basic balances on reactive processes

Basic Balances on Reactive Processes

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Motivation

• This topic was chosen because it is believed to be one of the core principles of chemical processes

• Students would first need to learn this foundation of process balances before building more specific skills

Introduction

• The knowledge of writing balances on reactive processes can be applied in many places: power plants, pharmaceutical plants, etc

• Usually, one would first need to draw a flowchart depicting the various processes

• Then, for most of the course, students will need to calculate the flow rates and compositions of all the flows in the flowchart

The General Balance Equation

Accumulation = In – Out + Generation - Consumption

• Accumulation: buildup within system

• In: enters through system boundaries

• Out: leaves through system boundaries

• Generation: produced within system

• Consumption: consumed within system

The Idea of Control Volume

• The irregular shape may represent any process unit (evaporator, reactor, condenser, etc.)

• We draw a rectangle around the unit to represent the desired control volume

• Then, the general balance equation is applied to this control volume– Whatever that is entering the rectangle is the “In”– Whatever that is exiting the rectangle is the “Out”– Whatever that is being produced/consumed in the rectangle is

the “Generation/Consumption”– Whatever that remains in the rectangle is the “Accumulation”

Process

Unitmin (kg CH4) mout (kg CH4)

Three Different Balances

Three different balances may be written to determine an unknown flow rate for a reactive process:

a) Molecular Species Balances

b) Atomic Species Balances

c) Extent of Reaction

Example 1

• Let us consider the dehydrogenation of ethane:

• Assuming 100 kmol/min of ethane is fed to the reactor and the molar flow rate of hydrogen in the product stream is 40 kmol/min

24262 HHCHC

Solution to Example 1

• Since most analyses would be done at steady-state (accumulation = 0), the general balance equation is reduced to the following:

In + Generation = Out + Consumption

• This example will now be solved using the three different balances introduced earlier:

a) Molecular Species Balances

b) Atomic Species Balances

c) Extent of Reaction

1a) Molecular Species Balances

• Usually used when a reaction is involved

• Thus, there should be some species that are being generated and some other species that are being consumed

• Once a generation/consumption term has been calculated, the other terms can be calculated directly from stoichiometric equation

1a) Molecular Species Balances

• H2: generation = output

• C2H6: input = output + consumption

• C2H4: generation = output

min/40min)/( 222HkmolgeneratedHkmolGenH

generatedHkmol

consumedHCkmolgeneratedHkmoln

2

6221

1

1

min

40100 min/60 621 HCkmoln

2

2

422

1

1

min

40n

generatedHkmol

generatedHCkmolgeneratedHkmol

min/40 422 HCkmoln

Process

Unit100 kmol C2H6/min 40 kmol H2/min

(kmol C2H6/min)

(kmol C2H4/min)1n

2n24262 HHCHC

1b) Atomic Species Balances

• Usually leads to most straightforward solution procedure, especially when more than one reaction is involved

• Since atomic species can neither be generated nor consumed, all atomic balances take the form of “input = output”

1b) Atomic Species Balances

• C:

• H:

• Solving simultaneously gives

Process


(kmol C2H6/min)

(kmol C2H4/min)1n

2n24262 HHCHC

21

42

422

62

621

62

62

100

1

2

min1

2

min1

2

min

100

nn

HCkmol

CkmolHCkmoln

HCkmol

CkmolHCkmoln

HCkmol

CkmolHCkmol

21

21

46520

)4()6()2(40)6(100

nn

nn

min/60 621 HCkmoln min/40 422 HCkmoln

1c) Extent of Reaction

• As for the molecular species balances, this method is usually used when a reaction is involved

• However, this method tends to be more useful when there are multiple reactions

• Recall:

– ν is the stoichiometric coefficient

• Reactants: -ve Products: +ve

– ξ is the extent of reaction

iii nn 0

1c) Extent of Reaction

• H2(ν=1):

• C2H6(ν=-1):

• C2H4(ν=1):

Process


(kmol C2H6/min)

(kmol C2H4/min)1n

2n24262 HHCHC

)(10min/40 2 Hkmol min/40kmol

6040100)1(1001 n

40)(102 n

min/60 621 HCkmoln

min/40 422 HCkmoln

Example 2

• Dehydrogenation of Propane

• The process is to be designed for a 95% overall conversion of propane. The reaction products 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 (moles recycled)/(mole fresh feed), and the single-pass conversion

26383 HHCHC

Solution to Example 2

• Basis: 100 mol fresh feed

• Product composition: n6/(n6+n7+n8)

• Recycle ratio: (n9+n10)/100

• Single-pass conversion: 100%*(n1-n3)/n1

• Need: n1, n3, n6, n7, n8, n9, n10

SeparatorReactor

Fresh feed

100 mol C3H8

n9 mol C3H8

n10 mol C3H6

Recycle

Product

n6 mol C3H8

n7 mol C3H6

n8 mol H2

n3 mol C3H8

n4 mol C3H6

n5 mol H2

n1 mol C3H8

n2 mol C3H6

Example 2: Discussion

• This time we have multiple units and a mixing point, therefore, we can draw a control volume around each unit, mixing point, and around the overall process

• “…95% overall conversion of propane.”

