, we must have velocity and acceleration as follows:

22 V V a a m1 p1 m1 p1 V V a a m p m p V V a a m p m p Geometric similarity is a necessary condition for kinematic similarity. Dynamic Similarity: This is the similarity of forces between model and prototype.the magnitude of forces at two similarly located points are in a fixed ratio. For systems involving fluids the forces may be due to viscosity, gravitation, pressure, inertia, surface tension, elasticity etc etc... It is generally accepted in fluid mechanics that the ratio of inertia forces is the most useful ratio. ( F1 ) ( F ) 1 m p ( Fv ) ( F ) ( Fp ) ( F ) ( F1 ),( Fv ) and( Fp ) m m m v m p p m p are the inertia,viscous and pressure forces at a point in the model. Dynamic similarity involving flow with viscous forces: They are numerous instances of fluid flow affected only by viscous pressure and inertia forces. A fluid flowing in a full pipe is such a case. For dynamic similarity the ratio of magnitude of any two forces must be the same at corresponding points (in a steady flow situation). The ratio of inertia force to net viscous force is chosen for review. The inertia force is the mass x acceleration. [Density (ρ ) x volume ( l ) x acceleration ( u / l )]. Note: The acceleration is chosen to be the characteristic velocity ( u ) divide by a particular time interval ( l/ u ) u / l. The magnitude of the inertia forces are therefore proportional ( ρ.l )( u / l ) ρ l u The magnitude of the shear stress resulting from viscosity is the product of the viscosity (µ )and the rate of shear ( u / l ) acting over an area proportional an area l. This is therefore proportional to ( µ ) ( u / l ) x ( l ) ( µ u l ) The ratio of inertia forces to viscous forces is therefore as follows:

23 This ratio is very important in fluid mechanics, mainly for problems involving flowing fluids, and it is called Reynolds number. The ratio for dynamic similarity between two flows past geometrically similar boundaries and affected by only viscous and inertia forces is the same if the fluids have the same Reynolds number. In the UK for pipe flow studies the characteristic length (l) is the diameter (D) and the characteristic velocity u is chosen as the mean velocity. Dynamic similarity involving flow with gravity forces... When considering forces with free surfaces e.g. flows over weirs, channel flows, or surface motion around ships, the most significant relationships is the ratio between the gravity forces and the inertia forces. These are summarized below. This ratio u /( lg ) 1/ is called the Froude number. Dynamic similarity exists between two flows which involve fluids subject to only gravity and inertial forces if the Froude number, based on corresponding velocities and lengths, is the same for both fluids. MODEL LAWS: The laws on which the models are designed for dynamic similarity are called model or similarity laws. They are 1. Reynolds model law. Froude model law. Euler model law 4. Weber model law 5. Mach model law.4.1. REYNOLDS MODEL LAW: Viscous force is predominant Example: Motion of air planes Flow of incompressible fluid in closed pipes Motion of submarines completely under water

24 r represent scale ratios Time scale ratio T r Acceleration scale ratio a r Force scale ratio f r m r a r Discharge scale ratio.4.. Froude Model Law: Gravitational force can be considered to be the only predominant force. Can be applied for: Free surface flows Flow of jet from an orifice Where waves are likely to be formed Fluids of diff mass densities. TIME SCALE RATIO T r T m T P Acceleration scale ratio: a V/t

25 Discharge scale ratio Q r : Q AV Force scale ratio: F ma Pressure intensity scale ratio: Energy scale ratio: Momentum scale ratio M r : Momentum mv Torque scale ratio TFL.4..EULER MODEL LAW: Pressure forces are the controlling forces Hence.4.4. WEBER MODEL LAW: This model is used when surface tension force predominates in addition to inertia forces.

26 ρ m ρ P for same fluid This model is generally used for Capillary rise in narrow passages Capillary waves in channels Capillary movement of water in soil...5. Mach Model law: Elastic forces are more predominant. Influences in the following: 1. Aero dynamic testing. Hydraulic model testing. Under water testing Density 1. An oil of sp gravity 0.9 & viscosity 0.0 poise is to be transported at the rate of500 l/s through a pipe of Tests are conducted on a 1 cm pipe using water at 0 C. If viscosity water at 0 C is 0.01 poise, find (i) velocity of flow in model (ii) Rate of flow in the model. Solution: Sp gravity of oil S P 0.9 Viscosity of oil µ P 0.0 Poise 0.0 x 0.00 Rate of flow Q 500 l/s.5 m /s D p 1. m D m 1 cm 0.1 m Viscosity of water at 0 C µm 0.01 poise 0.01 x 1/ Ns/m. (i) Velocity of flow in model V m :

27 Sp gravity S (ii) V m.067 X m/s. In a geometrically similar model of spill way the discharge 1 m length is 0. m /s. If the scale of the model is find the discharge /m run of the prototype. Solution: L r 6 By Froude law, Discharge/ meter length for model, 16. A 7. m high & 15 m long spillway discharges 94 m/s discharges under a head of.0m. If 1:9 scale model of this spillway is to be constructed, det model dimensions, head over spill way model & model discharge. If model experiences a force of 7500 N, determine force on the prototype.

