Chapter 5 Solutions: Warm-Ups P5.1
Chapter 5 Solutions: Warm-Ups P5.1
Chapter 5 Solutions: Warm-Ups P5.1
WARM-UPS
P5.1
Crystallization: differences in freezing point
Adsorption: differences in binding to solid
Extraction: differences in solubility in 2 immiscible liquids
Distillation: differences in volatility
Membrane filtration: differences in size
Absorption: difference in solubility of gas in liquid
P5.2
Flash vaporization, condensation, and distillation all involve vapor-liquid equilibrium
and the use of an energy separating agent. In flash vaporization, a liquid feed is heated to
evaporate part or all of the material. In condensation, a vapor feed is cooled to condense
part or all of the material. In distillation, a feed (liquid and/or vapor) is repeatedly
condensed and vaporized in multiple stages, resulting in two product streams of different
composition.
Adsorption, absorption, and solvent extraction all require the addition of a material
separating agent. In adsorption, a multicomponent fluid (gas or liquid) is contacted with a
solid separating agent; some components of the fluid stick to the solid whereas others do
not. In absorption, a multicomponent gas is contacted with a liquid; some components of
the gas dissolve in the liquid while others do not. In solvent extraction, a multicomponent
liquid is contacted with another immiscible liquid; some components of the first liquid
dissolve in the second while others do not.
P5.3
Antoine equation for water from 60 to 150 °C:
log10Pwsat = 7.96681−
At boiling point, P=Psat.
At P = 1 atm = 760 mmHg
Another way to find the temperature of saturated steam at this pressure is to use
Antoine’s equation
A third way will be learned in Chapter 6, when we are introduced to steam tables.
P5.5
Two data points are shown. The general shape comes from knowing that the vapor
pressure increases faster than linearly with temperature (which we can see for example
from the Antoine equation or from the water phase diagram).
4
Acetone
3.5
3
Saturation pressure (atm)
2.5
1.5
0.5
0
0 20 40 60 80 100
Temperature (•C)
P5.6
Refer to Figure 5.12, or to Table B.9. A saturated solution of benzene and naphthalene at
45°C contains about 48 mol% naphthalene and about 52 mol% benzene. The solid phase
formed upon cooling is pure naphthalene.
0.3−0.09
From inverse lever rule, nV = =1.90, or about 65.5% is
vaporized. nL 0.41−0.3
P5.8
The mixture is 20 wt% acetic acid, 60 wt% water, and 20 wt% MIBK. This composition
clearly falls in the two-phase (shaded) domain of Figure 5.15. To find the compositions
of the two liquid phases we follow the dashed tie line to the curve. The water-rich phase
contains about 21 wt% acetic acid, 4 wt% MIBK, and the remainder water. The extract
phase contains about 17 wt% acetic acid, 74 wt% MIBK, and the remainder water.
P5.9
ySO2 = 0.09. At 760 mm Hg, pSO2 = 0.09(760)=68.4. From table B.6 at 20°C this
corresponds to about 1.15 kg SO2/100 kg water, or xSO2 = 0.011. At 3040 mm Hg, pSO2 =
273.6. Evaluating data in table B.6, xSO2 = 0.057.
P5.10
Neither – Raoult’s and Henry’s law do not apply to liquid-solid equilibrium.
Henry’s – CO2 is a “noncondensable” gas
Neither – there are not two phases, so you can’t apply a phase equilibrium relationship.
Raoult’s – these compounds form both vapor and liquid phases, and are chemically
similar and relatively non-interacting.
Neither – acetic acid and water participate in hydrogen bonding which makes Raoult’s
law inappropriate.
P5.11
Because the air in the room is multicomponent, not single component. From Antoine
equation for water at 22°C, Pwsat =19.7 mm Hg. Maximum water content is if air is
saturated; from Raoult’s law this occurs at
xwPwsat =ywP
1(19.7)=yw(760)
yw = 0.026
P5.12
The Henry’s law constant for oxygen in water increases with increasing temperature
(e.g., 32,700 atm at 10°C and 47,500 atm at 30°C). Assuming the oxygen content of the
air is constant at 0.21 mol fraction, and the pressure is 1 atm, then the mole fraction of
oxygen dissolved in the water at 10°C is
yO P 0.21
xO2 = HO22 = 32700(1 atm atm) = 6.4 ×10−6
There is more oxygen dissolved in the cooler waters, so that is where the fish go.
P5.13
Colored dye partitions into the hexane layer, because like dissolves like, and oil is more
like hexane than like water. The colored dye is in the top layer, because hexane is less
dense than water.
P5.14
(1) Add stages. Multistaging reduces the solvent requirements but increases the cost of
building the equipment. (2) Add a chemical to the water that complexes with the H2S,
such as DEA or MEA. (3) Increase the operating pressure or decrease the operating
temperature.
P5.15
Antibiotics: extraction (difference in solubility in liquid, can’t use high temperatures,
want to selectively pluck out one component at low concentration from complex mixture)
Isopropanol-air: condensation (large difference in volatility)
Limestone: sedimentation (two phases so use mechanical, large difference in density)
Soybean oil: leaching
Colored impurities: adsorption (want to selectively remove component at low
concentration)
Methane: vaporization (methane is gas at ambient T and P, manure isn’t)
CO2 and H2: absorption (both gases at ambient, need to go too cold for distillation),
Ethylbenzene and styrene: distillation (boiling points are different and not much higher
than ambient)
Yeast: filtration (two phases so use mechanical) filtration
Potassium nitrate: crystallization (if cooled with form solid phase plus liquid solution)
Sedimentation and filtration are mechanical separations; others are equilibrium-based.
P5.16
For selection of separation technology, we consider the heuristics on p. 373-374. Given
the solubility information, it is possible that ethylene glycol is insoluble not just in
benzene but also in ethylbenzene, styrene, and naphthalene. (In fact the data in Table
B.14 supports the idea that ethylene glycol is only sparingly soluble in styrene and
ethylbenzene). If there are 2 liquid phases, we can simply mechanically separate, into one
phase that is mainly ethylene glycol and another phase that is a mixture of the 3
benzenelike compounds. We will next consider equilibrium-based separations (heuristic
2). All compounds have normal boiling points at un-exotic temperatures but all but one
have below-ambient melting points. Therefore, separations based on differences in
relative volatility (e.g., distillation) are better than crystallization (heuristics 4 and 5). The
differences in boiling points are large enough to make distillation feasible (heuristic 4),
so although we could separate based on differences in solubility, we don’t need to,
thereby avoiding addition of a foreign species (heuristic 6).
For sequencing of separation units, we consider the heuristics on p. 377-378. If the
mixture forms 2 liquid phases including one phase that is mostly ethylene glycol, then we
are left with distillation for the remaining 2 separations. We note that the order of relative
volatility is ethylbenzene > styrene > naphthalene. The most difficult separation
(approximated as the smallest difference between boiling point temperatures) is
ethylbenzene from styrene. Based on separating the components present in the greatest
quantities first (heuristic 2), we would first remove styrene, but based on saving difficult
separations for last (heuristic 3), we would first remove naphthalene. Since the
difference in quantity is relatively smaller than the difference in volatility, heuristic 3
probably wins out. Our block flow diagram might look like this:
EB
EG
EG
EB
S
N liquid -liquid distillation
phase
separator
distillation
S
EB
S
N
What if there is insufficient phase separation between ethylene glycol and the
benzenelike compounds to give the required purities? We might still use the liquid-liquid
phase separator to get an initial “cut”, then further purify the ethylene glycol product by
distillation (to remove the lower boiling styrene and ethylbenzene) and by crystallization
(to remove the naphthalene).
P5.17
Strategy: there is only a small amount of CHCl3, so use the Br2 to convert CHCl3 to
CHBr3. The boiling point differences are now much larger so the separation is easier.
Br2
Cl2
Br 2
Cl2
CHBr3
Br2,Cl Evapor Evapor
Reactor
CHCl3 ator ator
CHBr3
Br2
CHBr3
P5.18
Use M for methane and E for ethane. Flows are given in gmol/min, and compositions as
mol%. Use f, d, and b to indicate feed, distillate (or overhead), and bottoms streams,
respectively. The flow diagram is
d
90 % M
10 % E
70 M f
30 E distillation
b
2% M
98 % E
P5.19
Use N for sodium carbonate and W for water. Streams are numbered as shown on flow
diagram. Use c for crystal phase and l for liquid phase. All flows in lb/h.
W
2
crystals C
liquid L:
3000 N 1 17.7 % N 4 liquid:
7000 W Evaporator Filter
3 17.7 %N
Solids +
entrained
liquid
m ˙ c5= 2097
Combine and solve to find: m ˙ l5= 599
m ˙ l4 =
4504
3200 gmol/h
72 % M M
1 3
13 & CO2 CO2
12 % H2S H2S
3 % COS 0.3 % COS
4 H2S
COS
S
with M for methane and S for the solvent. We also know that the gas:solvent ratio is 3:1
and the solvent absorbs 97.2% of the H2S fed.
DOF analysis
Number of variables
˙ ˙
fR,H2S,4 = 0.972 = n n ˙HH22SS,4,1 = 0.12n H(23200S,4 )
From these calculations we find that the exit gas flow rate is 2739 gmol/h and the gas
contains 84.1 mol% CH4, 15.2 mol% CO2, 0.4 mol% H2S and 0.3 mol% COS. The liquid
stream flow rate is 1528 gmol/h; it contains 24.4 mol% H2S and 5.75 mol% COS in
addition to the solvent.
P5.21
Consider 3 components: dry air, H2O, and dry popcorn. The popcorn production is 50
kg/h total. At 10 wt% moisture, the popcorn product is therefore 45 kg/h dry popcorn and
5 kg/h water. Therefore, the wet popcorn, at 25% moisture, must contain 15 kg/h water
(plus 45 kg/hr dry popcorn), and 10 kg/h water are removed in the hot air stream, or
555.6 gmol/h.
Since the wet air is modeled as an ideal gas, mol fraction = volume fraction. A
steadystate material balance on water, with the entire process as the system, is
0.02n ˙gas,in + 555.6 = 0.15n ˙gas,out where the gas includes both dry air and water vapor.
The recycle flow rate is 4 times the gas feed rate, or 14532 gmol/h. The recycle gas
contains 15 mol% water – because there is simply a splitter on the exit air stream. From
this we can then do a material balance on the mixer, where the hot air feed is mixed with
the recycle stream.
Therefore, the moisture content of the gas to the dryer is 2252.5/18165, or 0.124.
P5.22
This is a semi-batch process. We can consider the ion exchanger as a separation unit that
achieves recovery of GMP and BLG by adding the separating agent, salt. The flow
diagram shows what is accomplished by the ion exchanger in the two steps of adsorption
and elution.
Whey
GMP
BLG
proteins
lactose water
salts
water Eluted
GMP
BLG
NaCl dried powder
Buffer water GMP
0.25 M NaCl ion exchanger dryer BLG
water NaCl
Unadsorbed
GMP
BLG
proteins
lactose salts
water
(a) During the first 30 minutes: consider the ion exchanger as the system. Whey is
pumped at a steady flow rate (150 mL/min) onto the ion exchanger unit, GMP and BLG
accumulate in the system, and unadsorbed materials leave the system at a steady flow
rate. There is no NaCl in this part of the process. We write integral material balance
equations because we are interested in what happens over a specified time interval. The
material balance equation for GMP is:
30 30
m m m
mGMP,sys,f − GMP,sys,0 GMP,sys,f − GMP,sys,0
∫ m˙
GMP,indt t0
The total quantity of GMP that exits in the “unadsorbed” stream is 5.4 – 4.8 or 0.6 g.
