Dimensional analysis and similitude

Dimensions and homogeneity, Rayleigh and Buckingham π methods, Re/Fr/Eu/We/Ma, similitude and the Reynolds and Froude model laws with scaling examples.

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Why it matters

Car bodies, pumps, turbochargers and spillways are developed on scale models in wind tunnels, water channels and test rigs long before the full-size item exists. Dimensional analysis tells you which groups of variables to test, and similitude tells you how to run the model and scale its results up. It also explains why the Moody chart and pump characteristic curves can be drawn once for a whole family of sizes.

Key ideas

  • Dimensions. Fluid-mechanics quantities can be written in terms of mass M, length L and time T (plus temperature θ for heat transfer). Examples: velocity LT⁻¹, force MLT⁻², pressure ML⁻¹T⁻², dynamic viscosity ML⁻¹T⁻¹, kinematic viscosity L²T⁻¹, power ML²T⁻³.
  • Dimensional homogeneity. Every term in a valid physical equation has the same dimensions. This is used to check equations and find the units of constants.
  • Rayleigh's method. Assume the dependent variable is a product of powers of the independent variables, then equate the exponents of M, L and T. It works well when there are few variables (up to about four).
  • Buckingham π theorem. If a problem involves n variables containing m fundamental dimensions, it can be written as a relation among (n − m) independent dimensionless π groups.
    • Choose m repeating variables that together contain all m dimensions and do not form a dimensionless group among themselves. Usually pick one geometric variable (D or L), one flow variable (V or N) and one fluid property (ρ).
    • Do not choose the dependent variable as a repeating variable.
    • Each remaining variable combined with the repeating set forms one π group.
  • Important dimensionless numbers (each a force ratio):
    • Reynolds number Re: inertia / viscous force. Governs pipe flow, boundary layers and drag at low speed.
    • Froude number Fr: inertia / gravity. Governs free-surface flows — ships, spillways, open channels.
    • Euler number Eu: pressure / inertia force. A dimensionless pressure drop or pressure coefficient.
    • Weber number We: inertia / surface-tension force. Governs sprays, droplets and fuel atomisation.
    • Mach number Ma: inertia / elastic (compressibility) force. Governs high-speed gas flow; compressibility matters above Ma ≈ 0.3.
  • Similitude between model (m) and prototype (p):
    • Geometric similarity: all length ratios equal (scale ratio L_r = L_p/L_m), including roughness ideally.
    • Kinematic similarity: velocity ratios (and flow directions) equal at corresponding points.
    • Dynamic similarity: ratios of all forces equal at corresponding points — achieved by matching the governing π groups. With geometric similarity, dynamic similarity gives kinematic similarity.
  • Model laws. Match the dominant force ratio: Reynolds model law for enclosed or fully submerged flows (pipes, submarines, low-speed aerodynamics); Froude model law for free-surface flows; Mach law for high-speed gas flow. Usually you cannot match Re and Fr at once with the same fluid, so the model is "distorted" and corrections are applied.
  • Turbomachines follow the same idea: geometrically similar pumps share the same flow coefficient Q/(ND³), head coefficient gH/(N²D²) and power coefficient P/(ρN³D⁵). These give the affinity laws used in the pump and turbine topics.

Formulas

Re = ρ·V·L/μ = V·L/ν ρ density (kg/m³), V velocity (m/s), L characteristic length (m), μ dynamic viscosity (Pa·s), ν kinematic viscosity (m²/s).

Fr = V/√(g·L) Eu = Δp/(ρ·V²) We = ρ·V²·L/σ Ma = V/c g = 9.81 m/s², Δp pressure difference (Pa), σ surface tension (N/m), c speed of sound (m/s).

Number of π groups = n − m n variables, m fundamental dimensions.

V_m = V_p·(L_p/L_m)·(ν_m/ν_p) Reynolds model law (Re_m = Re_p).

V_r = √L_r, Q_r = L_r^2.5, T_r = √L_r, F_r = L_r³ Froude model law with the same fluid and g; subscript r means prototype/model ratio; T is time.

F_p/F_m = (ρ_p/ρ_m)·(V_p/V_m)²·(L_p/L_m)² Force scaling when the force coefficient F/(ρV²L²) is matched.

Q/(N·D³), g·H/(N²·D²), P/(ρ·N³·D⁵) = constant Similar turbomachines; N speed (rev/s or rpm used consistently), D diameter (m), H head (m), P power (W).

