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# Chapter 9 Dim (continue)

## 9.2 Buckingham–$\pi$–Theorem

All the physical phenomena that is under the investigation have $n$ physical effecting parameters such that \begin{align} \label{dim:eq:bt:initial} F_1(q_1, q_2, q_3, \cdots, q_n) = 0 \end{align} where $q_i$ is the $i$'' parameter effecting the problem. For example, study of the pressure difference created due to a flow in a pipe is a function of several parameters such \begin{align} \label{dim:eq:bt:exPipe} \Delta P = f(L, \, D,\, \mu,\, \rho,\, U) \end{align} In this example, the chosen parameters are not necessarily the most important parameters. For example, the viscosity, $\mu$ can be replaced by dynamic viscosity, $\nu$. The choice is made normally as the result of experience and it can be observed that $\nu$ is a function of $\mu$ and $\rho$. Finding the important parameters is based on good fortune'' or perhaps intuition. In that case, a new function can be defined as \begin{align} \label{dim:eq:bt:exPipeG} F(\Delta P ,L, D, \mu, \rho, U) = 0 \end{align} Again as stated before, the study of every individual parameter will create incredible amount of data. However, Buckingham's methods suggested to reduce the number of parameters. If independent parameters of same physical situation is $m$ thus in general it can be written as \begin{align} \label{dim:eq:bt:initialDimless} F_2(\pi_1, \pi_2, \pi_3, \cdots, \pi_m) = 0 \end{align} If there are $n$ variables in a problem and these variables contain $m$ primary dimensions (for example M, L, T), then the equation relating all the variables will have (n-m) dimensionless groups. There are 2 conditions on the dimensionless parameters: beginNormalEnumerate change startEnumerate=1

1. Each of the fundamental dimensions must appear in at least one of the m variables
2. It must not be possible to form a dimensionless group from one of the variables within a recurring set. A recurring set is a group of variables forming a dimensionless group.
In the case of the pressure difference in the pipe (Equation \eqref{dim:eq:bt:exPipeG}) there are 6 variables or $n = 6$. The number of the fundamental dimensions is 3 that is $m = 3$ ([M], [L], [t]) The choice of fundamental or basic units is arbitrary in that any construction of these units is possible. For example, another combination of the basic units is time, force, mass is a proper choice. According to Buckingham's theorem the number of dimensionless groups is $n -m = 6-3 = 3$. It can be written that one dimensionless parameter is a function of two other parameters such as \begin{align} \label{dim:eq:bt:pipeDim} \pi_1 = f \left(\pi_2, \pi_3\right) \end{align} If indeed such a relationship exists, then, the number of parameters that control the problem is reduced and the number of experiments that need to be carried is considerably smaller. Note, the $\pi$–theorem does not specify how the parameters should be selected nor what combination is preferred.

### 9.2.1 Construction of the Dimensionless Parameters

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

### 9.2.2 Basic Units Blocks

In Thermo–Fluid science there are several basic physical quantities which summarized in Table 9.1. In the table contains two additional physical/basic units that appear in magnetohydrodynamics (not commonly use in fluid mechanics). Many (almost all) of the engineering dimensions used in fluid mechanics can be defined in terms of the four basic physical dimensions $M$, $L$ ,$t$ and $\theta$. The actual basic units used can be S.I. such as kilograms, meters, seconds and Kelvins/Celsius or English system or any other system. In using basic new basic physical units, $M$, $L$, $t$, and $\theta$ or the old system relieves the discussion from using particular system measurements. The density, for example, units are $Mass/Length^3$ and in the new system the density will be expressed as $M/L^3$ while in S.I. $kg/m^3$ and English system it $slug/ft^3$. A common unit used in Fluid Mechanics is the Force, which expressed in SI as Newton [$N$]. The Newton defined as a force which causes a certain acceleration of a specific mass. Thus, in the new system the force it will be defined as $M\,L\,t^{-2}$. There are many parameters that contains force which is the source reason why the old (or alternative) system use the force instead the mass. There many physical units which are dimensionless by their original definition. are the angle, strains, ratio of specific heats, $k$, friction coefficient, $f$ and ratio of lengths. The angle represented by a ratio of two sides of a triangle and therefor has no units nor dimensions. Strain is a ratio of the change of length by the length thus has no units. Quantities used in engineering can be reduced to six basic dimensions which are presented in Table 9.1. The last two are not commonly used in fluid mechanics and temperature is only used sometimes. Many common quantities are presented in the following Table 9.3.

