PowerPoint Presentation2 Introduction Process Feed Feedt Feed Product Product Product Outlet...

In t roduct ion to Chemica l Eng ineer ing

CHE-201

I N S T R U CT O R : D r . N a b e e l S a l i m A b o - G h a n d e r

Fundamentals of Material Balance

C h a p t e r 4

2

Introduction

Process

Feed

Feed

Feed

Inp

ut

Product

Product

Product

Ou

tlet

Material enters the process Material leaves the process

A Process is an operation or series of operations in which certain objectives are

achieved.

3

Let’s Consider the following situation:

4.1 Process classification

Goldsmith Shop

100 g gold 50 golden pieces

Each of which is 3 g

Is it possible?!!

Conservation of Mass:

Mass can neither be created nor destroyed.

4

Why do we need To c lass i fy chemica l Processes?


5

Classification of Chemical Processes:

1. Batch Processes.

2. Continuous Processes.

3. Semibatch Process.


6

Batch Process

Feed Product

Feed is charged at the beginning of the process and

the product is collected some time later

Batch Process

7

Continuous Process

Feed Product

Feed and product flow continuously throughout

the duration of the process

Continuous Process

8

Semibatch Process

Feed Product

Semibatch Process

Feed Product

Any process which are not batch or continuous

(Inputs without outputs OR outputs without inputs)

Semibatch Process

9

Chemical Processes can be also classified into two main categories:

1. Steady-State:

Process variables don’t vary with time, i.e.

2. Unsteady-State (Transient):

Process variables vary with time, i.e.

time

Pro

cess

Var

iab

les

time

Pro

cess

Var

iab

les

tfy

tfy


10


Summary:

All batch processes are considered to be unsteady-state if the changes

between the initial and final time is required to determine .

Continuous processes are operated in the unsteady state at the start-

up, then it is operated almost at steady state mode.

11

Example 4.1-1: Classification of processes:

1. A balloon is filled with air at a steady rate of 2 g/min.

2. A bottle of milk is taken from the refrigerator and left on the kitchen table.

3. Water is boiled in an open flask.

12

Let’s Consider the following situation:


Process min CH4/s mout CH4/s

in outm m

1. A chemical reaction is taking place.

2. Leakage exist.

3. Wrong measurement.

13

4.2 Balances

Input - Output + generation - consumption = Accumulation

enters through system boundaries

leaves through system boundaries

produced within system

consumed within system

Buildup within system

This equation can be applied for any conserved quantity such as total mass,

mass of a specific species, energy, momentum.

14

Example 4.2-2 The general Balance Equation

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.

4.2 Balances

15

4.2 Balances

Two types of balances may be written:

Aspects Differential balance Integral Balance

Indication Indicates what happens at a certain

moment of time

Indicates what happens between

two instants of time

Balanced quantity rate quantity

Unit of balanced

quantity Quantity/time Quantity

Process Continuous Batch with two instants of time

Mathematical

Model Algebraic equations Differential equations

16

4.2 Balances

Special Cases:

1. If the balanced quantity is the total mass, then:

generation = 0

consumption = 0

2. If the balanced substance is a nonreactive species, then:

generation = 0

consumption = 0

3. If the system is at steady state, then:

accumulation = 0


Input - Output = Accumulation

Input - Output = Accumulation

Input - Output + generation - consumption

17

4.2b Balances on continuous steady-state processes:

For continuous steady-state processes:

Accumulation = 0

Input - Output = consumption - generation

If the balance is on a nonreactive species:

generation = 0

consumption = 0

Input = Output


18

4.2b Balances on continuous steady-state processes:

Example 4.2-2: Material Balance 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/h and that of toluene in the bottom stream is 475

kgT/h. The operation is at steady state. Write balances on benzene and toluene to calculate

the unknown component flow rates in the output streams.

Distillation Process 500 kg B/h

500 kg T/h

450 kg B/h

m1 (kg T/h)

m2 (kg B/h)

475 kg T/h

19

4.2c Integral Balance on Batch Processes:


Input = 0

Output = 0

Accumulation =

final output - initial input = generation - consumption

If the balance is on a nonreactive species:

generation = 0

consumption = 0

initial input = final output

20

Example 4.2-3: Balances on a Batch Mixing Processes

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?

