CHAPTER 3
MATERIAL BALANCE PART I
Sem 1, 2015/2016
ERT 214 Material and Energy Balance / Imbangan
Bahan dan Tenaga
OBJECTIVES



Students will be able to identify types of
processes; batch, semi-batch and continuous
Students will be able to perform Degree of
Freedom Analysis
Students will be able to perform Material
Balance on Single Unit, Multiple Unit,
Nonreactive process under steady state
condition
Process Classification
Three type of process
1.
Batch process

Feed is charge to the process and product is removed when the
process is completed

No mass is fed or removed from the process during the operation

Used for small scale production

Operate in unsteady state
2.
Continuous process

Input and output is flow continuously throughout the duration of the
process

Operate in steady state

Used for large scale production
3.
Semi-batch process

Neither batch nor continuous

During the process a part of reactant can be fed or a part of product
can be removed.

Operate in unsteady state
1.
Steady state


2.
All the variables (i.e. temperatures, pressure, volume,
flow rate, etc) do not change with time (eg:
continuous process)
Minor fluctuation can be acceptable
Unsteady state or transient

Process variable change with time, in particular
mass flow rate. (eg; batch and semibatch)
Test Yourself
Define type and operation of process given below
 A balloon is filled with air at steady rate of 2
g/min
 A bottle of milk is taken from the refrigerator and
left on the kitchen
 Water is boiled in open flask
Answer
•
Semibatch and unsteady state
•
Batch and unsteady state
•
Semibatch and unsteady state
Balances
General Balance
A balance on a conserved quantity (total mass, mass
of a particular species, energy, momentum) in a
system ( a single process unit, a collection of units, or
an entire process) may be written in the following
way:
INPUT + GENERATION – OUTPUT – CONSUMPTION = ACCUMULATION
(Enters
through
system
boundaries)
(Produced
within
system)
(Leave
through
system
boundaries)
(Consumed
within
system)
(Buildup
within
system)
Differential vs. Integral Balances
Two types of balances may be written:
1.
Differential balances
 balances that indicate what is happening in a system at an instant
time.
 balance equation is a rate (rate of input, rate of generation, etc.) and
has units of the balanced quantity unit divided by a time unit
(people/yr, g SO2/s).
 usually applied to a CONTINUOUS process.
2.
Integral balances
 Balances that describe what happens between two instants of time.
 balance equation is an amount of the balanced quantity and has the
corresponding unit (people, g SO2).
 usually applied to a BATCH process, with the two instants of time being
the moment after the input takes place and the moment before the
product is withdrawn.
Simplified Rule for Mass Balance



If the balanced quantity is TOTAL MASS, set
generation = 0 and consumption = 0. Mass can
neither be created nor destroyed.
If the balanced substances is a NONREACTIVE
SPECIES (neither a reactant nor a product), set
generation = 0 and consumption = 0.
If a system is at STEADY STATE, set accumulation =
0, regardless of what is being balanced.
Balances on Continuous Steady State
Process
General Mass Balance Equation (GMBE)
INPUT + GENERATION – OUTPUT – CONSUMPTION = ACCUMULATION

Steady state: accumulation = 0
INPUT + GENERATION – OUTPUT – CONSUMPTION = 0

If balance on nonreactive species or total mass; balance
equation become (generation=consumption=0)
INPUT = OUTPUT
Non-reactive processes
Processes that undergo without chemical reaction
Depends on the physical / environmental factors like temperature, volume, pressure,
Example
Material Balances on a Continuous Distillation Process
One thousand kilograms per hour of a mixture of
benzene (B) and toluene (T) containing 50% benzene by
mass is separated by distillation into two fractions. The
mass flow rate of benzene in the top stream is 450 kg
Blh and that of toluene in the bottom stream is 475 kg T/h.
The operation is at steady state.
Write balances on benzene and toluene to calculate the
unknown component flow rates in the output streams.
The process can be depicted schematically as follows:
Benzene Balance
500 kg B/h = 450 kg B/h + m2
m2 = 50 kg B/h
Toluene Balance
Since the process is at steady state there can be
no buildup of anything in the system, so the
accumulation term equals zero in all material
balances. In addition, since no chemical
reactions occur, there can be no nonzero
generation or consumption terms. For all
balances, Equation 4.2-2
therefore takes the simple form input = output.
500 kg B/h = 475 kg B/h + m1
m1 = 25 kg T/h
Check the calculation:
Total Mass Balance
1000 kg /h = 450 + m1+m2+475
(all in kg/h)
ml = 25 kg T/h
M2= 50kg B/h
1000 kg/h = 1000 kg/h
Integral Balances on Batch Process

