Heat and Mass Transfer in Food Processing: Conduction, Convection, Radiation, Transients, Heat Exchangers, Diffusion, Balances and Water Activity

The first chapter of Section 7, Dairy and Food Engineering, of the GATE Agricultural Engineering (AG) paper. The section’s first sub-heading, Heat and Mass Transfer, is the engineering science beneath the unit operations and preservation methods of its other two, so the section is written as two chapters, and this one covers the science: steady-state conduction, convection and radiation; transient heat transfer in simple geometries; the working principles of heat exchangers; diffusive and convective mass transfer; simultaneous heat and mass transfer in agricultural processing; material and energy balances in food processing; and water activity with the sorption and desorption isotherms. It is written at the depth this paper examines — resistances, the lumped model, LMTD, Fick’s law, the Pearson square and water activity — rather than the design correlations of a chemical-engineering course.

1. Steady-state conduction: walls, cylinders and critical insulation

Fourier’s law q = −kA dT/dx gives the conduction rate through a material of thermal conductivity k. In steady one-dimensional conduction the heat flow is the same through every layer, so each layer acts as a thermal resistance and resistances in series add, exactly as in an electric circuit: a plane wall has R = L/(kA), a cylindrical shell R = ln(r₂/r₁)/(2πkL), and a surface film R = 1/(hA). A cold-store wall of 0.2 m brick (k = 0.8 W/m·K) and 0.05 m insulation (k = 0.04 W/m·K) has resistances per m² of 0.25 and 1.25 K·m²/W; under a 30 K difference it passes 30/1.5 = 20 W/m² — the thin insulation carries five-sixths of the total resistance.

Insulating a small pipe does not always reduce heat loss. Adding insulation to a cylinder raises the conduction resistance but enlarges the outer surface, lowering the convective resistance; the heat loss is greatest when the outer radius equals the critical radius r_c = k/h (2k/h for a sphere). For insulation of k = 0.2 W/m·K with h = 10 W/m²·K, r_c = 20 mm: on a pipe smaller than that, a thin layer increases the loss, which is useful for electric cables and a trap for steam lines.

⚠️ The critical radius is a maximum of heat loss, not a minimum
Below r_c, adding insulation increases the loss; only beyond r_c does more insulation reduce it. An option saying "insulation always reduces heat loss" is wrong for thin pipes.

2. Convection and radiation

Convection is described by Newton’s law of cooling, q = hA(T_s − T_∞), with the film coefficient h found from dimensionless groups: the Nusselt number Nu = hL/k, the Reynolds number Re, the Prandtl number Pr = c_pμ/k, and in natural convection the Grashof number Gr. For turbulent flow in tubes the Dittus–Boelter correlation Nu = 0.023 Re^0.8 Pr^n is used, with n = 0.4 when the fluid is heated and 0.3 when it is cooled; for fully developed laminar flow in a tube Nu is a constant, 3.66 at constant wall temperature and 4.36 at constant heat flux. Boiling and condensing give the highest coefficients, which is why steam is the preferred heating medium.

Radiation needs no medium. A black body emits σT⁴ per unit area, with the Stefan–Boltzmann constant σ = 5.67 × 10⁻⁸ W/m²·K⁴ and T in kelvin; a grey surface of emissivity ε exchanging with large surroundings loses q = εσA(T_s⁴ − T_sur⁴). A 1 m² surface of ε = 0.9 at 400 K facing surroundings at 300 K loses 0.9 × 5.67 × 10⁻⁸ × (2.56 × 10¹⁰ − 0.81 × 10¹⁰) = 893 W. Radiation dominates in baking ovens and infrared drying of grain, and must be counted in the heat load of any hot surface.

⚠️ Kelvin, and the difference of fourth powers
Using 127 °C and 27 °C, or raising the temperature difference (100 K) to the fourth power, gives nonsense. The fourth powers are taken separately, in kelvin, and then subtracted.

3. Transient heat transfer in simple geometries

When a food is heated or cooled, its temperature changes with time. If its internal resistance is small next to the surface resistance — Biot number Bi = hL_c/k < 0.1, with the characteristic length L_c = V/A (D/6 for a sphere) — its temperature stays uniform and the lumped-capacitance model applies: (T − T_∞)/(T_i − T_∞) = exp(−hAt/(ρVc)). A 10 mm pea-sized sphere (ρ = 1000 kg/m³, c = 4000 J/kg·K, k = 0.5 W/m·K) in water with h = 25 W/m²·K has Bi = 25 × 0.00167/0.5 = 0.083, so the model holds; hA/(ρVc) = 25/(1000 × 4000 × 0.00167) = 0.00375 s⁻¹, and the temperature difference halves in ln 2/0.00375 = 185 s.

