Dimensional Analysis: Similarity, the Buckingham Pi Theorem, Dimensionless Groups and Model Testing
2. The Buckingham Pi theorem and its procedure
Pi theorem. If a physical relation connects n variables (the dependent one included) and these contain j independent primary dimensions, the relation can be rewritten as a relation among n − j independent dimensionless groups Π₁, Π₂, …, Π₍ₙ₋ⱼ₎. Here j is the rank of the dimensional matrix. It is usually 3 (M, L, T), but is 2 if, say, only L and T occur, so count independent dimensions and not just the letters that appear.
- List all variables and write their M, L, T dimensions; find j.
- Choose j repeating variables. Together they must contain every primary dimension, no product of powers of them may be dimensionless, and the dependent variable must not be one of them. The usual choice is one geometric (D), one kinematic (V or N) and one fluid property (ρ).
- Take each remaining variable in turn and multiply it by the repeating variables raised to unknown powers; solve the powers that make the product dimensionless (one equation each for M, L and T).
- Check that each Π is dimensionless and write Π₁ = φ(Π₂, …). Groups may be inverted or multiplied together, so only their number is fixed, not their form.
Worked case, drag on a sphere. F = f(ρ, V, D, μ), so n = 5 and j = 3, giving 2 groups. Take ρ, V, D as repeating. For Π₁ = F ρᵃ Vᵇ Dᶜ: M gives 1 + a = 0, so a = −1; T gives −2 − b = 0, so b = −2; L gives 1 − 3a + b + c = 0, so 1 + 3 − 2 + c = 0 and c = −2. Hence Π₁ = F/(ρV²D²). The same steps for μ give Π₂ = μ/(ρVD) = 1/Re. The result is F/(ρV²D²) = φ(Re), which is why a single curve of drag coefficient against Reynolds number serves every sphere, speed and fluid.
3. Applications: pipe pressure drop, sphere drag and a pump
| System | Variables | n, j | Pi groups |
|---|---|---|---|
| Pipe pressure drop | Δp, ρ, V, D, L, ε, μ | 7, 3 | 4: Δp/(ρV²), ρVD/μ, L/D, ε/D |
| Sphere drag | F, ρ, V, D, μ | 5, 3 | 2: F/(ρV²D²), ρVD/μ |
| Pump (gH and P dependent; N is the speed in rev/s) | gH, P, Q, N, D, ρ, μ | 7, 3 | 4: gH/(N²D²), P/(ρN³D⁵), Q/(ND³), ρND²/μ |
Pipe. In fully developed flow Δp grows in proportion to L, so the group L/D is pulled out as a factor: Δp/(ρV²) = (L/D) φ(Re, ε/D). Writing φ = f/2 gives Δp = f (L/D)(ρV²/2), the Darcy-Weisbach relation, and shows that the friction factor f depends only on Re and the relative roughness ε/D.
Pump. The three coefficients are the head coefficient C_H = gH/(N²D²), the flow coefficient C_Q = Q/(ND³) and the power coefficient C_P = P/(ρN³D⁵). For geometrically similar pumps at dynamically similar operating points, and with Reynolds-number effects and the change of efficiency neglected, the affinity laws follow: Q₂/Q₁ = (N₂/N₁)(D₂/D₁)³, H₂/H₁ = (N₂/N₁)²(D₂/D₁)² and P₂/P₁ = (ρ₂/ρ₁)(N₂/N₁)³(D₂/D₁)⁵. At fixed D these reduce to Q ∝ N, H ∝ N² and P ∝ N³, so a 10% rise in speed raises power by about 33%.
4. The concept of similarity: geometric, kinematic and dynamic
A model test measures a dependent quantity on a small, cheap model and converts it to the full-size prototype. The conversion is valid only if the two flows are similar in three layers. Write the length scale as L_r = L_m/L_p (m = model, p = prototype).
- Geometric similarity: all corresponding length ratios are equal and all corresponding angles are equal. The scaling must include surface roughness (equal ε/D) and the position of nearby walls and free surfaces.
- Kinematic similarity: the velocity at corresponding points has the same direction and a constant ratio V_r, so the streamline patterns are geometrically similar. The time scale is then T_r = L_r/V_r. It is the ratio of velocities that is constant, not the velocities themselves.
- Dynamic similarity: the ratios of corresponding forces (inertia, viscous, gravity, pressure, surface tension, elastic) at corresponding points are equal; the force scale is F_r = ρ_r L_r² V_r².
The Pi theorem converts these force statements into a recipe: complete similarity holds when the geometry is similar and every independent Π group that does not contain the dependent variable has the same value in model and prototype. The dependent group (for example Δp/(ρV²) or F/(ρV²D²)) is then equal automatically. In the pipe case, equal L/D, ε/D and Re guarantee equal Δp/(ρV²). Geometric similarity alone is necessary but never sufficient.
5. Dimensionless groups and their physical significance
| Group | Definition | Ratio of | Where it governs |
|---|---|---|---|
| Reynolds Re | ρVL/μ = VL/ν | inertia force to viscous force | pipes, boundary layers, drag, transition to turbulence |
| Froude Fr | V/√(gL) | square root of inertia to gravity force | free-surface flows: ships, spillways, open channels, waves |
| Mach Ma | V/a, with a = √(γRT) for an ideal gas | flow speed to speed of sound | compressible flows; shocks at and above Ma = 1 |
| Euler Eu | Δp/(ρV²) (some texts use ½ρV²) | pressure force to inertia force | pressure coefficient, pressure drop |
| Weber We | ρV²L/σ | inertia force to surface-tension force | droplets, bubbles, thin films, small scales |
Reading the numbers. A large Re means inertia dominates and viscosity is confined to thin layers (boundary layers) or, in a pipe, the flow turns turbulent. Fr plays the role of a flow-speed to wave-speed ratio, since √(gL) is the speed of a long gravity wave in water of depth L: in an open channel Fr < 1 is subcritical, Fr = 1 critical and Fr > 1 supercritical. For gases a rule of thumb treats the flow as incompressible when Ma < 0.3, because density then changes by about 5% or less. Since a = √(γRT), a Mach number is never fixed by speed alone: for air at 300 K, a = √(1.4 × 287 × 300) ≈ 347 m/s.
6. Model testing and scaling laws
Often two groups matter and cannot both be matched. With the same fluid, Reynolds matching needs V_m/V_p = (ν_m/ν_p)(L_p/L_m) = 1/L_r, so a small model needs a high speed. Froude matching needs V_m/V_p = √L_r, a low speed. Both hold together only if ν_m/ν_p = L_r^(3/2): for L_r = 1/25 that is ν_m = ν_p/125, and no practical liquid exists. One therefore matches the dominant group. Ships, spillways and floating bodies, where gravity waves at a free surface govern, use Froude. Submerged bodies, aircraft at low speed, pipes, pumps and turbines use Reynolds; high-speed aircraft add Mach, which is why pressurised or cryogenic tunnels are used to raise Re.
| Quantity | Froude matching | Reynolds matching |
|---|---|---|
| Velocity | L_r^(1/2) | L_r⁻¹ |
| Time | L_r^(1/2) | L_r² |
| Flow rate | L_r^(5/2) | L_r |
| Force | L_r³ | 1 |
| Pressure | L_r | L_r⁻² |
| Power | L_r^(7/2) | L_r⁻¹ |
Worked example 1, Froude. A ship model at L_r = 1/36 is towed to represent a prototype speed of 12 m/s. Then V_m = 12 × √(1/36) = 12/6 = 2 m/s. A model wave-making drag of 8 N scales to F_p = 8 × 36³ = 373,248 N, about 373 kN, taking the same water density. Dynamic similarity also gives the prototype time as 6 times the model time.
Worked example 2, Reynolds. A water-going body (ν = 1.0 × 10⁻⁶ m²/s) at 3 m/s and length L_p is modelled at L_m = L_p/4 in air (ν = 1.5 × 10⁻⁵ m²/s). Re matching gives V_m = 3 × 15 × 4 = 180 m/s. At about Ma = 0.5 this air flow is no longer incompressible, so the test fails. A better choice is a water tunnel, where the model needs 3 × 4 = 12 m/s.
Key takeaways
- The Pi theorem gives n − j groups, with j the rank of the dimensional matrix (usually 3); repeating variables must cover M, L and T, be independent, and exclude the dependent variable. A variable that is the only one carrying a dimension drops out.
- Pipe: Δp/(ρV²) = (L/D)φ(Re, ε/D), the root of Darcy-Weisbach. Sphere: F/(ρV²D²) = φ(Re). Pump: C_H, C_Q, C_P and Re, giving the affinity laws Q ∝ ND³, H ∝ N²D², P ∝ ρN³D⁵.
- Complete similarity needs geometric similarity and equal values of every independent Π group; velocity and force ratios are constants of the scaling, not equalities.
- Re = inertia/viscous, Fr = √(inertia/gravity) = V/√(gL), Ma = V/a with a = √(γRT), Eu = pressure/inertia, We = inertia/surface tension. Gas flow is treated as incompressible below Ma ≈ 0.3.
- Match the dominant group. Froude (same fluid): V ∝ L^(1/2), F ∝ L³. Reynolds (same fluid): V ∝ 1/L, F unchanged. Both cannot hold in one fluid unless ν_m/ν_p = L_r^(3/2).
