ChemE 101 LE Practice Problem Set
Recycle and Purge
Department of Chemical Engineering
University of the Philippines Diliman
PROBLEM 1
Propane is dehydrogenated to form propylene in a catalytic reactor:
C 3H 8  C 3 H 6 + H 2
The process is to be designed for a 95% overall conversion of propane. The reaction products are separated into two
streams: the first, which contains H2, C3H6, and 0.555% of the propane that leaves the reactor, is taken off as a product;
the second stream, which contains the balance of the unreacted propane and 5% of the propylene in the first stream, is
recycled to the reactor. Calculate the composition of the product, the recycle to feed ratio (i.e., recycle ratio), and the
single-pass conversion.
Solution:
Diagram
Assume: steady-state
Basis: 100 mol propane fresh feed
From information on overall conversion:
mol C3 H8 into system - mol C3 H8 out of system
XA =
mol C3 H8 out of system
0.95 =
100 - x mol C3H8 in P
100
 C3 H8 in P = 5 mol C3 H8
This amount is 0.555% of the total propane leaving the reactor. The remaining 99.445% is recycled. Hence the amount of
propane in G1, G2, and R can be determined:
æ 100 ö÷
C3 H8 in G2 = 5 çç
çè 0.555 ÷÷ø = 900.901 mol
æ 99.445 ö÷
= 895.901 mol
C3 H8 in R = 5 çç
çè 0.555 ÷÷ø
Balance around mixing point:
C3 H8 bal:
100 + 895.901 = G1
 G1 = 995.901 mol C3 H8
1
Performing molecular balances around the reactor we can determine the amount of propylene and hydrogen formed.
Around the reactor:
C3 H8 bal:
C3 H6 bal:
995.901 = 900.901 + C3H8 consumed
 C3 H8 consumed = 95 mol consumed
æ 1 C3 H6 ö÷
÷ + C3 H6 in
C3 H6 out = 95 mol C3H8 consumed çç
çè 1 C H ÷÷ø
3
8
 C3 H6 out = 95 mol C3 H6 + y C3 H6 in
H2 bal:
æ 1 H2 ö÷
÷
H2 out = 95 mol C3 H8 consumed çç
çè 1 C3 H8 ÷÷ø
 H2 out = 95 mol H2
We can rewrite the diagram to reflect what we have solved so far:
We can perform balances for the overall system to determine the amount of propylene in P. We use the atomic method:
C bal:
æ3ö
æ3ö
æ3ö
100 C3 H8 çç ÷÷÷ = 5 C3 H8 çç ÷÷÷ + m C3 H6 çç ÷÷÷
çè 1 ø
çè 1 ø
èç 1 ø
 m = 95 mol C3H6
From the additional information, 5% of this amount is the amount of propylene in the recycle stream. We can now
determine the amount of propylene entering the reactor:
C3 H6 in R = C3 H6 in G1 = 0.05 (95) = 4.75 mol C3 H6
Answers to questions:
1. Product composition
2.
Comp
Mol
Mol fr
Mol %
C3H8
5
0.025641
2.56%
C3H6
95
0.487179
48.72%
H2
95
0.487179
48.72%
Total
195
1
100.00%
Recycle-to-feed ratio
mol R
895.901 + 4.75
=
= 9.00651
mol F
100
2
3.
