54 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 circle’s radius. 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 person is curious about why some insects can walk on water. The person discovers that a fluid property of importance in this problem is called surface tension, which is given the symbol \sigma_s and 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. A force is equivalent to mass times acceleration, which is MLT^{-2} in base units. Therefore, the base units of surface tension are

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

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:

  1. Energy, E
  2. Work, W
  3. Power, P
  4. 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), 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, all terms have the same fundamental dimensions of M L^{-1} T^{-2}.

Worked Example #7

In a particular fluids problem, the flow rate, Q, depends on a height, h, and acceleration under gravity, g. The relationship can be expressed as

    \[ Q = K \, h^{\alpha} \, g^{\beta} \]

where K is a constant with units of length. By satisfying dimensional homogeneity, determine the values of \alpha and \beta.

In terms of the dimensions

    \[ \left[ Q\right] = L \left[ h^{\alpha} \right] \left[ g^{\beta} \right] \]

and so for each parameter, then

    \[ \left[ Q \right] = L^3 T^{-1} \quad\quad \left[ K \right] = L \quad\quad \left[ h \right] = L \quad\quad \left[ g \right] = L T^{-2} \]

Therefore,

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

To obtain dimensional homogeneity, then

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

so \beta = 1/2 and \alpha = 3/2. Inserting the values gives the relationship as

    \[ Q = K \, h^{3/2} \, g^{1/2} = K \, \sqrt{g} \, h^{3/2} \]

Worked Example #8

Consider the drag of a sphere problem using the Buckingham \Pi method, 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 obtained.

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

    \[ D = \phi \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, d \right) \]

where D is the dependent variable. \varrho_{\infty}, V_{\infty}, \mu_{\infty} and d are the independent variables. In implicit form, the drag can be written as

    \[ \psi \left( \varrho_{\infty}, V_{\infty}, \mu_{\infty}, d, D \right) = 0 \]

Choose \mu_{\infty}, V_{\infty} and d as repeating variables. Following the Buckingham \Pi method, then the \Pi products for this 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, \alpha = -1, \beta = -1, and \gamma = -1. So, the first \Pi product is

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

or

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

which is still dimensionless, 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 vary 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 obtained this outcome by emphasizing 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 dimensionless 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 will be recognized as the Reynolds number.

Therefore, as a result of the dimensional analysis of the sphere, then

    \[ \phi_1 \left( \frac {D}{3 \pi\mu V_{\infty} d}, \frac{\varrho_{\infty} V_{\infty} d}{\mu_{\infty}} \right) = 0 \]

or

    \[ \phi_1 \left( C_{D_{\rm Stokes}}, Re \right) = 0 \]

or in explicit form

    \[ C_{D_{\rm Stokes}} = \phi_2 \left( Re \right) \]

Worked Example #9

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 dimensionless groupings that will describe this problem.

The power required for the propeller (the dependent variable), P, can be written in a 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 preceding equation can be written in an implicit form as

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

In this case, there are six variables (N = 6) and three fundamental dimensions (K = 3) comprising mass (M), length (L), and time (T). This means that N - K = 3 so three \Pi products must be determined.

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 dimensionless groupings to be determined.

Choose the variables \varrho_{\infty}, d, and n as the repeating variables, which are all linearly independent, which can be confirmed using the dimensional matrix (below). The dimensionless \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 so that each of the \Pi products must be dimensionless.

Now, the base dimensions of each of the variables can be written down. 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, it can quickly be confirmed 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 the base 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 dimensionless 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 latter 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.,  a dimensionless 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 dimensionless 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

    \[ C_P = f_0 \left( C_T,  J \right) \]

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

Worked Example #10

Consider the internal flow through a rough pipe. The objective is to determine the dimensionless 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, then in general functional form

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

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

The dimensional matrix is

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

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

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

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 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 dimensionless pressure drop or “head” drop. Usually, 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 dimensionless roughness of the pipe.

