Hello! Can somebody please help me with this problem? Thank you! :) Ammonia is b

Question

Hello! Can somebody please help me with this problem? Thank you! :)

Ammonia is being diffused through a stagnant gaseous mixture consisting of one third of nitrogen and two thirds of hydrogen by volume. The total pressure is 30 lbf / in ^ 2 abs (206.8 kN / m^2) and the temperature is 130 ° F (54 ° C). Calculate the rate of diffusion of the ammonia through a gas film of 0.5 mm thickness when the change in concentration through the film is 10 to 5% ammonia by volume. Answer: 2.05 (10 ^ -4) kmol / m2 . s

Explanation / Answer

DNH3-N2= 2.3*10-5m2/s at 298 K, at 54+273= 327K, DNH3-N2= 2.3*10-5*(327/298)1.5= 2.64*10-5 m2/s

DNH3-H2= 7*10-5 m2/s at 298K, at 327K, DNH3-H2= 7*10-5* (327/298)1.5= 8*10-5 m2/s

effective diffusivity, DAB = 1/ {(yb/DNH3-N2) + 1/yc/(NH3-H2)}

yb= mole fraction of N2= 2/3=0.667 and yc= 1/3= 0.333

effective diffusivity= 10-5 / {(0.667/2.64) +0.333/8)}=3.4*10-5 m2/s

PA1= partial pressure of NH3 on one side= 0.1*206.8= 20.68 KN/m2 and PA2= partial presure on the other side= 206.8*0.05= 10.34 KN/m2

PB1= partial presure of remaining gas= 206.8-20.68=186.12 KN/m2 and PB2= 206.8-10.34=196.46 KN/m2

pBM= mean partial pressure= (196.46-186.12)/ln(196.46/186.12) =191 KN/m2

NA= molar flux= DAB*Pt*(PA1-PA2) /RTZPBM

=3.4*10-5*206.8*(20.68-10.34)/(8314*327* 0.5*10-3*191)=9.15*10-5 kmol/m2.s

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