Download Mass Balance presentation

The accounting of all mass in an industrial
chemical process is referred to as a mass (or
material) balance.


‘day to day’ operation of process for
monitoring operating efficiency
Making calculations for design and
development of a process i.e. quantities
required, sizing equipment, number of items of
equipment
200 kg of a 40% w/w methanol/water solution is
mixed with 100 kg of a 70% w/w
methanol/water solution in a batch mixer unit.
What is the final quantity and composition?
Total initial mass = total final mass = 300 kg
Initial methanol mass = final methanol mass
80 + 70 = final methanol mass = 150 kg
Therefore final composition of batch is (150/300)
x 100 = 50 % by wt.
1000 kg of 8% by wt. sodium hydroxide (NaOH)
solution is required. 20% sodium hydroxide
solution in water and pure water are available.
How much of each is required?
Batch processes operate to a batch cycle and are
non-steady state. Materials are added to a
vessel in one operation and then process is
carried out and batch cycle repeated. Integral
balances are carried out on batch processes
where balances are carried out on the initial
and final states of the system.



Sequence of operations/steps repeated
according to a cycle
Batch cycle time
Batch size
Paul Ashall, 2008
3 steps
Add reactants etc
Start cycle t=0
reaction
Empty reactor
Next cycle
t, finish cycle
These processes are continuous in nature and
operate in steady state and balances are carried
out over a fixed period of time. Materials enter
and leave process continuously.
When there is no net accumulation or depletion of
mass in a system (steady state) then:
Total mass entering system = total mass leaving
system
or total mass at start = total final mass
Input + generation – output – consumption =
accumulation
Notes: 1. generation and consumption terms refer only to generation
of products and consumption of reactants as a result of chemical
reaction. If there is no chemical reaction then these terms are zero.
2. Apply to a system
3. Apply to total mass and component mass





System – arbritary part or whole of a system
Steady state/non-steady state
Accumulation/depletion of mass in system
Basis for calculation of mass balance (unit of
time, batch etc)
Component or substance
1000 kg of a 10 % by wt. sodium chloride solution
is concentrated to 50 % in a batch evaporator.
Calculate the product mass and the mass of
water evaporated from the evaporator.
F2
F1
F4
F3
Paul Ashall, 2008
Calculate E and x
evaporator feed E, composition x%
Fresh feed 1000kg, 15%
by wt sodium hydrogen carbonate
Recycle stream 300 kg, 10% satd. soln.



Streams
Operations/equipment sequence
Standard symbols

Process flow diagram
Recycle of unreacted material
reactor
Fresh feed
(reactants, solvents,
reagents, catalysts etc)
Separation &
purification
waste
product
Byproducts/coproducts
A 1000 kg batch of a pharmaceutical powder
containing 5 % by wt water is dried in a double
cone drier. After drying 90 % of the water has
been removed. Calculate the final batch
composition and the weight of water removed.
1000 kg of a 20% by wt mixture of acetone in
water is separated by multistage batch
distillation. The top product (distillate)
contains 95% by wt. acetone and the residue
still contains 2% acetone. Calculate the amount
of distillate.
It is often useful to calculate a mass balance using
molar quantities of materials and to express
composition as mole fractions or mole %.
Distillation is an example, where equilibrium data
is often expressed in mole fractions.






A mole is the molecular weight of a substance expressed
in grams
To get the molecular weight of a substance you need its
molecular formula and you can then add up the atomic
weights of all the atoms in the molecule
To convert from moles of a substance to grams multiply
by the molecular weight
To convert from grams to moles divide by the molecular
weight.
Mole fraction is moles divided by total moles
Mole % is mole fraction multiplied by 100
Benzene is C6H6. The molecular weight is (6x12) +
(6x1) = 78
So 1 mole of benzene is 78 grams
1 kmol is 78 kg
Paul Ashall, 2008
1000 kmol of an equimolar mixture of benzene
and toluene is distilled in a multistage batch
distillation unit. 90 % of the benzene is in the
top product (distillate). The top product has a
benzene mole fraction of 0.95. Calculate the
quantities of top and bottom products and the
composition of the bottom product.
1.
Read the problem and clarify what is to be
accomplished.
A train is approaching the station at 105 cm/s. A
man in one car is walking forward at 30 cm/s relative
to the seats. He is eating a foot-long hot dog which is
entering his mouth at the rate of 2 cm/s. An ant on
the hot dog is running away from the man’s mouth at
1 cm/s. How fast is the ant approaching the station?
Paul Ashall, 2008
2. Draw a sketch (flow diagram)
3. Label. Assign symbols to each variable.
4. Put down the known values
5. Select the basis.
6. List the symbols
7. Write down independent equations
8. Count the numbers
9. Solve the equations
10. Check the answer.
Paul Ashall, 2008