Standard System Old System
Name Letter Units Name Letter Units
Area $L^2$ [$m^2$] Area $L^2$ [$m^2$]
Volume $L^3$ [$m^3$] Volume $L^3$ [$m^3$]
Angular Velocity $\dfrac{1}{t}$ $[m^3]$ Angular Velocity $L^3$ $[m^3]$
Acceleration $\dfrac{L}{t^2}$ $\left[\dfrac{m}{sec^2}\right]$ Acceleration $\dfrac{L}{t^2}$ $\left[\dfrac{m}{sec^2}\right]$
Angular Acceleration $\dfrac{1}{t^2}$ $\left[\dfrac{1}{sec^2}\right]$ Angular Acceleration $\dfrac{1}{t^2}$ $\left[\dfrac{1}{sec^2}\right]$
Force $\dfrac{M\,L}{t^2}$ $\left[\dfrac{kg\,m}{sec^2}\right]$ Mass $\dfrac{F\,t^2}{L}$ $\left[\dfrac{N\,s^2}{m}\right]$
Density $\dfrac{M}{L^3}$ $\left[\dfrac{kg}{L^2}\right]$ Density $\dfrac{F\,t^2}{L^4}$ $\left[\dfrac{N\,s^2}{m^4}\right]$
Momentum $\dfrac{M L}{t}$ $\left[\dfrac{kg m}{sec}\right]$ Momentum ${F\,t}$ $\left[{N\,sec}\right]$
Angular Momentum $\dfrac{M L^2}{t}$ $\left[\dfrac{kg m^2}{sec}\right]$ Angular Momentum ${L\,F\,t}$ $\left[{m\,N\,sec}\right]$
Torque $\dfrac{M L^2}{t^2}$ $\left[\dfrac{kg m^2}{s^2}\right]$ Torque ${L\,F}$ $\left[{m\,N}\right]$
Absolute Viscosity $\dfrac{M }{M^1\,t^2}$ $\left[\dfrac{kg }{m\,s}\right]$ Absolute Viscosity $\dfrac{t\,F}{L^2}$ $\left[\dfrac{N\,s}{sec}\right]$
Kinematic Viscosity $\dfrac{L^2 }{t}$ $\left[\dfrac{m^2 }{s}\right]$ Kinematic Viscosity $\dfrac{L^2}{t}$ $\left[\dfrac{m^2 }{s}\right]$
Volume Flow Rate $\dfrac{L^3 }{t}$ $\left[\dfrac{m^3 }{sec}\right]$ Volume Flow Rate $\dfrac{L^3}{t}$ $\left[\dfrac{m^3 }{sec}\right]$
Mass Flow Rate $\dfrac{M }{t}$ $\left[\dfrac{kg }{sec}\right]$ Mass Flow Rate $\dfrac{F\,t}{L}$ $\left[\dfrac{N\,s }{m}\right]$
Pressure $\dfrac{M }{L\,t^2}$ $\left[\dfrac{kg }{m\,s^2}\right]$ Pressure $\dfrac{F}{L^2}$ $\left[\dfrac{N }{m^2}\right]$
Surface Tension $\dfrac{M }{t^2}$ $\left[\dfrac{kg }{s^2}\right]$ Surface Tension $\dfrac{F}{L}$ $\left[\dfrac{N }{m}\right]$
Work or Energy $\dfrac{M\,L^2 }{t^2}$ $\left[\dfrac{kg\,m^2 }{s^2}\right]$ Work or Energy ${F\,L}$ $\left[ N \,m\right]$
Power $\dfrac{M\,L^2 }{t^3}$ $\left[\dfrac{kg\,m^2 }{s^3}\right]$ Power $\dfrac{F\,L}{t}$ $\left[ \dfrac{N \,m}{sec}\right]$
Thermal Conductivity $\dfrac{M\,L }{t^3\,\theta}$ $\left[\dfrac{kg\,m }{s^3\,K}\right]$ Thermal Conductivity $\dfrac{F\,L}{t}$ $\left[ \dfrac{N} {s\,K}\right]$
Specific Heat $\dfrac{L^2 }{t^2\,\theta}$ $\left[\dfrac{m^2 }{s^2\,K}\right]$ Specific Heat $\dfrac{L^2\,T}{t^2}$ $\left[\dfrac{m^2 }{s^2\,K}\right]$
Entropy $\dfrac{M\,L^2 }{t^2\,\theta}$ $\left[\dfrac{kg\,m^2 }{s^2\,K}\right]$ Entropy $\dfrac{L\,F}{T}$ $\left[\dfrac{m\,N }{K}\right]$
Specific Entropy $\dfrac{L^2 }{t^2\,\theta}$ $\left[\dfrac{m^2 }{s^2\,K}\right]$ Specific Entropy $\dfrac{L^2}{t^2\,T}$ $\left[\dfrac{m^2 }{s^2\,K}\right]$
Molar Specific Entropy $\dfrac{M\,L^2 }{t^2\,\mathfrak{M}\,\theta}$ $\left[\dfrac{m^2 }{s^2\,K\,mol}\right]$ Molar Specific Entropy $\dfrac{F\,L^2}{T\,\mathfrak{M}}$ $\left[\dfrac{N\,m }{K\,mol}\right]$
Enthalpy $\dfrac{M\,L^2 }{t^2}$ $\left[\dfrac{kg\,m^2 }{s^2}\right]$ Enthalpy ${L\,F}$ $\left[{m\,N }\right]$
Specific Enthalpy $\dfrac{L^2 }{t^2}$ $\left[\dfrac{m^2 }{s^2}\right]$ Specific Enthalpy $\dfrac{L^2}{t^2}$ $\left[\dfrac{m^2 }{sec^2}\right]$
Thermo dynamic Force $\dfrac{M\,L }{t^2\, \mathfrak{M}}$ $\left[\dfrac{kg\,m }{s^2\, mol}\right]$ Thermo dynamic Force $\dfrac{N}{\mathfrak{M}}$ $\left[\dfrac{N}{mol}\right]$
Catalytic Activity $\dfrac{\mathfrak{M} }{t}$ $\left[\dfrac{mol }{sec}\right]$ Catalytic Activity $\dfrac{\mathfrak{M} }{t}$ $\left[\dfrac{mol}{sec}\right]$
Heat Transfer Rate $\dfrac{M\,L^2 }{t^3}$ $\left[\dfrac{kg\,m^2 }{sec^3}\right]$ Heat Transfer Rate $\dfrac{L\,F }{t}$ $\left[\dfrac{m\,N}{sec}\right]$

