Transport Phenomena and Rate Processes: Momentum, Heat and Mass Transfer, Kinetics and Polarization

Section 3 of the GATE Metallurgical Engineering (MT) paper, in one chapter. It takes the six headings in the syllabus’s order. Momentum transfer: the concept of viscosity, shell balances, Bernoulli’s equation, the mechanical energy balance, and flow past plane surfaces and through pipes. Heat transfer: conduction by Fourier’s law and one-dimensional steady conduction, convection, heat transfer coefficients and their correlations, and radiation with the Stefan-Boltzmann and Kirchhoff laws. Mass transfer: diffusion and Fick’s laws, mass transfer coefficients and correlations. Dimensional analysis: what each dimensionless number means and where it is used. The basic laws of chemical kinetics: first-order reactions, the rate constant, the Arrhenius relation, heterogeneous reactions and oxidation kinetics. And electrochemical kinetics: polarization. The examples are metallurgical — liquid steel in a nozzle, a refractory furnace wall, a cooling casting, a dissolving alloy addition, a scaling steel — because that is where this paper sets them.

1. Momentum transfer: viscosity, shell balances, Bernoulli and flow in pipes

Viscosity is the transport coefficient for momentum. Newton’s law of viscosity, τ_yx = −μ dv_x/dy, says the shear stress between fluid layers is proportional to the velocity gradient, which is the same as saying that x-momentum flows down its gradient in the y direction; μ is in Pa·s and the kinematic viscosity ν = μ/ρ, in m²/s, is the momentum diffusivity — the exact counterpart of the thermal diffusivity α = k/(ρc_p) and of the mass diffusivity D. Liquid metals have low kinematic viscosities, comparable to water’s, which is why flows in ladles, tundishes and moulds are almost always turbulent.

A shell balance is how every laminar result in this section is derived: take a thin shell of fluid, write (rate of momentum in) − (rate out) + (sum of forces) = 0 for steady flow, let the shell thickness go to zero to get a differential equation for the momentum flux, then substitute Newton’s law and apply boundary conditions (no slip at a wall, zero shear at a free surface or on the centreline). For a falling film of thickness δ on a vertical wall it gives a parabolic profile with v_max = ρgδ²/(2μ). For steady laminar flow in a tube of radius R and length L under a pressure drop ΔP it gives the Hagen-Poiseuille result: v(r) = (ΔP R²/4μL)(1 − r²/R²), the mean velocity is half the maximum, and Q = πΔP R⁴/(8μL).

For inviscid flow along a streamline, Bernoulli’s equation P/ρ + v²/2 + gz = constant holds. The mechanical energy balance extends it to real systems between two points: ΔP/ρ + Δ(v²/2) + gΔz + Σh_f = W_s, where Σh_f is the frictional loss per unit mass and W_s the shaft work done by a pump. The pipe friction loss is h_f = 4f (L/D)(v²/2) with the Fanning friction factor f (the Darcy factor is 4f); in laminar flow f = 16/Re, and in turbulent flow f depends on Re and wall roughness and is read from the Moody chart. Flow in a pipe is laminar below Re = ρvD/μ ≈ 2100. Over a flat plate a boundary layer grows from the leading edge; in laminar flow its thickness is δ ≈ 5x/√Re_x (Blasius), and transition to turbulence occurs near Re_x ≈ 5 × 10⁵. A classic metallurgical application is teeming from a ladle: Bernoulli between the melt surface and the nozzle gives the Torricelli discharge velocity v = √(2gh), so the teeming rate falls as the head falls.

⚠️ Fanning or Darcy?
Laminar pipe flow has f = 16/Re in the Fanning convention and f_D = 64/Re in the Darcy convention; the loss is 4f(L/D)(v²/2) in one and f_D(L/D)(v²/2) in the other. Mixing the factor from one book with the loss formula from another puts a factor of four into the answer.

2. Heat transfer: conduction, convection and radiation

Fourier’s law, q = −k dT/dx, makes the heat flux proportional to the temperature gradient. In one-dimensional steady conduction with no generation the temperature is linear through a plane wall and q = kΔT/L; it is convenient to write the wall as a thermal resistance R = L/(kA), so that layers in series add. A furnace wall of firebrick backed by insulating brick is the standard case: the heat flux is the overall temperature difference divided by the sum of the resistances, and the interface temperature follows from the drop across the first layer — which is how one checks that the insulating brick is not being run above its service limit. For a cylindrical shell the resistance is ln(r₂/r₁)/(2πkL). A convective film at a surface adds a resistance 1/(hA).

