ERT 416/3
CHAPTER 3: FUNDAMENTALS OF
MATERIAL AND ENERGY BALANCES
IN BIOPROCESS PLANT
MISS. RAHIMAH BINTI OTHMAN
(Email: rahimah@unimap.edu.my)
COURSE OUTCOMES
OUTLINES
IDENTIFY and APPLY engineering principle for
the design process; UNDERSTAND and APPLY
the material balances calculations in process
design in order to study plant operation and
troubleshooting; IDENTIFY the energy
requirement of the process using energy
balances. COMPARE the engineering
calculation done on paper with suitable
process simulators like Aspen HYSYS or Super
Pro Design.
OUTLINES
 Engineering principle for the
design process.
 The material balances
calculations in process design- to
study plant operation and
troubleshooting.
 The energy requirement of the
process using energy balances.
 Comparison between engineering
calculation with suitable process
simulators like Aspen HYSYS or
Super Pro Design.
Q UIZ 1 ( QUESTION 1)
The general equations for material and energy balances.
Material out = Material in + Generation – Consumption - Accumulation
[Mass] :
rate of mass accumulation  rate of mass in  - rate of mass out
[Component i ] :
[component i]  rate of component i in  - rate of companent i out  rate of component i produced
Total energy entering the system – Total energy leaving the system
= Change in the total energy of the system
∆(Energy of the system) = E in – E out =∆U+∆EK+ ∆EP
[Energy ] :
rate of energy  rate of energy in


accumulati
on

 by convection
net rate of heat

rate
of
energy
out
 
 



addition
t
o
the
system
 
 

