1. CHAPTER FOUR
4.1 a. Continuous, Transient
b. Input – Output = Accumulation
No reactions ⇒ Generation = 0, Consumption = 0
kg kg dn dn kg
6.00 − 3.00 = ⇒ = 3.00
s s dt dt s
c. 100 m3 1000 kg 1 s
.
t= = 333 s
1 m3 3.00 kg
4.2 a. Continuous, Steady State
b. k = 0 ⇒ C A = C A0 k = ∞ ⇒ CA = 0
c. Input – Output – Consumption = 0
Steady state ⇒ Accumulation = 0
A is a reactant ⇒ Generation = 0
FG m IJ C FG mol IJ = V FG m IJ C FG mol IJ + kVC FG mol IJ ⇒ C
3 3
CA0
V
H sK Hm K H sK Hm K
A0 3 A 3
HsK
A A =
1+
kV
V
4.3 a. b
mv kg / h g
100 kg / h 0.850 kg B / kg
0.550 kg B / kg Input – Output = 0
0.150 kg T / kg
Steady state ⇒ Accumulation = 0
0.450 kg T / kg
b
ml kg / h g No reaction ⇒ Generation = 0, Consumption = 0
0.106 kg B / kg
0.894 kg T / kg
(1) Total Mass Balance: 100.0 kg / h = mv + ml
(2) Benzene Balance: 0.550 × 100.0 kg B / h = 0.850mv + 0106ml
.
Solve (1) & (2) simultaneously ⇒ mv = 59.7 kg h, ml = 40.3 kg h
b. The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by
masses (kg). The balance equations are also identical (initial input = final output).
c. Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state,
the feed composition is incorrect, the flow rates are not what they are supposed to be, other
species are in the feed stream, measurement errors.
4- 1

2. 4.4 b. n (mol) b
0.500 n mol N 2 28 g N 2 1 kg g
= 0.014 n kg N 2 b g
0.500 mol N 2 mol mol N 2 1000 g
0.500 mol CH 4 mol
c. 100.0 g / s
nE =
b g
100 x E g C 2 H 6 1 lb m lb - mole C2 H 6 3600 s
b g
x E g C2 H 6 g s 453593 g 30 lb m C2 H 6
. h
xP bg C H gg
3 8 b
= 26.45x E lb - mole C 2 H 6 / h g
xB bg C H gg
4 10
d. b g
n1 lb - mole H 2 O s nO2 = 0.21n2 ( lb-mole O 2 / s )
Rn blb - mole DA sg
| 0.21 lb - moleO lb - mole DA U
2
| n1 ⎛ lb-mole H 2 O ⎞
S
| 0.79 lb - mole N lb - mole DAV
2
|
xH2O =
n1 + n2 ⎜ lb-mole ⎟
⎝ ⎠
T 2 W 0.21n2 ⎛ lb-mole O 2 ⎞
xO2 =
n1 + n2 ⎜ lb-mole ⎟
⎝ ⎠
e. n ( mol )
nN2O4 = n ⎡0.600 − yNO2 ⎤ ( mol N 2 O4 )
⎣ ⎦
0.400 mol NO mol
yNO2 ( mol NO 2 mol )
0.600 − yNO2 ( mol N 2 O 4 mol )
4.5 a. Basis: 1000 lbm C3H8 / h fresh feed
(Could also take 1 h operation as basis -
flow chart would be as below except
that all / h would be deleted.) b
n6 lb m / h g
0.02 lb m C3H8 / lb m
1000 lb m C3H 8 / h b
n7 lb m / h g 0.98 lb m C3H 6 / lb m
0.97 lb m C3H8 / lb m
Still
0.03 lb m C3H 6 / lb m
Compressor
b
n1 lb m C3H 8 / h g b
n1 lb m C3H 8 / h g
Reactor n blb
2 m C H / hg
3 6 b
n2 lb m C3H 6 / h g
n blb CH / h g
3
n blb
m
H / hg
4
b
n3 lb m CH 4 / h g
4 m 2
b
n4 lb m H 2 / h g Stripper
b
n5 lb m / h g
Note: the compressor and the off gas from Absorber
the absorber are not mentioned explicitly
in the process description, but their presence
should be inferred. b
n1 lb m C3H 8 / h g
b
n2 lb m C3H 6 / h g
n blb
5 m oil / h g
4- 2

3. 4.5 (cont’d)
b. Overall objective: To produce C3H6 from C3H8.
Preheater function: Raise temperature of the reactants to raise the reaction rate.
Reactor function: Convert C3H8 to C3H6.
Absorption tower function: Separate the C3H8 and C3H6 in the reactor effluent from the other
components.
Stripping tower function: Recover the C3H8 and C3H6 from the solvent.
Distillation column function: Separate the C3H5 from the C3H8.
4.6 a. 3 independent balances (one for each species)
b. 7 unknowns ( m1 , m3 , m5 , x2 , y2 , y4 , z4 )
– 3 balances
– 2 mole fraction summations
2 unknowns must be specified
c. y2 = 1 − x2
FG kg A IJ = m + b1200gb0.70g FG kg A IJ
A Balance: 5300 x2
H hK 3
H hK
F kg I
Overall Balance: m + 5300 G J = m + 1200 + m G J
F kg I
1
H hK 3
H hK 5
B Balance: 0.03m + 5300 x G
F kg BIJ = 1200 y + 0.60m FG kg BIJ
1
H hK 2
H hK 4 5
z4 = 1 − 0.70 − y4
4.7 a. 3 independent balances (one for each species)
b. bg
400 g 0.885 g H 2O mR g 0.995 g H 2O
= ⇒ mR = 356 g min
Water Balance:
min g min gb g
Acetic Acid Balance: b400gb0115g FGH g CH OOH IJK = 0.005m
.
min
3
R + 0.096mE
FG g CH OOH IJ
H min K
3
⇒ mE = 461g min
FG g IJ = m + m FG g IJ ⇒ m = 417 g min
Overall Balance: mC + 400
H min K H min KR E C
c.
b0115gb400g − b0.005gb356g FGH min IJK = b0.096gb461g FGH min IJK ⇒ 44 g min = 44 g min
.
g g
4- 3

