Question

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?

Answer 1

Answer:

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

Explanation:

Here Mass flux of water vapour is given as

                               $j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

• $j_{H_2O}$ is the mass flux of the water which is to be calculated.

• D is diffusion coefficient which is given as $0.25 cm^2/s$

• l is the thickness of the film which is 0.15 cm thick.
• $\bigtriangleup c$ is given as

                                $\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

• $P_{sat}$ 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

• $P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

• R is the universal gas constant as $8.314 kJ/kmol-K$

• T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

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

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

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

Putting this in the equation of mass flux equation gives

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

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

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

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