Worked Examples: Units, Conversion Factors & Dimensional Analysis

J. Gordon Leishman

doi:https://doi.org/10.15394/eaglepub.2022.1066.9

50 Worked Examples: Units, Conversion Factors & Dimensional Analysis

These worked examples have been fielded as homework problems or exam questions.

Worked Example #1

Determine the base dimensions of each of the following variables:

(a) Plane angle
(b) Specific volume
(c) Force
(d) Stress

(a) Plane angle: A plane angle $\theta$ is defined in terms of the lines from two points meeting at a vertex and is defined by the arc length of a circle subtended by the lines and the radius of that circle. The unit of plane angle is the radian. Because it is the ratio of an arc length to the radius, then the plane angle is dimensionless, i.e., a radian is one measurement unit that is already dimensionless, i.e.,

$\left[ \theta \right] = 1$

(b) Specific volume: The specific volume $SV$ is defined as the ratio of volume to mass, i.e., it is the reciprocal of the density. Therefore,

$\left[ SV \right] = \frac{L^3}{M} = L^3 M^{-1}$

(c) Force: Force $F$ is the product of mass times acceleration so

$\left[ F \right] = M L T^{-2}$

(d) Stress: Stress $\sigma$ has units of force per area, but a force is the product of mass times acceleration so

$\left[ \sigma \right] = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2}$

Worked Example #2

A student is curious about why some insects can walk on water. The student finds out that a fluid property of importance in this problem is called surface tension, which is given the symbol $\sigma_s$ , and it has dimensions of force per unit length. Write the dimensions of surface tension in terms of its base dimensions.

We are told that the units of surface tension, $\sigma_s$ , have dimensions of force per unit length. We also know that a force is mass times acceleration, which is $MLT^{-2}$ in base units. Therefore, we can write for the base units of surface tension that

$\left[ \sigma_s \right] = \left( MLT^{-2}\right) L^{-1} = MT^{-2}$

Worked Example #3

Write the primary dimensions of each of the following variables from the field of thermodynamics:

(a) Energy, $E$
(b) Specific energy, $e=E/m$
(c) Power, $P$

(a) Energy has units of force times distance, i.e.,

$\left[ E \right] = \left( MLT^{-2} \right) L = ML^{2} T^{-2}$

(b) Specific energy has units of energy per unit mass, i.e.,

$\left[ e \right] = ML^{2} T^{-2} M^{-1} = L^{2} T^{-2}$

(c) Power is the rate of doing work, so a force times distance per unit time, i.e.,

$\left[ P \right] = \left( MLT^{-2} \right) L T^{-1} = M L^{2} T^{-3}$

Worked Example #4

Determine the primary (base) dimensions of each of the following parameters from thermodynamics:

Energy, $E$
Work, $W$
Power, $P$
Heat, $Q$

1. Energy, $E$ , is the ability to do work and is measured in Joules (J) in the SI system and foot-pounds (ft-lb) in the USC system. Notice that “foot-pounds” is the USC unit and not “pounds-foot” or “pounds-feet.” Energy has the same units of work (force times distance), so $\left[ E \right] = \left( MLT^{-2} \right) L = ML^{2} T^{-2}$ which are the base units. Notice that the units of force are obtained from the product of mass and acceleration, i.e., [ $F$ ] = $( M) (LT^{-2})$ = $M L T^{-2}$ .

2. Work, $W$ , is also measured in Joules (J) in the SI system and “foot-pounds” (ft-lb) in the USC system. Work is equivalent to force times distance, so $\left[ W \right] = ML T^{-2} L$ $=$ $M L^{2} T^{-2}$ , which are the base units of work.

3. Power, $P$ , is the rate of doing work and is measured in Watts (W) in the SI system and foot-pounds per second (ft-lb/s or ft-lb s $^{-1}$ in the USC system. Power is equivalent to a force times distance per unit time (or force times velocity), so $\left[ P \right] = \left( MLT^{-2} \right) L T^{-1}$ $=$ $M L^{2} T^{-3}$ which are the base units of power. In practice, power is measured in terms of kiloWatts (kW) in SI and horsepower (hp) in USC, where one horsepower is equivalent to 550 ft-lb/s.

4. Heat $Q$ has units of energy (the ability to do work), which have the same units as work, i.e., units of Joules (J) in the SI system and foot-pounds (ft-lb) in the USC system. In base units then $\left[ Q \right] = \left( MLT^{-2} \right) L = ML^{2} T^{-2}$ , which are the base units of heat.

Worked Example #5

Given the average energy density of jet fuel is 43 to 45 Mega-Joules per kilogram (MJ/kg), then calculate the equivalent energy density in kilo-Watt-hours per kilogram (kWh/kg).

A Watt is a Joule per second. So, one Watt hour is the equivalent of 3,600 Joules per hour. Therefore, one kilo-Watt-hour (kWh) = 3.6 Mega-Joules (MJ). The energy density (in units of kWh/kg) = energy density (in units of MJ/kg)/3.6. Therefore, the energy density of jet fuel in kilowatt-hours per kilogram is approximately 11.9 to 12.5 kWh/kg.

Worked Example #6

Write down the Bernoulli equation and explain the meaning of each term. Verify that each of the terms in the Bernoulli equation has the same fundamental dimensions.

The Bernoulli equation can be written as

$p + \frac{1}{2} \varrho V^2 + \varrho g z = \mbox{constant}$

The first term is the local static pressure. The second term is dynamic pressure. The third term is the local hydrostatic pressure. The sum of the three terms is called total pressure. The Bernoulli equation is a surrogate for the energy equation in a steady, incompressible flow without losses or energy addition.

Each of the terms in the Bernoulli equation has units of pressure. In terms of fundamental dimensions, then

$\left[ p \right] = (M L T^{-2} ) (L^{-2}) = M L^{-1} T^{-2}$

$\left[ \varrho V^2 \right] = (M L^{-3}) (L^2 T^{-2}) = M L^{-1} T^{-2}$

$\left[ \varrho g z \right] = (M L^{-3}) (L T^{-2}) L = M L^{-1} T^{-2}$

So, we have proved that all terms have the exact fundamental dimensions of M L $^{-1}$ T $^{-2}$ .

Worked Example #7

Consider the drag of a sphere problem, which was previously performed using the repeating variables $V_{\infty}$ , $\varrho_{\infty}$ , and $d$ . Repeat the dimensional analysis process using $V_{\infty}$ , $\mu_{\infty}$ , and $d$ as the repeating variables. Show all of your work. Comment on the results you obtain.

The relationship between $D$ and the air properties may be written in a general functional form as

$D = f \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, d \right)$

where $D$ is the dependent variable. $\varrho_{\infty}$ , $V_{\infty}$ , $\mu_{\infty}$ and $d$ are the independent variables. We will now choose $\mu_{\infty}, V_{\infty}$ and $d$ as repeating variables. Following the Buckingham $\Pi$ method, then the $\Pi$ products for our problem are:

$\begin{eqnarray*} \Pi_{1} & = & g_{1} \left( \mu_{\infty}, V_{\infty}, d, D \right) \\ \Pi_{2} & = & g_{2} \left( \mu_{\infty}, V_{\infty}, d, \varrho_{\infty} \right) \end{eqnarray*}$

For the first $\Pi$ product

$\Pi_{1} = (\mu_{\infty})^{\alpha} (V_{\infty})^{\beta} (d)^{\gamma} D$

The values of the coefficients $\alpha$ , $\beta$ , and $\gamma$ must be obtained to make the equation dimensionally homogeneous. In terms of the dimensions of the problem, then

$\left[ \Pi_{1} \right] = 1 = M^{0}L^{0}T^{0} = \left( ML^{-1}T^{-1} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L \right)^{\gamma} \left( MLT^{-2} \right)$

For $\Pi_{1}$ to be dimensionless, then the powers or exponents of $M$ , $L$ , and $T$ must add to zero, i.e., we must have that

$\begin{eqnarray*} M: 0 & = & \alpha+1 \\ L: 0 & = & -\alpha+\beta+ \gamma+1 \\ T: 0 & = & -\alpha -\beta-2 \end{eqnarray*}$

By inspection we get $\alpha = -1$ , $\beta = -1$ , and $\gamma = -1$ . So, we get the first $\Pi$ product as

$\Pi_{1} = (\mu_{\infty})^{\alpha} (V_{\infty})^{\beta} (d)^{\gamma} D = \mu_{\infty}^{-1} V_{\infty}^{-1} d^{-1} \, D$

or

$\Pi_{1} = \frac{D}{\mu_{\infty} V_{\infty} d}$

which is non-dimensional, but it is a different grouping to what was obtained with $\varrho_{\infty}, V_{\infty}$ , and $d$ as the repeating variables.

This is an interesting outcome because this parameter is a form of Stoke’s Law, which Sir George Stokes determined in the 1840s. He found that the drag force $D$ on a sphere of radius $R$ moving through a fluid of viscosity $\mu$ at a very low speed $V$ is given by

$D = 6 \pi R \mu V$

Comments: This drag force is proportional to the sphere’s radius. This outcome is not obvious because, based on what we did before, it might be thought that drag would be proportional to the cross-section area, which would go as the square of the radius. The drag force is also directly proportional to the speed and $V^2$ . But this behavior ONLY occurs only at very low Reynolds numbers near unity. We have obtained this outcome here because we emphasized viscosity $\mu$ over density $\varrho$ in the repeating variables. When choosing the repeating variables, a general rule is that they must have an important effect on the dependent variable, in this case, the drag. So, by emphasizing viscosity in this case, we have ended up with a different non-dimensional grouping.

Therefore, in this case, then

$\Pi_{1} = C_{D_{\rm Stokes}} = \frac{D}{3 \pi \mu_{\infty} V_{\infty} d}$

For the second $\Pi$ product then

$\Pi_{2} = (\mu_{\infty})^{\alpha} (V_{\infty})^{\beta} (d)^{\gamma} \varrho_{\infty}$

so

$\left[ \Pi_{2} \right] = 1 = M^{0} L^{0} T^{0} = \left(M L^{-1} T^{-1} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L \right)^\gamma \left( ML^{-3} \right)$

and setting the sum of the powers to zero gives

$\begin{eqnarray*} M: 0 & = & \alpha+1 \\ L: 0 & = & -\alpha + \beta+ \gamma-3 \\ T: 0 & = & -\alpha -\beta \end{eqnarray*}$

In this case $\alpha = -1$ , $\beta = 1$ , and $\gamma = 1$ and so

$\Pi_{2} = (\mu_{\infty})^{\alpha} (V_{\infty})^{\beta} (d)^{\gamma} \varrho_{\infty} = \mu_{\infty}^{-1} V_{\infty}^{1} d^{1} \, \varrho_{\infty}$

i.e.,

$\Pi_{2} = \frac{\varrho_{\infty} V_{\infty} d}{\mu_{\infty}}$

which we recognize as the Reynolds number.

Therefore, as a result of our dimensional analysis on the sphere we can write

$\phi \left( \frac {D}{3 \pi\mu V_{\infty} d}, \frac{\varrho_{\infty} V_{\infty} d}{\mu_{\infty}} \right) = 0$

or

$\phi \left( C_{D_{\rm Stokes}}, Re \right) = 0$

or in explicit form

$C_{D_{\rm Stokes}} = f_{0} \left( Re \right)$

Worked Example #8

Based on experiments performed in a low-speed wind tunnel, it is determined that the power required at the shaft to drive a propeller forward is a function of the thrust the propeller produces, $F$ , the size of the propeller as characterized by its diameter $d$ , the rotational speed of the propeller in terms of revolutions per second $n$ , and the air density $\varrho_{\infty}$ , and the operating free-stream velocity $V_{\infty}$ . Find the appropriate non-dimensional groupings that will describe this problem.

The power required for the propeller (the dependent variable), $P$ , can be written in functional form as

$P = f_1(F, d, \varrho_{\infty}, n, V_{\infty})$

where $f_1$ is the function to be determined. The power would be given by $P = Q \. \Omega$ , where $Q$ is the torque and $\Omega$ is the angular velocity in radians per second.

This foregoing equation can be written in implicit form as

$f_2(F, d, n, V_{\infty}, P) = 0$

In this case we have 6 variables ( $N=6$ ) and 3 fundamental dimensions ( $K=3$ ) comprising mass ( $M$ ), length ( $L$ ), and time ( $T$ ). This means we have to determine $N - K = 3$ or three $\Pi$ products.

The functional dependence can also be written in the form

$f_3 (\Pi_1, \Pi_2, \Pi_3) = 0$

where $\Pi_1$ , $\Pi_2$ and $\Pi_2$ are the non-dimensional groupings to be determined.

Choose the variables $\varrho_{\infty}$ , $d$ , and $n$ as the repeating variables, which are all linearly independent, and we can confirm this using the dimensional matrix (below). The non-dimensional $\Pi$ products can each be written in terms of these repeating variables plus one other variable, that is

$\begin{eqnarray*} \Pi_1 & = & \varrho_{\infty}^\alpha d^\beta n^\gamma F \\ \Pi_2 & = & \varrho_{\infty}^\alpha d^\beta n^\gamma P \\ \Pi_3 & = & \varrho_{\infty}^\alpha d^\beta n^\gamma V_{\infty} \end{eqnarray*}$

where in each case the powers $\alpha$ , $\beta$ , and $\gamma$ are to be determined in such a way that each of the $\Pi$ products must be non-dimensional.

Now we can write down the dimensions of each of the variables. For this problem, we have

$\begin{eqnarray*} \left[ \varrho_{\infty} \right] & = & M L^{-3} \\ \left[ d \right] & = & L \\ \left[ V_{\infty} \right] & = & L T^{-1} \\ \left[ n \right] & = & T^{-1} \\ \left[ F \right] & = & M L T^{-2} \\ \left[ P \right] & = & (M L T^{-2}) (L T^{-1}) = M L^2 T^{-3} \end{eqnarray*}$

and so the dimensional matrix is

$\begin{array}{l|r|r|r|r|r|r} & \varrho_{\infty} & d & V_{\infty} & n & T & P \\ \hline \mbox{Mass: } M & 1 & 1 & 0 & 0 & 1 & 1 \\ \mbox{Length: } L & -3 & 0 & 1 & 0 & 1 & 2 \\ \mbox{Time: } T & 0 & 0 & -1 & -1 & -2 & -3 \end{array}$

By examining the determinants of the submatrix formed by each of the elected repeating variables then we can quickly confirm that they are indeed linearly independent.

Considering the first $\Pi$ product then

$\Pi_ 1 = \varrho_{\infty}^\alpha d^\beta n^\gamma F$

where $\alpha$ , $\beta$ , and $\gamma$ are to be determined. We know that

$\left[ \Pi_1 \right] = M^0 L^0 T^0 = 1$

so in this case, in terms of dimensions of the parameters, then

$(M L^{-3})^\alpha (L)^\beta (T^{-1})^\gamma (M L T^{-2}) = M^0 L^0 T^0$

Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

$\begin{eqnarray*} M: & 0 = & \alpha + 1 \\ L: & 0 = & -3\alpha + \beta +1 \\ T: & 0 = & -\gamma -2 \end{eqnarray*}$

These simultaneous equations have the solution that $\alpha = -1$ , $\beta = -4$ and $\gamma = -2$ . Therefore, the first $\Pi$ product can be written as

$\Pi_1 = \varrho_{\infty}^{-1} d^{-4} n^{-2} F = \frac{F}{\varrho_{\infty} n^2 d^4} = C_T$

which is a form of thrust coefficient, i.e., a non-dimensional measure of thrust.

