# 9.4.3: Examples for Dimensional Analysis

Example 9.20

The similarity of pumps is determined by comparing several dimensional numbers among them are Reynolds number, Euler number, Rossby number etc. Assume that the only numbers which affect the flow are Reynolds and Euler number. The flow rate of the imaginary pump is 0.25 [$$m^3/sec$$] and pressure increase for this flow rate is 2 [Bar] with 2500 [kw]. Due to increase of demand, it is suggested to replace the pump with a 4 times larger pump. What is the new estimated flow rate, pressure increase, and power consumption?

Solution 9.20

It provided that the Reynolds number controls the situation.  The density and viscosity remains the same and hence
$\label{pump1:Re} Re_m = Re_p \Longrightarrow U_m\,D_m = U_p\, D_p \Longrightarrow U_p = \dfrac{D_m}{D_P} U_m \tag{68}$
It can be noticed that initial situation is considered as the model and while the new pump is the prototype. The new flow rate, $$Q$$, depends on the ratio of the area and velocity as
$\label{pump1:Q} \dfrac{Q_p}{Q_m} = \dfrac{A_p \, U_p}{A_m \, U_m} \Longrightarrow Q_p = {Q_m}\dfrac{A_p \, U_p}{A_m \, U_m} = {Q_m}\dfrac{{D_p}^2 \, U_p}{{D_m}^2 \, U_m} \tag{69}$
Thus the prototype flow rate is
$\label{pump1:Qf} Q_p = Q_m\, \left(\dfrac{D_p}{D_m}\right)^3 = 0.25\times 4^3 = 16 \left[ \dfrac{m^3}{sec}\right] \tag{70}$
The new pressure is obtain by comparing the Euler number as
$\label{pump1:Eu} Eu_p = Eu_m \Longrightarrow \left(\dfrac{\Delta P} {\dfrac{1}{2}\rho\,U^2} \right)_p = \left(\dfrac{\Delta P} {\dfrac{1}{2}\rho\,U^2} \right)_m \tag{71}$
Rearranging equation (71) provides
$\label{pump1:Euarranged} \dfrac{\left(\Delta P\right)_p} {\left(\Delta P\right)_m} = \dfrac{\left(\cancel{\rho}\,U^2 \right)_p} {\left(\cancel{\rho}\,U^2 \right)_m} = \dfrac{\left(U^2 \right)_p} {\left(U^2 \right)_m} \tag{72}$
Utilizing equation (68)
$\label{pump1:DeltaP} \Delta P_p = \Delta P_m \left( \dfrac{D_p}{D_m} \right)^2 \tag{73}$
The power can be obtained from the following
$\label{pump1:power} \dot{W} = \dfrac{F\, \ell}{t} = F\, U = P \, A \, U \tag{74}$
In this analysis, it is assumed that pressure is uniform in the cross section. This assumption is appropriate because only the secondary flows in the radial direction (to be discussed in this book section on pumps). Hence, the ratio of power between the two pump can be written as
$\label{pump1:powerRatio} \dfrac{\dot{W}_p}{\dot{W}_m} = \dfrac{\left( P \, A \, U \right)_p} {\left( P \, A \, U \right)_m} \tag{75}$
Utilizing equations above in this ratio leads to
$\label{pump1:pRvalue} \dfrac{\dot{W}_p}{\dot{W}_m} = \overbrace{\left(\dfrac{D_p}{D_m}\right)^2}^{P_p/P_m} \overbrace{\left(\dfrac{D_p}{D_m}\right)^2}^{A_p/A_m} \overbrace{\left(\dfrac{D_p}{D_m}\right)}^{U_p/U_m} = \left(\dfrac{D_p}{D_m}\right)^5 \tag{76}$

Example 9.21

The flow resistance to flow of the water in a pipe is to be simulated by flow of air. Estimate the pressure loss ratio if Reynolds number remains constant. This kind of study appears in the industry in which the compressibility of the air is ignored. However, the air is a compressible substance that flows the ideal gas model. Water is a substance that can be considered incompressible flow for relatively small pressure change. Estimate the error using the averaged properties of the air.