A similar analysis leads to the conclusion that 3.6 g of BLG are fed to the system, 0.86 g
adsorb to the resin and accumulate in the system, and 2.74 g exit in the “unadsorbed”
stream.
(b) During the next 10 minutes, the high-salt buffer is pumped through the system and all
the material initially adsorbed to the resin is eluted, so that none remains in the system at
the end of the 10 minute interval. No protein enters the system, so the material balance is
10
− 4.8 =−1.5cGMP,eluted
cGMP,eluted = 3.2 g/L
The total quantity of GMP eluted is 4.8 g, and the total volume of the elution stream is
1.5 L. (The GMP is 2.67 times more concentrated in the elution stream than it was in the
whey.)
Similarly, we find that the total quantity of BLG eluted is 0.86 g, and the concentration of
BLG in the elution stream is 0.57 g/L. (The BLG is less concentrated in the elution
stream than it was in the whey.)
NaCl does not accumulate on the resin; rather it simply enters and leaves the system at a
steady state. The total quantity of NaCl that enters the system equals the quantity leaving
the system, and is equal to
10
∫ 0.25 moles NaClL× 58.5mole NaCl g NaCl × 0.15min Ldt = 21.9 g NaCl
0
(c) During drying: the dryer is the system, nothing accumulates inside the dryer, and all
water is removed while all NaCl, all GMP, and all BLG exits in the “dried powder”
stream. The material balance is simply IN = OUT, or the dried powder contains 4.8 g
GMP, 0.86 g BLG, and 21.9 g NaCl.
The total quantity of powder product is 27.56 g, and the purity is only (4.8/27.56) x 100%
or 17%. The majority of the impurity is salt. On a salt-free (protein-only) basis, the
product is (4.8/(4.8+0.86)) x 100% or 85% GMP.
The two product streams are the eluted and the unadsorbed materials. The
separation factor for the separation of GMP and BLG between these two streams is:
m
mGMP,eluted BLG,unadsorbed 4.8 g 2.74 g
× = × = 25.5
m
mGMP,unadsorbed BLG,eluted 0.6 g 0.86 g
P5.23
crystals C
liquid L
30 K 1 3
70 W Chiller Filter liquid
2
Solids +
entrained
liquid
From Figure 5.11, at 5°C, saturated solution is 15 wt% KNO3. If we call ES the entrained
solution, C the crystals in the filter cake, and L the filtrate liquid, and assume that the
composition of ES is the same as that of L, and that both are saturated solutions, then
material balances around the entire system yield:
Water: 70 = 0.85m ˙ L+ 0.85m ˙ ES
KNO3: 30 = 0.15m ˙ L+ 0.15m ˙ ES+ m ˙ C
ES 1
m˙ = from entrainment. This gives 3 equations in 3
unknowns. Also, we know that
m ˙ C 19
Solving:
m ˙ C =17.65
kg/h m ˙ ES =
0.93 kg/h m ˙ L
= 81.42 kg/h
Recycling won’t help. The salt concentration in the liquid is fixed by phase equilibrium,
and the solid:liquid ratio in the cake is fixed by the entrainment specification. Since these
are fixed, and we still need to satisfy the material balance equation, the purity, recovery,
and separation factor can’t change.
P5.24
The data in Table B.8 were plotted to generate a phase diagram:
100
anhydrous
80
Temperature (°C)
60
40
20
decahydrate heptahydrate
0
0 10 20 30 40 50 60
The curves show the salt form at different concentrations and temperatures. We
extrapolated the data a bit (light lines) so the curves would meet up.
30 g Na2SO4 = x g Na2SO4
70 g H2O 100 g H2O x=
42.9 g Na2SO4
(a) Reading from the diagram, at 42.9 g Na2SO4 per 100 g H2O, a solid phase of the
decahydrate form appear at about 32°C. Thus, the solution should be cooled to below
32°C to allow crystallization.
(b) If 50% of the Na2SO4 is crystallized, then 15 g of Na2SO4 out of the original 100
g solution is crystallized. The remaining solution therefore contains
30 g - 15 g Na2SO4 = 21.4 g Na2SO4
70 g H2O 100 g H2O
The solution must be cooled to 21°C.
(c) By reading from the diagram, or using the table, we find that at 50°C, a solid phase
appears if the salt concentration exceeds 46.7 g Na2SO4 per 100 g H2O, or
Case (a) and (c) are illustrated by the dashed lines on the figure:
100
anhydrous
80
Temperature (°C)
60
evaporate water
40 decrease T
20
decahydrate heptahydrate
0
0 10 20 30 40 50 60
To determine the quantity of water that must be evaporated we sketch a flow diagram and
complete material balance calculations:
water
1 3
3m ˙ 3= 94.3 g/h
2m ˙ 2 = 5.7 g/h
Since the water feed rate is 70 g/h, then 5.7/70 or 8.1% of the water fed must be
evaporated to just begin to see solid phase formation.
P5.25
The saturation pressures at 80°C are tabulated and plotted.
Compound Psat (mm Hg) Mw
Alkanes: n-C5 2724 72
n-C6 1068 86
n-C7 428 100
n-C8 175 114
Alcohols: n-C1 1350 32
n-C2 811 46
n-C3 381 60
n-C4 163 74
Water 355 18
Saturation pressure, mm Hg
Saturation pressure drops rapidly as molar mass increases. The alkanes have the highest
saturation pressure at a given molar mass. This is because attractive hydrogen bonding
interactions for the alcohols keep them in the liquid phase compared to the alkanes,
which have only weaker van der Waals and similar forces to hold together the liquid
phase. Water is a very anomalous compound due to its strong hydrogen bonding.
P5.26
The data are plotted.
Temperature, °C
(b) A 20% methanol mixture at 89.3°C clearly falls in the two-phase region, as shown:
Temperature, °C
At 89.3°C, the saturated vapor contains 36.5 mol% methanol (63.5 mol% water) and the
saturated liquid contains 8.0 mol% methanol (92 mol% water). By material balance on
methanol (using as a basis 100 gmol/h feed):
0.2( )= 0.365n ˙V + 0.08n ˙L
100
P5.27
1574.51
(a) log10(1500)= 7.14016−, or Tb = 173.1°C.
T+ 224.09
1424.255
(b) log10(1500)= 6.95719−, or Tb = 163.5°C. T+ 213.21
(c) At P = 2 bar = 1500 mm Hg from Eq. 5.12, at
bubblepoint 1500 = 0.5Pssat + 0.5Pebsat
1574.51 1424.255
7.14016− 6.95719−
1500 = 0.5×10 T+224.09 + 0.5×10 T+213.21
(Temperatures lie outside the range given in Table B.4, so the solution should be used
with caution.)
P5.28
For ethylene glycol, the Antoine equation is
( )
log10 Psat = 8.0908− 2088.9
T+ 203.5
( )
and for water we will use log10 Psat
= 7.96681−1668.21 T
+ 228
where T is in °C.
(a) 260°F is equivalent to 126.7°C, at which (from the Antoine equations) we find
P = 58.2 mm Hg for ethylene glycol and 1835 mm Hg for water. We want the
sat
Also,
xe+xw =1
Solving simultaneously, we find
xe = 0.605
(b) We return to the bubblepoint equation from part (a), except now we know the
molefractions and total pressure (15 psig, or 1535 mm Hg) but not the saturation
pressures:
0.605Pesat + 0.395Pwsat =1535 mm Hg
P5.29
1211.033
(a) log10(7600) = 6.90565− , or T = 179.6°C. T + 220.79
(b) From Antoine’s equation, at 202°C,
1211.033
benzene: log10Pbsat = 6.90565− , or Pbsat =10,997 mm Hg 202+
220.79
1344.8
toluene: log10Ptsat = 6.95464 − , or Ptsat = 5790 mm Hg 202+
219.48
(This temperature lies outside the range given in Table B.4, so the results should be
used with caution.) Material balances:
Benzene: 40=ybnV +xbnL Total:
100=nV +nL
yb= xb
nV = 32
P5.30
We’ll postulate that at any operating conditions N2 is noncondensable and only MEK will
appear in the liquid stream. The flow diagram is
vapor
MEK
N2
liquid
MEK
( )
log10 Psat = 6.97− 1210
T+ 216
(a) At 80°C, Psat for MEK is calculated to be 762 mm Hg. We have only one phase
equilibrium relationship, which we get from Raoult’s law:
x P
yMEKP= MEK MEKsat
yMEK = = 0.25
The vapor is 25 mol% MEK and 75 mol% N2. Liquid is pure MEK. From the material
balance equations we find the vapor and liquid flow rates:
100 = yN2n ˙V = 0.75n ˙V
(b) We maintain the pressure at 3050 mm Hg but increase the temperature until just
allthe liquid is gone – in other words, to the dewpoint temperature. Since all the MEK
and all the N2 goes to the vapor, then we know that yMEK = zMEK = 0.5. At the dewpoint we
postulate the existence of a tiny droplet of liquid, which in this case will be pure MEK.
x P
From Raoult’s law yMEKP= MEK MEKsat
0.5(3050)=PMEKsat =1525 mm Hg
We insert this value into the Antoine equation to find the dewpoint temperature:
1210
log10(1525)= 6.97− , or T=103.5°C
T+ 216
The flash tank must operate at temperatures of at least 103.5°C to ensure complete
vaporization of the feed stream.
(c) Now we maintain a constant temperature (80°C) and vary the pressure. Raoult’s
lawis
x P
yMEKP= MEK MEKsat
0.5P= 762 mm Hg, or P=1524 mm Hg
The maximum tank pressure at which full vaporization can occur at 80°C is 1524 mm
Hg.
P5.31
144 A
96 I
I
144 A
(a) I
Vapor is 5.9% isopropanol, 94.1 mol% air, 153 gmol/min total. Liquid condensate is
calculated from material balance on isopropanol: 96 – 9.03 = 87 gmol/min, 100%
isopropanol. Percent recovery is 87/96 x 100 = 90.5%.
Brrr!!
P5.32
The flow diagram is shown, with E for ethane and M for methane.
4
97 % M
3% E
90 % M
10 % E
col 2
3
2
70 M 1
mixer col 1
30 E
5
6 98 % E
2% M
Flows are given in gmol/min, compositions in mol%, and we assume steady-state
operation. We write material balances on all 3 process units. We will use mole fractions
and total molar flow rates as variables, because of the nature of the information provided.
We also have
xM5+xE5=1 xM2+xE2 =1
Alternatively, we can first group all 3 units into a single system and derive 2 equations in
2 unknowns:
(or, we use one of these balances plus a total mole balance equation). We solve by
substitution and elimination to find
n ˙4 = 71.6 gmol/min
n ˙6 = 28.4 gmol/min
This result inserted into the component balance equations on column 2 tells us
xM5= 0.85 xE5=
0.15
n ˙2= 200
gmol/min xM2 =
0.775 xE2 = 0.225
The flowrate of the desired product, stream 4, is 71.6 gmol/min. The recycled stream,
stream 5, is 85% methane and 15% ethane. The fractional recoveries of methane and
ethane are:
fRM4 = xM4n ˙4 = 0.97(71.6) = 0.992
xM1n ˙1 0.70(100)
P5.33
(a) The dewpoint temperature corresponds to the saturated vapor at yE = 0.3, or 92°C.
(b) Reading across from yE = 0.3, at 92°C, we find xE = 0.05
(c) The bubblepoint temperature corresponds to the saturated liquid at xE = 0.3, or 82°C.
