Similitude and physical modeling

Rules for choosing the repeating variables

1. The number of repeating variables equals the number of fundamental dimensions in the problem (usually 3: M, L, T), $m = 3$.
2. Together they must contain all the fundamental dimensions in the problem, and they must be dimensionally independent (no repeating variable, and no product of powers of them, can be dimensionless).
3. The dependent variable must not be chosen as a repeating variable.
4. The repeating variables should not themselves form a dimensionless $\pi$-group.
5. Pick variables that are easy to measure and control, one from each class: a geometric one (e.g. $D$ or $L$), a flow property (e.g. $V$), and a fluid property (e.g. $\rho$). Typical choice: $D, V, \rho$.
6. Avoid choosing two variables with the same dimensions (like $D$ and $L$).

Rules for forming the $\pi$ terms

1. Number of $\pi$ terms $= n - m$, where $n$ is the total number of variables.
2. Each $\pi$ term is formed from all repeating variables, each raised to an unknown power, times one non-repeating variable (raised to power 1).
3. The powers are found by equating the exponents of M, L and T to zero so that the group is dimensionless.
4. Each $\pi$ should contain only one non-repeating variable; the dependent variable appears in only one $\pi$ (called $\pi_1$).
5. Any $\pi$ may be inverted, raised to a power, or multiplied by a constant or another $\pi$ without changing its dimensionless nature. Dimensionless variables given in the problem (angles, ratios like $\epsilon/D$) are already $\pi$ terms.
6. The final relation is $\pi_1 = \phi(\pi_2, \pi_3, \dots)$.

Tides are gravity-dominated, so the Froude law applies: $\dfrac{V}{\sqrt{gL}}$ is the same in model and prototype. The time scale then follows from the length scale and the gravity ratio.

Time scale

Velocity scale $V_r = \sqrt{g_rL_r}$, and time $T = L/V$, so:

$$T_r = \frac{L_r}{V_r} = \sqrt{\frac{L_r}{g_r}}$$

Case 1: model tested on earth ($g_r = 1$)

$$T_r = \sqrt{\frac{1}{400}} = \frac{1}{20}$$

Model time for one prototype day $= 24 \times \dfrac1{20} = 1.2$ h $= 72$ min.

Case 2: model tested on the moon

The prototype is on earth, the model on the moon, so $g_r = g_m/g_p = 1/6$.

$$T_r = \sqrt{\frac{1/400}{1/6}} = \sqrt{\frac{6}{400}} = 0.1225$$

Model time for one prototype day $= 24 \times 0.1225 = 2.94$ h (about 2 h 56 min).

Answer: on earth, 1 prototype day corresponds to 1.2 h of model time ($T_r = 1/20$). On the moon, $T_r = 0.1225$ and 1 prototype day corresponds to about 2.94 h of model time; the model time is $0.1225\times$ the prototype time.

Rayleigh's method writes the dependent variable as a product of the independent variables, each raised to an unknown power. The powers are found by equating the dimensions on both sides.

Functional relation

$$V = K\, d^{a}\,\rho_s^{\,f}\,\rho_l^{\,b}\,\mu^{c}\,g^{e}$$

Dimensions

$V = LT^{-1}$, $d = L$, $\rho = ML^{-3}$, $\mu = ML^{-1}T^{-1}$, $g = LT^{-2}$.

There are 5 unknown powers and only 3 equations, so express three of them in terms of $c$ and $f$:

$$e = \frac{1-c}{2},\qquad b = -c - f,\qquad a = \frac12 - \frac{3c}{2}$$

Substitute

$$V = K\, d^{\,\frac12-\frac{3c}{2}}\,\rho_s^{\,f}\,\rho_l^{\,-c-f}\,\mu^{c}\,g^{\,\frac{1-c}{2}}$$

Group terms with the same power:

$$V = K\sqrt{gd}\left[\frac{\mu}{\rho_l\, d^{3/2}g^{1/2}}\right]^{c}\left[\frac{\rho_s}{\rho_l}\right]^{f}$$

Answer:

$$\frac{V}{\sqrt{gd}} = \phi\left[\frac{\mu}{\rho_l\,d^{3/2}g^{1/2}},\ \frac{\rho_s}{\rho_l}\right]$$

This says the dimensionless velocity (Froude number) depends on a viscous term (a form of inverse Reynolds number) and on the density ratio. The functions of these groups are found by experiment.