28 Solution: Height h p 7.m L p 15m Q p 94 m /s Head H p.0m L r 9 F m 7500 N Model dimension (h m,l m ): Head over model H m : Discharge through model Q m : Force on prototype F P : F P F M X N 4. In an aero plane model of size 1/10 of its prototype the pressure drop is 7.5 kn/m. The model is tested in water. Find corresponding Pressure drop in prototype. Solution:

29 Water is flowing through a pipe of ϕ 0 cm at a velocity of 4 m/s. Find velocity of oil flowing in another pipe of ϕ 10cm, if condition of dynamic similarity is satisfied between pipes. µ of water & oil is 0.01 poise& 0.05 poise sp gr of oil is 0.8. Given data: Pipe E 1 : d 1 0 cm 0. m V 1 4m/s µ Poise ρ Kg/m Pipe E : d 10 cm 0.1 m V 1? Solution: µ 0.05 Poise Sp. Gr. Of oil 0.8 Sp gr ρ 0.8 X Kg/m Pipes are dynamically similar, 6. An underwater missile, diameter m and length 10m is tested in a water tunnel to determine the forces acting on the real prototype. A 1/0 th scale model is to be used.

30 If the maximum allowable speed of the prototype missile is 10 m/s, what should be the speed of the water in the tunnel to achieve dynamic similarity? For dynamic similarity the Reynolds number of the model and prototype must be equal: So the model velocity should be As both the model and prototype are in water then, m p and m p so Note that this is a very high velocity. This is one reason why model tests are not always done at exactly equal Reynolds numbers. Some relaxation of the equivalence requirement is often acceptable when the Reynolds number is high. Using a wind tunnel may have been possible in this example. If this were the case then the appropriate values of the and ratios need to be used in the above equation. 7. A model aero plane is built at 1/10 scale and is to be tested in a wind tunnel operating at a pressure of 0 times atmospheric. The aero plane will fly at 500km/h. At what speed should the wind tunnel operate to give dynamic similarity between the model and prototype? If the drag measure on the model is 7.5 N what will be the drag on the plane? The equation for resistance on a body moving through air: For dynamic similarity Re m Re p, so

31 The value of density does not change much with pressure so ρ ρ The equation of state for an ideal gas is p ρ RT. As temperature is the same then the density of the air in the model can be obtained from m p So the model velocity is found to be The ratio of forces is found from So the drag force on the prototype will be.5. Types of Models: The hydraulic models, in general, are classified into the following two broad categories: Undistorted models Distorted models.5.1. Undistorted models: An undistorted model is one that is geometrically similar to its prototype. The conditions of similitude are completely satisfied for such models; hence the results obtained from the model tests are easily used to predict the performance of prototype body. In such

32 models the design and construction of the model and the interpretation of the model results are simpler..5.. Distorted models: A distorted model is one which is not geometrically similar to its prototype. In such a model different scale ratios for the linear dimensions are adopted. For example in case of a wide and shallow river it is not possible to obtain the same horizontal and vertical scale ratios, however, if these ratios are taken to be same then because of the small depth of flow the vertical dimensions of the model will become too less in comparison to its horizontal length. Thus in distorted models the plan form is geometrically similar to that of prototype but the cross section is distorted. A distorted model may have the following distortions: Geometric distortion Material distortion Distortion of hydraulic quantities Typical examples for which distorted models are required to be prepared are: rivers, dams across very wide rivers, harbours, etc. Merits of distorted Models: The necessary hydraulic similitude is obtained. Depth of flow is increased affording precise measurements. Height of waves is increased affording precise measurements. Viscous effects which are practically absent in the prototype can be practically eliminated in the model, for instance, by increasing the bed slope in an otherwise geometrically similar model, the velocity can be increased, thus decreasing viscous effects. Demerits of a distorted model: Pressure and velocity distributions are not truly reproduced Difficult to extrapolate and interpolate results obtained from distorted models The observer experiences an unfavorable psychological effect.5.. Scale effect in model: By model testing it is not possible to predict the exact behavior of the prototype. The behavior of the prototype as predicted by two models with different scale ratios is generally not the same. Such a discrepancy or difference in the prediction of behavior of the prototype is termed as scale effect. The magnitude of scale effect is affected by the type of the problem and the scale ratio used for the performance of experiments on models. The scale effect can be positive and negative and when applied to the results accordingly, the corrected results then hold good for prototype.

