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insens350 [35]
3 years ago
8

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

er 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
Engineering
1 answer:
Elodia [21]3 years ago
7 0

Answer:

1.25 cm/day

Explanation:

An air thickness , (l) = 0.15 cm

Air Temperature =

(T_a)=20^0C = (20+273)K\\(T_a)=293K

Mass Diffusion coefficient (D) = 0.25cm^2/sec

If the air pressure (P_a) = 0.5 P_{sat}

We are to determine how fast will the water (H_2O) level drop in a day.

From the property of air  at T = 20° C

P_{sat} = 2.34 from saturated water properties.

The mass flow of (H_2O)  can be calculated as:

H_2O = \frac{D}{\phi} \delta C

where:

\delta C = \frac{P_{sat}*P_a}{RT }

R(constant) = 8.314 kJ/mol.K

\delta C = \frac{2.34*0.5}{8.314*293 }

\delta C = 4.803*10^{-4}\\ \delta C =0.48*10^{-3} mol/m^3\\ \delta C = 0.48*10^{-6} mol/cm^3

Since 1 mole = 18 cm ³ of water

0.48*10^{-6}mol/cm^3 will be: (0.48*10^{-6}mol/cm^3 *18)cm^3/cm^3

\delta C = 8.64 * 10^{-6}

Again:

H_2O = \frac{D}{\phi} \delta C

= \frac{0.25}{0.15}*8.69*10^{-6} \frac{cm^2/sec}{cm}

=1.4481*10^{-5} \frac{cm^2/sec}{cm}

Converting the above value to cm/day: we have:

1.448*10^{-5}*3600*24\frac{cm}{s}*\frac{s}{yr}*\frac{yr}{day}

= 1.25 cm/day

∴ the rate at which  the water level drop in a day = 1.25 cm/day

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