Considering now the second $\Pi$ product then

$\Pi_2 = \varrho_{\infty}^\alpha d^\beta n^\gamma P$

where new values for $\alpha$ , $\beta$ , and $\gamma$ are to be determined. In terms of dimensions

$\left[ \Pi_2 \right] = (M L^{-3})^\alpha (L)^\beta (T^{-1})^\gamma (M L^2 T^{-3} ) = M^0 L^0 T^0 = 1$

Making this equation dimensionally homogeneous gives

$\begin{eqnarray*} M: & 0 = & \alpha + 1 \\ L: & 0 = & -3\alpha + \beta + 2 \\ T: & 0 = & -\gamma -3 \end{eqnarray*}$

These equations have the solution that $\alpha =-1$ , $\beta = -3$ and $\gamma$ = -5. Therefore, the second $\Pi$ product is

$\Pi_2 = \varrho_{\infty}^{-1} d^{-5} n^{-3} F = \frac{P}{\varrho_{\infty} n^3 d^5} = C_P$

which is a form of power coefficient, i.e., non-dimensional measure of power.

Finally, for the third $\Pi$ product then

$\Pi_3 = \varrho_{\infty}^\alpha d^\beta n^\gamma V_{\infty}$

and in terms of dimensions then

$\left[ \Pi_3 \right] = (M L^{-3})^\alpha (L)^\beta (T^{-1})^\gamma (L T^{-1} ) = M^0 L^0 T^0 = 1$

Making this final equation dimensionally homogeneous gives

$\begin{eqnarray*} M: & 0 = & -\alpha \\ L: & 0 = & -3\alpha + \beta + 1 \\ T: & 0 = & -\gamma -1 \end{eqnarray*}$

These latter equations have the solution that $\alpha = 0$ , $\beta = -1$ and $\gamma = -1$ . Therefore, the third $\Pi$ product is

$\Pi_3 = \varrho_{\infty}^{0} d^{-1} n^{-1} V_{\infty} = \frac{V_{\infty}}{n d} = J$

which is a non-dimensional airspeed called a tip speed ratio or advance ratio.

Therefore, for this propeller problem we have that

$\phi \left( \frac{F}{\varrho_{\infty} n^2 d^4}, \frac{P}{\varrho_{\infty} n^3 d^5} , \frac{V_{\infty}}{n d} \right) = 0$

or

$\frac{P}{\varrho_{\infty} n^3 d^5} = f_0 \left( \frac{F}{\varrho_{\infty} n^2 d^4}, \frac{V_{\infty}}{n d} \right) = 0$

or finally as

$C_P = f_0 \left( C_T, J \right)$

which allows us to evaluate the performance of the propeller in terms of power coefficient as a function of the thrust coefficient and the tip speed ratio.

Worked Example #9

Consider the internal flow through a rough pipe. The objective is to determine the non-dimensional groupings that will describe this problem. The dependencies include the average flow velocity $V_{\rm avg}$ , the diameter of the pipe, $d$ , the density of the fluid flowing through the pipe $\varrho$ , the viscosity of the fluid, $\mu$ , the roughness height of the pipe $\epsilon$ , and the pressure drop along the length of the pipe, $dp/dx$ .

Proceeding using the Buckingham $\Pi$ method, we can write in general functional form that

$\phi (V_{\rm avg}, d, \varrho, \mu, \epsilon, dp/dx ) = 0$

In this case we have 6 variables ( $N=6$ ) and 3 fundamental dimensions ( $K=3$ ) comprising mass ( $M$ ), length ( $L$ ), and time ( $T$ ). According to the Buckingham $\Pi$ Method then we have $N - K = 3$ , so we need to look for three $\Pi$ products.

The dimensional matrix is

$\begin{array}{l|r|r|r|r|r|r} & V_{\rm avg} & d & \varrho &\mu & \epsilon & dp/dx \\ \hline \mbox{Mass: } M & 0 & 0 & 1 & 1 & 0 & 1 \\ \mbox{Length: } L & 1 & 1 & -3 & -1 & 1 & -2 \\ \mbox{Time: } T & -1 & 0 & 0 & -1 & 0 & -2 \end{array}$

Choose $\varrho$ , $V_{\rm avg}$ and $d$ as the repeating variables, a common choice for fluid problems, and these variables are also linearly independent and contain all of the base dimensions.

Considering the first $\Pi$ product then

$\Pi_1 = \varrho^\alpha V_{\rm avg}^\beta d^\gamma \mu$

and in terms of dimensions, then

$\left[ \Pi_1 \right] = (ML^{-3})^\alpha (LT^{-1})^\beta (L)^\betta M L^{-1} T^{-1} = M^0 L^0 T^0$

Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

$\begin{eqnarray*} M: & 0 = & \alpha + 1 \\ L: & 0 = & -3\alpha -\beta + \gamma -1 \\ T: & 0 = & -\beta -1 \end{eqnarray*}$

These simultaneous equations have the solution that $\alpha = -1$ , $\beta = -1$ and $\gamma = -1$ . Therefore, the first $\Pi$ product can be written as

$\Pi_1 = \varrho^{-1} V_{\rm avg}^{-1} d^{-1} \mu = \frac{\mu}{\varrho V_{\rm avg} d}$

which we see is the reciprocal of the Reynolds number, but as discussed before we can just invert this grouping to get the first $\Pi$ product as

$\Pi_1 = \frac{\varrho V_{\rm avg} d}{\mu} = Re$

For the second $\Pi$ product then

$\Pi_2 = \varrho^\alpha V_{\rm avg}^\beta d^\gamma \epsilon$

and in terms of dimensions, then

$\left[ \Pi_2 \right] = (ML^{-3})^\alpha (LT^{-1})^\beta (L)^\gamma L = M^0 L^0 T^0 = 1$

and we can just reuse the exponents $\alpha$ , $\beta$ and $\gamma$ for convenience. Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

$\begin{eqnarray*} M: & 0 = & \alpha \\ L: & 0 = & -3\alpha +\beta +\gamma +1 \\ T: & 0 = & -\beta \end{eqnarray*}$

These simultaneous equations have the solution that $\alpha = 0$ , $\beta = 0$ and $\gamma = -1$ . Therefore, the second $\Pi$ product can be written as

$\Pi_2 = d^{-1} \, \epsilon = \frac{\epsilon}{d}$

which is a measure of the relative surface roughness.

Finally, for the third $\Pi$ product then

$\Pi_3 = \varrho^\alpha V_{\rm avg}^\beta d^\gamma dp/dx$

and in terms of dimensions, then

$\left[ \Pi_3 \right] = (ML^{-3})^\alpha (LT^{-1})^\beta (L)^\gamma ML^{-2}T^{-2} = M^0 L^0 T^0 = 1$

Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

$\begin{eqnarray*} M: & 0 = & \alpha + 1\\ L: & 0 = & -3\alpha +\beta + \gamma + 1 \\ T: & 0 = & -\beta -2 \end{eqnarray*}$

These simultaneous equations have the solution that $\alpha = -1$ , $\beta = -2$ and $\gamma = 1$ . Therefore, the third $\Pi$ product can be written as

$\Pi_3 = \varrho^{-1} V_{\rm avg}^{-2} d \, (dp/dx) = \frac{d \, (dp/dx)}{\varrho V^2}$

which is a non-dimensional pressure drop or “head” drop. Normally, this latter grouping is expressed in terms of a friction factor, i.e.,

$f = \frac{d \, (dp/dx)}{\frac{1}{2} \varrho V_{\rm avg}^2}$

Therefore, we get to the final result that

$f = \phi_1 \left( Re, \frac{\epsilon}{d} \right)$

which tells us that the frictional pressure drop along the pipe will be a function of the Reynolds number and the effective non-dimensional roughness of the pipe.

Worked Example #10

A solid rocket booster parachutes back to Earth and falls into the sea. The booster bobs around in the sea in an upright position at a frequency of $\omega$ .

Using dimensional analysis, show that the non-dimensional frequency of this motion, $k$ , is given by

$k = \omega \sqrt{\frac{d}{g}}$

You may assume that for this problem depends on the diameter of the booster $d$ , the mass of the booster, $m$ , the density of the water $\varrho$ , and acceleration under gravity, $g$ .

We are told that

$\omega = f(D, m, \varrho, g)$

or in implicit form then

$f_1 (d, m, \varrho, g, \omega) = 0$

Therefore, $N$ = 5, $K$ = 3 and so we will have 2 $\Pi$ products. The repeating variables must collectively include all the units of mass $M$ , length $L$ and time $T$ , so the best choice here is $d$ , $m$ and $g$ . Notice that if we were to choose $\varrho$ instead of $g$ , then collectively the repeating variables would not include time $T$ and so the Buckingham $\Pi$ method fails in this case. Of course, the dependent variable, $\omega$ , can never be used as a repeating variable.

Therefore, the $\Pi$ products are

$\Pi_1 = f_2(d, m, g, \omega)$

and

$\Pi_2 = f_3(d, m, g, \varrho)$

In this question we are only asked for one of the two $\Pi$ products, which is the one involving $\omega$ or the $\Pi_1$ product.

We first set up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|r|r} & d & m & g & \omega & \varrho \\ \hline \mbox{Mass} ~M: & 0 & 1 & 0 & 0 & 1 \\ \mbox{Length} ~L: & 1 & 0 & 1 & 0 & -3 \\ \mbox{Time} ~T: & 0 & 0 & -2 & -1 & 0 \end{array}$

So, the first $\Pi$ product is

$\Pi_{1} = (d)^{\alpha} (m)^{\beta} (g)^{\gamma} \omega$

where the values of the coefficients $\alpha$ , $\beta$ and $\gamma$ must be obtained to make the equation dimensionally homogeneous. In terms of the dimensions of the problem then

$\left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = \left( L \right)^{\alpha} \left( M \right)^{\beta} \left( L T^{-2} \right)^{\gamma} \left( T^{-1}\right)$

For $\Pi_{1}$ to be dimensionless, then the powers or exponents of $M$ , $L$ and $T$ must add to zero, i.e., we must have that

$\begin{eqnarray*} M: 0 & = & \beta \\ L: 0 & = & \alpha + \gamma \\ T: 0 & = & -2\gamma - 1 \end{eqnarray*}$

Therefore, $\alpha$ = 1/2, $\beta$ = 0, and $\gamma$ = -1/2, so the $\Pi_1$ product is

$\Pi_1 = (d)^{1/2} (m)^{0} (g)^{-1/2} \omega = \omega \sqrt{\frac{d}{g}}$

which is what we were asked to prove.

It is straightforward to show (not asked for) that the other $\Pi$ product is

$\Pi_2 = \frac{\varrho d^3}{m}$

For this second $\Pi$ product then

$\Pi_{2} = (d)^{\alpha} (m)^{\beta} (g)^{\gamma} \varrho$

In terms of the dimensions then

$\left[ \Pi_{2} \right] = M^{0}L^{0}T^{0} = \left( L \right)^{\alpha} \left(M \right)^{\beta} \left( L T^{-2} \right)^{\gamma} \left( ML^{-3}\right)$

For $\Pi_{2}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \beta + 1 \\ L: 0 & = & \alpha + \gamma - 3 \\ T: 0 & = & -2\gamma \end{eqnarray*}$

Therefore, $\beta$ = -1, $\gamma$ = 0, and $\alpha = 3$ , so the $\Pi_2$ product is

$\Pi_2 = (d)^{3} (m)^{-1} (g)^{0} \varrho = \frac{\varrho d^3}{m}$

which is a buoyancy similarity parameter, i.e., the ratio of the mass (or weight) of the water displaced to the mass (or weight) of the rocket booster.

Worked Example #11

The drag on the hull of a ship, $D$ can be written in general functional form as

$D = f \left( \varrho, V, l, g \right)$

where $\varrho$ is the density of the water, $V$ is the speed of the ship through the water, and $l$ is a length scale associated with the hull.Use the Buckingham $\Pi$ method to show that

$C_D = f(F\!r)$

where $C_D$ is a drag coefficient and the non-dimensional grouping $F\!r= V_{\infty}/\sqrt{g l}$ is known as the Froude number.

For this problem we are told that the drag $D$ can be written as

$D = f \left( \varrho, V, l, g \right)$

In implicit form then

$f_1 \left( \varrho, V, l, g, D \right) = 0$

So, we have $N=5$ and $K=3$ so there will be 2 $\Pi$ products.

Now we set up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|r|r} & \varrho & V & l & g & D \\ \hline \mbox{Mass} ~M: & 1 & 0 & 0 & 0 & 1 \\ \mbox{Length} ~L: & -3 & 1 & 1 & 1 & 1 \\ \mbox{Time} ~T: & 0 & -1 & 0 & -2 & -2 \end{array}$

We need to choose the repeating variables, for which the standard choice is $l$ , $V$ and $\varrho$ , which will all have a primary influence on drag. They also collectively include all of the fundamental dimensions mass, length and time, and they are clearly linearly independent just by inspection. So, for the first $\Pi$ product then

$\Pi_{1} = f_2 \left(l, V, \varrho, g \right)$

and for the second $\Pi$ product then

$\Pi_{2} = f_3 \left(l, V, \varrho, D \right)$

Continuing with the first $\Pi$ product then

$\Pi_{1} = (l)^{\alpha} (V)^{\beta} (\varrho)^{\gamma} g$

and terms of the dimensions then

$\left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = (L)^{\alpha} (L T^{-1} )^{\beta} (M L^{-3} )^{\gamma} L T^{-2}$

For $\Pi_{1}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \gamma \\ L: 0 & = & \alpha + \beta - 3\gamma + 1 \\ T: 0 & = & -\beta -2 \end{eqnarray*}$

Therefore, $\beta$ = -2, $\gamma$ = 0, and $\alpha = 1$ , so the $\Pi_1$ product is

$\Pi_1 = (l)^{1} (V)^{-2} (\varrho)^{0} g$

or

$\Pi_1 = \frac{l g}{V^2} \quad \mbox{or} \quad \frac{V^2}{l g} \quad \mbox{or} \quad \frac{V}{\sqrt{l g}}$

which is the Froude number, $F\!r$ .

For the second $\Pi$ product then

$\Pi_{2} = (l)^{\alpha} (V)^{\beta} (\varrho)^{\gamma} D$

and terms of the dimensions then

$\left[ \Pi_{2} \right] = M^{0}L^{0}T^{0} = (L)^{\alpha} (L T^{-1} )^{\beta} (M L^{-3} )^{\gamma} M L T^{-2}$

For $\Pi_{2}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \gamma + 1 \\ L: 0 & = & \alpha + \beta - 3\gamma + 1 \\ T: 0 & = & -\beta -2 \end{eqnarray*}$

Therefore, $\beta$ = -2, $\gamma$ = -1, and $\alpha = -2$ , so the $\Pi$ product is

$\Pi_2 = (l)^{-2}(V)^{-2} (\varrho)^{-1} D$

or

$\Pi_2 = \frac{D}{\varrho l^2 V^2}$

which is a drag coefficient, $C_D$ .

Therefore, finally we have that

$C_D = f_4 (F\!r)$

i.e., the drag coefficient on the hull is some function of the Froude number.

Worked Example #12

A tiny spherical particle of diameter $D$ is falling freely vertically at velocity $V$ in the atmosphere. The aerodynamic drag on the particle ${\cal {D}}$ can be written in general functional form as

${\cal{D}} = f( D, V, \mu )$

where $f$ is the function to be determined and $\mu$ is the coefficient of viscosity. Using the Buckingham $\Pi$ method, find the dimensionless similarity parameter that governs this freely falling behavior. Illustration of a tiny spherical particle falling freely in the vertical direction.