If both gravity and viscous forces are to be modelled, the Froude and Reynolds numbers must both be equal in model and prototype. The liquid in the model must then have a different viscosity.

Froude number equal

$$\frac{V_m}{\sqrt{g L_m}} = \frac{V_p}{\sqrt{g L_p}} \;\Rightarrow\; V_r = \sqrt{L_r}$$

(the same $g$ for both).

Reynolds number equal

$$\frac{V_mL_m}{\nu_m} = \frac{V_pL_p}{\nu_p} \;\Rightarrow\; \nu_r = V_rL_r$$

Combine

$$\nu_r = L_r^{1/2}L_r = L_r^{3/2} = \left(\frac15\right)^{1.5} = 0.0894$$ $$\nu_m = \nu_p\times 0.0894 = 4.645\times10^{-5}\times0.0894 = 4.155\times10^{-6}\ \text{m}^2/\text{s}$$

Answer: the model liquid must have a kinematic viscosity of $\nu_m \approx 4.15\times10^{-6}$ m²/s (about 0.0415 stokes). The dynamic viscosity then follows as $\mu_m = \rho_m\nu_m$ for the chosen liquid; if its density equals that of the oil, $\mu_m = 0.0894\,\mu_p$.

Use the Buckingham $\pi$ method with the variables $\Delta P, D, L, \rho, \mu, V$ and $\epsilon/D$.

Variables and dimensions

$\epsilon/D$ is already dimensionless, so it is a $\pi$ term by itself. For the other six variables: $n = 6$, fundamental dimensions $m = 3$ (M, L, T), so $n - m = 3$ more $\pi$ terms. Total $= 4$.

Repeating variables: $D$ (geometric), $V$ (kinematic), $\rho$ (fluid property).

$\pi_1$ (with $\Delta P$)

$$\pi_1 = \Delta P\,D^{a}V^{b}\rho^{c}$$

M: $1 + c = 0 \Rightarrow c = -1$. T: $-2 - b = 0 \Rightarrow b = -2$. L: $-1 + a + b - 3c = 0 \Rightarrow a = 0$.

$$\pi_1 = \frac{\Delta P}{\rho V^2}$$

$\pi_2$ (with $L$)

Same dimension as $D$, so $\pi_2 = \dfrac{L}{D}$.

$\pi_3$ (with $\mu$)

$$\pi_3 = \mu\,D^{a}V^{b}\rho^{c}:\quad c = -1,\; b = -1,\; a = -1 \;\Rightarrow\; \pi_3 = \frac{\mu}{\rho V D} = \frac1{Re}$$

Result

$$\frac{\Delta P}{\rho V^2} = \phi\left(\frac{\rho V D}{\mu},\ \frac LD,\ \frac\epsilon D\right)$$

Experiments show $\Delta P \propto L/D$, so

$$\Delta P = \frac{L}{D}\,\rho V^2\,\phi_1\left(Re,\ \frac\epsilon D\right)$$

This is the Darcy-Weisbach form, with friction factor $f$ related to $\phi_1$ by $\phi_1 = f/2$.

Use dimensional analysis with $C = f(\rho, \lambda, \sigma)$. Here $n = 4$ variables and $m = 3$ dimensions (M, L, T), so there is only one $\pi$ term.

Dimensions

$C = LT^{-1}$, $\rho = ML^{-3}$, $\lambda = L$, $\sigma = MT^{-2}$ (force per length).

Form of the relation

$$C = K\,\rho^{a}\lambda^{b}\sigma^{c}$$

From T: $c = \tfrac12$. From M: $a = -\tfrac12$. From L: $b = 1 + 3a = -\tfrac12$.

