Similarity — geometric, kinematic, dynamic; Reynolds similarity: Exercises

Recall The four tools you will reuse (peek if you forget)

Reynolds number $Re = \frac{ρ V L}{μ} = \frac{V L}{ν}$ — the ratio (inertia force)/(viscous force). Dimensionless.
Froude number $Fr = \frac{V}{g L}$ — the ratio that governs free surfaces and waves.
Reynolds matching $Re_{m} = Re_{p} \Rightarrow \frac{V _{m} L _{m}}{ν _{m}} = \frac{V _{p} L _{p}}{ν _{p}}$ .
Drag coefficient $C_{D} = \frac{F _{D}}{\frac{1}{2} ρ V ^{2} A}$ ; when Re is matched, $C_{D, m} = C_{D, p}$ . Here the subscript $m$ = model, $p$ = prototype (the real full-size thing), $r$ = ratio $L_{r} = L_{m} / L_{p}$ .

Figure — Similarity — geometric, kinematic, dynamic; Reynolds similarity

Level 1 — Recognition

L1-Q1

A submarine is tested fully submerged in a towing tank (no water surface broken). Which dimensionless number must you match for dynamic similarity — Reynolds or Froude? State why in one sentence.

Recall Solution

Reynolds. A fully submerged body has no free surface, so gravity waves play no role; the competition is inertia vs viscosity, which is exactly what $Re = ρ V L / μ$ measures. Froude only enters when a liquid surface can rise into waves.

L1-Q2

State whether each statement forces the next: (a) geometric similarity, (b) kinematic similarity, (c) dynamic similarity. Order them from weakest to strongest and say which one "implies" the others.

Recall Solution

Weakest → strongest: geometric → kinematic → dynamic.

Kinematic (matching streamline patterns) requires geometric (same shape).
Dynamic (matching force ratios) implies kinematic (matched forces produce matched motion). So achieve dynamic similarity and the other two come free. Dynamic is the king.

L1-Q3

Is $Re = \frac{ρ V L}{μ}$ dimensionless? Check by plugging SI units.

Recall Solution

$\frac{[ ρ ] [ V ] [ L ]}{[ μ ]} = \frac{( kg/m ^{3} ) ( m/s ) ( m )}{kg / ( m \cdot s )} = \frac{kg \cdot m ^{- 1} \cdot s ^{- 1}}{kg \cdot m ^{- 1} \cdot s ^{- 1}} = 1$ Yes — every unit cancels, so $Re$ is a pure number. That is why the same value in model and prototype means the same flow regime regardless of size.

Level 2 — Application

L2-Q1

A car prototype has length $L_{p} = 4.5$ m and drives at $V_{p} = 25$ m/s. A $1/6$ scale model ( $L_{m} = 0.75$ m) is tested in the same air. Find the wind-tunnel speed $V_{m}$ .

Recall Solution

Same fluid ⇒ $ρ, μ$ cancel in $Re_{m} = Re_{p}$ , leaving $V_{m} L_{m} = V_{p} L_{p}$ . $V_{m} = V_{p} \frac{L _{p}}{L _{m}} = 25 \times \frac{4.5}{0.75} = 25 \times 6 = 150 m/s$ The smaller model must be blown $6 \times$ faster to keep $ρ V L / μ$ unchanged.

L2-Q2

A pipe of diameter $D_{p} = 0.6$ m carries water ( $ν_{p} = 1.0 \times 1 0^{- 6} m^{2} / s$ ) at $V_{p} = 1.5$ m/s. A model pipe $D_{m} = 0.15$ m uses air ( $ν_{m} = 1.5 \times 1 0^{- 5} m^{2} / s$ ). Find $V_{m}$ .

