CHAPTER 2-MATERIAL BALANCE

FOR NON-REACTIVE SYSTEM

CHE243-MATERIAL AND ENERGY BALANCE AND SIMULATION

CHAPTER 2-MATERIAL BALANCE FOR

NON-REACTIVE SYSTEM

Process Classification

Fundamental Of Material Balance For Steady State

Flow Chart, Flow Chart Scaling, Basis Of Calculations And

Degree Of Freedom (DOF)

Material Balance Calculation on Single Unit Process and

Multiple Unit Processes

Recycle and Bypass Calculations

By the end of this topic, you should be able

to:

Briefly explain in your own words the meaning of the following terms: batch, semibatch, continuos, transient,steady state, recycle and its purpose, degree of freedom and etc.

Given a process description:

(a) Draw & fully label a flowchart

(b) Choosing a convenient basis of calculation

(c) For multi unit, identify the subsystems for which balances might be written

(d) Perform degree of freedom analysis for overall system and each subsytem

(e) Write in order the equations you would use to calculate specified variables

(f) Perform the calculations.

You should be able to do the computations for single unit & multiple unit processes that involve recycle, bypass or purge systems. Performed material balance for bypass and recycle.

PROCESS CLASSIFICATIONChemical processes may be classified as:

Batch process – The feed is fed into a vessel at beginning of theprocess and the vessel contents are removed sometime later. Nomass crosses the system boundaries between the time the feed ischarged and the time the product is removed.

Continuous process – The inputs and outputs flow continuouslythroughout the duration of the process.

Semibatch process – Any process that neither batch nor continuous

Steady State - Any process that have constant (do not change withtime)process variables such as flow rates, temperatures, pressures,volume etc

Unsteady State/Transient - If the process variables do change withtime.

By nature, batch and semi-batch process are unsteady stateoperation, while continuous process is steady state operation.

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.

EXAMPLE 1

Process Classification Remark

A balloon is filled with

air at a steady state rate

of 2 g/min

Volume, T, P change with

time.

Water is boiled in an

open flask

Volume, T, P change with

time.

Water is boiled in an

closed flask

T & P change with time.

Gasoline from car tank Volume change with time

Carbon dioxide and

steam are fed into reactor

to form carbon dioxide

and hydrogen

All process variables do

change with time.

The General Balance Equation

Input + Generation - Output - Consumption = Accumulation

Enters throughSystem

boundaries

Produced within system

Output Leave

through System

boundaries

ConsumedWithinsystem

Built upwithin system

BALANCE EQUATION FOR CONTINUOS

STEADY STATE PROCESS

For continuous steady-state operation for reactive system, the

accumulation team is equal to zero. Hence, general equation becomes,

Input + Generation - Output - Consumption = 0

For continuous steady-state operation for non reactive system, the

equation is simplified as,

Input = Output

BALANCE EQUATION FOR BATCH PROCESS

For batch processes, the input and output is zero, and the equation is

simplified as,

Generation - Consumption = Accumulation

For batch processes, accumulation is defined as,

Final Output - Initial Input = Accumulation

Hence for batch processes, the mass balance equation is given as,

Initial Input + Generation = Final Output + Consumption

Material Balance Calculations1. Read and try to understand on the process description. What type of

process unit 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 and lines with arrows to represent inputs and outputs.

3. Write all known stream variables i.e. inputs & outputs, on the flowchart.

4. Assign algebraic symbols to unknown stream variables.

5. Do the degree of freedom analysis. You must have equal number of equations related to unknown process variables (zero degree of freedom).

6. Take basis of calculation.

7. Write mass balance equation for the overall system and for specific component using selected Equation.

8. Perform mass balance for the process description. Always check the overall mass balance for Total Inputs = Total Outputs

FLOWCHARTS, FLOWCHART SCALING

AND BASIS OF CALCULATION

FLOWCHART

Organize the information – Draw a flowchart of the process,using boxes or other symbols to represent process units andlines with arrows to represent input and output. *Note: ReferFelder pp. 90

Several suggestion follow for labeling a flowchart in materialbalance calculation:

Write the values and units of all known stream variables at thelocation of the streams on the chart.

Assign algebraic symbols to unknown stream variables and writethese variable names and their associated units on the chart.

*Note: Refer Felder for Ex. 4.3-1 pp. 91-92.

FLOWCHARTS LABELLING

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:

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

2. Or directly as the amount or flow rate of each component

0.21 mol O2/mol0.79 mol N2/molT = 320˚C, P = 1.4 atm

400 mol/h

60 kmol N2/min40 kmol O2/min

0.6 kmol N2/kmol 0.4 kmol O2/kmol

100 kmol/min

FLOWCHARTS LABELLINGAssign algebraic symbols to unknown stream variables [such as m (kgsolution/min), x (lbm N2/lbm), and n (kmol C3H8)] and write these variablenames and their associated units on the flowchart.

If that the mass of stream 1 is half that of stream 2, label the masses ofthese streams as m and 2m rather than m1 and m2.

If you know that mass fraction of nitrogen is 3 times than oxygen, labelmass 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 onemust be 1 minus the sum of the others

If volumetric flow rate of a stream is given, it is generally useful to labelthe mass or molar flow rate of this stream or to calculate it directly, sincebalance are not written on volumetric qualities

mol/h

0.21 mol O2/mol0.79 mol N2/molT = 320˚C, P = 1.4 atm

n 400 mol/h

y mol O2/mol(1-y) mol N2/molT = 320˚C, P = 1.4 atm

CONSISTENT OF NOTATION

gasin fraction moles y

liquidin moles)or (massfraction component x

rate flow volumeV

volumeV

rate flowmolar n

moles n

rate flow mass m

mass m

EXAMPLE 1

A gas containing N2 and O2 is combined with propane in a batch

combustion chamber in which some (but not all) of the O2 and

C3 H8 react to form CO2 and H2O, and the product is then

cooled, condensing the water.

Draw the process flowchart.

Produce a flowchart for process described below…..

The catalytic dehydrogenation of propane is carried out in a continuouspacked bed reactor. One thousand kilograms per hour of pure propane ispreheated to a temperature of 670OC before it passes into the reactor.

The reactor effluent gas, which includes propane, propylene, methane andhydrogen, is cooled from 800OC to 110OC and fed to an absorption tower,where the propane and propylene are dissolved in oil.

The oil then goes to a stripping tower in which it is heated, releasing thedissolved gas; these gases are recompressed and sent to a distillation columnin 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 recycle toabsorption tower.

EXAMPLE 2

Continuous steady state processes (Felder pp.86)

1000 kg/h of mixture benzene (B) and toluene (T) containing 50% B in mass isseparated by distillation into two fractions.

The mass flow rate of B in the top stream is 450 kg B/h and that of T in thebottom stream is 475 kg T/h. The operation is at steady state.

Write balances on benzene and toluene to calculate the unknown componentflow rates in output stream.

EXAMPLE 3- Continuous steady state processes

hkgm

hkgm

B

T

/50

/25

.

.

Batch mixing process (Felder pp.87)

Methanol – water mixtures are contained in separate flasks. The first mixturecontains 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 themass and composition of product?

EXAMPLE 4- Batch processes

Mass of product =350 g Composition of product =0.529 g CH3OH/g

= 0.471 g H2O/g

A liquid mixture of benzene and toluene contains 55.0% benzene by mass.The mixture is to be partially evaporated to yield a vapor containing 85.0%benzene and residual liquid containing 10.6% benzene by mass.

(a) Suppose the process is to be carried out continuously andsteady state, with a feed rate of 100.0 kg/h of the 55% mixture.Let mv (kg/h) and ml (kg/h) be the mass flow rate of vapor andliquid product stream respectively. Draw and label processflowchart.

(b) Write and solve balances on total mass and on benzene todetermine the expected values of mv (kg/h) and ml (kg/h).

h

kgml 32.40

h

kg mv 68.59

Air Humidification and Oxygenation Process

An experiment on the growth rate of certain organisms requires anenvironment of humid air enriched in oxygen. Three input stream are fedinto an evaporation chamber to produce an output stream with the desiredcomposition. The input stream are:

Input Stream Input Description

A Liquid water, fed at a rate of 20.0 cm3/min

B Air (21 mole% O2, 79 mole% N2 )

CPure oxygen, with a molar flow rate one-fifth of the molar flow rate of stream B

EXAMPLE 5- Flowchart of an Air Humidification and Oxygenation Process

The output gas is analyzed and is found to contain 1.5 mole% water. Drawand label a flowchart of the process, and calculate:

(a) Molar flow rate for each stream input and output

(b) Mole fraction of O2 in output stream

StreamAmount

(mol/min)

A 1.11

B 60.74

C 12.15

Output 74.0

Mole fraction of O2.

336.0

649.0985.0

985.0

2

2

2

O fraction Mole

O fraction Mole

O fraction Mole y

FLOWCHART SCALING AND BASIS OF

CALCULATION

Basis of Calculation is an amount (mass or moles) or flow rate

(mass or molar) of one stream or stream component in a process.

*Note: Refer Felder pp. 95

If a stream amount or flow rate is given in a problem statement, it

is usually most convenient to use this quantity as a basis of

calculation. If no stream amounts or flow rates are known,

assume one, preferably that of a stream with a known

composition.

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.

Scaling-up – if final stream quantities are larger than the original quantities.

Scaling down – if final stream quantities are smaller than the original quantities. *Refer Felder for Ex. 4.3-2 pp. 94.

FLOWCHART SCALING AND 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 n1/n2.

Scaling Factor= Desired amount / Old amount

You cannot, however, scale masses or mass flow rates to molar quantities orvice versa by simple multiplication; conversions of this type must be carried outusing the methods as discussed in mass fraction and mol fraction section.

A 60 – 40 mixture (by moles) of A and B is separated into two fractions. Aflowchart of process is shown here.

100 mol

0.60 mol A/mol

0.40 mol B/mol

50 mol

0.95 mol A/mol

0.05 mol B/mol

12.5 mol A

37.5 mol B

It is desired to achieve the same separation with a continuous feed of 1250 lb-moles/h. Scale the factor accordingly.

EXAMPLE 5- Scale Up of a Separation Process Flowchart

1250 lbmole/h

0.60 lbmole A/lbmole

0.40 lbmole B/lbmole

625 lbmole/h

0.95 lbmole A/lbmole

0.05 lbmole B/lbmole

156.25 lbmole A/h

468.75 lbmole B/h

Final value after scale up…

A sulphuric acid (H2SO4) manufacturer required to produce H2SO4

solution with a certain concentration in order to fulfill the standardspecification. One of the process is to produce a 75.0 wt% H2SO4 solution(solution C) by diluting solution containing 90.0 wt% of H2SO4 (solutionB) with a dilute H2SO4 aqueous solution (solution A). For feed rate of 300lbm/h of solution A (55 wt% of H2SO4):

(a) Draw and label a flowchart for this process

(b) Calculate the flow rate of solution B and solution C.

A

B C300 lbm/h

0.55 lbm H2SO4/lbm

0.45 lbm H2O/lbm

m1 (lbm/h)

0.90 lbm H2SO4/lbm

0.10 lbm H2O/lbm

m2 (lbm/h)

0.75 lbm H2SO4/lbm

0.25 lbm H2O/lbm

Procedure to analysis whether you have enough information to

solve a given problem.

Before do any material balance calculation, use a properly

drawn and labeled flowchart to determine whether there is

enough information to solve a given problem.

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.

*Note: Refer Felder pp.99 and Ex-4.3-4 pp. 100.

4.0 DEGREE OF FREEDOM ANALYSIS

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 overspecified

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.

DEGREE OF FREEDOM ANALYSIS

1. Material balances.

For a nonreactive process, number of independent equation can be written is not more than number of molecules species (n ms) of the process

If benzene and toluene is involve in stream, we can write balance on benzene, toluene, total mass, atomic carbon and etc., but only TWO INDEPENDENT balance equation exist

2. An energy balance.

If the amount of energy exchanged between the system and its surroundings is specified or if it is one of the unknown process variables, an energy balance provides a relationship between inlet and outlet material flows and temperatures.

3. Process specifications

The problem statement may specify how several process are related.

i.e: Outlet flow rate is two times than flow rate stream 1 or etc.

4. Physical properties and laws

Two of the unknown variables may be the mass and volume of a stream material, in which case a tabulated specific gravity for liquids and solids or an equation of state for gases would provide an equation relating the variables

5. Physical constraints

For example, if the mole fractions of the three components of a stream labeled xA, xB, and xC, then the relation among these variables is xA + xB + xC = 1.

Instead label as xc, the las fraction should be 1-xA-xB

6. Stoichiometric relations

If chemical reactions occur in a system, stoichiometric equation provide a relationship between the quantities of reactant and the product

6 SOURCES OF EQUATION FOR BALANCE

5.0 GENERAL PROCEDURE FOR SINGLE – UNIT

PROCESS

1. Choose as basis of calculation an amount or flow rate of one of the process streams.

2. Draw a flowchart and fill in all unknown variables 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.

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 degree of freedom), write the equations in an efficient order (minimizing simultaneous equations) and circle the variables for which you will solve.

7. Solve the equations.

8. Calculate the quantities requested in the problem statement if they have not already been calculated.

9. 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.

Note: Refer Felder pp.101and Ex. 4.3-5 pp. 102.

A liquid mixture containing 45.0% benzene (B) and 55.0% toluene (T) by massis fed to a distillation column.

A product stream leaving the top of the column contains 95.0 mole% B, andbottom product stream contains 8.0% of the benzene fed to the column(92%benzene leaves with the overhead product.

The volumetric flow rate of the feed stream is 2000 L/h and the specificgravity of the feed mixture is 0.872.

Determine the mass flow rate of the overhead product stream and the massflow rate and composition (mass fractions) of the bottom product stream.Given molecular weight of Benzene is 78.11 kg/kmol and toluene is 92.13kg/kmol.

EXAMPLE 6 – MASS BALANCES ON A DC

Mass flow rate of overheadh

kg47.766

Mass flow rate of bottomh

kg53.977

Composition at bottom

kg

kg

kg

kg

T 936.0

B 064.0

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

6.0 MULTI – UNIT PROCESS

Unit 1

Product 1

Feed 1 Unit 2

Product 2

Product 3

Feed 2

Feed 3

A

B

C

D

E

Flowchart for a two – unit process is shown in Figure 4.4-1, Felder pp.105. Five boundaries drawn

about portions of process define systems on which balances may be written.

Boundary A encloses the entire process. Balance on this system are referred to as overall

balances.

Boundary B encloses a feed stream mixing points.

Boundary C encloses Unit 1

Boundary D encloses a stream splitting point

Boundary E encloses Unit 2

A labeled flow chart of continuous steady – state two – unit process is shownbelow. Each stream contains two components, A and B, in differentproportions. Three streams whose flow rates and/or compositions are notknown are labeled 1, 2, and 3. Calculate the unknown flow rates 1, 2 and 3.

Unit 1 Unit 2100.0 kg/h

0.50 kg A/kg

0.50 kg B/kg

40.0 kg/h

0.90 kg A/kg

0.10 kg B/kg

30.0 kg/h

0.30 kg A/kg

0.70 kg B/kg

1 2 3

30.0 kg/h

0.60 kg A/kg

0.40 kg B/kg

EXAMPLE 7 – MASS BALANCES ON TWO UNIT PROCESS

Flash distillation is used to separate some component in the hydrocarbonmixture. A hydrocarbon mixture containing 35.0 mole% pentane (A), 15.0mole% hexane (B) and balance octane (C) is fed to a flash distillationcolumn. The bottoms product contains 97.0 mole% C and no A, and 96.0%of the C in the feed is recovered in the bottom stream. The overheadproduct is fed to a second column. The overhead product from the secondcolumn contains 97.0 mol% of the A in the second column feed.Thecomposition of overhead product stream is 92.0 mole% A and the balanceB. Based on 100 mol/h of distillation column feed, calculate:

i) The molar flow rate and the composition of first column overheadproduct (10 marks)

ii) The molar flow rate and the composition of second column bottomproduct (6 marks)

Molar flow rate & composition of overheadFor First column

C/mol mol 039.0

B/mol mol 268.0

A/mol mol 693.0

515.50

C

B

A

x

x

x

h

molm

Molar flow rate & composition of bottomFor second column

C/mol mol 145.0

B/mol mol 778.0

A/mol mol 077.0

606.13

C

B

A

x

x

x

h

molm

A labeled flowchart of chemical process involving reaction, productseparation and recycle is shown in Figure 4.5-1, Felder pp. 110.