33 .6. Applications of Dimensional Parameters: ä Pipe Flow ä Pump characterization ä Model Studies and Similitude ä dams: spillways, turbines, tunnels ä harbors ä rivers ä ships Two marks questions with answers 1. What is dimensional analysis? It is mathematical technique used for designing and conducting the model test.. What are the fundamental dimensions? The fundamental dimensions are L, M, T length, mass, time.. What are secondary or derived quantities? Secondary or derived quantities are those posses more than one fundamental dimension.e.g. Velocity distance/ time (L/T) 4. What are dimensions of angular velocity? Angular velocity (angle covered in radians/time) (1/T)T What are the dimensions of discharge? Discharge Area* Velocity L²*L/TL³/TL³T Determine the dimension for kinematic viscosity. Kinematics viscosity µ/ ρ Where µ τ / (du/dy) Μ ((Force /area)/ (1/τ)) Μ ((mass* acceleration)/ (area*time)) Μ (M*L/T )/ (L *(1/T) ML -1 T -1 Ρ (MASS/VOLUME) M/L Kinematic viscosity (ML -1 T -1 )/ (ML - ) L T Determine the unit of specific weight? Specific weight (weight/volume) (force/volume) (M*LT - )/L ML - T What is meant by dimensional homogeneity? It means the dimensions of each term an both sides in an equation is equal. 9. What is an unit?

34 The quantitative measure of a physical quantity is done in terms of a unit. 10. What are the various kinds of units? 1) MKS system, ). CGS systems,).si systems. 11. Give an example of dimensional homogeneity? V gh LT -1 (L/T - )*LL/T LT -1 LT Check whether the following equation is dimensionally homogenous? h f FL/D*v /g LHS L RHSL/L((L /T )/(L/T ))L The equation is dimensionally homogeneous 1. What are the two methods used in dimensional analysis? 1. Rayleigh s method. Buckingham method 14. What is Rayleigh s method? It is a method used for determining the expression for a variable, which depends upon maximum or 4 variable only. 15. Give the mathematical expression for Raleigh s method. Xf(x 1, x, x ) Also x k x a 1 x b 1 x c When is Buckingham s π method used? When the variables are more than the number of fundamental dimensions (LMT) the Raleigh s method is more laborious. Then the Buckingham s π method is used. 17. How is Buckingham s method mathematically expressed? x 1 ( x 1, x, x n) 18. What are the variables associated with the geometric property? The variables associated with the geometric properties are a. length b. d c. Height etc. 19. What are the variables associated with flow property? Variables associated with flow properties are 1. Velocity. Acceleration 0. What are the variables with fluid property? 1. viscosity. density 1. What is a model? Model is a small-scale replica of the actual structure or machine.. What is model analysis? The study of models of actual machines is called model machine.. What are the advantages of dimensional and model analysis?

35 1. the performance of the machine can be predicted in advance. With the dimensional analysis, a relationship between the variables influencing a flow problem is obtained.. Merits of alternative designs can be predicted. 4. Useful information about the performance of prototype can be obtained in advance. 4. What is similitude? It is defined as the similarity between the model and its prototype in every respect. 5. What are the various types of similarities? The types of similarities are 1. geometric. kinematic. dynamic 6. When is a geometric similarity said to exist? When the ratio of all corresponding linear dimension in the model and prototype are equal a geometric similarity exists. 7. What is kinematic similarity? It means there is a similarity of motion between the model and prototype. 8. What is dynamic similarity? Dynamic similarity means the similarity of forces between the model and prototype. 9. What are dimensionless numbers? They are numbers, which are obtained by dividing the inertia force by viscous force or gravity force or pressure force or surface tension force or elastic force. 0. Why is it called dimensionless? As the ratio is between one force to another it will be dimensionless, hence it is called dimensionless. 1. Name some important dimensionless numbers. 1. Reynolds number. Froude s number. Euler s number 4. Weber s number 5. mach s number. What is Reynolds number? It is the ratio of the inertia force of a flowing fluid and the viscous force of the fluid.. What is Froude s number? It is the square root of the ratio of inertia force of a flowing fluid to the gravity. 4. What is Euler s number?