Single-pass conversion
XA =
mol C3 H8 into reactor - mol C3 H8 out of reactor
mol C3 H8 into reactor
XA =
995.901 - 900.901
´ 100 = 9.539%
995.901
PROBLEM 2
For an ammonia synthesis process, 100 moles per hour of fresh gas feed composed of 75.16% H2, 24.57% N2, and 0.27%
Ar is mixed with a recycle gas stream and enters the reactor with a composition of 79.52% H2. The purge stream leaving
the ammonia separator contains 80.01% H2 and no ammonia. The product ammonia contains no dissolved gases. Calculate:
1. The flow rate and composition of the recycle stream and the purge stream
2. The single-pass and overall conversion of the process
Solution:
Diagram
Assume: steady-state
Basis: 1 hour
Around overall system:
0.0027 (100) = Ar in U
Ar bal:
 Ar in U = 0.27 mol
H bal:
æ 2ö
æ 2ö
æ3ö
0.7516 (100)çç ÷÷÷ = 0.8001U çç ÷÷÷ + P çç ÷÷÷
çè 1 ø
çè 1 ø
èç 1 ø
N bal:
æ2ö
æ 2ö
æ 1ö
0.2457 (100)çç ÷÷÷ = (U - 0.8001U - 0.27 )çç ÷÷÷ + P çç ÷÷÷
çè 1 ø
çè 1 ø
èç 1ø
Solving the H and N balances simultaneously:
 U = 3.194 mol purge,
P = 48.403 mol product
Composition of the purge stream:
Comp
Mol
Mol fr
Mol %
H2
2.555519
0.8001
80.01%
N2
0.368481
0.115366
11.54%
Ar
0.27
0.084534
8.45%
Total
3.194
1
100.00%
3
Around the mixing point:
H2 bal:
0.7516 (100) + 0.8001R = 0.7952G1
OMB:
100 + R = G1
Solving the H2 bal and OMB simultaneously:
 R = 889.796 mol recycle,
G1 = 989.796 mol gross feed
N2 bal:
0.2457 (100) + 0.1154 (889.796) = N2 in G1
 N2 in G1 = 127.252 mol N2
Around the splitting point (recycle, purge, and stream I):
OMB:
I = 889.796 + 3.194
 I = 892.99 mol I
Around the separator:
OMB:
G2 = 892.99 + 48.403
 G2 = 941.393 mol gross product
H2 bal:
H2 in G2 = 0.8001 (892.99)
 H2 in G2 = 714.481 mol H2
N2 bal:
N2 in G2 = 0.1154 (892.99)
 N2 in G2 = 103.051 mol N2
Answers to questions:
1. Flow rate and composition of recycle and purge
R = 889.796 mol
h
Composition: 80.01% H2 , 11.54% N2 , 8.45% Ar
U = 3.194 mol
h
Composition: 80.01% H2 , 11.54% N2 , 8.45% Ar
2.
Single-pass and overall conversion
Single-pass X A
mol N2 into reactor - mol N2 out of reactor
´ 100
mol N2 into reactor
127.252 - 103.051
=
´ 100
127.252
= 19.018%
=
4
Overall X A
=
mol N2 into system - mol N2 out of system
´ 100
mol N2 into system
=
0.2457 (100) - 0.1154 (3.194)
´ 100
0.2457 (100)
= 98.500%
PROBLEM 3
A high-grade fuel oil with 0.8% ash is burned using wet air in a furnace. A sample of the flue gas is taken
and a portion is subjected to analysis in a standard Orsat apparatus. The analysis obtained is shown in
the table. At atmospheric conditions, the partial pressure of water in the flue gas is measured as 10.3
kPa. The wet air used in the combustion is supplied at 1.995 m3 per 100 mol dry flue gas at a total
pressure of 3.8 psig and 30℃. The partial pressure of water in the wet air is 2.35 kPa. The residue after
combustion is composed of ash and some of the sulfur in the fuel oil. For every cubic meter of dry flue
gas at STP, 1.2 grams of residue are recovered.
a. What is the complete (wet) analysis of the flue gas?
b. What percentage of the sulfur in the fuel oil that goes to the residue?
c. What is the composition of the high-grade fuel in weight percent?
d. What type of high-grade fuel oil was used based on the C/H weight ratio?
Parameter
C/H ratio
e.
No. 1 Fuel Oil
6.35
No. 2 Fuel Oil
6.93
No. 4 Fuel Oil
7.42
Low S, No. 6 Fuel Oil
8.31
Comp
SO2
CO2
O2
CO
H2
N2
%
0.07%
11.73%
4.47%
1.02%
0.85%
81.86%
High S, No. 6 Fuel Oil
7.62
What is the percent excess air used in the combustion process?
Solution:
We use the process conditions to determine the flow rates of the streams in the process.
