Dimensional analysis, Buckingham pi theorem and similitude

Why it matters

Most real flows are too complex to solve exactly, so engineers correlate experiments. Dimensional analysis tells you which dimensionless groups to correlate (friction factor against Reynolds number, drag coefficient against Reynolds number, power number against Reynolds number for agitators), and similitude tells you how to run a small model or a pilot plant so its results scale to the full plant. Every correlation in a chemical engineering data book rests on this idea.

Key ideas

Dimensional homogeneity. Every valid physical equation has the same dimensions in every additive term. Fluid mechanics uses three primary dimensions — mass M, length L, time T (add temperature θ for heat transfer). Examples: velocity [LT⁻¹], force [MLT⁻²], pressure [ML⁻¹T⁻²], viscosity μ [ML⁻¹T⁻¹], kinematic viscosity ν [L²T⁻¹], surface tension [MT⁻²], power [ML²T⁻³].

Rayleigh's method. Assume the dependent variable is a product of powers of the independent ones and equate exponents of M, L, T. It works well for up to about four variables.

Buckingham π theorem. If a physical problem involves n variables that together contain m primary dimensions (strictly, the rank of the dimensional matrix), the relation can be rewritten as a relation among n − m independent dimensionless π groups. Procedure:

1. List all n relevant variables, including the dependent one. Missing a variable is the most serious error; including an irrelevant one only adds a spurious group.
2. Choose m repeating variables that together contain all primary dimensions, are not themselves able to form a dimensionless group, and do not include the dependent variable. In flow problems the usual choice is a length (D), a velocity (V) and a fluid property (ρ).
3. Form each π group as the product of the repeating variables (raised to unknown powers) and one remaining variable; solve for the exponents.
4. Check each group is dimensionless and rearrange into familiar forms (any product, power or inverse of π groups is also valid).

Important dimensionless groups.

• Reynolds number Re = ρVL/μ: inertia/viscous force; laminar–turbulent transition, pipe friction, drag.
• Froude number Fr = V/√(gL): inertia/gravity; free-surface flows, ship waves, open channels.
• Euler number Eu = Δp/(ρV²): pressure/inertia; pressure coefficient.
• Weber number We = ρV²L/σ: inertia/surface tension; drops, bubbles, sprays.
• Mach number Ma = V/c: inertia/compressibility; gas flows.
• Power number N_P = P/(ρN³D⁵) for agitators; drag coefficient C_D and friction factor f are Euler-type groups.

Similitude.

• Geometric similarity: all lengths in the model are in the same ratio (scale) to the prototype, including roughness.
• Kinematic similarity: velocities at corresponding points are in a fixed ratio, so streamline patterns are alike.
• Dynamic similarity: ratios of all forces at corresponding points are equal, achieved by matching the governing dimensionless groups.
• Closed-conduit flows, flow past submerged bodies and agitated tanks are governed mainly by Re (model law: Re_m = Re_p). Free-surface flows are governed mainly by Fr. Matching both Re and Fr with the same fluid at a reduced scale is usually impossible, so the dominant one is matched and the other is corrected for (distorted or partial similarity).

Scale ratios under Froude similarity (length ratio L_r = L_p/L_m, same g): velocity ratio √L_r, time ratio √L_r, discharge ratio L_r^2.5, force ratio L_r³ (same fluid).

Formulas

• number of π groups = n − m — n variables, m primary dimensions (rank).
• Re = ρ·V·L/μ = V·L/ν — ρ (kg/m³), V (m/s), L characteristic length (m), μ (Pa·s), ν (m²/s).
• Fr = V/√(g·L), Eu = Δp/(ρ·V²), We = ρ·V²·L/σ, Ma = V/c — g (m/s²), Δp (Pa), σ (N/m), c speed of sound (m/s).
• V_m = V_p·(L_p/L_m)·(ν_m/ν_p) — Reynolds similarity.
• V_m = V_p·√(L_m/L_p) — Froude similarity.
• Q_p/Q_m = (L_p/L_m)^2.5 — discharge ratio under Froude similarity.
• Δp_p/Δp_m = (ρ_p/ρ_m)·(V_p/V_m)² — equal Euler numbers.

