3. Sulfur dioxide gas (SO2) is an agent that contributes significantly to enviro

Question

3. Sulfur dioxide gas (SO2) is an agent that contributes significantly to environmental problems, specifically acid rain. To reduce SO2 emissions, a coal burning plant employs a special reactor to oxidize SO2 with air to SO3. SO3 is then removed by scrubbing with liquid water, which leads to the formation of aqueous solution of H2SO4. The SO2 oxidation reactor takes place in the reactor. The reaction of SO3 with H2O to form H2SO4 takes place in the scrubber.

The fresh feed to the process has a flow rate of 100 lbmoles/h. This stream is 12 mol% SO2, remainder air. The single-pass conversion of SO2 in the reactor is 45%. The reactor outlet is fed to the scrubber, where the SO3 conversion is 100%. Pure liquid water enters the top of the scrubber. All the sulfuric acid and water leave the scrubber in the bottom stream.   The gas stream leaving the top of the scrubber (containing SO2, O2, and N2) is recycled (i.e. combined with the fresh feed). 2% of this stream is purged. The system is at steady state.

Calculate:
a) the molar flow rate of each species in each stream (in lbmoles/h)
b) the recycle ratio (moles recycled/moles fresh feed)
c) Overall conversion of SO2

Explanation / Answer

SO2+1/2O2--à SO3

Moles of air supplied =88 moles/hr

Moles of Nitrogen in the feed = 88*0.79= 69.52 moles/hr

Moles of oxygen   =69.52*0.21 =14.6 moles/hr

Under steady state conditions all this will be in the purge.

Flow rate of SO2 in the recycle= R

So2 in the feed = 100*12/100 =12 moles/hr

So total of SO2 entering the reactor = (R+12)

45% conversion takes place in the reactor = (R+12)*0.45

Moles of O2 consumed = (R+12)*0.45/2

Moles of SO2 remaining = (R+12)*0.55

This is recycled as well as purged

Purged SO2= R*2/100

Recycled SO2= R

Total SO2 remaining = R*(1+2/100)= 1.02R

Hence (R+12)*0.55= 1.02R

R+12 = 1.02R/0.55 =1.85R

0.85R= 12

R=12/0.85= 14.12

Moles of SO2 in the recycle = 14.12

Moles of O2 consumed = (14.12+12)*0.45/2 =5.877 moles/hr

Moles of SO2 in the purge= 14.12*2/100 = 0.2824 moles/hr

Moles of oxygen entering = moles of oxygen In the purge + moles of oxygen used for SO3 formation

14.6 = 5.877 + moles of oxygen in the purge

Moles of oxygen in the purge = 14.6-5.877 =8.7823 moles/hr

All the moles of nitrogen entering will be in the purge only

Hence 69.52 moles/hr of Nitrogen in the purge

Purge contains 8.7823 moles/hr O2, 0.2824 moles/ hr of SO2 and 69.52 moles/hr of nitrogen.

Total moles of purge = 8.7823+0.2824+69.52=78.5847

Composition: O2= 8.7823/(8.7823+0.2824+69.52) =0.111, SO2= 0.2824/78.5847 =0.0035 and N2= 0.885

0.0035 SO2 correspond to 14.12 moles of recycle

Hence moles of recycle = 14.12/0.0035=4034.3 moles/hr

Flow of each stream in recycle, SO2 : 4034.3*0.0035=14.12, O2= 0.111*4034.3 =448 and N2= 0.885*4034.3=3750.3 moles/hr

moles of SO2 entering =12 moles/hr

SO2 in purge = 0.2824

Moles of SO2 converted = 12-0.2824=11.72 moles/hr

Moles of SO3 formed = moles of SO2 converted = 11.72

Overall conversion = 100*11.72/12=97.7%

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