4. 4.7 (cont’d)
d.
H 2O CH 3COOH
some CH3COOH
CH3COOH
H 2O Extractor C4 H9OH Distillation
C 4 H9OH CH3COOH Column
C4 H 9OH
4.8 a.
X-large: 25 broken eggs/min
35
45 unbroken eggs/min
120 eggs/min
0.30 broken egg/egg
Large: n 1 broken eggs/min
0.70 unbroken egg/egg
n 2 unbroken eggs/min
b. 120 = 25 + 45 + n1 + n2 ( eggs min ) ⇒ n1 + n2 = 50 ⎫ n1 = 11
⎪
⎬ ⇒
( 0.30 )(120 ) = 25 + n1 ⎪ n2 = 39
⎭
c. n1 + n2 = 50 large eggs min
b
n1 large eggs broken/50 large eggs = 11 50 = 0.22 g
d. b g
22% of the large eggs (right hand) and 25 70 ⇒ 36% of the extra-large eggs (left hand)
are broken. Since it does not require much strength to break an egg, the left hand is probably
poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed.
4.9 a. b
m1 lb m strawberries g b
m3 lb m W evaporated g
015 lb m S / lb m
.
0.85 lb m W / lb m
1.00 lb m jam
c
m2 lb m S sugar h 0.667 lb m S / lb m
0.333 lb m W / lb m
b. 3 unknowns ( m1 , m2 , m3 )
– 2 balances
– 1 feed ratio
0 DF
c. Feed ratio: m1 / m 2 = 45 / 55 (1)
S balance: 0.15 m1 + m 2 = 0.667 (2)
Solve simultaneously ⇒ m1 = 0.49 lb m strawberries, m 2 = 0.59 lb m sugar
4- 4

5. 4.10 a.
300 gal
b g
m1 lb m
4 unknowns ( m1 , m2 ,V40 , m3 )
0.750 lb m C 2 H 5OH / lb m
0.250 lb m H 2O / lb m
b g
m3 lb m
– 2 balances
0.600 lb m C 2 H 5OH / lb m – 2 specific gravities
0.400 lb m H 2O / lb m 0 DF
b g
V40 gal
m b lb g
2 m
0.400 lb m C 2 H 5OH / lb m
0.600 lb m H 2 O / lb m
b. 300 gal 1ft 3 0.877 × 62.4 lb m
m1 = = 2195 lb m
7.4805 gal ft 3
Overall balance: m1 + m2 = m3 (1)
C2H5OH balance: 0.750m1 + 0.400m2 = 0.600m3 (2)
Solve (1) & (2) simultaneously ⇒ m2 = 1646 lb m, , m3 = 3841 lb m
1646 lb m ft 3 7.4805 gal
V40 = = 207 gal
0.952 × 62.4lb m 1ft 3
4.11 a. 3 unknowns ( n1 , n2 , n3 )
b
n1 mol / s g – 2 balances
0.0403 mol C3H 8 / mol 1 DF
0.9597 mol air / mol b
n3 mol / s g
0.0205 mol C3H 8 / mol
b
n2 mol air / s g 0.9795 mol air / mol
0.21 mol O 2 / mol
0.79 mol N 2 / mol
b. Propane feed rate: 0.0403n1 = 150 ⇒ n1 = 3722 mol / s b g
Propane balance: 0.0403n1 = 0.0205n3 ⇒ n3 = 7317 mol / s b g
Overall balance: 3722 + n2 = 7317 ⇒ n2 = 3600 mol / s b g
c. > . The dilution rate should be greater than the value calculated to ensure that ignition is not
possible even if the fuel feed rate increases slightly.
4- 5

6. 4.12 a.
b
m kg / h g
1000 kg / h 0.960 kg CH3OH / kg
2 unknowns ( m, x )
0.500 kg CH 3OH / kg 0.040 kg H 2O / kg – 2 balances
0.500 kg H 2O / kg 0 DF
673 kg / h
b
x kg CH 3OH / kg g
1 − x b kg H O / kgg
2
b. Overall balance: 1000 = m + 673 ⇒ m = 327 kg / h
b g b g b g
Methanol balance: 0.500 1000 = 0.960 327 + x 673 ⇒ x = 0.276 kg CH 3OH / kg
Molar flow rates of methanol and water:
673 kg 0.276 kg CH3OH 1000 g mol CH3OH
= 5.80 × 103 mol CH3OH / h
h kg kg 32.0 g CH3OH
673 kg 0.724 kg H 2O 1000 g mol H 2O
= 2.71 × 104 mol H 2O / h
h kg kg 18 g H 2O
Mole fraction of Methanol:
5.80 × 103
= 0176 mol CH 3OH / mol
.
5.80 × 103 + 2.71 × 104
c. Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the
system is not at steady state.
4.13 a. Product
Reactor effluent 1239 kg
Feed Reactor Purifier R = 583
2253 kg 2253 kg W aste
R = 388
b g
m w kg
R = 140
Analyzer Calibration Data
1
1.364546
x p = 0.000145R
xp 0.1
0.01
100 R 1000
4- 6

7. 4.13 (cont’d)
b g
b. Effluent: x = 0.000145 388 1.3645 = 0.494 kg P / kg
p
Product: x = 0.000145b583g
1.3645
p = 0.861 kg P / kg
Waste: x = 0.000145b140g
1.3645
p = 0123 kg P / kg
.
0.861b1239g
Efficiency = × 100% = 95.8%
0.494b2253g
c. Mass balance on purifier: 2253 = 1239 + mw ⇒ mw = 1014 kg
P balance on purifier:
b gb g
Input: 0.494 kg P / kg 2253 kg = 1113 kg P
Output: b0.861 kg P / kggb1239 kgg + b0123 kg P / kggb1014 kgg = 1192 kg P
.
The P balance does not close . Analyzer readings are wrong; impure feed; extrapolation
beyond analyzer calibration data is risky -- recalibrate; get data for R > 583; not at steady
state; additional reaction occurs in purifier; normal data scatter.
4.14 a.
b
n1 lb- mole/ h g
.
00100 lb- mole H2O/ lb- mole
.
09900 lb- mole DA/ lb- mole
b g
n3 lb- mole/ h
.
0100 lb- mole H2O/ lb- mole
b
n2 lb- mole HO/ h
2 g .
0900 lb- mole DA/ lb- mole
v dft / hi
2
3
4 unknowns ( n1 , n2 , n3 , v ) – 2 balances – 1 density – 1 meter reading = 0 DF
Assume linear relationship: v = aR + b
v2 − v1 96.9 − 40.0
Slope: a = = = 1.626
R2 − R1 50 − 15
Intercept: b = v a − aR1 = 40.0 − 1.626 15 = 15.61 b g
b g
v2 = 1.626 95 + 15.61 = 170 ft / h c 3
h
n2 =
170 ft 3 62 .4 lb m lb - mol
h ft 3 18.0 lb m
b
= 589 lb - moles H 2 O / h g
DA balance: 0.9900n1 = 0.900n3 (1)
Overall balance: n1 + n2 = n3 (2)
Solve (1) & (2) simultaneously ⇒ n1 = 5890 lb - moles / h, n3 = 6480 lb - moles / h
b. Bad calibration data, not at steady state, leaks, 7% value is wrong, v − R relationship is not
linear, extrapolation of analyzer correlation leads to error.
4- 7