Recall Solution

Different fluids ⇒ keep all three factors in Re-matching: $\frac{V _{m} D _{m}}{ν _{m}} = \frac{V _{p} D _{p}}{ν _{p}} \Rightarrow V_{m} = V_{p} \frac{D _{p}}{D _{m}} \frac{ν _{m}}{ν _{p}}$ $V_{m} = 1.5 \times \frac{0.6}{0.15} \times \frac{1.5 \times 1 0 ^{- 5}}{1.0 \times 1 0 ^{- 6}} = 1.5 \times 4 \times 15 = 90 m/s$

L2-Q3

For the car in L2-Q1, the model drag is measured as $F_{D, m} = 18$ N. Same fluid, Re matched. Find the prototype drag $F_{D, p}$ .

Recall Solution

Re matched ⇒ $C_{D, m} = C_{D, p}$ , so $\frac{F _{D, m}}{ρ V _{m}^{2} A _{m}} = \frac{F _{D, p}}{ρ V _{p}^{2} A _{p}}$ . With same fluid, $A = L^{2}$ , and $V_{m} = V_{p} / L_{r}$ where $L_{r} = L_{m} / L_{p} = 1/6$ : $F_{D, p} = F_{D, m} \frac{V _{p}^{2} A _{p}}{V _{m}^{2} A _{m}} = F_{D, m} \cdot \frac{V _{p}^{2}}{( V _{p} / L _{r} ) ^{2}} \cdot \frac{L _{p}^{2}}{L _{m}^{2}} = F_{D, m} \cdot L_{r}^{2} \cdot \frac{1}{L _{r}^{2}} = F_{D, m}$ $F_{D, p} = 18 N$ The higher model speed exactly cancels the smaller area — for same fluid + Re matched, forces are equal.

Level 3 — Analysis

L3-Q1

A 1/25 scale ship model must satisfy Froude similarity (it makes surface waves). If the prototype cruises at $V_{p} = 10$ m/s, find $V_{m}$ . Then compute the ratio of Reynolds numbers $Re_{m} / Re_{p}$ (same water), and comment on whether Re is also matched.

Recall Solution

Froude matching: $\frac{V _{m}}{g L _{m}} = \frac{V _{p}}{g L _{p}} \Rightarrow V_{m} = V_{p} \frac{L _{m}}{L _{p}} = V_{p} L_{r}$ with $L_{r} = 1/25$ . $V_{m} = 10 1/25 = 10 \times \frac{1}{5} = 2 m/s$ Reynolds ratio (same $ν$ ): $\frac{Re _{m}}{Re _{p}} = \frac{V _{m} L _{m}}{V _{p} L _{p}} = L_{r} \cdot L_{r} = L_{r}^{3/2} = (1/25)^{3/2} = \frac{1}{125}$ . So $Re_{m}$ is 125× too small — Re is not matched. Froude and Reynolds cannot both hold in the same fluid, so engineers match Froude (the dominant, wave-making force) and correct the viscous drag separately.

L3-Q2

Show algebraically why matching Reynolds ( $V \propto 1/ L$ ) and Froude ( $V \propto L$ ) at once with the same fluid is impossible unless $L_{m} = L_{p}$ .

Recall Solution

Re-matching (same $ν$ ): $V_{m} L_{m} = V_{p} L_{p} \Rightarrow V_{r} = 1/ L_{r}$ . Fr-matching (same $g$ ): $V_{m} / L_{m} = V_{p} / L_{p} \Rightarrow V_{r} = L_{r}$ . Both true simultaneously ⇒ $\frac{1}{L _{r}} = L_{r} \Rightarrow L_{r}^{3/2} = 1 \Rightarrow L_{r} = 1.$ $L_{r} = 1$ means the model is the same size as the prototype — no scaling at all. Hence you must pick one governing number. (To satisfy both with $L_{r} \neq = 1$ you would need a different fluid with a specially chosen $ν$ .)

L3-Q3

The parent note claims the non-dimensional Navier–Stokes equation carries only $1/ Re$ in front of the viscous term. Using this, explain in words why two geometrically similar flows with equal Re must have equal $C_{D}$ .