Assume: steady-state, ideal gas
Basis: 100 mol dry flue gas (P)
Using the pressure conditions of the flue gas:
Mol H2O =
10.3 kPa
´ 100 = 11.31557 mol H2O
101.325 kPa - 10.3 kPa
For wet air (A) :
æ N öæ P ö æT ö
æ 1 mol ÷ö çæ 3.8 + 14.7 psia ö÷ æ 273.15 K ÷ö
PV
PV
ç
= s s  N = V ççç s ÷÷÷ ççç ÷÷÷ çç s ÷÷÷ = 1.995 m 3 çç
çè 0.0224 m 3 ÷÷ø çèç 14.7 psia ÷ø÷ ççè 30 + 273.15 K ÷÷ø
÷ Ps ÷ø çèT ø
NT
N sTs
è Vs øè
 N = 100.9934 mol wet air
For residue (R ):
æ 0.0224 m 3 ÷ö
÷ = 2.24 m 3 P at STP
100 mol dry flue gas çç
çè 1 mol ÷÷ø
æ 1.2 g residue ö÷
2.24 m 3 P at STP çç
÷÷ = 2.688 g residue
çè
ø
1 m3
The composition of wet air is:
x H 2O =
2.35 kPa
= 0.01843
kPa
(3.8 + 14.7 psia )( 101.325
)
14.7 psia
x N 2 = 0.79 (1 - 0.01843) = 0.77544
xO2 = 0.20613
5
The diagram now becomes:
Let
C bal:
x = mol C in F
z = mol S in F
o = mol O in F
y = mol H in F
n = mol N in F
x ( 11 ) = 11.73 CO2 ( 11 ) + 1.02 CO ( 11 )
 x = 12.75 mol C
O bal:
o ( 11 ) + 100.9934 éë0.01843 H2O ( 11 ) + 0.20613 O2 ( 21 )ùû =
0.07 SO2 ( 21 ) + 11.73 CO2 ( 21 ) + 4.47 O2 ( 21 ) + 1.02 CO ( 11 ) + 11.31557 H2O ( 11 )
 o = 1.37872 mol O
H bal:
y ( 11 ) + 0.01843 (100.9934) H2O ( 21 ) = 0.85 H2 ( 21 ) + 11.31557 H2O ( 21 )
 y = 20.60852 mol H
N bal:
n ( 11 ) + 0.77544 (100.9934) N2 ( 21 ) = 81.86 N2 ( 21 )
 n = 7.09136 mol N
Let
a = mass ash in R
Ash mass balance:
Ash in F = a
S mass balance:
g
S in F = 0.07 ( 11 )(32 mol
) + 2.688 - a
g
Total mass of F = 12.75 (12 mol
) + 1.37872 (16 molg ) + 20.60852 (1 molg )
g
+7.09136 (14 mol
) + a + 0.07 (32) + 2.688 - a
 F = 299.87508
6
From additional information: Ash in F is 0.8%
a
 0.008 =
299.87508
 a = 2.39900 g ash in R
 2.688 - a = 0.289 g S in R
 0.07 (32) + 2.688 - a = 2.53000 g S in F
Answers to questions:
1. Complete analysis of flue gas
Comp
Mol
Mol fr
Mol %
SO2
0.07
0.000628843
0.06%
CO2
11.73
0.105376094
10.54%
O2
4.47
0.040156108
4.02%
CO
1.02
0.009163139
0.92%
H2
0.85
0.007635949
0.76%
N2
81.86
0.735386793
73.54%
H2O
11.31557
0.101653075
10.17%
Total
111.31557
1
100.00%
2.
% of S in fuel that goes to residue
% S in F that goes to R =
3.
S in R
0.289
=
´ 100 = 11.423%
S in F
2.53000
Composition of fuel in weight percent
Comp
Mol
MW
Mass
Mass fr
Mass %
C
12.75
12
153
0.510211
51.021%
H
20.60852
1
20.60852
0.068723
6.872%
2.53
0.008437
0.844%
S
N
7.09136
14
99.27904
0.331067
33.107%
O
1.37872
16
22.05952
0.073562
7.356%
2.399
0.008
0.800%
299.8761
1
100.000%
ash
Total
4.
Type of high-grade fuel based on C/H weight ratio
C
51.021
by weight =
= 7.424  No.4 Fuel Oil
H
6.872
5.
