Exercises — Reynolds number Re = ρvL - μ — laminar vs turbulent criterion

WHAT: We read off the fraction. $\mu$ (viscosity) is the denominator; $\rho$, $v$, $L$ are the numerator. WHY it matters (physical): Viscosity is neighbour-to-neighbour friction — a fast layer literally tugs a slow layer along and gets tugged back. That tugging spreads out velocity differences before they can curl into eddies. So a large $\mu$ physically irons flat any budding swirl; mathematically that shows up as $\mu$ in the denominator, driving $Re$ down toward the laminar picture. Viscosity is the peacemaker not by convention but because shear friction physically dissipates the energy a vortex would need. Answer: $\mu$ is in the denominator; it represents the calming/smoothing (viscous) force.

Compare to the table: $350<2000$, so the flow is laminar (smooth, layered) — the leftmost picture above. No calculation needed — just placement on the number line.

WHAT: Substitute SI units into $\rho vL/\mu$. $\frac{[{\text{kg/m}}^3][\text{m/s}][\text{m}]}{[\text{Pa⋅s}]}=\frac{\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-1}}{\text{Pa⋅s}}$ Now $\text{Pa}={\text{N/m}}^2=\text{kg}{\text{m}}^{-1}{\text{s}}^{-2}$, so $\text{Pa⋅s}=\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$. $\frac{\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}}{\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}}=1\Rightarrow \text{dimensionless.}$ Answer: $Re$ has no units.

WHAT: Plug SI numbers straight in. $Re=\frac{1000\times 0.30\times 0.010}{1.0\times 10^{-3}}=\frac{3.0}{1.0\times 10^{-3}}=3000$ WHY it comes out unitless — the deeper point: we aren't just lucky that units cancel. $Re$ is a non-dimensional ratio of the inertial push $\rho v^2{L}^2$ to the viscous drag $\mu vL$. When we feed in SI numbers the units must cancel because we are literally dividing a force by a force — and the surviving pure number is the physical answer: "$3000$ times more inertia than friction." Working in SI just lets us read that ratio off directly. Answer: $Re=3000$, which lies in $2000$–$4000$ ⇒ transitional (the middle picture: intermittent wobbles and the first vortices).

$Re=\frac{1.2\times 15\times 2.0}{1.8\times 10^{-5}}=\frac{36}{1.8\times 10^{-5}}=2.0\times 10^6$ WHY: Long body + fast air + tiny viscosity all push $Re$ enormously high. Such a large number means inertia utterly dominates ⇒ strongly turbulent boundary layer. Answer: $Re=2.0\times 10^6$.

WHAT: $\nu =\mu /\rho$. $\nu =\frac{1.0\times 10^{-3}}{1000}=1.0\times 10^{-6}{\text{m}}^2/\text{s}$ $Re=\frac{vL}{\nu }=\frac{0.30\times 0.010}{1.0\times 10^{-6}}=\frac{3.0\times 10^{-3}}{1.0\times 10^{-6}}=3000✓$ WHY this check matters (pedagogy): getting the same $3000$ two different ways is not busywork — it proves the two formulas are the same physics wearing different clothes. The form $\rho vL/\mu$ keeps density and viscosity separate; the form $vL/\nu$ has already fused them into one property $\nu$ (how fast momentum diffuses through the fluid). Whenever a problem hands you $\nu$ instead of $\rho$ and $\mu$, you now know you can use it directly without hunting for the missing density. Consistency checks like this are how you catch a slipped power of ten before it wrecks a whole answer. Answer: $Re=3000$, matching Exercise 2.1.

WHAT: Rearrange $Re=\rho vL/\mu$ for $v$. $v_c=\frac{Re\mu }{\rho L}=\frac{2000\times 1.0\times 10^{-3}}{1000\times 0.010}=\frac{2.0}{10}=0.20\text{m/s}$ WHY: Below $0.20\text{m/s}$ friction can still smear out every disturbance faster than inertia can grow it; exactly at $v_c$ the two rates balance, and above it inertia wins and the first vortices survive. So $v_c$ is the tipping speed where the leftmost picture starts turning into the middle one. Answer: $v_c=0.20\text{m/s}$.

