Module 5 Dimensional Analysis and Similitude_online2

Dimensional Analysis and Similitude

Dr. Om Prakash Singh

Asst. Prof., IIT Mandi

www.omprakashsingh.com

Dimensional Analysis and Similitude

• Many problems of interest in fluid mechanics cannot be solved using the integral

and/or differential equations.

• Wind motions around a football stadium, the air flow around the deflector on a

semitruck, the wave motion around a pier or a ship, and air flow around aircraft are

all examples of problems which are studied in the laboratory with the use of

models.

• A laboratory study with the use of models is very expensive, however, and to

minimize the cost, dimensionless parameters are used.

• In fact, such parameters are also used in numerical studies for the same reason.

• Once an analysis is done on a model in the laboratory and all quantities of interest

are measured, it is necessary to predict those same quantities on the prototype,

such as the power generated by a large wind machine from measurements on a

much smaller model.

• Similitude is the study that allows us to predict the quantities to be expected on a

prototype from measurements on a model.

• This will be done after our study of dimensional analysis that guides the model

study.

Dimensional Analysis

• Dimensionless parameters are obtained using a method called dimensional

analysis.

• It is based on the idea of dimensional homogeneity: all terms in an equation must

have the same dimensions.

• By simply using this idea, we can minimize the number of parameters needed in an

experimental or analytical analysis, as will be shown.

• Any equation can be expressed in terms of dimensionless parameters simply by

dividing each term by one of the other terms.

• For example, consider Bernoulli’s equation,

Now, divide both sides by gz2 . The equation can then be written as

Note the dimensionless parameters, V2/gz and p/γ z

(1)

(2)

Example for Dimensional Analysis

• Suppose the drag force is desired on an

object with a spherical front that is

shaped as shown in Fig.

• A study could be performed, the drag

force measured for a particular radius R

and length L in a fluid with velocity V,

viscosity µ, and density ρ.

• Gravity is expected to not influence the

force. This dependence of the drag force

on the other variables would be written

as Flow around an object

• To present the results of an experimental study, the drag force could be plotted

as a function of V for various values of the radius R holding all other variables

fixed.

• Then a second plot could show the drag force for various values of L holding all

other variables fixed, and so forth.

(3)

Example for Dimensional Analysis

• The plots may resemble those of Fig. 2 above. To vary the viscosity holding the

density fixed and then the density holding the viscosity fixed, would require a

variety of fluids leading to a very complicated study, and perhaps an impossible

study.

Fig. 2 Drag force versus velocity: (a) L, µ, ρ fixed; (b) R, µ, ρ fixed.

Example for Dimensional Analysis

• The actual relationship that would relate the drag force to the other variables

could be expressed as a set of dimensionless parameters, much like those of Eq.

(2), as

• The procedure to do this will be presented next.

• The results of a study using the above relationship would be much more organized

than the study suggested by the curves of Fig. 2.

• An experimental study would require only several different models, each with

different R/L ratios, and only one fluid, either air or water.

• Varying the velocity of the fluid approaching the model, a rather simple task, could

vary the other two dimensionless parameters.

• A plot of FD/ (ρV2 R2) versus ρVR/µ for the several values of R/L would then provide

the results of the study.

(4)

• Nondimensionalization of an equation by inspectional analysis is useful only when

one knows the equation to begin with.

• However, in many cases in real-life engineering, the equations are either not known or

too difficult to solve; often times experimentation is the only method of obtaining

reliable information.

• In most experiments, to save time and money, tests are performed on a

geometrically scaled model, rather than on the full-scale prototype.

• In such cases, care must be taken to properly scale the results.

• We introduce here a powerful technique called dimensional analysis. While

typically taught in fluid mechanics, dimensional analysis is useful in all disciplines,

especially when it is necessary to design and conduct experiments.

• You are encouraged to use this powerful tool on other subjects as well, not just in

fluid mechanics. The three primary purposes of dimensional analysis are

Dimensional analysis

• To generate nondimensional parameters that help in the design of experiments

(DoE) (physical and/or numerical) and in the reporting of experimental results

• To obtain scaling laws so that prototype performance can be predicted from

model performance

• To (sometimes) predict trends in the relationship between parameters

• Before discussing the technique of dimensional analysis, we first explain the

underlying concept of dimensional analysis—the principle of similarity.

• There are three necessary conditions for complete similarity between a

model and a prototype.

• The first condition is geometric similarity—the model must be the same

shape as the prototype, but may be scaled by some constant scale factor.

• The second condition is kinematic similarity, which means that the velocity at

any point in the model flow must be proportional (by a constant scale factor) to

the velocity at the corresponding point in the prototype flow

Dimensional analysis and similarity

Fig.: Kinematic similarity is achieved when, at all

locations, the velocity in the model flow is

proportional to that at corresponding locations

in the prototype flow, and points in the same

direction. In other words, ratio of velocity must

remain constant.

• The third and most restrictive similarity condition is that of dynamic similarity.

• Dynamic similarity is achieved when all forces in the model flow scale by a

constant factor to corresponding forces in the prototype flow (force-scale

equivalence).