Worked examples

Example 1 (standard) — Buckingham π for pipe pressure drop. Given: Δp in a pipe depends on D, L, V, ρ, μ and roughness ε. Find the π groups.

  1. Variables: Δp, D, L, V, ρ, μ, ε → n = 7. Dimensions M, L, T → m = 3. So 7 − 3 = 4 π groups.
  2. Repeating variables: D (L), V (LT⁻¹), ρ (ML⁻³).
  3. π₁ = Δp·D^a·V^b·ρ^c: M: 1 + c = 0 → c = −1; T: −2 − b = 0 → b = −2; L: −1 + a + b − 3c = 0 → a = 0. So π₁ = Δp/(ρV²).
  4. π₂ with μ (ML⁻¹T⁻¹): c = −1, b = −1, a = −1 → μ/(ρVD), i.e. 1/Re.
  5. π₃ = L/D and π₄ = ε/D (already lengths). Answer: Δp/(ρV²) = f(ρVD/μ, L/D, ε/D). Since Δp is proportional to L/D, this becomes the Darcy friction factor f(Re, ε/D) — the Moody chart.

Example 2 (GATE level) — car model in a water tunnel. Given: a car 1:5 scale model (L_p/L_m = 5) is tested in water to simulate the prototype at 25 m/s in air. ν_air = 1.5 × 10⁻⁵ m²/s, ν_water = 1.0 × 10⁻⁶ m²/s, ρ_air = 1.2 kg/m³, ρ_water = 1000 kg/m³. The model drag is 200 N. Find the model speed and the prototype drag.

  1. Reynolds law: V_m = V_p·(L_p/L_m)·(ν_m/ν_p) = 25 × 5 × (1.0 × 10⁻⁶/1.5 × 10⁻⁵) = 8.333 m/s.
  2. With Re matched, the drag coefficient is the same, so F_p/F_m = (ρ_p/ρ_m)·(V_p/V_m)²·(L_p/L_m)².
  3. F_p = 200 × (1.2/1000) × (25/8.333)² × 5² = 200 × 0.0012 × 9 × 25 = 54 N. Answer: V_m ≈ 8.33 m/s; prototype drag ≈ 54 N. In air, the same model would need 125 m/s (Ma ≈ 0.36), which brings in compressibility — the reason water tunnels are used.

Example 3 (Froude law). A 1:25 spillway model carries 0.2 m³/s. Prototype discharge Q_p = 0.2 × 25^2.5 = 0.2 × 3125 = 625 m³/s, and prototype velocities are √25 = 5 times the model's.

Common mistakes

  • Choosing the dependent variable, or a set that is itself dimensionless, as repeating variables.
  • Inverting the scale ratio — for Reynolds similarity in the same fluid, a smaller model needs a higher speed.
  • Using the Reynolds law for a free-surface model (use Froude) or Froude for a closed-pipe model.
  • Forgetting that kinematic viscosity, not dynamic viscosity, appears when the model fluid differs.
  • Writing Q_r = L_r² or L_r³ under Froude scaling; it is L_r^2.5.

For GATE ME

Typical questions: the number of π groups, identifying a dimensionless group or the dimensions of a quantity, the model velocity or discharge under the Reynolds or Froude law, force or power scaling, and the meaning of Re, Fr, We, Ma and Eu. Practise writing M-L-T dimensions quickly and checking the exponent algebra.

Quick check

  1. How many π groups for 6 variables involving M, L and T?
  2. What are the dimensions of dynamic viscosity?
  3. Which dimensionless number governs ship-model testing?
  4. Under Froude scaling, what is the velocity ratio for a 1:16 model?
  5. Water (ρ = 1000 kg/m³, μ = 0.001 Pa·s) flows at 2 m/s over a 0.5 m plate. What is Re_L? Answers: 1. 3. 2. ML⁻¹T⁻¹. 3. Froude number. 4. V_p/V_m = √16 = 4. 5. 1000 × 2 × 0.5/0.001 = 1.0 × 10⁶.

Try answering each one aloud before you open it.

  1. 1.What is dimensional analysis in the context of fluid mechanics?Concept

    Dimensional analysis is a method used in fluid mechanics to reduce the complexity of physical problems by expressing the variables involved in terms of their fundamental dimensions (such as mass, length, time). This helps in identifying dimensionless numbers that govern the behavior of the system, allowing for the comparison of different systems and the derivation of scaling laws.