Result

$$C = K\sqrt{\frac{\sigma}{\rho\lambda}}$$

where $K$ is a dimensionless constant (found by experiment or theory; the theory of capillary waves gives $K = \sqrt{2\pi}$).

Effect of doubling the surface tension

At fixed $\rho$ and $\lambda$, $C \propto \sqrt{\sigma}$:

$$\frac{C_2}{C_1} = \sqrt{\frac{2\sigma}{\sigma}} = \sqrt2 = 1.414$$

Answer: $C = K\sqrt{\sigma/(\rho\lambda)}$; if $\sigma$ is doubled, the speed increases by a factor $\sqrt2 \approx 1.41$ (about 41%).

River flow is controlled by gravity, so the Froude law is used. The slope scale is $S_r = \dfrac{\text{vertical scale}}{\text{horizontal scale}}$, and here it is to be unity.

Given

Prototype: $Q_p = 3500\ \text{m}^3/\text{s}$, depth $2.25$ m, width $1500$ m. Horizontal scale $L_r = 1/400$. $S_r = 1$.

Depth scale

$$S_r = \frac{d_r}{L_r} = 1 \;\Rightarrow\; d_r = L_r = \frac1{400}$$

So the model is geometrically undistorted. Model depth $= 2.25/400 = 5.625$ mm; model width $= 1500/400 = 3.75$ m.

Velocity and discharge scale (Froude law)

$$V_r = \sqrt{d_r} = \sqrt{1/400} = \frac1{20}$$ $$Q_r = A_rV_r = (L_rd_r)\sqrt{d_r} = L_r\,d_r^{3/2} = \left(\frac1{400}\right)^{5/2} = \frac1{3.2\times10^{6}} = 3.125\times10^{-7}$$

Model discharge

$$Q_m = 3500 \times 3.125\times10^{-7} = 1.094\times10^{-3}\ \text{m}^3/\text{s} \approx 1.09\ \text{L/s}$$

Answer: depth scale $d_r = 1/400$; discharge scale $Q_r = 1/(3.2\times10^6)$; model discharge $\approx 1.09$ L/s. Such a shallow model (5.6 mm deep) would suffer scale effects (surface tension, laminar flow), which is why real river models are usually distorted.

Relation

$Q$ depends on $d, H, \mu, \rho, g$. By Rayleigh's method:

$$Q = K\,d^{a}H^{b}\mu^{c}\rho^{e}g^{f}$$

Dimensions

$Q = L^3T^{-1}$, $d = H = L$, $\mu = ML^{-1}T^{-1}$, $\rho = ML^{-3}$, $g = LT^{-2}$.

There are 5 unknowns and 3 equations. Express in terms of $b$ and $c$:

$$e = -c,\qquad f = \frac{1-c}{2},\qquad a = \frac52 - \frac{3c}{2} - b$$

Substitute

$$Q = K\,d^{\,\frac52-\frac{3c}{2}-b}\,H^{b}\,\mu^{c}\,\rho^{-c}\,g^{\frac{1-c}{2}}$$

Collect terms with the same powers:

$$Q = K\,\sqrt g\,d^{5/2}\left(\frac Hd\right)^{b}\left(\frac{\mu}{\rho\,g^{1/2}d^{3/2}}\right)^{c}$$

Answer:

$$Q = \sqrt{g}\,d^{5/2}\ \phi\left[\frac Hd,\ \frac{\mu}{\rho\sqrt g\,d^{3/2}}\right]$$

Equivalently, $Q = d^2\sqrt{gH}\ \phi_1\!\left(H/d,\ Re\right)$. The dimensionless groups are the head ratio $H/d$ and a Reynolds-type number. For a given orifice, experiments give $Q = C_dA\sqrt{2gH}$.

Spillway flow is gravity-dominated, so the Froude law is used. Length scale $L_r = L_m/L_p = 1/16$.

