mass balance non reactive 1
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Mass Balance
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ECB 3013 MATERIAL AND ENERGY
BALANCES
The law of conservation of mass states that for any processunit (s), mass can neither be created nor destroyed but it
can change from one phase to another phase or converted
into other forms through reaction process.
The law concludes that:
Total mass INPUT = Total mass OUTPUT
FUNDAMENTAL OF MATERIAL BALANCES:
CONSERVATION OF MASS
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PROCESS CLASIFICATION
Chemical process may be classified as:
1. Batch process: The feed is charged into a process unit at the
beginning of the process and the output is removed from the process
unit at the end of the process.
2. Continuous process: The inputs and outputs flow continuouslythroughout the duration of the process.
3. Semi batch: Any process that is neither batch nor continuous.
If the process variables such as flow rates, temperatures, pressures,
volume etc do not change with time, the process is said at STEADYSTATE. On the other hand, if the process variables do change with
time, the process is said at UNSTEDY STATE or TRANSIENT. By
nature, batch and semi-batch process are unsteady state operation,
while continuous process is steady state operation.
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EXAMPLE 1
Process Classification Remark
A balloon is filled with air
at a steady state rate of 2
g/min
Water is boiled in an
open flaskWater is boiled in an
closed flask
Gasoline from car tank
Carbon dioxide andsteam are fed into
reactor to form carbon
dioxide and hydrogen
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EXAMPLE 1
Process Classification Remark
A balloon is filled with air
at a steady state rate of 2
g/min
Semi-batch, transient Volume, T, P change
with time.
Water is boiled in an
open flask
Semi-batch, transient Volume, T, P change
with time.Water is boiled in an
closed flask
Batch, transient T & P change with
time.
Gasoline from car tank Semi-batch, transient Volume change with
time
Carbon dioxide and
steam are fed into
reactor to form carbon
dioxide and hydrogen
Continuous, steady state All process variables
do change with time.
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MASS BALANCE EQUATION
Generally, the mass balance equation for any process unit is
given as:
Input + Generation - Output - Consumption = Accumulation
Enters
through
system
boundaries
Produced
within
system only
for reactive
system
Leaves
through
system
boundaries
Consumed
within
system only
for reactive
system
Buildup within
system only for
transient operation
[1]
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Each year 50,000 people move into a city, 75,000 people move
out, 22,000 are born and 19,000 die. Write a balance on the
population of the city.
Solution:
Let P denote to people:
Input + generation – output – consumption = accumulation
50,000 P/yr + 22,000 P/yr – 75,000 P/yr – 19,000 P/yr = A P/yr
A = -22,000 P/yr
Each year the city’s population decreases by 22,000 people.
EXAMPLE 2
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1) If the balanced quantity is total mass, set generation = 0 and
consumption = 0. Except in nuclear reactions, mass can
neither be created nor destroyed.
2) If the balanced substance is a nonreactive species (neither a
reactant nor a product), set generation = 0 and consumption
= 0.
3) If a system is at steady state, set accumulation = 0,
regardless of what is being balanced. By definition, in
steady-state system nothing can change with time, including
the amount of the balanced quantity.
Rules to simplify the material balance equation
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MASS BALANCE EQUATION
Balance on Continuous Steady-State Processes
For continuous steady-state operation for reactive system, the
accumulation term is equal to zero. Hence, EQ [1] becomes,
Input + Generation - Output - Consumption = 0 [2]
For continuous steady-state operation for non reactive system,
EQ [2] is simplified as,
Input = Output [3]
See example 4.2-2 in Text book
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MASS BALANCE EQUATION
Balances on Batch Processes
For batch processes, the input and output in Eq [1] is zero, and
the equation is simplified as,
For batch processes, accumulation is defined as,
Final Output - Initial Input = Accumulation [5]
[4]Generation - Consumption = Accumulation
Hence by equating Eq [4] and [5], for batch processes, the mass balance equation is given as,
Initial Input + Generation = Final Output + Consumption [6]
See example 4.2-3 in Text book
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ECB 3013 MATERIAL AND ENERGY
BALANCESSteps in Solving Material Balance
1. Read and try to understand on the process description. What type of processunit used and what type of process operation.
2. Draw a flowchart for the process description using boxes or other symbols to
represent process unit or unit operation (reactors, mixers, etc.) and lines with
arrows to represent inputs and outputs.
MASS BALANCE CALCULATION
DISTILLATION
COLUMN
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3. Write all known stream variables i.e. inputs & outputs, on the flowchart.
4. Assign algebraic symbols to unknown stream variables.(e.g., m1 , n1 , yCO etc.)
5. 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 or the other.