• As with geometric and kinematic similarity, the scale factor for forces can be

less than, equal to, or greater than one.

Dimensional analysis and similarity

All three similarity conditions must exist for complete similarity to be ensured.

In a general flow field, complete similarity between a model and

prototype is achieved only when there is geometric, kinematic, and

dynamic similarity.

Wind Tunnel Testing

We match the Reynolds numbers for the full

scale model and prototype.

which can be solved for the required wind

tunnel speed for the model tests Vm

For many objects, the drag coefficient levels off at

Reynolds numbers above some threshold value. This

fortunate situation is called Reynolds number

independence. It enables us to extrapolate to

prototype Reynolds numbers that are outside of the

range of our experimental facility.

While drag coefficient CD is a strong function

of the Reynolds number at low values of Re,

CD often levels off for Re above some value.

In other words, for flow over many

objects, especially “bluff” objects like trucks,

buildings, etc., the flow is Reynolds number

independent above some threshold value of

Re (Fig., typically when the boundary layer

and the wake are both fully turbulent.

The aerodynamic drag of a new sports car is to be predicted at a speed of

50.0 mi/h at an air temperature of 25°C. Automotive engineers build a one fifth

scale model of the car to test in a wind tunnel. It is winter and the wind tunnel is

located in an unheated building; the temperature of the wind tunnel air is only

about 5°C. Determine how fast the engineers should run the wind tunnel in

order to achieve similarity between the model and the prototype.

Problem

Ans: 221 mi/h

Hind: Reynolds number should be same both in model and prototype

Dimensions of fluid variables

• There are only three basic dimensions, since Newton’s second law can be used to

relate the basic dimensions.

• Using F, M, L, and T as the dimensions on force, mass, length, and time, we see that

F= ma demands that the dimensions are related by

(5)

• We choose to select the M-L-T system (F-L-T system can also be used) and use Eq.

(5) to relate F to M, L, and T.

• If temperature is needed, as with the flow of a compressible gas, an equation of

state, such as

could be expressed dimensionally as

where the brackets mean “the dimensions of.” Note that the product RT does

not introduce additional dimensions.

Dimensions of fluid variables

Symbols and Dimensions of Quantities of

Interest Using the M-L-T System

Buckingham π theorem

• The Buckingham π theorem is used to create the dimensionless parameters, given

a functional relationship such as that of Eq. (3). Write the primary variable of

interest as a general function, such as

• where n is the total number of variables.

• If m is the number of basic dimensions, usually 3, the Buckingham π theorem

demands that (n− m) dimensionless groups of variables, the π terms, are

related by,

• The π term π1 is selected to contain the dependent variable [it would be FD of Eq. (3)]

and the remaining π terms contain the independent variables.

• It should be noted that a functional relationship cannot contain a particular

dimension in only one variable; for example, in the relationship v=f(d, t, ρ) the

density ρ cannot occur since it is the only variable that contains the dimension M,

and M would not have the possibility of canceling out to form a dimensionless πterm.

Buckingham π theorem

Steps:

The steps that are followed when applying the Buckingham π theorem are:

1. Write the dependent variable as a function of the (n–1) independent

variables. This step requires knowledge of the phenomenon being studied. All

variables that influence the dependent variable must be included and all

variables that do not influence the dependent variable should not be

included. In most problems, this relationship will be given.

2. Identify the m repeating variables that are combined with the remaining

variables to form the π terms. The m variables must include all the basic

dimensions present in the n variables of the functional relationship, but they

must not form a dimensionless π term by themselves. Note that an angle is

dimensionless, so it is not a candidate to be a repeating variable.

3. Combine each of the (n– m) variables with the repeating variables to form the

π terms. Step 3 is carried out by either inspection or by an algebraic

procedure.

4. Write the π term containing the dependent variable as a function of the

remaining π terms.

Buckingham π theorem

Example

The method of inspection will be used in an example. To demonstrate the algebraic

procedure, let’s form a π term of the variables V, R, ρ, and µ. This is written as

In terms of dimensions, this is

Equating exponents on each of the basic dimensions provides the system of

equations:

Buckingham π theorem

The solution is

The π term is then written as

• This π term is dimensionless regardless of the value of d.

• If we desire V to be in the denominator, select d =1; if we desire V to be in

the numerator, select d = −1. Select d = −1 so that

Suppose that only one π term results from an analysis. That π term would then

be equal to a constant which could be determined by a single experiment.

Buckingham π theorem

• Finally, consider a very general functional relationship between a pressure change Δp,

a length l, a velocity V, gravity g, viscosity µ, a density ρ, the speed of sound c, the

surface tension σ , and an angular velocity Ω.

• All of these variables may not influence a particular problem, but it is interesting to

observe the final relationship of dimensionless terms.

• Dimensional analysis, using V, l, and ρ as repeating variables provides the relationship

Each term that appears in this relationship is an important parameter in certain

flow situations.