### 9.2.3.1 One Shot Method: Constructing Dimensionless Parameters

In this method, the solution is obtained by assigning the powers to the affecting variables. The results are used to compare the powers on both sides of the equation. Several examples are presented to demonstrate this method.

# Example 9.4 Fig. 9.3 Resistance of infinite cylinder.

The researcher intuition suggests that the resistance to flow, $R$ is a function of the radius $r$, the velocity $U$, the density, $\rho$, and the absolute viscosity $\mu$. Based on this limited information construct a relationship of the variables, that is } \begin{align} \label{CylinderR:first} R = f (r, U, \rho, \mu) \end{align}

# Solution

The functionality should be in a form of \begin{align} \label{CylinderR:form} R = f\left( \,r^a\, U^b\, \rho^c \mu^d \right) \end{align} The units of the parameters are provided in Table 9.3. Thus substituting the data from the table into equation \eqref{CylinderR:form} results in \begin{align} \label{CylinderR:withUnits} \overbrace{\dfrac{ML}{t^{2}} }^{R} = Constant \, \left(\overbrace{L}^{r} \right)^a\, \left(\overbrace{\dfrac{L}{t} }^{U}\right)^b\, \left(\overbrace{\dfrac{M}{L^{3} } }^{\rho}\right)^c\, \left(\overbrace{\dfrac{M}{L\,t } }^{\mu}\right)^d \end{align} From equation \eqref{CylinderR:withUnits} the following requirements can be obtained \begin{align} \label{CylinderR:Requirments} \begin{array}{lrcl} time, t & -2 &=& -b - d \ mass, M & 1 &=& c + d \ length, L& 1&=& a + b - 3c - d \end{array} \end{align} In equations \eqref{CylinderR:withUnits} there are three equations and 4 unknowns. Expressing all the three variables in term of $d$ to obtain \begin{align} \label{CylinderR:variables} \begin{array}{rcl} a & = & 2-d \ b & = & 2 - d \ c & = & 1 - d \end{array} \end{align} Substituting equation \eqref{CylinderR:variables} into equation qref{CylinderR:withUnits} results in \begin{align} \label{CylinderR:substiting} R = Constant\, r^{2-d}\, U^{2-d} \,\rho^{1-d}\, \mu^d = Constant\, \left(\rho\,U^2\,r^2\right) \left(\dfrac{\mu}{\rho\,U\,r} \right)^d \end{align} Or rearranging equation yields \begin{align} \label{CylinderR:subRearranging} \dfrac{R}{\rho\,U^2\,r^2} = Constant\,\left(\dfrac{\mu}{\rho\,U\,r} \right)^d \end{align} The relationship between the two sides in equation \eqref{CylinderR:subRearranging} is related to the two dimensionless parameters. In dimensional analysis the functionality is not clearly defined by but rather the function of the parameters. Hence, a simple way, equation \eqref{CylinderR:subRearranging} can be represented as \begin{align} \label{CylinderR:Rearranging} \dfrac{R}{\rho\,U^2\,r^2} = Constant\,f\left(\dfrac{\mu}{\rho\,U\,r} \right) \end{align} where the power of $d$ can be eliminated.

An example of a ship is be a typical example were more than one dimensionless is to constructed. Also introduction of dimensional matrix is presented.

# Example 9.5

The modern ship today is equipped with a propeller as the main propulsion mechanism. The thrust, $T$ is known to be a function of the radius, $r$, the fluid density, $\rho$, relative velocity of the ship to the water, $U$, rotation speed, $rpm$ or $N$, and fluid viscosity, $\mu$. Assume that no other parameter affects the thrust, find the functionality of these parameters and the thrust.

# Solution

The general solution under these assumptions leads to solution of \begin{align} \label{propeller:basic} T = C \, r^a \, \rho^{b}\, U^c \, N^d\, \mu^e \end{align} It is convenient to arrange the dimensions and basic units in table.

