Question

5) The following example illustrates an ammonia synthesis process in which an inert gas (I) is present, necessitating a purge stream. The fresh feed to the process contains nitrogen and hydrogen, as well as an inert gas (I). The feed is combined with a recycle stream containing the same three species, and the combined stream is fed to a reactor. The reactor effluent flows to a condenser. A liquid stream containing all of the ammonia formed in the reactor and a gas stream containing all the inerts and the unreacted nitrogen and hydrogen leave the condenser. The gas stream is split into two fractions with the same composition: one is removed from the process as a purge stream, and the other is the recycle stream, which is subsequently combined with the fresh feed. In every stream containing nitrogen and hydrogen, the two species are in stoichiometric proportion. Let yI0 be the mole fraction of inerts in the fresh feed, f the single-pass conversion of nitrogen in the reactor, and p the fraction of the gas leaving the condenser that is purged (mol purged/mol total). Draw and fully label a process flowchart, and using stoichiometric ratios, physical constraints and simple material balances in your head, try to introduce the fewest possible variables. In other words express as many quantities as possible in terms of yI0, f and p. Given yI0 = 0.01, f = 0.20, and p = 0.10, chose a basis anywhere in the problem and solve for the total moles fed to the reactor (nr), moles of ammonia produced (np), and overall nitrogen conversion (fov).

Answer 1

Let F= Moles of fresh feed of N2 and H2 and R= moles of Recycle containing N2 and H2

Mole fraction of feed of N2 and H2 = 1-y10   (1)

Total flow of N2, H2 and I= F/(1-y10) (2)

Moles of inerts in the feed = F*y10/(1-y10) (3)

Moles of N2 and H2 entering the reactor = F+R (4)

The reaction is N2+3H2-à2NH3   (5_

Percent conversion = (F+R)*f (6)

Moles of NH3 formed = (F+R)*f/2 (7)

Percent unconverted = (F+R)*(1-f) (8)

This is leaving as purge

Moles of N2 and H2 gas leaving as purge =(F+R)*(1-f)*P (9)

Under steady state conditions. Inerts in feed= inerts in purge

F*y10/(1-y10)= inerts in purge

Moles of N2 andd H2 recycled = (F+R)*(1-f)*(1-P) =R

(F+R)= R/(1-f)*(1-P) (11)

F= R*{1/(1-f)*(1-P)}-1) (12)

Basis : 1 moles/hr of N2 and H2 given y10=0.01,f =0.2 and P=0.1

From Eq.11, 1+R = R/(1-0.2)*(1-0.1) =1.4R

R =2.5 moles/hr of N2 and H2

Moles of NH3 formed, from Eq. 7, (1+2.5)*0.2/2= 0.35 moles/hr

Nitrogen in the feed = F*1/4 ( since feed contains 1 mole N2 and 3 moles H2)_=0.25 moles

As per the reaction, 1 mole of N2 gives 2 moles of NH3

0.25 moles gives 0.25*2=0.5 moles of NH3 (100% conversion)

Hence overall % conversion = 100*0.35/0.5= 70%

Inerts in feed = 1*0.01/(1-0.01)= 0.010

Total feed entering fresh = 1+0.010= 1.010 moles/hr ( including inerts)

Recycle = 2.5 moles/ hr of N2 and H2

Moles of N2 and H2 purged = 2.5*0.1= 0.25 moles/hr

So purge contains 0.010 moles/hr inerts and 0.25 moles/hr of N2 and H2

mole fraction inerts = 0.010/ (0.25+0.010)=0.038

this is also the mole fractino of inerts in the recylce, moles of inerts in recycle= 2.5*0.038= 0.095 moles

So total moles of recycle= 2.5+0.095= 2.595

total feed + recycle entering the reactor = 1.010+2.595 =3.605 moles/hr