Thermodynamic Properties of Moist Air for Drying Applications

Reference ID: MET-6D9F | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The thermodynamic properties of moist air are fundamental to Process Engineering, particularly in drying operations, HVAC system design, and industrial gas processing. Moist air is treated as a binary mixture of dry air and water vapor. Understanding the relationship between temperature, pressure, and humidity allows engineers to calculate the energy requirements for heating or cooling air streams and to determine the moisture-carrying capacity of air during drying processes. These calculations are essential for maintaining product quality, preventing condensation, and optimizing energy efficiency in industrial dryers.

Methodology & Formulas

The following formulas define the state of moist air at a constant total pressure P. The calculations rely on the saturation pressure of water vapor p_sat, which is determined using the Antoine equation:

\[ \log_{10}(p_{sat}) = A - \frac{B}{C + T} \]

where T is the dry-bulb temperature in °C. With the constants \(A = 8.07131\), \(B = 1730.63\), \(C = 233.426\), this equation yields the saturation pressure in mmHg. To work in kilopascals, the result must be converted:

\[ p_{sat}[\text{kPa}] = p_{sat}[\text{mmHg}] \cdot 0.133322 \]

The partial pressure of water vapor p_w is derived from the relative humidity φ (0 ≤ φ ≤ 1):

\[ p_{w} = \phi \cdot p_{sat} \]

The specific humidity ω, representing the mass of water vapor per unit mass of dry air, is calculated as:

\[ \omega = 0.622 \cdot \frac{p_{w}}{P - p_{w}} \]

The enthalpy h of the moist air stream is determined by the sensible heat of the dry air and the total enthalpy of the water vapor, referenced to 0 °C:

\[ h = c_{p,a} \cdot T + \omega \cdot (h_{g,0} + c_{p,v} \cdot T) \]

where c_p,a is the specific heat of dry air (≈ 1.005 kJ/(kg·K)), c_p,v is the specific heat of water vapor (≈ 1.82 kJ/(kg·K)), and h_g,0 is the enthalpy of vaporization at 0 °C (≈ 2501.3 kJ/kg).

Parameter	Condition / Regime	Constraint
Temperature Range	Empirical Validity	\( 0 \leq T \leq 100 \) °C
Relative Humidity	Physical Limit	\( 0 \leq \phi \leq 1.0 \)
Gas Behavior	Ideal Gas Assumption	\( P \approx 1 \) atm, \( T < 70 \) °C

To calculate the humidity ratio, also known as the mixing ratio, you must account for the partial pressures of water vapor and dry air. Follow these steps:

Measure the dry-bulb temperature and the relative humidity of the air stream.
Determine the saturation pressure of water vapor at the measured dry-bulb temperature using standard steam tables or the Antoine equation (ensuring correct unit conversion to kPa).
Calculate the partial pressure of water vapor by multiplying the saturation pressure by the relative humidity (as a fraction).
Apply the humidity ratio formula: ω = 0.622 P_v / (P_atm − P_v), where P_v is the partial pressure of water vapor and P_atm is the total atmospheric pressure.

The wet-bulb temperature is a direct indicator of the thermodynamic state of the air and its capacity to absorb moisture. It is essential because:

It represents the lowest temperature that can be reached by evaporative cooling.
It allows process engineers to determine the enthalpy of the air stream.
It helps identify the approach to saturation, which dictates the drying rate of the product.
It serves as a reliable control variable for adjusting heat input in closed-loop drying systems.

The dew point temperature is the threshold at which moist air becomes saturated and begins to condense into liquid water. For process engineers, monitoring this is vital to:

Prevent corrosion and microbial growth on internal ductwork surfaces.
Ensure that the exhaust air does not reach saturation before exiting the system.
Optimize the energy recovery process by avoiding condensation in heat exchangers.
Maintain consistent product quality by preventing moisture re-absorption during the cooling phase.

Worked Example: Heating of Moist Air for Drying

Scenario: An industrial drying process uses ambient air at 25.0 °C and 50 % relative humidity. The air is heated at constant total pressure to 40.0 °C without adding or removing moisture. This example calculates key thermodynamic properties before and after heating.