We are told that

${\cal{D}} = f( D, V, \mu )$

so in implicit form then

$f_1 (D, V, \mu, {\cal{D}} ) = 0$

We have $N$ = 4 and $K$ = 3 (by inspection, all of mass, length and time are involved in this problem) so we have just one $\Pi$ product.

Setting down the dimensional matrix gives

$\begin{array}{l|r|r|r|r|r} & D & V & \mu & {\cal{D}} \\ \hline \mbox{Mass} ~M: & 0 & 0 & 1 & 1 \\ \mbox{Length} ~L: & 1 & 1 & -1 & 1 \\ \mbox{Time} ~T: & 0 & -1 & -1 & -2 \end{array}$

For the $\Pi$ product then we have

$\Pi = (D)^{\alpha} (V)^{\beta} (\mu)^{\gamma} {\cal{D}}$

and in terms of dimensions then

$\left[ \Pi \right] = M^{0}L^{0}T^{0} = (L)^{\alpha} (L T^{-1})^{\beta} (M L^{-1} T^{-1})^{\gamma} M L T^{-2}$

For $\Pi$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \gamma + 1 \\ L: 0 & = & \alpha + \beta - \gamma + 1 \\ T: 0 & = & -\beta - \gamma - 2 \end{eqnarray*}$

and so $\gamma$ = -1, and $\beta = -1$ and $\alpha$ = -1.

Therefore, the $\Pi$ product is

$\Pi = (D)^{-1}(V)^{-1} (\mu)^{-1} {\cal{D}} = \frac{{\cal{D}}}{D V \mu}$

which is a viscous drag coefficient applicable, in this case, to what is known as a Stokes flow, which is a flow corresponding to Reynolds numbers near unity.

Worked Example #13

A flow experiment with a circular cylinder shows that at a certain condition a vortex shedding phenomenon at frequency $f$ appears in the wake downstream of the cylinder.

Use dimensional analysis to show that the non-dimensional parameter that governs this process, known as a Strouhal number $St$ , is given by

$St = \frac{ f d}{V}$

where $V$ is the flow speed and $d$ is the diameter of the cylinder.

In this case, we are told that the frequency of shedding $f$ will be a function of the diameter of the cylinder $d$ and flow velocity $V$ . Working with this information then we have

$f = \phi \left( d, V \right)$

or in implicit form then

$\phi_1 \left( d, V , f \right) = 0$

We see $N=3$ , but in this case $K=2$ because only length and time are involved in this group of variables (no mass). So, there is just the one $\Pi$ product to determine.

Setting up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|} & d & V & f \\ \hline \mbox{Length} ~L: & 1 & 1 & 0 \\ \mbox{Time} ~T: & 0 & -1 & -1 \end{array}$

So, we have

$\Pi_{1} = (d)^{\alpha} (V)^{\beta} f$

and terms of the dimensions then

$\left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = (L)^{\alpha} (L T^{-1} )^{\beta} T^{-1}$

For $\Pi_{1}$ to be dimensionless, then

$\begin{eqnarray*} L: 0 & = & \alpha + \beta \\ T: 0 & = & -\beta -1 \end{eqnarray*}$

Therefore, $\beta$ = -1 and $\alpha = 1$ , so the $\Pi_1$ product is

$\Pi_1 = (d)^{1}(V)^{-1} f = \frac{f d}{V}$

which is the Strouhal number $St$ , as called for in the question.

We could also have proceeded by recognizing that the frequency of shedding may also be a function of the flow density $\varrho$ and its viscosity $\mu$ . In this case, there is an expectation that Reynolds number may be involved. If we do this, then

$f = \phi \left( d, V , \varrho, \mu \right)$

or in implicit form then

$\phi_1 \left( d, V , \varrho, \mu f \right) = 0$

We now have $N=5$ and $K=3$ (mass is now involved) so there are now two $\Pi$ products to determine. Of course, there is an expectation that one of these has already been determined.

Setting up the dimensional matrix gives

$\begin{array}{l|r|r|r|r|r} & d & V & \varrho & \mu & f \\ \hline \mbox{Mass} ~M: & 0 & 0 & 0 & 1 & 0 \\ \mbox{Length} ~L: & 1 & 1 & 1 & -1 & 0 \\ \mbox{Time} ~T: & 0 & -1 & -3 & -1 & -1 \end{array}$

Proceeding as normal with the selection of the repeating variables (again, the standard choice is $d$ , $V$ and $\varrho$ ) then we have

$\Pi_{1} = \phi_2 \left(d, V, \varrho, f \right)$

and

$\Pi_{2} = \phi_3 \left(d, V, \varrho, \mu \right)$

For the first $\Pi$ product then

$\Pi_{1} = (d)^{\alpha} (V)^{\beta} (\varrho)^{\gamma} f$

and terms of the dimensions then

$\left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = (L)^{\alpha} (L T^{-1} )^{\beta} (M L^{-3} )^{\gamma} T^{-1}$

For $\Pi_{1}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \gamma \\ L: 0 & = & \alpha + \beta - 3\gamma \\ T: 0 & = & -\beta -1 \end{eqnarray*}$

Therefore, $\beta$ = -1, $\gamma$ = 0, and $\alpha = 1$ , so the $\Pi_1$ product is

$\Pi_1 = (d)^{1}(V)^{-1} f = \frac{f d}{V}$

which is the Strouhal number $St$ we have derived previously.

For the second $\Pi$ product, which we suspect will be a Reynolds number, then

$\Pi_{2} = (d)^{\alpha} (V)^{\beta} (\varrho)^{\gamma} \mu$

and terms of the dimensions then

$\left[ \Pi_{2} \right] = M^{0}L^{0}T^{0} = (L)^{\alpha} (L T^{-1} )^{\beta} (M L^{-3} )^{\gamma} M L^{-1} T^{-1}$

For $\Pi_{2}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \gamma + 1 \\ L: 0 & = & \alpha + \beta - 3\gamma \\ T: 0 & = & -\beta -1 \end{eqnarray*}$

Therefore, $\beta$ = -1, $\gamma$ = -1, and $\alpha = -1$ , so the $\Pi_1$ product is

$\Pi_2 = (d)^{-1}(V)^{-1}(\varrho)^{-1} \mu = \frac{\mu }{\varrho V d} \quad \mbox{or} \quad \frac{\varrho V d}{\mu }$

which indeed is the Reynolds number $Re$ .

Therefore, we have then that the frequency of shedding is expected to be a function of Reynolds number, i.e.,

$St = \phi_4 ( Re )$

Worked Example #14

Define Reynolds number and explain what it means. Show that the Reynolds number represents a ratio of the relative magnitude of inertial effects to viscous effects in the flow. Hint: Multiply both the numerator and denominator of the equation for the Reynolds number by a velocity and a length scale.

The Reynolds number is a non-dimensional grouping formed in terms of fluid density, $\varrho$ , a reference velocity, $V$ , a characteristic length scale, $L$ , and viscosity, $\mu$ , i.e.,

$Re = \frac{\varrho V L}{\mu}$

Reynolds number represents “the ratio of the relative effects of inertial effects to viscous effects,” which can be seen by writing

$Re = \frac{\varrho_{\infty} V_{\infty} c}{\mu_{\infty}} = \frac{\varrho_{\infty} V_{\infty} c (V_{\infty} c)}{\mu_{\infty} (V_{\infty} c)} = \frac{\varrho_{\infty} V_{\infty}^2 c^2}{\mu_{\infty} (V_{\infty}/c) c^2} \equiv \frac{\mbox{Inertial force}}{\mbox{Viscous force}}$

On the numerator, $\varrho_{\infty} V_{\infty}^2 c^2$ has units of force so in this case it represents an inertial force. The coefficient of viscosity, $\mu$ , is the shear force per unit area per unit velocity gradient, so the denominator is also a force but a viscous force. Hence, we see the significance of the Reynolds number as a relative measure of inertial effects to viscous effects in a fluid flow.

Worked Example #15

The distance traveled by a dimpled golf ball depends on its aerodynamic drag, $\cal{D}$ , which in turn on its flight speed $V$ , the density of the air $\varrho$ , the viscosity of the air $\mu$ , the diameter of the ball $D$ , and the diameter of the dimples on the ball $d$ , i.e., $\cal{D}$ = $\phi$ ( $V$ , $\varrho$ , $\mu$ , $D$ , $d$ ), where $\phi$ is some functional dependency. Use dimensional analysis (Buckingham $\Pi$ method) to determine the non-dimensional groupings that govern this problem.

The drag of the golf ball (the dependent variable) $\cal{D}$ , can be written in functional form as

${\cal{D}} = \phi (V, \varrho, \mu, D, d)$

where $\phi$ is a function to be determined. This equation can be written in implicit form as

$\phi_1(V, \varrho, \mu, D, d, {\cal{D}}) = 0$

In this case there are 6 variables ( $N=6$ ) and 3 fundamental dimensions ( $K=3$ ) so there are three $\Pi$ products.

The functional dependence can also be written in the form

$\phi_2 (\Pi_1, \Pi_2, \Pi_3) = 0$

where $\Pi_1$ , $\Pi_2$ and $\Pi_2$ are the non-dimensional groupings to be determined.

Choose the standard aerodynamic repeating variables $\varrho$ , $V$ and $D$ , which are all linearly independent. The non-dimensional $\Pi$ products can each be written in terms of these repeating variables plus one other variable, that is

$\begin{eqnarray*} \Pi_1 & = & \varrho^\alpha V^\beta D^\gamma\cal{D} \\ \Pi_2 & = & \varrho^\alpha V^\beta D^\gamma \\ \Pi_3 & = & \varrho^\alpha V^\beta D^\gamma \mu \end{eqnarray*}$

where in each case the values of the exponents $alpha$ , $\beta$ , and $\gamma$ are to be determined in such a way that each of the $\Pi$ products is non-dimensional.

Write down the dimensions of each of the variables. For this problem then

$\begin{eqnarray*} \left[ V \right] & = & L T^{-1} \\ \left[ \varrho \right] & = & M L^{-3} \\ \left[ \mu \right] & = & M L^{-1} T^{-1} \\ \left[ D \right] & = & L \\ \left[ d \right] & = & L \\ \left[ \cal{D} \right] & = & M L^{-1} T^{-2} \end{eqnarray*}$

and so the dimensional matrix is

$\begin{array}{l|r|r|r|r|r|r} & V & \varrho & \mu & D & d & {\cal{D}} \\ \hline \mbox{Mass: } M & 0 & 1 & 1 & 0 & 0 & 1 \\ \mbox{Length: } L & 1 & -3 & -1 & 1 & 1 & 1 \\ \mbox{Time: } T & -1 & 0 & -1 & 0 & 0 & -2 \end{array}$

Considering the first $\Pi$ product then

$\Pi_1 \right] = \varrho^\alpha V^\beta D^\gamma {\cal{D}$

and in terms of dimensions then

$\left[ \Pi_1 \right] = (M L^{-3})^\alpha (L T^{-1})^\beta (L)^\gamma (M L T^{-2}) = M^0 L^0 T^0$

Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

$\begin{eqnarray*} M: & 0 = & \alpha + 1 \\ L: & 0 = & -3\alpha + \beta + \gamma + 1 \\ T: & 0 = & -\beta - 2 \end{eqnarray*}$

These simultaneous equations have the solution that $\alpha = -1$ , $\beta = -2$ and $\gamma = -2$ . Therefore, the first $\Pi$ product can be written as

$\Pi_1 = \varrho^{-1} V^{-2} D^{-2} {\cal{D}} = \frac{ {\cal{D}} }{\varrho V^{2} D^{2}}$

which is a force coefficient.

Considering the second $\Pi$ product then

$\left[ \Pi_2 \right] = \varrho^\alpha V^\beta D^\gamma d$

and in terms of dimensions, then

$\left[ \Pi_2 \right] = (M L^{-3})^\alpha (L T^{-1})^\beta (L)^\gamma (L) = M^0 L^0 T^0$

Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

$\begin{eqnarray*} M: & 0 = & \alpha \\ L: & 0 = & -3\alpha + \beta + \gamma + 1 \\ T: & 0 = & -\beta \end{eqnarray*}$

These simultaneous equations have the solution that $a=0$ , $b=0$ and $c=-1$ . Therefore, the second $\Pi$ product can be written as

$\Pi_2 = \varrho^{0} V^{0} D^{-1} d = \frac{d}{D}$

which is a non-dimensional length scale, i.e., the ratio of the diameter of the dimples to the diameter of the golf ball.

Considering the third $\Pi$ product then

$\Pi_3 = \varrho^\alpha V^\beta D^\gamma \mu$

and in terms of dimensions, then

$\left[ \Pi_3 \right] = (M L^{-3})^\alpha (L T^{-1})^\beta (L)^\gamma (M L^{-1} T^{-1} ) = M^0 L^0 T^0$

Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

$\begin{eqnarray*} M: & 0 = & \alpha + 1 \\ L: & 0 = & -3\alpha + \beta + \gamma - 1 \\ T: & 0 = & -\beta - 1 \end{eqnarray*}$

These simultaneous equations have the solution that $\alpha = -1$ , $\beta = -1$ and $gamma = -1$ . Therefore, the third $\Pi$ product can be written as

$\Pi_3 = \varrho^{-1} V^{-1} D^{-1} \mu = \frac{\mu }{\varrho V D}$

or inverting the grouping (it is still non-dimensional)

$\Pi_3 = \frac{\varrho V D}{\mu}$

which is a Reynolds number.

Finally then

$\phi_2 (\Pi_1, \Pi_2, \Pi_3) = \phi_2 \left( \frac{ {\cal{D}} }{\varrho V^{2} D^{2}}, \frac{d}{D}, \frac{\varrho V D}{\mu} \right) = 0$

or just

$\frac{ {\cal{D}} }{\varrho V^{2} D^{2}} = \phi_3 \left( \frac{d}{D}, \frac{\varrho V D}{\mu} \right)$

Worked Example #16

The sound intensity $I$ from a jet engine is found to be a function of the sound pressure level $p$ (dimensions of pressure) and the fluid properties density, $\varrho$ , and speed of sound, $a$ , as well as the distance from the engine to an observer location, $r$ . Using the Buckingham $\Pi$ method, find a relationship for $I$ as a function of the other parameters. Show all of your work. Hint: Sound intensity, $I$ , is defined as the acoustic power per unit area emanating from a sound source.