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The dimensions of dynamic viscosity μ in terms of mass M, length L and time T are
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Answer: C — ML⁻¹T⁻¹, from τ = μ du/dy
From τ = μ du/dy: μ = τ/(du/dy) = ML⁻¹T⁻² divided by T⁻¹, which is ML⁻¹T⁻¹. The tempting L²T⁻¹ is the kinematic viscosity ν = μ/ρ, since dividing ML⁻¹T⁻¹ by ML⁻³ leaves L²T⁻¹. ML⁻¹T⁻² is the dimension of stress itself, and MLT⁻² that of force.The Froude number V/√(gL) is, in terms of forces, the square root of the ratio of
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Answer: A — inertia force to gravity force
The inertia force scales as ρV²L² and the gravity force as ρgL³, so their ratio is V²/(gL) = Fr². Inertia to viscous force is the Reynolds number, inertia to elastic force gives Ma², and inertia to surface tension is the Weber number.A smooth sphere of diameter D moves at speed V through a large, unbounded body of incompressible fluid of density ρ and viscosity μ. There is no free surface and V is far below the speed of sound. The drag force F obeys F/(ρV²D²) = φ(...). The argument of φ is
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Answer: B — the Reynolds number ρVD/μ alone
The variables are F, ρ, V, D, μ, so n = 5, j = 3 and there are 2 groups: F/(ρV²D²) and ρVD/μ. Gravity enters only as a buoyancy force and creates no waves without a free surface, so g, σ and the sound speed are not variables here. The Froude-number option is the tempting one but would need g in the list.A geometrically similar model, with L_m/L_p = 1/4, of a fully submerged body is tested in the same fluid at the same temperature, with the Reynolds number matched. The measured model drag is 25 N. The drag on the prototype is
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Answer: D — 25 N, because the force is the same at any scale
With ρ and μ unchanged, Re matching requires V_m/V_p = L_p/L_m = 4. Then F ∝ ρV²L² gives F_m/F_p = 4² × (1/4)² = 1, so F_p = 25 N. The options 100, 400 and 1,600 N come from scaling by L, L² or L³ and forgetting that the speed changed.A naval architect tests a ship model to estimate the wave-making resistance of the prototype. Dynamic similarity for this resistance is based on equal
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Answer: C — Froude numbers, because gravity waves govern the resistance
Wave making is a gravity effect at a free surface, so the Froude number V/√(gL) is matched. Reynolds matching with a small model would need an impossibly high speed, so the small viscous (friction) part is corrected separately. Mach number plays no role at ship speeds, and surface tension is negligible for large waves.Which statement about the Mach number is correct?
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Answer: A — It equals V/a, with a = √(γRT) for an ideal gas; above about 0.3 compressibility matters
Ma = V/a, and for an ideal gas a = √(γRT), so the same speed gives a different Mach number at a different temperature. The rule of thumb Ma < 0.3 for incompressible treatment rests on a density change of about 5% or less. The Reynolds-number option is wrong because Ma has nothing to do with viscous force, and the V/√(gL) option describes the Froude number.For the pipe-flow relation Δp = f(D, L, ε, ρ, μ, V), which set of three variables cannot be used as the repeating variables?
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Answer: D — D, L and ε
D, L and ε all have the dimension of length alone, so together they contain neither M nor T and L/D and ε/D are already dimensionless; they are not independent. Each of the other sets carries M, L and T independently (the 3 × 3 dimensional determinant is non-zero), so the usual set ρ, V, D is valid, and so are ρ, μ, D and the less usual V, D, μ.A 1:25 scale model (L_m/L_p = 1/25) of a ship hull is to be tested in the same liquid as the prototype with both the Froude number and the Reynolds number matched. This is not possible because
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Answer: B — Froude matching requires V_m/V_p = 1/5 while Reynolds matching requires V_m/V_p = 25
Froude matching gives V_m/V_p = √(1/25) = 1/5, while Reynolds matching in the same liquid gives V_m/V_p = L_p/L_m = 25. The two speeds conflict by a factor of 125 unless ν_m/ν_p = (1/25)^(3/2) = 1/125. Re does not involve g, and Froude matching does not depend on density, so the other statements are wrong.A simple pendulum has period T that may depend on its length ℓ, bob mass m, the acceleration g due to gravity and the amplitude θ₀. Dimensional analysis shows that
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Answer: C — T√(g/ℓ) is a function of θ₀ alone; m does not appear
Mass is the only variable containing M, so it cannot appear in any dimensionless group. With n = 5 and j = 3 there are two groups, θ₀ and T√(g/ℓ), so T√(g/ℓ) = φ(θ₀). Dimensional analysis cannot prove that φ is a constant; that is true only approximately, for small amplitudes, and needs the physics. The √(mℓ/g) option fails because it contains M.Which of the following statements about similarity of flows are correct?