Worked Example #11

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

Using dimensional analysis, show that the dimensionless frequency of this motion, k, is given by

    \[ k = \omega \sqrt{\frac{d}{g}} \]

You may assume that 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 the implicit form, then

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

Therefore, N = 5, K = 3, so there are two \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. If \varrho were to be chosen instead of g, the repeating variables would collectively not include time T, so the Buckingham \Pi method fails. 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, only one of the two \Pi products is asked for, which involves \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 \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 #12

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

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

where \varrho is the density of the water, V is the ship’s speed 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 dimensionless 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 two \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 primarily influence drag. They also collectively include all of the fundamental dimensions of mass, length, and time, and they are linearly independent 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, 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 #13

A tiny spherical particle of diameter D falls 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, mass, length, and time are all 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 #14

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


Use dimensional analysis to show that the dimensionless 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 the 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 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 the 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 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 usual with the selection of the repeating variables (again, the standard choice is d, V, and \varrho), 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 #15

Define Reynolds number and explain its meaning. 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 dimensionless 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 in 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, representing 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 #16

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 dimensionless groupings that govern this problem.

The drag of the golf ball (the dependent variable) \cal{D} can be written in a 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 an implicit form as

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

In this case, there are six variables (N = 6) and three 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 dimensionless groupings to be determined.

Choose the standard aerodynamic repeating variables \varrho, V, and D, which are all linearly independent. The dimensionless \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 so that each of the \Pi products is dimensionless.

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

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

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 dimensionless 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,n 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 dimensionless)

    \[ \Pi_3 = \frac{\varrho V D}{\mu} \]

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 #17

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 we could think of this as Watts per unit area in SI. 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 set 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 selecting the repeating variables, one choice is \varrho, a, and r. Therefore, the 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 #18

Consider a liquid in a cylindrical container where both the container and the liquid rotate as a rigid body (called solid-body rotation), as shown in the figure below.
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 selecting the repeating variables, one choice is \varrho, g, and R. Therefore, the 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 #19

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 a 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 dimensionless parameters that appear in your result. Show all of your work. Hint: Notice that there are three length scales in this problem, but choose h as the reference length scale for consistency.

We are asked to find the effects on the exit flow velocity V in terms of 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

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

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 selecting the repeating variables, the only choice, in this case, is \varrho, \mu, and h. (Note: Can you explain why?). Therefore, the 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 as 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 dimensionless groupings involved are such that

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

Worked Example #20

The AIAA Design Build & Fly (DBF) team must determine the factors influencing 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 dimensionless groupings governing 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 the variables gives N = 6 and K = 3 because this problem has three fundamental dimensions. 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 the drag coefficient. We usually 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 a matter of convenience and/or consistency with established conventions.

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 following established conventions or just because it is otherwise convenient.

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 the 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 the Reynolds number based on the 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 #21

The DBF team has observed that the banner in the previous problem begins to flutter at some critical airspeed, which results in a 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 dimensionless 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 a 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 dimensionless 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 dimensionless frequency or a structural reduced frequency.

As a result of the 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 dimensionless flutter speed of the banner will depend on the Reynolds number and its structural reduced frequency.

Worked Example #22

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 dimensionless groupings governing this problem.

The relationship between the drag on the sphere and the given variables can be written in a 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 that temperature is explicitly defined in this case, so there are four 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 drag, collectively include all the fundamental dimensions, and are linearly independent.

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 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 dimensionless 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 that 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 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 the 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 critical grouping that comes out of this problem is

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

Worked Example #23

A force F is applied at the tip of a cantilevered wing of length L and the 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 dimensionless groupings governing 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 dilemma because the choice of the repeating variables here is not apparent. Suppose we choose F, L_w, and E as the repeating variables. In that case, they will not be linearly independent. If we decide 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 dimensionless groupings.

The accepted solution to this 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, 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 dimensionless displacement is 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 dimensionless 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 dimensionless force or force coefficient.

Finally, all three \Pi groupings have been determined so that 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 #24

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, verify that the primary dimensions of \tau_p are units of time. Second, find a dimensionless form for the time constant 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 dimensionless 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 that a dimensionless form could be

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

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

Worked Example #25

Based on experiments performed with a wind turbine, it is determined that its power output is a function of 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 dimensionless groupings describing 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 two \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. In this case, a good choice is R, V_{\infty}, and \varrho_{\infty}, which will primarily influence power production from the turbine. They also collectively include all of the fundamental dimensions of mass, length, and time, and they are 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 an 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 #26

A sphere is located in a pipe through which a liquid flows. The drag force F_D on the sphere is assumed to be a function of the sphere’s diameter 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 the  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 have all of M, L, and T, so K = 3 (i.e., three fundamental dimensions,) so there are two \Pi products to determine.

