Dimensional Analysis

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[1]Dimensional analysis is the analysis of units with the intended purpose of creating variables or equations. Some applications are simple and just involve making sure the equation used yields the correct results but other applications are the creation of correct scales for experimentation. This article explains the basic concepts of dimensional analysis and briefly explains how it is used.

Contents

Textbook Readings

Cengel and Cimbala, Fluid Mechanics: Fundamentals and Applications, (2nd ed, Singapore, McGraw Hill Education, 2010), pp. 283 - 302.

Units and Dimensions

[2]At the heart of dimensional analysis are units, such as length or mass. There are seven primary units which are:

  1. Mass (kg)
  2. Length (m)
  3. Time (s)
  4. Temperature (K)
  5. Electric Current (A)
  6. Amount of Light (cd- candela)
  7. Amount of Matter (mol)

The idea is that all non-primary units can be formed by some combination of the seven primary units. For example, Force, which is given in Newtons (N) is given by:

F = ma
= kg.m/s2 = N

Note however that when dimensional analysis is carried out, it is not the units that we care about, it is the dimensions. That is, mass might be given in units of "kg" as convention in SI units, but in dimensional analysis we simply treat it as a dimension of mass since the units might be in mm, cm, m or km. Hence the symbols used in dimensional analysis for the seven primary units are:

  1. Mass (m)
  2. Length (L)
  3. Time (t)
  4. Temperature (T)
  5. Electric Current (I)
  6. Amount of Light (C)
  7. Amount of Matter (N)

When dimensional analysis is carried out, say for force, it is done with curly brackets which mean "dimension of" as in:

{Force} = {Mass.Length/Time2} = {mL/t2}

Dimensional Homogeneity

[3]An important aspect of dimensional analysis is that in term in an equation that is added must be of the same dimension as all other terms. As an example, consider Bernoulli's Equation for one point in a fluid. Remember that Bernoulli's equation is given by:

Bernouli.png

Now take each term separately:

  1. {P/ρ} = {Pressure/Density} = {Force per Unit Area/Mass per unit volume} = {(Mass.Length.Time2/Length2)/(Mass/Length3)} = {(L/t)2}
  2. {(1/2)V2} = {Velocity2} = {(Length/Time)2} = {(L/t)2}
  3. {gz} = {Gravity.Length} = {(Length/Time2).Length} = {(L/t)2}

Hence the constant must have units of {(L/t)2} as well. Note that this unit corresponds to the specific energy of the fluid (i.e. J/kg).

Dimensional Analysis and Similarity

[4]One very useful application of dimensional analysis is that of similarity. It is often that case that the equations for relationships between variables is not known and that only experiments yield reliable results. However in many situations, experiments become to expensive if real life models are used and hence engineers attempt experiments on scale models. The problem is that the model might be to scale, but the conditions of the experiments need to be to scale as well. For example, if in real life size experiments winds of 100 km/hr are needed, then under a scaled model, the winds might not need be so strong to get a reliable result. In essence there are three conditions that must be met in order to have a reliable experiment on scale models:

  1. Geometric Similarity (shape of model must be the exact same as the original but may be scaled up/down with a constant scale factor)
  2. Kinematic Similarity (the velocity at any point on the model must be the same as in the original but may be scaled up/down with a constant scale factor)
  3. Dynamic Similarity (all forces on the model must be the same as on the original, but may be scaled up/down by a constant scale factor

In order to achieve proper similarity all of these three conditions must be made.

Reynold's Number

[5]A very useful number used to achieve similarity is the Reynold's number which is dimensionless. Reynold's number is given by:

Reynold'sNumber.png

Where:

V = fluid velocity (m/s)
u = viscosity (kg/m.s)

The dimensional analysis shows the Reynold's number is dimensionless as:

  1. {ρVL} = {(Mass/Length3)(Length/Time)(Length)} = {mL2/(L3t)} = {m/Lt}
  2. {u} = {Mass/(Length.Time)} = {m/Lt}
  3. {Re} = {(m/Lt)/(m/Lt)} = {1} = Dimensionless

In order to achieve similarity between real size prototypes and scaled models, engineers equate the Reynold's number for the prototype with the Reynold's number for the model. Since the length is just a scale down of the real life object, it is already determined. Similarly, the viscosity and density are generally known since they are properties of the materials used. Then the velocity is the main unknown to be found in many situations.

The power of dimensional analysis is that all parameter values are ignored. All that is required is that the value of the non-dimensional parameters (such as Reynold's Number) to be equal between the real life and model.

Method of Repeating Variables and the Buckingham Pi Theorem

[6]Dimensional analysis is also used to generate dimensionless parameters (which help with achieving similarity) and the most popular method is that of the repeating variables. All that is needed for this method is to know what variables are related to each other (but not the actual relationship). For example we might know that energy is related to mass and velocity without knowing the exact equation.

The procedure is as follows:

  1. Find n, where n is the number of parameters
  2. List the primary dimensions (mentioned above) of each parameter
  3. Set the reduction j as the number of primary dimensions. Find k, where k = n - j and is the amount of Π's used later (Each Π is a dimensionless parameter)
  4. Choose j amount of repeating parameters. This is the most difficult part of the procedure- choose j repeating parameters, not dimensions
  5. Generate the Π's one at a time by grouping the j repeating parameters with one of the remaining parameters and with unknown exponents, forcing the product to be dimensionless
  6. Write the functional relationship between all the Π's with the dependent Π (usually the first one found)

For example, assume we want to find the relationship between energy, mass and velocity. Then the dependent variable is E and the procedure is:

  1. n = number of parameters = 3
  2. Primary dimensions = m, L, t
  3. j = 3. Therefore k = n - j = 0 which is pointless. Go back and guess j to be 1 less than the original guess. Therefore j = 2 => k = 1. Therefore the number of Π = 1.
  4. Choose j number of repeating parameters. Since j = 2 we need to choose two parameters: mass and velocity.
  5. Generate Π by combining all parameters with exponents: Π = Emavb
Dimensional Analysis:
{Π} = {Emavb} = {(m(L/t)2).(ma).((L/t)b)}
Now we require that Π be dimensionless, that is:
{t0} = {(t-2)(t-b)} => b = -2
{m0} = {m.ma} => a = -1
6. Therefore, Π = Em-1v-2}, or in other words, Π = E/(mv2)

References

"Textbook" refers to Cengel and Cimbala, Fluid Mechanics: Fundamentals and Applications, (2nd ed, Singapore, McGraw Hill Education, 2010).

  1. Textbook p. 283
  2. Textbook p. 284
  3. Textbook p. 285
  4. Textbook p. 291
  5. Textbook p. 293
  6. Textbook p. 295
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