Calculate the amount of water vapor

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SUMMARY

The discussion focuses on calculating the amount of water vapor in compressed air at different temperatures and pressures. At 25°C and 70% relative humidity, the water content is reduced from 22.8 g/m³ to 15.96 g/m³, resulting in a total of 3830.4 g/hr of water, with 608.08 g/hr in vapor form. At 20°C, the water content further decreases to 11.97 g/m³, yielding a total of 2872.8 g/hr, with 455.9 g/hr in vapor. The calculations assume the air is compressed to 9 bar absolute pressure, and the discussion highlights the importance of understanding the effects of temperature and pressure on water vapor density.

PREREQUISITES
  • Understanding of thermodynamics, specifically gas laws.
  • Knowledge of relative and absolute humidity concepts.
  • Familiarity with water vapor density calculations.
  • Experience with pressure conversions, particularly bar to atm.
NEXT STEPS
  • Research "Moist Air Properties" to understand the behavior of water vapor in air.
  • Learn about "Psychrometric Charts" for visualizing humidity and temperature relationships.
  • Study "Compressible Flow Dynamics" to grasp the effects of pressure on gas behavior.
  • Investigate "Condensation in Compressed Air Systems" for practical applications in engineering.
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Engineers, HVAC professionals, and anyone involved in thermodynamic calculations or compressed air systems will benefit from this discussion.

dbag123
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Homework Statement
Calculate the amount of water vapor when 4m^3/min or 25C air is cooled to 20C. How much vapor is condensed to water in 1 hour and how much of the vapor is still in the system.
Relevant Equations
pV/T=c
Hello
4m3/min is 240m3 /hr. Temperature is 293K, at 25C the amount of water is 22.8g/m3 and at 20C it is 17.1g/m3. Relative humidity is 70% and absolute pressure 9bar.

Solution:
@ 25C
relative humidity lowers the water content from 22.8 to 15,96g/m3 so in total 3830.4g/hr
240m3 or air is compressed to 1/9, so 26,67m3 and in that compressed air the amount of water vapor would be 608,08 g every hour
total amount of water 3830,4g/hr, in vapor its 608,08g and in solid water 3232,3g/hr

@20C
17,1g/m3 to 11,97g/m3 due to relative humidity. total amount of water is 2872.8g
then volume is compressed to 26.67m3 actually there is in compressed air 107.73g/m3 so the amount of vapor in compressed air would be 17.1g/m3*26.67m3 or 455.9g. and in form of water 2416.9g

How does this tell me the amount of water condensed or how much of the vapor is in form of liquid water during 1 hour from temperature of 25C to 20C?
 
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Is that 4m^3/min at 9 bars?
 
Chestermiller said:
Is that 4m^3/min at 9 bars?
the problem statement does not say that. i am assuming it is 4m^3/min at 1 atm

i have left some detail out sorry about that. the air is compressed to 9bar absolute due to the cooling
 
Last edited:
dbag123 said:
the problem statement does not say that. i am assuming it is 4m^3/min at 1 atm

i have left some detail out sorry about that. the air is compressed to 9bar absolute due to the cooling
This is not correct. You have moist air flowing at 9 bars, and the pressure does not change in the cooler. You have the correct water vapor density of 16.0 gm/m^3, so just multiply that by the volume flow rate of air. This will be the mass flow rate of water vapor entering. If there is any condensation at all, then the air coming out the cooler will be saturated with water vapor at 20 C. What would the gm/m^3 of water vapor be at the exit. Since the air cools from 298 to 293 K, its density increase and its volume flow rate has decreases by the exit. What would be the volume flow rate of air at the exit? What would be the gas flow rate of water vapor exiting the cooler, assuming the exit air was saturated?
 

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