Percent excess
Theo O2 = 12.75 C ( 11 OC2 ) + 20.60852 H ( 0.52 HO2 ) + 2.53 g S ( 132molg )( 11OS2 ) - 1.37872 ( 12 OO2 ) = 17.29183 mol O2
% excess air =
0.20613 (100.9934) - 17.29183
´ 100 = 20.391%
17.29183
7
PROBLEM 4
A process for methanol synthesis is shown in the figure below. The pertinent chemical reactions involved are:
CH 4 + H 2O  CO2 + H 2
(main reformer reaction)
(reformer side reaction)
CH 4 + H 2O  CO + H 2
(CO converter reaction)
CO + O2  CO2
(methanol synthesis reaction)
CO2 + H 2  CH 3OH + H 2O
Ten percent excess steam, based on the main reformer
reaction, is fed to the reformer, and conversion of methane
is 100%, with a 90% yield of CO2.
A stoichiometric quantity of oxygen is fed to the CO
converter, and the CO is completely converted to CO2.
Additional makeup CO2 is then introduced to establish a 3:1
ratio of H2 to CO2 in the feed stream to the methanol
reactor. Conversion in the methanol reactor is 55% on one
pass through the reactor.
The methanol reactor effluent is cooled to condense all the
methanol and water, with the noncondensable gases
recycled to the methanol reactor feed. The H2/CO2 ratio in the recycle stream is also 3:1.
Because the methane contains 1% nitrogen as an impurity, a portion of the recycle stream must be purged out to prevent
the accumulation of inerts. The purge stream analyzes 5% nitrogen.
On
1.
2.
3.
4.
the basis of 100 mol of methane feed (including N2), calculate the:
Moles of H2 lost in the purge
Moles of makeup CO2 required
Recycle to purge ratio
Amount (in kg) and strength (in weight percent) of methanol solution produced
Solution:
We can rewrite the diagram as shown in the next page:
8
The balanced reactions are:
CH 4 + 2H 2O  CO2 + 4H 2
(main reformer reaction)
CH 4 + H 2O  CO + 3H 2
(reformer side reaction)
CO + O2  CO2
(CO converter reaction)
CO2 + 3H 2  CH 3OH + H 2O
(methanol synthesis reaction)
1
2
Basis: 100 mol of methane feed
We can solve the amount of steam entering the reformer:
mol H2O
mol H2O fed = 99 mol CH4 ( 12 mol
CH4 ) (1.1) = 217.8 mol H2O
Using the definition of the yield:
x mol CO2 produced
mol CO2
99 mol CH 4 ( 11 mol
CH4 )
0.90 =
 x = 89.1 mol CO2 produced
Performing balances around the reformer:
N2 bal:
Let:
C bal:
0.01 (100) = N2 in R = 1 mol N2
x = mol CO out
y = mol H2 out
z = mol H2O out
0.99 (100) CH4 ( 11 ) = 89.1 CO2 ( 11 ) + x CO ( 11 )
 x = 9.9 mol CO
O bal:
217.8 H2O ( 11 ) = 89.1 CO2 ( 21 ) + 9.9 CO ( 11 ) + z H2O ( 11 )
 z = 29.7 mol H2O
H bal:
217.8 H2O ( 21 ) + 0.99 (100) CH 4 ( 41 ) = y H2 ( 21 ) + 29.7 H2O ( 21 )
 y = 386.1 mol H2
Since all of the CO entering the converter is completely converted to CO2, we can now determine the composition of stream
C:
C bal:
89.1 CO2 ( 11 ) + 9.9 CO ( 11 ) = x mol CO2 ( 11 )
 x = 99 mol CO2 in C
9
At stream C:
N 2 = 1 mol N2
H 2 = 386.1 mol H2
H 2O = 29.7 mol H2O
CO2 = 99 mol CO2
At stream N, it is expected that the ratio of H2 to CO2 is 3:1. Hence, the composition would be:
N 2 = 1 mol N2
H 2 = 386.1 mol H2
H 2O = 29.7 mol H2O
CO2 = ( 31 ) 386.1 = 128.7 mol CO2
The makeup CO2 can be computed by performing a balance around the mixing point before stream N:
CO2 bal:
M + 99 = 128.7
 M = 29.7 mol makeup CO2
