Consider moist air with absolute humidity of 0.02 (kg moisture)/(kg dry air) at 1 bar pressure. The vapour 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 ____________ °C (rounded off to the nearest integer).
Moist Air Dew Point Calculation
The dew point temperature is the temperature at which the air becomes saturated with water vapor, meaning the partial pressure of the water vapor ($P_v$) equals the saturation vapor pressure ($P^{sat}$) at that temperature.
Step 1: Calculate Partial Pressure of Water Vapor ($P_v$)
Use the formula for absolute humidity ($\omega$):
$ \omega = \frac{M_w}{M_a} \times \frac{P_v}{P_{total} - P_v} $
Where:
- $\omega = 0.02$ (kg moisture / kg dry air)
- $M_w = 18$ kg/kmol (Molecular weight of water)
- $M_a = 29$ kg/kmol (Molecular weight of dry air)
- $P_{total} = 1$ bar (Total pressure)
Rearranging the formula to solve for $P_v$:
$ P_v = \frac{\omega \times P_{total}}{\omega + \frac{M_w}{M_a}} $
Substitute the values:
$ P_v = \frac{0.02 \times 1}{0.02 + \frac{18}{29}} $
Calculate the ratio $\frac{M_w}{M_a}$:
$ \frac{18}{29} \approx 0.6207 $
Now calculate $P_v$:
$ P_v = \frac{0.02}{0.02 + 0.6207} = \frac{0.02}{0.6407} \approx 0.03121 \text{ bar} $
Step 2: Determine Dew Point Temperature ($T_{dp}$)
At the dew point, the partial pressure $P_v$ is equal to the saturation pressure $P^{sat}$.
Use the given saturation vapor pressure equation:
$ \ln P^{sat} = 12 - \frac{4000}{T - 40} $
Set $P^{sat} = P_v = 0.03121$ bar and solve for $T$ (which will be $T_{dp}$ in Kelvin):
$ \ln(0.03121) = 12 - \frac{4000}{T_{dp} - 40} $
Calculate $\ln(0.03121)$:
$ \ln(0.03121) \approx -3.467 $
Substitute this back into the equation:
$ -3.467 = 12 - \frac{4000}{T_{dp} - 40} $
Rearrange to solve for $T_{dp} - 40$:
$ \frac{4000}{T_{dp} - 40} = 12 - (-3.467) = 15.467 $
$ T_{dp} - 40 = \frac{4000}{15.467} \approx 258.6 \text{ K} $
Solve for $T_{dp}$ in Kelvin:
$ T_{dp} = 258.6 + 40 = 298.6 \text{ K} $
Step 3: Convert Dew Point Temperature to Celsius
Convert the dew point temperature from Kelvin to Celsius:
$ T_{dp}(°C) = T_{dp}(K) - 273.15 $
$ T_{dp}(°C) = 298.6 - 273.15 = 25.45 \text{ °C} $
Conclusion: Rounded Dew Point
Rounding the result to the nearest integer gives:
$ T_{dp} \approx 25 \text{ °C} $
This value lies between 23 and 27 °C.
Important Questions from Humidification
- Vapor pressure of water at various temperature is given in the table.
Temperature (in K) 284 289 294 299 310 Vapor pressure of water (in kPa) 1.28 1.80 2.50 3.40 6.40
An air and water vapor mixture at $100 \text{ kPa}$ with a relative humidity of $20\%$ has a dry-bulb temperature of $310 \text{ K}$. Assume latent heat of vaporization for water is $44 \text{ kJ mol}^{-1}$ and specific heat capacity of the air and water vapor mixture is $0.035 \text{ kJ mol}^{-1}\text{K}^{-1}$. Which one of the following is the closest to its wet-bulb temperature (in K)? - For what value of Lewis number, the wet-bulb temperature and adiabatic saturation temperature are nearly equal?
Partially saturated air at 1 bar and $50 ^{\circ}C$ is contacted with water in an adiabatic saturator. The air is cooled and humidified to saturation, and exits at $25^{\circ}C$ with an absolute humidity of 0.02 kg water per kg dry air. Use latent heat of vaporization of water as 2450 kJ. $kg^{-1}$, and average specific heat capacity for dry air and water, respectively as 1.01 kJ.$kg^{-1}K^{-1}$ and 4.18 kJ. $kg^{-1}K^{-1}$. If the absolute humidity of air entering the adiabatic saturator is $H \times 10^{-3}$ kg water per kg dry air, the value of $H$ is ________ (rounded off to two decimal places).
The dry-bulb temperature of air in a room is $30 \ \degree C$. The Antoine equation for water is given as
$ln \ P^{sat} = 12.00 - \frac{4000}{T-40}$
where $T$ is the temperature in $K$ and $P^{sat}$ is the saturation vapor pressure in $bar$. The latent heat of vaporization of water is $2000 \ kJ \ kg^{-1}$, the humid heat is $1.0 \ kJ \ kg^{-1} \ K^{-1}$, and the molecular weights of air and water are $28 \ kg \ kmol^{-1}$ and $18 \ kg \ kmol^{-1}$, respectively. If the absolute humidity of air is $Y' \ kg$ moisture per $kg$ dry air, then for a wet-bulb depression of $9 \ \degree C$, $1000 \times Y' = $ _________ (rounded off to one decimal place).The humidity of air at a dry-bulb temperature of $65^{\circ}C$ is $0.025 \; kg \; water/kg \; dry \; air$. The latent heat of vaporization of water at $0 ^{\circ}C$ is $2500 \; kJ/kg$. The psychrometric ratio of air is $0.95 \; kJ \; (kg \; dry \; air)^{-1} \; K^{-1}$. Considering $0 ^{\circ}C$ as reference temperature, the enthalpy of air (in kJ/kg) at its adiabatic saturation temperature of $35 ^{\circ}C$ is ________ (rounded off to two decimal places).
