Humidity Chart: Adiabatic Humidification

Organized by textbook: learncheme.com/
Calculates the outlet temperature and rate of water evaporation in a spray chamber for an adiabatic cooling process using the humidity charts. Made by faculty at Lafayette College and produced by the University of Colorado Boulder, Department of Chemical & Biological Engineering.
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[@SanjayYadav-fy4dg]

What minimum temperature of air can be achieved and what will be the required sprayed water temp.
Ex
If air initial temp 42 dec c
RH : 37 %
And i want achieve 13 deg C
Is it possible to achieve by adding water and what will be the water temp required

[@abhishekshukla6186]

Sir , Why doe adiabatic humidification occurs on the wet bulb line? Is it always the case or does it vary if we change the combination of fluids being used in the system (air/water ) ?

[@LearnChemE]

Thanks for your question! Adiabatic humidification actually occurs along the constant enthalpy line, because this is the definition of an adiabatic process. If you follow the yellow line to the "Enthalpy at Saturation" on the left side of the humidity chart, you'll see that the entire process occurs along the 65 kJ/kg dry air line.

[@dewan_lama]

@@LearnChemE thank you 😊 I also had this question ..Great explanation.

[@chrystaljanepante4775]

Good day. Is specific volume equal to humid volume? Does that mean that the volume of humid air is not significantly different from the volume of dry air? Thank you!

[@carultch]

Specific volume is the reciprocal of density. It accounts for all flavors of gas that are part of the mixture. Density is given by rho=P*M/(R*T), where M is the effective molar mass of the gas mixture, and P and T are pressure and temperature in the absolute scale. Its reciprocal v, is therefore v=R*T/(P*M).
To explore what humidity would do to either density or volume, calculate its effect on M, from realistic values of the water vapor mole fraction.
M = Xw*Mw + Xn*Mn + Xo*Mo
where X's are mole fractions, and M's are molar masses. The w/n/o refer to water, nitrogen, and oxygen respectively. The fixed data is the following:
Mn = 28 kg/kmol
Mo = 32 kg/kmol
Mw = 18 kg/kmol
Xo = 0.21*(1-Xw)
Xn = 0.78*(1-Xw)
Explore what happens to M as Xw changes from 0 to 6%, which is realistic for the molar composition of most air you'll experience at a survivable condition.
at Xw=0, M = 28.9 kg/kmol
at Xw=3%, M = 28.6 kg/kmol
at Xw=6%, M = 28.2 kg/kmol
You can see that the water vapor is a very small part in how it affects the molar mass, density, and specific volume of air.

[@AnamilM]

Thanks a lot for the explanation ! (Y)

[@muhandkhatm9719]

tanks man