Solution 9.21

For the first part, the Reynolds number is the single controlling parameter which affects the pressure loss. Thus it can be written that the Euler number is function of the Reynolds number.
$\label{airWater:ReEu} Eu = f (Re) \tag{77}$
Thus, to have a similar situation the Reynolds and Euler have to be same.
$\label{airWater:Re-Eu} Re_p = Re_m \qquad \qquad Eu_m = Eu_p \tag{78}$
Hence,
$\label{airWater:Rer} \dfrac{U_m}{U_p} = \dfrac{ll_p}{ll_m} \dfrac{\rho}{\rho_m} \dfrac{\mu_p}{\mu_m} \tag{79}$
and for Euler number
$\label{airWater:Eur} \dfrac{\Delta P_m}{\Delta P_p} = \dfrac{\rho_m}{\rho_p} \dfrac{U_m}{U_p} \tag{80}$
and utilizing equation (79) yields
$\label{airWater:ReEucombined} \dfrac{\Delta P_m}{\Delta P_p} = \left(\dfrac{ll_p}{ll_m}\right)^2 \left(\dfrac{\mu_m}{\mu_p}\right)^2 \left(\dfrac{\rho_p}{\rho_m}\right) \tag{81}$
Inserting the numerical values results in
$\label{airWater:numerical} \dfrac{\Delta P_m}{\Delta P_p} = 1 \times 1000 \times \tag{82}$
It can be noticed that the density of the air changes considerably hence the calculations produce a considerable error which can render the calculations useless (a typical problem of Buckingham's method). Assuming a new variable that effect the problem, air density variation. If that variable is introduced into problem, air can be used to simulate water flow. However as a first approximation, the air properties are calculated based on the averaged values between the entrance and exit values. If the pressure reduction is a function of pressure reduction (iterative process). to be continue

Example 9.22

A device operating on a surface of a liquid to study using a model with a ratio 1:20. What should be ratio of kinematic viscosity between the model and prototype so that Froude and Reynolds numbers remain the same. Assume that body force remains the same and velocity is reduced by half.

Solution 9.22

The requirement is that Reynolds
$\label{kinimaticViscosity:Re} Re_m = R_p \Longrightarrow \left(\dfrac{U\,\ell}{\nu} \right)_p = \left(\dfrac{U\,\ell}{\nu} \right)_m \tag{83}$
The Froude needs to be similar so
$\label{kinimaticViscosity:Fr} Fr_m= Fr_p \Longrightarrow \left(\dfrac{U}{\sqrt{g\,\ell}} \right)_p = \left(\dfrac{U\,\ell}{\nu} \right)_m \tag{84}$
dividing equation (83) by equation (84) results in
$\label{kinimaticViscosity:ReDRe1} \left(\dfrac{U\,\ell}{\nu} \right)_p / \left(\dfrac{U}{\sqrt{g\,\ell}} \right)_p = \left(\dfrac{U\,\ell}{\nu} \right)_m / \left(\dfrac{U}{\sqrt{g\,\ell}} \right)_m \tag{85}$
or
$\label{kinimaticViscosity:ReDRe} \left(\dfrac{\ell\, \sqrt{g\,\ell}} {\nu} \right)_p = \left(\dfrac{\ell\, \sqrt{g\,\ell}} {\nu} \right)_m \tag{86}$
The kinematic viscosity ratio is then
$\label{kinimaticViscosity:knRatio} \dfrac{\nu_p}{\nu_m} = \left(\dfrac{ll_m} {ll_p} \right)^{3/2} = \left(1/20\right)^{3/2} \tag{87}$
It can be noticed that this can be achieved using Ohnesorge Number like this presentation.

Example 9.23

In AP physics in 2005 the first question reads "A ball of mass $$M$$ is thrown vertically upward with an initial speed of $$U_0$$. Does it take longer for the ball to rise to its maximum height or to fall from its maximum height back to the height from which it was thrown? It also was mentioned that resistance is proportional to ball velocity (Stoke flow). Justify your answer.'' Use the dimensional analysis to examine this situation.