MASS BALANCE CALCULATION
DISTILLATION
COLUMN
0.55 kg T/kg
0.05 mol T/mol
mT3 kg T/kg
2000 L/h
0.45 kg B/kg
0.95 mol B/mol
mB3 kg B/kg 8% of B in feed
m 2 kg/h
m 1 kg/h
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6. Do the degree of freedom analysis. Count unknowns and identifyequations that relate them (zero degree of freedom). The equation:material balances, an energy balance, process specifications, physical
property relationships and laws, physical constraints and stoichiometricrelations.
7. Take basis of calculation. If no stream amount or flow rate is specified inthe problem statement, take as basis an arbitrary amount or flow rate ofthe stream with known composition (e.g., 100 kg or 100 kg/h or 100 molor 100 mol/h if all mass/mole fractions are known.)
8. Write mass balance equation for the overall system and for specificcomponent using selected Eq [1] to [6].
9. Perform mass balance for the process description. Always check theoverall mass balance for Total Inputs = Total Outputs
MASS BALANCE CALCULATION
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A liquid mixture of benzene (B) and toluene (T) containing
55% B by mass is fed continuously to a distillation column
with a feed rate of 100 kg/h. A product stream leaving the top
of the column (overhead product) contains 85% B and a
bottom product stream contains 10.6% B by mass. Determinethe mass flow rate of the overhead product stream and the
mass flow rate of the bottom product stream.
PROBLEM 4.3
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DISTILLATION
COLUMN
0.450 kg T/kg
0.150 kg T/kg
0.894 kg T/kg
100 kg/h
0.550 kg B/kg
0.850 kg B/kg
0.106 kg B/kg
m V kg/h
m L kg/h
STEPS 2, 3 & 4
EXAMPLE 3
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STEPS 5-DoF
DoF = unknown - independent equation
Unknown = 2
Independent equation = 2
Note: DoF must be zero to be solvable
mV
m L &
Material Balance for B and T
DoF = unknown - independent equation
= 2 - 2
= 0 (problem solvable)
EXAMPLE 3
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Take the basis of calculation = 100 kg/h of feed
Since this operation is at steady state and non-reactive
system, hence
Input = output
DC
0.450 kg T/kg
0.150 kg T/kg
0.894 kg T/kg
100 kg/h
0.550 kg B/kg
0.850 kg B/kg
0.106 kg B/kg
m V kg/h
m L kg/h
Total balance:
100 kg/h = m V m
L + A
Benzene balance:
100 (0.550)
h
Bkg
= 0.850 m V
0.106 m L
B+
STEPS 6 & 7
EXAMPLE 3
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Solve equation A and B simultaneously
The results are
m V
m L
= 59.7 kg/h
= 40.3 kg/h
EXAMPLE 3
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ECB 3013 MATERIAL AND ENERGY
BALANCES• Use to represent the process unit (reactors, mixers,
separation units)
• Lines with arrow - to represent inputs and outputs
Flowcharts
Flowchart scaling and basis of calculation
• Scaling up – final stream quantities are larger than the
original quantities
• Scaling down – final stream quantities are smaller than
the original quantities
• Basis of calculation – an amount (mass or moles) or
flow rate (mass or moles) of one stream
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Balancing a process
For non reactive process:
1. Maximum no of independent equations equals the number of
chemical species in the input and output
Ex: input has B and T,
• Independent equations involve mass or mole balance on Band T
• Balance on total mass or moles
2. Write balances first that involve the fewest unknown variables
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A stream containing four components (14 % w
/w A,36.1% w/w B, 23.6% w/w C and 26.3% w/w D) flows
at a rate of 984.0 kg/hr into a separator. The
separator produces two streams of differing
compositions. The upper product stream has a
composition of 16.5 % w/w A and 40.9 % w/w B with
C and D making up the remainder. The lower
product stream contains all four components but
only the weight percentages of B (20.9% w/w) and
C (37.6% w/w) are known. Draw and label theprocess and calculate the compositions and flow
rates of the two product streams.
PROBLEMS
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Degree of freedom
•
A way to identify that enough information isavailable for material balance calculation
• ndf = nunknowns – nindep eqns
– ndf =0, can be solved
– ndf >0, relations has been overlooked, problem is
underspecified
– ndf <0, flowchart is incompletely labeled, problem
is overspecified with redundant and inconsistentrelations
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Degree of freedom
•
Sources of relating unknown variables:1. Material balance
2. Energy balance
3. Process specification –
related4. Physical properties and laws
5. Physical constraint
6. Stoichiometric relations
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PROBLEMS
See problems 4.6, 4.7, 4.9, 4.10
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ECB 3013 MATERIAL AND ENERGY
BALANCESPROBLEM 4.20 OF TEXT BOOK
Wet air containing 4.0 mole% water vapor is passedthrough an adsorption column containing calcium chloride
pellets. The pellets adsorb 97.0% of the water and none
of the other constituents of the air. The column packing
was initially dry and had a mass of 3.40 kg. Following 5.0
hours of operation, the pellets are reweigh and found to
have a mass of 3.54 kg. Calculate the molar flow rate
(mol/h) of the feed gas and mole fraction of water vapor in
the product gas.
