The Reynolds Number from Nondimensionalisation

Derivation

The Reynolds Number from Nondimensionalisation

D-326 Home PU-307 Threads symmetry · force Depends on Navier-Stokes from the Newtonian Constitutive Law

Statement

For an incompressible Newtonian fluid, nondimensionalising the Navier–Stokes momentum equation with a single length scale \(L\), velocity scale \(U\) and the inertial pressure scale \(\rho U^2\) reduces the governing equation to \(\partial_{t^*}\mathbf{u}^* + (\mathbf{u}^*\cdot\nabla^*)\mathbf{u}^* = -\nabla^* p^* + \tfrac{1}{\mathrm{Re}}\nabla^{*2}\mathbf{u}^*\), in which the only dimensionless coefficient is the Reynolds number \(\mathrm{Re} = \rho U L/\mu = UL/\nu\). The Buckingham Pi theorem confirms that these four parameters form exactly one dimensionless group, so \(\mathrm{Re}\) is the sole governing parameter and two flows with equal \(\mathrm{Re}\) and matching geometry are dynamically similar.

Why it matters

The Reynolds number is the single most important parameter in fluid mechanics: it measures the ratio of inertial to viscous momentum transport and, once fixed together with the geometry, determines the entire flow field up to a rescaling. Nondimensionalisation shows this is not an empirical rule of thumb but a mathematical consequence of the structure of the momentum equation.

Dynamic similarity is what makes wind-tunnel and towing-tank testing possible: a scale model tested at the correct \(\mathrm{Re}\) reproduces the dimensionless force and pressure coefficients of the full-size vehicle. Without the collapse onto a single parameter, engineering prediction from models would require matching every dimensional quantity separately, which is impossible.

Assumptions

Incompressible, constant-density flow with \(\nabla\cdot\mathbf{u}=0\).If density varies, an additional continuity coupling and a Mach-number group appear; the momentum equation no longer closes on \(\mathrm{Re}\) alone.

Newtonian constitutive law, \(\boldsymbol{\tau}=2\mu\,\mathbf{S}\) with constant \(\mu\).For a shear-thinning or viscoelastic fluid the viscous term is not \(\mu\nabla^2\mathbf{u}\); extra dimensionless groups (e.g. Weissenberg, power-law index) enter and \(\mathrm{Re}\) is no longer sufficient.

A single, well-defined length scale \(L\) and velocity scale \(U\) characterise the flow.If the geometry imposes several disparate length scales (roughness, boundary-layer thickness, gap width), additional aspect-ratio groups survive nondimensionalisation and similarity requires matching each of them.

Inertial pressure scaling \(p_c=\rho U^2\) is appropriate.In creeping flow the pressure balances viscous stress, so \(p_c=\mu U/L\) is the correct scale; the nondimensional equation then reads \(0=-\nabla^* p^*+\nabla^{*2}\mathbf{u}^*\) and \(\mathrm{Re}\) multiplies the inertial terms instead. The physics is identical but the placement of \(\mathrm{Re}\) differs.

Body forces are either absent or absorbed into a modified pressure.A conservative body force \(\rho\mathbf{g}=-\rho\nabla\Phi\) folds into \(p\to p+\rho\Phi\). If it cannot be absorbed (free surfaces, buoyancy), a Froude or Grashof number appears alongside \(\mathrm{Re}\).

