This simulation shows how the Buckingham Pi theorem turns a list of dimensional variables into a smaller set of dimensionless groups, and how matching those groups lets a scale model predict the behavior of a full-size prototype.

In the Buckingham Pi view, choose a problem. The table is the dimensional matrix: the exponents of mass (M), length (L) and time (t) for each variable. Check three repeating variables. The simulation solves for the exponents that make each remaining variable dimensionless and identifies each group in terms of named groups such as the Reynolds number. Try different repeating variables: the groups change, but each is always a product of the same named groups. Choose variables that are not dimensionally independent (for example, D and L in the pipe problem) to see why that choice fails.

In the model testing view, the sphere case predicts the drag on a prototype sphere from a smaller model, usually in another fluid. Change the model velocity and compare the predicted prototype drag with the actual drag; the prediction is exact only when the Reynolds numbers match. The ship case shows why a model tested in water cannot match both the Froude number and the Reynolds number.

Buckingham Pi theorem

If a physical relation involves \(n\) variables and the dimensional matrix has rank \(r\) (here \(r = 3\), for M, L and t), the relation can be written in terms of \(n - r\) independent dimensionless groups:

\[ f(q_1, q_2, \ldots, q_n) = 0 \quad\Longrightarrow\quad \Pi_1 = \phi(\Pi_2, \ldots, \Pi_{n-r}) \]

Choose \(r\) repeating variables \(x_1, x_2, x_3\) that are dimensionally independent (none can be formed from the others). Each remaining variable \(q_j\) forms one group:

\[ \Pi_j = q_j\, x_1^{a_1} x_2^{a_2} x_3^{a_3} \]

The exponents make the powers of M, L and t zero. If \(\mathbf{R}\) is the 3 × 3 dimensional matrix of the repeating variables and \(\mathbf{d}_j\) is the dimension vector of \(q_j\):

\[ \mathbf{R}\,\mathbf{a} = -\mathbf{d}_j \]

If \(\mathbf{R}\) is singular, the repeating variables are not independent and no solution exists. The dependent variable (the one to be predicted) is usually not chosen as a repeating variable, so that it appears in only one group.

Named groups

\[ \mathrm{Re} = \frac{\rho V D}{\mu} \qquad \mathrm{Eu} = \frac{\Delta p}{\rho V^2} \qquad \mathrm{Fr} = \frac{V}{\sqrt{gL}} \qquad C_F = \frac{F}{\rho V^2 D^2} \] \[ \psi = \frac{\Delta p}{\rho N^2 D^2} \qquad \phi = \frac{Q}{N D^3} \qquad C_P = \frac{P}{\rho N^3 D^5} \qquad \mathrm{Re} = \frac{\rho N D^2}{\mu}\ \text{(pump)} \]

Similitude

A model and prototype are dynamically similar when all independent groups are equal. The dependent group is then also equal, so for sphere drag:

\[ \frac{F_p}{\rho_p V_p^2 D_p^2} = \frac{F_m}{\rho_m V_m^2 D_m^2} \quad\Longrightarrow\quad F_p = F_m \frac{\rho_p}{\rho_m}\left(\frac{V_p}{V_m}\right)^2\left(\frac{D_p}{D_m}\right)^2 \]

This prediction is exact only if \(\mathrm{Re}_m = \mathrm{Re}_p\), which requires

\[ V_m = V_p\,\frac{\nu_m}{\nu_p}\,\frac{D_p}{D_m} \]

The drag coefficient is \(C_D = F\big/\left(\tfrac{1}{2}\rho V^2 \tfrac{\pi}{4}D^2\right) = (8/\pi)\,C_F\). The simulation uses Morrison's correlation for a smooth sphere, valid for \(\mathrm{Re} < 10^6\):

\[ C_D = \frac{24}{\mathrm{Re}} + \frac{2.6\,(\mathrm{Re}/5)}{1 + (\mathrm{Re}/5)^{1.52}} + \frac{0.411\,(\mathrm{Re}/263{,}000)^{-7.94}}{1 + (\mathrm{Re}/263{,}000)^{-8}} + \frac{0.25\,(\mathrm{Re}/10^6)}{1 + \mathrm{Re}/10^6} \]

Where \(C_D\) changes slowly with Re (roughly \(10^3 < \mathrm{Re} < 2\times10^5\)), a mismatch in Re causes only a small error.

Ship models

Wave drag depends on Fr and friction drag depends on Re. Matching Fr with \(g\) fixed gives

\[ V_m = V_p\left(\frac{L_m}{L_p}\right)^{1/2} \qquad \frac{\mathrm{Re}_p}{\mathrm{Re}_m} = \frac{\nu_m}{\nu_p}\left(\frac{L_p}{L_m}\right)^{3/2} \]

so matching Re as well requires \(\nu_m = \nu_p\,(L_m/L_p)^{3/2}\). No practical liquid is that much less viscous than water, so towing tanks match Fr and correct the friction drag separately at each Re, for example with the ITTC-1957 line \(C_f = 0.075/(\log_{10}\mathrm{Re} - 2)^2\) (Froude's method).

Fluid properties (20 °C)

air: ρ = 1.204 kg/m³, μ = 1.825 × 10⁻⁵ Pa·s; water: ρ = 998 kg/m³, μ = 1.002 × 10⁻³ Pa·s; glycerin: ρ = 1260 kg/m³, μ = 1.41 Pa·s; mercury: ν = 1.15 × 10⁻⁷ m²/s.

Symbols

\(C_D\) = drag coefficient
\(C_F\) = force coefficient
\(C_P\) = power coefficient
\(D\) = diameter (m)
\(\mathrm{Eu}\) = Euler number
\(F\) = drag force (N)
\(\mathrm{Fr}\) = Froude number
\(g\) = gravitational acceleration (m/s²)
\(L\) = length (m); in the dimensional matrix, the length dimension
\(M\) = mass dimension
\(N\) = rotational speed (1/s)
\(n\) = number of variables
\(P\) = shaft power (W)
\(Q\) = volumetric flow rate (m³/s)
\(r\) = rank of the dimensional matrix
\(\mathrm{Re}\) = Reynolds number
\(t\) = time dimension
\(V\) = velocity (m/s)
\(\Delta p\) = pressure difference (Pa)
\(\varepsilon\) = wall roughness (m)
\(\mu\) = viscosity (Pa·s)
\(\nu\) = kinematic viscosity, \(\mu/\rho\) (m²/s)
\(\rho\) = density (kg/m³)
\(\phi\) = flow coefficient
\(\psi\) = head coefficient
\(\Pi\) = dimensionless group
subscripts \(m\), \(p\) = model, prototype

Reference: F. A. Morrison, An Introduction to Fluid Mechanics, Cambridge University Press, 2013.

This simulation was created in the Department of Chemical and Biological Engineering at University of Colorado Boulder for LearnChemE.com by John L. Falconer using Claude AI. It was prepared with financial support from the National Science Foundation (DUE 2336987 and 2336988). Address any questions or comments to LearnChemE@gmail.com.