4.2c Integral Balance on Batch Processes:

Mixer

200 g

0.400 g CH3OH/ g

0.600 g H2O/ g

150 g

0.700 g CH3OH/ g

0.300 g H2O / g

m (g)

x (g CH3OH/ g)

(1-x) (g H2O/ g)

Example 4.2-4: Integral Balance on a Semibatch Process is not included in the material

21

4.3 Flow Chart

1. Represent each process unit by a simple box or any other simple.

2. Feeds and Products are represented by lines and arrows.

Process Unit

Process Unit

Process Unit

Total mass (molar) flow rate

Mass (molar) composition

OR Component

Mass (molar) Flow rates

Other variables like temperature and pressure

Other variables like temperature and pressure

22

Example:

100 kmol/min

0.6 kmol N2/kmol

0.4 kmol O2/kmol

T = 320oC, P = 1.4 atm

OR 60 kmol N2

40 kmol O2

T = 320oC, P = 1.4 atm

10 lbm mixture

0.3 lbm CH4/lbm

0.4 lbm C2H4/lbm

0.3 lbm C2H6/lbm

T = 320oC, P = 1.4 atm

OR 3.0 lbm CH4

4.0 lbm C2H4

3.0 lbm C2H6

T = 320oC, P = 1.4 atm

4.3 Flow Chart

23

3. Assign algebraic symbols to unknown stream variables associating them with

units in parenthesis.

(mol/min)

0.21 mol O2/mol

0.79 mol N2/mol

n 400 mol/min

y (mol O2/mol)

(1-y)(mol N2/mol)

AND

4.3 Flow Chart

4. The following variables are normally used to describe:

Molar quantity (mole unit)

Volume (volume unit)

Mass (mass unit)

Molar flow rate (mol/time)

Volumetric flow rate (volume/time)

Mass flow rate (mass/time)

Temperature (temperature unit)

Pressure (pressure unit)

n

V

m

n

V

m

T

P

24

4.3a Flow Chart

Example 4.3-1: Flowchart of an Air Humidification and Oxygenation Process

An experiment on the growth rate of a certain organisms 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 a rate of 20.0 cm3/min

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

C: Pure oxygen, with a molar flow rate one-fifth of the molar flowrate of stream

B

The output gas is analyzed and is found to contain 1.5 mole% water. Draw and label a

flowchart of the process, and calculate all unknown stream variables.

25

Wet air containing 30.0 mole% water, flowing at a rate of 1 mole/s passes through a

condenser. Air leaving the condenser is 15.0 mole% water. Draw a fully labeled flowchart.

4.3a Flow Chart

26

A feed stock available at the rate of 1000 mol/h and consisting of (all in mol %):

20% Propane (C3)

30% Isobutane (i-C4)

20% Isopentane (i-C5)

30% Normal pentane (C5)

is to be separated into two fractions by distillation. The distillate (top product) is to

contain all of the propane feed to the unit and 80% of the isopentane fed to the

unit and is to consist of 40 mole% isobutane. The bottom stream is to contain all

the normal pentane fed to the unit. Draw the flowchart of the process

4.3a Flow Chart

27

Acetone can be recovered from a carrier gas by dissolving it in a pure water

stream in a unit called an absorber. In this process unit, 200 lbm/h of a stream

containing 20% acetone and 80% gas is treated with 1000 lbm/h of a pure

water stream to yield an acetone free overhead gas and an acetone-water

solution. Assume no carrier gas dissolves in water.

Draw a flowchart for the process.

4.3a Flow Chart

28

4.3b Flowchart Scaling and Basis of Calculation

Consider the following:

1 kg C6H6

1 kg C7H8

2 kg

0.5 kg C6H6/kg

0.5 kg C7H8/kg

10 kg C6H6

10 kg C7H8

20 kg

0.5 kg C6H6/kg

0.5 kg C7H8/kg

×10

Scaling Up

The final quantities are larger than the original quantities

29

1 kg C6H6

1 kg C7H8

2 kg

0.5 kg C6H6/kg

0.5 kg C7H8/kg

0.5 kg C6H6

0.5 kg C7H8

1.0 kg

0.5 kg C6H6/kg

0.5 kg C7H8/kg

×1/

2

Scaling Down

The final quantities are smaller than the original quantities

4.3 Flowchart Scaling and Basis of Calculation

30


Summary:

1. Balanced processes can be always scaled up or down by multiplying all steams of the

old process by a factor while maintaining stream compositions unchanged.

2. the scale factor is defined as :

3. When balanced processes are scaled up or down, compositions of all streams must

remain unchanged.

4. Changing balanced processes from batch to continuous by dividing quantities by unit

time will not change the original process.

streamoldingcorrespondtheof(flowrate)amount

streamnewof(flowrate)amountn

31


1 kg C6H6

1 kg C7H8

2 kg

0.5 kg C6H6/kg

0.5 kg C7H8/kg

1 kg C6H6/hr

1 kg C7H8/hr

2 kg/hr

0.5 kg C6H6/kg

0.5 kg C7H8/kg

32

Example 4.3-2: 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.

100.0 mol

0.6 mol A/mole

0.4 mol B/mole

12.5 mol A

37.5 mol B

50.0 mol

0.95 mol A/mole

0.05 mol B/mole


33

4.3c Balancing a Process

Condenser Humid Air

0.30 mole H2O/mole

0.70 mole air/mole

Humid Air

0.15 mole H2O/mole

0.85 mole air/mole

H2O (l)

Can we perform mass balance calculation and why?