Derivation

\[ \rho\left(\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u}\cdot\nabla)\mathbf{u}\right) = -\nabla p + \mu\nabla^2\mathbf{u} \]

Incompressible Navier–Stokes with the Newtonian constitutive result substituted; body forces absorbed into \(p\). A

\[ \mathbf{x}=L\,\mathbf{x}^*,\quad \mathbf{u}=U\,\mathbf{u}^*,\quad t=\frac{L}{U}\,t^*,\quad p=\rho U^2\,p^*,\quad \nabla=\frac{1}{L}\nabla^* \]

Introduce dimensionless starred variables through characteristic scales. The convective time \(L/U\) is chosen so that unsteady and convective terms share the same coefficient. A

\[ \frac{\partial \mathbf{u}}{\partial t} = \frac{U}{L/U}\frac{\partial \mathbf{u}^*}{\partial t^*} = \frac{U^2}{L}\frac{\partial \mathbf{u}^*}{\partial t^*} \]

Chain rule on the unsteady term: \(\mathbf{u}\) scales as \(U\), \(t\) as \(L/U\). A

\[ (\mathbf{u}\cdot\nabla)\mathbf{u} = U\,\mathbf{u}^*\cdot\frac{1}{L}\nabla^*\,(U\mathbf{u}^*) = \frac{U^2}{L}(\mathbf{u}^*\cdot\nabla^*)\mathbf{u}^* \]

Convective term carries two factors of \(U\) and one inverse length; identical prefactor \(U^2/L\) to step 3, confirming the time-scale choice. A

\[ \nabla p = \frac{1}{L}\nabla^*(\rho U^2 p^*) = \frac{\rho U^2}{L}\nabla^* p^*, \qquad \mu\nabla^2\mathbf{u} = \mu\frac{1}{L^2}\nabla^{*2}(U\mathbf{u}^*) = \frac{\mu U}{L^2}\nabla^{*2}\mathbf{u}^* \]

Pressure gradient uses \(p_c=\rho U^2\); the viscous term carries \(\nabla^2\sim L^{-2}\). A

\[ \rho\,\frac{U^2}{L}\left(\frac{\partial \mathbf{u}^*}{\partial t^*} + (\mathbf{u}^*\cdot\nabla^*)\mathbf{u}^*\right) = -\frac{\rho U^2}{L}\nabla^* p^* + \frac{\mu U}{L^2}\nabla^{*2}\mathbf{u}^* \]

Substitute steps 3–5 into step 1. Every term is now a dimensional prefactor times an \(O(1)\) starred quantity. B

\[ \frac{\partial \mathbf{u}^*}{\partial t^*} + (\mathbf{u}^*\cdot\nabla^*)\mathbf{u}^* = -\nabla^* p^* + \frac{\mu U/L^2}{\rho U^2/L}\,\nabla^{*2}\mathbf{u}^* \]

Divide through by the inertial prefactor \(\rho U^2/L\). Legal because it is a nonzero constant. Inertial and pressure terms reduce to coefficient unity. B

\[ \frac{\mu U/L^2}{\rho U^2/L} = \frac{\mu U}{L^2}\cdot\frac{L}{\rho U^2} = \frac{\mu}{\rho U L} = \frac{\nu}{UL} \equiv \frac{1}{\mathrm{Re}} \]

Simplify the viscous coefficient; \(\nu=\mu/\rho\) is the kinematic viscosity. Define \(\mathrm{Re}=\rho UL/\mu\). B

\[ \boxed{\;\frac{\partial \mathbf{u}^*}{\partial t^*} + (\mathbf{u}^*\cdot\nabla^*)\mathbf{u}^* = -\nabla^* p^* + \frac{1}{\mathrm{Re}}\,\nabla^{*2}\mathbf{u}^*\;} \]

The dimensionless momentum equation. Only one dimensionless coefficient survives — the whole flow depends on the dimensional data only through \(\mathrm{Re}\) and the boundary geometry. B

\[ \{\rho,\,U,\,L,\,\mu\}\ \text{have dimensions}\ \{ML^{-3},\,LT^{-1},\,L,\,ML^{-1}T^{-1}\};\quad n=4,\ k=\operatorname{rank}=3\ \Rightarrow\ n-k=1 \]

