Mass Balance - Answers to Practice Questions.pdf

mass balance problems its solutions

1. CAPE1020
Engineering Science 1
Mass Balance:
Answers to Practice Questions

2. 1. Mass Balance of Mixer
20 kg s
-1
of a solution containing 65% HNO
3
, 5% H
2
SO
4
and 30% H
2
O (by mass) is mixed
with 30 kg s
-1
of a solution containing 30% HNO
3
, 60% H
2
SO
4
and 10% H
2
O (by mass).
Calculate the composition of the resultant mixture.
Mixer
S
3
= S
1
+ S
2
= 50 kg s
-1
x
3,HNO3
= ?
x
3,H2SO4
= ?
x
3,H2O
= ?
S
1
= 20 kg s
-1
x
1,HNO3
= 0.65
x
1,H2SO4
= 0.05
x
1,H20
= 0.30
S
2
= 30 kg s
-1
x
2,HNO3
= 0.30
x
2,H2SO4
= 0.60
x
2,H20
= 0.10In = Out
Component balance
HNO
3
: (0.65)(20) + (0.3)(30) = (x
3,HNO3
)(50)
so (x
3,HNO3
) = 0.44
H
2
SO
4
: (0.05)(20) + (0.6)(30) = (x
3,H2SO4
)(50)
so (x
3,H2SO4
) = 0.38
H
2
O: x
3,H2O
= 1 - (x
3,HNO3
) + (x
3,H2SO4
)
= 1 – 0.44 + 0.38 = 0.18
Therefore
HNO
3
: 44 wt%
H
2
SO
4
: 38 wt%
H
2
O: 18 wt%

4. 2. Mass Balance of Evaporator
10 kg s
-1
of a 10% (by mass) NaClsolution (A) is concentrated to 50% in a continuous
evaporator. Calculate the production rate of concentrated solution (C) and the rate of
water removal (W) from the evaporator.
A = 10 kg/s
x
A,NaCl
= 0.10
W = ? kg/s
C = ? kg/s
x
C,NaCl
=
0.50
Component mass
balance on NaCl
(x
A,NaCl
)(A) = (x
C,NaCl
)(C)
(0.1)(10) = (0.5)(C)
C = 1/0.5 = 2 kg/s
Overall mass balance
A = C + W
W = A -C
= 10 –2
= 8 kg/s
evaporator

5. 3. Mass Balance of Evaporator
A 5% formaldehyde-in-water solution (by mass) is fed to a concentrating unit that
produces two product streams. One stream has 37% formaldehyde and the other stream
99.5% water, both by mass. How much of the streams is produced per 100 kg of feed.
evaporator
F = 100kg
x
F,form
= 0.05
x
F,H20
= 0.95
S
1
= ?. x
S1,form
= 0.37
S
2
= ?. x
S2,H2O
= 0.995
Component mass
balance on formaldehyde
(x
F,forml
)(F) = (x
S1,form
)(S
1
) + (x
S2,form
)(S
2
)
(0.05)(100) = (0.37)(S
1
) + (0.005)(S
2
)
5 = (0.37)(S
1
) + (0.005)(S
2
)
Rearranging
(S
1
) = 5 –0.005(S
2
) / 0.37 ----[1]
Component mass
balance on water
(x
F,H2O
)(F) = (x
S1,H2O
)(S
1
) + (x
S2,H2O
)(S
2
)
(0.95)(100) = (0.63)(S
1
) + (0.995)(S
2
)
95 = (0.63)(S
1
) + (0.995)(S
2
) ----[2]
Sub [1] into[2]
95 = (0.63)(5 –0.005(S
2
) / 0.37) + (0.995)(S
2
)
Multiply both sides by 0.37
35.15 = 3.15 –0.00315S
2
+ 0.368S
2
0.365S
2
= 32 so S
2
= 87.7 kg
Solve in [1]
(S
1
) = 5 – 0.005(87.7) / 0.37
(S
1
) = 12.33 kgSo S
1
= 12.33 kg and S
2
= 87.7 kg

