Influence of Air Temperature, Humidity, and Velocity on Drying Rate

Reference ID: MET-61DF | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The convective drying rate calculation is a fundamental process engineering analysis used to determine the moisture removal rate from a wetted surface, and it directly informs the design of continuous convective drying systems for solid foods. This calculation is critical in the design and optimization of industrial dryers, cooling towers, and food processing equipment. By modeling the steady‑state, constant‑rate drying period through experimental determination and modeling of drying curves, engineers can predict the specific energy and efficiency metrics for drying processes and time necessary to achieve target moisture levels, ensuring product quality and process efficiency.

Methodology & Formulas

The calculation follows a sequential approach based on heat and mass transfer analogies, as explained in the surface evaporation mechanism and heat‑mass transfer coupling. First, the mass velocity of the air is determined, which serves as the basis for calculating the convective heat transfer coefficient. Using the Lewis analogy (with the well‑established simplification \(Le \approx 1\) for air–water systems), the mass transfer coefficient is derived directly from the heat transfer coefficient, allowing for the calculation of the evaporation rate based on the humidity gradient between the surface and the bulk air, and can be further refined by incorporating a combined convection‑radiation heat transfer analysis for more accurate drying predictions.

The governing equations are defined as follows:

Mass Velocity: \( G = v \cdot \rho \)
Heat Transfer Coefficient: \( h = 20 \cdot G^{0.8} \)
Saturation Humidity: \( H_{s} = \dfrac{0.622 \cdot P_{sat}(T_{s})}{P - P_{sat}(T_{s})} \)
Mass Transfer Coefficient (Lewis analogy, air–water): \( k_{g} = \dfrac{h}{c_{p}} \)
Evaporation Rate: \( \dot{m}_{evap} = k_{g} \cdot A \cdot (H_{s} - H_{a}) \)
Heat Balance Check: \( \dot{m}_{evap} \cdot \lambda = h \cdot A \cdot (T_{a} - T_{s}) \)

Parameter	Condition/Threshold	Engineering Significance
Mass Velocity (\(G\))	\( 0.5 \leq G \leq 5.0 \)	Empirical validity range for the heat transfer correlation.
Temperature (\(T_{a}\))	\( 0 \leq T_{a} \leq 60 \)	Range where the simplified Lewis analogy (\(Le \approx 1\)) is sufficiently accurate for air–water systems.
Heat Balance	\( \text{Discrepancy} \leq 5\% \)	Ensures the assumed surface (wet‑bulb) temperature is physically consistent with the calculated evaporation rate.
Drying Regime	\( \dot{m}_{evap} / A < 0.04 \)	Recommended threshold to prevent surface crusting (case hardening) during the constant‑rate period. Above this flux, the surface may dry faster than moisture can diffuse from the interior.

Increasing the air temperature significantly accelerates the drying rate by providing more energy for moisture evaporation. This process impacts the system in the following ways:

It increases the vapor pressure of the moisture within the material, facilitating faster migration to the surface.
It lowers the relative humidity of the air, thereby increasing the driving force for mass transfer.
It reduces the viscosity of the liquid water, allowing for more rapid internal diffusion.

The humidity of the drying air determines the equilibrium moisture content of the material. When humidity is high, the air is closer to saturation, which results in:

A reduced concentration gradient between the material surface and the surrounding air.
A slower evaporation rate, particularly during the constant‑rate period of drying.
A higher final moisture content, as the material cannot dry below the equilibrium point dictated by the ambient humidity.

Air velocity directly influences the thickness of the boundary layer at the material surface. By increasing the velocity, process engineers can achieve the following:

Reduction of the stagnant air film resistance, which enhances both heat and mass transfer coefficients.
More efficient removal of the vaporized moisture from the immediate vicinity of the product.
Improved uniformity of drying across the entire surface area of the material.

