# 9.2.1: Construction of the Dimensionless Parameters

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In the construction of these parameters it must be realized that every dimensionless parameters has to be independent. The meaning of independent is that one dimensionless parameter is not a multiply or a division of another dimensional parameter. In the above example there are three dimensionless parameters which required of at least one of the physical parameter per each dimensionless parameter. Additionally, to make these dimensionless parameters independent they cannot be multiply or division of each other.
For the pipe problem above, $$\ell$$ and $$D$$ have the same dimension and therefore both cannot be chosen as they have the same dimension. One possible combination is of $$D$$, $$U$$ and $$\rho$$ are chosen as the recurring set. The dimensions of these physical variables are: $$D = [L^{1}]$$, velocity of $$U = [L\,t^{-1}]$$ and density as $$\rho = [M\,L^{-3}]$$. Thus, the first term $$D$$ can provide the length, $$[L]$$, the second term, $$U$$, can provide the time $$[t]$$, and the third term, $$\rho$$ can provide the mass $$[M]$$. The fundamental units, $$L$$, $$t$$, and $$M$$ are length, time and mass respectively. The fundamental units can be written in terms of the physical units. The first term $$L$$ is the described by $$D$$ with the units of [$$L$$]. The time, [$$t$$], can be expressed by $$D/U$$. The mass, [$$M$$], can be expressed by $$\rho\, D^3$$. Now the dimensionless groups can be constructed by looking at the remaining physical parameters, $$\Delta P$$, $$D$$ and $$\mu$$. The pressure difference, $$\Delta P$$, has dimensions of [$$M\,L^{-1}\,t^{-2}$$] Therefore, $$\Delta P\,M^{-1}\,L\,t^{2}$$ is a dimensionless quantity and these values were calculated just above this line. Thus, the first dimensionless group is

$\label{dim:eq:bt:pipeDim1} \pi_1 = \overbrace{\Delta P}^{ [M\,L^{-1}\,t^{-2} ] } \overbrace{\dfrac{1}{\rho \, D^3}}^ { [M^{-1}]} \overbrace{D} ^ { [L]} \overbrace{\dfrac{D^2}{U^2}} ^ { [t^{2}]} = \overbrace{\dfrac{\Delta P} {\rho \, U^2} }^ { unitless}$

The second dimensionless group (using $$D$$) is

$\label{dim:eq:bt:pipeDim2} \pi_2 = \overbrace{D}^{ [L]} \, \overbrace{\ell^{-1}}^{ [L^{-1}]} = \dfrac{D}{L}$ The third dimensionless group (using $$\mu$$ dimension of [$$M\,L^{–1}\,t^{-1}$$]) and therefore dimensionless is

$\label{dim:eq:bt:pipeDim3} \pi_3 = \mu\, \overbrace{\dfrac{1}{D^3\,\rho} }^{ [M^{-1}]} \overbrace{D}^{ [L]} \overbrace{ \dfrac{D}{U} }^{ [t]} = \dfrac{\mu}{D\,U\,\rho}$ This analysis is not unique and there can be several other possibilities for selecting dimensionless parameters which are "legitimately'' correct for this approach. There are roughly three categories of methods for obtaining the dimensionless parameters. The first one solving it in one shot. This method is simple and useful for a small number of parameters. Yet this method becomes complicated for large number of parameters. The second method, some referred to as the building blocks method, is described above. The third method is by using dimensional matrix which is used mostly by mathematicians and is less useful for engineering purposes. The second and third methods require to identification of the building blocks. These building blocks are used to construct the dimensionless parameters. There are several requirements on these building blocks which were discussed on page ??. The main point that the building block unit has to contain at least the basic or fundamental unit. This requirement is logical since it is a building block. The last method is mostly used by mathematicians which leads and connects to linear algebra. The fact that this method used is the hall mark that the material was written by mathematician. Here, this material will be introduced for completeness sake with examples and several terms associated with this technique.