Diffusive mass transfer 1) In the winter time (temp-5°C) a compound is diffusing

Question

Diffusive mass transfer 1) In the winter time (temp-5°C) a compound is diffusing from the sediment at the bottom of a very deep quiescent (no mixing) reservoir. The measured diffusion coefficient at that temperature is 1.0 x 105 cm'/s. It does not react with anything in the reservoir. The concentration at the sediment surface is 10 mg/L and is constant. For the following times a) What is the characteristic diffusion (penetration) distance? b) At what depth above the bottom will the concentration in the water be 8.0 mg/L? c) At what depth above the bottom will the concentration in the water be 9.0 mg/L at the given times for another non-reacting compound with a measured diffusion coefficient of 1.0 x 104 cm2/s. 100 10.000 100.000 Time (s) la Characteristic 1.000 diffusion distance (cm) Ib - Depth (cm) 1c - Depth (cm)

Explanation / Answer

using fick's second law of diffusion,

cx-co/cs-co=1-erf(x/2(Dt)^1/2 ) [for unsteady state diffusion]

Cs=surface concentration

cx=concentration at distance x

co=initial concentration at distance x=0

D=diffusion coefficient (cm^2/s)

x=position or lenght of diffusion (cm)

b)D=1.0*10^-5 cm2/s

for t=100 cm

cs=10mg/1000cm3=0.01 mg/cm3

cx=8mg/1000 cm3=0.008 mg/cm3

0.008/0.01=1-erf(x/2(Dt)^1/2)

0.8-1=-erf (x/2*(1.0*10^-5 *100)^1/2)

or,-0.2 =-erf(x/0.0632)

z=0.2 ,erf z=0.2227

x/0.0632=0.2227

x=0.014 cm

c) similarly, for t=100s

D=D=1.0*10^-5 cm2/s

cx=9mg/1000 cm3=0.009 mg/cm3

0.009/0.01=1-erf(x/2(Dt)^1/2)

0.9-1=-erf (x/2*(1.0*10^-5 *100)^1/2)

-0.1=-erf(x/0.0632)

z=0.1,erfz=0.1125

x/0.0632=0.1125

x=0.007 cm

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