6. 4. Mass Balance of Evaporator and Mixer
Liquid acetone (C
3
H
6
O) is fed at a rate of 400 litre min
-1
into a heated chamber, where it
evaporates into a nitrogen stream. The gas leaving the heater is diluted by another
nitrogen stream flowing at a rate of 419 m
3
(STP) min
-1
. The combined gases are then
compressed to a total pressure P
gauge
= 6.3 atmat a temperature of 325 ºC. The partial
pressure of acetone in this stream is p
a
=501 mm Hg. Atmospheric pressure is 760 mm
Hg. The density of liquid acetone is 0.792 g cm
-3
.
What is the molar composition of the stream leaving the compressor?
What is the volumetric flowrate of the nitrogen entering the evaporator if the temperature
and pressure of this stream are T = 27 ºC and P
gauge
=478 mm Hg?
A
1
= 400L/min
ρ = 792
kg/m
3
Evaporator Mixer
Compressor
N
1
T = 27 ºC
P
g
= 478 mm Hg
N
2
V̇ = 419 m3/min
(STP)
Pg = 6.3 atm
T = 325 ºC
Pa = 501 mm Hg
Calculation continued ….
A
1
N
1
A
1
N
3
A
1
N
4

7. To calculate composition of exit stream
Mass and molar flowrate of stream A
1
#
5
L
v r r
.
IEJ
H
I
7
s r r r.
H
y { t
GC
I
7
L
u s x
?
z
GC
IEJ
#
5
L
I=OO

N=PA
IKHA?QH=N

SAECDP
L
u s x
?
z
w z

L
w
?
v x

GIKH
IEJ
Composition of gas stream exiting
compressor
Acetone:
Nitrogen:
U
?
L sFU
?
Lr?{s
IKHAO KB 0
6
PKP=H IKHAO

Total moles of Nitrogen:
U
?
U
?
H#
5
L
r? { s
r? r {
H w? v xL w w? t s kmol min
Total molar flowrate of exit stream:
N
4
+ A
1
= 55.21+ 5.46 = 60.67 kmol/min
U
?
L
2
?
2
?????
L
w r s II *C
y?u =PIH y x r
II *C
=PI
L r? r {
IKHAO KB #
PKP=H IKHAO
To calculate volumetric flowrate of N
1
entering the
evaporator
N
2
= 419 m
3
/min
JL
??
??
L
54
1
??H85=
?
/
???
<?758
?
?????
H6;7
Lsz?vx GIKH IEJ
-1
N
1
= N
4
–N
2
= 55.21 –18.46 = 36.75 kmolmin
-1
Using ideal gas law
8L
???
?
where T = 300K and P = 1+(478/760) = 1.63 atm
i.e. 1.63*10
5
Pa
8L
7:?;9
????
???
H<758
?
??????
H744
5?:7H54
1
??
= 562 m
3
/min
Note: In chemical engineering calculations, pressure should be
absolute. See the calculation of the mole fraction y
A
(P
g
=6.3 atm,
but P
abs
=7.3 atm!) and the ideal gas law application (P
abs
=1bar).

8. Answer:
Water balance
In = Out
[(x
1,W
)(F) + (x
2,W
)(R)] * (%evaporated) = E + P
[(0.75)(2000) + (0.55)(R)] * (0.3) = 1500 + 0
450 + 0.165 R = 1500 Therefore R = 1050/0.165
= 6363.6 kg h
-1
5
.MassBalanceofEvaporator-Crystalliser
Anevaporator-crystalliserisfedwithafreshfeedof25wt%KNO
3
aqueoussolutionatarateof2000
kgh
-1
,toproduce500kgh
-1
ofsolidKNO
3
crystals.Thecrystallisationsolutioncontains45wt%KNO
3
isrecycledtojointhefreshfeed.Ofthewaterfedtotheevaporator-crystalliser,30wt%isevaporated.
If1500kgh
-1
ofwaterisevaporated,whatistheflowrateoftherecyclestream?
A.
6364kgh
-1
B.
567.8kgh
-1
C.
2985kgh
-1
D.
6042kgh
-1
Evaporator/
Crystalliser/
Dryer
Product, P = 500 kg h
-1
of
KNO3 solids
Feed,
F = 2000 kg h
-1
X
1,K
= 0.25
X
1,W
= 0.75
Recycle, R
x
2,K
= 0.45
x
2,W
= 0.55
Evaporation rate,
E = 1500 kg h
-1