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 ammonia, are
withdrawn. Between t0 and tf no ammonia enters or leaves through the reactor
boundaries.
General Mass Balance Equation (GMBE)
INPUT + GENERATION – OUTPUT – CONSUMPTION = ACCUMULATION
From GMBE: (input=0; output=0)
Generation – Consumption = Accumulation
For batch reactor:
Accumulation = Final output – Initial Input
Final GMBE for batch process
Initial input + Generation = Final output + Consumption
Example
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?
Observe that the input and output "streams" shown on the chart denote the initial and final states
for this batch process. Since no reactions are involved, the generation and consumption may be
omitted so that all balances have the simple form "input = output.“
Total Mass Balance: 200 g + 150 g = m
m = 350 g
Methanol Balance :
200 g
0.4 g CH3OH
+
150 g
g
m = 350 g, So….. X = 0.529 g CH3OH/g
Check the solution:
Water Balance :
input = output
(200)(0.6) + (150)(0.3) = (300)(1-0.529)
0.7 g CH3OH
g
=
mg
x g CH3OH
g
Example
Integral Balances on Semi-Batch Process
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.
We begin with a differential balance on air. Since we assume that air neither dissolves in the
liquid (accumulation = 0) nor reacts with hexane in the process unit (generation = consumption = 0).
the balance reduces to input = output:
Integral hexane balance, proceeding from time t = 0 to t = tf(min), the time we seek to calculate.
The balance has the form accumulation = -output (verify). The accumulation term, which is the total
change in the moles of liquid hexane in the system during time tf. Must be negative since hexane
is being lost from the system. Since the total number of moles of hexane evaporated occupied a
liquid volume of 10.0 m3and the specific gravity of liquid hexane is 0.659, the accumulation
term equals
The output term in the balance is the rate at which hexane is leaving the system [0.1n (kmol
C6H14/min)] times the total process time, tf (min). The balance (accumulation = -output) is therefore
Flowcharts


When you are given process information and asked
to determine something about the process,
ORGANIZE the information in a way that is EASY
for subsequent calculations.
The best way  draw a flowchart
 using
boxes or other symbols to represent process units
(reactors, mixers, separation units, etc.)
 lines with arrows to represent inputs and outputs.
Flowcharts



The flowchart of a process can help get material
balance calculations started and keep them
moving.
Flowchart must be fully labeled when it is first
drawn, with values of known process variables
and symbols for unknown variables being written
for each input and output stream.
Flowchart will function as a scoreboard for the
problem solution: as each unknown variable is
determined its value is filled in, so that the flowchart
provides a continuous record of where the solution
stands and what must still be done.
Flowcharts Labeling


Write the values and units of all known stream
variables at the locations of the streams on the
flowchart.
For example, a stream containing 21 mole% O2
and 79% N2 at 320˚C and 1.4 atm flowing at a
rate of 400 mol/h might be labeled as:
400 mol/h
0.21 mol O2/mol
0.79 mol N2/mol
T = 320˚C, P = 1.4 atm
Flowcharts Labeling

Process stream can be given in two ways
1.
As the total amount or flow rate of the stream and the fractions of each
component
100 kmol/min
60 kmol N2/min
40 kmol O2/min
2.
0.6 kmol N2/kmol
0.4 kmol O2/kmol
Or directly as the amount or flow rate of each component.
10 lbm
3.0 lbm CH4
4.0 lbm C2H4
3.0 lbm C2H6
0.3 lbm CH4/lbm
0.4 lbm C2H4/lbm
0.3 lbm C2H6/lbm
Flowcharts Labeling