• “…0.555% of propane that leaves the reactor,”– n6 = 0.00555*n3 ->

• “…5% of the propylene in the first stream,”– n10 = 0.05*n7 -> to be solved later

n6 = 0.05*100 = 5 mol C3H8

n3 = 5/0.00555 = 900 mol C3H8

2 a) Molecular Species Balances

Overall system

• C3H6: generation = out

• H2: generation = out

7

83

6383

1

1

min

95n

consumedHCmol

generatedHCmolconsumedHCmol

n7 = 95 mol C3H6

8

83

283

1

1

min

95n

consumedHCmol

generatedHmolconsumedHCmol

n8 = 95 mol H2

n6 = 5 mol C3H8

n7 = 95 mol C3H6

n8 = 95 mol H2

2.6 mol% C3H8

48.7 mol% C3H6

48.7 mol% H2


Separator: In = Out (no reaction)

• C3H8

– n3 = n6+n9 ->

• C3H6

– Recall: n10 = 0.05*n7 ->

– n4 = n7 + n10 ->

– Border is dashed because we did not need n4

• H2

– n5 = n8

n9 = n3-n6 = 900-5 = 895 mol C3H8

n10 = 0.05*95 = 4.75 mol C3H6

n4 = 95+4.75=99.75 mol C3H6

n5 = 95 mol H2


Mixing Point: In = Out (no reaction)

• C3H8

– 100+n9 = n1 ->

• C3H6

– n10 = n2 ->

• Recycle ratio = (n9+n10)/100

• Single-pass conversion = 100%*(n1-n3)/n1

n1 = 100+895 = 995 mol C3H8

n2 = 4.75 mol C3H6

feedfresh mol

recycle mol 9.00

9.6%

2 b) Atomic Species Balances

Overall System: In = Out

• C:

• H:

• Product composition:

)3()3(5)3()3()3(100 776 nnn n7 = 95 mol C3H6

)2()6(95)8(5)2()6()8()8(100 8876 nnnn n8 = 95 mol H2

n6 = 5 mol C3H8

n7 = 95 mol C3H6

n8 = 95 mol H2

2.6 mol% C3H8

48.7 mol% C3H6

48.7 mol% H2


Separator: In = Out

• Recall: n10 = 0.05*n7 ->

• C:

• H:

• We have 2 equations and 3 unknowns (n4, n5, n10)

• Since we know that no H2 is recycled, n5 = n8, thus, the unknowns can be found by solving the equations simultaneously

]1[)3(75.4)3()3(95)3(5)3()3(900

)3()3()3()3()3()3(

94

1097643

nn

nnnnnn

n10 = 0.05*95 = 4.75 mol C3H6

]2[)6(75.4)8()2(95)6(95)8(5)2()6()8(900

)6()8()2()6()8()2()6()8(

954

109876543

nnn

nnnnnnnn

n4 = 99.75 mol C3H6n5 = 95 mol H2 n9 = 895 mol C3H8


Mixing Point: In = Out

• C:

• H:

• Now, we have 2 equations and 2 unknowns (n1, n2)

• Recycle ratio = (n9+n10)/100

• Single-pass conversion = 100%*(n1-n3)/n1

]3[)3()3()3(75.4)3(895300

)3()3()3()3()3(100

21

21109

nn

nnnn

]4[)6()8()6(75.4)8(895800

)6()8()6()8()8(100

21

21109

nn

nnnn

n1 = 100+895 = 995 mol C3H8n2 = 4.75 mol C3H6

feedfresh mol

recycle mol 9.00

9.6%

Example 2: Further Discussion• As can be seen above, the problem can be solved

simply by analyzing only the overall system, separator, and mixing point

• Furthermore, since these units do not involve reactions, the extent of reaction method does not apply

• Also, for the separating and mixing sections, the molecular species balances method leads to simpler calculations as compared to the atomic species balances method

• It is also worthy to note here that you can use different methods at different points– E.g. It may be easier to solve the overall system balance

using atomic species balances, then use molecular species balances to solve for the other points

Problem 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 mole% CH4, 19.4% O2, and 72.8% N2. The percentage conversion of methane is 90.0%, and the gas leaving the reactor contains 8 mol CO2/mol CO. Calculate the molar composition of the product stream.

)2(22

)1(22

3

2224

224

OHCOOCH

OHCOOCH

Hints

• Basis = 100 mol Feed

Process

Unit0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

nH2O mol H2O

nO2mol O2

nN2mol N2

100 mol

nCH4mol CH4

nCO mol CO

nCO2mol CO2

Problem 2

• The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product stream is analyzed and found to contain 51.7 mole% C2H5Br and 17.3% HBr. The feed to the reactor contains only ethylene and hydrogen bromide.

• If the molar flow rate of the feed stream is 165 mol/s, what is the extend of reaction?

Problem 3

• Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carbon dioxide and water:

• The feed to the reactor contains equimolar amounts of methane and oxygen. Assume a basis of 100 mol feed/s. The fractional conversion of methane is 0.9. and the fractional yield of formaldehyde is 0.855.

• Calculate the two extents of reaction and the molar composition of the reactor output stream.

OHCOOCH

OHHCHOOCH

2224

224

22

References

• Felder, Richard M. Elementary principles of chemical processes. New York: John Wiley, 1999. Print.

basic balances on reactive processes

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