9. 6. Mass Balance of Distillation Column
An equimolar mixture of propane and butane is fed to a distillation column at the rate of
67 mols
-1
. 90% of the propane is recovered in the top product which has a propane
mole fraction of 0.95. Calculate the flowrate of the top and bottom products and the
composition of the bottom product.
S1,
F = 67 mol/s
x
1,P
= 0.5
x
1,B
= 0.5
S2, D = ?, 90% of
propane feed recovered
x
2,P
= 0.95
x
2,B
= 0.05
S3, B = ?
x
3,P
=
x
3,B
=
Balance on propane
(0.9)(x
1,P
)(F) = (x
2,P
)(D)
(0.9)(0.5)(67) = (0.95)(D)
So D = 31.736 mol/s
i.e. x
2,P
= 30.15 mols/s and
x
2,B
= 1.586 mols/s
Overall Balance to find B
F = D + B
B = F – D
B = 67 – 31.736
B = 35.264 mols/s
Balance on butane
(x
1,B
)(F) = (x
2,B
)(D) + (x
3,B
)(B)
(0.5)(67) = (0.05)(31.736) +
(x
3,B
)(35.264)
(x
3,B
) = (33.5) –(1.586) / (35.264)
(x
3,B
) = 0.905 i.e. 90.5%
(x
3,P
) = 1 -0.905 = 0.095 = 9.5%

10. 7.
MassBalanceofDistillationColumn
Adistillationcolumnisfedwithabinaryequimolarmixtureofethanolandwateratarateof400 kmol
h
-1
,andproducesatopandabottomproduct.If85mol%oftheethanolexitsinthetopproduct,which
contains90mol%ethanol,whatisthefractionofwaterinthebottomproduct?
A.
62mol% Note:theoriginaltaskwith60%
B.
71mol% ethanolinthedistillatecannot
C.
74mol% besolved.
D.
86mol%
Stream 1, F= 400 kmolh
-1
Distillation Column
x
1,ethanol
= 0.5
x
1,water
= 0.5
Stream 2
D, 85mol% of ethanol in Feed exits
x
2,ethanol
= 0.9
x
2,water
= 0.1
Stream 3, B = ?
x
3,ethanol, B
= ?
x
3,water, B
=?
Ans:
Balance on ethanol
(0.85)(x
1,ethanol
)(F) = (x
2,ethanol
)(D)
(0.85)(0.5)(400) = (0.9)(D)
So D = 188.9 kmolh
-1
Overall Balance to find B
F = D + B
B = F –D
B = 400 –188.9
B = 211.1 kmolh
-1
Balance on water
(x
1,water
)(F) = (x
2,water
)(D) +
(x
3,water
)(B)
(0.5)(400) = (0.1)(188.9) +
(x
3,water
)(211.1)
(x
3,water
) = (200 –18.89) / (211.1)
(x
3,water
) = 0.858 i.e. 86 mol%
Note: This solution is
mathematically correct, but
physically impossible!
Azeotrope at about 70% ethanol

11. 8. Mass Balance of Absorption Column
50 kg s
-1
of an acetone-water solution containing 10% acetone (by mass) is to be
produced in an absorption column. An acetone-air mixture containing 20% acetone (by
mass) is fed to the column and the acetone is scrubbed from the air by water fed to the
top of the column. The acetone is not completely removed and the exit gas stream
contains 3% acetone (by mass). Calculate the amount of acetone-air mixture fed to the
column.
50kg/s
10 wt% Ace
4
= 5 kg Ace/s
90 wt% W
4
= 45 kg W/s
Ace
1
= 20wt%
Air
1
= 80wt%
W
3
Ace
2
= 3 wt%
Air
2
= 97wt%
Absorber
Acetone balance
Ace
1
= 5 + Ace
2
---[1]
Air balance
Air
1
= Air
2
(80/20) Ace
1
= (97/3) Ace
2
(4) Ace
1
= (97/3) Ace
2
---[2]
Sub [1] into [2] and solve
(4) (5+Ace
2
)

= (97/3) Ace
2
20 + 4Ace
2
= (97/3) Ace
2
Ace
2
= 0.7 kg/s
Then solve for Ace
1
= 5.71 kg/s
Now Air
1
= Air
2
= (97/3) Ace
2
= (97/3)(0.7) = 22.82 kg/s
So stream 1 = Ace
1
+ Air
1
= 5.7 + 22.82 = 28.53 kg/s