Buckingham Pi theorem: four dimensional parameters spanning three independent dimensions \((M,L,T)\) yield exactly \(4-3=1\) independent dimensionless group. The dimensional matrix has rank 3, so the count is exact. C

\[ \Pi = \rho^{a}U^{b}L^{c}\mu^{d}:\quad M:\,a+d=0,\ \ L:\,-3a+b+c-d=0,\ \ T:\,-b-d=0 \]

Seek exponents making \(\Pi\) dimensionless. Solving with \(d=-1\): \(a=1,\ b=1,\ c=1\), giving \(\Pi=\rho U L/\mu=\mathrm{Re}\). Any dimensionless group is a power of this unique \(\Pi\). C

Result

\[ \mathrm{Re} = \frac{\rho U L}{\mu} = \frac{UL}{\nu} = \frac{\text{inertial force}}{\text{viscous force}} \]

Reading. The Reynolds number is the ratio of the characteristic inertial momentum flux \(\rho U^2/L\) to the characteristic viscous stress gradient \(\mu U/L^2\). When \(\mathrm{Re}\gg1\) inertia dominates and viscosity acts only in thin layers; when \(\mathrm{Re}\ll1\) viscosity dominates and inertia is negligible. Two geometrically similar flows with the same \(\mathrm{Re}\) obey the identical dimensionless equation with identical dimensionless boundary conditions, so their nondimensional fields \(\mathbf{u}^*,p^*\) coincide — this is dynamic similarity.

Units check. \([\rho U L/\mu] = (\mathrm{kg\,m^{-3}})(\mathrm{m\,s^{-1}})(\mathrm{m})/(\mathrm{kg\,m^{-1}s^{-1}}) = \mathrm{kg\,m^{-1}s^{-1}}/\mathrm{kg\,m^{-1}s^{-1}} = 1\). Dimensionless, as required.

Limiting cases

\(\mathrm{Re}\to0\) (creeping/Stokes flow): the \(1/\mathrm{Re}\) viscous term dominates; multiplying through by \(\mathrm{Re}\) gives \(0=-\nabla^*p^*+\nabla^{*2}\mathbf{u}^*\). Inertia drops out and the flow is reversible and linear.
\(\mathrm{Re}\to\infty\) (ideal/Euler limit): the \(1/\mathrm{Re}\) term formally vanishes, leaving the inviscid Euler equation. This is a singular limit: viscosity is retained in boundary layers of thickness \(\sim L/\sqrt{\mathrm{Re}}\), so d'Alembert's paradox does not actually apply.
\(\mathrm{Re}=O(1)\): inertial and viscous terms are comparable; no term may be dropped and the full nonlinear equation must be solved.
Fixed geometry, varying \(U\): \(\mathrm{Re}\propto U\), so increasing speed alone moves a flow from laminar toward turbulent along a single control axis.

Breaks when

Compressible flow (\(\mathrm{Ma}\gtrsim0.3\)): density is no longer constant, the continuity constraint changes, and a Mach number \(\mathrm{Ma}=U/c\) becomes a second independent governing parameter. Matching \(\mathrm{Re}\) alone no longer guarantees similarity — this is why high-speed wind-tunnel testing must match both \(\mathrm{Re}\) and \(\mathrm{Ma}\).
Non-Newtonian rheology: for viscoelastic or shear-dependent fluids the viscous term is not \(\mu\nabla^2\mathbf{u}\); nondimensionalisation produces extra groups (Weissenberg, Deborah, power-law index) and \(\mathrm{Re}\) is insufficient to characterise the flow.
Multiple length scales or free surfaces: when roughness, gap ratios, or gravity-driven interfaces impose additional scales, aspect-ratio groups or a Froude number \(\mathrm{Fr}=U/\sqrt{gL}\) survive; ship-hull testing cannot simultaneously match \(\mathrm{Re}\) and \(\mathrm{Fr}\) at model scale, forcing empirical corrections.
Continuum breakdown (Knudsen number \(\mathrm{Kn}\gtrsim0.1\)): in rarefied gases or microchannels the mean free path is comparable to \(L\); the Navier–Stokes equation itself fails, so no nondimensionalisation of it can be valid.

