Chemical Engineering problem. The fresh feed to an ammonia synthesis plant conta

Question

Chemical Engineering problem.

The fresh feed to an ammonia synthesis plant contains 24.30% N 2 , 75.20% H 2 and 0.50%
Ar. The reaction in the reactor is not complete so that recycle is necessary. In a specific
operation, the fresh feed is mixed with recycled gases and the mixture is sent to the reactor.
The gases leaving the reactor are sent to a cooler-condenser from which pure liquid
ammonia is obtained. The gas stream from the cooler-condenser also includes some
ammonia in gaseous form, and is recycled to the reactor. The unlimited accumulation of
argon in the system is prevented by incorporating a purge stream into the recycle line. Some
additional, partial information about some of the process streams is given below:
Reactor Feed – Ammonia 26.56%
Purge – Nitrogen 9.79%, ammonia 31.27%
Calculate, on the basis of 100 moles of fresh feed
a) The moles of purge, liquid ammonia product and the recycle streams and the
composition of the purge stream.
b) The reactor and overall conversion amounts – both based on the limiting reactant.
c) The overall plant yield, expressed as the percentage of total reactants in the fresh
feed that is converted to product.

Explanation / Answer

Since Recycle and purge will have the same composition

Making a material balance across the reactor

R*31.27 ( ammonia from Recycle stream)+ 0 (ammonia in the fresh feed)= (100+R)* 26.56

R*31.27= 100*26.56+R*26.56

R = 100*26.56/(31.27-26.56)= 563.9 moles

Purge stream contains 9.79 % Nitrogen, 31.27% NH3 and balance =100-(9.79+31.27) has to be Argon and NH3=58.94%

Let xAr= composition of Argon in the purge and xH2= composition of Argon in the purge ( Recycle also)

Nitrogen in the feed =24.3 moles, H2= 75.2 moles and Argon =0.5 moles

The reaction is N2+3H2--à 2NH3

Let L= Moles of liquid ammonia with drawn, P= purge

1 mole of nitrogen gives rise to 2 moles of Ammonia

Writing nitrogen balance (N)

Nitrogen entering = Nitrogen leaving in the purge + Nitrogen leaving in the form of ammonia

24.3*2= L ( Nitrogen in liquid ammonia)+ P*(9.79*2/100 ( nitrogen in the purge +31.27/100 Nitrogen in the form of ammonia)

48.6 =L + 0.5085P

L= 48.6- 0.5085P

Writing argon balance

100*0.5/100= P*xAr

0.5= P*xAr (3)

Writing Hydrogen balance

75.2*2= L*3 ( ammonia in the liquid + P*x2H ( hydrogen in the purge )

75.2*2= 3L + P*x2H (4)

Addition of Eq.3 and Eq.4 gives

150.4= 3L+P*(2xH+xAr)

But given that xH2+xAr= 0.5894 (58.94%)

150.9=3L +P*0.5984 (5)

Substitution of P value from Eq.2 in Eq.5 gives

75.7=3*(48.6-0.5085P)+0.5894P

(3*0.5085-0.5894)P= 3*48.6-75.7=70.1

P*0.9361= 70.1

P= 70.1/0.9361=74.88 moles/hr

L= 48.6-0.5085*74.88=48.6-38.08=10.52 moles/hr

From Eq.3

74.88*xAr= 0.5

xAr= 0.5/74.88=0.006677

% percentage of argon =0.67%

xH2= 0.5894-0.006677=0.582723=58.27%

Ammonia formed in the reactor = ammonia withdrawn as liquid + ammonia in the purge= 10.52+74.88*31.27/100 =23.414 moles/hr

Percentage conversion based on limiting reactant ( Nitrogen is the limiting reactant N2+3H2---> 2NH3 and 24.3 moles nitrogen requires 3*24.3= 72.9 moles of hydrogen where are hydrogen actually present =75.2)

1 mole of nitrogen upon 100% conversion gives 2 moles of ammonia

23.414 moles of ammonia correspond to 11.71 moles of N2

Hence percentage conversion = 100*11.71/24.3 =48.19%

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