When Bi exceeds 0.1, temperature varies inside the body and the one-term series solutions (Heisler charts) for a slab, a long cylinder and a sphere are used, in terms of Bi and the Fourier number Fo = αt/L². In canning, the slowest-heating point — the geometric centre for conduction-heated foods — governs the process. Freezing time is estimated by Plank’s equation, t = (ρλ/(T_f − T_∞))(Pa/h + Ra²/k), with λ the latent heat, a the thickness (or diameter), k the conductivity of the frozen food, and shape factors P = 1/2, R = 1/8 for a slab, P = 1/4, R = 1/16 for a cylinder and P = 1/6, R = 1/24 for a sphere. For a 5 cm slab with ρλ = 2.5 × 10⁸ J/m³, a 20 K driving difference, h = 20 W/m²·K and k = 1.0 W/m·K, t = 1.25 × 10⁷ × (0.00125 + 0.0003125) = 19 531 s = 5.43 h.

4. Working principles of heat exchangers

Food plants use plate heat exchangers (corrugated plates, high coefficients, easy cleaning, and regeneration — the pasteurised milk heating the incoming raw milk — for liquids like milk and juice), tubular and double-pipe exchangers (for viscous or particulate products and higher pressures), shell-and-tube exchangers, and scraped-surface exchangers whose rotating blades keep a very viscous or crystallising product (ice cream, margarine) moving off the wall. The rate is q = UAΔT_lm, where the log-mean temperature difference is LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) of the end differences. Hot fluid cooled from 80 to 50 °C against cold fluid heated from 20 to 40 °C in counterflow has end differences of 40 and 30 K and an LMTD of 34.8 K; in parallel flow the ends would be 60 and 10 K and the LMTD only 27.9 K, which is why counterflow needs less area. When outlet temperatures are unknown, the effectiveness–NTU method (ε as a function of NTU = UA/C_min and the capacity ratio) avoids iteration.

5. Mass transfer, simultaneous heat and mass transfer, and material and energy balances

Diffusive mass transfer follows Fick’s first law, J = −D_AB dC/dx: with D = 10⁻⁹ m²/s and a concentration difference of 10 kg/m³ across 1 mm, J = 10⁻⁵ kg/m²·s. Convective mass transfer from a surface to a moving fluid is N = k_c(C_s − C_∞), with k_c found through the Sherwood number Sh = k_cL/D and the Schmidt number Sc = μ/(ρD), by analogy with heat transfer (the Chilton–Colburn analogy, j_H = j_D). Simultaneous heat and mass transfer is the rule in agricultural processing: in drying, evaporative cooling, the wet-bulb thermometer, frying and freezing, heat flows in to supply the latent heat of the water moving out, and the surface settles at the temperature where the two rates balance.

Material and energy balances — input = output + accumulation, for total mass, for each component, and for energy — are the arithmetic of every food process: mixing, concentration, dilution, drying and blending. Standardising milk fat is the classic case, solved with the Pearson square: to make 4.5 per cent milk from 3 per cent milk and 40 per cent cream, the parts are (4.5 − 3) = 1.5 of cream to (40 − 4.5) = 35.5 of milk, so 1000 kg of standardised milk needs 1000 × 1.5/37 = 40.5 kg of cream. Energy balances give heat loads: heating 2000 kg/h of milk (c_p = 3.9 kJ/kg·K) from 10 to 70 °C takes 2000 × 3.9 × 60/3600 = 130 kW.

6. Water activity and sorption isotherms

Water activity a_w = p/p₀ is the vapour pressure of water in a food over that of pure water at the same temperature, equal at equilibrium to the equilibrium relative humidity/100. It, rather than moisture content, decides microbial growth, enzyme and chemical reactions and texture: most bacteria need a_w above about 0.91, most yeasts above about 0.88, most moulds above about 0.80, and practically no microorganism grows below about 0.6. Intermediate-moisture foods (jams, some dried fruits) sit around 0.6 to 0.85. Lowering a_w by drying, concentrating, or adding salt or sugar is therefore a preservation method in itself. A food whose water exerts 2.52 kPa where pure water at the same temperature exerts 3.17 kPa has a_w = 0.795.

A sorption isotherm plots equilibrium moisture content against a_w at constant temperature. Most foods give a sigmoid (type II) curve with three regions: tightly bound monolayer water at low a_w (the BET and GAB models estimate the monolayer), multilayer water in the middle, and free, capillary water at high a_w. The desorption isotherm (drying a wet food) lies above the adsorption isotherm (wetting a dry one) — hysteresis — so a food dried to a given a_w holds more water than one rewetted to it. Isotherms fix the safe storage moisture and the end-point of drying.