Knowns:

Total pressure, \(P = 101.325\) kPa
Initial dry-bulb temperature, \(T_{db1} = 25.0\) °C
Initial relative humidity, \(\phi_1 = 0.50\) (fraction)
Final dry-bulb temperature, \(T_{db2} = 40.0\) °C
Specific heat of dry air, \(c_{p,a} = 1.005\) kJ/(kg·K)
Specific heat of water vapor, \(c_{p,v} = 1.82\) kJ/(kg·K)
Reference enthalpy of vaporization at 0 °C, \(h_{g0} = 2501.3\) kJ/kg
Molecular weight ratio, \(M_w / M_a = 0.622\)

Step-by-Step Calculation:

Saturation pressure at initial temperature.
Using the Antoine equation (constants: \(A = 8.07131\), \(B = 1730.63\), \(C = 233.426\)) the saturation pressure in mmHg is: \[ \log_{10}(p_{sat1,mmHg}) = A - \frac{B}{C + T_{db1}} = 8.07131 - \frac{1730.63}{233.426 + 25.0} = 1.3753 \] \[ p_{sat1,mmHg} = 10^{1.3753} = 23.73 \text{ mmHg} \] Converting to kPa (\(1 \text{ mmHg} = 0.133322 \text{ kPa}\)): \[ p_{sat1} = 23.73 \cdot 0.133322 = 3.164 \text{ kPa} \]
Partial pressure of water vapor.
\[ p_{w1} = \phi_1 \cdot p_{sat1} = 0.50 \times 3.164 = 1.582 \text{ kPa} \]
Specific humidity (absolute humidity).
\[ \omega_1 = 0.622 \cdot \frac{p_{w1}}{P - p_{w1}} = 0.622 \cdot \frac{1.582}{101.325 - 1.582} = 0.00986 \text{ kg water/kg dry air} \]
Dew-point temperature.
The temperature at which \(p_{sat}(T_{dp}) = p_{w1} = 1.582\) kPa. Solving the Antoine equation in reverse (converting 1.582 kPa to 11.87 mmHg) gives: \[ T_{dp} = 13.91 \text{ °C} \]
Wet-bulb temperature (approximate).
Using the empirical formula of Stull (2011) for 1 atm and the initial state point: \[ T_{wb1} = 17.998 \text{ °C} \]
After heating to \(T_{db2} = 40.0\) °C (constant specific humidity, \(\omega_2 = \omega_1 = 0.00986\) kg/kg).
Saturation pressure at \(T_{db2}\) (mmHg to kPa): \[ \log_{10}(p_{sat2,mmHg}) = 8.07131 - \frac{1730.63}{233.426 + 40.0} = 1.7418 \] \[ p_{sat2,mmHg} = 10^{1.7418} = 55.21 \text{ mmHg} \quad\Rightarrow\quad p_{sat2} = 7.359 \text{ kPa} \] New relative humidity: \[ \phi_2 = \frac{p_{w1}}{p_{sat2}} = \frac{1.582}{7.359} = 0.2149 \quad (21.49\%) \]
Enthalpy before and after heating.
Enthalpy of moist air (kJ/kg dry air): \[ h_1 = c_{p,a} \cdot T_{db1} + \omega_1 \cdot (h_{g0} + c_{p,v} \cdot T_{db1}) \] \[ h_1 = 1.005 \cdot 25.0 + 0.00986 \cdot (2501.3 + 1.82 \cdot 25.0) = 50.24 \text{ kJ/kg dry air} \] \[ h_2 = 1.005 \cdot 40.0 + 0.00986 \cdot (2501.3 + 1.82 \cdot 40.0) = 65.58 \text{ kJ/kg dry air} \] Change in enthalpy: \[ \Delta h = h_2 - h_1 = 65.58 - 50.24 = 15.34 \text{ kJ/kg dry air} \]

Final Answer: The heating process reduces relative humidity from 50 % to 21.49 % and increases the enthalpy by 15.34 kJ per kg of dry air. The dew point remains at 13.9 °C, and the initial wet-bulb temperature is 18.0 °C.

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