12. 9. Mass Balance of Absorption Column
A gas mixture of CO
2
and N
2
is fed to an absorption column at a rate of 500 m
3
h
-1
at a temperature of
32
o
C and pressure of 700 kPa. The CO
2
in this stream has a partial pressure of 125 kPa, and 95% of
the CO
2
is removed and the remainder exits with the N
2
in the exit gas stream. The exit gas has a
temperature of 42
o
C and pressure of 600 kPa. What is the volumetric flow rate of the exit gas
stream?
A.
412 m
3
h
-1
B.
469 m
3
h
-1
C.
500 m
3
h
-1D.
531 m
3
h
-1
Absorption column
95% CO
2
removed
Stream 2
V
2
= ?
T
2
= 42°C
P
2
= 600 kPa
Stream 1
Gas mixture:
CO
2
+ N
2
V
1
= 500 m
3
T
1
= 32°C
P
1
= 700 kPa
P,co
2
= 125 kPa
1) Calculate total number of moles fed in:
J
5
L
2
5
8
5
46
5
L
y r rH s r
7
H w r r
z? u s vH: u tE t y u;
L??? o???
2) Calculate moles of CO
2
and N
2
entering and exiting
P,co
2
= 125 kPa IKHA BN=?PEKJ? U
??
.
?5
L
2
??
.
2
5
L
s t w
y r r
L r? s z
Moles of CO
2,1
= 0.18(n
1
) = 0.1*138 = 24.64 kmol
Moles of CO
2,2
= (1-0.95)(Moles of CO
2,1
) = 1.232 kmol
Moles of N
2,1
= (n
1
) -(Moles of CO
2,1
) = 138 –24.64 = 113.36 kmol
n
2
= (Moles of N
2,1
) + (Moles of CO
2,2
) = 114.6 kmol
2) Use ideal gas law to calculate volume at exit
8
6
L
J
6
46
6
2
6
L
s s v? xH s r
7
H z? u s vH v tE t y u
x r rH s r
7
L??? ?
?

13. 10.MassBalanceofExtractor
Asolutionofuraniuminnitricacidisfedtoanextractoratarateof100kgh
-1
,andasolvent(TBP)is
fedtotheextractoratarateof50kgh
-1
.Theprocesshastwoproducts:theextract,whichisuranium-
rich,andtheraffinate.Ifthefeedcontains40wt%uraniumand60wt%nitricacid,theextractcontains
15wt%nitricacidand55wt%TBP,andtheraffinatecontains60wt%nitricacid,whatisthefractionof
uraniumintheraffinate?
A.
19wt%
B.
24wt%
C.
29wt%
D.
34wt%
Stream 1
U in HNO3, F = 100 kg/h
TBP, S = 50 kg/h
Stream 2
x
1,U
= 40%
x
1,HNO3
= 60%
Stream 3
Extract, E
Stream 4, RaffinateR
x
3,HNO3
= 15%
x
3,TBP
= 55%
x
3,U
= 30%
x
4
,
HNO3
= 60%
x
4
,
TBP
= ?
x
4
,
U
= ?
Uranium balance
(x
1,U
)(F) = (x
3,U
)(E) + (x
4
,
U
)(R)
(0.4)(100) = (0.3)(E) +
(x
4
,
U
)(R)
40 = (0.3)(E) + (x
4
,
U
)(R)
HNO
3
balance
(x
1,HNO3
)(F) = (x
3,HNO3
)(E) +
(x
4
,
HNO3
)(R)
(0.6)(100) = (0.15)(E) + (0.6)(R)
60 = (0.15)(E) + (0.6)(150 -E)
Also we know (x
4
,
TBP
) = 0.4 - (x
4
,
U
) which can be substituted in eqns above
X
4,U
= (40 – 0.3 * 66.67)/83.33 = 0.24
0.15 E + 0.6 (150 – E) = 60
0.45 E = 30
Step 1: Draw a
concise process
flow diagram
Step 2: Mass balance overall
and on individual components
Overall balance
E + R = 150
E = 66.67 kg s
-1
R = 83.33 kg s
-1