Key takeaways

  • Resistances add in series: L/(kA) for a wall, ln(r₂/r₁)/(2πkL) for a cylinder, 1/(hA) for a film; r_c = k/h is a maximum of heat loss.
  • Dittus–Boelter Nu = 0.023Re^0.8Pr^n (n = 0.4 heating, 0.3 cooling); radiation q = εσA(T₁⁴ − T₂⁴) in kelvin.
  • Lumped model when Bi < 0.1: θ/θᵢ = exp(−hAt/(ρVc)); Plank’s equation estimates freezing time.
  • LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂); counterflow gives the larger LMTD and the smaller area.
  • Fick J = −D dC/dx; the Pearson square standardises fat; Q = ṁc_pΔT gives a heat load.
  • a_w = p/p₀ = ERH/100; bacteria need about 0.91, moulds about 0.80; desorption isotherms lie above adsorption.

Practice questions (13)

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. A 10 m² insulating panel 0.1 m thick has a thermal conductivity of 0.04 W/m·K. What is the steady heat flow through it under a temperature difference of 30 K, in W?

    Numerical answer — type the value.

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    Answer: 120

    q = kAΔT/L = 0.04 × 10 × 30/0.1 = 120 W. Multiplying by the thickness instead of dividing gives 1.2 W, which would make a thicker panel pass more heat.

  2. A cold-store wall consists of 0.2 m of brick (k = 0.8 W/m·K) and 0.05 m of insulation (k = 0.04 W/m·K). Neglecting surface films, what is the heat flow through it per m² for a temperature difference of 30 K, in W/m²?

    Numerical answer — type the value.

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    Answer: 20

    R_brick = 0.2/0.8 = 0.25 and R_ins = 0.05/0.04 = 1.25 K·m²/W, in series: R = 1.5, so q = 30/1.5 = 20 W/m². Adding the conductivities or averaging them treats the layers as parallel paths, which they are not.

  3. Insulation with a thermal conductivity of 0.2 W/m·K is to be put on a small pipe exposed to air with h = 10 W/m²·K. What is the critical radius of insulation, in mm?

    Numerical answer — type the value.

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    Answer: 20

    For a cylinder r_c = k/h = 0.2/10 = 0.02 m = 20 mm; at this outer radius the heat loss is a maximum. The spherical value 2k/h = 40 mm applies to a sphere, not a pipe.

  4. In a counterflow heat exchanger the hot fluid is cooled from 80 °C to 50 °C and the cold fluid heated from 20 °C to 40 °C. What is the log-mean temperature difference, in K (to one decimal place)?

    Numerical answer — type the value.

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    Answer: 34.8

    End differences: ΔT₁ = 80 − 40 = 40 K and ΔT₂ = 50 − 20 = 30 K. LMTD = (40 − 30)/ln(40/30) = 10/0.2877 = 34.8 K. The arithmetic mean, 35 K, is close here but always slightly high; pairing the ends as for parallel flow gives 27.9 K.

  5. A spherical food particle 10 mm in diameter (ρ = 1000 kg/m³, c = 4000 J/kg·K, k = 0.5 W/m·K) is placed in a fluid with h = 25 W/m²·K. Confirming that the lumped model applies, how long does it take for the particle–fluid temperature difference to fall to half its initial value, in s (to the nearest whole second)?

    Numerical answer — type the value.

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    Answer: 185

    L_c = V/A = D/6 = 1.667 mm; Bi = 25 × 0.001667/0.5 = 0.083 < 0.1, so lumped. hA/(ρVc) = h/(ρcL_c) = 25/(1000 × 4000 × 0.001667) = 0.00375 s⁻¹; t = ln 2/0.00375 = 184.8 ≈ 185 s. Taking L_c as the radius (5 mm) instead of V/A gives three times the time, 555 s.

  6. A surface of 1 m² with an emissivity of 0.9 is at 400 K and exchanges radiation with large surroundings at 300 K. What is the net radiative heat loss, in W (to the nearest whole number)? Take σ = 5.67 × 10⁻⁸ W/m²·K⁴.

    Numerical answer — type the value.

    Show answer

    Answer: 893

    q = εσA(T₁⁴ − T₂⁴) = 0.9 × 5.67 × 10⁻⁸ × (2.56 × 10¹⁰ − 0.81 × 10¹⁰) = 0.9 × 5.67 × 10⁻⁸ × 1.75 × 10¹⁰ = 893 W. Raising the 100 K difference to the fourth power gives only 5.1 W.