 by convection
 from the surroundings 


net rate of work performed



on the systemby the surroundings 
ASSIGNMENT 1( QUESTION 3)
Balances on a Batch Mixing Process
Two methanol-water mixtures are contained in separate flasks. The first mixture contains
40.0 wt % methanol, and the second contains 70.0 wt % methanol. If 200 g of the first
mixture is combined with 150 g of the second, what are the mass and composition of the
product?
Solution
Input = Output
Total mass balance : 200 g  150 g  m
m (g)  350 g
0.4 g CH 3OH
0.7CH 3OH
x(gCH 3OH )
Methanol balance : 200 g
 150 g
 m( g )
g
g
g
Known  m  350 g
 x  0.529 g CH 3OH/g
Verify!!
Water balance :
(200)(0.6)  (150)(0.3)  350 (1 - 0.529)
165 g H 2 O  165 g H 2 O
ASSIGNMENT
1 ( QUESTION 4)
Yield of a Crystallization Process
A salt solution weighing 10000 kg with 30 wt % Na2CO3 is cooled
to 293 K (20 °C). The salt crystallizes as the decahydrate. What
will be the yield of Na2CO3•10H2O crystals if the solubility is 21.5
kg anhydrous Na2CO3/100 kg of total water? Do this for the
following cases:
(a) Assume that no water is evaporated.
(b) Assume that 3% of the total weight of the solution is lost by
evaporation of water in cooling.
FIGURE3. Process flow for
crystallization
Q UESTION 4
Solution
The molecular weights are 106.0 for Na2CO3, 180.2 for
10H20, and 286.2 for Na2CO3 •10H2O. The process
flow diagram is shown in Fig. 3, with W being kg H2O
evaporated, S kg solution (mother liquor), and C kg
crystals of Na2CO3 •10H2O.Making a material balance
around the dashed line box for water for part
(a), where W = 0.
(13)
where (180.2)/(286.2) is wt fraction of water in the crystals.
Question 4
Making a balance for Na2CO3,
(14)
Solving the two equations simultaneously, C = 6370 kg of Na2CO3•10H2O
crystals and S = 3630 kg solution.
For part (b), W = 0.03(10000) = 300 kg H2O. Equation (13)
becomes
(15)
Equation (14) does not change, since no salt is in the W stream.
Solving Eqs. (14) and (15)simultaneously, C = 6630 kg of Na2CO3•10H2O
crystals and S = 3070 kg solution.
Q UIZ 1 ( QUESTION 5)
The diagram shows the main steps in a process for producing a
polymer. From the following data, calculate the stream flows for a
production rate of 10, 000 kg/h.
Filter wash water approx. 1 kg/ 1 kg polymer
Recovery column yield 98 % (percentage recovered)
Dryer feed ~ 5 % water, product specification 0.5 % H2O
Polymer losses in filter and dryer ~ 1 %
ANSWER
Only those flows necessary to illustrate the choice of system boundaries
and method of calculation are given in the solution.
Basis: 1 h
Take the first system boundary round the filter and dryer.
Take the next boundary round the reactor system; the feeds to the reactor can then be
calculated.
ANSWER
ANSWER
Now consider recovery system;
Composition of effluent is 22 kg monomer, 54,461 kg water.
Consider reactor monomer feed
OUTLINES
 Engineering principle for the
design process.
 The material balances
calculations in process design- to
study plant operation and
troubleshooting.
 The energy requirement of the
process using energy balances.
 Comparison between engineering
calculation with suitable process
simulators like Aspen HYSYS or
Super Pro Design.
W HAT
IS
M ATERIAL B ALANCE ?
Material balances (mass balances) are based on the fundamental “law of
conservation of mass (not volume, not moles)”. In particular, engineers
are concerned with doing mass balances around the specific processes.
 To describe the chemicals entering a process stream. Now, we must
learn how to:
1. Specify a process stream
2. Specify a process unit
3. Do a mass balance on a process unit
4. Do a mass balance on a sequence of process units.
Example 1:Input and Output Diagram
Water
Raw material
Energy
Auxiliary materials
Unit
Operation
Product
(expected)
byproduct
(usable)
Waste
Waste easily
assimilated by
the
environment
Inert waste
always available
toxic/dangerous
waste
Example 2: Input and Output
Raw
materials
Gaseous emissions
Plant, Process
or Unit
Operación
Catalyst
Air/Water
Energy
Recycle
Products
By-products
Wastewater
Liquid waste
Reusable residues in other
operation
Solid waste
Process Classification
1. Based on how the process varies with time.
a) Steady-state process is one that does not change with time. Every time we
take a snapshot, all the variables have the same values as in the first snapshot.
b) Unsteady-state (Transient) process is one that changes with time. Every time
we take a snapshot, many of the variables have different values than in the first
snapshot.
2. Based on how the process was built to operate.
a) A Continuous process is a process that has the feed streams and product
streams moving chemicals into and out of the process all the time. At every
instant, the process is fed and product is produced. Examples are an oil refinery,
a power grid and a steady salaried job.
b) A Batch process is a process where the feed streams are fed to the process to
get it started. The feed material is then processed through various process steps
and the finished products are created during one or more of the steps. The
process is fed and products result only at specific times. Examples are making a
batch of a product, like soup or a specialty chemical.
c) A Semi-batch process (also called semi-continuous) is a process that has some
characteristics continuous and batch processes. Some chemicals in the process
are handled batch-wise. Some chemicals are processed continuously.
System
Closed system
- Known as controlled mass
system (mass constant).
- Allow energy flow (in/out)
and volume is not
necessarily constant as a
controlled.
- Often undergo processes
during the internal energy
of the system changes.
∆(Energy of the system) + (∆Energy
of surroundings) = 0
∆(Energy of the system) = Q+W
∆(Energy of the system)
=∆U+∆EK+ ∆EP
Open system
- Known as controlled
volume system.
- Allow mass and energy
cross through the
boundary of open system
(= controlled surface).
Total Energy
E = U+ KE+ PE
Microscopic Energy
Internal energy, U
- The internal energy of a system
comes from the kinetic energy of
the molecules motion and potential
energy from the atomic bonds.
- Eg: chemical energy, nuclear, sensible
energy, latent energy.
Makroscopic energy
- Kinetic energy, KE
- Potential energy, PE
Thermodynamic
state and State
functions
The phase rule
(F = 2 – π +N)
The
reversible
process
Heat Capacity
At constant Volume,
Cv and Constant
Pressure, CP
CLOSED
SYSTEM
Equilibrium
(Eg: heat, mechanical,
phase, chemical)
Entalphy
(H = U +PV)
Constant-V &
Constant-P
Processes

dmCV
  (m) fs  0
dt



(2.25)

