fundamentals of material balancesteacher.buet.ac.bd/shoukat/che201handout3.pdf · balances on...

Material Balance-Part 2

Chapter 4

Fundamentals of Material Balances

April 5, 2009 ChE 201/[email protected] 2

Balances on Reactive Systems

• Stoichiometry– The theory of proportions in which chemical

species combine with one another.– Example: 2 SO2 + O2 2 SO3

• Stoichiometric Ratio– Ratio of stoichiometric coefficients – Example

2 mol SO3 produced1 mol O2 reacted

2 mol SO2 reacted2 mol SO3 produced


Terminology

• Limiting reactants– Exist less than stoichiometric proportion

• Excess reactants– Exist more than stoichiometric proportion

• Example2SO2 + O2 2 SO3

(30 mol) (10 mol)

Excess Limiting


Terminology

• Fractional excess • Percent excess• Example

– H2 + Br2 2HBr – H2 : 25 mol /hr– Br2 : 20 mol /hr– Fractional Excess H2 = (25 – 20 ) /20 = 0.25– Percent Excess = 25 %

s

s

nnn −

100s

s

n nn−

×


Terminology

• Fractional conversion– Chemical reactions are not always completed.– Factional conversion

• f = (moles reacted) / (moles fed) –– When fresh feed consists of more than one When fresh feed consists of more than one

material the conversion must be stated for a material the conversion must be stated for a single component, usually the limiting reactant.single component, usually the limiting reactant.


Balances of Atomic and Molecular Species

• Methods for solving mass balances with reactions– Using balances on molecular species– Using balances of atoms– Using the extent of reaction

• For multiple reactions, sometimes it is more convenient to use atomic balances

• Atomic species balances generally lead to the most straightforward solution procedure, especially when more than one reaction is involved.

• Extents of 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.


A simple Problem

The oxidation of ethylene to produce ethylene oxide proceeds as: 2 C2H4 + O2 = 2 C2H4OThe feed to the reactor contains 100 Kmol C2H4 and 200 Kmol O2. What is the limiting reactant? What is the percentage excess of the other reactant? If the reaction proceeds to a point where the fractional conversion of the limiting reactant is 50%, how much of each reactant and products is present at the end?


Multiple Reaction, Yield, Selectivity

• Multiple reaction : one or more reaction– Side Reaction : undesired reaction – Example: Production of ethylene

C2 H6 C2H4 + H2

Side Reactions C2 H6 + H2 2CH4

C2 H4 + C2 H6 C3H6 + CH4

– Design Objective• Maximize desired products (C2 H4 )• Minimize undesired products (CH4 , C3 H6 )


Terminology

– Yield and Selectivity are used to describe the degree to which a desired reaction predominates over competing side reactions.

moles of desired product formed = moles that would have been formed if therewere no side reactions and the limiting reactanthad reacted completelymoles of desired product formed

moles of reactant

YIELD

fedmoles of desired product formed

moles of reactant consumed

moles of desired product formedmoles of undesired product formed

SELECTIVITY =

Yield =

Yield =


Determination of yield and selectivityReaction:C2 H6 + 2.5O2 →

2CO + 3H2 OC2 H6 + 3.5O2 →

2CO2 +3H2 OUndesired product: CO80% reactor conversion

100 mol C2 H6

500 mol O2

REACTOR Products

20 mol C2H6

120 mol CO2

40 mol CO240 mol O2

240 mol H2O

Yield = 0.6 or 1.2 or 1.5Selectivity = 3.0


ProblemEthane is burned with air in a continuous steady- state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide, and water. The feed to the reactor contains 10% C2 H6 . The percentage conversion of ethane is 80%, and gas leaving the reactor contains 8 mol CO2 per mol of CO. Determine molar composition of product gas.

