Drying: drying rate curves and drying time

Moisture definitions, the drying-rate curve with constant- and falling-rate periods, predicting the constant rate from heat transfer, batch drying-time formulas, diffusion-controlled drying and dryer types.

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Why it matters

Almost every solid product – pharmaceuticals, fertilisers, polymers, food, ceramics, paper – is dried before sale or further processing, and dryers are among the largest energy users in a plant. The drying-rate curve tells you which mechanism controls, how long a batch takes, and whether hotter or faster air will actually help.

Key ideas

Moisture terms (dry basis).

  • Moisture content X = kg water per kg dry solid (dry basis keeps the denominator constant).
  • Equilibrium moisture X*: the moisture the solid holds in equilibrium with air of given humidity and temperature (from sorption isotherms). It cannot be removed with that air.
  • Free moisture = X − X*: the moisture that can be removed.
  • Bound moisture: water that exerts less than the vapour pressure of pure water (held in fine capillaries or adsorbed); unbound moisture: water in excess of that, behaving as free liquid.
  • Critical moisture content X_c: where the constant-rate period ends.

Drying-rate curve. From a batch test (mass versus time) one plots drying rate N (kg/(m²·h)) against X:

  1. Initial adjustment period – the solid heats or cools to the wet-bulb temperature.
  2. Constant-rate period – the surface is fully wet, behaves like a free water surface and stays at the wet-bulb temperature of the air (for convective drying). The rate is controlled by external heat and mass transfer: air temperature, humidity and velocity.
  3. Falling-rate period (X < X_c) – the surface dries out in patches (first falling-rate period, often linear in X), then the rate is controlled by internal movement of moisture by liquid diffusion or capillary flow (second falling-rate period). External air velocity matters little here; thickness and temperature of the solid matter a lot.
  4. Drying stops at X*.

Why the constant rate is "constant". At steady state the heat convected to the wet surface exactly supplies the latent heat of evaporation, as in a wet-bulb thermometer. So N_c can be predicted from a heat transfer coefficient alone.

Effect of variables.

  • Constant-rate period: higher air temperature, lower humidity (both increase T − T_w), and higher velocity (h ∝ G^0.8 for parallel flow) all shorten it.
  • Falling-rate period: thinner layers (time ∝ thickness² for diffusion control), higher solid temperature; air velocity has little effect.
  • Too-intense drying can cause case hardening – a dry, impermeable skin that traps moisture inside.

Dryer types. Tray (batch, small lots), tunnel and belt (continuous), rotary (granular, high capacity), spray (solutions and slurries to powders), fluidised bed (granules, uniform temperature), drum (thin films, pastes), freeze dryers (heat-sensitive biologicals), and vacuum dryers.

Formulas

Drying rate (per unit drying surface): N = −(L_s/A)·dX/dt

  • N – kg water/(m²·h); L_s – kg dry solid; A – drying area, m²; X – kg water/kg dry solid.

Constant-rate period: N_c = k_Y (Y_w − Y) = h (T − T_w)/λ_w t_c = (L_s/(A·N_c))·(X₁ − X_c)

  • h – W/(m²·K) (convert to kJ/(m²·h·K)); T – air dry-bulb; T_w – wet-bulb; λ_w – latent heat at T_w, kJ/kg; Y – humidity, kg/kg dry air.

Falling-rate period with rate linear in X down to X*: t_f = (L_s/(A·N_c))·(X_c − X*)·ln[(X_c − X*)/(X₂ − X*)]

General falling rate (graphical): t_f = (L_s/A)·∫ from X₂ to X_c of dX/N

Diffusion-controlled slab dried from both faces (half-thickness a, long times): t = (4a²/(π²·D_L))·ln[(8/π²)·(X₁ − X*)/(X₂ − X*)]

  • D_L – liquid diffusivity in the solid, m²/s; a in m; t in s.

Heat transfer coefficient for air flowing parallel to the surface (typical correlation): h = 0.0204·G^0.8 W/(m²·K), G in kg/(m²·h) – take from your data book.

Worked examples

Example 1 – total batch drying time (standard). Given: L_s = 50 kg dry solid; A = 2 m²; N_c = 1.5 kg/(m²·h); X₁ = 0.40, X_c = 0.20, X* = 0.02, X₂ = 0.05; falling rate linear in X.

  1. L_s/(A·N_c) = 50/(2 × 1.5) = 16.67 h (per unit X).
  2. Constant rate: t_c = 16.67 × (0.40 − 0.20) = 3.33 h.
  3. Falling rate: t_f = 16.67 × (0.20 − 0.02) × ln[(0.20 − 0.02)/(0.05 − 0.02)] = 16.67 × 0.18 × ln 6 = 3.00 × 1.792 = 5.38 h.
  4. Total = 3.33 + 5.38 = 8.71 h.

t_c = 3.33 h, t_f = 5.38 h, total ≈ 8.7 h. Note that the last small amount of water takes longest.

Example 2 – constant rate from heat transfer (GATE level). Given: tray of wet solid, L_s/A = 20 kg dry solid/m²; air at 60 °C with wet-bulb 30 °C; h = 25 W/(m²·K) (only convection); λ_w at 30 °C = 2430 kJ/kg. Dry from X = 0.35 to X_c = 0.15. Then estimate the time if the air mass velocity is doubled (h ∝ G^0.8).