( m) fs  m 3  m1  m 2
dmCV
  ( uA) fs  0
dt
H 
( 2.26)
u 2
g

z  Q  Ws
2 gc gc
(2.32a)
Energy Balance
at Steady-state
Flow Processes
Mass
Balance
OPEN
SYSTEM
Energy
Balance
Mass Balance
at Steady-state
Flow Processes

m
u2 A2 u1 A1 u A


V1
V1
V


d (mU ) cv
1 2
 [(U  u  zg ) m] fs  Q  work rate
dt
2
d



(mU )CV
1
 
   H  u 2  zg  m  Q  W
dt
2
  fs

(2.28)
(2.27)
Types of Balances
1. Differential Balance is a balance taken at a specific instant in time. It is
generally applied to a continuous process. If the process is at steady state, a
differential balance applied at any time gives the same result.
We will apply differential balances to steady-state continuous processes.
Each term in a differential balance represents a process stream and the mass
flow rate of the chemical(s) in that stream.
2. Integral balance is a balance taken at two specific instants in time. It
describes what has happened over the time period between the two points.
An integral balance is generally applied to the beginning and the end of a
batch process. It accounts for what happens to the batch of chemicals.
We will apply integral balances to batch processes.
Each term in an integral balance represents a process stream and the mass of
the chemical(s) in that stream.
Example: Integral Balances on Batch Processes
Ammonia is produced from nitrogen and hydrogen in a batch reactor.
At time t = 0 there are n0 mol of NH3 in the reactor, and at a later time tf the
reaction terminates and the contents of the reactor, which include nf mol of
ammonia, are withdrawn.
Between to and tf no ammonia enters or leaves through the reactor boundaries, so
the general balance equation is simply
generation = accumulation
input
  generation
 output
 consumption

 
 
 
(enters
through

(produced

(leaves
through

 
 
  (consumed within
systemboundaries)  within system)  systemboundaries) system)
accumulation 


 (buildup

 within system) 