Convection is described by Newton’s law of cooling, q = h(T_s − T_∞), where the heat transfer coefficient h lumps together the conduction through the thermal boundary layer and the flow that renews it. h is found from correlations in dimensionless form, Nu = hL/k = f(Re, Pr) for forced convection and f(Gr, Pr) for natural convection. For fully developed turbulent flow in a pipe the Dittus-Boelter correlation Nu = 0.023 Re^0.8 Pr^n (n = 0.4 for heating the fluid, 0.3 for cooling) is the one to know; fully developed laminar pipe flow has constant Nu, 3.66 at constant wall temperature and 4.36 at constant heat flux. When a body’s internal resistance is small compared with its surface resistance — Biot number Bi = h(V/A)/k < 0.1 — its temperature is uniform and it cools by the lumped-capacitance law, ρcV dT/dt = −hA(T − T_∞), so that (T − T_∞)/(T₀ − T_∞) = exp(−hAt/(ρcV)) and the instantaneous cooling rate is hA(T − T_∞)/(ρcV).

Every surface above absolute zero radiates. A black body emits E_b = σT⁴, with the Stefan-Boltzmann constant σ = 5.67 × 10⁻⁸ W/(m² K⁴); a real grey surface emits εσT⁴ with emissivity ε < 1. A small grey body in large surroundings at T_sur exchanges a net q = εσ(T_s⁴ − T_sur⁴) per unit area. Kirchhoff’s law states that at thermal equilibrium the absorptivity of a surface equals its emissivity, α = ε, at each wavelength; for a grey surface this holds over all wavelengths, which is what allows ε to be used for both emission and absorption. Because of the fourth power, radiation dominates heat loss from anything at steelmaking temperatures: a surface at 1600 °C emits (1873/1273)⁴ ≈ 4.7 times as much per unit area as one at 1000 °C.

3. Mass transfer: Fick’s laws, mass transfer coefficients and correlations

Fick’s first law, J = −D dC/dx, is the mass-transfer counterpart of Fourier’s and Newton’s laws: the diffusive flux (mol m⁻² s⁻¹) is proportional to the concentration gradient, with the diffusivity D in m²/s. Combining it with conservation of mass gives Fick’s second law, ∂C/∂t = D ∂²C/∂x² for constant D, the equation behind carburising, decarburisation and homogenisation, and solved with the error function in the physical-metallurgy chapter. Diffusivities in liquid metals are of order 10⁻⁹ m²/s and in gases of order 10⁻⁵ m²/s, which is why gas-phase transport is rarely the slow step in a gas-metal reaction.

Between a surface and a flowing fluid the flux is written with a mass transfer coefficient, J = k_m (C_s − C_b), exactly as convection is written with h. Three models relate k_m to D. The film model imagines a stagnant film of thickness δ, giving k_m = D/δ, so k_m ∝ D. Higbie’s penetration model assumes fluid elements that touch the surface for a time t_e before being swept away, giving k_m = 2√(D/(πt_e)), so k_m ∝ D^½ — the better model for a gas bubble rising through liquid steel or a stirred bath surface. The surface-renewal model gives the same square-root dependence. Correlations are written as Sh = k_mL/D = f(Re, Sc); for a sphere, Ranz-Marshall gives Sh = 2 + 0.6 Re^½ Sc^⅓, whose limit Sh = 2 is pure diffusion into a stagnant medium. The Chilton-Colburn analogy, j_D = j_H = f/2, carries a known friction factor or heat transfer coefficient over to mass transfer.

4. Dimensional analysis and the dimensionless numbers

Buckingham’s π theorem says that a relation among n variables involving m independent dimensions can be rewritten as a relation among n − m dimensionless groups. Its practical value is twofold: a correlation measured on a water model applies to liquid steel if the governing groups are matched (the basis of physical modelling of ladles and tundishes, usually by matching the Froude number), and every transport result can be read as one dimensionless number as a function of others.

