Determine the time necessary to reduce the center moisture content to 10 wt%, if

Question

Determine the time necessary to reduce the center moisture content to 10 wt%, if a 5-cm-thick slab of clay is placed on a belt passing through a continuous drier, thus restricting the drying to only one of the flat surfaces. (10.7 h). The initial moisture content is 15 wt% and the surface moisture content under the constant drying conditions is to be maintained at 4 wt%. The effective diffusivity of water through clay is 1.3 times 10^-4 cm^2/s. () As the volume of the slab is decreasing (shrinking) as it dries, it is best to use mass (weight) fractions on a dry basis in lieu of concentrations.

Explanation / Answer

Convert wt% to kg/m^3

Cx=10wt%

Density of water=1g/cm3

Density of clay=1.33g/cm3

Use eqn,

Cx=[C1/(C1/rho+C2/rho2)] *1000=[10/{(10/1g/cm3) +(90/1.33g/cm3)} ] *1000=(10/77.7 cm3/g)*1000=128.7kg/m3

Cs=4wt%= C1/C1/rho+C2/rho2 *1000=[4/(4/1g/cm3) +(96/1.33g/cm3) ] *1000=(4/76.9cm3/g)*1000=52.0 kg/m^3

Co=15wt%= C1/C1/rho+C2/rho2 *1000=[15/(15/1g/cm3) +(85/1.33g/cm3) ] *1000=(15/78.9)*1000=190.1kg/m3

From fick’s second law of diffusion,

Cx-Co/Cs-Co=1-erf(x/2(Dt)^1/2)

Cx= concentration at x=centre=5 cm/2=2.5 cm

C=co at t=o, 0<x<infinity

Cs=constant conc at the surface ,x=0

128.7-190.1/52-190.1=1-erf(2.5 cm/2(1.3*10^-4 cm2/s*t)^1/2

(61.4/138.1)-1=-erf(1.25 /(1.3*10^-4 cm2/s*t)^1/2)

0.55= erf(1.25 /(1.3*10^-4 cm2/s*t)^1/2)

Squaring both sides,

0.3025=1.56/(1.3*10^-4 cm2/s*t)

t=1.56 cm/(1.3*10^-4 cm2/s*0.3025)=39669.42 s=39669.42 s*1h/3600s=11.01 hr

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