Mass-transfer coefficients from naphthalene sublimation 2.2. Mass-transferc data
ID: 702528 • Letter: M
Question
Mass-transfer coefficients from naphthalene sublimation 2.2. Mass-transferc data In a laboratory experiment, air at 347 K and 1 atm is blown at high speed around a single naphthalene (CoH) sphere, which sublimates partially. When the experiment begins, the diameter of the sphere is 2.0 cm. At the end of the experi- ment, 14.32 min later, the diameter of the sphere is 1.85 cm. (a) Estimate the mass-transfer coefficient, based on the average surface area of the particle, expressing the driving force in terms of partial pressures. The density of solid naphthalene is 1.145 g/cm3 and its vapor pressure at 347 K is 670 Pa (Perry and Chilton, 1973). Answer: kG 0.012 mol/mag-kPa (b) Caleulate the mass-transfer coefficient, for the driving force in terms of molar concentrations. Answer: k 0.034 m/sExplanation / Answer
Volume of Naphthalene sublimate = 4/3*3.14*(d1^3-d2^3)/8=4/3*3.14*(2^3-1.85^3)*10^-6/8=8.73*10^-7 m3
Mass of Naphthalene sublimate=Volume of Naphthalene sublimate*Density=8.73*10^-7*1145=0.00099 kg
Napthalene Molecular weight = 128 g/mol
=> Moles of Naphthalene sublimate=.99/128=0.0077 moles
Initial area of sphere = 4*3.14*d1^2/4=3.14*(2*10^-2)^2=.001256 m2
Final area of sphere = 4*3.14*d2^2/4=3.14*(1.85*10^-2)^2=.00107 m2
=>Average Area = (.001256+.00107)/2=.001163 m2
Vapor pressure of Napthalene at 670 K = 670 Pa = .67 Kpa
Time to sublimate,t = 14.32 minute = 859.2 s
=> Kg=0.077/.001163/859.2/.67=.012 mol/m2/s/kPa
b) Using PV=nRT for Napthalene
n/V=Concentration of Napthalene in air
=>n/V=P/R/T=670*10^-5/.082/347=.235 mol/m3
Napthalene concentration in sphere = Napthalene sphere Volume*Density/molecular weight/volume of sphere
=4/3*3.14*(2*10^-2/2)^3*1145*1000/128/(4/3*3.14*(2*10^-2/2)^3)=8945 mol/m3
Total Napthalene transferred = .0077 mole
Time = 859.2 s
Average Area = .001163 m2
=>.0077/859.2=Kc*.001163*(8945-0.235)
=>kc=.034m/s