14. 11. Mass Balance of a system with Recycle
In a process producing KNO
3
salt, 1000 kg h
-1
of a feed solution containing 20 wt% KNO
3
is fed to an evaporator, which evaporates some water at 422 K to produce a 50% KNO
3
solution. This is then fed to a crystallizer at 311 K, where crystals containing 96% KNO
3
are removed. The saturated solution containing 37.5% KNO
3
is recycled to the
evaporator. Calculate the amount of recycle stream R in kg h
-1
and the product stream of
crystals P in kg h
-1
.
F, 1000kg/h
20wt% K
W
R
37.5wt% K
62.5wt% W
P, Crystal product
96wt% K
S, Solution
50wt% K
K = KNO
3
,
W= Water
Overall K Balance
(0.2)(1000) = (0.96)(P)
(P) = 208.3 kg/h
K (P) = (0.96)(208.3) = 199.97 kg/h
W (P) = (0.04)(208.3) = 8.33 kg/h
Crystallizer component balance
K: (0.5)(S) = (199.97) + 0.375R
W: (0.5)(S) = 8.33 + 0.625R
Subtract W from K
0 = 191.64 –0.25R, so R = 766.56 kg/h
Evaporator Crystalliser

15. 12. Mass Balance on Separation Units
A mixture containing 50% acetone and 50% water by mass is to be separated into its
components. The separation process consists of a two-stage extraction of acetone from
water into a solvent that dissolves acetone but is nearly immiscible with water. The
solvent used is methyl isobutyl ketone (MIBK). In the process for every 100 kg of
acetone-water fed to the first extractor, 100 kg of MIBK is fed to the first extractor and 75
kg to the second extractor. The extract from the first stage contains 27.5% acetone. The
second-stage raffinate is 43.1 kg and consists of 5.3% acetone, 1.6% MIBK and 93.1%
water and the second-stage extract contains 9.0% acetone, 88% MIBK and 3% water.
Calculate the masses and compositions of the unknown streams.
Extractor
1
Extractor
2
S
1
=
100kg
S
2
= 75kg
E
1
27.5wt% A
E
2
9wt% A
88wt% S
3wt% W
R
2
= 43.1
5.3wt% A
1.6wt% S
93.1wt% W
R
1
F
50wt% A
50wt% W
A = Acetone
S = Solvent
W = Water
F=Feed
E=Extract
Calccontinued..

16. Overall Balance
F + S
1
+ S
2
= E
1
+ E
2
+ R
2
275 = E
1
+ E
2
+ 43.1 ---[1]
A component balance
(0.5)(F) = (0.275)(E
1
) + (0.09)(E
2
) + (0.053)(43.1)
50 = 0.275E
1
+ 0.09E
2
+ 2.28 ---[2]
Solving [1] and [2] simultaneously gives
E
1
= 145.1 kg
E
2
= 86.8 kg
W component balance
(0.5)(100) = (E
1
,
W
)(145.1) + (0.03)(86.79) + (0.931)(43.1)
50 = (E
1
,
W
)(145.1) + 2.6 + 40.1
(E
1
,
W
) = 0.05 i.e. 5wt%
So (E
1
,
S
) = 1–0.275 –0.05 = 0.675 i.e. 67.5wt%
E
2
stream: A: (0.09)(86.8) = 7.81 kg
S: (0.88)(86.8) = 76.38 kg
W: (0.03)(86.8) = 2.6 kg

17. 13. Mass Balance on a Water Storage Tank
A 12.5-m
3
tank is being filled with water at a rate of 0.05 m
3
.s
-1
. At a moment when the
tank contains 1.20 m
3
of water, a bottom leak develops and gets progressively worse with
time. The rate of leakage can be approximated as 0.0025t (m
3
.s
-1
), where t is the time
from the moment the leak begins.
Draw a diagram of the tank showing the parameters given.
Write a mass balance on the tank and use it to obtain an expression for dV/dt, where Vis
the volume of water in the tank at any time. Provide an initial condition for the differential
equation.
Solve the balance equation to obtain an expression for V(t)and draw a plot of Vversus t.

18. Volume,
V
W
in
= 0.05m
3
/s
W
out
= 0.0025t m
3
/s
Tank
capacity =
12.5m
3
t = 0, V = 1.2m
3
Mass balance
RKHQIA
PEIA
LN=PA S
??
FN=PA S
???


@8
@P
L r? r wF r? r r t wP
Integrating over time from t=0 to the required
time t and volume V from the initial volume
V
0
=1.2 m
3
to the volume at time t, V
V = 1.2 + 0.05t –0.00125t
2
1) 2)
3)
Validity of the equation:
V = 0.05t –0.00125t
2
+ 1.2 0 ≤ t ≤ 57s
V = 0 > 57s