introduction to material balances
TRANSCRIPT
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Chapter 3:
Introduction to Material Balances
Learning Objectives
Upon completing this Chapter, you should be
able to understand what process flow sheet (PFS) or
process flow diagram (PFD) is
know a standard symbol for each of some
important process equipments
draw a simple process flow diagram (or a
block flow diagram or a flow chart) for the
given problem
make a necessary assumption/necessary
assumptions pertaining the given problem
set up an appropriate basis of calculation
(basis) of the given problem
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understand what steady-state process is
and how it affects the establishment/
set-up of material-balance equations understand what overall and species
balances are
establish overall- and species-balance
equations for the given problem
solve simple material-balance problems
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One of the mainresponsibilitiesof
chemical engineersis to create/construct/
analysechemical processes(or, at least,
to understandthe existing processes)
The layoutof a chemical processis
called process flow sheet (PFS) or
process flow diagram (PFD)
PFS or PFD can be for just a single
process unit of for the whole process,
either simple or complicated process
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Examples of PFS or PFD
PFD for an atmospheric
distillation of crude oil
(from Stoichiometry, 4thed; by Bhatt & Vora, 2004)
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PFD for a water-softeningby ion-exchange process
(from Stoichiometry, 4thed; by Bhatt & Vora, 2004)
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PFD for Ammonia Synthesis Plant
(from Introduction to Chemical Processes by Murphy, 2007)
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Normally, a PFS or a PFD comprises
all major process equipments/units
linesenteringor leavingthe
process/unit and/or linesconnecting
two or more process
equipments/units (these lines arecalled streams)
flow rateof each stream
compositionof each stream
operating conditionsof each stream
and/or unit/equipment (e.g., T, P)
energy/heat neededto be added to
and/or removed from any particularpart of the process or the entire
process
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Some important symbols of process
equipments are illustrated as follows
Mixer Packed-bed Batch Reactor
Reactor
Distillation Flash Drum Mixer-SettlerColumn
Pump Blower Turbine
Heat Exchanger Furnace
(from Introduction to Chemical Processes by Murphy, 2007)
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In order to be able to createor to
understandPFSor PFD, the knowledge
concerning MATERIAL & ENERGY
BALANCESis required
As a chemical engineer, you need to be
abletoperformmaterial and energy
balances for any particular process or for
the entire process efficiently/competently
We start our learning by doing MATERIAL
BALANCES, using an underlying knowledge
of law of conservation of mass
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Before starting learning material
balance, lets consider the following bank
account:
Date Description Deposit Withdrawal Balance
2/1/51
8/1/51
15/1/51
19/1/51
22/1/51
22/1/51
31/1/51
Beginning balance
Deposit from Mom
Deposit from Dad
ATM withdrawal
ATM withdrawal
Service charge
Closing balance
10,000
20,000
7,000
18,000
50
15,000
19,950
The data we obtain from the above
bank account are
The sum of deposits = 30,000
The sum of withdrawals = 25,050
The initialbalance = 15,000
Thefinalbalance = 19,950
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From the information in the previous
Page (Page 10), we can write the following
relationship:
The final balance The initial balance =
The sums of deposits The sums of withdrawals
(19,950) (15,000) =
(30,000) (25,050)
If we consider
the initial balance as the initial
mass of the process
the final balance as the final mass of
the process
the sums of deposits as the sums ofmass entering the process
the sums of withdrawals as the sums
of mass leaving the process
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Additionally, if we consider the
differencebetween thefinal balanceand
the initial balance, which is the amount
of money accumulatedin the bank
account as the amount of mass
accumulatedin the process, we can
establish the principle of material/mass
balanceas follows:
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Principles of MATERIAL BALANCE
In the case that there is NO Chemical
Rxn.
Total mass enteringa process/unit
Total mass leaving a process/unit
= Mass accumulation in a process/unit
However, the material balance
problemwill be more complicated(but
not too difficult believe me!) when thereis/are a Rxn./Rxns.in the process/unit,
as follows:
Total mass entering a process/unit
Total mass leaving a process/unit
+ Mass generating from a Rxn/Rxns
= Mass accumulation in a process/unit
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From the principles above, the following
equations can be written:
In the case that there is NO Rxn.
f isys sys sys
in out
m m m m m = =
(3.1)where
isysm = initialmass of a system
fsysm = final mass of a system
In the case that there is/are a
Rxn./Rxns.
.
sys
in out Rxn
m m m m + = (3.2)
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Normally, chemical processes are
continuous, the change in mass should,
therefore, be written in the rate form (i.e.
it changes with time)
Eqs. 3.1 & 3.2 can, then, be re-written,
as follows
sys
in out
dmdm dm
dt dt dt = (3.3)
.
sys
in out Rxn
dmdm dm dm
dt dt dt dt + =
(3.4)
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We can perform material balanceseither
for all species(called overall balance) or
for a specific/selected species(called
species balance)
In this level, we will carry out material
balances for steady-stateprocesses(do
you know what steady-state processes
are?)