Reaction: C2H6 + 5/2 O2 → 2CO + 3H2OC2H6 + 7/2 O2 → 2CO2 +3H2O


Concept of Purge- why needed?• Production of Ethylene Oxide• Reaction: 2C2 H4 + O2 → 2C2

H4

O•

Mixture of Ethylene and air stream is charged to the reactor •

Reactor effluent is charged to absorber and gas stream containing N2

, O2

and unreacted

ethylene is charged back to reactor

Reactor Absorber

Recycle

Fresh Feed

Products

Solvent


Production of Ethylene Oxide

• Problem: accumulation of N2

• Solution: allow purging of inert species

Reactor Absorber

Recycle

Fresh Feed

Products

Purge stream

Solvent

60 mol C2 H4 /s

30 mol O2 /s

113 mol N2 /s

100 mol C2 H4 /s

50 mol O2 /s

565 mol N2 /s

50 mol C2 H4 /s

50 mol C2 H4 O/s 25 mol O2 /s

565 mol N2 /s

50 mol C2 H4 /s

25 mol O2 /s

565 mol N2 /s

10 mol C2 H4 /s

5 mol O2 /s

113 mol N2 /s

50 mol C2 H4 O/s

solvent


Getting rid of undesired materials in recycle streamGetting rid of undesired materials in recycle stream..

ReactorProduct

SeparationUnit

Recycle

Reactants Products

Purging

Purging


Definition

• A Recycle Stream is a term denoting a process stream that returns material from downstream of a process unit back to the process unit.

• A Bypass stream - one that skips one or more stages of the process and goes directly to another downstream stage.

• A Purge stream – a stream bled off to remove an accumulation of inert or unwanted material that might otherwise build up in the recycle stream.


Example 4.7-3, page 139Recycle and Purge in the Synthesis of Methanol

Methanol is produced in the reaction of carbon dioxide and hydrogen: CO2 + 3H2 →

CH3 OH + H2 OThe fresh feed to the process contains hydrogen, carbon dioxide and 0.40 mole percent inerts(I). The reactor effluent passes to a condenser that removes essentially all of the methanol and water formed and none of the reactants or inerts. The latter substances are recycled to the reactor. To avoid buildup of the inerts in the system, a purge stream is withdrawn from recycle. The feed to the reactor contains 28.0 mole% CO2, 70 mole% H2 and 2% inerts. The single pass conversion of H2 is 60%. Calculate the molar flow rates and molar compositions of the fresh feed, the total feed to the reactor, the recycle stream, and the purge stream for a methanol production rate of 155 kmol methnol/hr.


Reaction: CO2

+ 3H2 → CH3

OH + H2

OBasis: 155 kmol

CH3

OH/h

REACTOR CONDENSER

Recycle

N8 kmol/h

Purge

N7 kmol/h

n1 kmol/h

155 kmol CH3 OH/h

155 kmol H2 O/h

x1C mol CO2 /mol

(0.996-x1C)mol H2 /mol

0.004 mol I/mol

0.28 mol CO2 /mol

0.70 mol H2 /mol

0.02 mol I/mol

n2 kmol/h

n3 kmol CO2 /mol

n4 kmol H2 /mol

n5 kmol I/mol

155 kmol CH3 OH/h155 kmol H2 O/h

N6 kmol/h

x6C mol CO2 /mol

x6H mol H2 /mol

(1-x6C-x6H)mol I/mol

x6C mol CO2 /mol

x6H mol H2 /mol

(1-x6C-x6H) mol I/mol

x6C mol CO2 /mol

x6H mol H2 /mol

(1-x6C-x6H) mol I/mol


Combustion Reaction

• Combustion– A rapid reaction of a fuel with oxygen– Fuels : coal, fuel oil, gas fuel, solid fuel, …– Complete combustion / incomplete combustion– Wet basis composition / dry basis composition

Remember:Orsat analysis yields dry basis composition


Terminology

• Theoretical oxygen : Amount of oxygen needed for complete combustion – all carbon in the fuel is oxidized to CO2 and – all the hydrogen is oxidized to H2 O

• Theoretical air : The quantity of air that contains theoretical oxygen

Air(theo) = 4.76 x O2(theo)

• Excess air : The amount by which the air fed to reactor exceeds the theoretical air

• Percent excess air

( ) %100air moles

air) (molesair) (moles

ltheoretica

altheroreticfed ×−


Composition of Flue or stack gas

• Wet Basis = => Dry Basis- Basis: 1 mole wet gas- basic idea, subtract the water and express the rest in %

• Dry Basis = => Wet Basis - Basis: 1 mole wet gas- need one extra information: How much water is there in one mole of wet gas? (Say, y mole fraction out of 1 mole wet gas)

- subtract water from one mole of wet gas and get the mole of dry gas. (dry gas=1-y)

- Now, dry gas fraction is (1-y)*yi, where yi is mole fraction of ith dry gas components


Questions

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