(i) Model dimensions

$$\begin{aligned} H_m &= \frac{12.5}{16} = 0.78125\ \text{m} \approx 0.78\ \text{m} \\ h_m &= \frac{2}{16} = 0.125\ \text{m} = 12.5\ \text{cm} \end{aligned}$$

(ii) Flow per metre length of the prototype

Model flume width $= 0.60$ m, $Q_m = 20$ L/s.

$$q_m = \frac{20}{0.6} = 33.33\ \text{L/s per m}$$

For Froude similarity, the discharge scale is $Q_r = L_r^{5/2}$, and the discharge per unit length scale is $q_r = L_r^{3/2}$:

$$q_p = \frac{q_m}{L_r^{3/2}} = 33.33\times16^{3/2} = 33.33\times64 = 2133\ \text{L/s per m}$$

Answer: (i) model spillway height $= 0.781$ m and head $= 0.125$ m; (ii) prototype discharge $\approx 2133$ L/s per metre length $= 2.13\ \text{m}^3/\text{s}$ per m.

Variables: $\tau_w, U, \delta, u', \rho, dp/dx$. So $n = 6$, $m = 3$ (M, L, T), giving $n - m = 3$ $\pi$ terms. Repeating variables: $\rho, U, \delta$ (as given).

Dimensions

$\tau_w = ML^{-1}T^{-2}$, $U = u' = LT^{-1}$, $\delta = L$, $\rho = ML^{-3}$, $dp/dx = ML^{-2}T^{-2}$.

$\pi_1$ (with $\tau_w$)

$$\pi_1 = \tau_w\,\rho^{a}U^{b}\delta^{c}$$

M: $1 + a = 0 \Rightarrow a = -1$. T: $-2 - b = 0 \Rightarrow b = -2$. L: $-1 - 3a + b + c = 0 \Rightarrow c = 0$.

$$\pi_1 = \frac{\tau_w}{\rho U^2}$$

$\pi_2$ (with $u'$)

It has the dimension of $U$, so:

$$\pi_2 = \frac{u'}{U}$$

$\pi_3$ (with $dp/dx$)

$$\pi_3 = \frac{dp}{dx}\rho^{a}U^{b}\delta^{c}$$

M: $1 + a = 0 \Rightarrow a = -1$. T: $-2 - b = 0 \Rightarrow b = -2$. L: $-2 - 3a + b + c = 0 \Rightarrow -2 + 3 - 2 + c = 0 \Rightarrow c = 1$.

$$\pi_3 = \frac{\delta}{\rho U^2}\,\frac{dp}{dx}$$

Result

$$\frac{\tau_w}{\rho U^2} = \phi\left(\frac{u'}{U},\ \frac{\delta}{\rho U^2}\frac{dp}{dx}\right)$$

The left side is the skin friction coefficient ($C_f/2$), and the second argument measures the pressure gradient.

Distorted vs undistorted models

An undistorted model is geometrically similar to the prototype: all lengths (horizontal and vertical) have the same scale ratio. A distorted model has different scale ratios in different directions, usually a larger horizontal reduction than vertical.

Buckingham's $\pi$ theorem: working principle

Statement: If a physical phenomenon involves $n$ variables with $m$ fundamental dimensions (M, L, T), the variables can be arranged into $(n - m)$ independent dimensionless groups ($\pi$ terms), and the relation becomes

$$\phi(\pi_1, \pi_2, \dots, \pi_{n-m}) = 0$$

Procedure

1. List all variables on which the phenomenon depends, and count $n$.
2. Write the dimensions of each variable (M, L, T) and find $m$.
3. Choose $m$ repeating variables (geometric, flow and fluid property, e.g. $D, V, \rho$) which together contain all dimensions and are independent.
4. Form $(n - m)$ $\pi$ terms, each containing the repeating variables raised to unknown powers and one non-repeating variable.
5. Find the powers by equating the exponents of M, L, T to zero.
6. Write $\pi_1 = f(\pi_2, \pi_3, \dots)$. Replace groups by standard ones (such as Reynolds or Froude number) if needed.

The result is found by experiment on models, with the $\pi$ terms equal in model and prototype.

Flow in a pressurised pipe is dominated by viscous forces, so the Reynolds model law is used: $Re_m = Re_p$.