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Example One thousand (1,000) kilograms
of a mixture of benzene (B) and toluene (T),
containing 40% by mass of B is to be
separated into two streams in a distillation
column. The top output stream of the
column contains 375 kg of B and the
bottom output stream contains 515 kg
of T.
(a)Perform the mass balance for B & T
(b)
Determine the composition of the
top and bottom streams
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Standard Procedure:
1)Make any necessary assumption(s)
For instance, in this case, we make an
assumptionthat the process is steady-
state
2)Draw a flow chart of the process/unit
Distilla tio n
co lumn
Input stream
1,000 kg mixture
40 wt% B
Top output stream
375 kg B
xkg T
Bottom output stream
ykg B
515 kg T
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3)Set a basis of calculations
In this case, we should set a basisas
1,000 kg of mixture
4)Determine the numbers of unknowns
In this case, there are 3 unknowns:
x = ?
y = ?
% of T of an input stream
5)
Establish material balance equations
In order to be able to solve for
unknowns, it is necessarythat the # of
Eqs.must be equalto the # of unknowns
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In this Example, since the input
stream consists of only 2 components, we
obtain the following equation:
wt% of B wt% of T 100+ =
Accordingly,
% T 100 % B
100 40
% T 60%
=
=
=
Note that one unknown is eliminated
(the # of unknowns are now only 2)
From the basis of calculationwe have set
(in Step 3) and from the percentages ofbenzene and toluene (the percentage of
toluene (%T) has been solved in Step 5), we
obtain the information that the input
streamcomprises:
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Benzene (B) = ( )40
1,000 kg 400 kg100
=
Toluene (T) = ( )60 1,000 kg 600 kg100
=
In a general form, let
Fm = mass of the input stream (feed)
Bm = mass of benzenein the input
stream
Tm = mass of toluenein the input
stream
By = mass fraction of benzenein the
input stream
Ty = mass fraction of toluenein the
input stream
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We can, then, write the following
equations
B B Fm y m=
T T Fm y m=
For the output streams, let
topm = mass of the topstream
bottomm = mass of the bottomstream
Since the process is steady-state(as we
made an assumption in Step 1) and has no
Rxn., the mass balance equation can be
written as follows
0sysin out
in out
m m m
m m
= =
=
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From the process flow chart (on Page
18), we obtain the facts that
F
in
m m=
and that
top bottom
out
m m m= +
Thus,
top bottomFm m m= + (3.5)
Eq. 3.5 is an example of an overall
mass balance equation
Substituting corresponding numerical
values into Eq. 3.5 yields
top bottom1,000 m m= + (3.6)
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In order to solve for 2 unknownsin Eq.
3.6 (i.e. topm & bottomm ), only 1 equationis
NOT enough
We needto have 2 equations; since we
have already got one, we, therefore, needanother equation
To obtain another equation, we need to
do a species balance
In this Example, we shall perform a
benzene balance, as follows
benzene benzenein out
m m
=
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Let
topBm = mass of benzenein the top
outputstream
bottomBm = mass of benzenein the
bottom outputstream
we can, then, write the following equation:
top bottomB B Bm m m= + (3.7)
ortop bottomB F B B
y m m m = + (3.8)
Eqs. 3.7 & 3.8 are the examples of
species balanceequations (in this case, it
is called benzene balance equations)
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Substituting numerical values into Eq.
3.8 gives
( ) ( ) top bottom0.4 1,000 400 B Bm m= = +
(3.9)
It is given that
topBm = 375 kg
Hence, from Eq. 3.9, we obtain
bottom
bottom
400 375
400 375
25 kg
B
B
m
m
y
= +
=
=
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We can also perform a toluene balance,
as follows
top bottomT T Tm m m= + (3.10)
top bottomT F T T x m m m = + (3.11)
It is given that
bottomTm = 515 kg
Thus, from Eq. 3.10, we obtain
top
top
600 515
600 515
85 kg
T
T
m
m
x
= +
=
=
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We can summarise our calculations as
illustrated in the following Table
SpeciesInput
[kg]
Output [kg]
Top Bottom
Benzene
Toluene
400
600
375
85
25
515
TOTAL 1,000 460 540
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Mole Balance
Performing mole balance is similartothat of the mass balance, but overall
mole balances are applicable ()
onlyfor theprocessesthat have no Rxns
Thus, in the case that there is NO Rxn.:
Overall balance
f isys sys sys
in out
n n n n n = =
(3.12)
Species balance
sys sys sys f ij j j j j
in out
n n n n n = =
(3.13)
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where
isysn = initialtotal # of moles of
all speciesin the system
fsysn = finaltotal # of moles of
all speciesin the system
sysijn = # of moles of species jin the
system at the initialstate
sysfjn = # of moles of species jin the
system at thefinalstate
Eqs. 3.12 & 3.13 can also be written in
the rate form (i.e.with respect to time),
as follows
Overall balance
sys
in out
dndn dn
dt dt dt
= (3.14)
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Species balance
sysjj j
in out
dndn dn
dt dt dt =
(3.15)
In the case that there is a Rxn/are
Rxns, mole balanceis still applicable,
but only for species balances(NOT for
an overall balance), as follows
.sysj j j j
in out Rxn
n n n n + = (3.16)
or, in the rate form
.
