95154977 Masgs Balance Reactive Systems
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Transcript of 95154977 Masgs Balance Reactive Systems
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Mass Balance on Reactive Processes
System
• Chapter 2.2
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CONTENT
• Stoichiometry• Limiting and Excess Reactant,
Fractional Conversion and Extent of Reaction
• Chemical Equilibrium• Multiple Reaction, Yield and Selectivity• Balance on Reactive System
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I. Stoichiometry• Stoichiometry – theory of proportions in which chemical
species combine with one another.• Stoichiometric equation of chemical reaction – statement
of the relative number of molecules or moles of reactants and products that participate in the reaction.
2 SO2 + O2 ---> 2 SO3• Stoichiometric ratio
– ratio of species stoichiometry coefficients in the balanced reaction equation
– can be used as a conversion factor to calculate the amount of particular reactant (or product) that was consumed (produced).
2 mol SO3 generated 2 mol SO2 consumed
2 mol SO2 consumed 1 mol O2 consumed33
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Test YourselfC4H8 + 6 O2 --------> 4 CO2 + 4 H2O
1. Is the stochiometric equation balance? Yes 2. What is stochiometric coefficient for CO2 ; 43. What is stochiometric ratio of H2O to O2 including it
unit–4 mol H2O generated/ 6 mol O2 consumed
1. How many lb-moles of O2 reacted to form 400lb-moles CO2 ; 600 lb-moles O2 reacted
1. 100 mol/min C4H8 fed into reactor and 50% is reacted. At what rate water is formed?
–200 mol/min water generated44
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II. LIMITING REACTANT & EXCESS REACTANT• The reactant that would run out if a reaction
proceeded to completion is called the limiting reactant, and the other reactants are termed excess reactants.
• A reactant is limiting if it is present in less than its stoichiometric proportion relative to every other reactant.
• If all reactants are present in stoichiometric proportion, then no reactant is limiting.
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%
stoich n
stoich n -
feed n
Excess Percentage
stoich n
stoich n -
feed n
Excess Fractional
111×=
=
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Example
C2H2 + 2H2 ------> C2H6Inlet condition: 20 kmol/h C2H2 and 50 kmol/h H2What is limiting reactant and fractional excess?
(H2:C2H2) o = 2.5 : 1(H2:C2H2) stoich = 2 : 1 H2 is excess reactant and C2H2 is limiting reactantFractional excess of H2 = (50-40)/40 = 0.25
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II. FRACTIONAL CONVERSION
• Fractional Conversion (f)
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%111fed mole
reacted moles f ,Conversion Percentage
fed mole
reacted moles f ,Conversion Fractional
×=
=
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II. EXTENT OF REACTION
• Extent of Reaction, ξ
ξ = extent of reactionni = moles of species i present in the
system after the reaction occurrednio = moles of species i in the system when
the reaction starts.vi = stoichiometry coefficient for species i
in the particular chemical reaction equation88
ξ
ξ
iioi
iioi
vnn
vnn
+=
+=or
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ExampleN2 + 3H2 ------------> 2NH3
Reactor inlet: 100 mol N2/s; 300 mol H2/s; 1 mol Ar/sIf fractional conversion of H2 0.6, calculate extent of reaction and the outlet composition.
Unreacted H2 or H2 outlet= (1-0.6) 300 = 120 mol H2/sSolve for extent of reaction : 60 mol/s
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ξ
ξξ
1
1
111
1111
1
1
1
==
−=−=
NH
Ar
N
H
n
n
n
n
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Test Yourself Page 1192 C2H4 + O2 ------->2 C2H4OThe feed to a reactors contains 100kmol C2H4 and 100 kmol O2.a) which is limiting reactant? C2H4b) Percentage of excess?
{ (100-50)/50} x 100% = 100%c) O2 out? C2H4 formed? Extent
of reaction?50kmol 100kmol C2H4 50kmol
d) if fractional conversion for limiting reactant is 50%, what is outlet composition and extent of reaction?50kmol C2H4; extent of reaction = 25 kmol;75 kmol O2 ;50 kmol C2H4O
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Test Yourself Page 1192 C2H4 + O2 ------->2 C2H4OThe feed to a reactors contains 100kmol C2H4 and 100kmol O2.
e) if reaction proceed to a point where 60kmol O2 left, what is fractional conversion for C2H4? Fractional conversion of O2 and extent of reaction?
fC2H4=0.8 fO2=0.4 extent of rxn=40 kmol
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Class DiscussionExample 4.6-1
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III. CHEMICAL EQUILIBRIUM• For a given set reactive species and
reaction condition, two fundamental question might be ask:
1. What will be the final (equilibrium) composition of the reaction mixture? – chemical engineering thermodynamics
2. How long will the system take to reach a specified state short of equilibrium? – chemical kinetics
• Irreversible reaction–. reaction proceeds only in a single direction
(from reactants to products)–. the concentration of the limiting reactant
eventually approaches zero.