$T$ $r$ $\rho$ $U$ $N$ $\mu$
$M$ $1$ $0$ $1$ $0$ $0$ $1$
$L$ $1$ $1$ $-3$ $1$ $0$ $-1$
$t$ $-2$ $0$ $0$ $-1$ $-1$ $-1$
Using the matrix results in \begin{align} \label{propeller:gov1} M\,L t^{-2} = L^a \, \left(Lt\right)^{b} \left( ML^{-3}\right)^c \left( t^{-t}\right)^d \left( ML^{-1} t^{-t}\right)^e \end{align} This matrix leads to three equations. \begin{align} \label{propeller:gov} \nonumber\text{Mass}, M & 1 =& c + e \\ \text{Length}, L & 1 =& a + b + -3c - e \\ \nonumber\text{time}, t & -2 =& - c - d - e \\ \end{align} The solution of this system is \begin{align} \label{propeller:sol1} \nonumber a = & 2 + d- e \\ b = & 2 - d - e \\ \nonumber c = & 1 - e \end{align} Substituting the solution \eqref{propeller:sol1} into equation qref{propeller:basic} yields \begin{align} \label{propeller:solIni} T = C \, r^{(2+d-e)} \, \rho^{(2-d-e)}\, U^{(1-e)} \, N^d\, \mu^f \end{align} Rearranging equation \eqref{propeller:solIni} provides \begin{align} \label{propeller:solIniF1} T = C \, \rho \, U^2 \, r^2 \left( \dfrac{\rho\,U\, r}{\mu} \right)^d \left( \dfrac{r\,N}{U} \right)^e \end{align} From dimensional analysis point of view the units under the power $d$ and $e$ are dimensionless. Hence, in general it can be written that \begin{align} \label{propeller:solIniF} \dfrac{T} {\rho \, U^2 \, r^2} = f \left( \dfrac{\rho\,U\, r}{\mu} \right) g \left( \dfrac{r\,N}{U} \right) \end{align} where $f$ and $g$ are arbitrary functions to be determined in experiments. Note the $rpm$ or $N$ refers to the rotation in radian per second even though $rpm$ refers to revolution per minute. It has to be mentioned that these experiments have to constructed in such way that the initial conditions and the boundary conditions are somehow eliminated.'' In practical purposes the thrust is a function of Reynolds number and several other parameters. In this example, a limited information is provided on which only Reynolds number with a additional dimensionless parameter is mentioned above.

# Example 9.6

The surface wave is a small disturbance propagating in a liquid surface. Assume that this speed for a certain geometry is a function of the surface tension, $\sigma$, density, $\rho$, and the wave length of the disturbance (or frequency of the disturbance). The flow–in to the chamber or the opening of gate is creating a disturbance. The knowledge when this disturbance is important and is detected by with the time it traveled. The time control of this certain process is critical because the chemical kinetics. The calibration of the process was done with satisfactory results. Technician by mistake releases a chemical which reduces the surface tension by half. Estimate the new speed of the disturbance.

# Solution

In the problem the functional analysis was defined as \begin{align} \label{surfaceWave:ini} U = f(\sigma,\, \rho,\, \lambda) \end{align} Equation \eqref{surfaceWave:ini} leads to three equations as \begin{align} \label{surfaceWave:gov1} \overbrace{\dfrac{L}{t} }^U = \left(\overbrace{\dfrac{M}{L^2} }^{\rho}\right)^a \left(\overbrace{\dfrac{M}{t^2} }^{\sigma}\right)^b \left(\overbrace{L }^{\lambda}\right)^c \end{align} \begin{align} \label{surfaceWave:gov} \begin{array}{rrl} \mbox{Mass}, M & a +b =& 0 \ \mbox{Length}, L & -2a+c =& 1 \ \mbox{time}, t & -2b =& - 1 \ \end{array} \end{align} The solution of equation set \eqref{surfaceWave:gov} results in \begin{align} \label{surfaceWave:solF} U = \sqrt{\dfrac{\sigma}{\lambda\,\rho}} \end{align} Hence reduction of the surface tension by half will reduce the disturbance velocity by $1/\sqrt{2}$.

# Example 9.7

Eckert number represent the amount of dissipation. Alternative number represents the dissipation, could be constructed as \begin{align} \label{dissaptionEc:dissNumber} Diss = \dfrac{\mu \left( \dfrac{dU}{d\ell}\right)^2}{\dfrac{\rho\,U^2}{\dfrac{\ell}{U}}} = \dfrac{\mu \left( \dfrac{dU}{d\ell}\right)^2\,\ell}{{\rho\,U^3}} \end{align} Show that this number is dimensionless. What is the physical interpretation it could have? Flow is achieved steady state for a very long two dimensional channel where the upper surface is moving at speed, $U_{up}$, and lower is fix. The flow is pure Couette flow i.e. a linear velocity. Developed an expression for dissipation number using the information provided.

# Solution

The nominator and denominator have to have the same units. \begin{align} \label{dissNumber:1units} \overbrace{\dfrac{M}{\cancel{L}\,t}}^{\mu}\, \overbrace{\dfrac{\cancel{L^2}}{t^2\,\cancel{L^2}}}^{\left(\dfrac{dU}{d\ell}\right)^2} \overbrace{\cancel{L}}^{\ell} & = \overbrace{\dfrac{M}{\cancel{L^3}}}^{\rho} \, \overbrace{\dfrac{\cancel{L^3}}{t^3}}^{U^3} \ \mbox{\Huge $\displaystyle\leadsto$} \qquad \dfrac{M}{t^3} & = \dfrac{M}{t^3} \end{align} The averaged velocity could be a represented (there are better methods or choices) of the energy flowing in the channel. The averaged velocity is $U/2$ and the velocity derivative is $dU/d\ell = constant = U/\ell$. With these value of the Diss number is \begin{align} \label{dissNumber:CouetteFlow} Diss = \dfrac{\mu \left( \dfrac{U}{\ell}\right)^2 \,\ell}{\rho\,\dfrac{U^3}{8}} = \dfrac{4\,\mu }{\rho\,\ell\,U} \end{align} The results show that Dissipation number is not a function of the velocity. Yet, the energy lost is a function of the velocity square $E \propto Diss\,\mu\,U$.