The dimensionless numbers this section uses
NumberDefinitionPhysical meaning
Reynolds, ReρvL/μInertial to viscous forces; laminar or turbulent
Prandtl, Prν/α = c_pμ/kMomentum to thermal diffusivity; very small for liquid metals
Schmidt, Scν/DMomentum to mass diffusivity
Nusselt, NuhL/k (k of the FLUID)Convective to conductive heat transfer across the fluid
Sherwood, Shk_mL/DConvective to diffusive mass transfer
Biot, BihL/k (k of the SOLID)Internal conduction resistance to surface resistance; lumped if < 0.1
Fourier, Foαt/L²Dimensionless time for transient conduction
Grashof, GrgβΔTL³/ν²Buoyancy to viscous forces; natural convection
Froude, Frv²/(gL)Inertial to gravity forces; matched in water models
Péclet, PeRe·Pr (or Re·Sc)Advective to diffusive transport
⚠️ Nu and Bi look identical
Both are hL/k. In the Nusselt number k is the conductivity of the fluid and the number describes the fluid side; in the Biot number k is that of the solid and the number compares the solid’s interior with its surface. A question that gives both conductivities is testing exactly this.

5. Chemical kinetics: first-order reactions, Arrhenius, heterogeneous reactions and oxidation

A first-order reaction has rate −dC/dt = kC, so ln(C₀/C) = kt and C = C₀e^(−kt); the rate constant k has units of s⁻¹, the half-life t½ = ln 2/k is independent of the starting concentration, and a plot of ln C against t is a straight line of slope −k. Decarburisation of liquid steel at low carbon, where the rate is controlled by carbon transport to the bubble or slag interface, behaves as first order in [%C]. The temperature dependence of k is the Arrhenius relation k = A exp(−E_a/RT): ln k against 1/T is a line of slope −E_a/R, and two measurements give E_a = R ln(k₂/k₁)/(1/T₁ − 1/T₂). The same form governs diffusivities, creep rates and the rate of any thermally activated process.

Most metallurgical reactions are heterogeneous — gas-solid (reduction of iron ore pellets, roasting), gas-liquid (degassing, decarburisation), liquid-liquid (slag-metal refining) — and proceed through steps in series: transport of the reactant through the fluid boundary layer, diffusion through a porous product layer if one forms, chemical reaction at the interface, and transport of products away. At steady state the flux through every step is the same, so resistances add: for a first-order interfacial reaction with rate constant k_r and a mass transfer coefficient k_m, the overall coefficient is 1/k = 1/k_m + 1/k_r. The slowest step is rate-controlling. The shrinking-core model of a dense particle reacting from the outside in gives distinct time laws: under chemical-reaction control, t/τ = 1 − (1 − X)^⅓; under product-layer diffusion control, t/τ = 1 − 3(1 − X)^⅔ + 2(1 − X); under gas-film control, t/τ = X — so plotting conversion data against each form identifies the controlling step. A high activation energy (above roughly 100 kJ/mol) points to chemical control; a low one, and sensitivity to gas velocity or particle size, points to transport control.

High-temperature oxidation of metals follows one of three rate laws depending on the scale. If the oxide is porous or cracks, oxygen reaches the metal freely and the thickness grows linearly, x = k_l t. If the oxide is dense and adherent, growth is controlled by diffusion of ions through it and the flux falls as the scale thickens: the parabolic law x² = k_p t (Wagner’s theory), which describes iron, nickel, chromia-forming and alumina-forming alloys. Very thin films at low temperatures follow a logarithmic law. The Pilling-Bedworth ratio, PBR = (volume of oxide formed)/(volume of metal consumed) = M_oxide ρ_metal/(n M_metal ρ_oxide) for an oxide MₙOₘ, predicts which: PBR < 1 (magnesium, alkali metals) gives a porous, non-protective oxide; PBR between about 1 and 2 (aluminium, nickel, chromium) a protective one; PBR well above 2 a scale that spalls under compressive stress.