Solution 9.23

The parameters that can effect the situation are (initial) velocity of the ball, air resistance (assuming Stokes flow e.g. the resistance is function of the velocity), maximum height, and gravity. Functionality of these parameters can be written as
$\label{dim:ex:APballBuckingham} t = f ( U,\, k,\, H,\, m,\, g) \tag{88}$
The time up and/or down must be written in the same fashion since fundamental principle of Buckingham's $$\pi$$ theorem the functionally is unknown but only dimensionless parameters are dictated. Hence, no relationship between the time up and down can be provided. However, Nusselt's method provides first to written the governing equations. The governing equation for the ball climbing up is
$\label{dim:eq:ballClimbing} m\,\dfrac{dU}{dt} = - m\,g - k\,U \tag{89}$
when the negative sign indicates that the positive direction is up. The initial condition is that
$\label{dim::eq:ballInitialConidtion} U(0) = U_0 \tag{90}$
The governing equation the way down is
$\label{dim:eq:ballDown} m\,\dfrac{dU}{dt} = - m\,g + k\,U \tag{91}$
with initial condition of
$\label{dim:eq:ballInitialConidtionDown} U(0) = 0 \tag{92}$
Equation (91) has no typical velocity (assuming at this stage that solution was not solved ever before). Dividing equation (91) by $$m\,g$$ and inserting the gravitation constant into the derivative results in
$\label{dim:eq:boatGovDimless2} \dfrac{dU}{d\,(g\,t)} = -1 + \dfrac{k\,U}{ m\,g} \tag{93}$
The gravity constant, $$g$$, could be inserted because it is constant. Equation suggests that velocity should be normalized by as dimensionless group, $$\left.{k\,U}\right/{ m\,g}$$. Without solving the equations, it can be observed that value of dimensionless group above or below one change the characteristic of the governing equation (positive slop or negative slop). Non–dimensioning of initial condition (90) yields
$\label{dim:eq:ballInitialConidtionD} \dfrac{k\,U(0)}{ m\,g} = \dfrac{k\,U_0}{ m\,g} \tag{94}$
In this case, if the value $$\left.{k\,U}\right/{ m\,g}$$ is above one change the characteristic of the situation. This exercise what not to solve this simple Physics mechanics problem but rather to demonstrate the power of dimensional analysis power. So, What this information tell us? In the case the supper critical initial velocity, the ball can be above critical velocity $$\dfrac{k\,U_0}{ m\,g} >1$$ on the up. However the ball never can be above the critical velocity and hence the time up will shorter the time done. For the initial velocity below the critical velocity, while it is know that the answer is the same, the dimensional analysis does not provide a solution. On the way up ball can start

Fig. Description of the boat crossing river.

Example 9.24

Two boats sail from the opposite sides of river (see Figure ). They meet at a distance $$ll_1$$ (for example 1000) meters from bank $$\mathbf{A}$$. The boats reach the opposite side respectively and continue back to their original bank. The boats meet for the second time at $$ll_2$$ (for example 500) $$[m]$$ from bank $$\mathbf{B}$$. What is the river width? What are the dimensional parameters that control the problem?

