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**Dimensional Analysis of a Fluid: Methods, Equations, Buckingham pi Theorem and Table! [with solved examples]**

**Dimensional Homogeneity of an Equation-Dimensional Analysis****: **

An equation is said to be dimensionally homogeneous if the dimensions of every term on each side of the equation are identical. Every equation representing a physical phenomenon derived from an analytical approach will satisfy this condition. Such equations are independent of the systems of units.

#### Methods

Given a number of quantities and also given that they are interrelated, the process of obtaining a relation between them by the use of dimensions is known as dimensional analysis. The relation between such quantities can be determined by two methods, namely, (i) Raleigh’s method. (ii) Buckingham’s method

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Suppose Q_{1}, Q_{2}, Q_{3}, Q_{4}, etc. are various inter-related quantities having different dimensions then a relation among them can be expressed in the form –

Q_{1} = KQ_{2}^{a}Q_{3}^{b}Q_{4}^{c}….

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where, K is non-dimensional factor.

Performing a dimensional analysis (i.e., writing down the dimensions of every quantity) and comparing the powers of corresponding fundamental units on both sides of the equation, the relation can be established.

For instance, consider the relation between the period of a simple pendulum t, its length I and the acceleration due to gravity g. We can express the relation in the form.

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**Buckingham’s Pi Theorem: **

This is a theorem which can be applied to form dimensionless constants from a set of variables.

**The theorem may be stated as follow: **

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If there are m variables influencing a phenomenon, which can be expressed fully in terms of n fundamental units, then it is possible to group these m variables to yield (m – n) dimensionless constants. Buckingham named these constants as π_{1}, π_{2}, π_{3y }….etc. and hence the theorem is sometimes called the Pi theorem.

**Method for Forming Dimensionless Constants****: **

**The following steps may be followed to determine the dimensionless constants, given a number of variables of different dimensions: **

(i) Compare the number of variables with the number of fundamental units involved and decide the number of dimensionless constants.

By Buckingham’s Pi Theorem

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Number of dimensionless constants = (Number of variables) – (number of fundamental units involved)

**See examples in the table below: **

(ii) Select the repeating variables. The number of repeating variables should be equal to the number of fundamental units involved. The repeating variables must contain, in one or the other, fundamental units involved, and must not among themselves reduce to a dimensionless parameter.

(iii) The repeating variables may preferable be selected as follows. The best choice would be to select one geometric characteristic (like I, d), one fluid characteristic (like ρ, μ) and one flow characteristic (like v). l or d, v and ρ would in most of the cases, be the best choice.

(iv) The repeating variables, each raised to an index, are grouped with a non-repeating variable to form a dimensionless constant. For example, if the given variables are F, D, v, ρ, g, μ then since there are 6 variables involving 3 fundamental units, we can frame 6 – 3 = 3 dimensionless constants. Since three fundamental units are involved we should select three repeating variables. Obviously, the best choice of repeating variables will be D, v and ρ.

The three dimensionless constants are given by –

Comparing the powers of the fundamental dimensions on both sides of each of the above relations, we can easily determine the indices provided to the variables. Once these indices are known, the dimensionless constants can be established. Now we can obtain the relation between the variables from a relation like –

π_{1 }= *f*(π_{2}, π_{3 }…)_{}