Example: Integral Balances on Semibatch and
Continuoius Processes
Air is bubbled through a drum of liquid hexane at a rate of 0.100 kmol/min. The
gas stream leaving the drum contains 10.0 mole% hexane vapor. Air may be
considered insoluble in liquid hexane. Use an integral balance to estimate the time
required to vaporize 10.0 m3 of the liquid.
SOLUTION
Example: Integral Balances on Semibatch and
Continuous Processes – Cont’
TEST YOURSELF?
Example: A process unit involves 3
chemical components. How many
mass balances can be written?
Solution:
We can write 4 balances. We can
write a total balance and 3
component balances.
Where am I? Where am I going ? How do
I get there?
To answer the first question, you need to
•Read, study and understand the problem.
•Draw a flow sheet for the process.
•Label it with all given information, including symbols for
the unknowns
•Note any special relationships.
To accomplish this step, you need to learn
•The information needed to specify a stream.
•How to use symbols to represent the required stream data.
•How to determine the mass of each component in a stream
(each mass will be a term in a mass balance)
EXAMPLE 1
Stream F contains 500 kg of O2 and 700 kg of CH4. Label the stream.
Solution: Note that the component masses must add to the total. The
total mass in F is 1200 kg. Thus,
Stream F F=1200 kg
m O2 = 500 kg
m CH4 = 700 kg
EXAMPLE 2
1200 kg of a mixture of O2 , N2 and CH4 are fed to a process. The stream
has 20% O2 by mass. Label the stream.
( Note the Mass of i in the stream is F xi )
Solution: The composition is partially known. Note that the fractional
compositions must add to 1.0. Thus, we can write two alternatives
Using fractional composition
Feed Stream F,
F=1200 kg
xO2 = 0.2
xN2 = ?
xCH4 = 1. – 0.2 - xN2 = 0.8 - xN2
or using component masses
Feed Stream F,
mO2 = 240 kg
F=1200 kg
mN2 = ?
mCH4 = 1200 - 240 - mN2 = 960 - mN2
Some Suggestions for Component Labeling
•If the stream composition is unknown (or if some of the component
masses are known) represent the component masses directly and use a
lower case letter for each chemical.
Eg. If stream F contains chemicals a, b and c, label the flow rates as;
F, aF, bF and cF=F- aF – Bf
•If the stream composition is known from fractional compositions, represent
the component masses directly and label as in 2.
•If stream composition is partially known with fractional compositions and
the total is known, represent the component masses indirectly and use
lower case x,y,z for each fractional composition.
•Avoid the creation of a product of two unknowns—this will result in a nonlinear equation.
FLOWCHARTS
CASE STUDY:
The catalytic dehydrogenation of propane is carried out in a continuous
packed-bed reactor. One thousand kilograms per hour of pure propane is
preheated to a temperature of 670 oC before it passes into the reactor. The
reactor effluent gas, which includes propane, methane, and hydrogen, is
cooled from 800 oC to 110 oC and fed to an adsorption tower, where the
propane and propylene are dissolved in oil. The oil then goes to a
stripping tower in which it is heated, releasing the dissolved gases; these
gases are recompressed and sent to a distillation column in which the
propane and propylene are separated. The propane stream is recycled
back to join the feed to the reactor preheater.The product stream from the
distillation column contains 98 % propylene, and the recycle stream is 97
% propane. The stripped oil is recycled to the adsorption tower. Draw
the flowchart for this system.
SOLUTION:
1. Write the values and units of all known stream variables at the locations of
the streams on the chart.
For example;
A stream containing 21 mole % O2, and 79 % N2 at 320 oC and 1.4 atm flowing at a
rate of 400 mol/h might be labeled.
 (kg
2. Assign algebraic symbols to unknown stream variables [such as m
solution/min), x (lbm N2/lbm), and n (kmol C3H8)] and write these variable
names and their associated units on the chart.
For example;
If you did not know the flow rate of the stream described in the first illustration of
step 1, you might label the stream.
EXAMPLE
SOLUTION
SOLUTION
FLOWCHART SCALING AND BASIS OF
CALCULATION
EXAMPLE
SOLUTION
DEGREE-OF-FREEDOM ANALYSIS
Equation:
ndf (= nunknowns – nindep eqns)
Three possibilities:
1. If ndf = 0, there are n independent equations in n unknowns and
the problem can in principle be solved.
2. If ndf > 0, there are more unknowns than independent equations
relating them, and at least ndf additional variable values must be
specified before the remaining variable values can be
determined. Either relations have been overlooked or the problem
is underspecified and has infinitely many solutions; in either
case, plugging into calculations is likely to be a waste of time.
3. If ndf < 0, there are more independent equations than unknowns.
Either the flowchart is incompletely labeled or the problem is
overspecified with rebundant and possibly inconsistent relations.
Again there is little point wasting time trying to solve it until the
equations and unknowns are brought into balance.
DEGREE-OF-FREEDOM ANALYSIS
1. Material
Balances
2. Energy
Balances
3. Process
Specifications
Source Of Equations Relating Unknown
Process Stream Variables
4. Physical
Properties and
Laws
5. Physical
Constrains
6. Stoichiometric
Relations
DEGREE-OF-FREEDOM ANALYSIS
EXAMPLE
Degree-of-Freedom Analysis
A stream of humid air enters a condenser in which 95 % of the water vapor in the air
is condensed. The flow rate of the condensate (the liquid leaving the condenser) is
