2) (55 pts.) The separation of methane M (CH) from carbon dioxide C (CO2) is ill

Question

2) (55 pts.) The separation of methane M (CH) from carbon dioxide C (CO2) is illustrated below using an absorber tower and a stripping tower. Methanol A (CHsOH) is used as the liquid in the absorber and is regenerated in the stripper using nitrogen gas (N2). Find all of the unknown molar flow rates n, n2, a a, and the molar composition of Stream 3, X3.c and X3. In Stream 4,90% of the CO2 is recovered with the nitrogen stripping gas. Notice that n, is the molar flow rate of only the COsin the exit stream from the stripping tower. nng h] 1o0 100 mal/h 0m Absorber pure N2

Explanation / Answer

Mole balance across absorber ( nf = 100mol/hr)

for CH4 balance, 100*.7 = n1*.99 this gives n1= 70.7 mol/hr

for CO2 balance, 100*.3+n2*.005 = n1*.01 + n3*x3,c

or 30+n2*.005 =.707 + n3*x3,c or 29.29+n2*.005 =n3*x3,c.........................(i)

Now moles balance across stripper:

n4 = .9*n3*x3,c

CO2 balance: n3*x3,c = n4+ n2*.005 = .9*n3*x3,c+ n2*.005 or n3*x3,c = n2*.05

using this in equation(i), we get, 29.29+n2*.005 = n2*.05 or n2 = 650.89 mol/hr

n3*x3,c = n2*.05 = 32.5 moles/hr now, n4 = .9*n3*x3,c so n4 = 29.29 mole/hr

now applying total mole balance across absorber,

100+n2= n1+ n3 or 100+650.89 = 70.7 + n3 so n3 =680.18 mol/hr

we know n3*x3,c = n2*.05 = 32.5 or x3,c = 32.5/680.18 = .04778

x3,a =1-x3,c = 0.9522

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