Recall Solution

Non-dimensionalising with $L$ , $V$ , $ρ V^{2}$ turns Navier–Stokes into an equation whose only free coefficient is $1/ Re$ . If two flows share the same shape (geometry ⇒ same boundary conditions in dimensionless coordinates) and the same Re, they solve the identical dimensionless equation with identical boundary conditions ⇒ identical dimensionless velocity and pressure fields. Since $C_{D}$ is computed by integrating those dimensionless fields over the (identical) dimensionless surface, $C_{D, m} = C_{D, p}$ automatically. This is the whole engine behind Drag Coefficient scaling.

Level 4 — Synthesis

L4-Q1

A spillway model at scale $1/40$ is Froude-scaled. (a) Find the velocity ratio $V_{r}$ . (b) Since discharge (volume flow rate) is $Q = V \times A$ with $A \propto L^{2}$ , find the discharge ratio $Q_{r} = Q_{m} / Q_{p}$ . (c) If the prototype discharge is $Q_{p} = 8000 m^{3} / s$ , find $Q_{m}$ .

Recall Solution

(a) Froude: $V_{r} = L_{r} = 1/40$ . (b) $Q_{r} = V_{r} A_{r} = L_{r} \cdot L_{r}^{2} = L_{r}^{5/2} = (1/40)^{5/2}$ . $Q_{r} = \frac{1}{4 0 ^{2.5}} = \frac{1}{4 0 ^{2} 40} = \frac{1}{1600 \times 6.3246} = \frac{1}{10119.3} = 9.882 \times 1 0^{- 5}$ (c) $Q_{m} = Q_{p} Q_{r} = 8000 \times 9.882 \times 1 0^{- 5} = 0.7906 m^{3} / s$ . So a real 8000 m³/s flood is reproduced by a manageable ~0.79 m³/s in the lab — the power of Froude scaling.

L4-Q2

A $1/8$ scale airfoil is tested in a pressurised wind tunnel where the air density is raised to $ρ_{m} = 4 ρ_{p}$ (viscosity $μ$ essentially unchanged, velocity kept equal to the prototype: $V_{m} = V_{p}$ ). Is Reynolds similarity achieved? If not, by what factor is $Re_{m}$ off?

Recall Solution

$\frac{Re _{m}}{Re _{p}} = \frac{ρ _{m} V _{m} L _{m} / μ _{m}}{ρ _{p} V _{p} L _{p} / μ _{p}} = \frac{ρ _{m}}{ρ _{p}} \cdot \frac{V _{m}}{V _{p}} \cdot \frac{L _{m}}{L _{p}} \cdot \frac{μ _{p}}{μ _{m}} = 4 \times 1 \times \frac{1}{8} \times 1 = \frac{4}{8} = 0.5$ Not matched: $Re_{m}$ is half the prototype's. The 4× density boost helps but the 1/8 size cut wins. To fully match at $V_{m} = V_{p}$ you would need $ρ_{m} = 8 ρ_{p}$ (density up by the same factor size went down). This is exactly why real pressurised tunnels raise density a lot.

L4-Q3

Combine ideas: for the airfoil of L4-Q2, suppose you instead keep density at $ρ_{m} = 4 ρ_{p}$ but are allowed to change speed. What $V_{m}$ (as a multiple of $V_{p}$ ) restores $Re_{m} = Re_{p}$ ? Then, if $C_{D}$ is now equal, express the model lift-or-drag force ratio $F_{r} = F_{m} / F_{p}$ .

Recall Solution

Speed for Re match: $\frac{ρ _{m} V _{m} L _{m}}{ρ _{p} V _{p} L _{p}} = 1 \Rightarrow 4 \cdot \frac{V _{m}}{V _{p}} \cdot \frac{1}{8} = 1 \Rightarrow \frac{V _{m}}{V _{p}} = 2$ . So $V_{m} = 2 V_{p}$ . Force ratio (any $C_{D}$ -type force, $F \propto ρ V^{2} A$ , $A \propto L^{2}$ ): $F_{r} = \frac{ρ _{m} V _{m}^{2} A _{m}}{ρ _{p} V _{p}^{2} A _{p}} = \frac{ρ _{m}}{ρ _{p}} (\frac{V _{m}}{V _{p}})^{2} (\frac{L _{m}}{L _{p}})^{2} = 4 \times 2^{2} \times (\frac{1}{8})^{2} = 4 \times 4 \times \frac{1}{64} = \frac{16}{64} = 0.25$ The model force is $1/4$ of the prototype force. (Notice we could not assume equal forces here because the fluid density differs — the Drag Coefficient is equal, but the raw force is not.)