The relationship between $I$ and the properties listed may be written in a general functional form as

$I = f \left( p, \varrho, a, r \right)$

or

$g \left( I, p, \varrho, a, r \right) = 0$

We are given a hint and told that sound intensity $I$ is defined as the acoustic power per unit area, so in SI we could think of this as Watts per unit area. We also know that power is the rate of doing work and so has units of force times displacement per unit time, i.e., when we come to setting up the problem then we will know that

$\left[ I \right] = \left( M L^{2} T^{-3} \right) L^{-2} = MT^{-3}$

Also, pressure is force per unit area so

$\left[ p \right] = \left( M L T^{-2} \right) L^{-2} = M L^{-1} T^{-2}$

In this problem we have $N=5$ and $K=3$ (by inspection mass, length and time are all involved) so there are two $\Pi$ products to determine. Setting up the dimensional matrix gives

$\begin{array}{l|r|r|r|r|r} & I & p & \varrho & a & r \\ \hline \mbox{Mass} ~M: & 1 & 1 & 1 & 0 & 0 \\ \mbox{Length} ~L: & 0 & -1 & -3 & 1 & 1 \\ \mbox{Time} ~T: & -3 & -2 & 0 & -1 & 0 \end{array}$

Proceeding with the selection of the repeating variables, then one choice is $\varrho$ , $a$ and $r$ . Therefore, our two $\Pi$ groups are

$\Pi_{1} = \phi_1 \left(\varrho, a, r, I \right)$

and

$\Pi_{2} = \phi_2 \left(\varrho, a, r, p \right)$

For the first $\Pi$ product then

$\Pi_{1} = (\varrho)^{\alpha} (a)^{\beta} (r)^{\gamma} I$

and terms of the dimensions then

$\left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = (M L^{-3})^{\alpha} ( LT^{-1})^{\beta} ( L )^{\gamma} \, MT^{-3}$

For $\Pi_{1}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \alpha + 1 \\ L: 0 & = & -3\alpha + \beta + \gamma \\ T: 0 & = & -\beta - 3 \end{eqnarray*}$

Therefore, $\alpha = -1$ , $\beta$ = -3, and $\gamma$ = 0, so the $\Pi_1$ product is

$\Pi_1 = (\varrho)^{-1} (a)^{-3} (r)^0 I = \frac{I}{\varrho a^3}$

For the second $\Pi$ product then

$\Pi_{2} = (\varrho)^{\alpha} (a)^{\beta} (r)^{\gamma} p$

and terms of the dimensions then

$\left[ \Pi_{2} \right] = M^{0}L^{0}T^{0} = (M L^{-3})^{\alpha} ( LT^{-1})^{\beta} ( L )^{\gamma} \, M L^{-1} T^{-2}$

For $\Pi_{2}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \alpha + 1 \\ L: 0 & = & -3\alpha + \beta + \gamma - 1\\ T: 0 & = & -\beta - 2 \end{eqnarray*}$

Therefore, $\alpha = -1$ , $\beta$ = -2, and $\gamma$ = 0, so the $\Pi_1$ product is

$\Pi_2 = (\varrho)^{-1} (a)^{-2} (r)^0 p = \frac{p}{\varrho a^2}$

Interestingly, the factor $1/\varrho a^2$ is related to the compressibility modulus of the medium in which the sound propagates. Therefore, we have in this case

$\frac{I}{\varrho a^3}= \phi_4 \left( \frac{p}{\varrho a^2} \right)$

Worked Example #17

Consider a liquid in a cylindrical container in which both the container and the liquid are rotating as a rigid body (this is called solid-body rotation).

The elevation difference $h$ between the center of the liquid surface and the rim of the liquid surface is a function of the angular velocity $\omega$ , the fluid density $\varrho$ , the gravitational acceleration $g$ , and the radius $R$ . Use the Buckingham $\Pi$ method to find the relationship between the height $h$ and the other parameters. Show all of your work.

In this problem we are asked to find the effects of the elevation difference $h$ between the center of the liquid surface and the rim of the liquid surface, which we are told is a function of the angular velocity $\omega$ , the fluid density $\varrho$ , the gravitational acceleration $g$ , and the radius $R$ , i.e.,

$h = f \left( \omega, \varrho, g, R \right)$

or

$f_1 \left( h, \omega, \varrho, g, R \right) = 0$

We have $N=5$ and $K=3$ (by inspection mass, length and time are all involved) so there are two $\Pi$ products to determine. Setting up the dimensional matrix gives

$\begin{array}{l|r|r|r|r|r|} & h & \omega & \varrho & g & R \\ \hline \mbox{Mass} ~M: & 0 & 0 & 1 & 0 & 0 \\ \mbox{Length} ~L: & 1 & 0 & -3 & 1 & 1 \\ \mbox{Time} ~T: & 0 & -1 & 0 & -2 & 0 \end{array}$

Proceeding with the selection of the repeating variables, then one choice is $\varrho$ , $g$ and $R$ . Therefore, our two $\Pi$ groups are

$\Pi_{1} = \phi_1 \left(\varrho, g, R, \omega \right)$

and

$\Pi_{2} = \phi_2 \left(\varrho, g, R, h \right)$

For the first $\Pi$ product then

$\Pi_{1} = (\varrho)^{\alpha} (g)^{\beta} (R)^{\gamma} \omega$

and terms of the dimensions then

$\left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = (M L^{-3})^{\alpha} ( LT^{-2})^{\beta} ( L )^{\gamma} \, T^{-1}$

For $\Pi_{1}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \alpha \\ L: 0 & = & -3\alpha + \beta + \gamma \\ T: 0 & = & -2\beta - 1 \end{eqnarray*}$

Therefore, $\alpha = 0$ , $\beta$ = -1/2, and $\gamma$ = 1/2, so the $\Pi_1$ product is

$\Pi_1 = (\varrho)^{0} (g)^{-1/2} (R)^{1/2} \omega = \omega \sqrt{ \frac{R}{g} }$

For the second $\Pi$ product then

$\Pi_{2} = (\varrho)^{\alpha} (g)^{\beta} (R)^{\gamma} \, h$

and terms of the dimensions then

$\left[ \Pi_{2} \right] = M^{0}L^{0}T^{0} = (M L^{-3})^{\alpha} ( LT^{-2})^{\beta} ( L )^{\gamma} \, L$

For $\Pi_{2}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \alpha \\ L: 0 & = & -3\alpha + \beta + \gamma + 1\\ T: 0 & = & -2\beta \end{eqnarray*}$

Therefore, $\alpha = 0$ , $\beta$ = 0, and $\gamma$ = -1, so the $\Pi_2$ product is

$\Pi_2 = (\varrho)^{0} (g)^{0} (R)^{-1} h = \frac{h}{R}$

Therefore, we have in this case

$\frac{h}{R} = f_3 \left( \omega \sqrt{ \frac{R}{g} } \right)$

Worked Example #18

A liquid of density $\varrho$ and viscosity $\mu$ flows by gravity through a hole of diameter $d$ in the bottom of a tank of diameter $D$ . At the start of the experiment, the liquid surface is at height $h$ above the bottom of the tank. The liquid exits the tank as a jet with average velocity $V$ straight down. Using the Buckingham $\Pi$ method, find a dimensionless relationship for $V$ as a function of the other parameters in the problem. Identify any established non-dimensional parameters that appear in your result. Show all of your work. Hint: Notice that there are three length scales in this problem, but for consistency choose $h$ as the reference length scale.

In this problem we are asked to find the effects on the exit flow velocity $V$ in terms the fluid density $\varrho$ , its viscosity $\mu$ , the diameter of the hole $d$ . the diameter of the tank $D$ , and the height of the liquid surface $h$ , i.e.,

$V = f \left( \varrho, \mu, d, D, h \right)$

or

$f_1 \left( V, \varrho, \mu, d, D, h \right) = 0$

In this problem we have $N=6$ and $K=3$ (by inspection mass, length and time are all involved) so there are three $\Pi$ products to determine. Setting up the dimensional matrix gives

$\begin{array}{l|r|r|r|r|r|r|} & V & \varrho & \mu & d & D & h \\ \hline \mbox{Mass} ~M: & 0 & 1 & 1 & 0 & 0 & 0 \\ \mbox{Length} ~L: & 1 & -3 & -1 & 1 & 1 & 1 \\ \mbox{Time} ~T: & -1 & 0 & -1 & 0 & 0 & 0 \end{array}$

Proceeding with the selection of the repeating variables, the only choice in this case is $\varrho$ , $\mu$ and $h$ . (Note: Can you explain why?). Therefore, our three $\Pi$ groups are

$\Pi_{1} = \phi_1 \left(\varrho, \mu, h, V \right)$

and

$\Pi_{2} = \phi_2 \left( \varrho, \mu, h, d \right)$

and

$\Pi_{3} = \phi_4 \left( \varrho, \mu, h, D \right)$

For the first $\Pi$ product then

$\Pi_{1} = (\varrho)^{\alpha} (\mu)^{\beta} (h)^{\gamma} \, V$

and terms of the dimensions then

$\left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = (M L^{-3})^{\alpha} ( M L^{-1}T^{-1})^{\beta} ( L )^{\gamma} \, LT^{-1}$

For $\Pi_{1}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \alpha + \beta \\ L: 0 & = & -3\alpha - \beta + \gamma + 1 \\ T: 0 & = & -\beta - 1 \end{eqnarray*}$

Therefore, $\beta$ = -1, $\alpha$ = 1, and $\gamma$ = 1, so the $\Pi_1$ product is

$\Pi_1 = (\varrho)^{1} (\mu)^{-1} (h)^{1} V = \frac{\varrho V h}{\mu}$

which we recognize is a Reynolds number.

For the second $\Pi$ product then

$\Pi_{2} = (\varrho)^{\alpha} (\mu)^{\beta} (h)^{\gamma} \, d$

and terms of the dimensions then

$\left[ \Pi_{2} \right] = M^{0}L^{0}T^{0} = (M L^{-3})^{\alpha} ( M L^{-1}T^{-1})^{\beta} ( L )^{\gamma} \, L$

For $\Pi_{2}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \alpha + \beta \\ L: 0 & = & -3\alpha - \beta + \gamma + 1 \\ T: 0 & = & -\beta \end{eqnarray*}$

Therefore, $\beta$ = 0, $\alpha$ = 0, and $\gamma$ = -1, so the $\Pi_2$ product is

$\Pi_2 = (\varrho)^{0} (\mu)^{0} (h)^{-1} d = \frac{d}{h}$

For the third $\Pi$ product then

$\Pi_{3} = (\varrho)^{\alpha} (\mu)^{\beta} (h)^{\gamma} \, D$

which quickly follows as per the $\Pi_2$ product as

$\Pi_3 = (\varrho)^{0} (\mu)^{0} (h)^{-1} d = \frac{D}{h}$

Therefore, in this case the non-dimensional groupings involved are such that

$\frac{\varrho V h}{\mu} = \phi_4 \left( \frac{d}{h}, \frac{D}{h} \right)$

Worked Example #19

The AIAA Design Build & Fly (DBF) team need to determine the factors that influence the aerodynamic drag on a rectangular banner being towed behind their airplane.

The size of the banner is determined by its length, $l$ , and height, $h$ . Use the Buckingham $\Pi$ method to determine the non-dimensional groupings that will govern this problem. You may also assume that the problem is governed by the airspeed of the airplane, as well as the density and viscosity of the air. Assume further that the banner remains flat and does not flutter in the flow behind the airplane.

The relationship between the drag on the banner $D$ and the air properties can be written in the general functional form as

$D = f \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, h, l \right)$

where we are told that the size of the banner is represented by its length, $l$ , and height, $h$ . Remember that $D$ is called the dependent variable and $\varrho_{\infty}$ , $V_{\infty}$ , $\mu_{\infty}$ , $h$ and $l$ are are the independent variables. The functional dependence of $D$ in implicit form is

$g \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, h, l, D \right) = 0$

Counting up the variables gives $N = 6$ and also $K = 3$ because there are 3 fundamental dimensions in this problem. Therefore, $N - K = 3$ and we will have three $\Pi$ products.

We must first find the dimensions of the variables. For each variable the dimensions are

$\begin{eqnarray*} \left[ D \right] & = & MLT^{-2} \\ \left[ \varrho_{\infty} \right] & = & ML^{-3} \\ \left[ V_{\infty} \right] & = & LT^{-1} \\ \left[ \mu_{\infty} \right] & = & ML^{-1}T^{-1} \\ \left[ h \right] & = & L \\ \left[ l \right] & = & L \end{eqnarray*}$

We can now set up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|r|r|r} & D & \varrho_{\infty} & V_{\infty} & \mu_{\infty} & h & l \\ \hline \mbox{Mass: } M & 1 & 1 & 0 & 1 & 0 & 0 \\ \mbox{Length: } L & 1 & -3 & 1 & -1 & 1 & 1 \\ \mbox{Time: } T & -2 & 0 & -1 & -1 & 0 & 0 \end{array}$

Choose $\varrho_{\infty}$ , $V_{\infty}$ and $l$ as the repeating variables, which will all have primary effects on the drag of the banner. These variables also collectively include all the fundamental dimensions of this problem and are linearly independent of each other.

Following the Buckingham $\Pi$ method, then the three $\Pi$ products are:

$\begin{eqnarray*} \Pi_{1} & = & g_{1} \left( \varrho_{\infty}, V_{\infty}, l, D \right) \\ \Pi_{2} & = & g_{2} \left( \varrho_{\infty}, V_{\infty}, l, \mu_{\infty} \right) \\ \Pi_{3} & = & g_{3} \left( \varrho_{\infty}, V_{\infty}, l, h \right) \end{eqnarray*}$

For $\Pi_{1}$ :

$\Pi_{1} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, D$

The values of the coefficients $\alpha$ , $\beta$ and $\gamma$ must now be obtained to make the equation dimensionally homogeneous. In terms of the dimensions of the problem then

$\left[ \Pi_{1} \right] = 1 = M^{0}L^{0}T^{0} = \left( ML^{-3} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L \right)^{\gamma} \left( MLT^{-2} \right)$

For $\Pi_{1}$ to be dimensionless, then the powers or exponents of $M$ , $L$ and $T$ must add to zero, i.e., we must have that

$\begin{eqnarray*} M: 0 & = & \alpha+1 \\ L: 0 & = & -3\alpha + \beta+\gamma+1 \\ T: 0 & = & -\beta-2 \end{eqnarray*}$

By inspection we get $\alpha = -1$ , $\beta = -2$ , and $\gamma = -2$ . Therefore, the first $\Pi$ product is

$\Pi_{1} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} D = \varrho_{\infty}^{-1} V_{\infty}^{-2} l^{-2} \, D$

or

$\Pi_{1} = \frac{D}{\varrho_{\infty} V_{\infty}^{2}\ l^2}$

i.e., a form of drag coefficient. We normally define aerodynamic force coefficients in terms of the dynamic pressure, i.e., $(1/2)\varrho_{\infty} V_{\infty}^2$ so that more conventionally we would write the force coefficient as

$\Pi_{1} = C_D = \frac{D}{ \frac{1}{2} \varrho_{\infty} V_{\infty}^{2} \, l^2}$

It would also be legitimate to write the drag coefficient as

$C_D = \frac{D}{ \frac{1}{2} \varrho_{\infty} V_{\infty}^{2} \, l \, h } = \frac{D}{ \frac{1}{2} \varrho_{\infty} V_{\infty}^{2} \, A }$

where the banner area $A = l \, h$ is used rather than $l^2$ . Ultimately, how we define $C_D$ is just matter of convenience and/or consistency with established convention.

For $\Pi_{2}$ :

$\Pi_{2} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, \mu_{\infty}$

so

$\left[ \Pi_{2} \right] = 1 = M^{0} L^{0} T^{0} = \left(ML^{-3} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L\right)^\gamma \left( ML^{-1}T^{-1} \right)$

and

$\begin{eqnarray*} M: 0 & = & \alpha+1 \\ L: 0 & = & -3\alpha + \beta+\gamma-1 \\ T: 0 & = & -\beta-1 \end{eqnarray*}$

Therefore, in this case $\alpha = -1$ , $\beta = -1$ , and $\gamma = -1$ , so

$\Pi_{2} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \mu_{\infty} = \varrho_{\infty}^{-1} V_{\infty}^{-1} l^{-1} \, \mu_{\infty}$

or

$\Pi_{2} = \frac{\mu_{\infty}}{\varrho_{\infty} V_{\infty} l}$

Inverting the grouping gives

$\Pi_{2} = \frac{\varrho_{\infty} V_{\infty} l}{\mu_{\infty}}$

which in the latter case is a Reynolds number based on the banner length. Notice that the grouping can be inverted if we want to, such as to follow established convention or just because it is otherwise convenient to do so.

