Calculate the amount of water vapor

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Homework Help Overview

The discussion revolves around calculating the amount of water vapor in air under varying conditions of temperature and pressure. The subject area includes thermodynamics and fluid dynamics, particularly focusing on the behavior of water vapor in compressed air systems.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between temperature, pressure, and water vapor content, questioning how changes in these variables affect condensation and the state of water (vapor vs. liquid). There are attempts to clarify the initial conditions regarding pressure and flow rates, with some participants suggesting different assumptions about the system.

Discussion Status

The discussion is active, with participants raising questions about the assumptions made regarding pressure and flow rates. Some guidance has been provided regarding the calculations of water vapor density and the implications of cooling on condensation, but there is no explicit consensus on the correct approach or final calculations.

Contextual Notes

There are uncertainties regarding the initial conditions of the air flow, specifically whether the flow rate is at 1 atm or 9 bar. Additionally, the problem statement lacks clarity on how to determine the amount of water condensed during the cooling process.

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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