Transport phenomena Study Guide

📖 Core Concepts Transport Phenomena – Study of how mass, energy, charge, momentum, and angular momentum move between a system and its surroundings. Conservation Principle – For any transported quantity, the sum of all fluxes entering and leaving a control volume must be zero (continuity). Constitutive Equation – Relates a flux to its driving force (e.g., Fourier’s law, Fick’s law, Newton’s law of viscosity). Driving Forces – Gradients that “push” transport: temperature gradient (heat), concentration gradient (mass), velocity gradient (momentum), pressure or chemical‑potential gradients. Analogy – Heat, mass, and momentum transfer share the same mathematical form: flux = diffusivity × gradient. This permits swapping dimensionless groups (Nu ↔ Sh, Pr ↔ Sc). Dimensionless Numbers – Collapse geometry, fluid properties, and flow regime into single parameters: Reynolds (Re) – inertial vs. viscous forces. Prandtl (Pr) – momentum vs. thermal diffusivity ($\nu/k$). Schmidt (Sc) – momentum vs. mass diffusivity ($\nu/D$). Nusselt (Nu) – convective vs. conductive heat transfer. Sherwood (Sh) – convective vs. diffusive mass transfer. Grashof (Gr) – buoyancy vs. viscous forces (natural convection). 📌 Must Remember Newton’s law of viscosity: $\tau{zx}= \mu \dfrac{\partial v{x}}{\partial z}= \rho \nu \dfrac{\partial v{x}}{\partial z}$ Fick’s first law: $J = -D\,\dfrac{\partial C}{\partial x}$ Fourier’s law (conduction): $\mathbf{q}= -k\,\nabla T$ Convective heat transfer: $Q = h\,A\,\Delta T$ Chilton–Colburn J‑factor (turbulent): $J = \dfrac{Nu}{Re\,Pr^{2/3}} = \dfrac{Sh}{Re\,Sc^{2/3}}$ Reynolds analogy (turbulent): $Pr \approx Sc \approx 1$ → turbulent diffusivities for heat, mass, momentum are equal. General correlation form (forced convection, circular tube): $Nu = C\,Re^{n}\,Pr^{m}$ 🔄 Key Processes Write a continuity (conservation) equation for the quantity of interest. Insert the appropriate constitutive law (Fourier, Fick, Newton) to replace the flux term. Non‑dimensionalize using characteristic length $L$, velocity $U$, etc., yielding dimensionless groups (Re, Pr, Sc, Nu, Sh). Select a correlation (e.g., $Nu=CRe^{n}Pr^{m}$) based on flow regime (laminar vs. turbulent) and geometry. Solve for the unknown coefficient (e.g., $h = Nu \,k/L$ for heat, $hm = Sh\,D/L$ for mass). 🔍 Key Comparisons Fourier vs. Fick vs. Newton – All are flux = diffusivity × gradient laws. Reynolds vs. Chilton–Colburn Analogy – Reynolds: assumes turbulent diffusivities are equal, no molecular contribution. Chilton–Colburn: adds the $Pr^{2/3}$ and $Sc^{2/3}$ correction, works for most turbulent engineering data. Forced vs. Natural Convection – Forced: $Nu = f(Re,Pr)$, external flow dominates. Natural: $Nu = f(Gr,Pr)$, buoyancy drives the flow. ⚠️ Common Misunderstandings “$Pr = 1$ always” – Only true for gases at certain conditions; liquids often have $Pr \neq 1$. Mixing up diffusivities – $\nu$ (kinematic viscosity) ≠ $k$ (thermal conductivity) nor $D$ (mass diffusivity). Using laminar correlations for turbulent flow – Leads to large errors in $h$, $hm$. Assuming the same boundary condition for all transports – Heat uses temperature difference, mass uses concentration difference; they are analogous but not interchangeable without the proper $h$ or $hm$. 