8. 4.15 a.
b
m kg / s g
100 kg / s 0.900 kg E / kg
0.600 kg E / kg 0100 kg H 2 O / kg
.
0.050 kg S / kg 3 unknowns ( m, xE , xS )
0.350 kg H 2 O / kg – 3 balances
b
m kg / s g 0 DF
b
x E kg E / kg g
x b kg S / kg g
S
1 − x − x b kg H O / kg g
E S 2
b. Overall balance: 100 = 2m ⇒ m = 50.0 kg / s b g
b g b g
S balance: 0.050 100 = xS 50 ⇒ xS = 0100 b kgS / kgg
.
b g b g b g
E balance: 0.600 100 = 0.900 50 + x E 50 ⇒ x E = 0.300 kg E / kg
kg E in bottom stream 0.300b50g kg E in bottom stream
= = 0.25
kg E in feed 0.600b100g kg E in feed
c. bg bg bg
x = aRb ⇒ ln x = ln a + b ln R
lnb x / x g lnb0.400 / 0100g .
b= 2 1
= = 1491
lnb R / R g lnb38 / 15g
.
2 1
lnbag = lnb x g − b lnb R g = lnb0100g − 1491lnb15g = −6.340 ⇒ a = 1764 × 10
1 1 . . . −3
x = 1764 × 10−3 R1.491
.
F x I F 0.900 IJ
1 1
R=G J =G
b 1.491
H a K H 1764 × 10 K
. −3
= 655
.
d. Device not calibrated – recalibrate. Calibration curve deviates from linearity at high mass
fractions – measure against known standard. Impurities in the stream – analyze a sample.
Mixture is not all liquid – check sample. Calibration data are temperature dependent – check
calibration at various temperatures. System is not at steady state – take more measurements.
Scatter in data – take more measurements.
4- 8

9. 4.16 a. 4.00 mol H 2SO 4 0.098 kg H 2SO 4 L of solution
L of solution mol H 2SO 4 1213 kg solution
.
b
= 0.323 kg H 2SO 4 / kg solution g
b.
v1 L bg 5 unknowns ( v1 , v2 , v3 , m2 , m3 )
– 2 balances
100 kg
bg
v3 L – 3 specific gravities
m b kg g
0.200 kg H 2SO 4 / kg 0 DF
3
0.800 kg H 2 O / kg
0.323 kg H 2SO 4 / kg
SG = 1139
.
0.677 kg H 2 O / kg
bg
v2 L SG = 1.213
m b kg g
2
0.600 kg H 2SO 4 / kg
0.400 kg H 2 O / kg
SG = 1.498
Overall mass balance: 100 + m2 = m3 U ⇒ m = 44.4 kg
V m = 144 kg 2
Water balance: 0.800b100g + 0.400m = 0.677m W
2 3 3
100 kg L
v1 = = 87.80 L20%solution
1139 kg
.
44.4 kg L
v2 = = 29.64 L 60% solution
1498 kg
.
v1 87.80 L 20%solution
= = 2.96
v2 29.64 L 60% solution
c. 1250 kg P 44.4 kg 60%solution L
= 257 L / h
h 144 kg P 1498 kgsolution
.
4.17 b g
m1 kg @$18 / kg
0.25 kg P / kg
0.75 kg H2O / kg 100 kg
.
b g
m2 kg @$10 / kg 017 kg P/ kg
.
012 kg P / kg
. 0.83 kg H2O / kg
0.88 kg H 2O / kg
Overall balance: m1 + m2 = 100
. (1)
Pigment balance: 0.25m1 + 012m2 = 017 100
. . . b g (2)
Solve (1) and (2) simultaneously ⇒ m = 0.385 kg 25% paint, m = 0.615 kg12% paint
Cost of blend: 0.385b$18.00g + 0.615b$10.00g = $13.08 per kg
1 2
Selling price: 110b$13.08g = $14.39 per kg
.
4- 9

10. 4.18 a. b gb
m1 kg H 2O 85% of entering water g
100 kg
0.800 kgS / kg
0.200 kg H 2O / kg
b g
m2 kgS
m b kg H Og
3 2
gb g b
85% drying: m1 = 0.850 0.200 100 = 17.0 kg H 2O
Sugar balance: m = 0.800b100g = 80.0 kgS
2
Overall balance: 100 = 17 + 80 + m3 ⇒ m3 = 3 kg H 2O
3 kg H 2O
xw = = 0.0361 kg H 2O / kg
b
3 + 80 kg g
m1 17 kg H 2O
= = 0.205 kg H 2O / kg wet sugar
m2 + m3 80 + 3 kg b g
b. 1000 tons wet sugar 3 tons H 2 O
= 30 tons H 2 O / day
day 100 tons wet sugar
1000 tons WS 0.800 tons DS 2000 lb m $0.15 365 days
= $8.8 × 107 per year
day ton WS ton lb m year
c.
xw =
1
10
b g
x w1 + x w 2 +...+ x w10 = 0.0504 kg H 2 O / kg
SD =
1
9
b 2
g
x w1 − x w +...+ x w10 − x w b
2
g
= 0.00181 kg H 2 O / kg
Endpoints = 0.0504 ± 3 0.00181 b g
Lower limit = 0.0450, Upper limit = 0.0558
d. The evaporator is probably not working according to design specifications since
x w = 0.0361 < 0.0450 .
4.19 a. c h
v1 m 3
m b kg H O g 5 unknowns ( v1 , v2 , v3 , m1 , m3 )
d i
1 2
SG = 1.00 v3 m 3
– 1 mass balance
m b kg suspension g
3 – 1 volume balance
v2 m d i 3 SG = 1.48 – 3 specific gravities
0 DF
400 kg galena
SG = 7.44
Total mass balance: m1 + 400 = m3 (1)
4- 10