(D) Reading across from xE = 0.3, at 82°C, we find yE = 0.57
P5.34
Assuming a basis of 100 gmol/min, the flow diagram is Vapor: 50
gmol/min
yE
yW
Liquid: 50 gmol/min
xE
xW
where we’ve used E for ethanol and W for water; compositions are given in mole
fractions.
Figure 5.13 (and Table B.11) provides the phase equilibrium relationships (relating yE to
xE, and yW to xW). Since the information is in graphical rather than equation form, we are
stuck with a trial-and-error approach: guess a temperature, find yE and xE from the graph,
insert values into the material balance equation and see if the equation is satisfied.
This is close enough – a temperature of about 85°C will vaporize 50% of the feed stream
and produce a vapor of 47 mol% ethanol and a liquid of 12.4 mol% ethanol.
P5.35
From the Antoine equation constants for benzene and toluene:
( )=
log10 Pb
sat
6.90565−
sat
Pb = 2840 mm Hg
( )
log10 Ptsat = 6.95464 −
Ptsat =1278 mm Hg
The minimum pressure to have two phases is the dewpoint (maximum is the
bubblepoint). Using Raoult’s law,
b
2840x
yb = 0.3=
P
1278xt
yt = 0.7 =
P
Using as a basis 100 gmol mixture fed to the distillation column, and using the
specifications of 96% recovery of toluene in the bottoms product and a 98% purity in this
product, we find by material balance that the distillate must contain 28.63 gmol benzene
and 2.8 gmol toluene while the bottoms must be 67.2 gmol toluene and 1.37 gmol
benzene. Assuming operation at 130 °C, from the Fenske equation
28.63
67.2 log
2.8
1.37
Nmin = 2840 = 7.78
log
1278
To convert moles H2S per mole MEA solution to mole fraction H 2S in the liquid phase,
n
where H2S is the moles of H2S per mole of MEA solution.
xH2S
The data are clearly highly nonlinear, and cannot be described by Henry’s law. This is
not surprising – MEA is a good solvent for H2S because there are strong noncovalent
(acid-base) interactions.
P5.37
(a) From the problem statement, and assuming no methane is absorbed into the water,
and no water evaporates into the gas, the flow diagram is
gas M
0.01 % S
feed liquid
98 M S
Separator
2 S W
solvent
W
where M = methane and S = H2S. From the process stream specification on the outlet
˙ ˙
n S,gas n S,gas
0.0001= =
˙ ˙ ˙
n S,gas + n M,gas n S,gas + 98
n ˙S,gas = 0.0098
By material balance, then n ˙S,liq =1.9902 kmol/h. For dilute solutions of H2S in water,
we can use Henry’s law to model the gas-liquid equilibrium. At 40°C, from App. B, HH2S
=
6
n ˙W ,liq = 2.97×10 kmol/h!
(b) At T = 10°C, by linear interpolation, HH2S =367 atm. With the change in temperature
and pressure we find
A big improvement.
P5.38
The flow diagram for the clean-up unit is
solvent S S
E
99 lb DEA
solvent
The solvent choices listed in Table B.15 for the DEA-water system are chloroform,
benzene, toluene, and xylene. Chloroform has the highest KD, indicating that DEA will
partition best into chloroform. In the absence of any other considerations (e.g., safety,
cost), chloroform is the best choice.
We’ll assume that the chloroform does not partition into the raffinate (aqueous) phase.
There is one equilibrium relationship relating the concentration of DEA in the extract and
raffinate phases (the exiting streams):
K x
xDEA,E = D DEA,R
99 = 2.2 1
P5.40
The flow diagram is shown, using as a basis 100 kg/h feed. Compositions are given as wt
%.
25 kg A F R 6% A
solvent extraction unit W
75 kg W
M
E A
W
MIBK
M
The raffinate R and extract E stream compositions are related by the phase equilibrium
information in Figure 5.15 (and Table B.13).
Reading from the phase envelope in Figure 5.15, we see that a saturated waterrich
solution containing 6 wt% acetic acid would also have ~2% MIBK and 92 wt% water.
Following along an imaginary tie-line that falls in-between the two dashed tie lines in
Figure 5.15, we estimate that this solution would be in equilibrium with an organic phase
of 5 wt% acetic acid, 92 wt% MIBK and 3 wt% water. Thus, we find
Raffinate: 6 wt% acetic acid, 2 wt% MIBK, 92 wt% water
Extract: 5 wt% acetic acid, 92 wt% MIBK, 3 wt% water
Then we find the solvent flow rate from the 3rd equation:
m ˙ S = 357 kg/h
P5.41
We can draw a flow diagram as
phenol
1 L water
adsorbed phenol
5 g carbon Separator
5 g cargon
1 g phenol
1 L water
To use the phase diagram, we need to convert to mmol: 1 g phenol (C6H5OH) = 10.638
mmol. Letting cP = phenol concentration (mmol/L) in the liquid, and qP = the phenol
adsorbed (mmol/g) on the carbon, the material balance equation is simply
Now we need to find, by trial and error, the point (cP, qP) that lies on the curve and that
satisfies the material balance equation. These two conditions are satisfied at cP ~ 0.6
mmol/L, where qP = 2 mmol/g.
% removal= ×100% = 95%
P5.42
The water leaving the vessel must contain 5 mmol phenol, and there will be 10 L, so the
concentration is 0.5 mmol/L. This is in equilibrium with the adsorbed phenol, which we
read directly off the chart as 1.8 mmol/g activated carbon. The total quantity of phenol to
be removed comes from a material balance: 200 – 5 = 195 mmol. Thus, the grams of
activated carbon required is 195/1.8, or 108.3 g.
P5.43
Since 99% of the MAb fed should be adsorbed, then 1%, or 3 nmoles/liter (nM) MAb,
are left in the liquid phase. There are a total of 300 x 2 or 600 nmoles MAb fed, so by
material balance there must be 594 nmoles adsorbed to the beads. This information is
shown on the flow diagram.
3 nM MAb
2 L
300 nM MAb
2L
nmoles/g adsorbed MAb. Therefore, the total grams of beads required is 594/60, or just
under 10 grams beads.
P5.44
Some possibilities: Sugar melts on a stove top (but it does brown) while salt does not.
The solubility of sugar in water is much higher than that of salt. Students may try other
solvents (ethanol seems to be a popular choice).
P5.45
Some possibilities:
(a) food manufacture: drying, crystallization (sugars)
(b) petroleum refining: distillation
(c) pharmaceutical manufacturing: crystallization (synthetic drugs),
solventextraction (antibiotics made by fermentation), adsorption chromatography
(protein therapeutics)
(d) photographic film manufacture: drying
(e) water purification: membrane filtration, reverse osmosis
P5.46
Technology Category Feed Product Property Separating
phase(s) phase(s) exploited agent
Drying of Equilibrium Moist solid Vapor and Differences Heat
solids solid in volatility
Adsorption Equilibrium Fluid Fluid + Differences Solid
solid in affinity adsorbent
for solid
From this, we propose the following: first, remove salt by ion exchange adsorption, then
concentrate lactic acid, lactose, and protein by reverse osmosis filtration (to remove
water), then, ultrafilter to separate proteins from lactic acid and lactose. Don’t want to
remove all the water yet, because solutions are easier to process than solids. If lactic acid
is not marketable, could try running the ion exchanger at a pH where the lactic acid is
ionized, so that some/all of it adsorbs as well. Alternatively, if lactic acid is a useful
product, could adjust pH to neutralize charge, then extract with ether (better than ethanol
because less soluble in water and lower boiling point.) Finally, remove remaining water
in driers.
4 W
W W P
P L
W dryer
L P 6
LA L
S
ion exchange reverse osmosis 9
P
1 3
L
5 ultrafiltration
2
W
P 10
L W
7
LA
S
dryer
L
W
11
L
Note the following. First, two ion exchange steps will be required, one to remove cations
and the other to remove anions. Second, the ultrafiltration filter will separate lactose from
protein, because the protein can’t move across the filter, but some lactose will stay with
the protein solution. Thus, the final dried protein solution will have some lactose in it.
Using as a basis 1000 kg whey fed to the process, we make these preliminary estimates
of flows (all in kg). We want to keep lactose soluble until the dryer (0.1 mg/ml max, or
~1 kg/10 kg water), and we want to avoid gelling of the protein (1 kg protein/kg water),
which limits the amount of water removal in earlier steps,
1 2 3 4 5 6 7 8 9 10 11
P 9 9 9 9 9
LA 2 2
S 5 5
The dried protein product is 91 wt% protein and 9 wt% lactose. The dried lactose product
is pure.
P5.48
Here is one reasonable scheme. Notice how much effort is expended to recover and
recycle Cl2!
In this scheme, fresh benzene (B) and chlorine (Cl) are mixed with recycled streams and
fed to Reactor 1. All the chlorine but only a fraction of the B react, since Cl is by far the
limiting reactant. B and monochlorobenzene (MCB) are easily separated from HCl by
cooling and condensing in Separator 1, because the boiling points of B (80°C) and MCB
are much higher than that of HCl (-85°C). Benzene and MCB are separated by
distillation in Separator 2; their relative volatilities are significantly different and
distillation is fairly easy.. The MCB is taken off as product and benzene is recycled.
HCl is mixed with air (oxygen and nitrogen) and fed to Reactor 2, where the HCl is
partially converted back to Cl2, with water as a byproduct. HCl and Cl2 are both quite
soluble in water whereas oxygen and nitrogen are relatively insoluble in water.
Therefore, water is used as an inexpensive solvent in Separator 3, which is an absorption
tower. (The Reactor 2 outlet will likely be cooled prior to feeding it to Separator 3.)
Oxygen and nitrogen gas is discharged to the atmosphere (or to a flare, just in case of a
process upset.) The HCl and Cl2 dissolved in water is sent to Separator 4, which is a
liquid-liquid contactor. Carbon tetrachloride is added, and much of the Cl2 partitions into
this phase. (Carbon tetrachloride is essentially immiscible with water – see Table B.15.)
The boiling point of carbon tetrachloride is 77°C, so it is readily separated from Cl2 by
heating in Separator 5. The chlorine is recycled back to the mixer, and the carbon
tetrachloride is recycled back to Separator 4. HCl and water (along with some Cl2 that
remains in the aqueous phase is fed back to Reactor 2. (Some purge of this stream, not
shown on the flow diagram, will be necessary to remove the water made by reaction. It
may not be necessary to add much if any water to Separator 3, as it can be recycled.)
P5.49
Notice the following features of this problem:
(1) Propylene and HCl are the major components of stream.
(2) HCl and Cl2 are both corrosive.
(3) Propylene, HCl, Cl2 all boil at very low temperatures, and are significantly different in
boiling point than the chlorinated hydrocarbons.
(4) Want to recycle propylene and Cl2 to reactor feed but not the HCl. Therefore we need
to separate Cl2 from HCl, but we don’t necessarily need to separate propylene and Cl2.
(5) HCl is highly soluble in water.
(6) 1,3-dichloropane and acrolein chloride need to be separated from the product allyl
chloride, but there is no need to separate these two byproducts from each other.
Removing the HCl early would be suggested by the heuristic “remove corrosives early”,
but so is Cl2. So it makes sense to apply the heuristic “do the easy separation first” and
the heuristic “separate large quantities first”.
Distillation to separate propylene and Cl2 from HCl doesn’t make sense because it would
have to operate at very low temperatures (violates “avoid very high or very low
temperatures and pressures” heuristic). Absorption into water takes advantage of the
differences in solubility.
