The rate of evaporation at high temperatures

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    Evaporation Rate
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SUMMARY

The discussion centers on calculating the evaporation rate of 2 liters of water in a venting system with an air temperature of 360 degrees Celsius. The user applies the Langmuir evaporation rate formula, resulting in a calculated evaporation time of approximately 108 seconds for complete evaporation. Key parameters include a surface area of 0.1 m², a vapor pressure of 18666 Pa, and an ambient pressure of 2.3392 Pa. The user expresses skepticism about the accuracy of this rapid evaporation time, prompting a request for validation or correction of their calculations.

PREREQUISITES
  • Understanding of Langmuir evaporation rate equations
  • Familiarity with thermodynamic principles, specifically vapor pressure
  • Knowledge of calculus for rate of change calculations
  • Basic concepts of gas laws and molecular weight
NEXT STEPS
  • Research advanced evaporation models beyond Langmuir, such as the Hertz-Knudsen equation
  • Explore the effects of varying ambient pressures on evaporation rates
  • Investigate the thermal dynamics of water at high temperatures and pressures
  • Learn about experimental methods to measure evaporation rates in controlled environments
USEFUL FOR

Students in thermodynamics, engineers working on cooling systems, and researchers studying phase changes in high-temperature environments will benefit from this discussion.

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To whom it may concern
I'm a University of Portsmouth student in the UK and have a conundrum for one of my projects. I'm trying to evaporate a reservoir of water at the bottom of a venting system to cool down the airflow in a gap between two membranes for a period of time. The reservoir is at the bottom of the void and the temperature of the air above the water is 360 degrees C. How long would it be for 2 litres of water to evaporate?
I have looked for a simple solution for this and have come across Langmuir evaporation rate but I'm struggling to make sense of it. If this the correct way forward?
According to my findings I need to use the equation
(mass loss rate)/(unit area)= (vapour pressure- ambient partial pressure)* sqrt((molecular weight)/(2*pi*Gas constant*temperature in Kelvin))
The Surface area (unit Area) of the water will be 0.1m²
the vapour pressure of water at 360 degrees C is 18666
the pressure at ambient (20degrees) is 2.3392
Molecular weight of water is 0.014kg/mole
Gas Constant (R) = 461.5 for water vapour
Temperature in kelvin = 360+273=633

Therefore this would mean through calculus that the rate of evaporation is 0.185 kg/m²/sec(3 d.p.). Meaning 2 litres of water with a surface area of 0.1m² and an air temperature above of 360C will evaporate dry in roughly 108 seconds.
This surely can not be right.
Can anyone advise on the errors of my ways?
Thank you in advance if you can.
 
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UOP Student said:
Meaning 2 litres of water with a surface area of 0.1m² and an air temperature above of 360C will evaporate dry in roughly 108 seconds.
This surely can not be right.
"Surely" because of an observation, or "surely" because it just doesn't sound right? I would be surprised if water could hang around for too long so high above its boiling point. Unfortunately, my oven doesn't get that hot, so I can't test it. But perhaps I will see how long it takes for 2 L of water at 561 K (288 C) to boil away.
 
Surely in the sense that this time seems far too short a time for 2 litres of water to last. Air entering the vent system is 20 degrees and the air 250mm up is 360. That figure seems just too quick to me.
 

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