Define each of the terms and give the units for each term in the following equat

Question

Define each of the terms and give the units for each term in the following equation: N_A = D_AB rho/L_chi_Bim (xhi_A0 - chi_AL) State the significance of the chi_Bim term. Ammonia gas diffuses through a 2 mm thick stagnant air film to a sulphuric acid stream, where it disappears immediately via a chemical reaction. The whole process takes place at 20 degree C and 1 atm pressure. The mole fraction of ammonia at the bottom layer of the stagnant air is 2 %. It is assumed that ammonia is an ideal gas, calculate rho for ammonia at 20 degree C. Calculate chi_Bim Given that D = 0.191 cm^2 s^-1 for ammonia diffusing into air, calculate the mass transfer rate of 1 kmol ammonia diffusing through the stagnant air. Calculate the mass of ammonia reacting with the sulphuric acid over a 72-hour period if the reaction vessel has a surface area of 4 m^2.

Explanation / Answer

NA= Flux due to diffusion its units are gm/cm2.s

DAB= Diffsivity = cm2/s, L= thickness(cm), p= density in g/cm3

X= mole fraction of diffusing component.

2.During diffusion, the concentration profiles are not linear. Hence logarithmic concentratino difference is a meausre of mean concentration difference.

3. density of ammoia at 20 deg,c and 1 atm can be calculated from ideal gas equation or refering to standard conditions.

From PV= nRT, P*Molar mass= p*RT, p= density of ammonia in g/L

1* 17= p* 0.0821L.atm/mole.K* (20+273)

Hence p= 0.706 g/L = 0.706* 10-3 g/cc

B is the second component, given XA1=0.2, XB1=1-0.2= 0.8, XA2= 0 and XB2= 1-0 =1 ( since the reaction is instantaneous)

Xbln= (XB2-XB1)/ ln(XB2/XB1)= (1-0.8)/ ln(1/0.8)= 0.896

Given L =2mm =2*10-3 m , p= 0.706 g/L= 0.706 kg/m3, D= 0.191 cm2/s = 0.191*10-4 m2/s

NA= 0.191*10-4*0.706 (0.2-0)/ (2*10-3*0.896)=0.0015 kg/m2.s

For 72 hrs= 72*60*60 seconds and A= 4 m2

Mass transferred = 0.0015* 72*60*60*4 kg=1555.2 kg

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