A closed vessel contains moist air at 20 degree celcius, the relative humidity being 30 %. What would be the relative humidity become if the vessel were cooled to 10 degree celcius ?
SVP of watyer at 20 deg celcius = 17.5 mm
SVP of water at 10 deg celcius = 9.2 mm
Relative humidity at 20 deg C is
= (Actual vapor pressure * 100)/Saturation vapor pressure at 20 deg ...
The relative humidity in a closed vessel is determined. The solution explains each step and provides all mathematical steps.
Please solve the problem using an atmospheric pressure of 85 kPa. Do not use a psychrometric chart but from first principles using formulae. I think you need to use steam tables:
In an air-conditioning unit 3.5 m^3/s of air at 27 degrees Celsius dry-bulb temperature, 50% relative humidity and standard atmospheric pressure enters the unit. The leaving condition of the air is 13 degrees Celsius dry-bulb temperature and 90 percent relative humidity. Using properties from the psychrometric chart:
a. Calculate the refrigerating capacity in kilowatts
b. Determine the rate of water removal from the air
c. Determine the mass of water vapor in 400 m^3 of air at 52 degrees Celsius and a relative humidity of 20% when the atmospheric pressure is 100 kPa and 85 kPa respectively.