Distillation would work well to separate allyl chloride from the chlorinated waste
products, because allyl chloride is more volatile than the byproducts.
P
P Cl2
HCl recycle to
CL2 reactor
Absorber
Water
HCl
P Cl2 ( little? )
HCl P ( little? )
Cl2 Flash to water
Allyl drum treatment
Acrolein
13 DCP Allyl
Allyl
Acrolein
13 DCP
Distillation
Acrolein
13 DCP
P5.50
We picked chloroform to extract diethylamine. The boiling point of chloroform is 61°C
while that of DEA is 58°C. Distillation might be an option, although these are pretty
close so many stages may be required. The melting points are –63.6 and –93°C,
respectively. These are pretty different, but are so low that it would be very expensive to
try to exploit crystallization as a separation method. Maybe there is another solvent we
could add, but that doesn’t seem to be getting us anywhere! Diethylamine is a strong
base. One possibility might be to use adsorption with an ion exchange resin (with acid
groups immobilized on the resin). Chloroform would not adsorb to this resin, so it could
be recycled back to the solvent extraction unit. The diethylamine could accumulate on
the resin until it is saturated, then recovered by flowing an elution solvent across the
resin. Now of course the DEA is dissolved in the new solvent…but it is at least more
concentrated than the initial contaminated water.
It might be worth a look at a solvent other than chloroform for the original extraction.
Benzene, for example, is almost as good at extracting DEA from water as chloroform (KD
= 1.8 vs 2.2), but it’s boiling point is much more different than DEA (80°C), so
distillation would be a relatively easy way to separate benzene from DEA. (One concern
with benzene is that it is not completely immiscible in water; we might be substituting
one pollution problem with another!) Still, this illustrates how, in separation processes
that require the addition of a foreign species, the choice of that species depends not only
on its performance in the main separation step, but also on the ease of recovering and
reusing the material.
P5.51
Scheme A: Flow rates are shown in molar units, with a basis of 100 gmol/min S fed. The
inert:oxygen ratio in the mix fed to the reactor is 476:100 which is greater than 7:3.
376 N2
Air 376 N2
100 O2 100 SO2
376 N2 Reactor Separator
100 SO2
100 S
Scheme B: Recylce of SO2 to the reactor is required to maintain the inert:oxygen ratio in
the mix fed to the reactor at 7:3.
376 N2 233 SO2
Air
100 O2 333 SO2
376 N2 Separator Reactor Splitter
100 O2
100 SO2
100 S
Scheme A has a lot going for it. It avoids an expensive separation (N2/O2) and adds a
much easier separation (N2/SO2). (SO2 differs significantly from N2 in its boiling point,
size, and its solubility in water, any of which difference could potentially be used as a
basis for separation.) It also uses nitrogen as the diluent, avoiding the need for recycle.
Finally, if not all the O2 is consumed by reaction, with the separator after the reactor you
can easily remove traces of O2 from SO2 in this scheme.
P5.52
This problem reviews material from Chapters 3 and 4 and adds some new material from
Chapter 5.
The ethane:oxygen reactor feed is maintained at 6:1 molar ratio, so oxygen is the limiting
reactant. This is done to reduce excess oxidation of ethanol to methanol, CO, and CO2.
Because air is the source of the oxygen, the nitrogen in air must be removed from the
system somewhere. It is expensive to separate nitrogen from other gases (CO2, CO), so it
makes more sense simply to purge this stream rather than try to recover all the ethane.
Separating the gases from the liquids first in Separator 1 makes a lot of sense. This is an
easy separation (following the heuristic – save the difficult separations for last), and it
splits the reactor outlet into roughly equal flows. The boiling points of acetaldehyde,
methanol and water are 20°C, 65°C, and 100°C, all at or slightly above ambient and
fairly widely separated. Distillation, since it operates at pressure and temperature at
which vapor and liquid phases can co-exist, and because it exploits differences in
relative volatility, makes sense. The sequence shown is reasonable; an equally reasonable
sequence would be to have distillation column separate into water + methanol and
acetaldehyde, followed by a methanol-water separation. This alternative sequence might
be preferable if methanol is sold as a byproduct (following the heuristic – take products
off as distillates one-by-one). More detailed analysis is required to make a final choice.
The flow diagram is sketched below (we assume that the separators are perfect).
Components are named as E (ethane), O (oxygen), N (nitrogen), CD (carbon dioxide),
CM (carbon monoxide), A (acetaldehyde), M (methanol), W (water). Streams are
numbered as shown.
N
CD N
CM CD
10 11
E CM
splitter
E
N 7
9 A
CD
CM
E E
O A
N 2 3 gas-liq 5
mixer reactor M
sep 1 dist
1 E
E 3
N
O CD
N CM
CD A 4 M
CM dist
W 2 8
A
M W
M
W
6
Reactions:
C2H6+O2→C2H4O+H2O (R1)
C2H6+ 3.5O2 → 2CO2+ 3H2O (R2)
C2H6+1.5O2 → CH3OH + CO + H2O (R3)
DOF Analysis
No. of stream variables 35 Counting components in each stream and summing:
3+5+7+3+2+1+1+1+4+4+4
No. of reaction variables 3
n ˙CM11 =
˙
0.5 n CD11
n ˙E11 = 0.10
n ˙E11+ n ˙CM11+ n ˙CD11+ 24452
Excluding the nitrogen material balance, we have 9 equations in 9 variables. We can use
an equation solving package like EES, or a spreadsheet with Solver, to find a solution.
˙ ˙
ξ 1= 468 ξ
˙
2 =ξ 3=1206
Purge gas: 31,190 gmol/h: 10% ethane, 78.4% nitrogen, 3.9% CO, 7.7% CO2.
The overall conversion of ethane is
ξ˙
fCE= 1+ξ˙ 2+ξ˙ 3= 468+1206+1206 = 0.48
n ˙E1 6000
The overall yield and selectivity for converting ethane to acetaldehyde are:
ξ˙ 1 468
sE→A= ˙ 1+ξ˙ 2+ξ˙ 3= 468+1206+1206 =
0.1625 ξ
P5.53
(a) Cellulose acetate is a polymer made from wood pulp and acetic anhydride. It is used
for example in textiles, photographic film, absorbent disposable materials, and filters. (b)
There are four separation steps. Extraction separates acetic acid from sulfuric acid and
water. It is used instead of distillation because acetic acid boils at a temperature in
between water and sulfuric acid, yet we want to separate it from these two in one step.
Extraction is placed first in the sequence to get rid of the corrosive acid early, in line with
the heuristics. The other 3 steps are all distillation: the solvent recovery column separates
ether from the acetic acid/water mix; the acid finishing separates acetic acid from water,
and the ether stripping separates ether from water.
(c) The flow diagram is shown, with W for water, S for sulfuric acid, A for acetic acid,
and E for diethyl ether.
DOF Analysis
No. of stream variables 31 Counting components in each stream and summing:
3+3+3+2+2+2+2+2+2+2+4+4
No. of reaction variables 0
DOF 31 – (11+1+18) = 1
If we choose a basis, the system is correctly specified. We’ll choose 1000 lb/h in stream 1
as the basis. Then we write the 19 material balance equations, the 11 composition
specifications, and the 1 performance specification equations, and solve. Sulfuric acid
balances are solved first. After that, acetic acid is the component that is the easiest to
work with first because we have the most information. Once the acetic acid flows are
calculated, there is sufficient stream composition information to get several of the water
and ether flows. The equations and solution are summarized.
698+ m ˙ W7 = m ˙ W2 m ˙
W10 + m ˙ W2 = m ˙ W3 + m ˙
W11
Water: m ˙ W3 = m ˙ W6 + m ˙ W4 m ˙ W4 =
m ˙ W5 + m ˙ W7 m ˙ W11 =
m ˙ W9 + m ˙ W12 m ˙ W8 +
m ˙ W9 + m ˙ W6 = m ˙ W10
m ˙ E8+ m ˙ E6+ m ˙ E9 =
m ˙ E10 m ˙ E10 = m ˙ E3+ m ˙ E11 Ether:
m ˙ E11= m ˙ E9+ m ˙ E12 m ˙ E3= m ˙ E6
S 2 2 2 2
W 698 962.8 279.5 267.5 2.7 12 264.8 0.007 0.65 12.6 696 695
A 300 430.4 401.2 401.2 270.8 130.4 29.2 29.2
sum 1000 1395 1672 669 274 100 395 1 54 1058 782 727
3
90.3% of acetic acid fed is recovered in concentrated product.
P5.54
Let’s look at the boiling points of the 4 components:
Pentane: 36°C
Benzene: 80°C
Toluene; 110.6°C
o-xylene: 144°C
Based on simply the differences in boiling points, the easiest separation is between
pentane and benzene, followed by toluene and o-xylene, and the most difficult (although
still not particularly difficult) is between benzene and toluene. It would be a little better
to calculate the saturation pressures from Antoine’s equation, and then compare the ratios
of saturation pressures as a measure of the relative volatility. Using the data in Table B.4,
and arbitrarily choosing a temperature of 100°C for comparison, we find saturation
pressures of:
Pentane: 4520 mm Hg
Benzene: 1350 mm Hg
Toluene; 556 mm Hg
o-xylene: 198 mm Hg
The ratio of saturation pressures is in the order of 3.35 (pentane/benzene) > 2.81
(toluene/xylene) > 2.43 (benzene/toluene) – same as what we concluded by looking at
differences in boiling point temperatures. The proposal from the supervisor follows this
sequence exactly.
Based on the heuristics of separating into equal amounts and removing the largest
quantities first, one might argue for doing the o-xylene – toluene separation first, then the
pentane-benzene separation, and finally the benzene-toluene separation.
The supervisor’s design is quite reasonable, but there is also a reasonable alternative.
Since he’s the boss, I’d probably just agree with him!
F
90 % T
2% B
3 8% X
1% T
99 % X
4
1000
n ˙2 = with all flows in gmol/h
n ˙3 1000
n ˙4 5980
The fractional recoveries of each component in their respective product stream are
0.98
fRP1 == 0.985
0.90
fRB2 == 0.90
0.9
fRT3= = 0.90
0.99
fRX4 == 0.987
P5.55
The flow diagram is shown.
liquid L
water
limestone
NaCl
2 g NaCl
230 g limestone
water
sludge
water
limestone
NaCl
In the clear liquid L, the maximum concentrations of limestone and NaCl are set at their
solubility limits of 0.015 and 300 g/l, respectively. If there is no limestone or no NaCl in
the solid phase, then the concentration may be below the solubility limit.
The limestone solubility limit is very low, so some of it will be in the solid phase, thus
we can pin the liquid concentration at 0.015 g/ml. From material balance:
230 g = 0.015 g/L ×0.722 L +mLS,S
mLS,S = 229.99 g limestone in the sludge
the concentration of limestone in the sludge is therefore 827 g/l.
For NaCl, let’s assume that it is at its solubility limit in the liquid
2 g = 360 g/L ×0.722 L +mNaCl,S mNaCl,S =−257.9 g?? the negative sign shows
us that our assumption was wrong, NaCl is below its solubility limit – none of it goes into
the solid phase; it is in the clear liquid as well as the entrained solution in the sludge.
To calculate the volume of entrained solution vs. solid phase in the sludge, I need the
density of limestone – 2.7 g/ml. The total volume of solid limestone is therefore
229.99/2.7, or 85.2 ml. The sludge volume is 278 ml, so the entrained liquid is 278 – 85.2
or 192.8 ml.