Worked examples

Example 1 (standard). The drag force F on a smooth sphere depends on diameter D, velocity V, fluid density ρ and viscosity μ. Find the dimensionless relation.

1. n = 5 variables (F, D, V, ρ, μ); m = 3 (M, L, T); so 5 − 3 = 2 π groups.
2. Repeating variables: D, V, ρ.
3. π₁ = F·D^a·V^b·ρ^c. Dimensions: MLT⁻² · L^a · L^bT⁻ᵇ · M^cL⁻³ᶜ. M: 1 + c = 0 → c = −1. T: −2 − b = 0 → b = −2. L: 1 + a + b − 3c = 0 → a = −2. So π₁ = F/(ρV²D²).
4. π₂ = μ·D^a·V^b·ρ^c. M: 1 + c = 0 → c = −1. T: −1 − b = 0 → b = −1. L: −1 + a + b − 3c = 0 → a = −1. So π₂ = μ/(ρVD) = 1/Re. F/(ρV²D²) = φ(Re) — the basis of the C_D–Re curve for spheres.

Example 2 (GATE level). A crude oil line (D = 0.5 m, ν = 4 × 10⁻⁵ m²/s, ρ = 850 kg/m³) carries oil at 2 m/s. It is studied with a geometrically similar 0.1 m model using water (ν = 1 × 10⁻⁶ m²/s, ρ = 1000 kg/m³). (a) What water velocity gives dynamic similarity? (b) If the model shows a pressure drop of 1.2 kPa over its scaled length, what is the prototype pressure drop over the corresponding length?

1. Pipe flow without a free surface → match Re: V_m = V_p·(D_p/D_m)·(ν_m/ν_p).
2. V_m = 2 × (0.5/0.1) × (1 × 10⁻⁶ / 4 × 10⁻⁵) = 2 × 5 × 0.025 = 0.25 m/s.
3. With Re matched, Eu is equal: Δp_p = Δp_m·(ρ_p/ρ_m)·(V_p/V_m)².
4. Δp_p = 1200 × (850/1000) × (2/0.25)² = 1200 × 0.85 × 64 = 65 280 Pa. (a) 0.25 m/s; (b) ≈ 65.3 kPa.

Example 3 (Froude scaling). A 1:25 spillway model carries 0.2 m³/s. Prototype discharge = 0.2 × 25^2.5 = 0.2 × 3125 = 625 m³/s; a model velocity of 2 m/s corresponds to 2 × √25 = 10 m/s in the prototype.

Common mistakes

• Choosing the dependent variable (or a set like ρ, μ, ν that is itself dimensionally dependent) as repeating variables.
• Omitting a relevant variable such as roughness ε, which loses the ε/D group.
• Matching Froude number for a pipe flow, or Reynolds number for a ship's wave resistance.
• Writing the dimensions of μ as ML⁻¹T⁻² (that is pressure); μ is ML⁻¹T⁻¹.
• Applying velocity scale √L_r but forgetting that discharge scales as L_r^2.5 and force as L_r³.

For GATE CH

Expect questions on the number of π groups, identifying a given group (Re, Fr, We, power number), the dimensions of a property, and model-prototype scaling under Reynolds or Froude similarity. Agitator scale-up with the power number and equal tip speed or equal power per volume is a favourite application in chemical engineering. Practise solving exponent equations for M, L and T quickly.

Quick check

1. A problem has 7 variables and 3 primary dimensions. How many π groups?
2. State the dimensions of kinematic viscosity.
3. Which group governs a ship model's wave resistance?
4. Under Froude similarity at 1:16 scale, what is the velocity ratio V_p/V_m?

Answers: 1. 4. 2. L²T⁻¹. 3. Froude number. 4. √16 = 4.