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Answer: A — Geometric similarity requires equal ratios for all corresponding lengths, including the roughness height; C — Dynamic similarity requires equal ratios of the corresponding forces at corresponding points
A scale model needs all lengths, angles and roughness in a fixed ratio, and dynamic similarity means that the force ratios agree at corresponding points. Kinematic similarity needs a constant ratio of velocities, not equal velocities, since a smaller model generally runs at a different speed. Geometric similarity is necessary but never sufficient: the flows must also match in Re, Fr or Ma as relevant.Which of the following pairs of a dimensionless group with the ratio it represents are correct?
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Answer: A — Reynolds number: inertia force to viscous force; B — Weber number: inertia force to surface-tension force; D — Euler number: pressure force to inertia force
Re = ρVL/μ compares inertia with viscous force, We = ρV²L/σ compares inertia with surface tension, and Eu = Δp/(ρV²) compares pressure force with inertia. The Froude number compares inertia with gravity, not with surface tension, so that pairing is wrong.Which of the following combinations of the variables are dimensionless? (V velocity, D length, ρ density, μ viscosity, σ surface tension, g gravitational acceleration, N rotational speed in rev/s, Q volume flow rate)
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Answer: B — ρV²D/σ; C — Q/(ND³); D — μ/(ρVD)
Substitute the dimensions. ρV²D/σ = (ML⁻³ L²T⁻² L)/(MT⁻²) = 1. Q/(ND³) = L³T⁻¹/(T⁻¹L³) = 1. μ/(ρVD) = ML⁻¹T⁻¹/(ML⁻³ LT⁻¹ L) = 1. But gD/V = LT⁻² L/(LT⁻¹) = LT⁻¹, which is a speed, so it is not dimensionless; gD/V² would be.The pressure rise Δp of a pump depends on the volume flow rate Q, the rotational speed N, the impeller diameter D, the fluid density ρ, the fluid viscosity μ and the impeller blade roughness height ε (a length). The number of independent dimensionless groups, including the dependent one, is ____.
Numerical answer — type the value.
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Answer: 4
The variables are Δp, Q, N, D, ρ, μ, ε, so n = 7. M, L and T all occur and the dimensional matrix has rank j = 3. Number of groups = n − j = 7 − 3 = 4: Δp/(ρN²D²), Q/(ND³), ρND²/μ and ε/D.A pump with impeller diameter 0.30 m runs at 1450 rpm and delivers a head of 24.0 m at its best-efficiency point. A geometrically similar pump of diameter 0.45 m, handling the same liquid, is run at 960 rpm at the corresponding point. Using the affinity laws and ignoring Reynolds-number and efficiency changes, the head of the larger pump, in metres to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 23.7
H₂/H₁ = (N₂/N₁)²(D₂/D₁)². N₂/N₁ = 960/1450 = 0.6621 and D₂/D₁ = 0.45/0.30 = 1.5. H₂ = 24.0 × 0.6621² × 1.5² = 24.0 × 0.4383 × 2.25 = 23.67 m, which is 23.7 m to one decimal place. Using only the speed ratio or only the diameter ratio gives 10.5 m or 54.0 m instead.A 1:36 scale model of a ship (length ratio L_m/L_p = 1/36) is towed in fresh water, and Froude similarity is used. The model wave-making drag is 8.0 N. Assuming the model and the prototype fluids have the same density, the prototype wave-making drag, in kN to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 373.2
Under Froude matching with the same density, F ∝ L³, so F_p = F_m (L_p/L_m)³ = 8.0 × 36³ = 8.0 × 46,656 = 373,248 N = 373.2 kN. The common slips are to scale by 36 (giving 0.29 kN) or by 36² (giving 10.4 kN).A submerged body of length L_p moves in water (kinematic viscosity 1.0 × 10⁻⁶ m²/s) at 3.0 m/s. A model of length L_p/4 is tested in air (kinematic viscosity 1.5 × 10⁻⁵ m²/s) with the Reynolds number matched. The required air speed, in m/s, is ____.
Numerical answer — type the value.
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Answer: 180
V_m L_m/ν_m = V_p L_p/ν_p, so V_m = V_p (ν_m/ν_p)(L_p/L_m) = 3.0 × (1.5 × 10⁻⁵/1.0 × 10⁻⁶) × 4 = 3.0 × 15 × 4 = 180 m/s. At about 340 m/s sound speed this is Ma ≈ 0.5, so the air flow would be compressible and the test is not usable as it stands.An airliner cruises at 900 km/h at an altitude where the air temperature is 220 K. Treat air as an ideal gas with γ = 1.4 and R = 287 J/kg·K. The flight Mach number, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.84
V = 900/3.6 = 250 m/s. a = √(γRT) = √(1.4 × 287 × 220) = √88,396 = 297.3 m/s. Ma = 250/297.3 = 0.841, so 0.84. Using the sea-level value a = 340 m/s would give 0.74, and forgetting the conversion to m/s would give an absurd answer.