WHAT: Only $L$ changes; $\rho ,v,\mu$ fixed. Since $Re\propto L$, $\frac{R{e}_B}{R{e}_A}=\frac{L_B}{L_A}=3.$ WHY it looks like this (physical): widening the pipe gives inertia a bigger cross-section to carry momentum through while the shearing friction still only acts over the wall region — so the inertia-to-friction balance tips upward. The figure below plots $Re$ against diameter: it is a straight line through the origin, so the two marked points A and B sit at heights in exact $1:3$ proportion, and the same speed that was laminar in the narrow pipe can be turbulent in the wide one.

Figure: $Re$ versus pipe diameter at fixed $\rho ,v,\mu$ (water, $v=0.30\text{m/s}$). The line is straight because $Re\propto L$; pipe B (three times A's diameter) sits at three times A's height. Dashed lines mark the laminar limit ($Re=2000$) and turbulent onset ($Re=4000$).

Answer: $Re$ becomes $3\times$ larger.

WHAT: With $v,L$ equal, $Re\propto \rho /\mu$. $\frac{R{e}_{\text{water}}}{R{e}_{\text{honey}}}=\frac{{\rho }_w/{\mu }_w}{{\rho }_h/{\mu }_h}=\frac{1000/10^{-3}}{1400/10}=\frac{1.0\times 10^6}{140}\approx 7143$ WHY: Honey's molecules cling to each other far harder, so a swirl started in honey is smothered almost instantly — its momentum diffuses sideways before it can roll into an eddy. That physical stickiness is the huge $\mu$ in the denominator; it swamps the modest density difference, leaving water's $Re$ thousands of times bigger. Answer: $R{e}_{\text{water}}/R{e}_{\text{honey}}\approx 7.1\times 10^3$ (so honey is overwhelmingly more laminar).

WHAT (a): First compute $Q$. $Q=\frac{\pi R^4\Delta P}{8\mu \ell }=\frac{\pi (5.0\times 10^{-3}{)}^4(8.0)}{8(1.0\times 10^{-3})(1.0)}$ Step — the fourth power. $R^4=(5.0\times 10^{-3}{)}^4$. Raise the mantissa and the exponent separately: $5.0^4=625$ and $(10^{-3}{)}^4=10^{-12}$, so $R^4=625\times 10^{-12}=6.25\times 10^{-10}{\text{m}}^4$. (The pipe radius enters to the fourth power — this is why a slightly wider pipe carries dramatically more flow.) Step — combine the top. Numerator $=\pi \times 6.25\times 10^{-10}\times 8.0=\pi \times 5.0\times 10^{-9}$. Step — divide. Denominator $=8\times 10^{-3}$, so $Q=\frac{\pi \times 5.0\times 10^{-9}}{8.0\times 10^{-3}}=\pi \times 6.25\times 10^{-7}.$ Step — multiply out $\pi$. Multiplying the pure number $6.25\times 10^{-7}$ by $\pi \approx 3.1416$ gives $6.25\times 3.1416\approx 19.63$, i.e. $Q\approx 19.63\times 10^{-7}=1.963\times 10^{-6}{\text{m}}^3/\text{s}$. (We keep $\pi$ symbolic until the last step so rounding never piles up.) WHAT (b): Average speed, then $Re$ with $L=D=0.010\text{m}$: $v=\frac{Q}{\pi R^2}=\frac{1.963\times 10^{-6}}{\pi (5.0\times 10^{-3}{)}^2}=\frac{1.963\times 10^{-6}}{\pi \times 2.5\times 10^{-5}}=\frac{1.963\times 10^{-6}}{7.854\times 10^{-5}}\approx 0.025\text{m/s}$ $Re=\frac{\rho vD}{\mu }=\frac{1000\times 0.025\times 0.010}{1.0\times 10^{-3}}=\frac{0.25}{1.0\times 10^{-3}}=250$ WHY the check matters: Poiseuille's law is only valid in laminar flow (it assumes those tidy parallel layers from the left-hand picture). Here $Re=250<2000$ ⇒ laminar ⇒ the law was legitimately used. Had $Re$ come out $>2000$, our whole $Q$ calculation would rest on a broken assumption. Answer: $v\approx 0.025\text{m/s}$, $Re\approx 250$ — laminar, so Poiseuille's law is valid.