Assign algebraic symbols to unknown stream variables [such as m (kg
solution/min), x (lbm N2/lbm), and n (kmol C3H8)] and write these variable
names and their associated units on the flowchart.
n mol/h
0.21 mol O2/mol
0.79 mol N2/mol
T = 320˚C, P = 1.4 atm
400 mol/h
y mol O2/mol
(1-y) mol N2/mol
T = 320˚C, P = 1.4 atm
Example : Flowchart
An experiment on the growth rate of certain organism requires
an environment of humid air enriched in oxygen. Three input
streams are fed into an evaporation chamber to produce an
output stream with the desired composition.
A: Liquid water fed at rate of 20 cm3/min
B: Air (21% O2 and 79% N2)
C: Pure O2 with a molar flow rate one-fifth of the molar flow
rate of stream B
The output gas is analyzed and is found to contain 1.5 mole%
water.
Draw and label the flowchart of the process, and calculate all
unknown stream variables.
0.200n1 mol O2/min
n1 mol air/min
0.21 mol O2 /mol
0.79 mol N2 /mol
n 3 mol/min
Evaporation
0.015 mol H2O /mol
y mol O2 /mol
(0.985-y) mol N2/mol
20 cm3 H2O (l)/min
n 2mol H2O/min
Notes on the Labeling:
1. Since the one known flow rate (20 cm3 H20/min) is given on a per minute basis, it is most
convenient to label all stream flow rates on this basis.
2. Once the variable name (n1 is chosen for the air flow rate, the given information about the
ratio of the air and O2flow rates may be used to label the O2 flow rate 0.200 nl.
3. The mole fractions of the components of any stream must add up to 1. Since the mole
fraction of H20 in the outlet stream is known to be 0.015, once the mole fraction of O2 is
labeled y, that of N2 must be 1 - 0.015 - y) = (0.985 - y) (mol N2/mol).
0.200n1 mol O2/min
n1 mol air/min
0.21 mol O2 /mol
0.79 mol N2 /mol
n 3 mol/min
Evaporation
0.015 mol H2O /mol
y mol O2 /mol
(0.985-y) mol N2/mol
20 cm3 H2O (l)/min
n 2 mol H O/min
2
The quantity n2 may be calculated from the given volumetric flow rate and the density of liquid
water (conversion unit):
The three remaining unknowns (nl' n3, and y) may be determined from balances, all of which have
the simple form input = output for this nonreactive steady-state process. The balances are easily
written by referring to the flowchart.
Flowcharts Labeling




If that the mass of stream 1 is half that of stream 2, label the
masses of these streams as m and 2m rather than m1 and m2.
If you know that mass fraction of nitrogen is 3 times than
oxygen, label mass fractions as y g O2/g and 3y g N2/g
rather than y1 and y2.
When labeling component mass fraction or mole fraction, the
last one must be 1 minus the sum of the others
If volumetric flow rate of a stream is given, it is generally
useful to label the mass or molar flow rate of this stream or to
calculate it directly, since balance are not written on volumetric
qualities
Consistent on Notation
m  mass
  mass flow rate
m
n  moles
n  molar flow rate
V  volume
  volume flow rate
V
x  component fraction (mass or moles) in liquid
y  moles fraction in gas
Flowchart Scaling & Basis of
Calculation

Flowchart scaling – procedure of changing the
values of all stream amounts or flow rates by a
proportional amount while leaving the stream
compositions unchanged. The process would still be
balance.
1) Scaling-up – if final stream quantities are larger
than the original quantities.
2) Scaling down – if final stream quantities are
smaller than the original quantities.
Flowchart Scaling & Basis of
Calculation
1 kg C6H6
1 kg C7H8
300 kg C6H6
300 kg C7H8
300 lbm/h
300 lbm/h
x 300
2 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
600 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
kg
kg/h
Replace kg with lbm
600 lbm/h
0.5 lbm C6H6/lbm
0.5 lbm C7H8/lbm
Flowchart Scaling & Basis of
Calculation