6. Electrochemical kinetics: polarization

When a current flows through an electrode its potential moves away from the equilibrium (Nernst) value; the displacement η = E − E_eq is the overpotential, and the phenomenon is polarization. Three kinds add. Activation polarization is the energy barrier of the charge-transfer step itself; the Butler-Volmer equation describes it, and at high overpotential it reduces to the Tafel equation η = β log(i/i₀), where i₀ is the exchange current density (the equal and opposite forward and back currents at equilibrium) and β the Tafel slope, typically about 0.1 V per decade. A large i₀ means a reaction that polarizes little — hydrogen evolution on platinum — and a small i₀ one that polarizes strongly — hydrogen on zinc or lead. Concentration polarization appears when the reacting species cannot be supplied fast enough by diffusion; the current cannot exceed the limiting current density i_L = nFDC_b/δ, reached when the surface concentration falls to zero. Resistance (ohmic) polarization is the IR drop through the electrolyte and any surface film.

Polarization is what turns the Pourbaix diagram’s “can it corrode?” into “how fast?”. In mixed-potential theory a corroding metal carries an anodic reaction (M → Mⁿ⁺ + ne⁻) and a cathodic one (hydrogen evolution or oxygen reduction) on the same surface; with no external current they must balance, and the potential at which the anodic and cathodic polarization curves intersect is the corrosion potential E_corr, the current there the corrosion current i_corr. Anything that raises i₀ of the cathodic reaction, or its limiting current (more dissolved oxygen, more stirring), raises the corrosion rate. Passivation shows as an anodic curve that, above a critical current density, suddenly drops to a small passive current — the basis of anodic protection and of stainless steel’s behaviour.

ℹ️ The three analogies to carry through the paper
τ = −μ dv/dy, q = −k dT/dx and J = −D dC/dx are one law in three costumes, with ν, α and D all in m²/s. Nu, Sh and the friction factor are the matching dimensionless fluxes, which is why the Chilton-Colburn analogy can pass a result from one to the other.

Key takeaways

  • Shell balance plus Newton’s law gives the laminar results: Hagen-Poiseuille Q = πΔPR⁴/(8μL), mean velocity half the maximum, Fanning f = 16/Re below Re ≈ 2100.
  • Plane-wall resistances L/(kA) add in series; convection adds 1/(hA); a body with Bi < 0.1 cools by the lumped law with rate hA(T − T∞)/(ρcV).
  • Radiation: q = εσ(T⁴ − T_sur⁴) with σ = 5.67 × 10⁻⁸ W/(m² K⁴); Kirchhoff: α = ε at equilibrium.
  • First order: ln(C₀/C) = kt, t½ = ln 2/k; Arrhenius from two points E_a = R ln(k₂/k₁)/(1/T₁ − 1/T₂); resistances in series add and the slowest step controls.
  • Oxidation is linear for porous scales and parabolic (x² = k_p t) for protective ones; Tafel η = β log(i/i₀) and limiting current i_L = nFDC/δ describe polarization.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Liquid steel (density 7000 kg/m³, viscosity 0.006 Pa·s) flows at a mean velocity of 0.5 m/s through a nozzle of 50 mm bore. What is the Reynolds number (to the nearest whole number)?

    Numerical answer — type the value.

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    Answer: 29167

    Re = ρvD/μ = 7000 × 0.5 × 0.05/0.006 = 175/0.006 = 29 167. Check with the kinematic viscosity: ν = 0.006/7000 = 8.57 × 10⁻⁷ m²/s, and vD/ν = 0.025/8.57 × 10⁻⁷ = 29 167 — well above 2100, so the flow is turbulent. Using the radius instead of the bore halves it.
  2. What is the Fanning friction factor for fully developed laminar flow in a pipe at a Reynolds number of 1000?

    Numerical answer — type the value.

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    Answer: 0.016

    From Hagen-Poiseuille, the wall shear stress gives f = τ_w/(½ρv²) = 16/Re, so f = 16/1000 = 0.016. The Darcy factor is four times this, 64/Re = 0.064, and quoting it answers a different convention from the one asked.
  3. A furnace wall consists of 230 mm of firebrick (k = 1.5 W/(m K)) backed by 115 mm of insulating brick (k = 0.15 W/(m K)). The inner face is at 1200 °C and the outer face at 100 °C. Neglecting contact resistance, what is the temperature at the firebrick-insulation interface, in °C (to the nearest whole number)?

    Numerical answer — type the value.

