Dimensional Analysis and Hydraulic Similitude s1

NPTEL Course Developer for Fluid Mechanics Dr. Niranjan Sahoo

Module 05; Lecture 37 IIT Guwahati

DIMENSIONAL ANALYSIS AND HYDRAULIC SIMILITUDE

Introduction and Objective

· Many practical real flow problems in fluid mechanics can be solved by using equations and analytical procedures.

· However, solutions of some real flow problems depend heavily on experimental data. Based on the measurements, refinements in the analysis are made. Hence, there is an essential link in this iterative process.

· Sometimes, the experimental work in the laboratory is not only time-consuming, but also expensive. So, the main goal is to extract maximum information from fewest experiments.

· In this regard, dimensional analysis is an important tool that helps in correlating analytical results with experimental data.

· Also, some dimensionless parameters and scaling laws are introduced in order to predict the prototype behavior from the measurements on the model.

Dimensional Analysis

· The analytically derived equations in engineering applications are correct for any system of units and consequently each group of terms in the equation must have the same dimensional representation. This is the law of dimensional homogeneity.

· In many instances, the variables involved in physical phenomena are known, while the relationship among the variables is not known. Such a relationship can be formulated between a set of dimensionless groups of variables and the groups numbering less than the variables. This procedure is called dimensional analysis. This procedure requires less experimentation and the nature of experimentation is considerably simplified.

The following examples will make the things clear.

Example I

Consider a steady flow of an incompressible Newtonian fluid through a long, smooth walled, horizontal circular pipe. It is desired to measure the pressure drop per unit length of the pipe without the use of experimental data.

§ The first step is to list out the variables that affect the pressure drop per unit length . These variables may be pipe diameter , fluid density , fluid viscosity and mean velocity at which the fluid is flowing through the pipe. Thus, the relationship can be expressed as,

(1)

At this point, the nature of the function is unknown and the experiments are to be performed to determine the nature of the function.

§ In order to perform the experiments in a systematic and meaningful manner, it is necessary to change one variable at a time keeping the others constant and measure the corresponding pressure drop. The series of tests would result the data that can be represented in graphical form as shown in Fig. 1 (a-d).

Fig. 1: Illustration of factors affecting the pressure drop in a pipe flow.

Referring to the Fig. 1(c), it would be difficult to vary fluid density while holding viscosity constant. Moreover, it would be rather impossible to obtain a general functional relationship between , , , and for any similar pipe system.

§ A simple approach to this problem is to collect two non-dimensional combinations of the variables (i.e. dimensionless products/dimensional groups) such that

(2)

Now, the working variables are reduced to only two instead of five. The necessary experiment would simply consist of varying the dimensionless product and determining the corresponding value . The results of the experiment could then be represented by a single, universal curve as illustrated in Fig. 2. It would be valid for any combination of smooth-walled pipe and incompressible Newtonian fluid.

Fig. 2: Illustrative plot of pressure drop in a pipe flow using dimensionless parameters.

Example II

Consider a problem to determine the drag force on a smooth sphere moving with certain velocity in a viscous fluid.

§ The first step is to list out the variables that affect the drag force (say ). These variables are diameter of sphere , velocity of the sphere , fluid density and viscosity of the fluid . It can be represented by some unknown function as,

(3)

§ Next step is to determine this relationship experimentally by varying one parameter at a time in the right-hand side parentheses of Eq. (3) by keeping the others constant. These results may be represented in graphical form as shown in Fig. 3. It would also lead to similar kinds of situations, which are extremely time-consuming and expensive investigation.

Fig. 3: Illustration of factors affecting the drag force on a sphere; (a) Vary density holding viscosity constant; (b) Vary viscosity holding density constant;

In order to eliminate these difficulties, the drag force can be expressed as functional relation between two dimensionless groups as,

(4)

where the nature of the function is not known. However, by experiment, a single curve may be obtained by relating the dimensionless groups as shown in Fig. 4.

Fig. 4: Illustrative plot of drag force on a sphere using dimensionless parameters.

Thus, the dimensionless combination of variables helps in building up the functional relationship for the experimental data, which is independent of the system of units (which is clear from the above examples). This method is known as dimensional analysis and the basis for its application to wide variety of problems is found in the Buckingham pi theorem.

Principle of Dimensional Homogeneity

It is stated as, “If an equation truly expresses a proper relationship between variables in a physical process, then it will be dimensionally homogeneous”. It means each of its additive terms will have the same dimension. For example,

§ Displacement of a free falling body is, . Each term in this equation has the dimension of length and hence it is dimensionally homogeneous.

§ Bernoulli’s equation for incompressible flow is, . Each term in this equation including the constant has dimension of velocity squared and hence it is dimensionally homogeneous.

Dimensional variables: These are the quantities, which actually vary during a given case and can be plotted against each other. In the first example, are the variables where as in the second example, the variables are .

Dimensional constants: These are normally held constant during a given run. But, they may vary from case to case. In above examples, are the dimensional constants.