sysjj j j
in out Rxn
dndn dn dn
dt dt dt dt + =
(3.17)
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Example An experiment on the growth rate of
organisms requires an environment of humid
air enriched in oxygen
Three input streams are fed into an
evaporation chamber:
liquid water, fed at the volumetric flow
rate of 20 cm3/min
air(21.0 vol% O2& 79 vol% N2)
pure O2, with the molar flow rate of
one-fifth (1/5) of the stream of air
to produce an output stream with the desired
composition
The output gas is analysed and is found to
contain 1.5 mol% of water
Calculate all unknowns
(Data: Density of water is 1.0 g/cm3; and
MW of water = 18.02, of O2= 32.00, & of
N2= 28.02)
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Assumption: Steady-state process
Flow chart:
Since the process is steady-state, (as we
made an assumption),
0sysdn
dt =
Eq. 3.14:
sys
in out
dndn dn
dt dt dt = (3.14)
Evaporation
chamber
Air
21 vol% O2
79 % N2
Pure O2
1
3
2
4
H2O (l)
20 cm3/min
Output stream
1.5 mol% H2O
x % N2
y % O2
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is then reduced to
0in out
dn dn
dt dt
=
orin out
dn dn
dt dt = (3.18)
From the flow chart (see Page 33), we
can write an overallmole balance equation,
as follows
2 31 4dn dndn dn
dt dt dt dt
+ + =
(3.19)
Let i idn
n
dt
=
Thus, Eq. 3.19 can be re-written as
follows
[ ]1 2 3 4n n n n + + = (3.20)
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It is given that
2 1 1
10.2
5
n n n= = (3.21)
Combining Eq. 3.21 with Eq. 3.20 gives
( )[ ]1 1 3 40.2n n n n + + =
and
1 3 41.2n n n+ = (3.22)
It is also given that the volumetricflow
rate of stream 3 = 20 cm3/min
Thus, the massflow rate of stream 3 can
be calculated as follows
3
3
cm g g20 1.0 20
min cm min
=
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which can be converted to the molarflow
rate as follows
20 g 1 g-mol= 1.11 g-mol/min
min 18.02 g
Thus,
3n = 1.11 g-mol/min
Performing species balances:
H2O balance:
( )23 H O 44
n x n= (3.23)
where
( )2H O 4x = mole fraction of H2O in
stream 4
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Substituting corresponding numerical
values into Eq. 3.23:
( )23 H O 44n x n= (3.23)
results in
4
4
1.51.11
10074.0 g-mol/min
n
n
=
=
Substituting 3n & 4n into Eq. 3.22:
1 3 41.2n n n+ = (3.22)
gives
1 3 4
1
1.2
1.2 1.11 74.0
n n n
n
+ =
+ =
1 60.7 g-mol/minn =
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Thus, the flow rate of stream 2 can be
computed, using Eq. 3.21, as follows
( )
2 115
160.7
5
n n=
=
2 12.1 g-mol/minn =
N2balance:
( ) ( )
( ) ( )
2 2N 1 N 41 4
79 60.7 g-mol/min 74.0 g-mol/min100 100
64.8 mol%
x n x n
x
x
=
=
=
(note that, for gases, vol% = mol%)
In other words, mol% of N2in the output
stream (stream 4) is 64.8%
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Thus, mol% of O2in the output stream
(stream 4) or the value of ycan be
computed as follows
2 2 2mol% O 100 mol% H O mol% N
100 1.5 64.8
33.7 %mol
y
=
=
=
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Finally, we can summarise our
calculations in the following Table:
INPUT OUTPUT
Stream 1*
1 60.7n =
Stream 2*
2 12.1n =
Stream 3*
3 1.11n =
Stream 4*
4 74.0n =
Composition
(mol% orvol%)
Composition
(mol% orvol%)
Composition
(mol% orvol%)
Composition
(mol% orvol%)
O2 21O2 100 H2O 100
H2O 1.5
N2 79 O2 33.7
N2 64.8
* Flow rateof each stream is in the
unitof g-mol/min