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    Answer: 1017

    Resistances per unit area: 0.23/1.5 = 0.1533 and 0.115/0.15 = 0.7667 m² K/W, total 0.92. Flux q = 1100/0.92 = 1195.7 W/m². Interface T = 1200 − 1195.7 × 0.1533 = 1200 − 183.3 = 1016.7 °C. Check from the cold side: 100 + 1195.7 × 0.7667 = 100 + 916.7 = 1016.7 °C. Splitting the temperature drop in proportion to thickness rather than resistance gives 466.7 °C.
  4. A small grey body of emissivity 0.8 at 1000 K sits in large surroundings at 300 K. What is its net radiative heat loss, in kW/m² (to the nearest whole number)? Take σ = 5.67 × 10⁻⁸ W/(m² K⁴).

    Numerical answer — type the value.

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    Answer: 45

    q = εσ(T_s⁴ − T_sur⁴) = 0.8 × 5.67 × 10⁻⁸ × (10¹² − 8.1 × 10⁹) = 0.8 × 5.67 × 10⁻⁸ × 9.919 × 10¹¹ = 44 993 W/m² ≈ 45 kW/m². The surroundings take back only 0.8% of the emission (8.1 × 10⁹ against 10¹²). Using the temperature difference to the fourth power, (700)⁴, gives 10.9 kW/m², the classic slip.
  5. A steel ball 10 mm in diameter (ρ = 7800 kg/m³, c = 500 J/(kg K), k = 40 W/(m K)) at 800 °C is dropped into oil at 30 °C with h = 100 W/(m² K). What is its initial cooling rate, in K/s (to one decimal place)?

    Numerical answer — type the value.

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    Answer: 11.8

    V/A for a sphere is d/6 = 1.667 × 10⁻³ m, so Bi = h(V/A)/k = 100 × 1.667 × 10⁻³/40 = 0.004 < 0.1 and the lumped law applies. dT/dt = h(T − T∞)/(ρc(V/A)) = 100 × 770/(7800 × 500 × 1.667 × 10⁻³) = 77 000/6500 = 11.8 K/s. Using the radius (5 mm) as V/A instead of d/6 gives 3.9 K/s.
  6. Kirchhoff’s law of radiation states that, for a surface in thermal equilibrium with its surroundings:

    1. Its emissivity equals its absorptivity
    2. Its emissive power is proportional to T⁴
    3. Its reflectivity equals its transmissivity
    4. The wavelength of peak emission is inversely proportional to T
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    Answer: A — Its emissivity equals its absorptivity

    Kirchhoff’s law is α = ε at equilibrium, which is why a good absorber is a good emitter and a black body (α = 1) is the perfect emitter. The T⁴ dependence is the Stefan-Boltzmann law and λ_max ∝ 1/T is Wien’s displacement law — both true, but not Kirchhoff’s; reflectivity and transmissivity obey no such equality.
  7. Which of the following statements about dimensionless numbers are correct?

    1. The Prandtl number is the ratio of momentum diffusivity to thermal diffusivity
    2. The Biot number uses the thermal conductivity of the fluid
    3. The Schmidt number is ν/D
    4. The Sherwood number is the mass-transfer counterpart of the Nusselt number
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    Answer: A — The Prandtl number is the ratio of momentum diffusivity to thermal diffusivity; C — The Schmidt number is ν/D; D — The Sherwood number is the mass-transfer counterpart of the Nusselt number

    Pr = ν/α (A) and Sc = ν/D (C) are both ratios of momentum diffusivity to another diffusivity, and Sh = k_mL/D is the mass-transfer form of Nu = hL/k (D). B is the trap: Bi = hL/k uses the conductivity of the SOLID, which is what lets it compare internal with surface resistance; the fluid’s k belongs in Nu.
  8. The rate constant of a reaction doubles when the temperature is raised from 800 K to 900 K. What is its activation energy, in kJ/mol (to one decimal place)? Take R = 8.314 J/(mol K).

    Numerical answer — type the value.

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    Answer: 41.5

    E_a = R ln(k₂/k₁)/(1/T₁ − 1/T₂) = 8.314 × 0.6931/(1/800 − 1/900) = 5.763/(1.389 × 10⁻⁴) = 41 490 J/mol = 41.5 kJ/mol. Check: exp(−41 490/(8.314 × 900))/exp(−41 490/(8.314 × 800)) = exp(41 490 × 1.389 × 10⁻⁴/8.314) = exp(0.693) = 2.00. Using ΔT = 100 K in place of the reciprocal difference gives a meaningless 0.058.
  9. A first-order reaction has a rate constant of 0.02 min⁻¹. How long, in minutes, does it take to reach 90% conversion (to one decimal place)?