For $\Pi_{3}$ :

$\Pi_{3} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, h$

so

$\left[ \Pi_{3} \right] = 1 = M^{0}L^{0}T^{0} = \left( ML^{-3} \right)^{\alpha} \left( LT^{-1} \right) ^{\beta} \left( L \right)^{\gamma} L$

giving

$\begin{eqnarray*} M: 0 & = & \alpha \\ L: 0 & = & -3\alpha + \beta+\gamma+1 \\ T: 0 & = & -\beta \end{eqnarray*}$

Therefore, in this case $\alpha = 0$ , $\beta = 0$ , and $\gamma = -1$ , i.e.,

$\Pi_{3} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, h = \varrho_{\infty}^{0} V_{\infty}^{0} l^{-1} \, h = \frac{h}{l}$

so

$\Pi_{3} = \frac{h}{l}$

or we can again invert this grouping (for convenience) giving

$\Pi_{3} = \frac{l}{h} = A \! R$

which is a length to height ratio or what would be called an aspect ratio $A\!R$ .

As a result of our dimensional analysis then

$\phi \left( \frac {D}{\frac{1}{2} \varrho_{\infty} V_{\infty}^{2} \, A }, \frac{\varrho_{\infty} V_{\infty} l }{\mu_{\infty}}, \frac{l}{h} \right) = 0$

or

$\phi \left( C_{D}, Re, A\!R\right) = 0$

Finally, in explicit form the drag coefficient can be written as a function of Reynolds number based on banner length and the aspect ratio of the banner, i.e.,

$C_{D} = f_{0} \left( Re, A\!R \right)$

Note: Try this problem again using $\varrho_{\infty}$ , $\mu_{\infty}$ and $h$ as the repeating variables. What happens to the groupings?

Worked Example #20

The DBF team have observed that banner in the previous problem begins to flutter at some critical airspeed, which results in much higher drag on the banner. The flutter speed of the banner appears to depend on the length of the banner, $l$ , and its structural characteristics, which can be expressed in terms of a natural frequency, $\omega_n$ . By extending the steps in the previous question, use the Buckingham $\Pi$ method to determine the non-dimensional groupings that will govern the flutter speed of the banner.

The relationship between the flutter speed $V_f$ and the expected dependencies can be written in general functional form as

$V_f = f \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, l, \omega_n \right)$

or in implicit form as

$g \left( V_f, \varrho_{\infty}, V_{\infty}, \mu_{\infty}, l, \omega_n \right) = 0$

Hence, in this problem $N = 6$ , $K = 3$ , $N - K = 3$ , and so we will have three $\Pi$ products.

For each variable the dimensions are

$\begin{eqnarray*} \left[ V_f \right] & = & LT^{-1} \\ \left[ \varrho_{\infty} \right] & = & ML^{-3} \\ \left[ V_{\infty} \right] & = & LT^{-1} \\ \left[ \mu_{\infty} \right] & = & ML^{-1}T^{-1} \\ \left[ l \right] & = & L \\ \left[ \omega_n \right] & = & T^{-1} \end{eqnarray*}$

Now we can set up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|r|r|r} & V_f & \varrho_{\infty} & V_{\infty} & \mu_{\infty} & l & \Omega_n \\ \hline \mbox{Mass: } M & 0 & 1 & 0 & 1 & 0 & 0 \\ \mbox{Length: } L & 1 & -3 & 1 & -1 & 1 & 0 \\ \mbox{Time: } T & -1 & 0 & -1 & -1 & 0 & -1 \end{array}$

Again, as in most aerodynamic problems, we choose $\varrho_{\infty}$ , $V_{\infty}$ and $l$ as the repeating variables. Following the Buckingham $\Pi$ method then the $\Pi$ products are:

$\begin{eqnarray*} \Pi_{1} & = & g_{1} \left( \varrho_{\infty}, V_{\infty}, l, V_f\right) \\ \Pi_{2} & = & g_{2} \left( \varrho_{\infty}, V_{\infty}, l, \mu_{\infty} \right) \\ \Pi_{3} & = & g_{3} \left( \varrho_{\infty}, V_{\infty}, l, \omega_n \right) \end{eqnarray*}$

For $\Pi_{1}$ :

$\Pi_{1} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, V_f$

In terms of the dimensions of the problem then

$\left[ \Pi_{1} \right] = 1 = M^{0}L^{0}T^{0} = \left( ML^{-3} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L \right)^{\gamma} L T^{-1}$

For $\Pi_{1}$ to be dimensionless we must have

$\begin{eqnarray*} M: 0 & = & \alpha \\ L: 0 & = & -3\alpha+\beta+\gamma+1 \\ T: 0 & = & -\beta-1 \end{eqnarray*}$

By inspection we get $\alpha = 0$ , $\beta = -1$ , and $\gamma = 0$ . Therefore, the first $\Pi$ product is

$\Pi_{1} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} V_f = \varrho_{\infty}^{0} V_{\infty}^{-1} l^{0} \, V_f$

or

$\Pi_{1} = \frac{V_f}{V_{\infty}}$

which is a speed ratio or a non-dimensional flutter speed.

For $\Pi_{2}$ :

$\Pi_{2} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, \mu_{\infty}$

In terms of the dimensions of the problem then

$\left[ \Pi_{2} \right] = 1 = M^{0}L^{0}T^{0} = \left( ML^{-3} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L \right)^{\gamma} M L^{-1} T^{-1}$

For $\Pi_{2}$ to be dimensionless we must have

$\begin{eqnarray*} M: 0 & = & \alpha+1 \\ L: 0 & = & -3\alpha-\beta+\gamma-1 \\ T: 0 & = & -\beta-1 \end{eqnarray*}$

By inspection, we get $\alpha = -1$ , $\beta = -1$ , and $\gamma = -1$ . Therefore, the second $\Pi$ product is

$\Pi_{2} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \mu_{\infty} = \varrho_{\infty}^{-1} V_{\infty}^{-1} l^{-1} \, \mu_{\infty}$

So

$\Pi_{2} = \frac{\mu_{\infty} }{\varrho_{\infty} V_{\infty} l }$

or

$\Pi_{2} = \frac{\varrho_{\infty} V_{\infty} l }{\mu_{\infty} }$

and, once again, a Reynolds number comes into the problem.

For $\Pi_{3}$ :

$\Pi_{3} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \, \omega_n$

In terms of the dimensions of the problem then

$\left[ \Pi_{3} \right] = 1 = M^{0}L^{0}T^{0} = \left( ML^{-3} \right)^{\alpha} \left( LT^{-1} \right)^{\beta} \left( L \right)^{\gamma} T^{-1}$

For $\Pi_{3}$ to be dimensionless we must have

$\begin{eqnarray*} M: 0 & = & \alpha \\ L: 0 & = & -3\alpha + \beta+\gamma \\ T: 0 & = & -\beta-1 \end{eqnarray*}$

By inspection we get $\alpha = 0$ , $\beta = -1$ , and $\gamma = 1$ . Therefore, the third $\Pi$ product is

$\Pi_{3} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (l)^{\gamma} \omega_n = \varrho_{\infty}^{0} V_{\infty}^{-1} l^{1} \ \omega_n$

or

$\Pi_{3} = \frac{ \omega_n \, l }{V_{\infty} }$

which is a form of structural non-dimensional frequency or what we might call a structural reduced frequency.

As a result of our dimensional analysis then

$\phi \left( \frac{V_f}{V_{\infty}}, \frac{\varrho_{\infty} V_{\infty} l }{\mu_{\infty} }, \frac{ \omega_n \, l }{V_{\infty} } \right) = 0$

or
or in explicit form

$\frac{V_f}{V_{\infty}} = f_{0} \left( \frac{\varrho_{\infty} V_{\infty} l }{\mu_{\infty} }, \frac{ \omega_n \, l }{V_{\infty} } \right)$

Therefore, the dimensional analysis tells us that the non-dimensional flutter speed of the banner will depend on the Reynolds number and its structural reduced frequency.

Worked Example #21

A spherical projectile of diameter $d$ is moving supersonically. The drag $D$ is assumed to depend on the free-stream velocity $V_{\infty}$ , the free-stream density $\varrho_{\infty}$ , the free-stream viscosity $\mu_{\infty}$ , and the free-stream temperature $T_{\infty}$ , as well as the heat capacities at constant volume and constant pressure, $C_{\cal V}$ and $C_p$ , respectively. Use the Buckingham $\Pi$ method to determine the non-dimensional groupings that will govern this problem.

The relationship between the drag on the sphere and the given variables can be written general functional form as

$D = f \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, T_{\infty}, C_{\cal V}, C_p, d \right)$

or in implicit form as

$g \left( D, \varrho_{\infty}, V_{\infty}, \mu_{\infty}, T_{\infty}, C_{\cal V}, C_p, d \right) = 0$

Hence, $N = 8$ , $K = 4$ , $N - K = 4$ , and so we will have four $\Pi$ products. Notice in this case that temperature in explicitly defined so there are 4 fundamental dimensions.

For each variable the units are

$\begin{eqnarray*} \left[ D \right] & = & MLT^{-2} \\ \left[ \varrho_{\infty} \right] & = & ML^{-3} \\ \left[ V_{\infty} \right] & = & LT^{-1} \\ \left[ \mu_{\infty} \right] & = & ML^{-1}T^{-1} \\ \left[ T_{\infty} \right] & = & \theta \\ \left[ C_{\cal V} \right] & = & L^{2} T^{-2} \theta^{-1} \\ \left[ C_p \right] & = & L^{2} T^{-2} \theta^{-1} \\ \left[ d \right] & = & L \end{eqnarray*}$

Now we can set up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|r|r|r|r|r|} & D & \varrho_{\infty} & V_{\infty} & \mu_{\infty} & T_{\infty} & C_{\cal V} & C_p & d \\ \hline \mbox{Mass: } M & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ \mbox{Length: } L & 1 & -3 & 1 & -1 & 0 & 2 & 2 & 1 \\ \mbox{Time: } T & -2 & 0 & -1 & -1 & 0 & -2 & -2 & 0 \\ \mbox{Temperature: } \theta & 0 & 0 & 0 & 0 & 1 & -1 & -1 & 0 \end{array}$

Choose $\varrho_{\infty}$ , $V_{\infty}$ , $T_{\infty}$ and $d$ as the repeating variables. They are not unique but have primary dependencies on the drag and collectively include all the fundamental dimensions as well as being linearly independent of each other.

Following the Buckingham $\Pi$ method then the $\Pi$ products are:

$\begin{eqnarray*} \Pi_{1} & = & g_{1} \left( \varrho_{\infty}, V_{\infty}, T_{\infty}, d, D \right) \\ \Pi_{2} & = & g_{2} \left( \varrho_{\infty}, V_{\infty}, T_{\infty}, d, C_{\cal V} \right) \\ \Pi_{3} & = & g_{3} \left( \varrho_{\infty}, V_{\infty}, T_{\infty}, d, C_p \right) \\ \Pi_{4} & = & g_{4} \left( \varrho_{\infty}, V_{\infty}, T_{\infty}, d , \mu_{\infty} \right) \end{eqnarray*}$

For $\Pi_{1}$ :

$\Pi_{1} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (T_{\infty})^{\gamma} (d)^{\delta} \, D$

In terms of the dimensions of the problem then

$\left[ \Pi_{1} \right] = 1 = M^{0}L^{0}T^{0} \theta^{0} = \left( ML^{-3} \right)^{\alpha} \left( L T^{-1}\right)^{\beta} \left( \theta \right)^{\gamma} \left(L\right)^{\delta} \, M L T^{-2}$

For $\Pi_{1}$ to be dimensionless we must have

$\begin{eqnarray*} M: 0 & = & \alpha+ 1 \\ L: 0 & = & -3\alpha +\beta +\delta +1\\ T: 0 & = & -\beta - 2 \\ \theta: 0 & = & \gamma \end{eqnarray*}$

By inspection we get $\gamma = 0$ , $\alpha = -1$ , $\beta = -2$ and $\delta = -2$ . Therefore, the first $\Pi$ product is

$\Pi_{1} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (T_{\infty})^{\gamma} (d)^{\delta} \, D = (\varrho_{\infty})^{-1} (V_{\infty})^{-2} (T_{\infty})^{0} (d)^{-2} \, D$

and

$\Pi_{1} = \frac{D}{\varrho_{\infty} V_{\infty}^2 \, d^2}$

which is a drag coefficient, or we just write it in the conventional way that

$\Pi_{1} = \frac{D}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 \, d^2} = C_D$

For $\Pi_{2}$ :

$\Pi_{2} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (T_{\infty})^{\gamma} (d)^{\delta} \, C_{\cal V}$

In terms of the dimensions of the problem then

$\left[ \Pi_{2} \right] = 1 = M^{0}L^{0}T^{0} \theta^{0} = \left( ML^{-3} \right)^{\alpha} \left( L T^{-1}\right)^{\beta} \left( \theta \right)^{\gamma} \left(L\right)^{\delta} \ L^{2} T^{-2} \theta^{-1}$

For $\Pi_{2}$ to be dimensionless we must have

$\begin{eqnarray*} M: 0 & = & \alpha \\ L: 0 & = & -3\alpha +\beta +\delta +2\\ T: 0 & = & -\beta - 2 \\ \theta: 0 & = & \gamma-1 \end{eqnarray*}$

By inspection we get $\alpha = 0$ , $\gamma=1$ , $\beta = -2$ and $\delta = 0$ . Therefore, the second $\Pi$ product is

$\Pi_{2} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (T_{\infty})^{\gamma} (d)^{\delta} \, C_{\cal V} = (\varrho_{\infty})^{0} (V_{\infty})^{-2} (T_{\infty})^{1} (d)^{0} \, C_{\cal V}$

and

$\Pi_{2} = \frac{C_{\cal V} \, T_{\infty} }{V_{\infty}^2 }$

or more conventionally this ratio is written as

$\Pi_{2} = \frac{V_{\infty}^2 }{C_{\cal V} \, T_{\infty} }$

Notice that $C_{\cal V} \, T_{\infty}$ is the internal energy per unit mass of the free-stream flow, so this non-dimensional grouping represents a ratio of kinetic energy to internal energy.

For $\Pi_{3}$ the process will be identical to that for $\Pi_{2}$ , which is redundant. But we also see that both $C_{\cal V}$ and $C_p$ have the same units so we can write immediately that

$\Pi_{3} = \frac{C_p}{C_{\cal V}}$

which is the familiar ratio of specific heats. This ratio would have been a product of the dimensional analysis if either $C_{\cal V}$ or $C_p$ had been used as a repeating variable.

For $\Pi_{4}$ :

$\Pi_{4} = (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (T_{\infty})^{\gamma} (d)^{\delta} \ \mu_{\infty}$

In terms of the dimensions of the problem then

$\left[ \Pi_{2} \right] = 1 = M^{0}L^{0}T^{0} \theta^{0} = \left( ML^{-3} \right)^{\alpha} \left( L T^{-1}\right)^{\beta} \left( \theta \right)^{\gamma} \left(L\right)^{\delta} \ M L^{-1} T^{-1}$

For $\Pi_{4}$ to be dimensionless, then we must have

$\begin{eqnarray*} M: 0 & = & \alpha + 1\\ L: 0 & = & -3\alpha +\beta +\delta -1\\ T: 0 & = & -\beta - 1 \\ \theta: 0 & = & \gamma \end{eqnarray*}$

By inspection we get $\alpha = -1$ , $\gamma=0$ , $\beta = -1$ and $\delta = 1$ . Therefore, the fourth $\Pi$ product is

$\Pi_{4}= (\varrho_{\infty})^{\alpha} (V_{\infty})^{\beta} (T_{\infty})^{\gamma} (d)^{\delta} \, \mu_{\infty} = (\varrho_{\infty})^{-1} (V_{\infty})^{-1} (T_{\infty})^{0} (d)^{1} \ \mu_{\infty}$

and

$\Pi_{4} = \frac{\mu_{\infty}}{\varrho_{\infty} V_{\infty} d}$

or just

$\Pi_{4} = \frac{\varrho_{\infty} V_{\infty} d}{\mu_{\infty}}$

which is the Reynolds number.

