12.2 Dimensionless Numbers and Flow Regimes

12.2 Dimensionless Numbers and Flow Regimes

The NCEES Fluid Mechanics specification lists "Dimensionless numbers (e.g., Reynolds number, Froude number, Mach number)" and "Laminar and turbulent flow" as two of its twelve sub-topics. Every dimensionless group is a ratio of two competing forces, and knowing which forces a group compares tells you immediately what it predicts.

The Groups and What They Compare

Reynolds Number and the Flow Regimes

$Re = \frac{\rho V D}{\mu} = \frac{VD}{\nu}$

where $\nu = \mu/\rho$ is the kinematic viscosity (m²/s) and $\mu$ is the dynamic (absolute) viscosity (Pa·s).

For flow in a circular pipe:

The critical value depends on the geometry, and using the pipe value everywhere is a classic error:

• Circular pipe: $Re_{\text{crit}} \approx 2{,}100$ (based on diameter)
• Flat plate boundary layer: $Re_{\text{crit}} \approx 5\times10^5$ (based on distance from the leading edge)
• Flow around a sphere: transition near $Re \approx 2\times10^5$

For non-circular ducts, replace $D$ with the hydraulic diameter:

$D_h = \frac{4A}{P}$

where $A$ is the flow area and $P$ the wetted perimeter. For a circular pipe running full, $D_h = 4(\pi D^2/4)/(\pi D) = D$ ✓, confirming the definition.

Velocity Profiles: The Consequence of Regime

Laminar flow in a pipe has $V_{\max} = 2V_{\text{avg}}$ exactly. Turbulent mixing flattens the profile because eddies transport momentum across the section, so the centerline velocity is only about 20% above the average. This matters for pitot-tube measurement: a single centerline reading must be corrected to obtain the average velocity, and the correction differs by a factor of nearly two between regimes.

Worked Example: Regime Determination

Water at 20 °C ($\rho = 998$ kg/m³, $\mu = 1.002\times10^{-3}$ Pa·s) flows at 0.08 m/s in a 40 mm pipe. Determine the regime, then find the velocity at which transition begins.

$Re = \frac{\rho V D}{\mu} = \frac{998(0.08)(0.040)}{1.002\times10^{-3}} = \frac{3.194}{1.002\times10^{-3}} = 3{,}187$

Transitional — an awkward regime where neither the laminar formula nor the Moody chart is reliable, and one that designers avoid.

Velocity at the laminar limit $Re = 2{,}100$:

$V = \frac{Re\,\mu}{\rho D} = \frac{2{,}100(1.002\times10^{-3})}{998(0.040)} = \frac{2.104}{39.92} = 0.0527\ \text{m/s}$

So laminar flow requires $V < 5.3$ cm/s in a 40 mm water line. Almost all practical water flow is turbulent, because the required velocity is so low — a useful sanity check when a problem's answer implies otherwise.

Note the strong temperature sensitivity: heating the water to 60 °C drops $\mu$ to about $4.67\times10^{-4}$ Pa·s, more than doubling $Re$ at the same velocity. Viscosity of a liquid falls with temperature while that of a gas rises.

Froude Number and Open-Channel Classification

$Fr = \frac{V}{\sqrt{gy}} \qquad\text{(rectangular channel, depth } y)$

The Froude number compares the flow velocity to the speed at which a shallow-water surface wave travels, $c = \sqrt{gy}$. That comparison determines whether a disturbance can propagate upstream:

Why the control section matters: in subcritical flow, downstream conditions propagate upstream and set the depth — which is why a weir or a downstream gate controls an upstream water surface. In supercritical flow they cannot, so upstream conditions govern. A hydraulic jump — an abrupt, highly turbulent rise in depth with substantial energy loss — is the only way flow transitions from supercritical to subcritical, and stilling basins are designed to force it deliberately.

Critical depth for a rectangular channel of width $b$ carrying discharge $Q$, with $q = Q/b$:

$y_c = \left(\frac{q^2}{g}\right)^{1/3}$

At critical depth the specific energy is minimized:

$E_{\min} = \frac{3}{2}y_c$

Mach Number and Compressibility

$Ma = \frac{V}{c}, \qquad c = \sqrt{kRT}$

with $k = c_p/c_v$ (1.4 for air) and $R$ the specific gas constant (287 J/kg·K for air).

The $Ma = 0.3$ threshold is the practical dividing line for this exam. For air at 20 °C, $c = \sqrt{1.4(287)(293)} = 343$ m/s, so $Ma = 0.3$ corresponds to about 103 m/s (230 mph). Below that, treating air as incompressible introduces less than 5% error — which is why ventilation, HVAC, and most fan problems use incompressible relations even though air is a gas.

Dynamic Similarity in Model Testing

Two flows are dynamically similar when they are geometrically similar and the governing dimensionless group matches. The essential judgment is which group to match:

$Re_m = Re_p \;\Rightarrow\; \frac{V_m L_m}{\nu_m} = \frac{V_p L_p}{\nu_p} \qquad\qquad Fr_m = Fr_p \;\Rightarrow\; \frac{V_m}{\sqrt{gL_m}} = \frac{V_p}{\sqrt{gL_p}}$

The classic conflict: a ship model must match Froude for wave drag, giving $V_m = V_p\sqrt{L_m/L_p}$ — a slower model. But matching Reynolds for viscous drag requires $V_m = V_p(L_p/L_m)$ with the same fluid — a much faster model. Both cannot be satisfied at once in the same fluid. Naval practice matches Froude and corrects the viscous component analytically. If an exam item asks which to match, identify the dominant force: a free surface means Froude.

Worked Example: Spillway Model Scaling

A spillway model is built at 1:25 scale. The prototype discharge is 900 m³/s. Find the model velocity ratio and the required model discharge.

Free-surface, gravity-driven flow → match Froude number:

$\frac{V_m}{\sqrt{gL_m}} = \frac{V_p}{\sqrt{gL_p}} \;\Rightarrow\; \frac{V_m}{V_p} = \sqrt{\frac{L_m}{L_p}} = \sqrt{\frac{1}{25}} = \frac{1}{5}$

The model runs at one-fifth the prototype velocity. Discharge scales as velocity × area, and area scales as length squared:

$\frac{Q_m}{Q_p} = \frac{V_m}{V_p}\times\frac{A_m}{A_p} = \frac{1}{5}\times\frac{1}{625} = \frac{1}{3{,}125} = \left(\frac{1}{25}\right)^{5/2}$

$Q_m = \frac{900}{3{,}125} = \boxed{0.288\ \text{m}^3/\text{s}}$

The $L^{5/2}$ discharge scaling is the signature of Froude similarity and is worth recognizing directly. Note also that the model's Reynolds number is only $\frac{1}{5}\times\frac{1}{25} = \frac{1}{125}$ of the prototype's — so viscous effects are badly exaggerated in the model, which is exactly why models must be built large enough to remain turbulent.