Solution 9.24

The original problem was constructed so it was suitable to the 11 years old author's daughter who was doing her precalculus. However, it appears that this question can be used to demonstrate some of the power of the the dimensional analysis. Using the Buckingham's method it is assumed that diameter is a function of the velocities and lengths. Hence, the following can be written
$\label{dim:eq:govBuck} D = f(ll_1, \, ll_2,\,U_A,\,U_B) \tag{95}$
Where $$D$$ is the river width. Hence, according basic idea the following can be written
$\label{dim:eq:govR} D = {ll_1}^a\,{ll_1}^b\,{U_A}^c\,{U_B}^d \tag{96}$
The solution of equation (96) requires that
$D = [L]^a[L]^b\left[\dfrac{L}{T}\right]^c\left[\dfrac{L}{T}\right]^d \tag{97}$
The time has to be zero hence it requires that
$\label{dim:eq:boatTime} 0 = c+d \tag{98}$
The units length requires that
$\label{dim:eq:boatLength} 1 = a +b + c+ d \tag{99}$
Combined equation (98) with equation (99) results in
$\label{dim:eq:boatCombined} 1=a+b \tag{100}$
It can noticed that symmetry arguments require that $$a$$ and $$b$$ must be identical. Hence, $$a=b=\sqrt{1/2}$$ and the solutions is of the form $$D= f(\sqrt{ll_1\,ll_2})$$. From the analytical solution it was found that this solution is wrong. Another approach utilizing the minimized Buckingham's approach reads
$\label{dim:eq:boatMinGov} D = f (ll_1,\,U_A) \tag{101}$
In the standard form this leads to
$\label{dim:eq:boadMinLead} D = [L]^a\,\left[\dfrac{L}{T}\right]^b \tag{102}$
Which leads to the requirements of $$b=0$$ and $$a=1$$. Which again conflict with the actual analytical solution. Using Nusselt's method requires to write the governing equation. The governing equations are based equating the time traveled to first and second meeting as the following
$\label{dim:eq:boatNusseltGov1} \dfrac{ll_1}{U_A} = \dfrac{D-ll_1}{U_B} \tag{103}$
At the second meeting the time is
$\label{dim:eq:boatNusseltGov2} \dfrac{D+ll_2}{U_A} = \dfrac{2\,D-ll_2}{U_B} \tag{104}$
Equations (103) and (104) have three unknowns $$D$$, $$U_A$$ and $$U_B$$. The non–dimensioning process can be carried by dividing governing equations by $$D$$ and multiply by $$U_B$$ to become
$\label{dim:eq:boatNusseltGov1d} \overline{ll_1} = \left( 1 - \overline{ll_1} \right) \,\dfrac{U_A}{U_B} \tag{105}$
$\label{dim:eq:boatNusseltGov2d} 1+\overline{ll_2} = \left( 2 - \overline{ll_2} \right) \,\dfrac{U_A}{U_B} \tag{106}$
Equations (105) and (106) have three unknowns. However, the velocity ratio is artificial parameter or dependent parameter. Hence division of the dimensionless governing equations yield one equation with one unknown as
$\label{dim:eq:boatNusseltSsol} \dfrac{\overline{ll_1}}{1+\overline{ll_2} } = \dfrac{1 - \overline{ll_1}}{ 2 - \overline{ll_2}} \tag{107}$
Equation determines that $$\overline{ll_1}$$ is a function of $$\overline{ll_2}$$. It can be noticed that $$D$$, $$ll_1$$ and $$ll_2$$ are connected. Hence, knowing two parameters leads to the solution of the missing parameter. From dimensional analysis it can be written that the
$\label{dim:eq:boatSolNusselt} \overline{ll_2} = f ( \overline{ll_1} ) = \dfrac{2\,\dfrac{\overline{ll_1} }{1-\overline{ll_1} } - 1 } { 1+ \dfrac{\overline{ll_1}}{1-\overline{ll_1} } } \tag{108}$
It can be concluded that river width is a function of implicit of $$\overline{ll_1}$$ and $$\overline{ll_2}$$. Clearly the Nusselt's technique provided write based to obtain the dimensionless parameters. A bit smarter selection of the normalizing parameter can provide explicit solution. An alternative definition of dimensionless parameters of $$\widetilde{D} = D/ll_1$$ and $$\widetilde{ll_2} = ll_2/ll_1$$ can provide the need path. Equation (107) can be converted quadratic equation for $$D$$ as
$\label{dim:eq:boatSolNusseltNew} \dfrac{1}{\widetilde{D} - \widetilde{ll_2} } = \dfrac{\widetilde{D} - 1}{ 2\, \widetilde{D} - \widetilde{ll_2} } \tag{109}$
Equation (109) is quadratic which can be solved analytically. The solution can be presented as
$\label{dim:eq:boatSolPresentation} D = ll_1 \, f \left(\dfrac{ll_2}{ll_1} \right) \tag{110}$

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.