11. 4.19 (cont’d)
Assume volume additivity:
b g
m1 kg m3
+
400 kg m 3 m kg
= 3
m3
(2)
b g
1000 kg 7440 kg 1480 kg
Solve (1) and (2) simultaneously ⇒ m1 = 668 kg H 2O, m3 = 1068 kg suspension
668 kg m3
v1 = = 0.668 m 3 water fed to tank
1000 kg
b. Specific gravity of coal < 1.48 < Specific gravity of slate
c. The suspension begins to settle. Stir the suspension. 1.00 < Specific gravity of coal < 1.48
4.20 a.
b
n1 mol / h g b
n2 mol / h g
0.040 mol H 2O / mol b
x mol H 2O / mol g
0.960 mol DA / mol b
1 − x mol DA / mol g
b
n3 mol H 2O adsorbed / h g
97% of H 2O in feed
Adsorption rate: n3 =
b354 − 3.40g kg
. mol H 2O
= 1556 mol H 2O / h
.
5h 0.0180 kg H 2O
b g
97% adsorbed: 156 = 0.97 0.04n1 ⇒ n1 = 401 mol / h
. .
Total mole balance: n1 = n2 + n3 ⇒ n2 = 401 − 1556 = 38.54 mol / h
. .
Water balance: 0.040 ( 40.1) = 1.556 + x ( 38.54 ) ⇒ x = 1.2 × 10−3 ( mol H 2 O/mol )
b. The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction
will reach that of the inlet stream, i.e. 4%.
4.21 a. 300 lb m / h
0.55 lb m H 2SO 4 / lb m
0.45 lb m H 2O / lb m
b
mC lb m / h g
b
mB lb m / h g 0.75 lb m H 2SO 4 / lb m
0.90 lb m H 2SO 4 / lb m 0.25 lb m H 2O / lb m
010 lb m H 2O / lb m
.
Overall balance: 300 + mB = mC (1)
b g
H2SO4 balance: 0.55 300 + 0.90mB = 0.75mC (2)
Solve (1) and (2) simultaneously ⇒ mB = 400 lb m / h, mC = 700 lb m / h
4- 11

12. 4.21 (cont’d)
500 − 150
b.
mA − 150 =
70 − 25
b g
RA − 25 ⇒ mA = 7.78 RA − 44.4
800 − 200
mB − 200 =
60 − 20
b g
RB − 20 ⇒ mB = 15.0 RB − 100
ln 100 − ln 20
ln x − ln 20 =
10 − 4
b g
Rx − 4 ⇒ ln x = 0.2682 Rx + 1923 ⇒ x = 6.841e 0.2682 Rx
.
300 + 44.4 400 + 100
mA = 300 ⇒ RA = = 44.3, mB = 400 ⇒ RB = = 333,
.
7.78 15.0
1 FG
55 IJ
x = 55% ⇒ Rx =
0.268
ln
H
6.841
= 7.78
K
c. Overall balance: mA + mB = mC
H2SO4 balance: 0.01xmA + 0.90mB = 0.75mC = 0.75 mA + mB ⇒ mB = b g b0.75 − 0.01xgm A
015
.
⇒ 15.0 RB − 100 =
d
0.75 − 0.01 6.841e 0.2682 Rx
i b7.78 R A − 44.4 g
015
.
d
⇒ RB = 2.59 − 0.236e 0.2682 Rx
iR A + 135e0.2682 Rx − 813
. .
b g b g − 813 = 333
Check: RA = 44.3, Rx = 7.78 ⇒ RB = 2.59 − 0.236e e 0.2682 7.78
j44.3 + 135e
.
0.2682 7.78
. .
4.22 a.
b
n A kmol / h g
010 kmol H 2 / kmol
. 100 kg / h
0.90 kmol N 2 / kmol b
nP kmol / h g
0.20 kmol H 2 / kmol
b
nB kmol / h g 0.80 kmol N 2 / kmol
0.50 kmol H 2 / kmol
0.50 kmol N 2 / kmol
b g b
MW = 0.20 2.016 + 0.80 28.012 = 22.813 kg / kmol g
100 kg kmol
⇒ nP = = 4.38 kmol / h
h 22.813 kg
Overall balance: n A + nB = 4.38 (1)
H2 balance: 010n A + 0.50 nB = 0.20 4.38
. b g (2)
Solve (1) and (2) simultaneously ⇒ n A = 3.29 kmol / h, n B = 110 kmol / h
.
4- 12

13. 4.22 (cont’d)
b. mP
nP =
22.813
mP
Overall balance: n A + nB =
22.813
x m
H2 balance: x A n A + x B nB = P P
22.813
⇒ nA =
b
mP x B − x P g nB =
b
mP x P − x A g
b
22.813 x B − x A g b
22.813 x B − x A g
c. Trial XA XB XP mP nA nB
1 0.10 0.50 0.10 100 4.38 0.00
2 0.10 0.50 0.20 100 3.29 1.10
3 0.10 0.50 0.30 100 2.19 2.19
4 0.10 0.50 0.40 100 1.10 3.29
5 0.10 0.50 0.50 100 0.00 4.38
6 0.10 0.50 0.60 100 -1.10 5.48
7 0.10 0.50 0.10 250 10.96 0.00
8 0.10 0.50 0.20 250 8.22 2.74
9 0.10 0.50 0.30 250 5.48 5.48
10 0.10 0.50 0.40 250 2.74 8.22
11 0.10 0.50 0.50 250 0.00 10.96
12 0.10 0.50 0.60 250 -2.74 13.70
The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot
blend a 10% H2 mixture with a 50% H2 mixture and obtain a 60% H2 mixture.
d. Results are the same as in part c.
4.23 Venous blood
Arterial blood
195.0 ml / min
200.0 ml / min
1.75 mg urea / ml
1.90 mg urea / ml
Dialysate
Dialyzing fluid
b
v ml / min g
c b mg urea / mlg
1500 ml / min
a. Water removal rate: 200.0 − 195.0 = 5.0 ml / min
. b . g b g
Urea removal rate: 190 200.0 − 175 195.0 = 38.8 mg urea / min
b. v = 1500 + 5.0 = 1505ml / min
38.8 mg urea/min
c= = 0.0258 mg urea/ml
1505 ml/min
c. b2.7 − 11g mg removed
. 1 min 10 3 ml 5.0 L
= 206 min (3.4 h)
ml 38.8 mg removed 1L
4- 13