As a result of our dimensional analysis then

$\phi \left( \frac{D}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 \, d^2}, \frac{V_{\infty}^2 }{C_{\cal V} \, T_{\infty} }, \frac{C_p}{C_{\cal V}}, \frac{\varrho_{\infty} V_{\infty} d}{\mu_{\infty}} \right) = 0$

or in explicit form

$\frac{D}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 \, d^2} = f_0 \left( \frac{V_{\infty}^2 }{C_{\cal V} \, T_{\infty} }, \frac{C_p}{C_{\cal V}}, \frac{\varrho_{\infty} V_{\infty} d}{\mu_{\infty}} \right)$

But the important grouping that comes out of this problem is

$\Pi_{2} = \frac{V_{\infty}^2 }{C_{\cal V} \, T_{\infty} }$

Worked Example #22

A force $F$ is applied at the tip of a cantilevered wing of length $L$ and second moment of area $I$ . The modulus of elasticity of the material used for the wing is $E$ . When the force is applied, the tip deflection is $z_d$ . Use the Buckingham $\Pi$ method to find the non-dimensional groupings that will govern this problem.

The relationship between the force and the tip deflection can be written in the general functional form as

$z_d = f \left( F, L_w, I, E \right)$

or in implicit form as

$g \left(z_d, F, L_w, I, E \right) = 0$

where we have used $L_w$ to denote the length of the wing to avoid confusion with the dimensions of length. Hence, $N = 5$ , $K = 3$ , $N - K = 2$ , and so we will have two $\Pi$ products.

For each variable the units are

$\begin{eqnarray*} \left[ z_d \right] & = & L \\ \left[ F \right] & = & M L T^{-2} \\ \left[ L_w \right] & = & L \\ \left[ I \right] & = & L^4 \\ \left[ E \right] & = & M L^{-1} T^{-2} \end{eqnarray*}$

Now we can set up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|r|r} & z_d & F & L_w & I & E \\ \hline \mbox{Mass: } M & 0 & 1 & 0 & 0 & 1 \\ \mbox{Length: } L & 1 & 1 & 1 & 4 & -1 \\ \mbox{Time: } T & 0 & -2 & 0 & 0 & -2 \end{array}$

This problem poses a bit of a dilemma because we see that the choice of the repeating variables here is not at all obvious. If we choose $F$ , $L_w$ and $E$ as the repeating variables they will not be linearly independent, and if we choose $F$ , $I$ and $E$ as the repeating variables (try it!) then the solution to the problem becomes indeterminate, i.e., we cannot uniquely solve for the non-dimensional groupings.

The accepted solution to this type of dilemma is to reduce the number of repeating variables by one and create a third $\Pi$ grouping. If we now choose $L_w$ and $E$ as repeating variables, which are linearly independent and include all of the fundamental dimensions, then following the Buckingham $\Pi$ method then the three $\Pi$ products will be:

$\begin{eqnarray*} \Pi_{1} & = & g_{1} \left( L_w, E, z_d \right) \\ \Pi_{2} & = & g_{2} \left( L_w, E, I \right) \\ \Pi_{3} & = & g_{3} \left( L_w, E, F \right) \end{eqnarray*}$

For $\Pi_{1}$ :

$\Pi_{1} = (L_w)^{\alpha} (E^{\beta}) \, z_d$

In terms of the dimensions of the problem then

$\left[ \Pi_{1} \right] = 1 = M^{0}L^{0}T^{0} = \left( L \right)^{\alpha} \left( M L^{-1} T^{-2} \right)^{\beta} L$

For $\Pi_{1}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \beta \\ L: 0 & = & \alpha-\beta+1 \\ T: 0 & = & -2\beta \end{eqnarray*}$

By inspection we get $\alpha = -1$ and $\beta = 0$ , so the first $\Pi$ product is

$\Pi_{1} = (L_w)^{\alpha} (E^{\beta}) \, z_d = (L_w)^{-1} (E^0) \, z_d$

so

$\Pi_{1} = \frac{ z_d}{L_w}$

i.e., a non-dimensional displacement and this seems to be an expected if not obvious grouping.

For $\Pi_{2}$ :

$\Pi_{2} = (L_w)^{\alpha} (E^{\beta}) \, I$

In terms of the dimensions of the problem then

$\left[ \Pi_{2} \right] = 1 = M^{0}L^{0}T^{0} = \left( L \right)^{\alpha} \left( M L^{-1} T^{-2} \right)^{\beta} L^4$

For $\Pi_{2}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \beta \\ L: 0 & = & \alpha-\beta+4 \\ T: 0 & = & -2\beta \end{eqnarray*}$

By inspection we get $\alpha = -4$ and $\beta = 0$ , so the second $\Pi$ product is

$\Pi_{2} = (L_w)^{\alpha} (E^{\beta}) \, I = (L_w)^{-4} (E^0) \, I$

and so

$\Pi_{2} = \frac{ I}{L_w^4}$

which is a non-dimensional form of the second moment of area.

For $\Pi_{3}$ :

$\Pi_{3} = (L_w)^{\alpha} (E^{\beta}) \, F$

In terms of the dimensions of the problem then

$\left[ \Pi_{3} \right] = 1 = M^{0}L^{0}T^{0} = \left( L \right)^{\alpha} \left( M L^{-1} T^{-2} \right)^{\beta} M L T^{-2}$

For $\Pi_{3}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \beta + 1 \\ L: 0 & = & \alpha-\beta+1 \\ T: 0 & = & -2\beta -2 \end{eqnarray*}$

By inspection we get $\alpha = -2$ and $\beta = -1$ , so the third $\Pi$ product is

$\Pi_{3} = (L_w)^{\alpha} (E^{\beta}) \, I = (L_w)^{-2} (E^{-1}) \, F$

so

$\Pi_{3} = \frac{ F}{E \, L_w^2}$

which is a form of non-dimensional force.

Finally, we have all three $\Pi$ groupings and so we can write the result in functional form as

$\phi \left( \frac{ z_d}{L_w}, \frac{ I}{L_w^4}, \frac{ F}{E \, L_w^2} \right) = 0$

or

$\frac{z_d}{L_w} = f_0 \left( \frac{ I}{L_w^4}, \frac{ F}{E \, L_w^2} \right)$

Worked Example #23

The ERAU wind tunnel uses tiny oil-based aerosol particles of characteristic size, $D_p$ , and density, $\varrho_p$ , to make flow measurements using a method called Particle Image Velocimetry (PIV). The characteristic time required for the aerosol particle to adjust to a sudden change in flow speed is called the particle relaxation time $\tau_p$ , which is given by the equation

$\tau_p = \frac{\varrho_p D_p^2}{18 \mu_{\infty}}$

where $\mu_{\infty}$ is the viscosity of the flow. First, proceed to verify that the primary dimensions of $\tau_p$ are units of time. Second, find a non-dimensional form for the time constant that is based on a characteristic flow velocity, $V$ , and a characteristic length, $L$ . Comment on your result. Do you see anything interesting?

We are given that

$\tau_p = \frac{\varrho_p D_p^2}{18 \mu_{\infty}}$

and told that the units of $\tau_p$ are time. For each variable the units are

$\begin{eqnarray*} \left[ \varrho_p \right] & = & M L^{-3} \\ \left[ D_p \right] & = & L \\ \left[ \mu_{\infty} \right] & = & M L^{-1} T^{-1} \\ \end{eqnarray*}$

so

$\left[ \tau_p \right] = \frac{ (M L^{-3} ) (L^2)}{M L^{-1} T^{-1}} = \frac{ M L^{-1}}{M L^{-1} T^{-1}} = T$

which confirms that the units of $\tau_p$ are indeed time.

We are asked to find a non-dimensional form of $\tau_p$ based on a characteristic flow velocity, $V$ , and a characteristic length, $L$ . We see that the ratio $L/V$ has units of time so as non-dimensional form could be

$\hat{\tau}_p = \frac{\varrho_p D_p^2}{18 \mu_{\infty}}\ \frac{V}{L} = \frac{\varrho_p V D_p}{18 \mu_{\infty}} \, \frac{D_p}{L} = \frac{1}{18} \left( \frac{\varrho_p V D_p}{ \mu_{\infty}} \right) \left( \frac{D_p}{L} \right)$

which is interesting because it involves a Reynolds number based on particle diameter as well as the ratio of the particle diameter to the length scale. Therefore, the higher the Reynolds number and/or the bigger the particle then the longer it will take for the particle to adjust to any changes in the flow conditions.

Worked Example #24

Based on experiments performed with a wind turbine, it is determined that its power output is a function the size of the wind turbine as characterized by its radius $R$ , the rotational angular velocity of the turbine $\Omega$ , the wind speed $V_{\infty}$ , and the air density $\varrho_{\infty}$ . Using the Buckingham $\Pi$ method, determine the non-dimensional groupings that will describe this problem.

For this problem we are told that the power output $P$ can be written as

$P = f \left( R, \Omega, V_{\infty}, \varrho_{\infty} \right)$

In implicit form then

$f_1 \left( R, \Omega, V_{\infty}, \varrho_{\infty}, P \right) = 0$

So, we have $N=5$ and again $K=3$ so there will be 2 $\Pi$ products.

We first set up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|r|r} & R & \Omega & V_{\infty} & \varrho_{\infty} & P \\ \hline \mbox{Mass} ~M: & 0 & 0 & 0 & 1 & 1 \\ \mbox{Length} ~L: & 1 & 0 & 1 & -3 & 2 \\ \mbox{Time} ~T: & 0 & -1 & -1 & 0 & -3 \end{array}$

Notice the base units of power are M L $^{2}$ T $^{-3}$ .

Now we need to choose the repeating variables. A good choice in this case is $R$ , $V_{\infty}$ and $\varrho_{\infty}$ , which will have a primary influence on power production from the turbine. They also collectively include all of the fundamental dimensions mass, length and time, and they are clearly linearly independent just by inspection.

For the first $\Pi$ product then

$\Pi_{1} = f_2 \left(R, V_{\infty}, \varrho_{\infty}, P \right)$

and for the second $\Pi$ product then

$\Pi_{2} = f_3 \left(R, V_{\infty}, \varrho_{\infty}, \Omega \right)$

Continuing with the first $\Pi$ product then

$\Pi_{1} = (R)^{\alpha} (V_{\infty})^{\beta} (\varrho_{\infty})^{\gamma} P$

and terms of the dimensions then

$\left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = (L)^{\alpha} (L T^{-1} )^{\beta} (M L^{-3} )^{\gamma} M L^{2} T^{-3}$

For $\Pi_{1}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \gamma + 1 \\ L: 0 & = & \alpha + \beta - 3\gamma + 2 \\ T: 0 & = & -\beta -3 \end{eqnarray*}$

Therefore, $\beta$ = -3, $\gamma$ = -1, and $\alpha = -2$ , so the $\Pi_1$ product is

$\Pi_1 = (R)^{-2} (V_{\infty})^{-3} (\varrho_{\infty})^{-1} P$

or

$\Pi_1 = \frac{P}{\varrho_{\infty} R^2 V_{\infty}^{3}}$

which is a form of power coefficient, i.e.,

$C_P = \frac{P}{\varrho_{\infty} R^2 V_{\infty}^{3}}$

Considering now the second $\Pi$ product then

$\Pi_{2} = (R)^{\alpha} (V_{\infty})^{\beta} (\varrho_{\infty})^{\gamma} \Omega$

and terms of the dimensions then

$\left[ \Pi_{2} \right] = M^{0}L^{0}T^{0} = (L)^{\alpha} (L T^{-1} )^{\beta} (M L^{-3} )^{\gamma} T^{-1}$

For $\Pi_{2}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \gamma \\ L: 0 & = & \alpha + \beta - 3\gamma \\ T: 0 & = & -\beta -1 \end{eqnarray*}$

Therefore, $\beta$ = -1, $\gamma$ = 0, and $\alpha = -1$ , so the $\Pi_2$ product is

$\Pi_2 = (R)^{1} (V_{\infty})^{-1} (\varrho_{\infty})^{0} \Omega$

or

$\Pi_2 = \frac{\Omega R} {V_{\infty}} \quad \mbox{or} \quad \frac{V_{\infty}}{\Omega R}$

which is a form of advance ratio or tip speed ratio, i.e.,

$J = \frac{V_{\infty}}{\Omega R}$

Therefore, we see based on the information given that the power output of the wind turbine in terms of a power coefficient $C_P$ is related to the wind speed in the form of a tip speed ratio $J$ , i.e.,

$\frac{P}{\varrho_{\infty} R^2 V_{\infty}^{3}} = \phi \left( \frac{V_{\infty}}{\Omega R} \right)$

or

$C_P = \phi (J)$

Worked Example #25

A sphere is located in a pipe through which a liquid is flowing. The drag force $F_D$ on the sphere is assumed to be function of the diameter of the sphere $d$ , the pipe diameter, $D$ , the average flow velocity $V$ , and the fluid density $\varrho$ .

1. Write down the functional expression for the drag force $F_D$ in terms of the parameters given above.
2. Write down the dimensional matrix for this problem in terms of base units MLT.
3. Determine the relevant $\Pi$ groups for this problem.
4. If the drag force on the sphere with $D$ = 0.1 m and $d$ = 0.07 m in a specific liquid flowing at an average flow speed of 3 m/s is 600 N, what would the drag force be on a sphere with $D$ = 0.4 m and $d$ = 0.28 m at 6.7 m/s using the same liquid? Assume that $\varrho$ = 900 kg m $^{-3}$ .

1. In explicit form then

$F_D = \phi \left( d, D, V, \varrho \right)$

or in implicit form then

$\phi_1 \left( F_D, d, D, V, \varrho \right) = 0$

2. Setting up the dimensional matrix for this problem gives

$\begin{array}{l|r|r|r|r|r|r|} & F_D & d & D & V & \varrho \\ \hline \mbox{Mass}~M: & 1 & 0 & 0 & 0 & 1 \\ \mbox{Length} ~L: & 1 & 1 & 1 & 1 & -3 \\ \mbox{Time} ~T: & -2 & 0 & 0 & -1 & 0 \\ \end{array}$

3. We see in this $N=5$ (five variables) and by inspection we clearly have all of M, L and T so $K=3$ (i.e., all three fundamental dimensions in terms M, L and T) so there are two $\Pi$ products to determine.