Suppose you have balanced a process and the amount or flow rate of one
of the process streams is n1.You can scale the flow chart to make the
amount or flow rate of this stream n2 by multiplying all stream amounts or
flow rate by the ratio n2/n1.
Scaling Factor= Desired amount / Old amount

You cannot, however, scale masses or mass flow rates to molar quantities
or vice versa by simple multiplication; conversions of this type must be
carried out using the methods as discussed in mass fraction and mol
fraction section.
Example : Scale Up of a
Separation Process Flowchart
A 60-40 mixture (by moles) of A and B is separated into two fractions.
A flowchart of the process is shown here.
It is desired to achieve the same separation with a continuous feed of
1250 lb-moles/h. Scale the flowchart accordingly.
desired
Balancing a Process
3.0 kg/min of benzene and 1.0 kg/min of toluene are mixed
3 kg C6H6/min
m (kg/min)
1 kg C7H8/min
x (kg C6H6/kg)
(1-x) (kg C7H8/kg)
Two unknown quantities – m and x, associated with process, so two equations are needed to
calculate them.
For NONREACTIVE STEADY STATE process, input = output.
3 possible balance can be written – Balance on total mass, benzene, and toluene – any two of
which provide the equations needed to determine m and x.
For example,
Total Mass Balance:
3.0 kg/min + 1.0 kg/min = m kg/min = 4.0 kg/min
Benzene Balance:
3.0 kg C6H6/min = 4.0 kg/min (x kg C6H6/kg)
x = 0.75 kg C6H6/kg

Balancing a Process


Which balance to be used when a choice exists
and the order in which these balanced should be
written?
Rules of thumb for NONREACTIVE process
1.
2.
The maximum number of independent equations that
can be derived by writing balances on a nonreactive
system equals the number of chemical species in the
input and output streams.
Write balances first that involve the fewest unknown
variables.
Non-reactive processes
Processes that undergo without chemical reaction
Depends on the physical / environmental factors like temperature, volume, pressure,
Basis of Calculation




Balanced process can always be scaled. Mean that material balance
calculation can be performed on the basis of any convenient set of
stream amount or flow rate and the results can afterward be scaled
to any desired extent.
A basis of calculation is an amount (mass or moles) of flow rate
(mass or molar) of one stream or stream component in a process.
All unknown variables are determined to be consistent with the basis.
If a stream amount or flow rate is given in problem, choose this
quantity as a basis
If no stream amount or flow rate are known, assume one stream with
known composition. If mass fraction is known, choose total mass or
mass flow rate as basis. If mole fraction is known, choose a total
moles or molar flow rate as basis
Example : Balance on Mixing Unit
An aqueous solution of NaOH contains 20.0% NaOH by mass. It is desired to produce an
8.0% NaOH solution by diluting a stream of the 20% solution with a stream of pure water.
Calculate the ratios (liters H20/kg feed solution) and (kg product solution/kg feed solution).
Solution:
Choose a basis of calculation-an amount or flow rate of one of the feed or product streamsand then draw and label the flowchart. We will arbitrarily choose a basis of 100 kg of the 20%
feed solution. (We could also have chosen a flow rate of 100 Ibm/min of the 8% product solution
or 10 tons of diluent water. The final results do not depend on the basis chosen since we are only
asked to find ratios of stream amounts.) The flowchart appears as follows:
input = output.
Total mass balance
100 kg + ml = m2.
NaOH Balance
Total mass balance (input = output).
Diluent water volume.
Although we are not given the temperature or pressure at which the mixing is done, the density of
liquid water is approximately constant at 1.00 kg/liter. So,
Ratios requested in problem statement. ratios (liters H20/kg feed solution) and (kg product
solution/kg feed solution).
Degree-of-Freedom



Before doing any material balance calculation, use a properly drawn
and labeled flowchart to determine whether there is enough
information to solve a given problem.
The procedure for doing so is referred to as degree-of-freedom
analysis.
Procedure to perform a degree-of-freedom analysis:
a) draw and completely label a flowchart
b) count the unknown variables on the chart (n unknowns)
c) count the independent equations (n indep. eq.)
d) Find number of degree-of-freedom (ndf)
ndf= n unknowns - n indep. eq.
Number of Degree-of-Freedom
Three possibilities number of degree-of-freedom (n df)
1. If ndf = 0
 the problem can in principle be solved.