14. 4.24 a. b
n1 kmol / min g
20.0 kg CO 2 / min
b
n3 kmol / min g
b
n2 kmol / min g 0.023 kmol CO 2 / kmol
0.015 kmol CO 2 / kmol
20.0 kg CO 2 kmol
n1 = = 0.455 kmol CO 2 / min
min 44.0 kg CO 2
Overall balance: 0.455 + n2 = n3 (1)
CO2 balance: 0.455 + 0.015n2 = 0.023n3 (2)
Solve (1) and (2) simultaneously ⇒ n2 = 55.6 kmol / min, n3 = 561 kmol / min
.
b. 150 m
u= = 8.33 m / s
18 s
1 561 kmol
. m3 1 min s
A = πD 2 = ⇒ D = 108 m
.
4 min 0123 kmol 60 s 8.33 m
.
4.25 b
Spectrophotometer calibration: C = kA ====> C μg / L = 3.333 A
A = 0.9
g
C =3
. . . b gb g
Dye concentration: A = 018 ⇒ C = 3333 018 = 0.600 μg / L
0.60 cm 3 1L 5.0 mg 103 μ g
Dye injected = = 3.0 μ g
103 cm 3 1L 1 mg
b g bg
⇒ 3.0 μ g V L = 0.600 μ g / L ⇒ V = 5.0 L
4.26 a. 1000 L B / min b
n3 kmol / ming
b
m2 kg B / min g y b kmol SO / kmolg
3 2
1 − y b kmol A / kmolg
V d m / min i
3 3
m b kg / min g
1
n b kmol / min g 4
x b kg SO / kgg
1
y b kmol SO / kmolg 4 2
1 − x b kg B / kgg
1 2
1 − y b kmol A / kmolg
1
4
4- 14

15. 4.26 (cont’d)
8 unknowns ( n1 , n3 , v1 , m2 , m4 , x4 , y1 , y3 )
– 3 material balances
– 2 analyzer readings
– 1 meter reading
– 1 gas density formula
– 1 specific gravity
0 DF
b. Orifice meter calibration:
d i
A log plot of V vs. h is a line through the points h1 = 100, V1 = 142 and h2 = 400, V2 = 290 . d i
ln V = b ln h + ln a ⇒ V = ah b
b=
d h = lnb290 142g = 0.515
ln V2 V1
lnbh h g lnb400 100g
2 1
ln a = ln V − b ln h = lnb142g − 0.515 ln 100 = 2.58 ⇒ a = e
1 1
2 .58
= 13.2 ⇒ V = 13.2h 0.515
Analyzer calibration:
ln y = bR + ln a ⇒ y = aebR
b=
b
ln y 2 y1 g = lnb01107 0.00166g = 0.0600
. U
|
R2 − R1 90 − 20 |
|
ln a = ln y1 − bR1 = lnb0.00166g − 0.0600b20g = −7.60V ⇒ y = 5.00 × 10 −4
e 0.0600 R
E |
|
a = 5.00 × 10 −4 |
W
c. h1 = 210 mm ⇒ V1 = 13.2 210 b g = 207.3 m min
0.515 3
ρ feed gas =
b12.2g b150 + 14.7g 14.7 batmg = 0.460 mol / L = 0.460 kmol / m 3
b75 + 460g 18 bKg
.
E
207.3 m 3 0.460 kmol
n1 = = 95.34 kmol min
min m3
b g
R1 = 82.4 ⇒ y1 = 500 × 10−4 exp 0.0600 × 82.4 = 0.0702 kmol SO2 kmol
.
R3 = 116 ⇒ y3 = 500 × 10−4
. . expb0.0600 × 116g = 0.00100 kmol SO
. 2 kmol
1000 L B 130 kg
.
m2 = = 1300 kg / min
min LB
4- 15

16. 4.26 (cont’d)
b gb g b g
A balance: 1 − 0.0702 95.34 = 1 − 0.00100 n3 ⇒ n3 = 88.7 kmol min
SO2 balance: b0.0702gb9534g( 64.0 kg / kmol) = b0.00100gb88.7g( 64) + m x
. 4 4 (1)
B balance: 1300 = m4 (1 − x4 ) (2)
Solve (1) and (2) simultaneously ⇒ m4 = 1723 kg / min, x4 = 0.245 kg SO2 absorbed / kg
SO2 removed = m4 x4 = 422 kg SO 2 / min
d. Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a
higher rate of transfer of SO2 from the gas to the liquid phase.
4.27 a. d
V2 m 3 / min i b
n3 kmol / min g
m b kg B / min g y b kmolSO / kmol g
3 2
1 − y b kmol A / kmol g
2
3
d
V1 m / min 3
i R3
n b kmol / min g
1 b
m4 kg / min g
y b kmolSO / kmol g
1 2 x b kgSO kg g
4 2
1 − y b kmol A / kmol g
1 1 − x b kg B / kg g
4
P , T1 , R1 , h1
1
b. 14 unknowns ( n1 ,V1 , y1 , P , T1 , R1 , h1 ,V2 , m2 , n3 , y3 , R3 , m4 , x4 )
1
– 3 material balances
– 3 analyzer and orifice meter readings
– 1 gas density formula (relates n1 and V1 )
– 1 specific gravity (relates m2 and V2 )
6 DF
b g b
A balance: 1 − y1 n1 = 1 − y3 n3 g (1)
x4 m4
SO2 balance: y1n1 = y3n3 + (2)
64 kgSO 2 / kmol
b
B balance: m2 = 1 − x4 m4 g (3)
Calibration formulas: y1 = 5.00 × 10−4 e0.060 R1 (4)
−4 0.060 R3
y3 = 5.00 × 10 e (5)
V1 = 13.2h10.515 (6)
Gas density formula: n1 =
b
12.2 P + 14.7 / 14.7
1 g
bT + 460g / 18
1 .
V1 (7)
Liquid specific gravity: SG = 130 ⇒ V2 =
.
m2 kgb gm3
(8)
h 1300 kg
4- 16