Failure modes

Wrong length scale. Using an arbitrary length (e.g. total pipe length instead of hydraulic diameter, or chord instead of momentum thickness) gives a numerically different \(\mathrm{Re}\); comparisons across sources are meaningless unless the reference scale is stated.
Confusing \(\mu\) and \(\nu\). Dividing by dynamic viscosity where kinematic viscosity is required (or vice versa) mis-scales \(\mathrm{Re}\) by a factor of \(\rho\) — a common numerical error, especially in gases.
Believing \(\mathrm{Re}\to\infty\) recovers inviscid flow uniformly. The high-\(\mathrm{Re}\) limit is singular; dropping the viscous term everywhere loses the boundary layer and predicts zero drag (d'Alembert's paradox).
Treating the laminar–turbulent transition \(\mathrm{Re}\) as a universal constant. The critical \(\mathrm{Re}\) depends on geometry and disturbance level (pipe \(\sim2300\), flat-plate boundary layer \(\sim5\times10^5\)); it is not a property of the fluid.
Assuming equal \(\mathrm{Re}\) alone gives similarity. Similarity also requires geometric similarity and matching of every other relevant group (\(\mathrm{Ma},\mathrm{Fr},\) roughness ratio). Matching one group while ignoring others is a frequent modelling error.

Discussion

The power of nondimensionalisation is that it converts a problem with four dimensional inputs into a problem with one dimensionless input. Before scaling, one might imagine that doubling the velocity, halving the size, or changing the fluid each produces qualitatively new behaviour. After scaling, all of these are seen to act only through their combined effect on \(\mathrm{Re}\): a flow at \(U=1\,\mathrm{m/s}\) around a \(1\,\mathrm{m}\) cylinder in water behaves identically (in dimensionless terms) to a flow at \(15\,\mathrm{m/s}\) around a \(1\,\mathrm{m}\) cylinder in air, because both have \(\mathrm{Re}\approx10^6\).

Physically, \(\mathrm{Re}\) compares two rates of momentum transport. Inertia advects momentum at rate \(\rho U^2/L\) per unit volume; viscosity diffuses it at rate \(\mu U/L^2\). Their ratio, \(\rho U L/\mu\), can equivalently be read as the ratio of the viscous diffusion time \(L^2/\nu\) to the convective time \(L/U\). Large \(\mathrm{Re}\) means momentum is carried downstream long before it can diffuse across the flow, which is why high-\(\mathrm{Re}\) flows organise into thin shear layers, wakes and eventually turbulence.

The Buckingham Pi count is what guarantees uniqueness. With three independent dimensions and four parameters, exactly one dimensionless group can be formed, and the derivation identifies it as \(\mathrm{Re}\) rather than leaving it as an abstract \(\Pi\). This is the bridge between the symmetry/dimensional-analysis viewpoint and the force-balance viewpoint: the same number arises whether one scales the differential equation or merely counts dimensions, because both reflect the same underlying scale invariance of the equations under \((\mathbf{x},t,\mathbf{u},p)\to(L\mathbf{x}^*,\,(L/U)t^*,\,U\mathbf{u}^*,\,\rho U^2 p^*)\).

The high-\(\mathrm{Re}\) limit is mathematically a singular perturbation: \(1/\mathrm{Re}\) multiplies the highest-order derivative \(\nabla^{*2}\mathbf{u}^*\). Setting it to zero reduces the differential order and cannot satisfy the no-slip condition, so a boundary layer of nondimensional thickness \(\mathrm{Re}^{-1/2}\) forms in which the neglected term is restored to leading order. This is the origin of Prandtl's boundary-layer theory, and it explains why the inviscid outer solution and the viscous inner solution must be matched asymptotically rather than one simply replacing the other. The nonuniformity of the limit is also, ultimately, why turbulence — a genuinely multiscale phenomenon — is not captured by naive large-\(\mathrm{Re}\) expansions.

