Question

Consider moist air with absolute humidity of 0.02 (kg moisture)/(kg dry air) at 1 bar pressure. The vapor pressure of water is given by the equation: \[ \ln P_{{sat}} = 12 - \frac{4000}{T - 40} \] where \( P_{{sat}} \) is in bar and \( T \) is in K. The molecular weight of water and dry air are 18 kg/kmol and 29 kg/kmol, respectively. The dew temperature of the moist air is __________ ℃ (rounded off to the nearest integer).

Hint:

The dew point temperature corresponds to the temperature at which the water vapor in air condenses into liquid. You can calculate it using the vapor pressure equation, where the dew point is the temperature at which the vapor pressure equals the partial pressure of water vapor.

Answer 1

Correct Answer: 23

Step 1: Define the known values:
The absolute humidity is 0.02 kg of moisture per kg of dry air. The molecular weight of water is \( M_{{water}} = 18 \, {kg/kmol} \) and for dry air \( M_{{air}} = 29 \, {kg/kmol} \). The total pressure is 1 bar, and the vapor pressure equation is: \[ \ln P_{{sat}} = 12 - \frac{4000}{T - 40} \] Step 2: Calculate the partial pressure of the water vapor:
The absolute humidity is defined as the mass of water vapor per unit mass of dry air. For this, we first calculate the mole fraction of water vapor. Using the given absolute humidity: \[ y_{{H}_2{O}} = \frac{{moles of water vapor}}{{moles of dry air}} = \frac{{absolute humidity} \times 1000}{M_{{water}}} \div \frac{1}{M_{{dry air}}} \] Substituting the values: \[ y_{{H}_2{O}} = \frac{0.02 \times 1000}{18} \times \frac{29}{1000} = 0.0322 \] Now, the partial pressure of water vapor \( P_{{H}_2{O}} \) is: \[ P_{{H}_2{O}} = y_{{H}_2{O}} \times P_{{total}} = 0.0322 \times 1 = 0.0322 \, {bar} \] Step 3: Solve for the dew temperature \( T \):
At the dew point, the vapor pressure equals the partial pressure of the water vapor: \[ P_{{sat}} = P_{{H}_2{O}} = 0.0322 \, {bar} \] Substitute this into the vapor pressure equation: \[ \ln 0.0322 = 12 - \frac{4000}{T - 40} \] Taking the natural logarithm of 0.0322: \[ \ln 0.0322 = -3.442 \] Now solve for \( T \): \[ -3.442 = 12 - \frac{4000}{T - 40} \] \[ \frac{4000}{T - 40} = 12 + 3.442 = 15.442 \] \[ T - 40 = \frac{4000}{15.442} = 259.4 \] \[ T = 259.4 + 40 = 299.4 \, K \] Convert to Celsius: \[ T_{{dew}} = 299.4 - 273.15 = 26.25 \, {°C} \] After rounding to the nearest integer, the dew temperature is 26°C.