🧠 Mental Models / Intuition “Flux = Slope × Conductivity” – Visualize a hill: steeper slope (larger gradient) → faster flow; conductivity is the “smoothness” of the hill. Analogy Switch – Think of a heat‑transfer problem; replace $k \rightarrow D$, $T \rightarrow C$, $h \rightarrow hm$, $Nu \rightarrow Sh$, $Pr \rightarrow Sc$ → you instantly have the mass‑transfer version. Dimensionless Numbers as “knobs” – Turn Re to increase inertia (more turbulence), turn Pr/Sc to change relative diffusion rates. 🚩 Exceptions & Edge Cases Radiation heat transfer – Not covered by Fourier’s law; must be added separately. Highly compressible or rarefied gases – Continuum assumptions (Newton, Fourier, Fick) break down; slip flow or free‑molecular regimes apply. Non‑Newtonian fluids – Viscosity $\mu$ is not constant; Newton’s law must be modified. Strong temperature dependence of $k$, $\mu$, $D$ – Use property evaluated at film temperature, not bulk. 📍 When to Use Which | Situation | Choose | Reason | |-----------|--------|--------| | Laminar flow in a flat plate | $Nu = 0.664\,Re^{1/2}Pr^{1/3}$ (heat) → $Sh = 0.664\,Re^{1/2}Sc^{1/3}$ (mass) | Valid for $Re < 5\times10^5$, linear velocity profile. | | Turbulent flow in a smooth pipe | Chilton–Colburn $J$ factor or $Nu = 0.023\,Re^{0.8}Pr^{0.4}$ | Empirical fit for $Re>10^4$, captures enhanced mixing. | | Natural convection over a vertical plate | Correlation using $Gr\,Pr$ (e.g., $Nu = C (Gr\,Pr)^n$) | Buoyancy dominates; $Re$ is not defined. | | Non‑Newtonian fluid | Use rheological model (e.g., Power‑law) to compute an apparent viscosity, then plug into Newton’s law form | Viscosity varies with shear rate. | | Mass transfer with strong pressure driving force | Use pressure‑driven diffusion term in the flux expression ($J = -D\,\nabla C + \frac{D}{RT}\,C\nabla P$) | Simple Fick’s law ignores pressure effect. | 👀 Patterns to Recognize “$Nu = f(Re,Pr)$” → heat‑transfer correlation; replace $Pr$ with $Sc$ for mass. Boundary‑layer similarity – If the momentum equation yields a $Re$‑dependent solution, the thermal or concentration boundary layer will have the same functional form with $Pr$ or $Sc$ as the extra exponent. “$J = Nu/(Re\,Pr^{2/3})$ constant” – When you see a turbulent correlation, check whether the J‑factor is roughly constant across your data; if so, the Chilton–Colburn analogy applies. Linear vs. quadratic dependence on $Re$ – Laminar: $Re^{1/2}$; turbulent: $Re^{0.8\text{–}0.9}$. Spotting this quickly tells you which regime you’re in. 🗂️ Exam Traps Mistaking $Pr$ for $Sc$ – A heat‑transfer problem will never use $Sc$; swapping them yields a wrong answer. Using $Nu = C\,Re^{n}\,Pr^{m}$ for natural convection – Natural convection requires $Gr$; a forced‑convection form will be penalized. Neglecting the $2/3$ exponent in the J‑factor – Plugging $Nu/(Re\,Pr)$ instead of $Nu/(Re\,Pr^{2/3})$ gives a value that is too high for turbulent flow. Assuming constant properties – For large temperature differences, $k$, $\mu$, $D$ change; exam problems may explicitly state “variable properties” – using constant‑property correlations will be wrong. Confusing shear stress $\tau$ with pressure drop – $\tau = \mu \, dv/dz$ is a local gradient; pressure drop $\Delta P$ relates to integrated wall shear via $ \Delta P = 2L \tauw /R$ for pipe flow. --- Study tip: Memorize the three core flux laws (Fourier, Fick, Newton) and the generic correlation form $Nu = C\,Re^{n}\,Pr^{m}$. Whenever you see a new transport problem, map it onto one of these templates, swap the appropriate dimensionless groups, and you’re already halfway to the answer. Good luck!