Common misconceptions. \(\mathrm{Re}\) is not a material property of the fluid (it depends on \(U\) and \(L\)); a "high" \(\mathrm{Re}\) is not synonymous with "turbulent" (transition depends on geometry and disturbances); and a low \(\mathrm{Re}\) does not mean the fluid is viscous in an absolute sense — water at small scales (a swimming bacterium) lives at \(\mathrm{Re}\ll1\) despite its modest viscosity.

Worked examples

\[ \text{Water at }20^\circ\mathrm{C}:\ \rho=998\ \mathrm{kg\,m^{-3}},\ \mu=1.00\times10^{-3}\ \mathrm{Pa\,s};\quad D=0.05\ \mathrm{m},\ U=2\ \mathrm{m\,s^{-1}} \]

Pipe flow. Use the diameter \(D\) as the length scale \(L\) and the mean velocity as \(U\). A

\[ \mathrm{Re}=\frac{\rho U D}{\mu}=\frac{(998)(2)(0.05)}{1.00\times10^{-3}} \]

Substitute numbers into \(\mathrm{Re}=\rho U L/\mu\) after writing the symbolic form. A

\[ \mathrm{Re}=\frac{99.8}{1.00\times10^{-3}}=9.98\times10^{4} \]

Arithmetic. A

\[ \mathrm{Re}\approx1.0\times10^{5} \]

Reading. Well above the pipe transition value \(\sim2300\), so the flow is fully turbulent. Any geometrically identical pipe flow with \(\mathrm{Re}=1.0\times10^5\) — whatever the fluid or diameter — has the same dimensionless velocity profile and friction factor.

Units check. \((\mathrm{kg\,m^{-3}})(\mathrm{m\,s^{-1}})(\mathrm{m})/(\mathrm{Pa\,s})=1\), since \(\mathrm{Pa\,s}=\mathrm{kg\,m^{-1}s^{-1}}\).

\[ \text{Model test: full-scale wing chord }L_1=2\ \mathrm{m},\ U_1=50\ \mathrm{m\,s^{-1}}\ \text{in air};\ \ \text{model chord }L_2=0.4\ \mathrm{m} \]

Dynamic similarity requires \(\mathrm{Re}_2=\mathrm{Re}_1\). Same fluid (air), so \(\nu\) cancels. B

\[ \frac{U_1 L_1}{\nu}=\frac{U_2 L_2}{\nu}\ \Rightarrow\ U_2 = U_1\frac{L_1}{L_2} \]

Set the Reynolds numbers equal and solve symbolically for the required tunnel speed. B

\[ U_2 = 50\times\frac{2}{0.4} = 50\times5 = 250\ \mathrm{m\,s^{-1}} \]

Substitute numbers. B

\[ U_2 = 250\ \mathrm{m\,s^{-1}} \]

Reading. To match \(\mathrm{Re}\) on a \(1/5\)-scale model in the same air, the tunnel must run five times faster. But \(250\ \mathrm{m\,s^{-1}}\) gives \(\mathrm{Ma}\approx0.73\), so compressibility now matters: this illustrates exactly why \(\mathrm{Re}\)-matching alone breaks down and pressurised or heavy-gas tunnels are used to keep \(\mathrm{Ma}\) low while raising \(\rho\).

Units check. \(U_2=[\mathrm{m\,s^{-1}}]\cdot[\mathrm{m}]/[\mathrm{m}]=\mathrm{m\,s^{-1}}\). Consistent.