  1. N_c = h (T − T_w)/λ_w = 25 × 30/(2430 × 10³) = 3.086 × 10⁻⁴ kg/(m²·s) = 1.111 kg/(m²·h).
  2. t_c = (L_s/A)(X₁ − X_c)/N_c = 20 × 0.20/1.111 = 3.60 h.
  3. Doubling G: h increases by 2^0.8 = 1.741, so N_c and the time scale by that factor: t_c = 3.60/1.741 = 2.07 h.

N_c = 1.11 kg/(m²·h), t_c = 3.6 h, falling to about 2.07 h at double air velocity.

Common mistakes

  • Using wet-basis moisture in the dry-basis formulas (convert with X = x/(1 − x)).
  • Forgetting to subtract X* in the falling-rate formula.
  • Taking the solid temperature as the air temperature during the constant-rate period – it is the wet-bulb temperature.
  • Expecting higher air velocity to speed up a diffusion-controlled falling-rate period.
  • Mixing W, kJ/h and kJ/s when computing N_c from h.
  • Using the total area of solids when only the top surface is exposed (or vice versa).

For GATE CH

Expect batch drying-time calculations (constant-rate and linear falling-rate), N_c from heat transfer data, conversions between wet and dry basis, ratio questions on how drying time changes with air velocity, temperature or thickness, and conceptual questions on critical, equilibrium, bound and free moisture.

Quick check

  1. Convert 20 % moisture (wet basis) to dry basis.
  2. What temperature does the surface reach during the constant-rate period?
  3. Which period is affected little by air velocity?
  4. For diffusion control, how does drying time change if slab thickness is doubled?

Answers: 1. X = 0.2/0.8 = 0.25; 2. the wet-bulb temperature of the air; 3. the falling-rate (internally controlled) period; 4. it becomes four times longer.

Interview questions

All Mass Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is drying in the context of mass transfer?Concept

    Drying is the process of removing moisture from a solid material by evaporation. It involves the transfer of mass (moisture) from the interior of the solid to its surface and then to the surrounding environment. This process is crucial in various industries to ensure product stability and quality.

  2. 2.Explain the concept of a drying rate curve.Concept

    A drying rate curve plots the drying rate per unit area, N = −(L_s/A)dX/dt, against the free moisture content of the solid, obtained by differentiating batch weight-loss data. After a short warm-up it shows a constant-rate period, where the surface is fully wet, sits at the wet-bulb temperature and the rate is set by external heat and mass transfer. Below the critical moisture content comes the falling-rate period, first as the surface dries in patches and then as internal diffusion or capillary flow controls, until the equilibrium moisture content is reached. The curve is used to compute batch drying times.

  3. 3.What factors influence the drying rate of a material?Concept

    Several factors influence the drying rate, including the temperature and humidity of the drying air, the airflow rate, the surface area of the material, the initial moisture content, and the properties of the material itself, such as porosity and thermal conductivity. These factors determine how quickly moisture can be removed from the material.

  4. 4.Why is the constant rate period important in drying processes?Application

    The constant rate period is important because it represents the phase where the drying rate is at its maximum and remains steady. During this period, the surface of the material is saturated with moisture, and the drying process is controlled by external conditions such as air temperature and velocity. Understanding this period helps optimize drying conditions for efficiency.

  5. 5.What happens if the drying temperature is too high?Application

    If the drying temperature is too high, it can lead to overheating and damage to the material being dried. This can cause degradation of the material's quality, such as discoloration, loss of nutrients, or structural damage. Additionally, it may lead to uneven drying, where the surface dries too quickly, trapping moisture inside.

  6. 6.How does air humidity affect the drying process?Application

    Air humidity affects the drying process by influencing the rate of moisture evaporation. High humidity levels reduce the drying rate because the air is already saturated with moisture, making it less effective at absorbing additional moisture from the material. Conversely, low humidity levels enhance the drying rate by allowing more moisture to evaporate into the air.

  7. 7.A batch of 20 kg dry solid is dried from 0.5 to 0.1 kg water/kg dry solid entirely in the constant-rate period, at 1.0 kg/(m²·h) over a drying area of 2 m². Calculate the drying time.Numerical

    Water removed = 20 × (0.5 − 0.1) = 8 kg. Evaporation rate = N_c × A = 1.0 × 2 = 2 kg/h. Time = 8/2 = 4 h, equivalently t = (L_s/(A·N_c))(X₁ − X₂) = (20/2)(0.4) = 4 h. This holds only if the critical moisture content is at or below 0.1; otherwise the falling-rate period must be added.

  8. 8.What is the significance of the falling rate period in drying?Application

    The falling rate period is significant because it indicates the phase where the drying rate decreases as the material becomes drier. During this period, the drying process is controlled by internal diffusion of moisture to the surface. Understanding this period is crucial for optimizing drying processes, as it often requires different conditions than the constant rate period.

  9. 9.Explain how porosity of a material affects its drying rate.Application

    Porosity affects the drying rate by influencing the ease with which moisture can move through the material. High porosity allows for easier movement of moisture to the surface, potentially increasing the drying rate. Conversely, low porosity can restrict moisture movement, slowing down the drying process. The material's structure and pore size distribution are key factors in this behavior.

  10. 10.A drying process operates at a constant rate of 0.015 kg/(m²·h). If the drying area is 3 m² and 1.5 kg of moisture must be removed in the constant-rate period, calculate the drying time.Numerical

    Evaporation rate = 0.015 kg/(m²·h) × 3 m² = 0.045 kg/h. Time = 1.5 kg/0.045 kg/h = 33.3 h. Keeping the rate in per-hour units avoids the common slip of reporting seconds; check that the result is physically sensible for the dryer and material.

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