Thus, the total liquid (clear liquid plus entrained solution ) is 192.8 + 722 = 915 ml. The
NaCl concentration is therefore 2 g/915 ml, or 2.19 g/L. Therefore the sludge contains
2.19 x 0.193 L or 0.42 g NaCl, and the clear liquid has 1.58 g NaCl.
= 86,500 (approx.)
P5.56
Let’s start by considering phase behavior of potassium nitrate – water system. If the
solution is cooled to 5°C, from Figure 5.11 we find that the saturated solution is 0.15 kg
KNO3/kg solution. At 90°C, the solubility is very high – about 0.65 kg KNO3/kg solution.
The vacuum evaporator alone won’t due the trick, because its maximum salt is still below
the solubility limit. But, if we cool the solution we will get some crystals forming.
Can we accomplish the necessary separation with just the cooler, rotary drum filter, and
drum dryer? The flow diagram is
water
4
cake
+ ES
1 cooler plus
300 K dryer crystals
700 W filter 100 % K
3 5
filtrate
0.15 kg K/kg soln
For material balance purposes, we’ve lumped together the cooler plus rotary drum filter.
The entrained solution (ES) is 0.15 kg KNO3/kg solution, and there is 1 kg solution
entrained per 19 kg cake (which is pure KNO3). The material balance equation on KNO3
around the cooler/filter is
300 = 0.15m ˙ 2+ m ˙ C3+ 0.15m ˙ ES3 where we have distributed the flow in
stream 3 between cake C and entrained solution
ES. We also know from the performance specification on the filter that
m ˙ C3=19m ˙ ES3
and the material balance on water gives:
700 = 0.85m ˙ 2+ 0.85m ˙ ES3
with flows all in kg/h. Assuming all the water from stream 3 is removed in the dryer, then
a material balance around the dryer yields
What if we evaporate some of the water first, to raise the feed solution to the cooler/filter
up to 50 wt% KNO3?
water
6
4
cake
+ ES
1 cooler plus
300 K evap dryer crystals
700 W filter 100 % K
7 50 % K 3 5
filtrate
0.15 kg K/kg soln
The material balance equations around the cooler/filter are slightly modified:
300 = 0.15m ˙ 2+ m ˙ C3+ 0.15m ˙ ES3
m ˙ C3=19m ˙ ES3
300 = 0.85m ˙ 2+ 0.85m ˙ ES3
Let’s go shopping!
P5.57
(a) Product purity is ×100% = 84.6wt%if just the water is removed. Component
recovery is 100%.
(b)
Liquid
W
S
W
2 4 7
5
11 % ASA 3 6 8
evaporator crystallizer dryer ASA
2% S 1 S
87 % W Filter cake +
W 35% ASA
entrained
solution
ASA
S
W
We’ve used S for sodium acetate and W for water. We’ve denoted two streams 5 and 6 as
the solid phase and the entrained liquid solution, respectively. The entrained solution is
assumed to be at the solubility limit of ASA, that is, 35% ASA. The filter cake contains
no salt. Given 100 kg in the feed to the evaporator, there are 11 kg ASA fed to the
process (mA3=11 kg), and half is recovered as dried product in stream 8 (mA8 = 5.5 kg).
From this we write a material balance equation around the dryer
5.5 =m5+ 0.35m6 (the solid phase is pure ASA since sodium acetate is
considered infinitely soluble)
From the entrainment specification, m5=4m6
Therefore m5=
5.056
m6=1.264
Since half of the ASA leaves the dryer as dried product, the other half leaves the system
in stream 4, as liquid filtrate: mA4 = 5.5
This is 35 wt% ASA, so
5.5
m4 = =15.71
0.35
The 2 kg sodium acetate is at the same concentration in stream 4 and stream 5
Since there is 1.264 kg entrained solution fed to the dryer, there is 0.118(1.264) or 0.149
kg sodium acetate fed to the dryer, all of which ends up in the dried product. Therefore,
the final product purity is better in the alternative process:
×100% = 97.36 wt% ASA
5.5
The total feed to the evaporator is 100 kg, therefore the water removed in the evaporator
must be 100 – 22.03 or 78 kg.
(c) As we’ve seen from parts (a) and (b), product purity increases as component recovery
decreases. The impurity is due to the sodium acetate in the entrained solution. In terms of
our stream variables:
A8
m ×100%
product purity (%) =
mA8+mS8
where we use A for acetylsalicylic acid and S for sodium acetate. The fractional recovery
is
mA8
fRA= mA1
mA1 m m
product purity (%) = fRA ×100% fRA A1+ S8
From the phase equilibrium information on ASA and the entrainment specification, plus
a material balance on ASA,
mA8 = m5+ 0.35m6= 4.35m6
Also
mA1= mA3= 0.35m4 +mA8= 0.35m4 + fRAmA1 m4 =
mS1 = mS8
Now we can write purity in terms of fractional recoveries and feed quantities:
RAmA1
product purity (%) = f ×100%
mS1
fRA)
0.35 fRA
We check the expression against the two points that we previously calculated:
fRA=1.0 ⇒ product purity = 84.6% check!
fRA= 0.5 ⇒ product purity = 97.4%
Component recovery
At 90% recovery, we calculate a purity of 92.2wt%, while 10% recovery gives us 98.4 wt
% purity. The purity drops slowly as recovery increases to about 50%, then precipitously
as recovery increases further.
(d) The best change to make would be to install a separation device upstream that could
remove the sodium acetate, if possible. Alternatively, the quantity of entrained solution is
rather high, installing a better filter would improve purity (or improve recovery at equal
purity).
P5.58
The labeled flow diagram, with components indicated by S for sodium carbonate, and W
for water, is shown. Flow rates are in lb/h. We will distinguish between crystals and
entrained solution in stream 5 by subscripts C5 and ES5, respectively.
W
3000 S 2 4 5 crystals
7000 W mixer evaporator crystallizer/filter entrained
1 solution
6
S
W
The mass fraction of salt in the entrained and recycle solution is 0.177. Considering the
entire process as the system operating at steady-state, we write the material balance
equations:
The product stream includes the crystals (100% sodium carbonate) plus entrained
solution (17.7wt% sodium carbonate), so the product purity is
×100% = 81.7wt%
From the performance specification on the evaporator, that 40% of the water fed to the
This is the total feed to the evaporator, which includes both fresh (7000 lb/h) and
recycled. Therefore, the recycle flow rate of water must be m ˙ W4 =15820−
7000 = 8820 lb/h
Since the recycle stream is 82.3 wt% water, then the total recycle flow rate is
W4 8820
m˙ = =10716 lb/h
m ˙ 4=
0.823 0.823
(b) Increasing the temperature at which the evaporator operates (and removing
morewater in the evaporator) might help. Most salt solubilities increase with temperature,
so this would increase the concentration of salt in the entrained solution, thus reducing
the amount of water in the product. Improvements in the filter operation that reduce the
amount of entrained solution would help product purity. Alternatively, installation of a
dryer to remove the remaining water would also increase purity. Recovery remains at
100%.
P5.59
(This is a hard problem – students may need some suggestions to get them thinking in the
right direction.)
First, solvent-extraction is useless – benzene B and naphthalene N are chemically very
similar.
Let’s next examine crystallization. We can go as low as 12°C. The phase diagram in
Figure 5.12 is in mol fraction, so we need to convert: 60 wt% benzene/40 wt%
naphthalene is equivalent ot 71 mol% B/29 mol% N. From the phase diagram in Figure
5.12, a 29 mol% N solution cooled to 12°C would phase separate into a pure N solid
phase and a liquid phase of about 21 mol% N (or 30 wt% N). We have a drum filter that
entrains 1 lb solution per 10 lb solids. Material balances around such a filter, along with
this performance specification and the phase equilibrium information, are
600 = 0.7m ˙ L+ 0.7m ˙ ES (benzene) 400
= 0.3m ˙ L+ 0.3m ˙ ES+ m ˙ C
(naphthalene) m ˙ C =10m ˙ ES
where L, C, and ES indicate liquid, crystal, and entrained solution flows, in lb/h,
respectively. Solving by substitution and elimination, we find:
m ˙ C =143 lb/h
m ˙ ES =14.3
lb/h m ˙ L =
842.7 lb/h
The filtrate (liquid from filter) is only 70 wt% B, much below our target of 99%.
What about the distillation column? To prevent formation of tarry materials, the bottoms
temperature can be no greater than 150°C. The higher the pressure, the higher the
temperature, so we want to operate at the minimum allowable pressure – 1 atm. What we
need to do is calculate the bubblepoint composition if the temperature is fixed at 150°C.
This sets the composition of the bottoms product from the distillation column.
From Antoine equations, we calculate that at 150°C
PBsat = 5.74 atm
But, what if we still operated the distillation column, and produced a distillate that met
our purity specifications for benzene (99%), and an impure bottoms product of 85 mol%
N? Can the distillation column available handle that separation? (Reminder: the vapor
phase in equilibrium with the bottoms product is NOT the same as the distillate if we
have a multistage column!)
From the Fenske equation, a distillation column producing a 99% benzene
distillate and a 85% naphthalene bottoms requires
0.99 0.85
0.01 0.15
Nmin = log 5.74 ≈ 2 log
0.1585
Now we could feed the bottoms from the distillation column to the crystallizer/filter
instead of the raw feed, and recycle the filtrate solution back to the distillation column
feed:
distillate
99 % B
1% N
1000 lb/h
60 % B solid +
40 % N entrained
dist col
mixer solution
4 stages
The maximum operating temperature of the evaporator is 250°C, but we may still want to
keep the temperature at or below 150°C to avoid tarry polymer formation. Can we get the
separation we need in the evaporator? To find out, we plug the saturation pressures of
benzene and naphthalene into the equation, calculate P, and compare to 300 mm Hg (0.39
atm)
We can operate the evaporator at 0.49 atm and 150°C, well within its operating range,
and obtain the purity we need. We’ll recycle the bit of vapor from the evaporator back to
the feed. The final design is
distillate
99 % B
1% N
1000 lb/h
60 % B
40 % N
dist col
mixer evaporator
4 stages solid +
entrained
solution
P5.60
We have a couple of considerations based on the solubility data. Na2SO4 is pretty soluble
in water. At 40°C, the solubility is at a maximum of 48.8 g/100 g water, and the solid in
equilibrium with this saturated solution is anhydrous Na2SO4. At cooler temperatures, we
get lower solubility – this means more crystals produced, but the crystals are not
anhydrous and any entrained solution is lower in salt concentration. So the dryer has to
work harder if we operate the filter at a lower temperature. We want to remove as much
water as possible in the evaporator compared to the dryer, and recover as much of the salt
as possible. One way to accomplish this is to skip the cooler all together, and recycle the
filtrate back to the evaporator. We should operate the vacuum evaporator at 70°C, by
interpolation of the data in App. B we estimate that the solubility of Na2SO4 at this
temperature is 44.5 g/100 g (30.8 wt%). The proposed flow diagram, with flows in kg/h
and compositions in wt%, is shown.
W
W
100 S 60 % S solids
900 W 40 % W ES crystals
mixer evaporator filter dryer
30.8 % S
69.2 % W
Process flow calculations proceed using the usual combination of material balances,
performance specifications, and phase equilibrium. We’ll simply summarize the results:
Flows Fresh To Water To Filtrate Cake Water Crystals
(kg/h) feed evap from filter (recycle) (solids from
evap + soln) dryer
Na2SO4 100 139.9 139.9 39.9 100 100
P5.61
At 40°C, from Table B.8, the solubility of K2SO4 in water is 14.76 g salt/100 g water, or
12.9 g salt per 100 g solution (0.129 mass fraction).