measured and found to be 225 L/h. Dry air may be taken to contain 21 mole%
oxygen, with the balance nitrogen. Calculate the flow rate of the gas stream leaving
the condenser and the mole fractions of oxygen, nitrogen and water in this stream.
SOLUTION
DEGREE-OF-FREEDOM ANALYSIS
Suppose we had given an additional piece of information- for example, that the
entering air contains 10.0 mole% water. The flowchart would then appear as
follows;
GENERAL PROCEDURE FOR SINGLE-UNIT
PROCESS MATERIAL BALANCE
CALCULATIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
Choose as a basis of calculation an amount or flow rate of one of the process
streams.
Draw flowchart and fill in all known variable values, including the basis of
calculation. Then label unknown stream variables on the chart.
Express what the problem statement asks you to determine in terms of the
labeled variables.
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.
Do the degree-of-freedom analysis.
If the number of unknowns equals the number of equations relating them (i.e:
ndf =0), write the equations in an efficient order (minimizing simultaneous
equations) and circle the variables for which you will solve.
Solve the equations.
Calculate the quantities requested in the problem statement if they have not
already been calculated.
If the stream quantity or flow rate ng was given in the problem statement and
another value nc was either chosen as a basis or calculated for this stream,
scale the balanced process by the ratio ng/nc to obtain the final result.
EXAMPLE
Material Balances on a Distillation Column
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 2000 L/h and the specific
gravity of the fed 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
We will explicitly illustrate the implementation of the steps of the procedure just outlined.
1. 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.
2. Draw and label the flowchart.
NOTIFICATION:
BALANCES ON MULTIPLE-UNIT
PROCESSES
BALANCES ON MULTIPLE-UNIT
PROCESSES
EXAMPLE
SOLUTION
SOLUTION
SOLUTION
RECYCLE AND BYPASS
EXAMPLE
Material and Energy Balances on an Air Conditioner
Fresh air containing 4.00 mole% water vapor is to be cooled and dehumidified to a
water content of 1.70 mole% H2O. A stream of fresh air is combined with a recycle
stream of previously dehumidified air and passed through the cooler. The blended
stream entering the unit contains 2.30 mole% H2O. In the air conditioner, some of
the water in the feed stream is condensed and removed as liquid. A fraction of the
dehumidified air leaving the cooler is recycled and the remainder is delivered to a
room. Taking 100 mol of dehumidified air delivered to the room as a basis of
calculation, calculate the moles of fresh feed, moles of water condensed, and
moles of dehumidified air recycled.
SOLUTION
SOLUTION
ASSIGNMENT 2
An Extraction- Distillation Process
A mixture containing 50.0 wt% acetone and 50.0 wt% water is to be separated
into two streams-one enriched in acetone, the other in water. The separation
process consists of extraction of the acetone from the water onto methyl isobutyl
ketone (MIBK), which dissolves acetone but is nearly immiscible in water. The
description the follows introduces some of the terms commonly used in reference
to liquid extraction processes. The process is shown schematically below.
ASSIGNMENT 2-CONT’
The acetone (solute)-water(diluent) mixture is first contacted with the MIBK
(solvent) in a mixer that provides good contact between the two liquid phases. A
portion of the acetone in the feed transfers from the aqueous (water) phase to the
organic (MIBK) phase in this step. The mixture passes into a settling tank, where
the phases separate and are separately withdrawn. The phase rich in the diluent
(water, in the process) is referred to as the raffinate, and the phase rich in the
solvent (MIBK) is the extract. The mixer-settler combination is the first stage of this
separation process.
The raffinate passes to a second extraction stage where it is contacted with a
second stream of pure MIBK, leading to the transfer of more acetone. The two
phases are allowed to separate in a second settler, and the raffinate from this stage
is discarded. The extracts from the two mixer-settler stages are combined and feed
to a distillation column. The overhead effluent is rich in acetone and is the process
product. The bottom effluent is rich in MIBK and in a real process would be treated
further and recycled back to the first extraction stage, but we will not consider
recycle in this example.
ASSIGNMENT 2-CONT’
In a pilot-plant study, for every 100 kg of acetone-water fed to the first extraction
stage, 100 kg MIBK is fed to the first stage and 75 kg is fed to the second stage.
The extraction from the first stage is found to contain 27.5 wt% acetone. (All
percentages in the remainder of this paragraph are weight percents.) The secondstage raffinate has a mass of 43.1 kg and contains 5.3% acetone, 1.6 % MIBK, and
93.1 % water, and the second extract contains 9.0% acetone, 88.0 % MIBK, and
3.0% water. The overhead product from the distillation column contains 2.0% MIBK,
1.0% water, and the balance acetone.
Taking a basis of calculation of 100 kg acetone-water feed, calculate the masses
and compositions (component weight percentages) of the Stage 1 raffinate and
extract, the Stage 2 extract, the combined extract, and the distillation overhead and
bottoms products.
WHAT NEXT???
PROCESS SCHEME OF CITRIC
ACID PRODUCTION
MASS
BALANCE
SUMMARY OF MASS BALANCE
ENERGY BALANCE