### 9.2.3.2 Building Blocks Method: Constructing Dimensional Parameters

Note, as opposed to the previous method, this technique allows one to find a single or several dimensionless parameters without going for the whole calculations of the dimensionless parameters.

# Example 9.8

Assume that the parameters that effects the centrifugal pumps are

 $Q$ Pump Flow Rate $rpm$ or $N$ angular rotation speed $D$ Rotor Diameter $\rho$ liquid density (assuming liquid phase) $B_T$ Liquid Bulk Modulus $\mu$ liquid viscosity $\epsilon$ Typical Roughness of pump surface $g$ gravity force (body force) $\Delta P$ Pressure created by the pump
Construct the functional relationship between the variables. Discuss the physical meaning of these numbers. Discuss which of these dimensionless parameters can be neglected as it is known reasonably.

# Solution

The functionality can be written as \begin{align} \label{pumpScaling:fun} 0 = f \left( D,\, N,\,\rho,\,Q,\,B_T,\, \mu ,\, \epsilon,\,g,\,\Delta P \right) \end{align} The three basic parameters to be used are $D$ [L], $\rho$ [M], and $N$ [t]. There are nine (9) parameters thus the number of dimensionless parameters is $9-3=6$. For simplicity the $RPM$ will be denoted as $N$. The first set is to be worked on is $Q,\,D,\,\rho,\,N$ as \begin{align} \label{pumpScaling:Q:ini} \overbrace{\dfrac{L^3}{t} }^Q = \left(\overbrace{ {L} }^{D}\right)^a \left(\overbrace{\dfrac{M}{L^3} }^{\rho}\right)^b \left(\overbrace{\dfrac{1}{t} }^{N}\right)^c \end{align} \begin{align} \label{pumpScaling:Q:gov} \left. \begin{array}{rrl} \text{Length}, L & a - 3b =& 3 \\ \text{Mass}, M & b =& 0 \\ \text{time}, t & -c =& - 1 \end{array} \right\} \Longrightarrow \pi_1 = \dfrac{Q}{N\,D^3} \end{align} For the second term $B_T$ it follows \begin{align} \label{pumpScaling:BT:ini} \overbrace{\dfrac{M}{L\,t^2} }^{B_T} = \left(\overbrace{ {L} }^{D}\right)^a \left(\overbrace{\dfrac{M}{L^3} }^{\rho}\right)^b \left(\overbrace{\dfrac{1}{t} }^{N}\right)^c \end{align} \begin{align} \label{pumpScaling:BT:gov} \left. \begin{array}{rrl} \text{Mass}, M & b =& 1 \\ \text{Length}, L & a - 3b =& -1 \\ \text{time}, t & -c =& - 2 \end{array} \right\} \Longrightarrow \pi_2 = \dfrac{B_T}{\rho\,N^2\,D^2} \end{align} The next term, $\mu$, \begin{align} \label{pumpScaling:mu:ini0} \overbrace{\dfrac{M}{L\,t} }^{\mu} = \left(\overbrace{ {L} }^{D}\right)^a \left(\overbrace{\dfrac{M}{L^3} }^{\rho}\right)^b \left(\overbrace{\dfrac{1}{t} }^{N}\right)^c \end{align} \begin{align} \label{pumpScaling:mu:gov1} \left. \begin{array}{rrl} \text{Mass}, M & b =& 1 \\ \text{Length}, L & a - 3b =& -1 \\ \text{time}, t & -c =& - 1 \end{array} \right\} \Longrightarrow \pi_3 = \dfrac{\rho\,N^2\,D^2}{\mu} \end{align} The next term, $\epsilon$, \begin{align} \label{pumpScaling:mu:ini1} \overbrace{L}^{\epsilon} = \left(\overbrace{ {L} }^{D}\right)^a \left(\overbrace{\dfrac{M}{L^3} }^{\rho}\right)^b \left(\overbrace{\dfrac{1}{t} }^{N}\right)^c \end{align} \begin{align} \label{pumpScaling:mu:gov2} \left. \begin{array}{rrl} \text{Mass}, M & b =& 0 \\ \text{Length}, L & a - 3b =& 1 \\ \text{time}, t & -c =& 0 \end{array} \right\} \Longrightarrow \pi_4 = \dfrac{\epsilon}{D} \end{align} The next term, $g$, \begin{align} \label{pumpScaling:mu:ini2} \overbrace{\dfrac{L}{t^2}}^{g} = \left(\overbrace{ {L} }^{D}\right)^a \left(\overbrace{\dfrac{M}{L^3} }^{\rho}\right)^b \left(\overbrace{\dfrac{1}{t} }^{N}\right)^c \end{align} \begin{align} \label{pumpScaling:mu:gov3} \left. \begin{array}{rrl} \text{Mass}, M & b =& 0 \\ \text{Length}, L & a - 3b =& 1 \\ \text{time}, t & -c =& -2 \end{array} \right\} \Longrightarrow \pi_5 = \dfrac{g}{D\,N^2} \end{align} The next term, $\Delta P$, (similar to $B_T$) \begin{align} \label{pumpScaling:mu:ini} \overbrace{\dfrac{L}{t^2}}^{\Delta P} = \left(\overbrace{ {L} }^{D}\right)^a \left(\overbrace{\dfrac{M}{L^3} }^{\rho}\right)^b \left(\overbrace{\dfrac{1}{t} }^{N}\right)^c \end{align} \begin{align} \label{pumpScaling:mu:gov} \left. \begin{array}{rrl} \text{Mass}, M & b =& 1 \\ \text{Length}, L & a - 3b =& -1 \\ \text{time}, t & -c =& -2 \end{array} \right\} \Longrightarrow \pi_6 = \dfrac{\Delta P}{\rho \,N^2\,D^2} \end{align} The first dimensionless parameter $\pi_1$ represents the dimensionless flow rate. The second number represents the importance of the compressibility of the liquid in the pump. Some argue that this parameter is similar to Mach number (speed of disturbance to speed of sound. The third parameter is similar to Reynolds number since the combination $N\,D$ can be interpreted as velocity. The fourth number represents the production quality (mostly mode by some casting process The fifth dimensionless parameter is related to the ratio of the body forces to gravity forces. The last number represent the effectiveness'' of pump or can be viewed as dimensionless pressure obtained from the pump.