4. Use $\varrho$ , $V$ and $d$ as the repeating variables, which the students should hopefully use if they were paying attention. This choice includes all the fundamental dimensions and it is obvious that they are all linearly independent. Following the Buckingham $\Pi$ method then the two $\Pi$ products are

$\begin{eqnarray*} \Pi_{1} & = & g_{1} \left( \varrho, V, d, F_D \right) \\ \Pi_{2} & = & g_{2} \left( \varrho, V, d, D \right) \\ \end{eqnarray*}$

So, we have for the first $\Pi$ product that

$\Pi_{1} = (\varrho)^{\alpha} (V)^{\beta} d^{\gamma} \, F_D$

and terms of the dimensions then

$\left[ \Pi_{1} \right] = M^{0}L^{0}T^{0} = (ML^{-3})^{\alpha} (L T^{-1} )^{\beta} (L)^{\gamma} \, M L T^{-2}$

For $\Pi_{1}$ to be dimensionless then

$\begin{eqnarray*} M: 0 & = & \alpha + 1 \\ L: 0 & = & -3\alpha + \beta + \gamma + 1\\ T: 0 & = & -\beta -2 \end{eqnarray*}$

Therefore, $\beta = -2$ , $\alpha = -1$ , and $\gamma = -2$ so the $\Pi_1$ product is

$\Pi_1 = (\varrho)^{-1} (V)^{-2} d^{-2} F_D = \frac{F_d}{\varrho V^2 d^2} = C_{F}$

which is a force coefficient.

For the second $\Pi$ product then

$\Pi_{2} = (\varrho)^{\alpha} (V)^{\beta} (d^{\gamma}) D$

and terms of the dimensions then

$\left[ \Pi_{2} \right] = M^{0}L^{0}T^{0} = (ML^{-3})^{\alpha} (L T^{-1} )^{\beta} (L)^{\gamma} \, L$

For $\Pi_{2}$ to be dimensionless, then

$\begin{eqnarray*} M: 0 & = & \alpha \\\ L: 0 & = & -3\alpha + \beta + \gamma + 1\\ T: 0 & = & -\beta \end{eqnarray*}$

Therefore, $\beta$ = 0 and $\alpha$ = 0, and $\gamma$ = -1 so the $\Pi_2$ product is

$\Pi_2 = (\varrho)^{0} (V)^{0} (d^{-1}) D = \frac{D}{d}$

which is a non-dimensional length. Therefore, we have

$\phi_2 \left( \Pi_1, \Pi_2\right) = 0$

and so finally in explicit form then

$C_F = \phi_3 \left( \frac{D}{d} \right)$

5. To examine dimensional similitude then for both cases for dimensional similitude then the force coefficients must be the same so

$C_F = \frac{F_{D_{1}}}{\varrho V_1^2 d_1^2} = \frac{F_{D_{2}}}{\varrho V_2^2 d_2^2}$

so by rearrangement then

$F_{D_{2}} = F_{D_{1}} \left( \frac{V_2^2}{V_1^2} \right) \left( \frac{d_2^2}{d_1^2} \right) = 600 \left( \frac{6.7^2}{3.0^2} \right) \left( \frac{0.28^2}{0.07^2} \right) = 47,882.7~\mbox{N}$

and also confirming for geometric similarity gives

$\left(\frac{D_1}{d_1} \right) \left(\frac{d_2}{D_2} \right) = \left(\frac{0.1}{0.07} \right) \left(\frac{0.28}{0.4} \right) = 1$

Worked Example #26

The singing sound produced by power lines in the wind are called as Aeolian tones, which is caused by vortex shedding behind the lines. The frequency of the sound, $f$ , is a function of the diameter of the wires, $d$ , the wind speed, $V$ , the density of the air, $\varrho$ , and its dynamic viscosity, $\mu$ .
1. Write down the functional relationship for the frequency in terms of the other parameters given, in both implicit and explicit form.
2. How many base dimensions and $\Pi$ groupings are involved in this problem
3. Write down the dimensional matrix for the problem.
4. Use the Buckingham $\Pi$ method to determine the dimensionless parameter(s) that describes this problem.
5. Rewrite the functional relationship in term of the non-dimensional parameter(s).

1. The frequency of the sound can be written explicitly as

$f = \phi_1 (d, V, \varrho, \mu)$

or implicitly as

$\phi_2 (d, V, \varrho, \mu, f) = 0$

2. The number of variables is 5 so $N = 5$ and the number of base dimensions (mass, length and time are all involved) is 3, so $K = 3$ . This means that there are $N - K = 2 \Pi$ groupings.

3. The dimensional matrix is

$\begin{array}{l|r|r|r|r|r|r} & d & V & \varrho & \mu & f \\ \hline M & 0 & 0 & 1 & 1 & 0 \\ L & 1 & 1 & -3 & -1 & 0 \\ T & 0 & -1 & 0 & -1 & -1 \\ \end{array}$

4. Choose the variables as $\varrho$ , $V$ , and $d$ , which are a standard choice for aerodynamic problems. We can now proceed to find the two $\Pi$ groups, i.e.,

$\Pi_1 = g_1(d,V,\varrho,\mu)$

and

$\Pi_2 = g_2(d,V,\varrho,f)$

For $\Pi_1$ then

$\Pi_1 = g_1(d, V, \varrho, \mu)$

Raising the repeating variables to unknown powers gives

$\left[ \Pi_1 \right] = d^{\alpha} \, V^{\beta} \, \varrho^{\gamma} \, \mu = 1$

In terms of dimensions then

$M^0 L^0 T^0 = 1 = L^{\alpha} \, (LT^{-1})^{\beta} \, (ML^{-3})^{\gamma} \, ML^{-1}T^{-1}$

For $\Pi_1$ to be dimensionless, the powers of each base dimension must add to zero, i.e.

$\begin{eqnarray*} M: 0 & = & \gamma + 1\\ L: 0 & = & \alpha + \beta - 3 \gamma - 1\\ T: 0 & = & -\beta - 1\\ \end{eqnarray*}$

Solving the equations gives $\alpha = -1$ , $\beta = -1$ , and $\gamma = -1$ so

$\Pi_1 = d^{-1} \, V^{-1} \, \varrho^{-1} \, \mu = \frac{\mu} {\varrho Vd}$

which is the inverse of the well known Reynolds number so we can invert (still having a dimensionless grouping) giving

$\Pi_1 = \frac {\varrho Vd} {\mu} = Re$

Solving for $\Pi_2$ gives

$\Pi_2 = g_2(d, V, \varrho, f)$

so

$\left[ \Pi_2 \right] = d^{\alpha} \, V^{\beta} \, \varrho^{\gamma} \, f = 1$

In terms of dimensions then

$M^0 L^0 T^0 = 1 = L^{\alpha} \, (LT^{-1})^{\beta} \, (ML^{-3})^{\gamma} \ T^{-1}$

For $\Pi_2$ to be dimensionless, the powers of each base dimension must add to zero, i.e.

$\begin{eqnarray*} M: 0 & = & \gamma\\ L: 0 & = & \alpha + \beta - 3 \gamma\\ T: 0 & = & -\beta - 1\\ \end{eqnarray*}$

Solving the equations gives $\alpha = 1$ , $\beta = -1$ , and $\gamma = 0$ so

$\Pi_2 = d^{1} \, V^{-1} \, \varrho^{0} \, f = \frac{f d} {V}$

which is a Strouhal number, i.e.,

$\Pi_2 = \frac{f d} {V} = St$

5. Therefore, based on the foregoing analysis, we have that

$\phi_3(\Pi_1, \Pi_2) = 0$

or

$\phi_3 (\frac{\varrho V d} {\mu}, \frac{f \, d} {V}) = 0$

or

$St = \phi_4 (Re)$

so the Strouhal number is a function of the Reynolds number.

Worked Example #27

Using Worked Example #25 as a basis, it is desired to replicate the physics of the singing sound and study it in a low-speed wind tunnel. For the actual power wires they have a diameter of 2.2 cm and are known to sing at wind speeds between 25 mph and 70 mph. How would you develop a wind tunnel test plan to study this problem? The equivalent wire available for the wind tunnel test is 1.1 cm in diameter and the wires is strung across the width of the test section. The tunnel can reach a maximum flow speed of 75 ft/s. Is it possible to obtain dynamic similarity of this problem in the wind tunnel test? If not, why not and what other consideration might be given to the wind tunnel test?

Based on the previous problem then the two relevant similarity parameters in this case are the Reynolds number and the Strouhal number, and the Strouhal number is a function of the Reynolds number. The actual power wires the Reynolds number based on diameter $d$ = 2.2 cm will be

$Re = \frac{\varrho \, V \, d}{\mu} = \frac{1.225 \times 70.0 \times 0.447 \times 0.022}{0.000017894} = 471,256$

using the highest wind speed of 70 mph and MSL ISA values for air. Notice that 70 mph is 102.67 ft/s.

In the wind tunnel, the wire available is only 1.1 cm in diameter, i.e., $d/2$ . So, to get the same Reynolds number the flow speed will need to be twice, i.e., 140 mph or 205.3 ft/s, but this is significantly less than the maximum flow speed of the wind tunnel. Even if the wire to be used in the tunnel was 2.2 cm in diameter, the required flow speed to match the Reynolds number would be higher than is attainable.

One solution would be to use a wire of diameter of say 3.3 cm in the wind tunnel, which would need a flow speed of

$V = 70.0 \frac{2.2}{3.33} = 46.67 \mbox{mph} = 68.45 \mbox{ft/s}$

to match the Reynolds numbers and this is easily achievable, and it is a factor of 0.67 of the actual wind speed.

Therefore, if the Reynolds number is matched by increasing the wire diameter the question is can we also match the Strouhal number? In this case we would get the same sound frequency if

$\frac{d_1} {V_1} = \frac{d_2} {V_2} = 1$

but in this case we have

$\frac{d_2} {V_2} = \frac{3 d_1/2} {0.67 V_1 } = 2.24 \frac{d_1} {V_1}$

So even though the Reynolds number could be matched we would not get the same frequency of the Aeolian sounds. Nevertheless, by matching Reynolds number in the wind tunnel we would get the same Strouhal number so the frequency obtained in this case would be higher by a factor of 2.24. In principle then, it would be possible to study the singing sound behavior of the wires in the wind tunnel it is just that the frequencies so obtained would be higher.

This is another example of the challenges in sub-scale testing to study real problems. But it can be seen that with a little ingenuity the problem can be studied by matching, or by matching as closely as possible, the similarity parameters that govern the physics.

Worked Example #28

A Covid-19 particle has a density $\varrho_p$ and characteristic size $D_p$ . It is carried along in the air of density $\varrho$ and viscosity $\mu$ . In still air, the particles slowly settle out only very slowly under the action of gravity and reach a terminal settling speed $V$ . It can be assumed that $V$ depends only on $D_p$ , $\mu$ , $g$ , and the density difference $\varrho_e = (\varrho_p - \varrho)$ .

Write down the functional dependency of $V$ and the other variables $D_p$ , $\mu$ , $g$ . and $\varrho_e$ in both explicit and implicit forms.
How many base dimensions and $\Pi$ groupings are involved in this problem?
Write down the dimensions of each of the variables involved.
Form the dimensional matrix for this problem.
Choose $D_p$ , $\mu$ and $g$ as the repeating variables, so then proceed to find the $\Pi$ grouping involving $V$ .

1. We can let $(\varrho_p - \varrho) = \Delta \varrho$ . The relationship between the settling velocity $V$ and other properties can be written in the general functional form as

$V = f \left( D_p, \mu, g, \Delta \varrho \right)$

and in implicit form

$g \left( V, D_p, \mu, g, \Delta \varrho \right) = 0$

2. $N = 5$ and $K = 3$ ; there are 3 fundamental dimensions in this problem. Therefore, $N - K = 2$ , so we will have two $\Pi$ products.

3. For each variable the dimensions are

$\begin{eqnarray*} \left[ V \right] & = & LT^{-1} \\ \left[ D_p \right] & = & L \\ \left[ \mu \right] & = & M L^{-1} T^{-1} \\ \left[ g \right] & = & L T^{-2} \\ \left[ \Delta \varrho \right] & = & M L^{-3} \end{eqnarray*}$

4. We can now set up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|r|r|r} & V & D_p & \mu & g & \Delta \varrho \\ \hline \mbox{Mass: } M & 0 & 0 & 1 & 0 & 1 \\ \mbox{Length: } L & 1 & 1 & -1 & 1 & -3 \\ \mbox{Time: } T & -1 & 0 & -1 & -2 & 0 \end{array}$

5. We are told to choose $D_p$ , $\mu$ , and $g$ as the repeating variables, which will all have primary effects on the settling velocity. They also collectively include all the fundamental dimensions and are linearly independent of each other. We are asked just to find the grouping that involves $V$ so

$\Pi_{1} = (D_p)^{\alpha} (\mu)^{\beta} (g)^{\gamma} \, V$

In terms of the dimensions of the problem then

$\left[ \Pi_{1} \right] = 1 = M^{0}L^{0}T^{0} = \left( L \right)^{\alpha} \left( M L^{-1} T^{-1} \right)^{\beta} \left( L T^{-2} \right)^{\gamma} \left(LT^{-1} \right)$

For $\Pi_{1}$ to be dimensionless, then the powers or exponents of $M$ , $L$ and $T$ must add to zero, i.e., we must have that

$\begin{eqnarray*} M: 0 & = & \beta \\ L: 0 & = & \alpha-\beta+\gamma+1 \\ T: 0 & = & -\beta-2\gamma-1 \end{eqnarray*}$

By inspection we get $\beta = 0$ , $\gamma = -1/2$ and $\alpha = -1/2$ so

$\Pi_{1} = (D_p)^{\alpha} (\mu)^{\beta} (g)^{\gamma} \, V = (D_p)^{-1/2} (\mu)^{0} (g)^{-1/2} \, V$

or

$\Pi_{1} = \frac{V}{\sqrt{D_p \, g}}$

Worked Example #29

A weir is an obstruction in an open channel water flow. The volume flow rate $\overbigdot{Q}$ over the weir depends on acceleration under gravity $g$ , the width of weir $b$ (into the screen), and the water height $H$ above the weir.

Write out the functional relationship in explicit and implicit form.
How many fundamental (base) dimensions are involved?
Write out the dimensional matrix for this problem.
Determine the non-dimensional $\Pi$ groupings that govern this problem.

1. The explicit form of the relationship is

$\overbigdot{Q} = f( g, b, H )$

and in implicit form then

$g( \overbigdot{Q}, g, b, H) = 0$

2. In this problem we see that there are actually only two base dimensions, length $L$ and time $T$ , so the procedure should be easier.

The units involved are

$\begin{eqnarray*} \left[ \overbigdot{Q} \right] & = & L^3 T^{-1} \\ \left[ g \right] & = & L T^{-2} \\ \left[ b \right] & = & L \\ \left[ H \right] & = & L \end{eqnarray*}$

3. We can now set up the dimensional matrix, i.e.,

$\begin{array}{l|r|r|r|r|r|r} & \overbigdot{Q} & g & b & H \\ \hline \mbox{Length: } L & 3 & 1 & 1 & 1 \\ \mbox{Time: } T & -1 & -2 & 0 & 0 \end{array}$

4. We will have two $\Pi$ products as the number of the basic quantities $K = 2$ and number of variables in this problem $N = 4$ . We will use $g$ and $H$ as repeating variables, which collectively include all the fundamental dimensions and are linearly independent of each other. Note: We cannot choose both $H$ and $b$ repeating variables because they have the same units and, therefore, they are not linearly independent of each other.