Rotary Vacuum Filtrafion

Specific Enthalpy for liquid at 20 oC = 83.90 kJ / kg

Specific Enthalpy for liquid at 22 oC = 83.90 kJ / kg
Net Enthalpy (kJ/batch)
= Output Enthalpy – Input Enthalpy
= (2016830.989 - 2037730.989) kJ / batch
= - 20900 kJ / batch
SUMMARY OF ENERGY
BALANCE
PART 1
•The importance of the process or product selected from Malaysia Bioprocess
Industry point of view.
•Information of the chosen process with general data on the plant capacity, total
cost in RM (latest currency rate conversion if in $USD, etc.), location of similar type
of units and capacity in Malaysia or outside Malaysia.
•Write down at least 3 different alternatives of the process scheme for the
production of the desired bio-product.
•Choice of the best process supported by data, safety and environmental issues and
economical indicators, market analysis, flexibility and controllability.
•Process chemistry, reactions, kinetics, thermodynamics data and other physical
and chemical properties data (MSDS Sheet of all the raw materials, products and
by-products in the process).
•Choose the capacity of the plant based on the literature background and provide
the justification. Give the accurate reference and complete description of the
process. References must be quoted in the standard format.
•Selection of the plan site location.
•Conceptual design process.
•Analysis for the most economical load of the plant.
•The general profitability analysis of the plant.
•Environmental effects and risks of the plant.
Graf jumlah ACF melawan tahun bagi muatan 15000 tan/tahun
0
0
2
4
6
8
10
12
14
16
18
20
-5000000
aliran tunai (RM)
-10000000
i=10%
i=15%
i=20%
-15000000
-20000000
-25000000
-30000000
tahun
Rajah 1 Graf aliran tunai tahunan melawan tahun bagi muatan 15000 tan/tahun
Graf jumlah ACF lawan tahun bagi muatan 25000 tan/tahun
30000000
20000000
aliran tunai (RM)
10000000
0
0
2
4
6
8
10
12
14
16
18
-10000000
-20000000
-30000000
tahun
Rajah 2 Graf aliran tunai tahunan melawan tahun bagi muatan 25000 tan/tahun
20
i=10%
i=15%
i=20%
Graf jumlah ACF melawan tahun bagi muatan 35000 tan/tahun
50000000
40000000
30000000
aliran tunai (RM)
20000000
i=10%
i=15%
i=20%
10000000
0
0
2
4
6
8
10
12
14
16
18
-10000000
-20000000
-30000000
tahun
Rajah 3 Graf aliran tunai tahunan melawan tahun bagi muatan 35000 tan/tahun
20
Graf jumlah ACF melawan tahun bagi muatan 100000 tan/tahun
200000000
aliran tunai (RM)
150000000
100000000
i=10%
i=15%
i=20%
50000000
0
0
2
4
6
8
10
12
14
16
18
-50000000
tahun
Rajah 4 Graf aliran tunai tahunan melawan tahun bagi muatan 100000 tan/tahun
20
PROFITABILITY ANALYSIS
AT DIFFERENT FEED LOAD
15000
17.63E+06
LOAD (ton/tahun)
25000
35000
13.12E+06
18.16E+06
NPV pada i = 0.1
(RM/tahun)
-14.00E+06
22.27E+06
45.20E+06
187.75E+06
NPV pada i = 0.15
(RM/tahun)
-16.54E+06
10.40E+06
27.43E+06
126.50E+06
NPV pada i = 0.2
(RM/tahun)
-17.61E+06
3.29E+06
16.46E+06
90.48E+06
DCFRR (%) pada i = 0.1
-
0.1125
0.2232
0.6645
DCFRR (%) pada i = 0.15
-
-
0.1942
0.6435
DCFRR (%) pada i = 0.20
-
-
-
0.5751
14
1
2
0.5
Jumlah modal pelaburan,
FCI (RM/tahun)
Tempoh bayar balik
100000
4.06E+06
T HANK YOU
Prepared by,
MISS RAHIMAH OTHMAN