In practice, the roughness is similar to similar size pump and can be neglected. However, if completely different size of pumps are compared then this number must be considered. In cases where the compressibility of the liquid can be neglected or the pressure increase is relatively insignificant, the second dimensionless parameter can be neglected.

A pump is a device that intends to increase the pressure. The increase of the pressure involves energy inserted to to system. This energy is divided to a useful energy ( pressure increase) and to overcome the losses in the system. These losses has several components which includes the friction in the system, change order of the flow and ideal flow'' loss. The most dominate loss in pump is loss of order, also know as turbulence (not covered yet this book.). If this physical phenomenon is accepted than the resistance is neglected and the fourth parameter is removed. In that case the functional relationship can be written as \begin{align} \label{pumpScaling:finalFun} \dfrac{\Delta P }{N^2, D^2} = f \left( \dfrac{Q}{N\,D^3}\right) \end{align}

### 9.2.3.3 Mathematical Method: Constructing Dimensional Parameters

under construction please ignore for time being In the progression of the development of the technique the new evolution is the mathematical method. It can be noticed that in the previous technique the same matrix was constructed with different vector solution (the right hand side of the equation). This fact is the source to improve the previous method. However, it has to be cautioned that this technique is overkill in most cases. Actually, this author is not aware for any case this technique has any advantage over the building block'' technique. In the following hypothetical example demonstrates the reason for the reduction of variables. Assume that water is used to transport uniform grains of gold. The total amount grains of gold is to be determined per unit length. For this analysis it is assumed that grains of gold grains are uniformly distributed. The following parameters and their dimensions are considered: \begin{center} \begin{longtable}{|l|c|c|l|} \caption[Units of the Pendulum] {Units and Parameters of gold grains \label{dim:tab:gold}} \ \hline \multicolumn{1}{|c|}{ {|c|}{ { Parameters } & \multicolumn{1}{c|}{ {c|}{ { Units } & \multicolumn{1}{c|}{ {c|}{ { Dimension } & \endfirsthead \multicolumn{4}{c} {\bfseries ablename\ \thetable{} – continued from previous page} \ \hline \endhead \hline \multicolumn{4}{|r|}{{Continued on next page}} \ \hline \endfoot \hline \hline \endlastfoot grains amount & q & $M/L$ & total grains per unit length \ cross section area & A & $L^2$ & pipe cross section \ grains per volume & gr & $grains/L^3$ & count of grain per V \ grain weight & e & $M/grain$ & count of grain per V \ \end{longtable} \end{center} Notice that $grains$ and $grain$ are the same units for this discussion. Accordingly, the dimensional matrix can be constructed as \begin{center} \begin{longtable}{c|cccc} \caption[gold grain dimensional matrix] { gold grain dimensional matrix } \ \multicolumn{1}{c|}{} & \multicolumn{1}{p{0.5in}}{\centering{\bf q}} & \multicolumn{1}{p{0.5in}}{\centering{\bf A}} & \multicolumn{1}{p{0.5in}}{\centering{\bf gr}} & \multicolumn{1}{p{0.5in}}{\centering{\bf e}} \ \hline \endfirsthead \multicolumn{5}{c} {\bfseries ablename\ \thetable{} – continued from previous page} \ \multicolumn{1}{c|}{} & \multicolumn{1}{p{0.5in}}{\centering{\bf q}} & \multicolumn{1}{p{0.5in}}{\centering{\bf A}} & \multicolumn{1}{p{0.5in}}{\centering{\bf gr}} & \multicolumn{1}{p{0.5in}}{\centering{\bf e}} \ \hline \endhead \multicolumn{5}{|r|}{{Continued on next page}} \ \hline \endfoot \endlastfoot M & 1 & 0& 0& 1 \ L & 1 & 2 & 3 & 0 \ grain & 0 & 0 & 1 & -1 \ \end{longtable} \end{center} In this case the total number variables are 4 and number basic units are 3. Thus, the total of one dimensional parameter. End ignore section

