(B.Sc. II Semester Industrial Chemistry) By Wasi ur Rahman ...
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Material Balance (B.Sc. II Semester Industrial Chemistry) By Wasi ur Rahman Department of Chemistry AMU Aligarh
Transcript of (B.Sc. II Semester Industrial Chemistry) By Wasi ur Rahman ...
Material Balance (B.Sc. II Semester Industrial Chemistry)
By
Wasi ur Rahman
Department of Chemistry
AMU Aligarh
Flow Chart
1. Represent each process unit by a simple box or any other simple.
2. Feeds and Products are represented by lines and arrows.
Process Unit
Process Unit
Process Unit
Total mass (molar) f low rate
Mass (molar) composition
OR
Other variables like temperature and pressure
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Component Mass (molar)
Flow rates
Other variables like temperature and pressure
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3. Assign algebraic symbols to unknown stream variables associating them with
units in parenthesis.
n (mol/min)
0.21 mol O2/mol
0.79 mol N2/mol
400 mol/min
y (mol O2/mol)
(1-y)(mol N2/mol)
AND
Flow Chart
4. The following variables are normally used to describe:
Molar quantity (mole unit)
Volume (volume unit)
Mass (mass unit)
Molar flow rate (mol/time)
Volumetric flow rate (volume/time)
Mass flow rate (mass/time)
Temperature (temperature unit)
Pressure (pressure unit)
n
V
m
n
V
m
T
P
Flowchart Scaling and Basis of Calculation
Consider the following:
1 kg C6H6
1 kg C7H8
2 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
10 kg C7H8
20 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
×10
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Scaling Up
The final quantities are larger than the original quantities
10 kg C6H6
1 kg C6H6
1 kg C7H8
2 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
0.5 kg C7H8
1.0 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
×1/
2
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Scaling Down
The final quantities are smaller than the original quantities
0.5 kg C6H6
Flowchart Scaling and Basis of Calculation
4.3 Flowchart Scaling and Basis of Calculation
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Summary:
1. Balanced processes can be always scaled up or down by multiplying all steams of the
old process by a factor while maintaining stream compositions unchanged.
2. the scale factor is defined as :
n amount (flowrate) of new stream
amount (flowrate) of the corresponding old
stream
3. When balanced processes are scaled up or down, compositions of all streams must
remain unchanged.
Balancing Processes
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Consider the following system of equations:
2x 5y 10
x 3y 3
The solution set is (15, -4)
The equations in the above system is recognized as independent equations, i.e. can’t be
derived algebraically from each other. This results in a unique solution set
Consider the following system of equations:
2x 5y 10
4x 10 y 20
The equations in the above system is recognized as dependent equations, i.e. Eq(2) =
2×Eq(1). This results in infinitely many solutions!
Balancing Processes
3.0 kg C6H6/min
1.0 kg C7H8/min
m (kg/min )
x (kg C6H6/kg)
(1-x) (kg C7H8/kg)
Total mass balance:
C6H6 balance:
C7H8 balance:
3.0 1.0 m
3.0 x m
1.0 1 x m
(1)
(3)
(2)
Conclusion:
For two components, we can write two balance equations and for “n” components we can
write “n” balance equations plus one total mass balance equation = (n+1) equations.
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Balancing Processes
Dependant and Independent Equations:
Addition of (2) and (3) yields:
3.0 1.0 xm 1 xm
Substation of (2) and (1) yields:
3.0 1.0 3.0 m xm
Substation of (3) and (1) yields:
3.0 1.01.0 m 1 xm
Total mass balance equation
Toluene balance equation
Benzene balance equation
Conclusion:
Although we can write three balance equations, we can only solve for two variables as
they are not linearly independent.
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Balancing Processes
Summary:
Total number of balance equation which can be written on physical systems equals
number of components +1
Number of independent balance equations which can be written on physical systems
equals the number of the chemical components.
Try always to write balances which involve the fewest number of unknowns
variables.
Degree-of-freedom Analysis
Is there a way which can save
my time and tell me if the
mass balance problem is
solvable or not?
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Degree-of-freedom Analysis
Solvable problems mean that enough information is available so that all variable
can be uniquely determined.
Degree of Freedom Analysis:
is an analysis which checks if enough information is available so that mass balance
problems are solvable or not before attempting to derive any equations.
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DF nunknowns nindep.equations
DF = 0
DF > 0
DF < 0
Correctly specified system
Underspecified system, i.e. more unknowns are there than number of
equations.
Overspecified system, i.e. more information is available than required.
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Degree-of-freedom Analysis
Sources of Equations
Sources of equations relating unknown process stream variables include
the following:
1. Material balances
2. An energy balance
3. Process specifications
4. Physical properties and laws.
5. Physical constraints.
General Procedure for Single-Unit Process Material Balance Calculations
1. Choose as a basis of calculation an amount or flow rate of one of the process
streams.
2. Draw a flowchart and fill in all known variable values, including the
basis of calculation. Then label unknown stream variables on the chart.
3. Express what the problem statement asks you to determine in terms of the
labeled variables. You will then know which unknowns you have to
determine in order to solve the problem.
4. If you are given mixed mass and mole units for a stream (such as a total
mass flow rate and component mole fractions or vice versa), convert all
quantities to one basis.
5. Do the degree-of-freedom analysis.
6. If the number of unknowns equals the number of equations relating them (i.e., if
the system has zero degrees of freedom), write the equations in an efficient
order (minimizing simultaneous equations) and circle the variables for which you
will solve.
