In: Chemistry
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.
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%