### 9.2.4 Similarity and Similitude

One of dimensional analysis is the key point is the concept that the solution can be obtained by conducting experiments on similar but not identical systems. The analysis here suggests and demonstrates that the solution is based on several dimensionless numbers. Hence, constructing experiments of the situation where the same dimensionless parameters obtains could, in theory, yield a solution to problem at hand. Thus, knowing what are dimensionless parameters should provide the knowledge of constructing the experiments. In this section deals with these similarities which in the literature some refer as analogy or similitude. It is hard to obtain complete similarity. Hence, there is discussion how similar the model is to the prototype. It is common to differentiate between three kinds of similarities: geometric, kinetics, and dynamic. This characterization started because historical reasons and it, some times, has merit especially when applying Buckingham's method. In Nusselt's method this differentiation is less important.

#### Geometric Similarity

One of the logical part of dimensional analysis is how the experiences should be similar to actual body they are supposed to represent. This logical conclusion is an add–on and this author is not aware of any proof to this requirement based on Buckingham's methods. Ironically, this conclusion is based on Nusselt's method which calls for the same dimensionless boundary conditions. Again, Nusselt's method, sometimes or even often, requires similarity because the requirements to the boundary conditions. Here this postulated idea is adapted. Under this idea the prototype area has to be square of the actual model or \begin{align} \label{dim:eq:AprotypeAmodel} \dfrac{A_{p}} {A_{m}} = \left(\dfrac{{ll_1}_{prototype}}{{ll_1}_{model}} \right)^2 = \left(\dfrac{{ll_2}_{p}}{{ll_2}_{m}} \right)^2 \end{align} where $ll_1$ and $ll_2$ are the typical dimensions in two different directions and subscript $p$ refers to the prototype and $m$ to the model. Under the same argument the volumes change with the cubes of lengths. In some situations, the model faces inability to match two or more dimensionless parameters. In that case, the solution is to sacrifice the geometric similarity to minimize the undesirable effects. For example, river modeling requires to distort vertical scales to eliminate the influence of surface tension or bed roughness or sedimentation.

#### Kinematic Similarity

The perfect kinetics similarity is obtained when there are geometrical similarity and the motions of the fluid above the objects are the same. If this similarity is not possible, then the desire to achieve a motion picture'' which is characterized by ratios of corresponding velocities and accelerations is the same throughout the actual flow field. It is common in the literature, to discuss the situations where the model and prototype are similar but the velocities are different by a different scaling factor. The geometrical similarity aside the shapes and counters of the object it also can requires surface roughness and erosion of surfaces of mobile surfaces or sedimentation of particles surface tensions. These impose demands require a minimum on the friction velocity. In some cases the minimum velocity can be $U_{min} = \sqrt{\tau_w/\rho}$. For example, there is no way achieve low Reynolds number with thin film flow.