WHAT: Use particle diameter $L=2r=2.0\times 10^{-4}\text{m}$. $R{e}_p=\frac{1000\times 0.02\times 2.0\times 10^{-4}}{1.0\times 10^{-3}}=\frac{4.0\times 10^{-3}}{1.0\times 10^{-3}}=4.0$ WHY: Stokes assumes viscous drag dominates; that holds only for $R{e}_p\lesssim 1$. Here $R{e}_p=4>1$, so inertial effects around the grain are not negligible — Stokes' law would over- or mis-predict the drag. Answer: $R{e}_p=4.0$; Stokes' law is not strictly valid (it's already outside the safe $R{e}_p\lesssim 1$ window).

WHAT: Solve for $v$ at $Re=1.0\times 10^5$. $v=\frac{Re\mu }{\rho D}=\frac{1.0\times 10^5\times 1.0\times 10^{-3}}{1000\times 0.010}=\frac{100}{10}=10\text{m/s}$ WHY the rule bends: $R{e}_c\approx 2000$ is not a law of nature; it is the point where the disturbances that actually exist get amplified. In an ultra-smooth, vibration-free pipe there is almost nothing to amplify, so flow stays laminar to far higher $Re$. Geometry and disturbance level set the threshold, not the fluid. Answer: $v=10\text{m/s}$; the $2000$ figure is a practical guideline for ordinary rough pipes, not a hard ceiling.

WHAT: Set $R{e}_1=R{e}_2$ using $Re=vL/\nu$. $\frac{v_1{L}_1}{{\nu }_{\text{air}}}=\frac{v_2{L}_2}{{\nu }_{\text{water}}}\Rightarrow v_2=v_1\frac{L_1}{L_2}\frac{{\nu }_{\text{water}}}{{\nu }_{\text{air}}}$ Plug in, one factor at a time: $\frac{L_1}{L_2}=\frac{2.0}{0.20}=10$ and $\frac{{\nu }_{\text{water}}}{{\nu }_{\text{air}}}=\frac{1.0\times 10^{-6}}{1.5\times 10^{-5}}=0.0667$. Then $v_2=50\times 10\times 0.0667=50\times 0.667=33.3\text{m/s}.$ WHY this works: if two flows share the same $Re$, their inertia-to-viscosity balance is identical, so the shape of the flow (where it separates, where it turns turbulent) is the same — that's why wind-tunnel and water-tunnel models are trustworthy. Answer: $v_2\approx 33.3\text{m/s}$.

WHAT: Stokes prediction: $C_d^{\text{Stokes}}=\frac{24}{R{e}_p}=\frac{24}{4.0}=6.0$ Percent error against measured $6.9$: $\text{error}=\frac{|6.0-6.9|}{6.9}\times 100\%=\frac{0.9}{6.9}\times 100\%\approx 13\%$ WHY the gap: at $R{e}_p=4$ we are already past the pure-Stokes regime; inertial effects add extra drag the $24/Re$ formula ignores, so it underestimates $C_d$ by about $13\%$. This is the same warning as Exercise 4.2, now quantified. Answer: Stokes predicts $C_d=6.0$; error $\approx 13\%$ low.