The dimensionless term with its common name is listed as follows:

Buckingham π theorem

Flows with Free Surfaces: Froude number

• For the case of model testing of flows with free surfaces (boats and ships, floods, river

flows, aqueducts, hydroelectric dam spillways, interaction of waves with piers, soil erosion,

etc.), complications arise that preclude complete similarity between model and prototype.

• For example, if a model river is built to study flooding, the model is often several hundred times

smaller than the prototype due to limited lab space. If the vertical dimensions of the model were

scaled proportionately, the depth of the model river would be so small that surface tension

effects (and the Weber number) would become important, and would perhaps even

dominate the model flow, even though surface tension effects are negligible in the

prototype flow.

• In addition, although the flow in the actual river may be turbulent, the flow in the model river

may be laminar, especially if the slope of the riverbed is geometrically similar to that of the

prototype. To avoid these problems, researchers often use a distorted model in which the

vertical scale of the model (e.g., river depth) is exaggerated in comparison to the horizontal

scale of the model (e.g., river width). In addition, the model river bed slope is often made

proportionally steeper than that of the prototype. These modifications result in incomplete

similarity due to lack of geometric similarity.

• Model tests are still useful under these circumstances, but other tricks (like deliberately

roughening the model surfaces) and empirical corrections and correlations are required to

properly scale up the model data.

Flows with Free Surfaces: Froude number

In many practical problems involving free surfaces, both

the Reynolds number and Froude number appear as relevant

independent π groups in the dimensional analysis (Fig.). It is

difficult (often impossible) to match both of these

dimensionless parameters simultaneously. For a free-

surface flow with length scale L, velocity scale V, and

kinematic viscosity ν, the Reynolds number is matched

between model and prototype when

The Froude number is matched between model and

prototype when

To match both Re and Fr, we solve above

simultaneously for the required length scale factor

Lm/Lp

In many flows involving a liquid with

a free surface, both the Reynolds

number and Froude number are

relevant nondimensional

parameters. Since it is not always

possible to match both Re and Fr

between model and prototype, we

are sometimes forced to settle for

incomplete similarity.

Eliminating the ratio Vm/Vp from above Eq., we see that the equired ratio of

kinematic viscosities to match both Re and Fr:

Flows with Free Surfaces: Froude number

Thus, to ensure complete similarity (assuming geometric similarity is

achievable without unwanted surface tension effects as discussed

previously), we would need to use a liquid whose kinematic viscosity

satisfies above Eq.

In the late 1990s the U.S. Army Corps of

Engineers designed an experiment to

model the flow of the Tennessee River

downstream of the Kentucky Lock and

Dam (Fig. ). Because of laboratory

space restrictions, they built a scale

model with a length scale factor of

Lm/Lp = 1/100. Suggest a liquid that

would be appropriate for the experiment.

Problem

Ans: νm = 1.00 x 10-9 m2/s

Note on previous problem

• We need to find a liquid that has a viscosity of 1.00 x 10-9 m2//s. A quick

glance through the property table yields no such liquid. Hot water has a

lower kinematic viscosity than cold water, but only by about a factor of

3. Liquid mercury has a very small kinematic viscosity, but it is of order

10-7 m2/s—still two orders of magnitude too large to satisfy .

• Even if liquid mercury would work, it would be too expensive and too

hazardous to use in such a test. What do we do? The bottom line is that

we cannot match both the Froude number and the Reynolds number in

this model test.

• In other words, it is impossible to achieve complete similarity between

model and prototype in this case. Instead, we do the best job we can

under conditions of incomplete similarity. Water is typically used in such

tests for convenience.

Buckingham π theorem

• Not all of the above numbers would be of interest in a particular flow; it is highly

unlikely that both compressibility effects and surface tension would influence the

same flow.

• These are, however, the primary dimensionless parameters in our study of fluid

mechanics.

• The Euler number is of interest in most flows used to characterize losses in the flow

(pressure drop by kinetic energy per unit volume) where a perfect frictionless flow

corresponds to an Euler number of 1; the Froude number in flows with free surfaces

in which gravity is significant (e.g., wave motion), it the the ratio of a characteristic

velocity to a gravitational wave velocity; the Reynolds number in flows in which

viscous effects are important, the Mach number in compressible flows, the Weber

number in flows affected by surface tension (e.g., sprays with droplets), it is a

measure of the relative importance of the fluid's inertia compared to its surface

tension; and the Strouhal number in flows in which rotation or a periodic motion

plays a role.

• Each of these numbers, with the exception of the Weber number (surface tension

effects are of little engineering importance), will appear in flows studied in other

cases.

• Note: The Froude number is often defined as V2/lg; this would not influence the

solution to problems.

Problem

The pressure drop Δp over a length L of pipe is assumed to depend on the average

velocity V, the pipe’s diameter D, the average height e of the roughness elements of

the pipe wall, the fluid density ρ, and the fluid viscosity π. Write a relationship

between the pressure drop and the other variables using Buckingham π theorem.

Ans:

Problem

The speed V of a weight when it hits the floor is assumed to depend on gravity g, the

height h from which it was dropped, and the density ρ of the weight. Use dimensional

analysis and write a relationship between the variables.

Ans:

A simple experiment would show that C= 2

End