  7. Milk with 3 per cent fat is to be standardised to 4.5 per cent fat by adding cream with 40 per cent fat. How much cream is needed to make 1000 kg of standardised milk, in kg (to one decimal place)?

    Numerical answer — type the value.

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    Answer: 40.5

    Pearson square: cream parts = 4.5 − 3 = 1.5, milk parts = 40 − 4.5 = 35.5, total 37. Cream = 1000 × 1.5/37 = 40.5 kg. Check: (959.5 × 0.03 + 40.5 × 0.40)/1000 = 0.045. Dividing 1.5 by 35.5 instead of by 37 gives the cream per kg of milk, not per kg of mixture.

  8. At 25 °C the water in a food exerts a vapour pressure of 2.52 kPa, and pure water at the same temperature exerts 3.17 kPa. What is the water activity of the food (to three decimal places)?

    Numerical answer — type the value.

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    Answer: 0.795

    a_w = p/p₀ = 2.52/3.17 = 0.795, equivalent to an equilibrium relative humidity of 79.5 per cent — above the limit for most moulds, below that for most bacteria. Inverting the ratio gives 1.26, impossible since a_w cannot exceed 1.

  9. A solute diffuses through a 1 mm layer with a diffusivity of 1 × 10⁻⁹ m²/s under a concentration difference of 10 kg/m³ across the layer. What is the diffusion flux, in units of 10⁻⁵ kg/(m²·s)?

    Numerical answer — type the value.

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    Answer: 1

    J = D ΔC/Δx = 10⁻⁹ × 10/10⁻³ = 10⁻⁵ kg/(m²·s), i.e. 1 in the units asked. Leaving the thickness in millimetres gives 10⁻⁸, a thousand times too small.

  10. A slab of food 5 cm thick (ρλ = 2.5 × 10⁸ J/m³ for its latent heat) freezes at −1 °C in air at −21 °C with h = 20 W/m²·K; the frozen food has k = 1.0 W/m·K. Using Plank’s equation with P = 1/2 and R = 1/8, what is the freezing time, in hours (to two decimal places)?

    Numerical answer — type the value.

    Show answer

    Answer: 5.43

    t = (ρλ/ΔT)(Pa/h + Ra²/k) = (2.5 × 10⁸/20)(0.5 × 0.05/20 + 0.125 × 0.05²/1.0) = 1.25 × 10⁷ × (0.00125 + 0.0003125) = 19 531 s = 5.43 h. Dropping the conduction term Ra²/k gives 4.34 h, the time for a food with no internal resistance.

  11. In the Dittus–Boelter correlation Nu = 0.023 Re^0.8 Pr^n for turbulent flow in a tube, n is:

    1. 0.4 when the fluid is being heated and 0.3 when it is being cooled
    2. 0.3 when the fluid is being heated and 0.4 when it is being cooled
    3. 0.8 in both cases
    4. 1/3 in both cases
    Show answer

    Answer: A — 0.4 when the fluid is being heated and 0.3 when it is being cooled

    The exponent is 0.4 for heating and 0.3 for cooling of the fluid, reflecting how viscosity near the wall changes with the direction of heat flow. Swapping them is the standard trap; 0.8 is the Reynolds-number exponent, and 1/3 belongs to other correlations such as Sieder–Tate.

  12. Which of the following statements about water activity are correct?

    1. Pure water has a water activity of 1
    2. Most bacteria need a higher water activity for growth than most moulds
    3. The desorption isotherm of a food usually lies above its adsorption isotherm
    4. Intermediate-moisture foods have water activities below 0.3
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    Answer: A — Pure water has a water activity of 1; B — Most bacteria need a higher water activity for growth than most moulds; C — The desorption isotherm of a food usually lies above its adsorption isotherm

    a_w = p/p₀ is 1 for pure water; bacteria generally need about 0.91 against about 0.80 for moulds; hysteresis puts desorption above adsorption. Intermediate-moisture foods sit around 0.6 to 0.85 — a_w below 0.3 describes dried foods such as milk powder — so the fourth statement is false.

  13. The lumped-capacitance model of transient heating is valid when the Biot number is:

    1. less than about 0.1
    2. greater than 10
    3. exactly 1
    4. greater than 0.1
    Show answer

    Answer: A — less than about 0.1

    Bi = hL_c/k compares internal conduction resistance with surface convection resistance; below about 0.1 the inside is nearly isothermal and one temperature describes the body. At large Bi the surface resistance is negligible and temperature varies strongly inside — the opposite case, needing Heisler charts.