(a) The labeled flow diagram is shown, with components potassium sulfate K and
waterW. Stream 5 includes both crystals C (pure K) and entrained solution ES (saturated
solution of K and W); flows are in lb/h.
W
2000 K 40 % K
1 60 % W Crystals
8 000 W
mixer evaporator crystallizer/filter Entrained
3 5 solution
4
K
W
Solving
m ˙ C5=1975 lb/h Crystal production rate
m ˙ ES5=197 lb/h
The balance equation for water (again with the entire process as system), is
8000 = m ˙ 2+ 0.871m ˙ ES5= m ˙ 2+ 0.871(197)
Water evaporation rate
m ˙ 2= 7828 lb/h
Now let’s lump the mixer + evaporator together as our system. Material balance
equations are
2000+ 0.129m ˙ 4 = 0.4m ˙ 3 (K balance)
8000+ 0.871m ˙ 4 = 7828+ 0.6m ˙ 3 (W balance)
(b) The lower temperature would produce a filtrate with a lower concentration of
salt.However, without the water evaporation step this could not be recycled – because the
water must be removed somewhere. This solution would result in a significantly lower
recovery of salt in the filter cake product. Not a good idea.
(c) If we assume that the operation of the crystallizer/filter remains at 40°C, and
theentrained solution remains at 1 lb/10 lb crystals, then the purity of the product must
stay the same. This is because the salt concentration in the saturated liquid solution
remains at 0.129 g/g. But, if less water is evaporated, then the feed rate to the process
must decrease.
Let’s look at the case where the water evaporation rate is 6000 lb/h. Now we allow the
feed rate to vary. The balances around the entire system are:
0.8m ˙ F = m ˙ 2+ 0.871m ˙ ES5= 6000+ 0.871m ˙ ES5
0.2m ˙ F = m ˙ C5+ 0.129m ˙ ES5
Solving, we find that the feed and production rates at the new lower evaporation rate are
m ˙ F = 7665 lb/h Feed rate m ˙ C5=1510 lb/h Crystal production rate m ˙
ES5=151 lb/h
The feed rate drops linearly with the drop in the water evaporation rate, as does the
crystal production rate. Further calculations would show that the recycle rate also drops
linearly with water evaporation rate. This is because all the equations involved are linear.
(d) If the fresh feed rate is maintained constant while the water evaporation rate
decreases, then by overall material balance some of the filtrate must be split off and
discarded – it can’t all be recycled. It seems like a dumb idea – the net effect of what the
superintendent is proposing is to simply feed material in and then discard it. (Of course,
one would be more polite in responding to the superintendent!)
P5.62
The crystallizer operates at –50°F (-45°C). From Table B.10 we see that this temperature
produces a liquid phase containing 16.5 mol% p-xylene/83.5 mol% m-xylene, and a
solid p-xylene phase. The process is shown, with flows in gmol/h and compositions in
mol%
solids ( P )
300 P cooler/crystalli
ES (16.5 % P,
700 M zer/filter
83.5 % M)
Filtrate
16.5% P
83.5% M
The material balance equations are (with S for solids, ES for entrained solution, and F for
filtrate)
300 = n ˙S+ 0.165n ˙ES + 0.165n ˙F (p-xylene)
700 = 0.835n ˙ES + 0.835n ˙F (m-xylene)
If p-xylene is the more valuable product, the addition of the isomerization reactor is a
great idea. It allows eventually almost all of the feed to be converted to the desired
product. This is achieved through the coupling of reaction and separation. Since this is an
isomerization, the reaction is reversible, and reaction alone could not achieve complete
conversion of m-xylene to p-xylene.
P5.63
The flow diagram for the flash drum is
H
V EB
0.2 % S
50 H F
30 EB flash drum
20 S
EB
L S
where flows are in kgmol/h and compositions in mol%. We have made an assumption
that hydrogen H does not condense at any reasonable operating conditions for the drum,
and therefore appears only in the vapor (V) phase. We have assumed that ethylbenzene
EB and styrene S are in both vapor and liquid phases. Since the flash drum operates as a
single equilibrium stage, V and L are in equilibrium. Since EB and S are chemically
similar compounds, Raoult’s law is an appropriate model for phase equilibrium:
xEBPEBsat =yEBP
xSPSsat =ySP
We do not write an equilibrium expression for hydrogen – all of it is in the vapor phase.
We do write material balance equations for all three components:
50 = yHn ˙V
30 = yEBn ˙V + xEBn ˙L
20 = ySn ˙V + xSn ˙L
The pressure is specified at 760 mm Hg, and the styrene content of the vapor stream yS =
0.002. We also know that mole fractions in each stream must sum to 1: yH +yEB+yS =1
xEB+xS =1
Finally, we have two Antoine equations that relate saturation pressures to temperature:
log10PEBsat = 6.95719− 1424.255
213.21+T
Note that there is a short-cut we can take to solve this problem. Very little ethylbenzene
and styrene are in the vapor phase, and their saturation pressures are not that different, so
to a good approximation we can say that the EB:S ratio in the feed is the same as that in
the liquid product. With this approximation, we estimate that xEB = 0.6, xS = 0.4, and we
can immediately use these values in the Raoult’s law equations (plus the Antoine
expressions) to find the desired operating temperature and the mole fraction ethylbenzene
the vapor phase.
P5.64
For methanol and water, Antoine equations are
log10Pmsat = 7.97328−
1515.14
T+ 232.85
log10Pwsat = 7.96681−1668.21
T+
228
Our procedure will be to set P = 1 atm, pick a temperature, calculate the bubblepoint
composition by satisfying xmPmsat +xwPwsat =P
then calculating the vapor composition in equilibrium with that liquid at the calculated T
from Raoult’s law. We can do this easily by recognizing that the bubblepoint equation
can be re-arranged to
xm = PmPsat−−PwPsatwsat
Calculations are then carried out in a spreadsheet, results are tabulated and plotted below.
(Note that we choose the temperature range from the methanol boiling point to the water
boiling point.)
T Pmsat Pwsat xm xw ym yw
65.00 769.76 187.61 0.98 0.02 1.00 0.00
67.50 848.59 209.62 0.86 0.14 0.96 0.04
75.30 1138.65 292.83 0.55 0.45 0.83 0.17
84.40 1575.52 423.48 0.29 0.71 0.61 0.39
91.20 1984.47 550.29 0.15 0.85 0.38 0.62
96.40 2352.27 667.36 0.05 0.95 0.17 0.83
100.00 2637.91 759.98 0.00 1.00 0.00 1.00
Temperature (°C)
The solid lines are the data and the dashed lines are the calculations from Raoult’s law.
This system exhibits significant deviations from Raoult’s law. This is not surprising
because methanol and water are both good hydrogen bonders. Raoult’s law should NOT
be used to model methanol-water vapor-liquid equilibrium.
P5.65
Water and methanol are the most condensable. If these were pure at 200°C and 4000 kPa,
would they be liquid or vapor? We can estimate this from the Antoine equation (although
the accuracy is not great, as 200°C is outside the range of the constants in Table B.4).
1668.21
Water: log10Pwsat = 7.96681− Pwsat =11725 mm Hg at 200 °C
T+ 228
4000 kPa = 39.5 atm = 30,000 mm Hg. Since P > Psat for both compounds, if these
compounds were pure they would be liquid.
From the Gibbs energy and enthalpy of formation (Table B.3) assuming all gas phase, we
calculate:
ˆ
ΔG R1=−25.05 kJ/gmol
ˆ
ΔH R1=−90.41 kJ/gmol
ˆ
ΔG R2 =+28.51 kJ/gmol
ˆ
ΔG R2 =+41.14 kJ/gmol
ym 1= (
ξ˙ 1 700−2ξ˙ 1)2 2 12
˙
ξ 1=123.8
˙
The solution is ξ 2 =
0.14
The mole fraction of methanol in the outlet is 123.8/452.3 or 0.274. The mole fraction of
water in the outlet is 0.14/452.3 or 0.0003. Is this above the dewpoint of the mixture? (we
can assume the other components are not condensable). We’ll use Raoult’s law as a crude
)
measure: if the mixture is at its dewpoint then xw = ywsatP = 0.000311725(30000 =
0.0008 Pw
xm = ymsatP = 0.27429709(30000) =
0.277 Pm
Since xw + xm << 1, then this mixture is NOT condensing, it is above its dewpoint, and the
mixture is all vapor, consistent with our original assumption. (One could calculate the
dewpoint temperature of the mixture and see how far we are from having a liquid phase
present.)
P5.66
The flow diagram is
O2
0.20 = yPWWsat1P
sat
From Antoine’s equation we calculate that at 20°C PW =17.5 mm Hg. Since the total
pressure is 1 atm (760 mm Hg)
0.20
yW1 == 0.0046
yW4 = = 0.062
The volumetric flow rate of warm humid air serves as our basis. To convert to a molar
flow rate, we use the ideal gas law
)
PV ˙ (1 atm (2000 ft3/min)(28.32 L/ft3)
n ˙4 = = = 2226 gmol/min
RT (0.082057 atm L/gmol K)(273+ 37)
Now working back:
n ˙N1= 0.538n ˙4 = 0.538(2226)=1197.6 gmol/min
n ˙N1 = 318.5gmol/min
n ˙O1=
3.76
+ n ˙O1
n ˙1= n ˙N1 =1523 gmol/min
0.9954 n ˙W1= 0.0046n ˙1= 7 gmol/min n
˙2= 0.40(2226)− 318.5 = 572 gmol/min n ˙3=
0.062(2226)−0.0046(1523)=131 gmol/min
Now that we have all the molar flow rates, we convert to mass flow rates (through molar
masses) and volumetric flow rates (through ideal gas law for air and O2, using liquid
density of 1 g/cm3 for liquid water).
Results:
Air in: 1523 gmol/min, 43.8 kg/min, 1290 ft3/min
Pure oxygen in: 572 gmol/min, 18.3 kg/min, 486 ft3/min (at 20°C and 1 atm)
Liquid water in: 131 gmol/min, 2.36 kg/min, 0.083 ft3/min
P5.67
Antoine equations for isopentane (i) and n-pentane (n) are
(a) The pressure is 350 kPa = 2625 mm Hg. For equal-sized vapor and liquid products
(assuming an arbitrary basis of 1000 gmol/s)
V = 500
L =500
Raoult’s law is appropriate as a model of vapor-liquid equilibrium for this similar
compounds.
yiP=xiPisat
ynP=xnPnsat
Also
yi +yn =1 xi
+xn =1
Solving this set of equations simultaneously in EES, Excel, or other tool, we find
T= 74.1°C
(at which Pisat = 2991 mm Hg and Pnsat = 2404 mm Hg. Notice that one saturation
pressure lies above the total pressure and the other lies below. This is a good check on
the calculated temperature.)
(b) A single stage flash cannot achieve the desired recoveries. The column flow
diagramshould be
distillate
398 iC5
12 nC5
bottoms
2 iC5
588 nC5
where distillate and bottoms flows are calculated from the specified component
recoveries of 99.5% for isopentane and 98% for n-pentane. At this flows,
Saturation pressures at 60°C are 2039.6 and 1654.9 mm Hg for isopentane and n-pentane,
respectively. From the Fenske equation
0.9707 0.9966
log
0.0293
0.00339
Nmin = 2039.6 = 44 stages
log1654.9
(c) The distillate is almost pure isopentane, so the temperature should be roughly
thesaturation pressure of isopentane at 350 kPa. From Antoine’s equation:
1020.012
log10(2625)= 6.78967−
233.097+T T=
69.5°C
P5.68
The desired operation is
Vapor
0.9 M
0.9 B
1 M
flash drum
4 B
Liquid
0.1 M
3.1 B
with flows in gmol/min and calculated from the recovery and purity specifications.
