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## Main Question or Discussion Point

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.

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.