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Chemical Equilibrium• Reversible reaction
– reactants form products for forward reaction and products undergo the reverse reactions to reform the reactants.
– Equilibrium point is a rate of forward reaction and reverse reaction are equal
• However the discussion to get the chemical equilibrium point is not covered in this text- learn in chemical engineering thermodynamic
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Class Discussion Example 4.6-2CO + H2O <----> CO2 + H2
nco = 1-ξnH2O = 2- ξn CO2 = ξnH2 = ξntotal = 3
K=yCO2 yH2 / y CO y H2O=1yY CO2 = ξ/3yH2 = ξ/3y CO = (1- ξ)/3y H2O = (2- ξ)/3 1515
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MULTIPLES REACTION, YIELD & SELECTIVITY
• Some of the chemical reaction has a side reaction which is formed undesired product- multiple reaction occurred.
• Effects of this side reaction might be:1. Economic loss2. Less of desired product is obtained for a
given quantity of raw materials3. Greater quantity of raw materials must be fed
to the reactor to obtain a specified product yield.
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selectivity =moles of desired product
moles of undesired product
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YIELD
• 3 definition of yield with different working definition
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Yield =Moles of desired product formed
Moles that would have been formed if there were no side reaction and the limiting
reactant had reacted completely
Yield =Moles of desired product formed
Moles of reactant fed
Yield =Moles of desired product formed
Moles of reactant consumed
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EXTENT OF REACTION FOR MULTIPLE REACTION
• Concept of extent of reaction can also be applied for multiple reaction
• only now each independent reaction has its own extent.
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ijjjiioi vnn ∑+= ξ
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Class DiscussionExample 4.6-3
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BALANCE OF REACTIVE PROCESSES
• Balance on reactive process can be solved based on three method:
1. Atomic Species Balance
1. Extent of Reaction
1. Molecular Species Balance
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Atomic Species Balance
– No. of unknowns variables– No. of independent atomic species
balance– No. of molecular balance on indep.
nonreactive species– No. of other equation relating the
variable =============================
No. of degree of freedom =============================
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Extent of Reaction
No. of unknowns variables+ No. of independent chemical reaction- No. of independent reactive species- No. of independent nonreactive species- No. of other equation relating the variable
=============================No. of degree of freedom
=============================
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Molecular Species Balance
No. of unknowns variables+ No. of independent chemical reaction- No. of independent molecular species balance- No. of other equation relating the variable
=============================No. of degree of freedom=============================
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Application of Method
C2H6 -------> C2H4 + H2
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40 kmol H2/min
n1 kmol C2H6/min
n2 kmol C2H4/min
Reactor
100 kmol C2H6/min
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Method 1: Atomic Species Balance
• All atomic balance is INPUT=OUTPUT• Degree-of-freedom analysis
2 unknowns variables (n1, n2)
- 2 independent atomic species balance (C, H)
- 0 molecular balance on indep. nonreactive species
- 0 other equation relating the variable
=============================
0 No. of degree of freedom
=============================
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Method 1: Atomic Species Balance
• Balance on atomic C (input= output)
200=2n1 + 2n2100=n1 + n2 [1]
• Balance on atomic H (input = output)100(6)=40(2) + 6n1+4n2520 = 6n1 + 4n2 [2]
Solve simultaneous equation, n1= 60 kmol C2H6/min; n2= 40
kmol C2H4/min
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100 kmol C2H6 2 knol C
=
n1 kmol C2H6 2 kmol C
+ n2(2)1 kmol C2H6 1 kmol C2H6
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Method 2: Extent of Reaction
• Degree-of-freedom analysis
2 unknowns variables (n1,n2)
+ 1 independent chemical reaction
- 3 independent reactive species (C2H6, C2H4, H2)
- 0 independent nonreactive species
- 0 other equation relating the variable
=============================
0 No. of degree of freedom
=============================2727
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Method 2: Extent of Reaction
• Write extent of reaction for each species
C2H6 : n1 = 100-ξC2H4 : n2= ξH2 : 40= ξ
Solve for n1 and n2 (ξ =40)
n1= 60 kmol C2H6/min; n2= 40 kmol C2H4/min
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Method 3: Molecular Species Balance
• Degree-of-freedom analysis
2 unknowns variables (n1, n2)
+1 independent chemical reaction
- 3 independent molecular species balance (C2H6, C2H4, H2)
- 0 other equation relating the variable
=============================
0 No. of degree of freedom
=============================
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Method 3: Molecular Species Balance
H2 balance (Gen=Output):H2 Gen= 40 kmol H2/min
C2H6 Balance (input=output + cons.):100 kmol C2H6/min = n1 kmol C2H6/min
+ 40 kmol H2 gen X (1 kmol C2H4 gen/1
kmol H2 gen)n1= 60 kmol C2H6/min
C2H4 balance (Gen.=Ouput):40 kmol H2 gen x (1 kmol C2H4 gen./ 1
kmol H2 gen) = n2n2= 40 kmol C2H4/min
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CLASS DISCUSSION
EXAMPLE 4.7-1
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Product Separation & Recycle
Overall ConversionReactant input to Process – reactant output
from ProcessReactant input to Process
Single Pass ConversionReactant input to Reactor – reactant output
from ReactorReactant input to Reactor
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75 mol B/min100 mol A/min75 mol A/minReactor Product
Separation Unit
25 mol A/min75 mol B/min
25 mol A/min
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Purging
• To prevent any inert or insoluble substance build up and accumulate in the system
• Purge stream and recycle stream before and after the purge have a same composition.