This system forms an azeotrope and is clearly nonideal. Raoult’s law cannot be used to
model its behavior!
(a) Reading from the diagram, the vapor contains 50% methanol at ~61°C and
is inequilibrium with a liquid of ~13 mol% methanol. From the inverse
lever rule or from completing the material balance equations:
1= 0.5n ˙V + 0.13n ˙L
4 = 0.5n ˙V + 0.87n ˙L we find that if we meet purity specifications by
operating at 61°C, n ˙V = 0.95,n ˙L = 4.05 and fRMV = 0.5(0.95)1= 0.475, well below
the desired fractional recovery of 0.90.
(b) To ensure that we meet the recovery specification, we require thatyM n ˙V =
0.90 xM n ˙L = 0.10
and that n ˙V + n ˙L
=5
˙
We resort to trial-and-error: guess yM, find xM from the phase diagram, calculate n V and
n ˙L, and see if the total mass balance equation is satisfied. Some iterations are tabulated:
T °C yM xM n ˙V n ˙L n ˙V + n ˙L
66.44 0.371 0.050 2.43 2 4.43 (too low)
70.67 0.267 0.026 3.37 3.38 7.21 (too high)
67.5 ~0.345 ~0.044 2.61 2.27 4.88 (a little low)
~67.6°C would produce the desired recovery, but the purity of the vapor would be only
~34 mol% methanol.
P5.69
(Note to instructor: Students are likely to need some guidance to make appropriate
assumptions, which are necessary to solve this problem.) From
Table B.20, we find that the boiling point temperatures are
Ethylbenzene: 136.2°C p-
xylene: 138.4°C m-
xylene: 139.1°C o-
xylene: 144.4°C
Our heuristics would argue for first separating OX, then EB, then finally PX from MX.
But we will consider several different configurations. We’ll assume that the only
impurities in each product are the compounds with adjacent volatilities. Where there are
2 impurities we simply assumed they were in equal quantities.
Configuration 1:
98 % EB
PX
EB
300 EB 95 % PX
150 PX MX
200 MX col 1
350 OX
PX
90 % MX
OX
col 2
col 3
MX
95% OX
Minimum stages are calculated from the Fenske equation, using the compositions in the
table and separation factors calculated from the ratio of the saturation pressures (1.059
for EB/PX, 1.02 for PX/MX, and 1.155 for MX/OX). Results are summarized in the
table, assuming the actual number of stages equals twice the minimum.
Nmin N Q $ (relative)
300 EB col 3
150 PX
200 MX col 1
350 OX
col 2 PX
90 % MX
OX
MX
95 % OX
Nmin N Q $ (relative)
col 3
col 2
EB
95 % PX
MX
300 EB
PX
150 PX
90 % MX
200 MX col 1
OX
350 OX
MX
95 % OX
OX 350 10 340 10
Nmin N Q $ (relative)
P5.70
We’ll use Raoult’s law to model the phase equilibrium behavior for both benzene and
toluene.
ybP=xbPbsat ytP=xtPtsat
Also
yb +yt =1 xb
+xt =1
Adding together the two Raoult’s law equations and then substituting:
ybP+ytP=xbPbsat +xtPtsat
The material balance equations for benzene and total moles are:
zbn ˙F = ybn ˙V + xbn
˙L n ˙F = n ˙V + n ˙L
Combining:
V V
˙L zb +1 = yb n ˙
+ xb
n˙
n ˙L n ˙L
n˙
(zb− yb) V = xb− zb
n ˙L
We set P = 100 atm and n ˙F =100. The minimum temperature is the bubblepoint
temperature, and the maximum is the dewpoint. We calculate saturation pressures from
Antoine’s equation
(i) zb = 0.2
T n ˙V n ˙L xb yb
absorber stripper
MEA
900 M H2S
650 N
100 H2S 1
with all flows given in kmol/h, and H2S flows calculated from the stream composition
specifications and an overall material balance.
Each unit operates as a single equilibrium stage. The gas leaving the absorber is in
equilibrium with the liquid, stream 1, leaving the absorber. The mole fraction of H2S in
the gas leaving the absorber is 0.001. The absorber pressure is 10,000 mm Hg, therefore,
pH2S =yH2SP= 0.001(10000)=10 mm Hg
From the data, at 40°C (the operating temperature of the absorber) the liquid in
equilibrium with this gas is at 0.374 moles H2S per mole MEA, or
n ˙H2S,1 = 0.374
˙
n MEA,1
˙ ˙
n MEA,1 = n MEA,2 and from the H2S
We have 4 equations in 4 unknowns that we can solve to find that the MEA flow rate is
1043 kmol/h.
If the N2 flow rate decreased while solvent flows and operating T and P remained
constant, the molar flow rate of H2S would have to decrease because yH2S would remain
constant. Not as much H2S would be removed from the MEA solution, so not as much
could be picked up in the absorber. The mole fraction of H2S in the gas leaving the
absorber would increase above its specification. (Think of what would happen if the
nitrogen stream was shut off entirely!)
P5.72.
Differences in solubility in water could be used to separate Cl2 and HCl from the decanes.
However, the decanes have similar insolubility in water, and are chemically similar so
their solubility in other solvents is also likely to be similar and therefore not a basis for
separation. Their melting points are below ambient, making crystallization unattractive.
Boiling points are all above ambient, but are not unusually high, so distillation is a
reasonable separation technology.
Configuration 1:
1 D
2 % MCD
400 D
500 MCD D
100 DCD 2
95 % MCD
col 1
DCD
col 2
D
MCD
DCD
3
MCD
DCD
The two key components in column 1 are decane and MCD, while the key components in
column 2 are MCD and DCD.
DOF analysis
Number of variables
Configuration 2:
1
D
2 % MCD
D
MCD
DCD
col 2
400 D
500 MCD
100 DCD 2 D
col 1 95 % MCD
DCD
3
MCD
DCD
Key components in column 1 are MCD and DCD, and in column 2 the key components
are decane and MCD.
Flow rates are calculated from stream composition and system performance
specifications as well as material balance equations.
P5.73
(a) If the air is insoluble and the water nonvolatile, then the single absorption stage is
sketched as
95 gmol/s air
0.5 % CO2
S
water
water absorption
CO2
L
F
95 gmol/s air
5 gmol/s CO2
where the streams are labeled solvent S, feed F, vapor V and liquid L. V and L are
x H
in equilibrium, which Henry’s law describes. yCO2P= CO2 CO2
The Henry’s law constant for CO2 at 40°C in water is 2330 atm, the operating pressure is
26 atm, and yCO2 = 0.005:
yCO P 0.005
CO 2, V
= 0.005 = n ˙COn ˙ 2,V + 95 ⇒ n ˙CO2,V = 0.48
gmol/s
gmol/s The water feed rate is 81,000 gmol/s, to treat 100 gmol/s gas!
(b) Is air insoluble? Model N2 and O2 dissolved in water using Henry’s law, assume N2
mol fraction in gas stream is same as in air.
= × −4
0.79(26) 1.97 10
=
x yN2P 2
N = HN2
104,000
A similar calculation for O2 reveals that about 8 gmol/s O2 leaves in the liquid stream –
about 40% of the O2 in the air feed stream!
The assumption that air is insoluble is not good in this case, because there is so much
water.
Is water nonvolatile? The saturation pressure of water at 40°C is calculated from Antoine
equation to be 0.073 atm. Raoult’s law is used to estimate the mole fraction of water
vapor
xW PWsat 1(0.073)
yW = = = 0.0028
P 26
Relatively insignificant.
(c) We return to the assumptions of part (a). We use Henry’s law to model the phase
behavior, and we combine that with a material balance equation on CO2 and the
stream composition specifications. n ˙CO
Given T, we find HCO2 from the data in App. B. Given T and P we use the above equation
to find nW,L at 0, 10, 20, 30, 40 and 50°C, and at 1, 10 and 100 atm. Results are plotted
(Notice the log scale). Water usage decreases with increasing pressure and decreasing
temperature. Within this range, I would choose 0°C and 100 atm.
Water usage, gmol/s
Temperature (°C)
(d)
To use the KSB equation, fractional
recovery = = 0.904
P 100
K= xCO2 = = = 0.137 at 0°C and 100 atm yCO2
H
CO2 728
(0.137n ˙
)N +1 (
0.904 = S /n ˙F ˙ S / n ˙F−) N0.137+1−n 1˙S /n ˙F)
(0.137n
The ratio of solvent to feed flow is calculated for different number of stages; results are
summarized
No. of stages 1 2 5 10
Solvent/feed 69 19.2 8.9 7.2
There is an enormous advantage to increasing from 1 to 2 stages. Increasing stages much
above 5 does not seem very helpful.
P5.74
The flow diagram is
980 H2
W
0.4 H2S
absorption
W 980 H2
20 H2S
19.6 H2S
with a basis chosen as 1000 gmol/min and flows calculated from stream composition and
system performance specifications. We’ll use Henry’s law to model the phase
H
equilibrium. At 10°C, H2S = 367 atm, therefore
(0.0545n ˙
)N +1
0.98 = S /n ˙F ˙ S / n ˙−F()0.0545N+1−1n ˙S /n ˙F)
(0.0545n
We find the solution in Excel at various values of N; the results are plotted:
Water flow/hydrogen flow
Number of stages
19.6 0.4
(367)= (20)
19.6+ m ˙ W 980.4
m ˙ W = 882,000 (versus 857,000 from KSB)
For N = 2,
980 2 H2980 H2
0.4 H2S20 H2S
stage 1 stage 2
1
W
W
19.6 H2S
The procedure continues to N = 10 in a similar manner. The assumption that the water
flow is constant simplifies the solution somewhat. Solutions from the stagewise
calculations are very similar to those obtained from the KSB equation, because the
assumptions that went into the KSB equation are good for this situation – but the
solutions are obtained with much less tedium with the KSB equation.
P5.75
(a) The exit liquid stream is in equilibrium with the entering gas stream. From
materialbalance and the system performance specification that 80% of the CO2 fed is
recovered in the aqueous stream, we calculate that the flow rate of CO2 in the aqeous
stream is 2 gmol/min. From this, the equilibrium constant is
KD= moles CO2/moles liquid = 2/(50+ 2) = 0.385 moles
CO2/moles gas 2.5/(22.5+ 2.5)
absorber
2.5 C 50 S
22.5 M 4 C
1
where C = carbon dioxide, M = methane, and S = the solvent (with additives). We’ve
assumed that none of the methane dissolves in the solvent and that none of the solvent is
vaporized. All flows are in gmol/min.
The % recovery is
2.208
×100% = 88.3%
2.5
25 S 25 S
splitter
Gas out
22.5 M
C
5
Gas out
absorber 2
absorber 1 22.5 M
C 3
2.5 C 25 S 25 S
22.5 M C C
4
1 2
The exit liquid from absorber 1 (stream 2) is in equilibrium with the entering gas stream
(stream 1)
C2
2.5 n˙
xC2 = 0.385yC1 = 0.385 = 0.0385 =
2.5+ 22.5 n ˙C2 + 25
n ˙C2 =1.0 gmol/min
Frannie’s idea is much better; Zooey’s idea is actually worse than the base case!