Given

Prototype: $D_p = 2$ m, $Q_p = 5\ \text{m}^3/\text{s}$, oil with $\rho_p = 800\ \text{kg/m}^3$ (s.g. 0.8), $\mu_p = 0.042$ poise $= 0.0042\ \text{N s/m}^2$.

Model: $D_m = 0.2$ m, water $\rho_m = 1000\ \text{kg/m}^3$, $\mu_m = 0.01$ poise $= 0.001\ \text{N s/m}^2$.

Prototype velocity

$$V_p = \frac{Q_p}{\frac\pi4D_p^2} = \frac{5}{\frac\pi4(2)^2} = 1.5915\ \text{m/s}$$

Model velocity

$$\frac{\rho_mV_mD_m}{\mu_m} = \frac{\rho_pV_pD_p}{\mu_p} \;\Rightarrow\; V_m = V_p\,\frac{\rho_p}{\rho_m}\,\frac{D_p}{D_m}\,\frac{\mu_m}{\mu_p}$$ $$V_m = 1.5915\times\frac{800}{1000}\times\frac{2}{0.2}\times\frac{0.001}{0.0042} = 3.03\ \text{m/s}$$

Model discharge

$$Q_m = \frac\pi4D_m^2V_m = \frac\pi4(0.2)^2(3.0315) = 0.0952\ \text{m}^3/\text{s}$$

Answer: model velocity $\approx 3.03$ m/s; model discharge $\approx 0.0952\ \text{m}^3/\text{s}$ (95.2 L/s).

A distorted model is a scale model in which the scale ratios are not the same in all directions, i.e. the horizontal scale $L_r$ differs from the vertical scale $h_r$. Geometric similarity is therefore not complete. The distortion is $h_r/L_r$ (greater than 1).

Importance (why it is used)

1. Space and cost: for large and flat prototypes such as rivers, harbours, estuaries and tidal basins, an undistorted model would be either too big or so shallow that the depth would be a few millimetres.
2. Avoids scale effects: a larger vertical scale gives enough depth to keep the flow turbulent, avoiding surface tension and viscosity effects, and keeps the Reynolds number high.
3. Measurable quantities: depths, velocities and slopes become large enough to measure accurately.
4. Movable beds: it gives enough slope and tractive force to study scour, silting and sediment transport.
5. Better visualisation of flow patterns, waves and currents.

The disadvantages are that the results need correction factors and that velocity distributions, bed forces and wave patterns are not exactly reproduced.

Flow in a pressurised pipe is dominated by viscous forces, so the Reynolds model law is used: $Re_m = Re_p$.

Given

Prototype: $D_p = 2$ m, $Q_p = 5\ \text{m}^3/\text{s}$, oil of s.g. 0.92, so $\rho_p = 920\ \text{kg/m}^3$, $\mu_p = 0.04$ poise $= 0.004\ \text{N s/m}^2$.

Model: $D_m = 0.2$ m, water $\rho_m = 1000\ \text{kg/m}^3$, $\mu_m = 0.01$ poise $= 0.001\ \text{N s/m}^2$.

Prototype velocity

$$V_p = \frac{Q_p}{\frac\pi4D_p^2} = \frac{5}{\frac\pi4(2)^2} = 1.5915\ \text{m/s}$$

Model velocity

$$\frac{\rho_mV_mD_m}{\mu_m} = \frac{\rho_pV_pD_p}{\mu_p} \;\Rightarrow\; V_m = V_p\,\frac{\rho_p}{\rho_m}\,\frac{D_p}{D_m}\,\frac{\mu_m}{\mu_p}$$ $$V_m = 1.5915\times\frac{920}{1000}\times\frac{2}{0.2}\times\frac{0.001}{0.004} = 3.66\ \text{m/s}$$

Model discharge

$$Q_m = \frac\pi4D_m^2V_m = \frac\pi4(0.2)^2(3.6606) = 0.115\ \text{m}^3/\text{s}$$

Answer: model velocity $\approx 3.66$ m/s; model discharge $\approx 0.115\ \text{m}^3/\text{s}$ (115 L/s).