#### Dynamics Similarity

The dynamic similarity has many confusing and conflicting definitions in the literature. Here this term refers to similarity of the forces. It follows, based on Newton's second law, that this requires similarity in the accelerations and masses between the model and prototype. It was shown that the solution is a function of several typical dimensionless parameters. One of such dimensionless parameter is the Froude number. The solution for the model and the prototype are the same, since both cases have the same Froude number. Hence it can be written that \begin{align} \label{dim:eq:FrSame} \left( \dfrac{U^2}{g\,\ell} \right)_m = \left( \dfrac{U^2}{g\,\ell} \right)_p \end{align} It can be noticed that $t \sim \ell /U$ thus equation \eqref{dim:eq:FrSame} can be written as \begin{align} \label{dim:eq:FrSame1} \left( \dfrac{U}{g \, t } \right)_m = \left( \dfrac{U}{g\,t } \right)_p \end{align} and noticing that $a \propto U /t$ \begin{align} \label{dim:eq:FrSame2} \left( \dfrac{a}{g } \right)_m = \left( \dfrac{a}{g } \right)_p \end{align} and $a \propto F / m$ and $m = \rho \, \ell^3$ hence $a = F / \rho \, \ell^3$. Substituting into equation \eqref{dim:eq:FrSame2} yields \begin{align} \label{dim:eq:FrSame3} \left( \dfrac{F}{\rho\,\ell^3} \right)_m = \left( \dfrac{F}{\rho\,\ell^3 } \right)_p \Longrightarrow \dfrac{F_p}{F_m} = \dfrac { \left( \rho\,\ell^3\right)_p }{ \left( \rho\,\ell^3\right)_m} \end{align} In this manipulation, it was shown that the ratio of the forces in the model and forces in the prototype is related to ratio of the dimensions and the density of the same systems. While in Buckingham's methods these hand waiving are not precise, the fact remains that there is a strong correlation between these forces. The above analysis was dealing with the forces related to gravity. A discussion about force related the viscous forces is similar and is presented for the completeness. The Reynolds numbers is a common part of Navier–Stokes equations and if the solution of the prototype and for model to be same, the Reynolds numbers have to be same. \begin{align} \label{dim:eq:ReSame} Re_m = Re_p \Longrightarrow \left( \dfrac{\rho\,U \,\ell}{\mu} \right)_m = \left( \dfrac{\rho\,U \,\ell}{\mu} \right)_p \end{align} Utilizing the relationship $U \propto \ell/t$ transforms equation qref{dim:eq:ReSame} into \begin{align} \label{dim:eq:ReSame1} \left( \dfrac{\rho\,\ell^2}{\mu\,t} \right)_m = \left( \dfrac{\rho\,\ell^2}{\mu\,t} \right)_p \end{align} multiplying by the length on both side of the fraction by $\ell\,U$as \begin{align} \label{dim:eq:ReSame2} \left( \dfrac{\rho\,\ell^3\,U}{\mu\,t\,\ell\,U} \right)_m = \left( \dfrac{\rho\,\ell^3\,U}{\mu\,t\,\ell\,U} \right)_p \Longrightarrow \dfrac{\left( \rho\,\ell^3 \,U /t \right)_m }{\left( \rho\,\ell^3\,U /t \right)_p } = \dfrac{\left( \mu\,\ell\,U \right)_m }{ \left( \mu\,\ell\,U \right) _p} \end{align} Noticing that $U/t$ is the acceleration and $\rho\,\ell$ is the mass thus the forces on the right hand side are proportional if the $Re$ number are the same. In this analysis/discussion, it is assumed that a linear relationship exist. However, the Navier–Stokes equations are not linear and hence this assumption is excessive and this assumption can produce another source of inaccuracy. While this explanation is a poor practice for the real world, it common to provide questions in exams and other tests on this issue. This section is provide to this purpose.

# Example 9.9

The liquid height rises in a tube due to the surface tension, $\sigma$ is $h$. Assume that this height is a function of the body force (gravity, $g$), fluid density, $\rho$, radius, $r$, and the contact angle $\theta$. Using Buckingham's theorem develop the relationship of the parameters. In experimental with a diameter 0.001 [m] and surface tension of 73 milli-Newtons/meter and contact angle of $75^\circ$ a height is 0.01 [m] was obtained. In another situation, the surface tension is 146 milli-Newtons/meter, the diameter is 0.02 [m] and the contact angle and density remain the same. Estimate the height.

# Solution

It was given that the height is a function of several parameters such \begin{align} \label{tubesH:ini} h = f \left( \sigma, \rho, g, \theta, r \right) \end{align} There are 6 parameters in the problem and the 3 basic parameters [$L,\, M,\, t$]. Thus the number of dimensionless groups is (6-3=3). In Buckingham's methods it is either that the angle isn't considered or the angle is dimensionless group by itself. Five parameters are left to form the next two dimensionless groups. One technique that was suggested is the possibility to use three parameters which contain the basic parameters [M, L, t] and with them form a new group with each of the left over parameters. In this case, density, $\rho$ for [M] and $d$ for [L] and gravity, $g$ for time [t]. For the surface tension, $\sigma$ it becomes \begin{align} \label{tubesH:surfaceTg} {\left[ \overbrace{M\,L^{-3}}^{\rho}\right]}^a\, {\left[ \overbrace{L}^{r} \right]}^b\, {\left[ \overbrace{L\,t^{-2}}^{g} \right]}^c \, {\left[ \overbrace{M\,t^{-2}}^{\sigma} \right]}^1 = M^0 \, L^0 \, t^0 \end{align} Equation \eqref{tubesH:surfaceTg} leads to three equations which are \begin{align} \label{tubesH:surfaceTgEq} \begin{array}{rrl} \text{Mass}, M & a + 1 = & 0 \\ \text{Length}, L & -3a + b + c =& 0 \\ \text{time}, t & -2c - 2 =& 0 \\ \end{array} \end{align} the solution is $a=-1\quad b=-2 \quad c=-1$ Thus the dimensionless group is $\dfrac{\sigma}{\rho\, r^2 \,g}$. The third group obtained under the same procedure to be $h/r$. In the second part the calculations for the estimated of height based on the new ratios. From the above analysis the functional dependency can be written as \begin{align} \label{tubesH:functionality} \dfrac{h}{d} = f \left( \dfrac{\sigma}{\rho\, r^5 \,g}\,,\, \theta \right) \end{align} which leads to the same angle and the same dimensional number. Hence, \begin{align} \label{tubesH:a} \dfrac{h_1}{d_1} = \dfrac{h_2}{d_2} = f \left( \dfrac{\sigma}{\rho\, r^2 \,g}\,,\, \theta \right) \end{align} Since the dimensionless parameters remain the same, the ratio of height and radius must be remain the same. Hence, \begin{align} \label{tubesH:b} h_2 = \dfrac{h_1\, d_2}{d_1} = \dfrac{0.01\times 0.002}{0.001} = 0.002 \end{align}