Worked Example: Convective Drying Rate of a Wet Surface

Scenario: A flat, fully wetted plate of dimensions 0.5 m × 0.5 m is being dried in the steady‑state, constant‑rate period. Air at 40 °C and 20 % relative humidity (absolute humidity \(H_{a} = 0.0093\) kg water/kg dry air) flows parallel to the surface at a velocity of 2.5 m/s. The plate surface is at the wet‑bulb temperature of 22 °C. The air density is 1.127 kg/m³, the specific heat capacity is 1005 J/(kg·K), and the latent heat of vaporization of water is \(2.407 \times 10^{6}\) J/kg. The objective is to compute the evaporation rate and verify the heat balance.

Knowns:

Air temperature: \(T_{a} = 40.0\) °C
Relative humidity: \(\text{RH} = 0.2\) (20 %)
Air velocity: \(v = 2.5\) m/s
Air density: \(\rho = 1.127\) kg/m³
Surface (wet‑bulb) temperature: \(T_{s} = 22.0\) °C
Absolute humidity of bulk air: \(H_{a} = 0.0093\) kg/kg
Plate width \(w = 0.5\) m, length \(L = 0.5\) m
Plate area: \(A = w \cdot L = 0.25\) m²
Atmospheric pressure: \(P = 101.325\) kPa
Specific heat of air: \(c_{p} = 1005.0\) J/(kg·K)
Latent heat of vaporization: \(\lambda = 2.407 \times 10^{6}\) J/kg

Step‑by‑step calculation:

Compute mass velocity \(G\)
\[ G = v \cdot \rho = 2.5 \times 1.127 = 2.818 \; \text{kg/(m}^{2}\!\cdot\!\text{s)} \]
Compute heat transfer coefficient \(h\)
\[ h = 20 \cdot G^{0.8} = 20 \times (2.818)^{0.8} = 45.81 \; \text{W/(m}^{2}\!\cdot\!\text{K)} \]
Determine saturation humidity \(H_{s}\) at the surface temperature
Using the Antoine‑type approximation for water vapor pressure:

\[ P_{\text{sat}}(T_{s}) = 0.611 \cdot \exp\!\left(\frac{17.27 \cdot 22.0}{22.0 + 237.3}\right) = 2.645 \; \text{kPa} \]

\[ H_{s} = \frac{0.622 \cdot P_{\text{sat}}(T_{s})}{P - P_{\text{sat}}(T_{s})} = \frac{0.622 \times 2.645}{101.325 - 2.645} = 0.0167 \; \text{kg/kg} \]
Compute mass transfer coefficient \(k_{g}\) using the Lewis analogy for air–water
For air–water systems the Lewis number \(Le \approx 1\); therefore the mass transfer coefficient is obtained directly from the heat transfer coefficient:

\[ k_{g} = \frac{h}{c_{p}} = \frac{45.81}{1005.0} = 0.0456 \; \text{kg/(m}^{2}\!\cdot\!\text{s)} \]
Calculate evaporation rate \(\dot{m}_{evap}\)
\[ \dot{m}_{evap} = k_{g} \cdot A \cdot (H_{s} - H_{a}) = 0.0456 \times 0.25 \times (0.0167 - 0.0093) \]

\[ \dot{m}_{evap} = 8.40 \times 10^{-5} \; \text{kg/s} \]
Verify heat balance
Heat required for evaporation:

\[ Q_{\text{req}} = \dot{m}_{evap} \cdot \lambda = 8.40 \times 10^{-5} \times 2.407 \times 10^{6} = 202.1 \; \text{W} \]

Heat supplied by convection:

\[ Q_{\text{sup}} = h \cdot A \cdot (T_{a} - T_{s}) = 45.81 \times 0.25 \times (40.0 - 22.0) = 206.1 \; \text{W} \]

The discrepancy ratio is:

\[ \left| \frac{Q_{\text{req}} - Q_{\text{sup}}}{Q_{\text{sup}}} \right| = 0.0194 \quad (1.94\%) \]

This is within the acceptable 5 % tolerance, confirming consistency between the assumed wet‑bulb temperature and the calculated evaporation rate.

Final Answer: The total evaporation rate is \(8.40 \times 10^{-5}\) kg/s, corresponding to a drying flux of approximately \(3.36 \times 10^{-4}\) kg/(s·m²) of surface area.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle