chapter 5: extent of reactionsfalhuthali.kau.edu.sa/getfile.aspx?id=238734&lng=ar&fn... ·...

27

Chapter 5: Extent of Reaction

𝜉 =𝑛𝐴,𝑜𝑢𝑡 − 𝑛𝐴,𝑖𝑛

𝜐𝐴Extent of reaction:

𝑛𝐴,𝑜𝑢𝑡 = 𝑛𝐴,𝑖𝑛 + 𝜐𝐴𝜉nN2,out = nN2,in − Consumption

𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 𝑖𝑛𝑝𝑢𝑡 - 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

0 = 𝑖𝑛𝑝𝑢𝑡 - 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

Steady state system:

𝑛1,𝑁2𝑛1,𝐻2

ReactorInput (1) Output (2)

𝑛2,𝑁𝐻3

N2 + 3H2 2NH3

𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑖𝑛𝑝𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

nNH3,out = nNH3,in + Generation

28

Chapter 5: Species Mole Balances, SMB


𝜐𝐴

𝜉 =𝑛𝐴,𝑓𝑖𝑛𝑎𝑙 − 𝑛𝐴,𝑖𝑛𝑖

𝜐𝐴

Extent of reaction (Batch process):

Extent of reaction (Continuous process):

𝜉 =−𝑓 ∙ 𝑛𝑖𝑛,𝑙𝑖𝑚𝑒𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡

𝜐𝐿𝑅

f =Conversion factor of the limited reactant

𝑓 = 𝑓𝐷𝐶 =𝑚𝑜𝑙𝑒𝑠(𝑚𝑎𝑠𝑠) 𝑜𝑓 𝑙𝑖𝑚𝑖𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑟𝑒𝑎𝑐𝑡𝑒𝑑

𝑚𝑜𝑙𝑒𝑠(𝑚𝑎𝑠𝑠) 𝑜𝑓 𝑙𝑖𝑚𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑓𝑒𝑑× 100%

𝑛𝐴,𝑜𝑢𝑡 = 𝑛𝐴,𝑖𝑛 + 𝜐𝐴𝜉

𝑛𝐴,𝑓𝑖𝑛𝑎𝑙 = 𝑛𝐴,𝑖𝑛𝑖 + 𝜐𝐴𝜉

Species Material Balances


𝜐𝐴

29

Chapter 5: Species Mole Balances

Example 5.7: Reaction in Which the Fraction Conversion is Specified:

The chlorination of methane occurs by the following reaction

CH4 + Cl2 CH3Cl + HCl

You are asked to determine the product composition if the conversion of

the limiting reactant is 67%, and the feed composition in mole % is given as:

40% CH4, 50%Cl2,and 10% N2.

Assumptions: The reactor is Open, Steady state process

Solution:

𝑛1,Cl2𝑛1,CH

4

𝑛1,N2

1 2

𝑛2,Cl2𝑛2,CH

4

𝑛2,N2

Reactor

Species moles % feed

CH4 40%

Cl2 50%

N2 10%

fLR 67%

30


CH4 + Cl2 CH3Cl + HCl

𝑛1,Cl2𝑛1,CH

4

𝑛1,N2

1 2

𝑛2,Cl2𝑛2,CH

4

𝑛2,N2𝑛2,HCl𝑛2,CH3Cl

Reactor

31


Example 10.2: A Reaction in Which the Fraction Conversion is to Be

Calculated:

H2S is toxic in very small quantities and is quite corrosive to process

equipment.

A proposed process to remove H2S is by reaction with SO2:

2H2S(g) + SO2(g) 3S(s) + 2H2O(g)

In a test of the process, a gas stream containing 20% H2S and 80% CH4

was combined with a stream of pure SO2.

The process produced 5000 kg of S(s), and in the product gas the ratio of

SO2 to H2S was equal to 3, and the ratio of H2O to H2S was 10.

You are asked to determine the fractional conversion of the limiting

reactant, and the feed rates of the H2S and SO2 streams.

Assumptions: The reactor is Open, Steady state process.