7. Start with equations that only involve one unknown variable, then pairs
of simultaneous equations containing two unknown variables, and so on.
8. Solve the equations.
9. Calculate the quantities requested in the problem statement if they have not
already been calculated.
Solved Example
Problem
A liquid mixture containing 45.0% benzene (B) and 55.0% toluene (T) by mass is
fed to a distillation column. A product stream leaving the top of the column (the
overhead product) contains 95.0 mole% B, and a bottom product stream contains
8.0% of the benzene fed to the column (meaning that 92% of the benzene leaves
with the overhead product). The volumetric flow rate of the feed stream is 2000
L/h and the specific gravity of the feed mixture is 0.872. Determine the mass flow
rate of the overhead product stream and the mass flow rate and composition (mass
fractions) of the bottom product stream.
Solution Choose a basis. Having no reason to do otherwise, we choose the given feed
stream flow rate (2000 L/h) as the basis of calculation.
Draw and label the flowchart
Choose a basis. Having no reason to do otherwise, we choose the given feed stream flow rate (2000 L/h) as the basis of calculation. Draw and label the flowchart
Feed
2000 L/h
m•
1(kg/h)
Overhead Product
m•
2(kg/h)
0.45 kg B/kg
0.55 kg T/kg
0.95 mol B/mol
0.05 mol T/mol
Bottom Product
m•
B3(kg B/h) (8% of B in feed)
m•
T3(kg T/h)
yB2(kg B/kg)
(1 – yB2)(kg T/kg)
Note several points about the flowchart labeling:
◆A volumetric flow rate is given for the feed stream, but mass flow rates and fractions will be
needed for balances. The mass flow rate of the stream should therefore be considered an unknown
process variable and labeled as such on the chart. Its value will be determined from the known
volumetric flow rate and density of the feed stream.
◆Since mass balances will be written, the given component mole fractions in the overhead product
stream will have to be converted to mass fractions. The mass fractions are accordingly labeled as
unknowns.
◆We could have labeled the mass flow rate and mass fractions of the bottom stream as we did the
overhead. However, since we have no information about either the flow rate or composition of this
stream, we have instead labeled the component flow rates (following the rule of thumb given in
Step 2 of the general procedure).
◆Every component mass flow rate in every process stream can be expressed in
terms of labeled quantities and variables. (Verify this statement.) For example,
the flow rates of toluene (kgT/h) in the feed, overhead, and bottom streams are,
respectively, 0.55m , m (1 — y ), andmT3. The flowchart is therefore labeled
completely.
◆The 8%–92% benzene split between the product streams is not a stream flow rate
or composition variable; nevertheless, we write it on the chart to remind
ourselves that it is an additional relation among the stream variables and so
should be included in the degree-of-freedom analysis.
3. Write expressions for the quantities requested in the problem statement. In terms of
the quantities labeled on the flowchart, the quantities to be determined are m (the overhead product mass flow rate), m3 = mB3 + mT3 (the bottom product mass flow rate), xB = mB3/m3
(the benzene mass fraction in the bottom product), and xT = 1 — xB (the toluene mass
fraction). Once we determine m2 , mB3 , and mT3 , the problem is essentially solved.
4. Convert mixed units in overhead product stream
Basis: 100 kmol overhead 95.0 kmol B, 5.00 kmol T
(95.0 kmol B) × (78.11 kg B/kmol B) = 7420 kg B, (5.00 × 92.13) = 461 kg T
(7420 kg B) +(461 kg T) = 7881 kg mixture
yB2 = (7420 kg B)/(7881 kg mixture) = 0.942 kg B/kg (write on chart)
5. Perform degree-of-freedom analysis. 4 unknown m1, m2, mB2, mT3
—2 material balances (since there are two molecular species in
this nonreactive process)
—1 density relationship (relating the mass flow rate to the given
volumetric flow rate of the feed)
—1 specified benzene split (8% in bottom–92% in overhead)
0 degrees of freedom
The problem is therefore solvable.
6. Write system equations and outline a solution procedure. The variables for which each equation will
be solved are circled.
◆Volumetric flow rate conversion. From the given specific gravity, the density of the feed stream is
0.872 kg/L. Therefore,
m = 2000 kg
L )(0.872 ) ( L
h
◆ Benzene split fraction. The benzene in the bottom product stream is 8% of the benzene in the feed
stream. This statement translates directly into the equation mB3 = 0.08(0.45m1)
There are two unknowns remaining on the flowchart (m2 and mT3 ), and we are allowed to
write two balances. Balances on total mass and on toluene each involve both unknowns, but a benzene
balance only involves m2, so we begin with that one.
Benzene balance 0.45m1 = m2 yb2 +m B3
Toluene balance 0.55m1 = (1 — yB2 )m2 + mT3
7. Do the algebra. The four equations may be solved manually.
Each newly calculated variable value should be written on the flowchart for ease of
reference in the remainder of the solution.
The results are m1 = 1744 kg/h, mB3 = 62.8 kg benzene/h, m2 = 766 kg/h , and
mT3 = 915 kg toluene/h.
A total mass balance (which is the sum of the benzene and toluene balances) may
be written as a check on this solution:
m1 = m2 + mB3 + mT3 1744 kg/h = (766 + 62.8 + 915) kg/h = 1744 kg/h
8. Calculate additional quantities requested in the problem statement. m3 = mB3 + mT3 = 62.8 kg/h + 915 kg/h = 978 kg/h
yB3=mB3/mT3=62.8 kg B/ 978 kg/h = 0.064 kg B/kg
yT3 = 1 — yB3 = 0.936 kg T/kg