17. 4.27 (cont’d)
c. T1 75 °F y1 0.07 kmol SO2/kmol
P1 150 psig V1 207 m3/h
h1 210 torr n1 95.26 kmol/h
R1 82.4
Trial x4 (kg SO2/kg) y3 (kmol SO2/kmol) V2 (m3/h) n3 (kmol/h) m4 (kg/h) m2 (kg/h)
1 0.10 0.050 0.89 93.25 1283.45 1155.11
2 0.10 0.025 1.95 90.86 2813.72 2532.35
3 0.10 0.010 2.56 89.48 3694.78 3325.31
4 0.10 0.005 2.76 89.03 3982.57 3584.31
5 0.10 0.001 2.92 88.68 4210.72 3789.65
6 0.20 0.050 0.39 93.25 641.73 513.38
7 0.20 0.025 0.87 90.86 1406.86 1125.49
8 0.20 0.010 1.14 89.48 1847.39 1477.91
9 0.20 0.005 1.23 89.03 1991.28 1593.03
10 0.20 0.001 1.30 88.68 2105.36 1684.29
V2 vs. y3
3 .5 0
3 .0 0
2 .5 0
V 2 (m /h)
2 .0 0
3
1 .5 0
1 .0 0
0 .5 0
0 .0 0
0 .0 0 0 0.0 2 0 0 .0 4 0 0 .06 0
y 3 ( k m o l S O 2 /k m o l)
x4 = 0 .1 0 x4 = 0 .20
For a given SO2 feed rate removing more SO2 (lower y3) requires a higher solvent feed
rate ( V2 ).
For a given SO2 removal rate (y3), a higher solvent feed rate ( V2 ) tends to a more dilute
SO2 solution at the outlet (lower x4).
d. Answers are the same as in part c.
4.28 Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3
Overall mass balance ⇒ m3
Mass balance - Unit 1 ⇒ m1
A balance - Unit 1 ⇒ x1
Mass balance - mixing point ⇒ m2
A balance - mixing point ⇒ x2
C balance - mixing point ⇒ y2
4- 17

18. 4.29 a. 100 mol / h b
n2 mol / h g b
n4 mol / h g
0.300 mol B / mol b g
x B 2 mol B / mol 0.940 mol B / mol
0.250 mol T / mol Column 1 x b mol T / molg Column 2 0.060 mol T / mol
T2
0.450 mol X / mol 1 − x − x b mol X / molg
B2 T2
n b mol / h g
3 b
n5 mol / h g
0.020 mol T / mol b g
x B5 mol B / mol
0.980 mol X / mol x b mol T / molg
T5
1 − x − x b mol X / molg
B5 T5
Column 1 Column 2:
4 unknowns ( n2 , n3 , x B 2 , xT 2 ) 4 unknowns ( n3 , n4 , n5 , y x )
–3 balances – 3 balances
– 1 recovery of X in bot. (96%) – 1 recovery of B in top (97%)
0 DF 0 DF
Column 1
b gb g
96% X recovery: 0.96 0.450 100 = 0.98n3 (1)
Total mole balance: 100 = n2 + n3 (2)
b g
B balance: 0.300 100 = x B 2 n2 (3)
T balance: 0.250b100g = x T 2 n2 + 0.020n3 (4)
Column 2
97% B recovery: 0.97 x B 2 n2 = 0.940n4 (5)
Total mole balance: n2 = n4 + n5 (6)
B balance: x B 2 n2 = 0.940 n4 + x B5 n5 (7)
T balance: xT 2 n2 = 0.060n4 + xT 5 n5 (8)
b. (1) ⇒ n3 = 44.1 mol / h (2) ⇒ n2 = 55.9 mol / h
(3) ⇒ x B 2 = 0.536 mol B / mol (4) ⇒ x T 2 = 0.431 mol T / mol
(5) ⇒ n4 = 30.95 mol / h ( 6) ⇒ n5 = 24.96 mol / h
(7) ⇒ x B5 = 0.036 mol B / mol (8) ⇒ x T 5 = 0.892 mol T / mol
b
0.940 30.95 g
× 100% = 97%
Overall benzene recovery:
b g
0.300 100
0.892b24.96g
× 100 = 89%
0.250b100g
Overall toluene recovery:
4- 18

19. 4.30 a.
100 kg / h b
m3 kg / h g b g
m4 kg / h b
m10 kg / h g
0.035 kg S / kg b
x3 kg S / kg g 4 x b kg S / kgg
10 0.050 kg S / kg
0.965 kg H 2O / kg
1 b
1 − x3 kg H 2O / kg g
4
1 − x b kg H O / kgg
4 2
0.950 kg H 2O / kg
b
0100mw kg H 2O / h
. g b
0100mw kg H 2O / h
. g b
0100mw kg H 2O / h
. g
b
mw kg H 2O / h g
b. Overall process
100 kg/h m10 ( kg / h)
0.035 kg S/kg 0.050 kg S/kg
0.965 kg H2O/kg 0.950 kg H2O/kg
mw ( kg H 2 O / h)
b g
Salt balance: 0.035 100 = 0.050m10
Overall balance: 100 = mw + m10
H2O yield: Yw =
b
mw kg H 2O recovered g
b
96.5 kg H 2O in fresh feed g
First 4 evaporators
100 kg/ h b
m4 kg / h g
0.035 kg S/ kg b
x 4 kg S/ kg g
0.965 kg H2 O / kg b g
1 − x4 kg H2 O / kg
4 × 0100m b kg H O / hg
. w 2
b
Overall balance: 100 = 4 0100 mw + m4
. g
b g
Salt balance: 0.035 100 = x4 m4
c. Yw = 0.31
x 4 = 0.0398
4- 19