    Numerical answer — type the value.

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    Answer: 115.1

    ln(C₀/C) = kt with C/C₀ = 0.1: t = ln 10/0.02 = 2.3026/0.02 = 115.1 min. Check with half-lives: t½ = ln 2/0.02 = 34.66 min, and 90% conversion takes log₂ 10 = 3.32 half-lives = 115.1 min. Using ln(0.9) instead — 10% conversion — gives 5.3 min.
  10. A steel oxidises at constant temperature by the parabolic law, forming a 2 µm scale in 1 h. How long, in hours, will it take for the scale to reach 6 µm?

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    Answer: 9

    x² = k_p t gives k_p = 2²/1 = 4 µm²/h, so t = 6²/4 = 9 h. Ratio check: tripling the thickness takes 3² = 9 times as long. A linear law would give 3 h — the parabolic scale slows down because each ion must diffuse through the whole scale already formed.
  11. Aluminium (M = 26.98 g/mol, ρ = 2.70 g/cm³) oxidises to Al₂O₃ (M = 101.96 g/mol, ρ = 3.95 g/cm³). What is the Pilling-Bedworth ratio (to two decimal places)?

    Numerical answer — type the value.

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    Answer: 1.29

    PBR = M_oxide ρ_metal/(n M_metal ρ_oxide) with n = 2 metal atoms per formula unit: 101.96 × 2.70/(2 × 26.98 × 3.95) = 275.3/213.1 = 1.29. As volumes: one mole of Al₂O₃ occupies 101.96/3.95 = 25.81 cm³ and consumes 2 × 26.98/2.70 = 19.99 cm³ of Al, ratio 1.29 — between 1 and 2, so the oxide is protective. Forgetting n = 2 gives 2.58.
  12. For a heterogeneous gas-solid reaction described by the shrinking-core model, which of the following statements are correct?

    1. At steady state the fluxes through the gas film, the product layer and the interface are equal
    2. If the interfacial reaction controls, t/τ = 1 − (1 − X)^⅓
    3. A strong dependence of rate on gas velocity indicates chemical-reaction control
    4. The overall resistance is the sum of the resistances of the steps in series
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    Answer: A — At steady state the fluxes through the gas film, the product layer and the interface are equal; B — If the interfacial reaction controls, t/τ = 1 − (1 − X)^⅓; D — The overall resistance is the sum of the resistances of the steps in series

    Steps in series carry the same flux at steady state (A), so their resistances add (D). Under interfacial control the rate is proportional to the core area, which integrates to t/τ = 1 − (1 − X)^⅓ (B). C is reversed: sensitivity to gas velocity means the gas-film boundary layer is controlling; a chemically controlled rate is insensitive to flow and strongly temperature-dependent.
  13. An electrode reaction has an exchange current density of 10⁻⁶ A/cm² and a Tafel slope of 0.12 V per decade. What activation overpotential, in V, is needed to drive a current density of 10⁻³ A/cm²?

    Numerical answer — type the value.

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    Answer: 0.36

    Tafel: η = β log(i/i₀) = 0.12 × log(10⁻³/10⁻⁶) = 0.12 × 3 = 0.36 V. Each decade of current costs one Tafel slope, and the current rises by three decades. Using the natural logarithm gives 0.83 V.
  14. Cu²⁺ ions (bulk concentration 10 mol/m³, diffusivity 1 × 10⁻⁹ m²/s) are deposited through a diffusion boundary layer 0.1 mm thick. What is the limiting current density, in A/m² (to one decimal place)? Take F = 96 485 C/mol.

    Numerical answer — type the value.

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    Answer: 19.3

    i_L = nFDC_b/δ = 2 × 96 485 × 1 × 10⁻⁹ × 10/(1 × 10⁻⁴) = 1.930 × 10⁻³/10⁻⁴ = 19.3 A/m². Equivalently the molar flux is D C_b/δ = 10⁻⁴ mol/(m² s), times nF = 192 970 C/mol. Leaving out n = 2 gives 9.6 A/m²; leaving δ in mm gives a value a thousand times too small.