Fluid Mechanics III 1. Dimensional analysis and similarity Similarity The real world is non-dimensional. The proposition the Eiffel Tower is tall has no sense unless we state what is the object we compare it with. The Eiffel Tower could be tall compare to your house, but it is not if we compare it with Everest. The proposition the Eiffel Tower is tall compare to my house is equivalent to the proposition the ratio of the height of the Eiffel Tower to the height of my house is large, where the non-dimensional ratio of two heights is used. The real world is non-dimensional If all possible non-dimensional parameters of one system equal to the corresponding non-dimensional parameters of another system these two systems have identical behaviour. This is the case of complete similarity. The similarity includes geometric similarity, kinematic similarity and dynamic similarity. The geometric similarity means that all possible length ratios of one system are the same as the corresponding length ration of another system. It follows that angles in a system are equal to the corresponding angles of a similar system. The kinematic similarity means that ratios of velocities are the same in both systems. For example, to achieve a kinematic similarity between a model and a prototype of a propeller in a fluid flow the ratios of fluid velocity U to the velocity of a blade end ω R for the model and the prototype must be the same. The dynamic similarity means that ratios of forces of different origins are the same in both systems. For example, Reynolds number represents the ratio of inertia force to viscous force. If Reynolds numbers 1
2 of two systems are the same, viscous forces produce in them identical effects. If system behaviour is known from theory or experiment, the results represented in non-dimensional form can be transferred to any similar system. The complete similarity between model and prototype is very difficult (if at all possible) to achieve. More common is partial similarity when only similarity of the most important effects is ensured. For example, for ship models the most important drag component is due to generation of surface waves, which are driven by gravity. The non-dimensional parameter describing the relative importance of gravity forces is Froude number: F r = U 2 /(L g). Then the main requirement for the model is to have the same Froude number as a prototype. It is often impossible to get the similarity of Reynolds numbers Re = U L/ν in the same experiment, and corrections to experimental results should be made to account for viscous effects. Equations of motion in non-dimensional form See lecture slides for theory and examples π-theorem Let a system is described by n dimensional variables v 1, v 2,..., v n, and their dimensions are constructed by using k basic dimensions d 1,..., d k. If thermal or electromagnetic effects are not considered there are three basic dimensions (k = 3) in fluid problems: length [L], mass [M] and time [T ]. All other dimensions are products of powers of basic dimensions. For example, the dimension of density is [M L 3 ], dimension of velocity is [M T 1 ], etc. The behaviour of the system can be described by an equation connecting the system variables f(v 1, v 2,..., v n ) = 0, (1) where f is some function. If we are interested in the behaviour of a specific system parameter, say v 1, and consider other variables as control parameters, we could write v 1 = F (v 2,..., v n ). (2) However, the explicit form of the solution (2) is not always possible, and then the more general implicit form (1) should be used. The original dimensional variables can be expressed by using a new set of n independent variables.
3 If we use basic dimensions d 1,..., d k as first k of these new variables, then other m = n k of them will be non-dimensional. Therefore, such system can be described by n k non-dimensional parameters. This is the so-called π-theorem. The solution can be then written either in the implicit form φ(π 1, π 2,..., π m ) = 0; m = n k or in the explicit form π 1 = Φ(π 2,..., π m ). The explicit representation could not always be possible. For all similar systems these non-dimensional forms are identical, with the same functions φ and Φ. Example: Solution for the velocity profile of a laminar flow with mean velocity U in a circular pipe of radius R is: ( u ( ) r 2 U = 2 1. R) This form of the solution is valid for circular pipes of any radius and any mean flow velocity as long as flow is laminar (Re d 2000). The laminar solution helps to guess the general form of a velocity profile in a smooth circular pipe, which can be written as How to use π-theorem u U = Φ( r R, Re ) Determine what dimensional parameters specify behaviour of a system and write down their dimensions. Define the basic dimensions. For fluid problems it is usually length [L], time [T ] and mass [M]. Find the number of non-dimensional parameters: m = n k. Specify primary parameters, that is operational parameters, output parameters and control parameters of the system. These are usually parameters which can easily be changed or measured in model experiment or during prototype operation. The number of these parameters should be m. These, for example, could be such parameters as flow rate,
4 head, rotation speed, etc. Other are secondary parameters or repeating variables. They could appear in more then one non-dimensional groups. These are fluid properties or geometrical parameters of the system. Construct the non-dimensional groups π 1,..., π m as products of unknown powers of dimensional parameters. Find the powers from the condition that all groups are non-dimensional, that is the total power of each dimension should be 0. You can use the following rules: Each of the primary parameters should appear only in one nondimensional group and usually have power 1. If there are more then m primary parameters you should chose which of them is less important for a particular problem, and treat it as a secondary parameter. If there are less then m primary parameters treat the most important secondary parameter as a primary one. Specify the general form of the solution and use non-dimensional groups to ensure the similarity between model and prototype. Example: Parameters influencing the operation of a pump are: 1 Primary: Flow rate Q [L 3 T 1 ] Head gh [L 2 T 2 ] Power P [ML 2 T 3 ] Speed of rotation N [T 1 ] Secondary (repeating variables): Rotor diameter D [L] Density ρ [ML 3 ] Viscosity µ [ML 1 T 1 ] Operating the pump we would like to provide a certain head for a certain flow rate, and to know what power is required to operate it. Therefore, the pump characteristics can be represented in the following general form: gh = f 1 (Q, N, D, ρ, µ) and P = f 2 (Q, N, D, ρ, µ). 1 Note that H and g only appear in the combination but not separately. Why?
5 Knowing these two characteristics we can specify the pump efficiency as η = ρ g H P = f 3(Q, N, D, ρ, µ), where f 3 can be expressed from f 1 and f 2. The total number of dimensional parameters is n = 7, and the number of basic dimensions is k = 3. Thus, the number of non-dimensional groups is m = n k = 4, and we have 4 primary parameters. If we consider viscous effects one of the non-dimensional parameters should be Reynolds number, which in the case of pump can be represented as Re = ρ ND2 µ This excludes µ from further consideration, and we should treat one of the primary parameters as a repeating parameter. Usually pumps operate on the constant rotation speed N, which is specified by the choice of a specific engine. Therefore, we will treat N as a repeating parameter. Now we have 3 primary parameters: Q, gh, P ; and 3 repeating variables: N, D, ρ. Let each of the non-dimensional groups be proportional to one of the primary parameters. Then we have: Non-dimensional flow rate: Dimensions: π 1 = Q N α D β ρ γ. [1] = [L 3 T 1 ][T 1 ] α [L] β [ML 3 ] γ. The power of each of the basic dimensions should be 0, which gives the following equations for unknown powers: L : 3 + β 3γ = 0 T : 1 α = 0 M : γ = 0. Then we have γ = 0, α = 1 and β = 1, and the corresponding nondimensional group is called discharge number or flow coefficient: K Q = Q ND 3. Applying the same approach to gh and P we obtain the head coefficient K H = gh N 2 D 3
6 and power coefficient K P = P ρ N 3 D 5. Then the universal pump characteristic, which can be applied to all pipes of similar designs, are: K H = φ 1 (K Q, Re); K P = φ 2 (K Q, Re). The efficiency is already non-dimensional, and can be found as Reading: η = K HK Q K P. Massey,B.S. Mechanics of Fluids, 8 th edition Chapter 1: Fundamental Concepts 1.2 Notation, dimensions, units and related matters Chapter 5: Physical Similarity and Dimensional Analysis Douglas,J.F., et al. Fluid Mechanics, 5 th edition Chapter 8: Dimensional Analysis Chapter 9: Similarity