Process description
Flowsheet
Label
Assign algebraic symbols to unknowns
(compositions, concentrations, quantities)
Select basis
Write mass balance equations (overall, total,
component, unit)
Solve equations for unknowns
Paul Ashall, 2008
Paul Ashall, 2008
1000 kg of a 20% by wt mixture of acetone in
water is separated by multistage batch
distillation. The top product (distillate)
contains 95% by wt. acetone and the residue
still contains 2% acetone. Calculate the amount
of distillate.
wash water/solvent
solid
feed suspension
waste water
filtrate
5000 kg DM water
F1
Impurity 55 kg
Water 2600 kg
API 450 kg
Water 7300 kg
Impurity 50 kg
API 2kg
Water 300 kg
API 448 kg
Impurity 5 kg
water/evaporated solvent
feed
product
A+B
A+B
S
S+B
A – feed solvent; B – solute; S – extracting solvent
feed
raffinate
E1
solvent
extract
Paul Ashall, 2008
The next step in the synthesis of aspirin in water
is to extract the aspirin with chloroform in a batch
process. A 195-kg particular batch containing 0.11
kg aspirin/kg water is extracted with 596 kg
chloroform. Extraction coefficient is give by
y=1.72x where y = kg aspirin/kg chloroform and
x = kg aspirin/kg water in raffinate. Calculate
amount and composition of extract and raffinate
Paul Ashall, 2008
F = 195 kg; xf = 0.11 kg API/kgwater
S = 596 kg chloroform
y = 1.72x, where y is kgAPI/kg chloroform in extract and x is kg API/kg
water in raffinate.
Total balance 195 + 596 = E + R
API balance 19.5 = 175.5x1 + 596y1
19.5 = 175.5x1 + 596.1.72x1
x1 = 0.0162 and y1 = 0.029
R is 175.5 kg water + 2.84 kg API
and E is 596 kg chloroform + 17.28 kg API
Note: chloroform and water are essentially immiscible
exit gas stream
feed solvent
feed gas stream
waste solvent stream
Paul Ashall, 2008
A crystallizer contains 1000 kg of a saturated solution of potassium
chloride at 80 deg cent. It is required to crystallise 100 kg KCl from
this solution. To what temperature must the solution be cooled?
Paul Ashall, 2008
T deg cent
Solubility gKCl/100 g
water
80
51.1
70
48.3
60
45.5
50
42.6
40
40
30
37
20
34
10
31
0
27.6
At 80 deg cent satd soln contains (51.1/151.1)x100
% KCl i.e. 33.8% by wt
So in 1000 kg there is 338 kg KCl & 662 kg water.
Crystallising 100 kg out of soln leaves a satd soln
containing 238 kg KCl and 662kg water i.e.
238/6.62 g KCl/100g water which is 36 g
KCl/100g. So temperature required is approx
27 deg cent from table.



Overall balance
Unit balances
Component balances
Paul Ashall, 2008
W2
F1
E
C
F
P3
R4
E – evaporator; C – crystalliser; F – filter unit
F1 – fresh feed; W2 – evaporated water; P3 – solid product; R4 – recycle of
saturated solution from filter unit
Paul Ashall, 2008







Process description
Flowsheet
Label
Assign algebraic symbols to unknowns
(compositions, concentrations, quantities)
Select basis
Write mass balance equations (overall, total,
component, unit)
Solve equations for unknowns
Paul Ashall, 2008







Stoichiometric quantities
Limiting reactant
Excess reactant
Conversion
Yield
Selectivity
Extent of reaction
Paul Ashall, 2008
Refers to quantities of reactants and products
in a balanced chemical reaction.
aA + bB
cC + dD
i.e. a moles of A react with b moles of B to give c
moles of C and d moles of D.
a,b,c,d are stoichiometric quantities

Paul Ashall, 2008

In practice a reactant may be used in excess of
the stoichiometric quantity for various reasons.
In this case the other reactant is limiting i.e. it
will limit the yield of product(s)
Paul Ashall, 2008
A reactant is in excess if it is present in a quantity
greater than its stoichiometric proportion.
% excess = [(moles supplied – stoichiometric
moles)/stoichiometric moles] x 100
Paul Ashall, 2008
Paul Ashall, 2008


Fractional conversion = amount reactant
consumed/amount reactant supplied
% conversion = fractional conversion x 100
Note: conversion may apply to single pass reactor
conversion or overall process conversion
Paul Ashall, 2008
Yield = (moles product/moles limiting reactant
supplied) x s.f. x 100
Where s.f. is the stoichiometric factor =
stoichiometric moles reactant required per
mole product
Paul Ashall, 2008
Paul Ashall, 2008
Selectivity = (moles product/moles reactant
converted) x s.f. x100
OR
Selectivity = moles desired product/moles
byproduct
Paul Ashall, 2008
Extent of reaction = (moles of component leaving
reactor – moles of component entering
reactor)/stoichiometric coefficient of component
Note: the stoichiometric coefficient of a component in
a chemical reaction is the no. of moles in the
balanced chemical equation ( -ve for reactants and
+ve for products)
Paul Ashall, 2008
A
B
i.e. stoichiometric coefficients a = 1; b = 1
100 kmol fresh feed A; 90 % single pass
conversion in reactor; unreacted A is
separated and recycled and therefore overall
process conversion is 100%
R
F
reactor
separation
Paul Ashall, 2008
P

Elementary Principles of Chemical Processes, R. M.
Felder and R. W. Rousseau, 3rd edition, John
Wiley, 2000
Paul Ashall, 2008