Level 5 — Mastery

L5-Q1

Using only the Buckingham Pi Theorem idea, the drag on a smooth sphere depends on $F_{D} = f (ρ, V, D, μ)$ . (a) How many dimensionless $Π$ -groups arise? (b) Write them and show the physics reduces to $C_{D} = ϕ (Re)$ .

Recall Solution

(a) Variables $n = 5$ ( $F_{D}, ρ, V, D, μ$ ); fundamental dimensions $k = 3$ (mass, length, time). By Buckingham, number of $Π$ -groups $= n - k = 5 - 3 = 2$ . (b) A natural choice: $Π_{1} = \frac{F _{D}}{ρ V ^{2} D ^{2}} (a drag coefficient), Π_{2} = \frac{ρ V D}{μ} = Re .$ The theorem says $Π_{1} = ϕ (Π_{2})$ , i.e. $C_{D} = ϕ (Re)$ . So matching Re fixes $C_{D}$ — precisely the Reynolds-similarity claim, now proven by dimensional analysis rather than asserted.

L5-Q2 (the "gotcha")

An engineer scales a $1/10$ ship by Reynolds (thinking "it's underwater, so Re") and computes $V_{m} = V_{p} / L_{r} = 10 V_{p}$ . The prototype cruises at $V_{p} = 8$ m/s, so she sets $V_{m} = 80$ m/s in the water channel. Identify the physical error and give the correct governing number and correct $V_{m}$ .

Recall Solution

Error: a ship rides on the water surface and makes waves ⇒ gravity waves dominate ⇒ Froude, not Reynolds. Reynolds is for fully submerged / closed-pipe flow. Correct scaling (Froude): $V_{m} = V_{p} L_{r} = 8 \times 1/10 = 8 \times 0.31623 = 2.53 m/s$ . So the correct channel speed is about 2.53 m/s, not 80 m/s. Her Reynolds answer over-speeds the model by a factor $\frac{80}{2.53} \approx 31.6 = L_{r}^{- 3/2}$ — physically impossible in a water channel and wrong physics.

L5-Q3 (open synthesis)

A drag test on a $1/12$ scale bridge pier is done in a wind tunnel (same-fluid-type air, Re matched) and gives $C_{D} = 1.2$ . The real pier faces a river current of water, $ρ_{p} = 1000 kg/m^{3}$ , $V_{p} = 3$ m/s, projected area $A_{p} = 6 m^{2}$ . Assuming Re is high enough that $C_{D}$ is Reynolds-independent (constant $1.2$ ), find the actual drag force on the pier.

Recall Solution

When $C_{D}$ has plateaued (high-Re regime), it is the same number regardless of fluid, so we apply it directly to the real conditions: $F_{D, p} = \frac{1}{2} C_{D} ρ_{p} V_{p}^{2} A_{p} = \frac{1}{2} \times 1.2 \times 1000 \times 3^{2} \times 6$ $= 0.6 \times 1000 \times 9 \times 6 = 32400 N = 32.4 kN .$ The model (air) only measured the dimensionless $C_{D}$ ; the prototype force uses the prototype's own $ρ, V, A$ . This is the everyday workflow: measure $C_{D}$ on a model, apply it to reality.

Recall Final self-check

Which number for a submerged submarine? ::: Reynolds (no free surface). Which number for a ship on the surface? ::: Froude (gravity waves). Same fluid, Re matched — how do $V_{m}$ and $V_{p}$ relate for scale $L_{r}$ ? ::: $V_{m} = V_{p} / L_{r}$ . Same fluid, Re matched — how do model and prototype forces compare? ::: equal ( $F_{m} = F_{p}$ ). Can you match Re and Fr together in one fluid? ::: only if $L_{r} = 1$ (no scaling).