Since we are now entering into the methanol reactor-recycle scheme, it is best if we start our calculations by performing a
balance around the entire reactor-recycle system:
Let
x = CH3OH in P
y = H2O in P
z = CO2 in E
m = H2 in E
N2 bal:
1 N2 = 0.05E
E = 20 mol
C bal:
128.7 CO2 ( 11 ) = z ( 11 ) + x ( 11 )
H bal:
386.1 H2 ( 21 ) + 29.7 H2O ( 21 ) = m ( 21 ) + x ( 41 ) + y ( 21 )
O bal:
128.7 CO2 ( 21 ) + 29.7 H2O ( 11 ) = z ( 21 ) + x ( 11 ) + y ( 11 )
From physical constraint: m = 20 - 1 mol N2 - z = 19 - z
Solving all four equations simultaneously, we get:
x = 123.95 mol CH3OH in P
y = 153.65 mol H2O in P
z = 4.75 mol CO2 in E
m = 14.25 mol H2 in E
The amount of methanol leaving the condenser is also the amount of methanol in stream G2 (leaving the reactor). Using
the definition of conversion:
mol CO2
0.55 (CO2 fed) = 123.95 mol CH3OH ( 1 1mol
CH3OH )
\ CO2 fed = 225.36 mol CO2
10
Since the recycle stream maintains a ratio of 3:1 for H2 and CO2, as well as the fresh feed stream, the gross feed must also
maintain the same ratio. Hence, the amount of H2 can also be computed:
H 2 fed to reactor = 225.36 (3) = 676.08 mol H2
Around the mixing point:
x = CO2 in R
y = H2 in R
Let
CO2 bal:
128.7 + x = 225.36  x = 96.66 mol CO2
H2 bal:
386.1 + y = 676.08  y = 289.98 mol H2
The 3:1 ratio is maintained. With the purge and recycle having the same composition, the composition of N2 in R is also
5%. Using the information on the H2:CO2 ratio, the composition of the recycle is:
5% N2
% H2
= (100 - 5) 34 = 71.25%
% CO2 = (100 - 5) 14 = 23.75%
The amount of recycle is:
R=
96.66
= 406.989 mol
0.2375
Answers to questions:
1. 14.25 mol H2
2.
29.7 mol makeup CO2
3.
Recycle to purge ratio
R
406.989
=
= 20.349
20
P
4. Amount and strength of methanol solution
The total amount of methanol solution and its strength are also computed:
Component Mole
Mass (g)
Mass
%
CH3OH
123.95 mol 3966.4 g
58.92%
H2O
Total
153.65
mol
2765.7 g
41.08%
6732.1 g = 6.732
kg
11
PROBLEM 5
A synthetic-ammonia unit with a single converter is converting a fresh feed gas containing 75.8% H2, 23.7% N2, and 0.5%
inerts. The gas leaves the converter containing 18.6% NH3 and 70.0% H2. This converted gas is fed to a cooler under
conditions of temperature and pressure such that most of the ammonia leaving the converter is condensed to a liquid and
removed. On reduction of pressure all the dissolved gases are evolved from this liquid ammonia. These are compressed and
returned to the system just before the cooler. The residual gases from the cooler are recycled to the inlet of the converter,
except for a purge provided in the recycle line. The uncondensed ammonia leaving the cooler in these gases is found to be
8% of that leaving the converter. Calculate the amount and composition of the purged gases per 100 mols of feed gas, and
determine the net mol of residual gas recycled to the converter inlet per 100 mol of feed gas.
Hints:
1. Lump the cooler, rectifier, and dissolved gas stream into a single unit. No need to determine the flow rate and
composition of the dissolved gas stream and the intermediate stream between the cooler and rectifier.
2. For easier analysis, start with a basis of 100 mol of gases leaving the converter, then scale-up your answer later by using
the 100 mol of feed gas.