20. 4.31 a.
2n1 mol b g
0.97 mol B / mol Condenser
0.03 mol T / mol
b g
n1 mol b g
n1 mol (89.2% of Bin feed )
0.97 mol B / mol 0.97 mol B / mol
100 mol
0.03 mol T / mol 0.03 mol T / mol
0.50 mol B / mol
Still
0.50 mol T / mol
b gb
n4 mol 45% of feed to reboiler g
y b mol B / molg
B
1 − y b mol T / molg
B
b g
n3 mol
b g
n2 mol
Reboiler x b mol B / molg
z b mol B / molg
B
B
1 − x b mol T / molg
1 − z b mol T / molg
B
B
Overall process: 3 unknowns ( n1 , n3 , x B ) Still: 5 unknowns ( n1 , n2 , n4 , y B , z B )
– 2 balances – 2 balances
– 1 relationship (89.2% recovery) 3 DF
0 DF
Condenser: 1 unknown ( n1 ) Reboiler: 6 unknowns ( n2 , n3 , n4 , x B , y B , z B )
– 0 balances – 2 balances
1 DF – 2 relationships (2.25 ratio & 45% vapor)
3 DF
Begin with overall process.
b. Overall process
b gb g
89.2% recovery: 0.892 0.50 100 = 0.97 n1
Overall balance: 100 = n1 + n3
b g
B balance: 0.50 100 = 0.97 n1 + x B n3
Reboiler
e j = 2.25
yB / 1 − yB
/ b1 − x g
Composition relationship:
xB B
Percent vaporized: n4 = 0.45n2 (1)
Mole balance: n2 = n3 + n4 (2)
(Solve (1) and (2) simultaneously.)
B balance: z B n2 = x B n3 + y B n4
4- 20

21. 4.31 (cont’d)
c. B fraction in bottoms: x B = 0100 mol B / mol
.
Moles of overhead: n1 = 46.0 mol Moles of bottoms: n3 = 54.0 mol
b1 − x gn × 100% = b1 − 010gb54.02g × 100% = 97%
B 3 .
0.50b100g 0.50b100g
Recovery of toluene:
4.32 a.
b
m3 kg H 2O g
Bypass Mixing point
Basis: 100 kg
100 kg b g
m1 kg Evaporator b g
m4 kg b g
m5 kg
0.12 kg S / kg 012 kg S / kg
. 0.58 kg S / kg 0.42 kg S / kg
0.88 kg H2O / kg 0.88 kg H2O / kg 0.42 kg H 2O / kg 0.58 kg H2O / kg
b g
m2 kg
012 kg S / kg
.
0.88 kg H2O / kg
Overall process: 2 unknowns ( m3 , m5 ) Bypass: 2 unknowns ( m1 , m2 )
– 2 balances – 1 independent balance
0 DF 1 DF
Evaporator: 3 unknowns ( m1 , m3 , m4 ) Mixing point: 3 unknowns ( m2 , m4 , m5 )
– 2 balances – 2 balances
1 DF 1 DF
b g
Overall S balance: 012 100 = 0.42m5
.
Overall mass balance: 100 = m3 + m5
Mixing point mass balance: m4 + m2 = m5 (1)
Mixing point S balance: 0.58m4 + 012m2 = 0.42m5
. (2)
Solve (1) and (2) simultaneously
Bypass mass balance: 100 = m1 + m2
b. m1 = 90.05 kg, m2 = 9.95 kg, m3 = 714 kg, m4 = 18.65 kg, m5 = 28.6 kg product
.
m2
Bypass fraction: = 0.095
100
c. Over-evaporating could degrade the juice; additional evaporation could be uneconomical; a
stream consisting of 90% solids could be hard to transport.
4- 21

22. 4.33 a.
b
m4 kg Cr / h g
b
m1 kg / h g b
m2 kg / h g b g
m5 kg / h b g
m6 kg / h
0.0515 kg Cr / kg 0.0515 kg Cr / kg Treatment x b kg Cr / kgg
5 x b kg Cr / kgg
6
0.9485 kg W / kg 0.9485 kg W / kg Unit 1 − x b kg W / kgg
5 1 − x b kg W / kgg
6
b
m3 kg / h g
0.0515 kg Cr / kg
0.9485 kg W / kg
b. b
m1 = 6000 kg / h ⇒ m2 = 4500 kg / h maximum allowed value g
Bypass point mass balance: m3 = 6000 − 4500 = 1500 kg / h
b gb g
95% Cr removal: m4 = 0.95 0.0515 4500 = 220.2 kg Cr / h
Mass balance on treatment unit: m5 = 4500 − 220.2 = 4279.8 kg / h
Cr balance on treatment unit: x5 =
b
0.0515 4500 − 220.2 g
= 0.002707 kg Cr / kg
4779.8
Mixing point mass balance: m6 = 1500 + 4279.8 = 5779.8 kg / h
Mixing point Cr balance: x6 =
b g
0.0515 1500 + 0.0002707 4279.8 b
= 0.0154 kg Cr / kg
g
5779.8
c. m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h) x5 m 6 (kg/h) x6
1000 1000 0 48.9 951 0.00271 951 0.00271
2000 2000 0 97.9 1902 0.00271 1902 0.00271
3000 3000 0 147 2853 0.00271 2853 0.00271
4000 4000 0 196 3804 0.00271 3804 0.00271
5000 4500 500 220 4280 0.00271 4780 0.00781
6000 4500 1500 220 4280 0.00271 5780 0.0154
7000 4500 2500 220 4280 0.00271 6780 0.0207
8000 4500 3500 220 4280 0.00271 7780 0.0247
9000 4500 4500 220 4280 0.00271 8780 0.0277
10000 4500 5500 220 4280 0.00271 9780 0.0301
4- 22

23. 4.33 (cont’d)
m 1 vs. x 6
0.03500
0.03000
x 6 (kg Cr/kg)
0.02500
0.02000
0.01500
0.01000
0.00500
0.00000
0 2000 4000 6000 8000 10000 12000
m 1 (kg/h)
d. Cost of additional capacity – installation and maintenance, revenue from additional
recovered Cr, anticipated wastewater production in coming years, capacity of waste lagoon,
regulatory limits on Cr emissions.
4.34 a.
b
175 kg H 2O / s 45% of water fed to evaporator g
b
m1 kg / s g b
m4 kg K 2SO 4 / s g b
m6 kg K 2SO 4 / s g Filter cake
0196 kg K 2SO 4 / kg
. b
m5 kg H 2O / s g Evaporator b
m7 kg H 2O / s g Crystallizer
Filter b
10 m2 kg K 2SO 4 / s g
0.804 kg H 2O / kg b g
m2 kgsoln / s
R0.400 kg K SO / kg U
S0.600 kg H O / kg V
2 4
Filtrate T 2 W
b
m3 kg / s g
0.400 kg K 2SO 4 / kg
0.600 kg H 2 O / kg
Let K = K2SO4, W = H2 Basis: 175 kg W evaporated/s
Overall process: 2 unknowns ( m1 , m2 ) Mixing point: 4 unknowns ( m1 , m3 , m4 , m5 )
- 2 balances - 2 balances
0 DF 2 DF
Evaporator: 4 unknowns ( m4 , m5 , m6 , m7 ) Crystallizer: 4 unknowns ( m2 , m3 , m6 , m7 )
– 2 balances – 2 balances
– 1 percent evaporation 2 DF
1 DF
Strategy: Overall balances ⇒ m1 , m2 U verify that each
| chosen subsystem involves
% evaporation ⇒ m5 |
V
Balances around mixing point ⇒ m , m | no more than two
Balances around evaporator ⇒ m , m | unknown variables
3 4
W 6 7
4- 23