4. Use \varrho, V, and d as the repeating variables. 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 dimensionless 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, for both cases, 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 #27

The singing sounds produced by power lines in the wind are called Aeolian tones, 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 in 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 terms of the dimensionless 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 five, so N = 5, and the number of base dimensions (mass, length, and time are all involved) is 3, so K = 3. This means 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 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 preceding 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 #28

Using Worked Example #27 as a basis, it is desired to replicate the physics of the singing sound and study it in a low-speed wind tunnel. The actual power wires 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 are strung across the test section’s width. The tunnel can reach a maximum flow speed of 75 ft/s.  Is it possible to obtain the 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, the two relevant similarity parameters in t, in this case, the Reynolds number and the Strouhal number, and the Strouhal number is a function of the Reynolds number. For 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 used in the tunnel were 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 a 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, 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 the 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, it would be possible to study the singing sound behavior of the wires in the wind tunnel. It is just that the frequencies obtained would be higher.

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

Worked Example #29

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).

  1. Write down the functional dependency of V and the other variables D_p, \mu, g, and \varrho_e in both explicit and implicit forms.
  2. How many base dimensions and \Pi groupings are involved in this problem?
  3. Write down the dimensions of each of the variables involved.
  4. Form the dimensional matrix for this problem.
  5. Choose D_p, \mu, and g as the repeating variables, then 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, 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, primarily affecting the settling velocity. They also collectively include all the fundamental dimensions and are linearly independent. We are asked 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 #30

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

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

1. The explicit form of the relationship is

    \[ Q = f( g, b, H ) \]

and in implicit form, then

    \[ g( Q, g, b, H) = 0 \]

2. In this problem, there are only two base dimensions, length L and time T, so the procedure should be more straightforward.

The units involved are

    \begin{eqnarray*} \left[ 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.,

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

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. Note: We cannot choose both H and b repeating variables because they have the same units and are not linearly independent.

For the first \Pi product then

    \[ \Pi_{1} = (g)^{\alpha} (H)^{\beta} Q \]

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

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

or

    \[ \Pi_{1} =\frac{Q} {g^{1/2} \, H^{5/2}} \]

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

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

or

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

Worked Example #31

Sir Geoffrey I. Taylor (1886–1975) was a British physicist and engineer. He used dimensional analysis to estimate an explosion’s blast wave propagation characteristics. 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 dimensionless 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 includes the dimensions of mass, length, and time, and so we have just one \Pi product. Energy is the ability to do work, so the units of energy will be the same as the units of work, 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 we get that \alpha = -1/5 from the second equation. 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 the 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 derived then, the radius would increase by 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 #32

An ocean surface wave is a sinusoidal-like disturbance propagating along the ocean’s surface, as shown in the figure below. The speed of the wave, U, is found to be a function of the surface tension of the seawater, \sigma, the density of the seawater, \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 preceding 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 dimensionless 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 #33

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 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.

  1. Write down the functional relationship for the volumetric flow rate Q in terms of the other parameters, in both implicit and explicit form.
  2. Write down the base units of the parameters involved in this problem.
  3. How many base dimensions and dimensionless groupings are involved in this problem?
  4. Write out the dimensional matrix for this problem.
  5. Choose the repeating variables and explain your choice.
  6. Determine the dimensionless grouping(s) for the parameters involved.
  7. Write down the final dimensionless functional relationship(s).

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

    \[ Q =  f_1 (\Delta p, \varrho, D, d) \]

where f_1 is some function, and implicit form as

    \[ f_2 (Q, \Delta p, \varrho, D, d) = 0 \]

where f_2 is some other function.

2. The dimensions of the parameters involved are:

  • \left[ 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 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

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

5. The only possible choice of repeating variables are \Delta p, \varrho, D ,and d. 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 that we will choose D, and the repeating variables will be \Delta p, \varrho, and D.

6. The \Pi groups will be formed from:

    \[ \Pi_1 = g_1(\Delta p, \varrho, D, Q) \]

and

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

We now have to solve for the dimensionless groupings.

For \Pi_1 then:

    \[ \Pi_1 = g_1(\Delta p, \varrho, D, Q) \]

so

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

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

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

or

    \[ \Pi_1 = Q \, \sqrt{ \frac{\varrho}{\Delta p \, D^{4}}} \]

As a check to see if this is a dimensionless 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 dimensionless.

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 evident that the ratio of one length to another is dimensionless.

7. Finally, the dimensionless relationship between the volumetric flow rate Q and the other parameters is

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

License

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Introduction to Aerospace Flight Vehicles Copyright © 2022, 2023 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