P5.76
The proposed block flow diagram is shown, along with the flow rates (in kgmol/h)
calculated from the specifications.
3.75 CO2 2
21.25 CO2
75 N2 MEA
CO2
absorber stripper
MEA
25 CO2 CO2
75 N2 1
The absorber should be operated at highest allowable P – 20 atm - and lowest allowable
T – 40°C since that’s the data available. The gas leaving the absorber is in equilibrium
with the liquid leaving the absorber (stream 1).
From this, we estimate that 724 mm Hg CO2 is in equilibrium with 0.698 moles
CO2/mole MEA at the absorber conditions, or
˙
n CO2,1
= 0.698.
˙
n MEA,1
The stripper should be operated at the lowest possible pressure (1 atm) and the highest
possible temperature (100°C). At these conditions, pCO2 =yCO2P= 760 mm Hg
P5.77
The flow diagram is
B
S
Raffinate
10000 kg/h R
BA
BA solvent extraction
B
W F
W
Extract
B
E BA
with components labeled as B (bezene), W (water) and BA (benzoic acid), and streams
labeled as feed F, solvent S, raffinate R and extract E.
m ˙ BAF
= 0.02 and m ˙ WF+ m ˙ BAF
=10000 m ˙ WF
so m ˙ WF = 9804 kg/h and m ˙ BAF =196 kg/h
We know that all the water exits in the raffinate, therefore from a mass balance
m ˙ WR= m ˙ WF = 9804 kg/h
Benzoic acid concentrations in the raffinate and in the extract are related by the
equilibrium distribution coefficient
P5.78
We’ll use the sour gas composition listed on p. 452 and the MEA data given on p. 456. If
there is some CO2 in the sour gas, we would need to consider its solubility in MEA,
which complicates this problem.) Some water will evaporate and leave in the sour gas
stream. We’ll assume that the hydrogen, as a nonpolar compound, has negligible
solubility in the MEA solution. (A 15.3 wt% solution of MEA is a 5 mol% solution.)
We’ll also assume that none of the MEA evaporates, and that any small change in the
MEA concentration in the liquid does not affect its solubility behavior. We’ll use 40°C
and 20 atm, as in the case study.
We’ll use A for MEA, W for water, H for hydrogen, and S for hydrogen sulfide.
sweet gas
sour gas
H2
98 H2
1 3 5
stage1 stage2
2 4 6
0.04 mol% H2S
2 H2S
H2O sour solventSolvent
MEA5% MEA
H2O H2O
H2S
The material balance equations are straightforward. We’ll use 100 kgmol/h sour gas as
the basis.
Stage 1
2+ n ˙S4 = n ˙S2+ n ˙S3 (hydrogen sulfide)
98 = n ˙H3 (hydrogen)
n ˙A4 = n ˙A2 (MEA)
n ˙W4 = n ˙W2+ n ˙W3 (water)
Stage 2
n ˙S3= n ˙S4 + n (hydrogen sulfide)
˙S5
n ˙H3= n ˙H5 (hydrogen)
n ˙A6 = n ˙A4 (MEA)
n ˙W3+ n ˙W6 = n ˙W4 + n ˙W5 (water)
n ˙S5 = 0.0004
n ˙S5+ n ˙H5+ n ˙W5
n ˙A6 =
0.05 n ˙A6+ n ˙W6
For water, we use Raoult’s law (with the saturation pressure calculated from Antoine’s
equation to be 0.0728 atm at 40°C and P = 20 atm).
x
yW 3= W 2PWsat or n ˙W 3 = 0.0036n ˙W 2
xW 4PWsat n˙
yW 5= or W5 = 0.0036n ˙W 4
P n ˙S5+ n ˙H5+ n ˙W 5 n ˙W 4 + n ˙A4 + n ˙S4
For H2S dissolved in 5 mol% MEA, we have phase equilibrium data but no simple
equation. We have 2 choices: one is to guess a partial pressure, find the solubility from
the data, and see if the material balance equations are satisfied. If not, we make a new
guess and repeat. An alternative is to fit an empirical equation to the data and use that
fitted equation. The equation is simply a mathematical tool for correlating data and does
not describe any underlying theory, nor should it be used outside of the range of the data.
Let’s plot the data, and see if we can find a reasonable equation. (We did this in
Kaleidagraph, see App. A for information on this graphing software. Just for fun we did
this for both H2S and CO2. ) After a few false starts we came up with 2 empirical
equations:
20 3
(
pH2S =yH2SP=804 XH2S +176 XH2S ) ( )
9.6
Now we can describe the phase equilibrium for our particular case using these empirical
equations
Now we are finally ready to proceed. (It wouldn’t be a bad idea to check a DOF analysis
first, just to make sure the problem is solveable!) Solution requires the use of an
equation-solving program like EES, or a spreadsheet like EXCEL with a SOLVER.
From the solution, we find that the MEA required drops to only 2.35 kgmol/h! (Total
solution flow is 47 kgmol/h).
Liquid from stage 2: 0.76 kgmol/h H2S, 2.35 kgmol/h MEA, 44.6 kgmol/h H2O
Liquid from stage 1: 1.96 kgmol/h H2S, 2.35 kgmol/h MEA, 44.3 kgmol/h H2O
P5.79
(a) Given 95% recovery of penicillin in the extract, the flow diagram for a single-
stageextraction is
9500 g penicillin
amyl acetate
5000 L broth F
500 g penicillin
2 g/L Penicillin mixer/settler
5000 L broth
R
Amyl acetate
where we indicate streams as feed F, solvent S, extract E and raffinate R. We also assume
that the quantity of penicillin removed is small enough that the total volume of broth
doesn’t change. The concentration of penicillin P in the raffinate x is
500 g
x= = 0.1g/L
5000 L
This is in equilibrium with the penicillin concentration y in the extract phase;
y=KDx= 9(0.1)= 0.9g/L
(b) With a 7000 Liter tank and 5000 L broth, we can add at most 2000 L solvent
eachstep. In the first extraction,
y g/L penicillin
amyl acetate
5000 L broth F
x g/L penicillin
2 g/L Penicillin mixer/settler
5000 L broth
R
2000 L
Amyl acetate
The material balance equation for penicillin is
5000x+ 2000y= 2(5000)=10000 g
This is coupled to the phase equilibrium equation; y=9x
We solve to find x = 0.435 g/L, and y = 3.91 g/L. The percent penicillin recovery in the
solvent phase is
3.91
×100% = 78.2%
If we remove the penicillin-loaded solvent, and then add back 2000 L fresh solvent, we
can recover more penicillin. There is less penicillin in the system, and the material
balance equation is
5000x+ 2000y= 0.435(5000)= 2175 g
The phase equilibrium equation remains y=9x
We solve to find x = 0.0948 g/L, and y = 0.853 g/L. The percent penicillin recovery in the
solvent phase in this step is
0.853
×100% = 78%
3.91
×100% = 95.3%
In two steps, we meet our performance objective, and we use only 4000 L total solvent
(vs. 10,555 L for a single-step extraction). Furthermore, the penicillin concentration in
the solvent is higher in the two-step process than in the single-step process, making
further purification less costly.
It makes no sense at all to drain the broth and leave the solvent in the tank – the broth
after the first step of equilibration still has enough penicillin in it to warrant a second
contact with fresh solvent.
P5.80
(a)
For single-stage contacting, using the flows and compositions given, with all flows in
kg/h:
288 W
12 A
ketone -
rich
200 A phas e
single 300 K
300 K stage A
water -
rich
phase
288 W
A
The material balance plus phase equilibrium relationship allows us to calculate the
quantity of alcohol A in each of the phases:
200+12 = m ˙ AI + m ˙ AII
xAI m ˙ AI (m ˙ AI + 300)
The solution is m ˙ AI =
32 kg/h m ˙ AII
=180 kg/h
The fractional removal of alcohol from the ketone-rich phase is
fRAII = = 0.84
(b)
Crosscurrent with 3 stages, with equal distribution of wash water among the 3 stages:
96 W 96 W 96 W
4 A 4 A 4 A
A A A
200 A 300 K 300 K 300 K
stage 1 stage 2 stage 3
300 K 2 4 6
1 5
3
96 W 96 W 96 W
A A A
We write 1 material balance equation and 1 phase relationship for each stage:
Stage 1:
200+ 4 = m ˙ A2+ m ˙ A1
KD = 4 = xAII = m ˙ A
1
(m ˙ A1+ 96)
xAI m ˙ A2 (m ˙ A2+ 300)
Stage 2:
m ˙ A2+ 4 = m ˙ A4 + m ˙ A3
KD = 4 = xAII = m ˙ A
3
(m ˙ A3+ 96)
xAI m ˙ A4 (m ˙ A4 + 300)
Stage 3:
m ˙ A4+ 4 = m ˙ A6+ m ˙ A5
fRAII = = 0.95
(c)
300 K 300 K 300 K
200 A 2 A 4 A 6
300 KA
1 stage 1 3 stage 2 5 stage 3
288 W288 W
288 288 12 A
A
W W
A A
We write one material balance equation and one phase relationship for each stage:
Stage 1:
200+ m ˙ A3= m ˙ A2+ m ˙ A1
xAI m ˙ A4 (m ˙ A4 + 300)
Stage 3:
m ˙ A4 +12 = m ˙ A6+ m ˙ A5
KD = 4 = xAII = m ˙ A
5
(m ˙ A5+ 96)
xAI m ˙ A6 (m ˙ A6+ 300)
207.6 m ˙ A2
= 35 m ˙
A3= 42.7 m
˙ A4 =10 m ˙
A5=17.6 m˙
A6= 4.4
fRAII = = 0.978
P5.81
Ima’s flow diagram must look like this:
Broth
picro
water
Solvent Extract
Elixir crystallizer/filte Solids
mixer settler Picro
r Pure Picro
Elixir
Raffinate Filtrate
Picro Elixir
Water Picro
There are 3 components: Picro P, water (and other water-soluble components) W, and
Elixir E.
(a) Start with a balance on the mixer-settler. Assuming no change in volume of the
brothdue to the loss of Picro, the Picro mass balance is
0.001053(10000)= 0.00001053(10000)− m ˙
P,ex m ˙ P,ex =10.425 g/day
where “ex” indicates the extract stream. From the distribution coefficient
KD = 45000 = m ˙ P,ex m ˙ ex 10.425 m ˙ ex
=
m ˙ P,raff m ˙ raff 0.00001053
m ˙ ex = 22 g/day
which is the total flow of both Picro and Elixir, Therefore, the Elixir flow rate must be
11.575 g/day.
(b) Now we need to estimate the liquid-solid phase equilibrium graph. We plot the
twopure melting point temperatures and the eutectic temperature, and connect the dots
with smooth curves. The dashed lines indicate the temperature of the crystallization unit
and the mass fraction Picro of the feed to the crystallizer, (10.425/22 or 0.474).
Operation at 10°C produces a pure Picro solid plus a saturated solution of about 32 wt%
Picro.
80
60
Temperature (°C)
40
20
The overall fractional recovery of Picro as crystals from the fermentation broth is
fRP= = 0.475
(c) The mixer-settler unit is very effective at extracting Picro, recovering 99% of
thematerial in the broth. There is not much opportunity for improving this step. Most of
the losses appear in the filtrate. Purchasing a new crystallizer that could operate at lower
temperatures would improve this a bit, but we are limited to a minimum wt% of >20%
Picro in the filtrate. Better would be an alternative method to partially remove some of
the Elixir from the filtrate (perhaps by evaporation?) and then recycle the concentrated
filtrate back to the crystallizer.