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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

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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<h2>Further explanation </h2>

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Equation (4) becomes :

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$A=0.125\ m^2$

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4 years ago

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a.

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b.

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P1/yH20 + V1²/2g + z1 = P3/yH20 + V3²/2g + z3

Substitute in the values

33.03 + 20 = P3/yH20 + 55

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