Solution:
The diagram is as shown:
fresh feed
H2 75.8%
N2 23.7%
Inerts 0.5%
catalytic
converter
recycle gas
converter gas
NH3 18.6%
H2 70.0%
cooler
rectifier
pure
liquid
NH3
purge gas
Basis: 100 mol converter gases
NH 3 balance around cooler+rectifier system:
0.186 (100 mol) = mol liq. NH 3 in product + mol gaseous NH 3
mol liq. NH 3 = (0.92)(0.186)(100 mol) = 17.112 mol liq. NH 3
mol gaseous NH 3 = (0.08)(0.186)(100 mol) = 1.488 mol gaseous NH 3
Since the cooler-rectifier system separates the dissolved gases from the liquid ammonia, the composition of the stream
leaving the cooler-rectifier system is:
1.488 mol gaseous NH 3
70 mol H 2
(100 - 18.6 - 70) mol N 2 and inerts
12
In terms of fractions:
0.01795 gaseous NH 3
0.84451 H 2
0.13753 (N 2 + inerts)
If we let x be the fraction of inerts, the compositions will be:
0.01795 gaseous NH 3
0.84451 H 2
x inerts
0.13753 - x N 2
These percentages are constant for the gross recycle, purge, and recycle streams. Now that we know the composition of the
purge stream, we can perform overall atomic balances to determine the flowrates of the fresh feed and the purge:
H bal:
æ2ö
æ3ö
æ3ö
æ 2ö
0.758F çç ÷÷÷ = 17.112 çç ÷÷÷ + 0.017952U çç ÷÷÷ + 0.844513U çç ÷÷÷
çè 1 ø
çè 1 ø
çè 1 ø
çè 1 ø
 1.516F - 1.742882U = 51.336
N bal:
æ2ö
æ 1ö
æ 1ö
æ2ö
0.237F çç ÷÷÷ = 17.112 çç ÷÷÷ + 0.017952U çç ÷÷÷ + (0.137535 - x )U çç ÷÷÷
èç 1 ø
èç 1ø
èç 1ø
èç 1 ø
 0.474F - 0.293022U + 2xU = 17.112
inert bal:
0.005F = xU
 0.005 - xU = 0
Solving simultaneously:
F = 37.01572 mol fresh feed
U = 2.742489 mol purge gas
xU = 0.185079
 x inerts = 0.067486, x N 2 = 0.070049
The complete composition for the purge is:
1.795%
NH 3
84.451%
H2
7.005%
N2
6.749%
inerts
At the splitting point, an overall mole balance can be carried out:
mol gross recycle = mol purge gas + mol recycle gas
1.488 + 70 + (100 - 18.6 - 70) = 2.742 + R
R = 80.146 mol recycle
Since we need values for the purge and recycle gas for 100 moles of fresh feed, we can scale-up the values via ratio and
proportion:
æ 2.74 mol purge gas ö÷
mol purge gas = ççç
÷ (100 mol fresh feed) = 7.402 mol purge gas
è 37.016 mol fresh feed ÷ø
13
æ 80.146 mol recycle ö÷
mol recycle = ççç
÷ (100 mol fresh feed) = 216.517 mol recycle
è 37.016 mol fresh feed ÷ø
PROBLEM 6
Acetone is produced by the catalytic
dehydrogenation of isopropyl alcohol.
Air and vaporized isopropyl alcohol are
reacted in contact with heated brass or
copper catalyst at 500°C and 300 kPa.
The hot reaction gases containing
acetone, isopropyl alcohol, and
hydrogen pass through a water
condenser and then into a water
scrubber where the noncondensable
gases are completely removed. The
condenser product and solution from
the scrubber are fed to the
fractionating column. Pure acetone is
distilled off at the overhead while an
isopropyl alcohol-water mixture is
removed from the bottom of the column. This isopropyl alcohol mixture is sent to another column where it is separated
into a 91 mol% alcohol constant boiling solution (an azeotrope) at the top and pure water at the bottom. The 91%
isopropyl alcohol is recycled without further purification. The reactions involved are:
(CH 3 )2 CHOH + 12 O2  CH 3COCH 3 + H 2O
(CH 3 )2 CHOH  CH 3COCH 3 + H 2
The air fed is one-half the theoretical amount needed to react with the fresh isopropyl alcohol. The conversion of isopropyl
alcohol per pass in the reactor is 60%. The overall conversion to acetone is 100%. Take a basis of 1000 kg of acetone
product.
1. Calculate the amount of nitrogen and hydrogen that issues out of the reactor.
2. How much 91% alcohol-water mixture is recycled?
3. If the scrubber utilizes 500 kg of water, what will be the mass composition of the liquid leaving the scrubber (just before
it enters the first fractionating column)?