For the first $\Pi$ product then

$\Pi_{1} = (g)^{\alpha} (H)^{\beta} \overbigdot{Q}$

In terms of the dimensions of the problem then

$\left[ \Pi_{1} \right] = 1 = L^{0}T^{0} = \left( L T^{-2} \right)^{\alpha} \left( L \right)^{\beta} \left( L ^3T^{-1} \right)$

For $\Pi_{1}$ to be dimensionless, then the powers or exponents of $L$ and $T$ must add to zero, i.e., we must have that

$\begin{eqnarray*} L: 0 & = & \alpha + \beta + 3 \\ T: 0 & = & -2 \alpha -1 \end{eqnarray*}$

By inspection we get $\alpha = -1/2$ and $\beta = -5/2$ so the grouping is

$\Pi_{1} = (g)^{\alpha} (H)^{\beta} (g) \overbigdot{Q} = (g)^{-1/2} (H)^{-5/2} \, \overbigdot{Q}$

or

$\Pi_{1} =\frac{\overbigdot{Q}} {g^{1/2} \, H^{5/2}}$

For the second $\Pi$ product then

$\Pi_{2} = (g)^{\alpha} (H)^{\beta} b$

In terms of the dimensions of the problem then

$\left[ \Pi_{2} \right] = 1 = L^{0}T^{0} = \left( L T^{-2} \right)^{\alpha} \left( L \right)^{\beta} \left( L \right)$

For $\Pi_{2}$ to be dimensionless, then the powers or exponents of $L$ and $T$ must add to zero, i.e., we must have that

$\begin{eqnarray*} L: 0 & = & \alpha + \beta + 1 \\ T: 0 & = & -2 \alpha \end{eqnarray*}$

By inspection we get $\alpha = 0$ and $\beta = -1$ so the grouping is

$\Pi_{2} = (g)^{0} (H)^{-1} b = \frac{b}{H}$

Therefore, we have that

$\phi_1 \left( \frac{\overbigdot{Q}} {g^{1/2} \, H^{5/2}}, \frac{b}{H} \right) = 0$

or

$\left( \frac{\overbigdot{Q}} {g^{1/2} \, H^{5/2}} \right) = \phi_2 \left( \frac{b}{H} \right)$

Worked Example #30

Sir Geoffrey I. Taylor (1886–1975) was a British physicist and engineer. He used dimensional analysis to estimate the blast wave propagation characteristics from an explosion. Taylor assumed that the radius of the wave $R$ was a function of the energy released $E$ during the explosion, the density of the air $\varrho$ , and the time $t$ . Using the Buckingham Pi method, recreate Taylor’s steps and find the non-dimensional grouping that governs this behavior. How does the radius of the blast wave change by a doubling of: (i) $E$ and (ii) time?

The relationship may be written in a general functional form as

$R = \phi \left( E, \varrho, t \right)$

or in implicit form then

$\phi_1 \left(R, E, \varrho, t \right) = 0$

We have $N = 4$ and $K = 3$ because the problem clearly includes the dimensions of mass, length and time, and so we have just one $\Pi$ product. Energy is the ability to to work and so the units of energy will be the same as the units of work, which is equivalent to a force times a distance, i.e., $M L^2 T^{-2}$ . Setting up the dimensional matrix gives

$\begin{array}{l|c|r|r|r} & R & E & \varrho & t \\ \hline \mbox{Mass} ~M: & 0 & 1 & 1 & 0 \\ \mbox{Length} ~L: & 1 & 2 & -3 & 0 \\ \mbox{Time} ~T: & 0 & -2 & 0 & 1 \end{array}$

Following the Buckingham $\Pi$ method, then the $\Pi$ product is

$\Pi = (E)^{\alpha} (\varrho)^{\beta} (t)^{\gamma} \, R$

where the specific values of the coefficients $\alpha$ , $\beta$ and $\gamma$ must be obtained to make the equation dimensionally homogeneous. In terms of the dimensions of the problem then

$\left[ \Pi \right] = 1 = M^{0}L^{0}T^{0} = \left( M L^2 T^{-2} \right)^{\alpha} \left( M L^{-3} \right)^{\beta} \left(T \right)^{\gamma} \, L$

For $\Pi$ to be dimensionless, then the powers or exponents of $M$ , $L$ and $T$ must add to zero, i.e., we must have that

$\begin{eqnarray*} M: 0 & = & \alpha + \beta \\ L: 0 & = & 2\alpha - 3\beta + 1 \\ T: 0 & = & -2\alpha + \gamma \end{eqnarray*}$

In this case we see that $\alpha = -\beta$ so from the second equation we get that $\alpha = -1/5$ . This further gives that $\beta = 1/5$ , and $\gamma = -2/5$ . Therefore, we get that

$\Pi = (E)^{\alpha} (\varrho)^{\beta} (t)^{\gamma} \, R = E^{-1/5} \varrho^{1/5} t^{-2/5} R$

or

$\Pi = \frac{ R \, \varrho^{1/5}}{E^{1/5} \, t^{2/5}}$

In final form we can write

$R = \phi_2 \left( \frac{E^{1/5} \, t^{2/5}}{\varrho^{1/5}} \right)$

In the second part of the question we are asked about how the radius of the blast wave changes by a doubling of $E$ . According to the relationship we have derived then the radius would increase by a factor of $2^{1/5}$ or about 1.15. By doubling the time then the radius of the wave would increase by a factor of $2^{2/5}$ or about 1.32.

Worked Example #31

An ocean surface wave is a sinusoidal-like disturbance propagating along the surface of the ocean, as shown in the figure below. The speed of the wave, $U$ , is found to be a function of the surface tension of the sea water, $\sigma$ , the density of the sea water, $\varrho$ , and the wavelength, $\lambda$ , of the wave.

1. Write down the functional relationship for the speed of the wave in terms of the given parameters, in both implicit and explicit form.
2. How many base dimensions and $\Pi$ groupings are involved in this problem?
3. Write down the dimensions of all the parameters in base units. Notice: The units of surface tension are force per unit length.
4. Create the dimensional matrix for this problem.
5. Determine the dimensionless parameter(s) that describe this problem.

1. In explicit form, the speed of the wave is

$U = f(\sigma,\, \varrho, \lambda)$

where $f$ is some function. In implicit form then

$g( \sigma,\, \varrho,\, \lambda, U) = 0$

where $g$ is some other function.

2. The base dimensions and $\Pi$ groupings involved in this problem are:

$\begin{eqnarray*} \mbox{Number~of~variables:} \, N & = & 4\\ \mbox{Number~of~base~dimensions:} \, K & = & 3\\ \mbox{Number~of~ $\Pi$~ groups:} \, N - K & = & 4 - 3 = 1\\ \end{eqnarray*}$

3. Below are the dimensions of all the parameters in this problem in terms of base units:

The speed of the wave, $U$ , has base units of

$\left[ U \right] = L T^{-1}$

Surface tension, $\sigma$ , is given in the question as a “force per unit length” so it has base units of

$\left[ \sigma \right] = ( M L T^{-2}) L^{-1} = M T^{-2}$

Density, $\varrho$ , has base units of

$\left[ \varrho \right] = M \, L^{-3}$

Wavelength, $\lambda$ , has base units of

$\left[ \lambda \right] = L$

4. From the previous results, then the dimensional matrix is

$\begin{array}{l|r|r|r|r|r|r} & \sigma & \varrho & \lambda & U \\ \hline M & 1 & 1 & 0 & 0 \\ L & 0 & -3 & 1 & 1 \\ T & -2 & 0 & 0 & -1 \\ \end{array}$

5. The wave speed $U$ cannot be a repeating variable so the only possible choice of the repeating variables is $\sigma$ , $\varrho$ , and $\lambda$ . The $\Pi$ group will be

$\Pi_1 = g_1 (\sigma, \, \varrho, \, \lambda, \, U)$

Therefore, we have that

$\left[ \Pi_1 \right] = 1 = M^ 0 \, L^0 \, T^0 = \sigma^{\alpha} \, \varrho^{\beta} \, \lambda^{\gamma} \, U$

In terms of dimensions then

$\left[ \Pi_1 \right] = 1 = M^0 \, L^0 \, T^0 = ( MT^{-2})^{\alpha} \, (ML^{-3})^{\beta} \ (L)^{\gamma} \, LT^{-1}$

or alternatively the foregoing can be written as

$M^0 \, L^0 \, T^0 = M^{(\alpha + \beta)} \, L^{(-3\beta + \gamma + 1)} \, T^{(-2\alpha - 1)}$

For this equation to be mathematically balanced on the left and right sides, i.e., to be dimensionally homogeneous, then

$\begin{eqnarray*} M: \;\; 0 & = & \alpha + \beta \\ L: \;\; 0 & = & -3\beta + \gamma + 1 \\ T: \;\; 0 & = & -2\alpha - 1 \\ \end{eqnarray*}$

Solving the foregoing equations gives $\alpha = -1/2$ , $\beta = 1/2$ , $\gamma = 1/2$ so that the resulting $\Pi$ grouping is

$\Pi_1 = \sigma^{-1/2} \ \varrho^{1/2} \, \lambda^{1/2} \, U$

or

$\Pi_1 = U \sqrt{\frac{\varrho \lambda} {\sigma }}$

As a final check, it is easy to show that this grouping is non-dimensional because

$\left[ \Pi_1 \right] = L T^{-1} \left( \frac{ (M L^{-3}) L }{M T^{-2}} \right)^{1/2} = L T^{-1} \left( L^{-2} T^{2} \right)^{1/2} = L T^{-1} L^{-1} T^{1} = 1$

Worked Example #32

The flow rate in a pipe is to be measured with an orifice plate, as shown in the figure below. The static pressure before and after the plate is measured using two pressure gauges. The volumetric flow rate $\overbigdot{Q}$ is found to be a function of the measured pressure drop across the plate, $p_1 - p_2 = \Delta p$ , the fluid density, $\varrho$ , the pipe diameter, $D$ , and the orifice diameter, $d$ .

Write down the functional relationship for the volumetric flow rate $\overbigdot{Q}$ in terms of the other parameters given, in both implicit and explicit form.
Write down the base units of the parameters involved in this problem.
How many base dimensions and dimensionless groupings are involved in this problem?
Write out the dimensional matrix for this problem.
Choose the repeating variables and explain your choice.
Determine the non-dimensional grouping(s) for the parameters involved.
Write down the final dimensionless functional relationship(s).

1. The volumetric flow rate of water can be written in explicit form as

$\overbigdot{Q} = f_1 (\Delta p, \varrho, D, d)$

where $f_1$ is some function, and implicit form as

$f_2 (\overbigdot{Q}, \Delta p, \varrho, D, d) = 0$

where $f_2$ is some other function.

2. The dimensions of the parameters involved are:

$\left[ \overbigdot{Q} \right] = L^3 T^{-1}$
$\left[ \Delta p \right] = ( M L T^{-2} ) L^{-2} = M L^{-1} T^{-2}$
$\left[ \varrho \right] = M L^{-3}$
$\left[ D \right] = L$
$\left[ d \right] = L$

3. The number of base dimensions and groupings are involved in this problem:

Number of variables: $N = 5$ .
Number of base dimensions (mass, length, and time are all involved): $K = 3$ .
Number of $\Pi$ groups: $N - K = 5 - 3 = 2$ .

4. The dimensional matrix is

$\begin{array}{l|r|r|r|r|r|r|r} & \overbigdot{Q} & \Delta p & \varrho & D & d \\ \hline M & 0 & 1 & 1 & 0 & 0 \\ L & 3 & -1 & -3 & 1 & 1 \\ T & -1 & -2 & 0 & 0 & 0 \\ \end{array}$

5. The only possible choice of repeating variables are: $\Delta p$ , $\varrho$ , $D$ and $d$ . Obviously, we cannot choose both $D$ and $d$ as repeating variables because they are not linearly independent. But we should choose one or the other, so we will choose $D$ and so the repeating variables will be $\Delta p$ , $\varrho$ , and $D$ .

6. The $\Pi$ groups will be formed from:

$\Pi_1 = g_1(\Delta p, \varrho, D, \overbigdot{Q})$

and

$\Pi_2 = g_2( \Delta p, \varrho, D, d)$

We now have to solve for the dimensionless groupings.

For $\Pi_1$ then:

$\Pi_1 = g_1(\Delta p, \varrho, D, \overbigdot{Q})$

so

$\left[ \Pi_1 \right] = 1 = M^0 L^0 T^0 = ( \Delta p)^{\alpha} \, ( \varrho)^{\beta} \, (D)^{\gamma} \, \overbigdot{Q}$

and inserting the dimensions for each parameter gives

$\left[ \Pi_1 \right] = 1 = M^0 L^0 T^0 = (M L^{-1} T^{-2} )^{\alpha} \, (M L^{-3})^{\beta}\, (L)^{\gamma} \, L^3 T^{-1}$

or

$\left[ \Pi_1 \right] = 1 = M^0 L^0 T^0 = M^{(\alpha+\beta)} \, L^{(-\alpha -3\beta + \gamma+3)} \, T^{(-2\alpha - 1)}$

For $\Pi_1$ to be dimensionless the powers must add to zero, i.e.,

$\begin{eqnarray*} M: 0 & = & \alpha+\beta \\ L: 0 & = & -\alpha -3\beta + \gamma + 3 \\ T: 0 & = & -2\alpha - 1 \end{eqnarray*}$

Solving the equations we get: $\alpha = -1/2$ , $\beta = 1/2$ , and $\gamma = -2$ . This means, therefore, that $\Pi_1$ is

$\Pi_1 =\Delta p^{-1/2} \, \varrho^{1/2} \, D^{-2} \, \overbigdot{Q}$

or

$\Pi_1 = \overbigdot{Q} \, \sqrt{ \frac{\varrho}{\Delta p \, D^{4}}}$

As a check to see if this is a non-dimensional grouping, we can substitute the units of the parameters so that

$\left[ \Pi_1 \right] = L^3 T^{-1} \left( \frac{M L^{-3}}{M L^{-1} T^{-2} L^4} \right)^{-1/2} = L^3 T^{-1} \left( \frac{L^{-6}}{T^{-2}} \right)^{-1/2} = L^3 T^{-1} L^{-3} T = 1$

so confirming that the grouping is indeed nondimensional.

For $\Pi_2$ then:

$\Pi_2 = g_1(\Delta p, \varrho, D, d)$

so

$\left[ \Pi_2 \right] = 1 = M^0 L^0 T^0 = ( \Delta p)^{\alpha} \, ( \varrho)^{\beta} \, (D)^{\gamma} \, d$

and inserting the dimensions for each parameter gives

$\left[ \Pi_2 \right] = 1 = M^0 L^0 T^0 = (M L^{-1} T^{-2} )^{\alpha} \, (M L^{-3})^{\beta}\, (L)^{\gamma} \, L$

or

$\left[ \Pi_2 \right] = 1 = M^0 L^0 T^0 = M^{(\alpha+\beta)} \, L^{(-\alpha -3\beta + \gamma+1)} \, T^{(-2\alpha)}$

For $\Pi_2$ to be dimensionless the powers must add to zero, i.e.,

$\begin{eqnarray*} M: 0 & = & \alpha+\beta \\ L: 0 & = & -\alpha -3\beta + \gamma + 1 \\ T: 0 & = & -2\alpha \end{eqnarray*}$

Solving the equations we get: $\alpha = 0$ , $\beta = 0$ , and $\gamma = -1$ . This means, therefore, that $\Pi_2$ is

$\Pi_2 = \Delta p^{0} \, \varrho^{0} \, D^{-1} \, d = \frac{d}{D}$

In this case, it is obvious that the ratio of one length to another is non-dimensional.

7. Finally then, the non-dimensional relationship between the volumetric flow rate $\overbigdot{Q}$ and the other parameters is

$\overbigdot{Q} \, \sqrt{ \frac{\varrho}{\Delta p \, D^{4}}} = \phi \left( \frac{d}{D} \right)$

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Introduction to Aerospace Flight Vehicles by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.9