2. If ndf > 0
 there are more unknowns than independent equations relating to them
 at least ndf additional variable values must be specified before remaining
variable values can be determined.
 Either relations have been overlooked or the problem is underspecified.
3. If ndf < 0
 there are more independent equations than unknowns.
 Either the flowchart is incompletely labeled or the problem is over specified
with redundant and possibly inconsistent relations.

There is little point wasting time trying to solve material balance for n df > 0 or n df
<0.
Sources of Equation for Balance
Material balances.

An energy balance.

Process specifications
(when problem statement specify process variables)

Physical properties and Laws
(eg: equation of state for gases etc)

Physical constrains
(eg:if the mole fractions of the three components of a stream are
labeled XA, XB, and Xc, then a relation among these variables is XA
+XB +Xc =1)

General Procedure for Single Unit Process
Material Balance Calculation
1.
2.
3.
4.
5.
6.
7.
8.
9.
Choose as basis of calculation an amount or flow rate of one of the process streams.
Draw a flowchart and fill in all unknown variables 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., if the
system has zero degree of freedom), 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 a 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.
BALANCE ON MULTIPLE UNIT
PROCESSES
Balances on Multiple Unit Processes



In real chemical industries, more than one unit processes exist such
as a separation unit after reactor and so on.
Need to know term called SYSTEM in order to solve material
problem
SYSTEM:
 Any portion of process that can be enclosed within a
hypothetical box (or boundary)
 It can be the entire process, an interconnected of process unit, a
single unit, a point which two or more stream come together
into one stream or etc.
 The inputs and outputs to a system are the process streams that
are intersect to the system boundary
System of Multiple Unit Processes
C
A
D
B
E
Balances on Multiple Unit Processes



Solving material balances in multiple unit process is
basically the same as single unit processes
In multiple unit, must isolate and write balance on
several subsystems to obtain enough equation to
determine all unknowns stream variables
Always perform degree-of-freedom analysis
before solving a material balance of system.
Example : Balance on Two Unit
Process
A labeled flowchart of a continuous steady-state two-unit process is shown below. Each
stream contains two components, A and B, in different proportions. Three streams whose
flow rates and/or compositions are not known are labeled 1,2, and 3. Calculate the
unknown flow rates and compositions of streams 1,2, and 3.
Basis-Given Flow Rates
The systems about which balances might be written are shown on the following representation of the
flowchart:
The outer boundary encompasses the entire process and has as input and output streams all of the
streams that enter and leave the process. Two of the interior boundaries surround individual
process units, and the third encloses a stream junction point.
RECYCLE & BYPASS
RECYCLE
BYPASS
Recycle stream
Bypass stream
Eg: Unused reactant, recovery catalyst etc
Fraction of the feed to process unit is
diverted around the unit and
combined with the output stream
from the unit.
Example: RECYCLE PROCESS
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% H20. 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% H20. 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.
The labeled flowchart for this process, including the assumed basis of calculation, is shown below.
Dashed lines depict the four subsystems about which balances might be written-the overall
process, the recycle-fresh feed mixing point, the air conditioner, and the recycle-product gas
splitting point. The quantities to be determined are n1, n3, and n5.
All balances have the form input = output, and each additive term in each equation has the units
(mol of the balanced quantity).
THANK YOU
MID TERM EXAM 1
WEEK 8
(26 – 30 Oct 2015)
TUESDAY, 27 Oct 2015
DKD3 (10 am-12.00 pm)