Solution:
MW: Alcohol = 60, Acetone = 58
Reactions:
(CH 3 )2 CHOH + 12 O2  CH 3COCH 3 + H 2O
(CH 3 )2 CHOH  CH 3COCH 3 + H 2
Basis:
1000 kg acetone = 17.241 kmol acetone
We can rewrite the entire diagram to include the expected components for each stream (see next page)
14
We can use the information on the overall conversion of alcohol. Performing a balance around the entire system:
C bal:
17.241 kmol acetone ( 31 ) = x kmol alcohol ( 31 )
\ x = 17.241 kmol alcohol
We can now compute the amount of fresh feed entering the system:
mol O2 in air = 17.241 kmol alcohol
(
1 mol O
2
2
1 mol alcohol
)(0.5) = 4.31025 kmol O
2
mol N2 in air = 4.31025 kmol O2 ( 79
21 ) = 16.21475 kmol N2
Still performing balances around the entire system:
Let
W1 = water entering the scrubber
W2 = water leaving the column
nH 2 = H2 in S1
O bal:
4.31025 O2 ( 21 ) + 17.241 alcohol ( 11 ) + W1 ( 11 ) = 17.241 acetone ( 11 ) + W2 ( 11 )
\ W2 -W1 = 8.6205
H bal:
17.241 alcohol ( 81 ) + W1 ( 21 ) = nH 2 ( 21 ) + W2 ( 21 ) + 17.241 acetone ( 61 )
\ 17.241 (8) - 17.241 (6) = 2nH 2 + 2 (W2 -W1 )
Solving the two equations: nH 2 = 8.6205 kmol H2
The amounts of N2 and H2 issuing out of the reactor are:
16.21475 kmol N2
8.6205 kmol H2
15
The H2 produced may be used to determine how much of the alcohol was consumed by the second reaction. Performing
molecular balances around the reactor:
H 2 bal:
H 2 gen by Rxn 2 = H 2 out
H 2 gen by Rxn 2 = 8.6205 kmol H2
Acetone bal: Acetone gen by Rxn 1 + Acetone gen by Rxn 2 = Acetone out
acetone
Acetone gen by Rxn 1 + 8.6205 kmol H2 ( 1 mol
1 mol H2 ) = 17.241 kmol
Acetone gen by Rxn 1 = 8.6205 kmol acetone
Alcohol bal: Alcohol in = Alcohol out + Alcohol cons by Rxn 1 + Alcohol cons by Rxn 2
mol alcohol
1 mol alcohol
Alcohol in = Alcohol out + 8.6205 kmol acetone ( 11 mol
acetone ) + 8.6205 H2 ( 1 mol H2 )
Alcohol in = Alcohol out + 17.241
Using the single-pass conversion of 60%: Alcohol out = 0.4 (Alcohol in )
Solving the alcohol balance:
Alcohol in = 0.4 (Alcohol in ) + 17.241
\ Alcohol in = 28.735 kmol alcohol
\ Alcohol out = 11.494 kmol alcohol
This amount of alcohol is a tie-component of both the scrubber and the two distillation columns. Since this amount of
alcohol constitutes 91% of the recycled mixture:
Recycle stream =
11.494
= 12.631 kmol 91% isopropyl mixture
0.91
The total amount of water fed to the reactor is obtained by performing a material balance around the mixing point before
the reactor:
H2O bal:
H 2O out = (0.09)(12.631) = 1.137 kmol H2O
The amount of water leaving the reactor is:
H2O bal:
H 2O out = H 2O in + H 2O gen by Rxn 1
mol H2O
H 2O out = 1.137 kmol + 8.6205 kmol acetone ( 1 1mol
acetone ) = 9.757 kmol H 2O
The amount of water leaving the scrubber is:
H2O bal:
H 2O out = H 2O in from water feed + H 2O in from reactor effluent
H 2O out = 27.78 kmol + 9.757 kmol = 37.537 kmol H2O
The composition of the liquid leaving the scrubber is:
Comp
Mol
MW
Mass
Mass fr
Mass %
H2O
37.537
18
675.666
0.28566
28.57%
Acetone
17.241
58
999.978
0.422773
42.28%
Alcohol
11.494
60
689.64
0.291568
29.16%
Total
66.272
2365.284
1
100.00%
16