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)/(1 hmol n
)/(2 hmol n
x mol H2O / mol(1 - x ) mol DA / mol
)/( 23 hadsorbed O H mol n
0.040 mol H2O / mol
0.960 mol DA / mol
97.0% of H2O in feed
Adsorption Unit
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Need H2O
Adsorption Rate
Problem Statement: xnnn ,,, 321
4 unknowns :3 independent equations
Degree of Freedom:
DoF = Unks. - IE = 4 – 3
DoF = + 1
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Overall mass balance
INPUT=OUTPUT + ACCUMULATION
321 nnn 1
Component mass balances
321040.0 nn xn H2O :
21 )1(960.0 n xn -DA :
2
3
Water in CaCl2 pellets
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Take the basis of calculation= 100 mol/hr of feed stream
mol/h12.96
[3]insubsituteand0.12/x,
/88.3497.0
ascalculated becanthatknow
)1(96
4
100
2
2
3
3
2
32
2
32
321
-
n
n Hence
hmol n
nwe
n x
DA
nn x
O H
nn
nnn
[1]
[2]
[3]
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Stream INPUT OUTPUT
H2O
(mol/h)
DA
(mol/h)
H2O
(mol/h)
DA
(mol/h)
n1 4 96 - -
n2 - - 0.12 96
n3 - - 3.88 -
TOTAL 4 96 4 96
Take the basis of calculation= 100 mol/hr of feed stream
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Stream INPUT OUTPUT TOTALH2O
(mol/h)
DA
(mol/h)
H2O
(mol/h)
DA
(mol/h)
n1 1.6 38.4 - - 40
n2 - - 0.048 38.4 38.448
n3 - - 1.552 - 1.552
TOTAL 1.6 38.4 1.6 38.4
AdsorptionRatehO H mol O H kg
O H mol
h
kg n /56.10180.0
1
5
)40.354.3(2
2
23
-
Based on the given problem,
Scale down factor = 1.56/3.88 = 0.40. hence the new INPUT/OUTPUT values
will be multiplied by a factor of 0.4 to meet the process description.
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Adsorption
RatehO H mol
O H kg
O H mol
h
kg n /56.1
0180.0
1
5
)40.354.3(2
2
23
-
hmol nnnn /54.3856.11.402321 - Total balance :
Given: 97% H2O (of the input stream) is adsorbed. Therefore
hmol nn /1.4056.1)040.0(97.0 11
H2O balance : x = 1.25×10-3
Note : DA balance may also be used to calculate x.
Take basis calculation of 0.14 kg of water vapor adsorbed in 5 hour operation
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Gas absorption of gas scrubbing is a commonly usedmethod for removing environmentally undesired species
from waste gases in chemical manufacturing and
combustion processes. A waste gas containing 10 mol%
SO2 (a precursor of acid rain) and several other species
(collectively as A) is fed to scrubbing tower where it
contacts a solvent (B) that absorbs SO2. The ratio of
liquid stream to gas stream fed to the column is 5. The
solvent feed rate to the tower is 1000 mol/min. The
absorption unit has an efficiency of 98%. Assume that theabsorption of A and evaporation of B in the scrubbing can
be neglected, calculate the mole fraction of SO2 in the
liquid and gas streams.
LOOK LIKE PROBLEM 4.26 OF TEXT BOOK
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10 mol% SO2
90 mol% A
G1=200 mol/min
y1 mol% SO2y2 mol% A
G2
1000 mol/min B
L1=1000 mol/min
L2
x1 mol% SO2
x2 mol% S
Gasabsorption
unit
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Degree of Freedom
Problem Statement:G2, L2, x1, x2, y1 and y2 = 6 unknown
4 independent equations, 1 overall, 3 specific equations
Degree of Freedom:
DoF = Unks – IE= 6-4 =+2
The remaining 2 independent equations can be obtained from the column
efficiency.
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Overall Balance
G1 + L1 = G2 + L2
200 + 1000 = G2 +
L2
1200= G2 + L2
Solvent Balance
1000 = (1-x1)L2
L2=1000 +
(0.98)20
L2=1019.6mol/min
Hence, x1=0.019
Gas A Balance
0.9(200) = (1-y1)G2
180 = (1-y1)G2
G2=Gas A +
Remaining SO2 ingas A
G2=180 + (0.02)20
=180.4 mol/min
Hence, y1=2.21x10-3
Take basis of calculation L1= 1000 mol/min solvent
SO2 Balance
0.1(200) =
y1G2+x1L2
20 = y1G2 + x1L2