34


Calculation can be made for any basis of any convenient set of stream

amounts or flow rates and the results can afterward be scaled to any desired

extent.

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

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

Choosing basis of calculation is considered to be the first step in balancing a

process.

If no stream amounts or flow rates are known, assume one, preferably that of

known compositions.

35

4.3c Balancing Processes

Consider the following system of equations:

1052 yx

33 yx

The solution set is (15, -4)

The equations in the above system is recognized as independent equations, i.e. can’t be

derived algebraically from each other. This results in a unique solution set

Consider the following system of equations:

1052 yx

20104 yx

The equations in the above system is recognized as dependent equations, i.e. Eq(2) =

2×Eq(1). This results in infinitely many solutions!

36


3.0 kg C6H6/min

1.0 kg C7H8/min

m (kg/min )

x (kg C6H6/kg)

(1-x) (kg C7H8/kg)

Total mass balance:

C6H6 balance:

C7H8 balance:

m 0.10.3

mx0.3

mx 10.1

(1)

(3)

(2)

Conclusion:

For two components, we can write two balance equations and for “n” components we can

write “n” balance equations plus one total mass balance equation = (n+1) equations.

37


Dependant and Independent Equations:

Addition of (2) and (3) yields:

mxxm 10.10.3

Substation of (2) and (1) yields:

xmm 0.30.10.3

Substation of (3) and (1) yields:

mxm 10.10.10.3

Total mass balance equation

Toluene balance equation

Benzene balance equation

Conclusion:

Although we can write three balance equations, we can only solve for two variables as

they are not linearly independent.

38


Summary:

Total number of balance equation which can be written on physical systems equals

number of components +1

Number of independent balance equations which can be written on physical systems

equals the number of the chemical components.

Try always to write balances which involve the fewest number of unknowns

variables.

39


Example 4.3-3: Balances on a Mixing Unit

An aqueous solution of sodium hydroxide 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 H2O/kg feed solution) and (kg product

solution/kg feed solution)

40

4.3d Degree-of-freedom Analysis

Is there a way which can save

my time and tell me if the

mass balance problem is

solvable or not?

41


Solvable problems mean that enough information is available so that all variable

can be uniquely determined.

Degree of Freedom Analysis:

is an analysis which checks if enough information is available so that mass balance

problems are solvable or not before attempting to derive any equations.

equationsindepunknowns nnDF .

DF = 0 Correctly specified system

DF > 0 Underspecified system, i.e. more unknowns are there than number of

equations.

DF < 0 Overspecified system, i.e. more information is available than required.

42


Sources of Equations:

1. Material balances

2. An energy balance

3. Process specifications

4. Physical properties and laws.

5. Physical constraints.

43


Example 4.3-4 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 flowrate of the condensate (the liquid leaving the condenser) is

measured and found to be 225 L/h. dry air may be taken to contain 21.0% oxygen, with

the balance nitrogen. Calculate the flowrate of the gas stream leaving the condenser and

the mole fractions of oxygen, nitrogen, and water in this stream.

44

Condenser 0.30 mole H2O/mole

0.70 mole air/mole

0.15 mole H2O/mole

0.85 mole air/mole

H2O (l)

m3 (kg/s)

m2 (kg/s)

m1 (kg/s)


Perform Degree of Freedom on the above system?

45

4.3d General Procedure for Single-Unit Process Material

Balance Calculations

2000 L/h

m1 (kg/h)

0.45 kg B/kg

0.55 kg T/kg

0.95 mol B/mole

0.05 mol T/mole

m2 (kg/h)

mB3 (kg/h)

mT3 (kg/h)

(8% of B in feed)

Bottom Product

Overhead Product

46

Unit

1

Unit

2

A

D

E

B

Product 1 Product 2 Feed 3

Product 3 Feed 1

Feed 2

C

Flowchart of a two-unit process

4.4 Balances on Multiple-Unit Processes

47


Mixing Points:

The resulting stream after mixing has different flowrate and compositions than the

constituting streams.

The total balance equations which can be written is number of chemical species +1.

The number of independent equations equals the number of chemical species.

stream 2

stream 1 stream 1

stream 2

stream 3 stream 3

48


Splitting Points:

The resulting streams after mixing have different flowrate but the same compositions as

the main stream.

The total balance equations which can be written is only ONE.

stream 2

stream 1 stream 1

stream 2

stream 3 stream 3

49


Procedure to solve multiple unit systems:

1. Draw a fully labeled flowchart.

2. Divide the multiple systems into subsystems.

3. Perform degree of freedom on each subsystems.

4. Solve systems of zero degree of freedom.

50

Example 4.4-1 Two-Unit Process

A labeled flowchart of a continuous steady-state two-unit process is shown below. Each

stream contains two components, A, B, in different proportions. Three streams whose

flow rates and/or compositions are not known are labeled 1,2, and 3. Calculate the

unknown flowrates and composition of streams 1,2, and 3?


Unit I

Unit II

100.0 kg/h

0.500 kg A/kg

0.500 kg B/kg

30.0 kg/h

0.300 kg A/kg

0.700 kg B/kg

30.0 kg/h

0.600 kg A/kg

0.400 kg B/kg

40.0 kg/h

0.900 kg A/kg

0.100 kg B/kg

1 2 3

51

Extractor Extractor

Distillation

Unit

100 kg

100 kg M 75 kg M

43.1 kg

mA2 (kg A)

mM2 (kg M)

mW2 (kg W)

m1 (kg)

0.275 (kg A/kg)

xM1 (kg M/kg)

(0.725-xM1 )(kg W/kg)

0.053 kg A/kg

0.016 kg M/kg

0.931 kg W/kg

mA4 (kg A)

mM4 (kg M)

mW4 (kg W)

m5 (kg)

0.97 kg A/kg

0.02 kg M/kg

0.01 kg W/kg

mA6 (kg A)

mM6 (kg M)

mW6 (kg W)

m3 (kg)

0.09 (kg A/kg)

0.88 (kg M/kg)

0.03 (kg W/kg)

0.5 kg W/kg

0.5 kg A/kg


52

4.5 Recycle and by-pass

Process Unit

Consider the following:

Feed Product

Process Unit

Feed Product 1 Process Unit

Product 2

Impractical as large

number of units

are sometimes needed

to be built

53


Process Unit

Fresh feed Product

A recycle stream is simply a stream which is split off from the outlet of a unit

and sent back as the input of an upstream unit

Feed

Recycle stream

54

Example 4.5-1 Material balance on an Air Conditioner.

Fresh air containing 4.00 % 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.


55

AIR COND.

DA = Dry Air

W = Water


0.960 mol DA/mole

0.040 mol W/mole

0.983 mol DA/mole

0.017 mol W/mole 0.983 mol DA/mole

0.017 mol W/mole

n3 [mol W(l)]

n5 (mol)

0.983 mol DA/mole

0.017 mol W/mole

n2 (mol)

0.977 mol DA/mole

0.023 mol W/mole

n1(mol) n4(mol) 100 mol

56

The flowchart of a steady-state process to recover crystalline chromate (K2CrO4) from an aqueous solution of this salt is

shown below:

Forty-five hundred kilograms per hour of a solution that is one-third K2CrO4 by mass is joined by a recycle stream containing

36.4% K2CrO4, and the combined stream is fed into an evaporator. The concentrated stream leaving the evaporator contains

49.4% K2CrO4; this stream is fed into a crystallizer in which it is cooled (causing crystals of K2CrO4 to come out of solution)

and then filtered. The filter cake consists of K2CrO4 crystals and a solution that contains 36.4 % K2CrO4 by mass; the crystal

account for 95% of the total mass of the filter cake. The solution that passes through the filter, also 36.4% K2CrO4, is the

recycle stream.

1. Calculate the rate of evaporation, the rate of production of crystalline K2CrO4, the feed rates that the evaporator and

crystallizer must be designed to handle, 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 of the recycling?

EVAPORATOR

CRYSTALIZER AND FILTER

H2O

4500 kg/h

33.3% K2CrO4

49.4% K2CrO4

Filtrate

36.4% K2CrO4 solution

Filter cake

K2CrO4 solid

36.4 % K2CrO4 solution

(the crystals constitute 95% by

mass of the filter cake)


57

EVAPORATOR CRYSTALIZER

AND FILTER

Fresh feed

4500 kg/h 0.494 kg K/kg

0.506 kg W/kg

Filtrate (recycle)

m6 (kg /h)

m5 (kg soln/h)

0.364 kg K/kg

0.636 kg W/kg

Filter cake

m4 (kg K(s)/h)

0.364 kg K/kg

0.636 kg W/kg

x1 (kg K/kg)

(1-x1) (kg W/kg)

m1 (kg /h)

m2 (kg W(v)/h)

m3 (kg /h)


58

EVAPORATOR CRYSTALIZER

AND FILTER

Fresh feed

4500 kg/h 0.494 kg K/kg

0.506 kg W/kg

m5 (kg /h)

0.364 kg K/kg

0.636 kg W/kg

m4 (kg soln/h)

0.364 kg K/kg

0.636 kg W/kg

m3 (kg K(s)/h)

m1 (kg W(v)/h)

m2 (kg /h)


59


Process Unit

Fresh feed Product Feed

Recycle stream

Process Unit

Fresh feed Product Feed

Bypass stream

60


Reasons for installing recycle lines:

1. Recovery of catalyst.

2. Dilution of a process stream.

3. Control of a process variable

4. Circulation of working fluid.

61

4.6 Chemical Reaction Stoichiometry

Stoichiometry:

is the theory of the proportions in which chemical species combine with one another.

Stoichiometric equation:

is a statement of the relative number of molecules or moles of reactants and products

that participate in the reaction.

322 22 SOOSO

It tells the reactive species and relative amounts in mole, k-mole, or lb-mole.

It must be balanced to be used.

Atoms can’t be generated nor created in chemical reactions.

62

4.6 Chemical Reaction Stoichiometry

Stoichiometric Ratio:

is defined for two species participating in a reaction in the ratio of their stoichiometric

coefficients in the balanced reaction equations.

322 22 SOOSO

consumedOmol1

generatedSOmol2

2

3

generatedSOmoleslb2

consumedSOmoleslb2

3

2

It can be used as conversion factor to calculate the amount of a particular reactant (or

product) that was consumed or (produced), given a quantity of another reactant of product

the participated in the reaction.

Example:

Calculate the amount of oxygen required to produce 1600 kg/h of SO3?

63

4.6b Limiting and Excess Reactants, Fractional Conversion,

and Extent of Reaction

322 22 SOOSO

Consider:

SO2 O2 SO3

Case 1:

4 moles 2 moles 4 moles

Stoichiometric proportion:

For any two chemical species A and B, they are at their stoichiometric proportion if their

feed ratio equals the stoichiometic ratio obtained from the balanced reactions equation.

64



322 22 SOOSO

Consider:

SO2 O2 SO3

Case 2:


Comment:

If the reaction goes to completion, only 2 moles of SO3 are produced, 2 moles of SO2

consumed and all oxygen are consumed.

We say that: O2 is the limiting reactant

SO2 is excess reactant

Limiting reactant:

The reactant that would run out if a reaction proceed to completion. It presents in less

than its stoichiometric proportion relative to every other reactant.

65



322 22 SOOSO

Consider:

SO2 O2 SO3

Case 2:


Comment:

If the reaction goes to completion, only 2 moles of SO3 are produced, 4 moles of SO2

consumed and 2 moles of oxygen are consumed.

We say that: O2 is excess reactant

SO2 is the limiting reactant

Limiting reactant:

The reactant that would run out if a reaction proceed to completion. It presents in less

than its stoichiometric proportion relative to every other reactant.

66



Fractional Excess:

Fractional Conversion:

stoichA

stoichAfeedA

n

nnAofexcessfractional

fedmoles

reactedmolesf

67



Reactor Flowchart:

Reactor Feed Product

Product stream of a reactor is composed of a matrix of:

1. Unconverted reactants.

2. Products formed.

3. Inert materials

68



Reactor Flowchart:

Reactor Feed n (mole/s)

Reactor Feed

yA (mole A/mole)

yB (mole B/mole)

yC (mole C/mole)

(1-yA -yB- yC) (mole D/mole)

nA (mole A/s)

nB (mole B/s)

nC (mole C/s)

nD (mole D/s)

DCBA

69



Techniques to balance reactive processes:

1. Extent of chemical reaction.

2. Molecular balance.

3. Atomic balance.

70



Extent of chemical reactions:


Assumptions:

Steady state, i.e. accumulation =0

Generation = consumption i

This can be reduced to:

iioi nn

: feed molar flowrate of component i.

: product molar flowrate of component i.

: stoichiometric number s obtained from the balanced stoichiometric equation and

defined to be negative for reactants and positive for products.

ion

i

in

71

Acrylonitrile is produced in the reaction of propylene, ammonia, and oxygen:

The feed contains 10.0% propylene, 12.0% ammonia, and 78.0% air. A fractional conversion

of 30.0% of the limiting reactant is achieved. Taking 100 mol of feed as a basis, determine

which reactant is limiting, the percentage by which each of the other reactants is excess, and

the molar amounts of all product gas constituents for a 30.0% conversion of the limiting

reactant.

OHNHCONHHC 2332363 32

3



72

OHNHCONHHC 2332363 32

3



Reactor 100 mole

n1 (mole C3H6)

n2(mole NH3)

n3(mole O2)

n4 (mole C3H3N)

n5 (mole H2O)

n6 (mole N2)

0.10 mole C3H6 /mole

0.12 mole NH3 /mole

0.78 mole Air /mole

0.21 mole O2 / mole Air

0.79 mole N2 / mole Air

73

Consider the ammonia formation reaction:

Suppose the feed to a continuous reactor consists of 100 mol/s of nitrogen, 300 mol/s of

hydrogen, and 1 mol of argon (an inert gas), calculate the molar flowrates of all

components knowing that the fractional conversion of hydrogen is 0.6?

322 23 NHHN



74

322 23 NHHN



Reactor 100 mole N2/s

300 mole H2/s

1.0 mole Ar /s

n1 (mole N2/s)

n2(mole H2/s)

n3(mole NH3/s)

n4 (mole Ar/s)

Fractional conversion of hydrogen is 0.60

1

2

3

4

100

300 3

0.0 2

1.0 0

n

n

n

n

76

4.6c Chemical Equilibrium:

Progress of Reaction Progress of Reaction

Ch

emic

al S

pec

ies

Ch

emic

al S

pec

ies

Irreversible reaction Reversible reaction

dDcCbBaA dDcCbBaA

Objective:

We are interested in knowing the molar quantities (compositions) at equilibrium.

77


Example 4.6-2: Calculation of an Equilibrium Composition

If water-gas shift reaction:

Proceed to equilibrium at a temperature T(K), the mole fraction of the four reactive species satisfy

the relation:

Where K(T) is the reaction equilibrium constant. At T=1105 K, K=1.00. Suppose the feed to a

reactor contains 1.00 mol of CO, 2.00 mol H2O, and no CO2 or H2, and the reaction mixture comes

to equilibrium at 1105K. Calculate the equilibrium composition and the fractional conversion of the

limiting reactant.

gHgCOgOHgCO 222

TKyy

yy

OHCO

HCO

2

22

78


gHgCOgOHgCO 222

TKyy

yy

OHCO

HCO

2

22

Reactor 1.0 mole CO

2.0 moles H2O

n1 (mole CO)

n2(mole H2O)

n3(mole CO2)

n4 (mole H2)

79

4.6d Multiple Reactions, Yield, and Selectivity:

Consider:

Process: Dehydrogenation of Ethane to produce ethylene

24262 HHCHC

4262 2CHHHC

4636242 CHHCHCHC

Main reaction

Side reactions

Two major mathematical quantities are needed to be defined as they measure the

performance of the production of the process.

80


completelyreactedhadreactantlimiting

theandreactionssidenowerethere

ifformedbeenhavewouldthatmoles

formedproductdesiredofmolesYield

formedproductundersiredofmoles

formedproductdesiredofmolesySelectivit

High values of selectivity and yield signify that the undesired side reactions have been

successfully suppressed relative to the desired reaction.

R

jjijioi nn

1

Extent of Chemical Reaction:

81

Example 4.6-3 Yield and Selectivity in a Dehydrogenation Reactor

The reactions:

Take place in a continuous reactor at steady state. The feed contains 85.0 mole%, ethane (C2H6) and

the balance inert (I). The fractional conversion of ethane is 0.501, and the fractional yield of ethylene

is 0.471. calculate the molar composition of the product gas and selectivity of ethylene to methane

production.

24262 HHCHC

4262 2CHHHC


82

24262 HHCHC

4262 2CHHHC


Reactor 0.85 mole C2H6/mol

0.15 mol I/ mol

n1 (mole C2H6)

n2(mole C2H4)

n3(mole H2)

n4 (mole CH4)

n5 (mol I)

f = 0.501

Y = 0.471

83

4.7 BALANCES ON REACTIVE PROCESSES:

Molecular Balance:

Atomic Balance:

For each chemical species participating in a chemical reaction, a generation or

consumption term is defined.

Atoms can’t be destroyed nor created in a chemical reaction.

84

4.7 BALANCES ON REACTIVE PROCESSES:

Example:

Ethane is dehydrogenated according to the following stoichiometric equation:

In a steady-state continuous reactor, one hundred kmol/min of ethane is fed to the

reactor. If the conversion on the reactor is obtained to be 0.45, calculate the molar

flowrate of all components constituting the product stream using the molecular balance

and atomic balance techniques?

24262 HHCHC

85

Reactor 100 kmol C2H6/min

n1 (kmole C2H6/min)

4.7a BALANCES ON REACTIVE PROCESSES:

n2 (kmole C2H4/min)

n3 (kmole H2/min)

Fractional conversion is 0.45

86

4.7b Independent Equations, Independent Species and Independent

Reactions:

n1 (mol O2/s)

3.67n1 (mol N2/s)

n2 (mol CCl4(l)/s)

n3 (mol O2/s)

3.67n3 (mol N2/s)

n4 (mol CCl4(v)/s)

n5 (mol CCl4(l)/s)

O2-balance:

N2-balance:

CCl4-balance:

31 nn

31 67.367.3 nn

542 nnn

If two molecular species are in the same ratio to each other wherever they appear in a

process and this ratio is incorporated in the flowchart labeling, balances on those

species will not be independent equations.

87

4.7b Independent Equations, Independent Species and Independent

Reactions:

BA 2

CB

CA 2

Chemical reaction equations:

Chemical reactions are independent if the stoichiometric equation of any one of them

cannot be obtained by adding and subtracting multiples of the stoichiometic equations

of the others.

1

2

3

88

4.7c Molecular Species Balances:

Degree of Freedom is obtained by:

No. of unknown labeled variables

+ No. of independent chemical equations.

- No. of independent molecular species balances.

- No. of other equations relating unknown variables

= No. of degrees of freedom

89

Reactor 100 kmol/min

n1 (kmole C2H6/min)

4.7c Molecular Species Balances:

n2 (kmole C2H4/min)

n3 (kmole H2/min)


90

4.7d Atomic Species Balances:



- No. of independent atomic species balances.

- No. molecular balances on independent nonreactive species



91


n1 (kmole C2H6/min)

4.7d Atomic Species Balances:

n2 (kmole C2H4/min)

n3 (kmole H2/min)


92

4.7e Extent of Reaction:



+ No. of independent chemical equations.

- No. of independent reactive species

- No. of independent nonreactive species.



93


n1 (kmole C2H6/min)


n2 (kmole C2H4/min)

n3 (kmole H2/min)


94

Summary of Balances on Reactive Systems:

Atomic species balances generally lead to the most straightforward solution

procedure, especially when more than one reaction is involved.

Extent of chemical reaction are convenient for chemical equilibrium problems and

when equation-solving software is to be used.

Molecular species balances require more complex calculations than either of the

other two approaches and should be used only for simple systems involving one

reaction.

95

Example 4.7-1 Incomplete combustion of Methane

Methane is burned with air in a continuous steady-state combustion reactor to yield a

mixture of carbon monoxide, carbon dioxide, and water. The reactions taking place are:

The feed to the reactor contains 7.80% CH4, 19.4% O2, 72.8% N2. the percentage

conversion of methane is 90.0%, and the gas leaving the reactor contains 8.0 mol

CO2/mol CO. Carry out a degree-of-freedom analysis on the process. Then calculate the

molar composition of the product stream using molecular species balances, atomic

species balances, and extent of reaction.

OHCOOCH 224 22

3

OHCOOCH 2224 22


96

REACTOR PRODUCT

SEPARATION

UNIT

75 mol

A/ min

100 mol

A/ min

75 mol B/ min

25 mol A/ min

75 mol

B/ min

25 mol A/ min

4.7f Product Separation and Recycle:

97


Two types of conversions are defined:

processtoinputreactant

processfromoutputreacrantprocesstoinputreactantConversionOverall

reactortoinputreactant

reactorfromoutputreacrantreactortoinputreactantConversionPassSingle

98

REACTOR PRODUCT

SEPARATION

UNIT

75 mol

A/ min

100 mol

A/ min

75 mol B/ min

25 mol A/ min

75 mol

B/ min

25 mol A/ min


99

Example 4.7-2 Dehydrogenation of Propane

Propane is dehydrogenated to form propylene in a catalytic reactor:

The process is designed for a 95% overall conversion of propane. The reaction products are

separated are separated into two streams: the first, which contains H2, C3H6, and 0.555% of

the propane that leaves the reactor, is taken off as product: the second stream, which

contains the balance of the unreacted propane and 5% of the propylene in the first stream, is

recycled to the reactor. Calculate the composition of the product, the ratio (mole

recycled)/(mole fresh feed), and the single-pass conversion.

26383 HHCHC


100


REACTOR SEPARATOR

100 mol

C3H8

n1 (mol C3H8)

n2 (mol C3H6)

Recycle

n9 (mol C3H8)

n10 (mol C3H6)

n3 (mol C3H8)

n4 (mol C3H6)

n5 (mol H2)

n6 (mol C3H8)

n7 (mol C3H6)

n8 (mol H2)

Product

101

4.7g Purging

REACTOR ABSORBER 60 mol C2H4/s

30 mol O2/s

113 mol N2/s

100 mol C2H4/s

50 mol O2/s

565 mol N2/s

Product

Fresh feed

50 mol C2H4/s

25 mol O2/s

565 mol N2/s

50 mol C2H4O/s

50 mol C2H4O/s

Solvent

Solvent

Recycle

40 mol C2H4/s

20 mol O2/s

452 mol N2/s

10 mol C2H4/s

5 mol O2/s

113 mol N2/s

Purge stream

OHCOHC 42242 2

102

Example 4.7-3 Recycle and Purge in the Synthesis of Methanol

Methanol is produced in the reaction of carbon dioxide and hydrogen:

The fresh feed to the process contains hydrogen, carbon dioxide, and 0.400 mole% inert (I). The

reactor effluent passes through a condenser to remove essentially all of the methanol and water

formed and none of the reactants or inert.. The later substances are recycled to the reactor. To

avoid buildup of the inert in the system, a purge stream is withdrawn from the recycle.

The feed to the reactor (not the fresh feed to the process) contains 28.0 mole% CO2, 70.0 mole%

H2, and 2.00 mole% inert. The single-pass conversion of hydrogen is 60.0%. Calculate the molar

flow rates and molar composition of the fresh feed, the total feed to the reactor, the recycle stream,

and purge stream for a methanol production rate of 155 kmol CH3OH/h.

OHOHCHHCO 2322 3

4.7g Purging

103

4.7g Purging

REACTOR CONDENSER 0.280 mol CO2/mol

0.700 mol H2/mol

0.020 mol I/mol

Product

n1 (mole CO2)

n2 (mole H2)

2.0 mol I

n3 (mole

CH3OH)

n4 (mole H2O)

x5C (mol CO2/mol)

x5H (mol H2/mol)

(1- x5C- x5H)

np (mol)

no

x0C (mol CO2/mol)

(0.996- x0C)mol H2/mol

0.00400 mol I/mol

nr (mol)

n5 (mol)

x5C (mol CO2/mol)

x5H (mol H2/mol)

(1- x5C- x5H)

x5C (mol CO2/mol)

x5H (mol H2/mol)

(1- x5C- x5H)

OHOHCHHCO 2322 3

104

4.8 Combustion Reactions

Combustion: the rapid reaction of fuel with oxygen.

gasesofmatrixoxygenfuel

They are run to produce tremendous amount of energy used to run turbine and

boilers.

Fuel can be:

1. Coal (carbon, some hydrogen and sulfur and various noncombustible materials.

2. Fuel oil (high molecular weight of hydrocarbons, some sulfur).

3. Gaseous fuel (natural gas such as methane).

4. Liquefied petroleum gas (propane and/or butane).

Elements of the matrix of gas product are CO2, CO, H2O, SO2 and at high temperature

NO is produced.

105


Combustion reactions can be divided into two main categories:

1. Complete combustion.

2. Incomplete combustion (Partial).

Complete Combustion Incomplete combustion

Gas Matrix

(Products)

Production of CO2 and

steam

Production of CO and

steam

Energy High energy Low energy

Oxygen required More Less

106


Example 4.8-1: Classify each of the following reactions as complete or incomplete:

22 COOC

COOC 22

1

OHCOOHC 22283 435

OHCOOHC 2283 432

7

OSCOOCS 2222 23

107


Flowchart of a combustion reactor:

Combustion

Reactor

Fuel

Air

0.21 mol O2/mol

0.79 mol N2/mol

Gaseous products

Unburned fuel

CO

CO2

SO2

H2O

Noncombustible material

Compositions of gaseous products can be given in two ways:

1. Composition on a dry basis.

2. Composition on a wet basis.

Flue gas

Stack gas

108

Example 4.8-1: Composition on Wet and Dry Bases

a. A stack gas contains 60.0 mole% N2, 15.0% CO2, 10.0% O2, and the balance

H2O. Calculate the molar composition of the gas on a dry basis.

b. An orsat analysis (a technique for stack analysis) yields the following dry basis

composition:

N2 65%

CO2 14%

CO 11%

O2 10%

A humidity measurement shows that the mole fraction of H2O in the stack gas is

0.0700. Calculate the sack gas composition on a wet basis.


109

4.8b Theoretical and Excess Air

100%

airmoles

airmolesairmoles:AirExcessPercent

ltheoretica

ltheoreticafed

Theoretical Oxygen:

The moles (batch) or molar flowrate (continuous) of O2 needed for complete

combustion of all the fuel fed to the reactor, assuming that all carbon in the fuel is

oxidized to CO2, hydrogen to H2O, and sulfur to SO2.

Theoretical Air:

The amount of air containing the theoretical oxygen.

Excess Air:

The amount by which the air fed to the reactor exceeds the theoretical air.

ltheoreticafed airmolesairexcessoffraction1airmoles

110

4.8b Theoretical and Excess Air

Example 4.8-2 Theoretical and Excess Air

One hundred mol/h of butane (C4H10) and 5000 mol/h of air are fed to a combustion

reactor. Calculate the percent excess air.

111

4.8c Material Balances on Combustion Reactors

Example 4.8-3 combustion of Ethane

Ethane is burned with 50% excess air. The percentage conversion of the ethane is 90%;

of the ethane burned, 25% reacts to form CO and the balance reacts to form CO2.

calculate the molar composition of the stack gas on a dry basis and the mole ratio of

water to dry stack gas.

112

4.8c Material Balances on Combustion Reactors

Example 4.8-4 Combustion of a Hydrocarbon of Unknown Composition

A hydrocarbon gas is burned with air. The dry-basis product gas composition is 1.5

mole% CO, 6.0% CO2, 8.2% O2, and 84.3% N2. there is no atomic oxygen in the

fuel. Calculate the ratio of hydrogen to carbon in the fuel gas and speculate on what

the fuel might be. The calculate the percent excess air fed to the reactor.

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