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ProductFresh FeedReactor Product
Separation Unit
Recycle Purge
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CLASS DISCUSSION
EXAMPLE 4.7-2 & 4.7-3
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Balance on Reactive Processes System: COMBUSTION REACTIONS
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COMBUSTION REACTIONS• Combustion – Rapid reaction of a fuel with
oxygen. This reaction releases tremendous quantities of energy that can be manipulated to boil water to produce steam.
• Combustion releases products such as CO, CO2 and SO2 and as chemical engineers, we are tasked to monitor and analyze the production of these noxious gases.
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COMBUSTION REACTIONS• Combustion fuels could be coal (carbon,
some hydrogen, and sulfur and various noncombustible materials), fuel oil (mostly high molecular weight hydrocarbons, some sulfur), gaseous fuel (such as natural gas, which is primarily methane) or it could be liquefied petroleum gas, which is usually propane or butane.
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COMBUSTION CHEMISTRYFrom Fuel Perspective
• Fuel contains carbonaceous material that will form either CO2 or CO, Hydrogen forming H2O and Sulfur forming SO2.
From O2 Source Perspective• For economic reason, AIR is the source of
oxygen in most combustion reactions. Dry air has the following average molar composition:
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N2 78.03%O2 20.99%Ar 0.94%CO2 0.03%H2, He, Ne, Kr, Xe 0.01%
Average MolecularWeight = 29.0
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• However, in most combustion calculations, it is acceptable to simplify this composition to 79% N2, 21% O2
• For combustion reaction, generally we have TWO type of expressions to express the mole composition of a gas, that is Composition On A Wet Basis and Composition On A Dry Basis.
• Composition On A Wet Basis is commonly used to denote mole fractions of a gas that contains water.
• Composition On A Dry Basis can also be used to denote mole fractions of the same gas that contains water but by excluding the presence of water in the calculation.
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• Example: A gas contains 33.3 mole% CO2, 33.3%
N2 and 33.3% H2O on wet basis is deemed to have a composition of 50.0 mol% CO2 and 50.0 mol% H2O on dry basis.
• It is important to have a grasp of knowledge on the correct technique to calculate these two type of compositions or to convert from dry basis to wet basis or vice versa.
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• The MAIN REASON is whenever flow rate of a product gas leaving the stack (stack gas or flue gas refers to product gas that leaves a combustion furnace) is measured, the measurement is for the total flow rate that also involved the product H2O, while on the other hand, common techniques for analyzing stack gases provide compositions on a dry basis.
• The procedure to convert Dry Basis to Wet Basis or vice versa follows exactly the same procedure outlined earlier on for converting Mass Compositions to Mole Compositions or vice versa in previous chapter.
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EXAMPLE 4.8-1:
A stack gas contains 60.0 mole% N2, 15.0% CO2, 10.0% O2 and the balance O2. Calculate the molar composition of the gas on a dry basis.
SOLUTION• Basis: 100 mol Wet Gas 60.0 mol N2
15.0 mol CO210.0 mol O285.0 mol dry gas (DG)
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60.0/85.0 = 0.706 mol N2/mol DG15.0/85.0 = 0.176 mol CO2/mol DG10.0/85.0 = 0.118 mol O2/mol DG
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Theoretical and Excess Air• Theoretical Oxygen: The moles (batch
process) or molar flow rate (continuous process) of O2 needed for complete combustion of all the fuel fed to the reactor, assuming that all the carbon in the fuel is oxidized to CO2 and all the hydrogen is oxidized to H2O.
• Theoretical Air: The quantity of air that contains the theoretical oxygen.
• Excess Air: The amount by which the air fed to the reactor exceeds the theoretical air.
• Percent Excess Air:4343
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Material Balances on Combustion Process
• The procedure for writing and solving material balances for a combustion reactor is the same as that for any other reactive system. However, be extra cautious with these:
• When you draw and label flow chart, be sure that the outlet stream (the stack gas) includes:
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Unreacted fuel unless you are told otherwise
Unreacted O2
H2O and CO2, also CO if the problem statement says so
N2 if the fuel burned With AIR and Not PURE O2
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• To calculate the O2 feed rate from a specified percent oxygen excess, first is to calculate the theoretical O2 from the fuel feed rate and the reaction stoichiometry for COMPLETE COMBUSTION, then calculate the oxygen feed rate by multiplying the theoretical O2 by (1+fractional excess of O2).
• Atomic balances are usually most convenient for use in the calculation.
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EXAMPLE 4.8-3
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