The fresh feed to an ammonia production process contain 24.75 mole% N_2, 74.25 m

Question

The fresh feed to an ammonia production process contain 24.75 mole% N_2, 74.25 mole% H_2, and the balance inserts (I). The feed is combined with a recycle stream containing the same species, and the combined stream is fed to a reactor in which a 25% single-pass conversion of N_2 is achieved. The products pass through a condenser in which essentially all the ammonia is removed, and the remaining gases are recycled. However, to prevent buildup of the inserts in the system, a purge stream must be taken off. The recycle stream contains 12.5 mole% inserts. Calculate the overall conversion of nitrogen, the ratio (moles purge gas/mole of fresh feed), and the ratio (mole fed to reactor/mole fresh feed). [92.9%; 0.0800 mole purge gas/mole fresh feed; 4.07 mole fed reactor/mole fresh feed]

Explanation / Answer

Basis : Feed flow rate = 100 mol/hr. It contains   Inerts = 100- (24.75+74.25)= 1 mol/hr

Since inerts do not participate in the reaction. At steady state

Mole of inerts in = moles of inerts out in the purge

Let P= molar flow rate of purge

Purge balance ( Overall )

P*0.125= 1

P= 1/0.125= 8 moles/hr

Ratio of Product/ Feed = 8/100 =0.08 moles of Product/ moles of Feed

Composition of Recycle ( same as purge)= 12.5% I, and N2 and H2= 100-12.5= 87.5%

Molar ratio of N2: H2 in the feed = 24.75: 74.25 =1:3

The molar ratio of reactants N2: H2= 1:3

The reaction is N2+3H2--à2NH3. 4 moles of mixture gives 2 moles of NH3

Moles of mixture of N2 and H2 entering = 24.75+74.25= 99 moles

Moles of N2 and H2 entering in recycle = 100-12.5= 87.5

Let R = Recycle flow rate

Hence feed of recycle and fresh feed = R*0.875+99

Percentage conversion is 25%

So moles of N2 and H2 converted = R*0.875+99)

Moles of N2 and H2 converted = (R*0.875+99)*0.25

Moles of N2 and H2 unconverted= (R*0.875+99)*0.75

This must be equal to R*0.875+10 ( Recycle + purge)

(R*0.875+99)*0.75= R*0.875+8

R*0.875*0.75+99*0.75 = R*0.875+8

R*0.875*(1-0.75)= 99*0.75-8

R= 302.86

Feed to reactor = 302.86+100= 402.86 moles/hr

Fresh feed = 100 moles/hr

Moles of feed to reactor/moles of fresh feed =402.86/100 = 4.03 moles/mole of feed.

moles of NH3 formed = (R*0.875+99)*0.25/2=45 moles/ hr

moles of N2 utilized for producing 45 moles/hr= 22.5 moles/hr

Moles of N2 supplied = 24.25 moles/hr

% conversion of N2 = 100{(22.5/24.25}=92.78%

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