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 problem describes an adiabatic saturator process, which involves cooling and humidifying air adiabatically until it reaches saturation. This scenario is analyzed using the principles of thermodynamics and energy balance.
Adiabatic Saturator Energy Balance
For an adiabatic saturator, the energy balance per unit mass of dry air entering the system is established between the inlet state and the outlet state. The general form of the energy balance equation is:
$ c_p T_1 + \omega_1 (h_{fg} + c_w T_1) + (\omega_2 - \omega_1) c_w T_2 = c_p T_2 + \omega_2 (h_{fg} + c_w T_2) $
Where the terms represent:
- $T_1$: Inlet temperature of air ($50 ^{\circ}C$)
- $\omega_1$: Inlet absolute humidity ($H \times 10^{-3}$ kg water / kg dry air)
- $T_2$: Outlet temperature of air ($25^{\circ}C$)
- $\omega_2$: Outlet absolute humidity ($0.02$ kg water / kg dry air)
- $c_p$: Specific heat capacity of dry air ($1.01$ kJ. $kg^{-1}K^{-1}$)
- $c_w$: Specific heat capacity of water ($4.18$ kJ. $kg^{-1}K^{-1}$)
- $h_{fg}$: Latent heat of vaporization of water ($2450$ kJ. $kg^{-1}$)
Applying the Energy Balance Equation
Substitute the given values into the energy balance equation:
$ 1.01 \times 50 + \omega_1 (2450 + 4.18 \times 50) + (0.02 - \omega_1) \times 4.18 \times 25 = 1.01 \times 25 + 0.02 (2450 + 4.18 \times 25) $
Calculating the Humidity Ratio ($\omega_1$)
Simplify and solve the equation for $\omega_1$ step-by-step:
$ 50.5 + \omega_1 (2450 + 209) + (0.02 - \omega_1) \times 104.5 = 25.25 + 0.02 (2450 + 104.5) $
$ 50.5 + 2659 \omega_1 + 2.09 - 104.5 \omega_1 = 25.25 + 0.02 (2554.5) $
$ 52.59 + (2659 - 104.5) \omega_1 = 25.25 + 51.09 $
$ 52.59 + 2554.5 \omega_1 = 76.34 $
Isolate the term with $\omega_1$:
$ 2554.5 \omega_1 = 76.34 - 52.59 $
$ 2554.5 \omega_1 = 23.75 $
Solve for $\omega_1$:
$ \omega_1 = \frac{23.75}{2554.5} \approx 0.009297318 \text{ kg water / kg dry air} $
Determining the Value of H
The problem states the inlet absolute humidity is $\omega_1 = H \times 10^{-3}$. Use this to find $H$:
$ H \times 10^{-3} = 0.009297318 $
$ H = 0.009297318 \times 1000 $
$ H = 9.297318 $
Rounding the value of $H$ to two decimal places gives $9.30$.
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)? 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).- For what value of Lewis number, the wet-bulb temperature and adiabatic saturation temperature are nearly equal?
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).