Pure constants: have no dimensions. But, they arise from mathematical manipulation. In above examples, arises from mathematical manipulation. The other common dimensionless constants are, .

Buckingham pi Theorem

· It states that if an equation involving variables is dimensionally homogeneous, it can be reduced to a relationship among independent dimensionless products, where is the minimum number of reference dimensions required to describe the variable.

· The dimensionless products are frequently referred to as pi terms and the theorem is named accordingly after famous scientist Edgar Buckingham (1867-1940). It is based on the idea of dimensional homogeneity.

· Mathematically, if a physically meaningful equation involving variables is assumed,

(5)

such that the dimensions of the variables on the left side of the equation are equal to the dimensions of any term on the right side of equation, then, it is possible to rearrange the above equation into a set of dimensionless products (pi terms), so that

(6)

where is a function of through .

· The required number of pi terms is less than the number of original variables by , where is determined by the minimum number of reference dimensions required to describe the original list of variables. These reference dimensions are usually the basic dimensions (Mass, Length and Time) .

Determination of pi Terms

Several methods can be used to form dimensionless products or pi terms that arise in dimensional analysis. But, there is a systematic procedure called method of repeating variables that allows in deciding the dimensionless and independent pi terms. For a given problem, following distinct steps should be followed.

Step I: List out all the variables that are involved in the problem.

§ The ‘variable’ is any quantity including dimensional and non-dimensional constants in a physical situation under investigation. Typically, these variables are those that are necessary to describe the “geometry” of the system (diameter, length etc.), to define fluid properties (density, viscosity etc.) and to indicate the external effects influencing the system (force, pressure etc.).

§ All the variables must be independent in nature so as to minimize the number of variables required to describe the system.

Step II: Express each variable in terms of basic dimensions.

Typically, for fluid mechanics problems, the basic dimensions will be either or . Dimensionally, these two sets are related through Newton’s second law so that e.g. or . It should be noted that these basic dimensions should not be mixed.

Step III: Decide the required number of pi terms.

It can be determined by means of Buckingham pi theorem which indicates that the number of pi terms is equal to , where is the number of variables in the problem (determined from Step I) and is the number of reference dimensions required to describe these variables (determined from Step II).

Step IV: Select the number of repeating variables.

§ Amongst the original list of variables, select those variables that can be combined to form pi terms.

§ The required number of repeating variables is equal to the number of reference dimensions.

§ Each repeating variable must be dimensionally independent of the others i.e. they cannot themselves be combined to form dimensionless product.

§ Since there is a possibility of repeating variables to appear in more than one pi term, so dependent variables should not be chosen as one of the repeating variable.

Step V: Formation of pi terms

Essentially, the pi terms are formed by multiplying one of the non-repeating variables by the product of the repeating variables each raised to an exponent that will make the combination dimensionless. It usually takes the form of where the exponents , and are determined so that the combination is dimensionless.

Step VI: Repeat the Step V for each of the remaining non-repeating variables. The resulting set of pi terms will correspond to the required number obtained from Step III.

Step VII: Checking of pi terms

Make sure that all the pi terms must be dimensionless. It can be checked by simply substituting the basic dimension () of the variables into the pi terms.

Step VIII: Final form of relationship among pi terms

Typically, the final form among the pi terms can be written in the form of Eq. (6) where would contain the dependent variable in the numerator. The actual functional relationship among pi terms is determined from experiment.

Illustration of Pi Theorem

In order to illustrate various steps in the pi theorem, following examples are considered.

Example 1 (Pressure drop in a pipe flow)

§ According to Step I, all the pertinent variables involved in the experimentation of pressure drop per unit length of the pipe, must be noted and should be expressed in the form,

(7)

where is the pipe diameter, is the fluid density, is the viscosity of the fluid and is the mean velocity at which the fluid is flowing through the pipe.

§ Next step is to express all the variables in terms of basic dimensions i.e. . It then follows that

, , , and (8)

§ Third step is to apply pi theorem to decide the number of pi terms required. Since there are five variables (including the dependent variable ) and three reference dimensions, so . So, only two pi terms are required.

§ The repeating variables to be used to form pi terms (Step IV) need to be selected from the list . It is to be noted that the dependent variable should not be used as one of the repeating variable. Since, there are three reference dimensions involved, so we need to select three repeating variable. These repeating variables should be dimensionally independent i.e. dimensionless product can not be formed from this set. In this case, are chosen as the repeating variables.

§ Now, first pi term is formed between the dependent variable and the repeating variables. It is written as,

(9)

Since this combination need to be dimensionless, it follows that

(10)

The exponents must be determined by equating the exponents for each of the terms i.e.

(11)

The solution of this algebraic equations gives . Therefore,

(12)

§ The process is repeated for remaining non-repeating variables with other additional variable so that,

(13)

and

(14)

Equating the exponents,

(15)

The solution of this algebraic equation gives . Therefore,

(16)

§ Now, the correct numbers of pi terms are formed as determined in step 3. In order to make sure about the dimensionality of pi terms, they are written as,

§ Finally, the result of dimensional analysis is expressed among the pi terms as,

(17)