  2. 2.Explain the concept of similitude in turbomachinery.Concept

    Similitude in turbomachinery refers to the concept of creating a scaled model of a machine that behaves in the same way as the full-scale machine. This is achieved by ensuring geometric, kinematic, and dynamic similarity between the model and the prototype. Similitude allows engineers to predict the performance of a full-scale machine based on tests conducted on a smaller, more manageable model.

  3. 3.What are dimensionless numbers, and why are they important in fluid mechanics?Concept

    Dimensionless numbers are ratios of forces or other quantities that have no units. They are important in fluid mechanics because they allow for the comparison of different fluid systems and help in understanding the relative importance of different physical effects. Examples include the Reynolds number, which indicates the relative importance of inertial versus viscous forces, and the Mach number, which compares the speed of an object to the speed of sound.

  4. 4.Why is the Reynolds number used in the study of fluid flow?Application

    The Reynolds number is used in the study of fluid flow to predict the flow regime, whether it is laminar or turbulent. It is a dimensionless number that represents the ratio of inertial forces to viscous forces in a fluid. A low Reynolds number indicates laminar flow, where viscous forces dominate, while a high Reynolds number indicates turbulent flow, where inertial forces are more significant.

  5. 5.What happens if geometric similarity is not maintained in a model test of a turbomachine?Application

    If geometric similarity is not maintained in a model test of a turbomachine, the results may not accurately predict the performance of the full-scale machine. Geometric similarity ensures that all linear dimensions of the model are proportionally scaled from the prototype, which is crucial for maintaining the correct flow patterns and pressure distributions. Without it, the model may exhibit different flow characteristics, leading to incorrect conclusions about the prototype's behavior.

  6. 6.How does the Mach number affect the design of turbomachinery?Application

    The Mach number measures compressibility. Below about Ma 0.3, density changes are small and the flow can be treated as incompressible. Between about 0.3 and 0.8, the flow is subsonic but compressible, and density changes through a stage must be included. As the relative Mach number at compressor blade tips approaches 1, shocks form and losses rise sharply, and the passage chokes at a maximum mass flow. Designers limit tip speed, use thin, swept or transonic blade profiles, and match Mach number when testing models.

  7. 7.Calculate the Reynolds number for a fluid with a density of 1.2 kg/m³, a velocity of 10 m/s, a characteristic length of 0.5 m, and a dynamic viscosity of 0.001 Pa·s.Numerical

    Reynolds number (Re) is calculated using the formula: Re = (ρ·V·L) / μ, where ρ is the density, V is the velocity, L is the characteristic length, and μ is the dynamic viscosity. Substituting the given values: Re = (1.2 kg/m³ · 10 m/s · 0.5 m) / 0.001 Pa·s = 6000. Therefore, the Reynolds number is 6000.

  8. 8.What is the significance of the Froude number in fluid mechanics?Concept

    The Froude number Fr = V/√(gL) is the ratio of inertia to gravity forces, and it governs flows with a free surface. In open channels, with L the flow depth, Fr < 1 is subcritical (tranquil) flow, where surface waves can travel upstream. Fr > 1 is supercritical (shooting) flow, and Fr = 1 is critical flow. Ship hulls, spillways and harbour models are tested under the Froude model law, giving V_r = √L_r and Q_r = L_r^2.5.

  9. 9.Explain how dynamic similarity is achieved in model testing of fluid systems.Concept

    Dynamic similarity is achieved in model testing of fluid systems by ensuring that the dimensionless numbers governing the flow are the same for both the model and the prototype. This typically involves matching the Reynolds number, Froude number, or other relevant dimensionless numbers. By doing so, the forces and flow patterns in the model will accurately represent those in the full-scale system, allowing for reliable predictions of the prototype's behavior.

  10. 10.A model turbine runs at 3000 rpm and the prototype at 1500 rpm. If the model is one-quarter the size of the prototype, what is the ratio of flow velocities at corresponding points?Numerical

    For geometrically similar turbomachines with similar velocity triangles, every velocity scales with the blade speed, so V ∝ ND. V_m/V_p = (N_m·D_m)/(N_p·D_p) = (3000 × 0.25)/(1500 × 1) = 0.5. The model velocities are half those in the prototype; equivalently, the prototype velocities are twice the model's.

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