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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of water in 20 !C air is about 0.25 cm2/sec. If the air out of the film is fifty percent saturated, how fast will the water level drop in a day?

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danialamin

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

                               [tex]j_{H_2O}=\frac{D}{l} \bigtriangleup c[/tex]

where

  • [tex]j_{H_2O}[/tex] is the mass flux of the water which is to be calculated.
  • D is diffusion coefficient which is given as [tex]0.25 cm^2/s[/tex]
  • l is the thickness of the film which is 0.15 cm thick.
  • [tex]\bigtriangleup c[/tex] is given as

                                [tex]\bigtriangleup c= \frac{P_{sat}-P_a}{RT}[/tex]

In this

  • [tex]P_{sat}[/tex] is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa
  • [tex]P_a[/tex] is the air pressure which is given as 0.5 times of [tex]P_{sat}[/tex]
  • R is the universal gas constant as [tex]8.314 kJ/kmol-K[/tex]
  • T is the temperature in Kelvin scale which is [tex]20+273= 293K[/tex]

By substituting values in the equation

                                    [tex]\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3[/tex]

Converting [tex]\bigtriangleup c[/tex] into [tex]cm^3/cm^3[/tex]

As 1 mole of water 18 [tex]cm^3[/tex] so

                               [tex]\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3[/tex]

Putting this in the equation of mass flux equation gives

                            [tex]j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s[/tex]

For calculation of water level drop in a day, converting mass flux as

                     [tex]j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day[/tex]

So the water level will drop by about 1.24 cm in 1 day.