24. 4.34 (cont’d)
Overall mass balance: m1 = 175 + 10m2 + m2 U
|
0196m1 = 10m2 + 0.400m2
V
|
Overall K balance: .
W
Production rate of crystals = 10m2
45% evaporation: 175 kg evaporated min = 0.450m5
W balance around mixing point: 0.804m1 + 0.600m3 = m5
Mass balance around mixing point: m1 + m3 = m4 + m5
K balance around evaporator: m6 = m4
W balance around evaporator: m5 = 175 + m7
m4
Mole fraction of K in stream entering evaporator =
m4 + m5
b. Fresh feed rate: m1 = 221 kg / s
Production rate of crystals = 10m2 = 416 kg K s s
. bg
b
m3 kg recycle s g=
352.3
= 160
kg recycle
Recycle ratio:
b
m1 kg fresh feed s g
220.8
.
kg fresh feed
c. Scale to 75% of capacity.
Flow rate of stream entering evaporator = 0.75(398 kg / s) = 299 kg / s
46.3% K, 53.7% W
d. Drying . Principal costs are likely to be the heating cost for the evaporator and the dryer and
the cooling cost for the crystallizer.
4- 24

25. 4.35 a. Overall objective: Separate components of a CH4-CO2 mixture, recover CH4, and discharge
CO2 to the atmosphere.
Absorber function: Separates CO2 from CH4.
Stripper function: Removes dissolved CO2 from CH3OH so that the latter can be reused.
b. The top streams are liquids while the bottom streams are gases. The liquids are heavier than
the gases so the liquids fall through the columns and the gases rise.
c.
b
n1 mol / h g b g
n5 mol N 2 / h
0.010 mol CO 2 / mol
n b mol CO / h g
6 2
0.990 mol CH 4 / mol
100 mol / h Absorber b
n2 mol / h g Stripper
0.300 mol CO 2 / mol 0.005 mol CO 2 / mol b
n5 mol N 2 / h g
0.700 mol CH 4 / mol 0.995 mol CH 3OH / mol
b g
n3 mol CO 2 / h
n b mol CH OH / h g
4 3
Overall: 3 unknowns ( n1 , n5 , n6 ) Absorber: 4 unknowns ( n1 , n2 , n3 , n4 )
– 2 balances – 3 balances
1 DF 1 DF
Stripper: 4 unknowns ( n2 , n3 , n4 , n5 )
– 2 balances
– 1 percent removal (90%)
1 DF
Overall CH4 balance: b0.700gb100g bmol CH / hg = 0.990n 4 1
Overall mole balance: 100b mol / h g = n + n 1 6
Percent CO2 stripped: 0.90 n3 = n6
Stripper CO2 balance: n3 = n6 + 0.005n2
Stripper CH3OH balance: n4 = 0.995n2
d. n1 = 70.71 mol / h , n2 = 6510 mol / h, n3 = 32.55 mol CO 2 / h, n4 = 647.7 mol CH3OH / h,
.
n6 = 29.29 mol CO 2 / h
30.0 − 0.010n1
Fractional CO2 absorption: f CO 2 = = 0.976 mol CO 2 absorbed / mol fed
30.0
4- 25

26. 4.35 (cont’d)
Total molar flow rate of liquid feed to stripper and mole fraction of CO2:
n3
n3 + n4 = 680 mol / h, x3 = = 0.0478 mol CO 2 / mol
n3 + n4
e. Scale up to 1000 kg/h (=106 g/h) of product gas:
b g b
MW1 = 0.01 44 g CO 2 / mol + 0.99 16 g CH 4 / mol = 16.28 g / mol g
bn g = d10 × 10 g / hib16.28 g / molg = 6142 × 10 mol / h
1 new. 6
. 4
bn g = b100 mol / hg (6142 × 10 mol / h) / (70.71 mol / h) = 8.69 × 10
feed new . 4 4
mol / h
f. Ta < Ts The higher temperature in the stripper will help drive off the gas.
Pa > Ps The higher pressure in the absorber will help dissolve the gas in the liquid.
g. The methanol must have a high solubility for CO2, a low solubility for CH4, and a low
volatility at the stripper temperature.
4.36 a. Basis: 100 kg beans fed
e
m kg C H
5 6 14 j Condenser
e
m kg C H j 300 kg C 6 H14 b g
m2 kg b g
m4 kg b
m6 kg oil g
1 6 14
Ex b
x2 kg S / kg g F b g
y4 kg oil / kg Ev
y b kg oil / kgg
2
1 − y b kg C H
4 6 14 / kg g
1 − x − y b kg C H
2 2 6 14 / kg g
13.0 kg oil
87.0 kg S
b g
m3 kg
0.75 kg S / kg
b
y3 kg oil / kg g
b
0.25 − y3 kg C 6 H14 / kg g
Overall: 4 unknowns ( m1 , m3 , m6 , y3 ) Extractor: 3 unknowns ( m2 , x2 , y2 )
– 3 balances – 3 balances
1 DF 0 DF
Mixing Pt: 2 unknowns ( m1 , m5 ) Evaporator: 4 unknowns ( m4 , m5 , m6 , y4 )
– 1 balance – 2 balances
1 DF 2 DF
Filter: 7 unknowns ( m2 , m3 , m4 , x2 , y2 , y3 , y4 )
– 3 balances
– 1 oil/hexane ratio
3 DF
Start with extractor (0 degrees of freedom)
Extractor mass balance: 300 + 87.0 + 13.0 kg = m2
4- 26