36 It is defined as the square root of the ratio of the inertia force of a flowing fluid to the pressure force. 5. What is Weber s number? It is defined as the square root of the ratio of the inertia force of the flowing fluid to the surface tension force. 6. What is mach s number? It is defined as the square root of the ratio of the inertia force of the following fluid to the elastic force. 7. What are model laws or similarity laws? The laws on which the models are designed for dynamic similarity are called model laws. 8. Which are the models based on Reynolds number? 1. pipe flow. Resistance experienced by airplanes, submarines and fully submerged bodies. 9. Give some applications of Froude s model law. 1. Free surface flows such as weirs, sluices, channels etc.. Flow of jet from an orifice or nozzle.. Where waves are likely to be formed on surface. 4. Where fluids of different densities flow one over the other. 40. When the Euler s model law applied? Euler s model law is applied for fluid flow problems where flow occurs in a closed pipe in which case turbulence is fully developed so that viscous force are absent and gravity amend surface tension force is absent. 41. What is the application of Weber s model law? It is applied where surface tension effects predominate in addition to inertia force e.g. Capillary rise in narrow passage. Capillary movements of water in soil. Capillary waves in channels. Flow over weirs for small heads. 4. Give some of the applications of mach model lane. Applications of mach model lane are: 1. aerodynamic testing. Underwater testing of torpedoes.. Water hammer problems. 4. What are the two types of hydraulic models? They are 1. Undistorted. Distorted 44. What are undistorted models?

37 Undistorted models are those models, which are geometrically similar to their prototypes. 45. What are distorted models? When a model is not geometrically similar to its prototype it is called distorted model. EXERCISE PROBLEMS 1. Check whether the equation is dimensionally homogeneous. Find the unit conversion factor to make it dimensionally homogeneous. P P 0 (1/) ρ u ρ gz (P-pressure, - ρ density, u-velocity, z-height above datum). The centripetal acceleration of a particle in circular motion is dependent on velocity u and radius r. using dimensional analysis determines the functional relation.. In flow over a smooth flat plate, the boundary layer thickness δ is found to depend on the free stream velocity u, fluid density and viscosity and the distance x from the leading edge. Express the correlation in the form of dimensionless groups. 4. In flow over a smooth flat plate, the wall shear τ w in the boundary layer depends on the free stream velocity, density and viscosity of the fluid and the distance from the leading edge. Determine the dimensionless parameters to express the relation between the variables. 5. The volume flow Q over a weir depends on the upstream height h, the width of the weir b, and acceleration due to gravity. Obtain a relationship between the variables in terms of dimensionless parameters. 6. Obtain a relationship for the torque T to rotate a disk of diameter D in a fluid of viscosity τ at an angular speed ω over a plate, with clearance h. 7. The pressure P, developed in a jet pump is found to depend on the jet diameter d, diffuser diameter D, the velocity u of the jet, the volume flow Q and the density and viscosity of the fluid. Determine the dimensionless parameters to organize experimental results. 8. Players use spin in ball plays like tennis, golf etc. As the ball moves the spin rate will decrease. If the aerodynamic torque T on the ball in flight depends on the forward speed u, density and viscosity of air, the ball diameter D, angular velocity of spin ω, and the

38 roughness height e on the ball surface, determine the dimensionless parameters to correlate situation. 9. The power required to drive a propeller in a gas medium depends upon the forward speed u, the rotational speed N, diameter D, density and viscosity of the gas and the speed of sound c in the medium. Obtain dimensionless parameters to correlate experimental results. 10. An airship is to operate in air at 0 o C and 1 bar at 0 m/s speed. A model of scale 1/0 is used for tests in a wind tunnel, the test speed being 75 m/s. Determine the pressure of the tunnel for dynamic similarity. The air temperatures are equal. If the drag force on the model was 50 N. Determine the drag on the prototype. 11. One fifth scale model of an automobile is tested in a towing water tank. Determine the ratio of speeds of the model and prototype. Assume 0oC in both cases. It is found that the coefficient of drag remains constant for the model after speeds of 4 m/s and the drag at this speed was 18 N. Estimate the drag on the prototype when operating at 90 kmph. 1. An open channel of rectangular section of width 7 m carries water to a depth of 1 m and a flow rate of m /s. A model to have Froude number similarity is to be designed. The discharge scale is 1/1000. Determine the depth of flow in the model. 1. A 1/50 scale model of a ship is tested in a towing tank to determine the wave drag on the ship s hull. The ship is to designed to cruise at 18 knots (knot 185 m). Determine the velocity with which the model is to be towed. Also determine the ratio of drag values on model and prototype. Neglect viscous drag. 14. A model of an aero plane of 1/0 size is to be tested in a pressurized wind tunnel at the same speed as that of the prototype to get over compressibility effects. If the temperatures are the same, determine the pressure in the wind tunnel in atm. The aero plane is to be operated at 0.8 atm. 15. The drag force on a sphere submerged in water at 0 o C, when moved at 1.5 m/s was measured as 10 N. An enlarged model of :1 scale was tested in a pressurized wind tunnel at a pressure of 1.5 MN/m and temperature of 0 o C. Determine the velocity for dynamic similarity. Also determine the drag force on the model. Kinematic viscosity of water m/s. Viscosity of air kg/ms. Density of air 17.8 kg/m.

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