*Note: Refer Felder pp.110 and Ex-4.5-1 pp. 110.

There are several reason for using recycle in chemical process:

a) Recovery of catalyst

b) Dilution of a process stream

c) Control process variables

d) Circulation of a working fluid.

*Note: Refer Felder Ex.4-5.2 pp. 112

7.0 RECYCLE AND BYPASS

A procedure that has several features in common with recycle isbypass, in which a fraction of feed to a process unit is diverted aroundthe unit and combined with the output stream.

By varying the fraction of feed that is bypassed, we can vary thecomposition and properties of product.

Process

unit

Feed

Bypass stream

Product

Fresh air containing 4.00 mole% water vapor is to be cooled anddehumidified to a water content of 1.70 mole% H2O. A stream of fresh air iscombined with a recycle stream of previously dehumidified air and passedthrough 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 condensedand removed as liquid. A fraction of the dehumidified air leaving the cooleris recycled and the remainder is delivered to a room.

Taking 100 mol of dehumidified air delivered to the room as a basis ofcalculation, calculate the moles of fresh feed, moles of water condensed, andmoles of dehumidified air recycled.

*Note: refer felder pp.110

EXAMPLE 8 – RECYCLE

Moles of fresh feed… mol 4.102

Moles of water condensed mol 4.2

Moles of dehumidified air recycled mol 8.290

The flowchart of a steady state process to recover crystalline potassiumchromate (K2Cr2O7) from an aqueous solution of this salt is shown below.

*Note: Refer felder pp.113

Evaporator4500 kg/h

33.3% C

49.4% C

Filter cake

C (solid crystal)

36.4% C solution

(the crystal

constitute 95% by

mass of the filter

cake)

Crystallizer

and filter

Filtrate

36.4% C

solution

C = K2Cr2O7

H2O

EXAMPLE 9 – RECYCLE & BYPASS

4500 kg/h of a solution that is 1/3 K2Cr2O7 (C) by mass is joined by a recyclestream containing 36.4% C, and the combined stream is fed into anevaporator. The concentrated stream leaving the evaporator contains 49.4%C; this stream is fed into a crystallizer in which it is cooled (causing crystalsof C to come out of solution) and then filtered. The filter cake consists of Ccrystals and a solution that contains 36.4% C by mass; the crystals account for95% of the total mass of the filter cake. The solution that pass through thefilter, also 36.4%, in the recycle stream.

1) Calculate the rate of evaporation, the rate of production of crystalline C,the feed rates that the evaporator and the crystallizer must be designed tohandle, and the recycle ratio (mass of recycle)/(mass of fresh feed)

2) Suppose that the filtrate were discarded instead of being recycled.Calculate the production rate of crystals. What are the benefits and costs ofrecycling?

EXAMPLE 9 – RECYCLE & BYPASS

Rate of evaporation…h

kg740.2950

Rate of production of crystallineh

kg797.1471

Feed rate to evaporator..h

kg231.10151

Recycle ratio…eedkg fresh f

kg recycle.2561

Fresh orange juice contains 12.0 wt% solids and the balance water,

and concentrated orange juice contains 42.0 wt% solids. Initially a

single evaporation process was used for the concentration, but

volatile constituents of the juice escaped with the water, leaving the

concentrate with a flat taste. The current process overcomes the

problem by bypassing the evaporator with a fraction of fresh juice.

The juice that enters the evaporator is concentrated to 58 wt%

solids, and the evaporator product stream is mixed with the bypassed

fresh juice to achieve the desired final concentration. Draw and label

the flowchart. Calculate the amount of product (42% concentrate)

produced per 100 kg fresh juice fed to the process and the fraction of

the feed that bypasses the evaporator.

EXERCISE