Problems

A microorganism of length \(L=10\ \mu\mathrm{m}\) swims at \(U=30\ \mu\mathrm{m\,s^{-1}}\) in water (\(\nu=1.0\times10^{-6}\ \mathrm{m^2 s^{-1}}\)). Find \(\mathrm{Re}\) and state which term of the momentum equation dominates.
Solution
\(\mathrm{Re}=UL/\nu=(30\times10^{-6})(10\times10^{-6})/(1.0\times10^{-6}) = 3.0\times10^{-10}/1.0\times10^{-6}=3.0\times10^{-4}\). Since \(\mathrm{Re}\ll1\), the \(1/\mathrm{Re}\) viscous term dominates: this is creeping (Stokes) flow, inertia is negligible, and swimming is time-reversible (the "scallop theorem" regime).
Glycerine has \(\rho=1260\ \mathrm{kg\,m^{-3}}\) and \(\mu=1.5\ \mathrm{Pa\,s}\). A sphere of diameter \(D=0.02\ \mathrm{m}\) falls at \(U=0.1\ \mathrm{m\,s^{-1}}\). Compute \(\mathrm{Re}\).
Solution
\(\mathrm{Re}=\rho U D/\mu=(1260)(0.1)(0.02)/1.5 = 2.52/1.5 = 1.68\). Order unity: neither inertia nor viscosity is negligible, so Stokes drag (\(\mathrm{Re}\ll1\)) is only marginally valid and a correction is needed.
Show, starting from \(\mathrm{Re}=\rho U L/\mu\), that \(\mathrm{Re}\) equals the ratio of the viscous diffusion time \(t_\nu=L^2/\nu\) to the convective time \(t_c=L/U\).
Solution
\(t_\nu/t_c=(L^2/\nu)/(L/U)=(L^2/\nu)(U/L)=UL/\nu=\mathrm{Re}\). Thus a large \(\mathrm{Re}\) means viscous diffusion across the flow is slow compared with convection along it, consistent with the reading of \(\mathrm{Re}\) as an inertia-to-viscosity ratio.
A ship model is built at \(1/25\) scale and tested in a towing tank in water. If the full-scale ship travels at \(U_1=10\ \mathrm{m\,s^{-1}}\), what model speed matches \(\mathrm{Re}\)? Comment on whether \(\mathrm{Fr}\) can be matched simultaneously.
Solution
\(\mathrm{Re}\)-matching: \(U_2=U_1 L_1/L_2 = 10\times25 = 250\ \mathrm{m\,s^{-1}}\), which is physically impossible in a tank. Froude matching (relevant for wave drag) needs \(U_2=U_1\sqrt{L_2/L_1}=10/\sqrt{25}=2\ \mathrm{m\,s^{-1}}\). The two requirements conflict, so ship testing matches \(\mathrm{Fr}\) and corrects the viscous (\(\mathrm{Re}\)-dependent) drag empirically. This is the classic limitation of single-parameter similarity.
Air at \(\rho=1.2\ \mathrm{kg\,m^{-3}}\), \(\mu=1.8\times10^{-5}\ \mathrm{Pa\,s}\) flows over a flat plate at \(U=15\ \mathrm{m\,s^{-1}}\). At what distance \(x\) from the leading edge does \(\mathrm{Re}_x=\rho U x/\mu\) reach the transition value \(5\times10^5\)? Estimate the boundary-layer thickness there using \(\delta\sim x/\sqrt{\mathrm{Re}_x}\).
Solution
Solve \(\mathrm{Re}_x=\rho U x/\mu=5\times10^5\) for \(x\): \(x=\mathrm{Re}_x\,\mu/(\rho U)=(5\times10^5)(1.8\times10^{-5})/[(1.2)(15)] = 9.0/18 = 0.50\ \mathrm{m}\). Boundary-layer thickness: \(\delta\sim x/\sqrt{\mathrm{Re}_x}=0.50/\sqrt{5\times10^5}=0.50/707\approx7.1\times10^{-4}\ \mathrm{m}\approx0.71\ \mathrm{mm}\). The thin layer (\(\delta/x\sim1.4\times10^{-3}\)) confirms the singular nature of the high-\(\mathrm{Re}\) limit.