Solution:

32


Species moles % feed

CH4 80%

H2S 20%

𝑚4,S 5000 kg

𝑛3,SO2/𝑛3,H2S 3

𝑛3,H2O/𝑛3,H2S 10

𝑛1𝑦1,H2S=0.20

𝑦1,CH4=0.80

𝑛1,H2S𝑛1,CH4

1 3

𝑛3,H2S𝑛3,CH4

𝑛3,H2O𝑛3,SO2

Reactor

2

4

𝑛2,SO2

n4,s= 156.25 mol

𝑚4,S=5000 kg

Product

2H2S(g) + SO2(g) 3S(s) + 2H2O(g)

Moles, Kmol MW Mass, Kg

Inp

ut

n1 H2S 114.6 34 3896.4

n1 CH4 458.48 16 7335.68

n2 SO2 83.3 64 5331.2

Sum input 656.38 Kmol 16563.3 Kg

Ou

tpu

t

n3 H2S 10.42 34 354.28

n3 CH4 458.48 16 7335.68

n3 H2O 104.2 18 1875.6

n3 SO2 31.2 64 1996.8

n4 S 156.25 32 5000

Sum out 760.55 Kmol 16562.4 Kg

33

Chapter 5: Process Involving Multiple Reactions

Example 5.8: Material Balances Involving Two Ongoing Reactions

Formaldehyde (CH2O) is produced industrially by the catalytic oxidation of

methanol (CH3OH) according to the following reaction:

CH3OH + ½O2 CH2O + H2O (1)

Unfortunately, under the conditions used to produce formaldehyde an

undesired reaction occurs, that is:

CH2O + ½ O2 CO + H2O (2)

Assume that methanol and twice the stoichiometric amount of air needed

for complete conversion of the CH3OH to the desired products (CH2O and

H2O) are fed to the reactor.

Also assume that 90% conversion of the methanol results, and that a 75%

yield of formaldehyde occurs based on the theoretical production of CH2O

by Reaction 1.

Determine the composition of the product gas leaving the reactor.

Solution:

34


𝑦2,O2 21%

𝑦2,N2 79%

𝐶𝑜𝑛𝑣. , 𝑓CH3OH 90%

𝑌𝑖𝑙𝑒𝑑CH2O 75%

𝑛1,CH3OH

1 3

𝑛3,CH3OH𝑛3,CH2O𝑛3,H2O𝑛3,CO𝑛3,O2𝑛3,N2

Formaldehyde

Reactor

𝑛2(Air)

𝑦2,O2=0.21

𝑦2,N2 =0.79

𝑛2,O2𝑛2,N2

Product

2

• Assume that CH3OH

requires twice the

stoichiometric amount of

Air are fed to the reactor.

CH3OH + ½O2 CH2O + H2O (1)

CH2O + ½ O2 CO + H2O (2)

35

Example 5.9: Analysis of a Bioreactor

A bioreactor is a vessel in which biological conversion is carried out. The

following overall reactions occurs:

Reaction 1: C6H12O6 2C2H5OH + 2CO2

Reaction 2: C6H12O6 2C2H3CO2H + 2H2O

In a batch process, a tank is charged with 4000 kg of a 12% solution of

glucose(C6H12O6) in water. After fermentation, 120 kg of CO2 are produced

and 90 kg of unreacted glucose(C6H12O6) remains in the solution. What are

the weight (mass) percent of ethanol(C2H5OH) and propenoic

acid(C2H3CO2H) in the solution at the end of the fermentation process?

Assume that none of the glucose(C6H12O6) is assimilated(digested) into the

bacteria.

Solution:


36


C6H12O6 2C2H5OH + 2CO2 (1)

C6H12O6 2C2H3CO2H + 2H2O (2)

𝑚𝑖𝑛𝑖,𝑠𝑜𝑙𝑛=4000 kg

𝑛𝑖𝑛𝑖,𝐻2𝑂𝑛𝑖𝑛𝑖,𝐶6𝐻12𝑂6

𝑛𝑓𝑖𝑛𝑎𝑙𝑛𝑓𝑖𝑛𝑎𝑙,𝐻2𝑂𝑛𝑓𝑖𝑛𝑎𝑙,𝐶6𝐻12𝑂6𝑛𝑓𝑖𝑛𝑎𝑙,𝐶2𝐻3𝐶𝑂2𝐻𝑛𝑓𝑖𝑛𝑎𝑙,𝐶𝑂2𝑛𝑓𝑖𝑛𝑎𝑙,𝐶2𝐻5𝑂𝐻

𝑚𝑖𝑛𝑖,𝑠𝑜𝑙𝑛=4000 kg Soln𝑥𝑖𝑛𝑖,𝐶6𝐻12𝑂6 =0.120𝑥𝑖𝑛𝑖,𝐻2𝑂 =0.88

𝑚𝑓𝑖𝑛𝑎𝑙,𝐶𝑂2 =120 kg CO2

Unreacted C6H12O6

𝑚𝑓𝑖𝑛𝑎𝑙,𝐶6𝐻12𝑂6 =90 kg

C6H12O6 H2O CO2 C2H5OH C2H3CO2HMW(g/mol) 180 18 44.0 46 72

37

Chapter 5: Element Material Balances, EMB

Input (atoms)= Output (atoms)

CO2 + H2O H2CO3

𝑛𝐴,𝑜𝑢𝑡 = 𝑛𝐴,𝑖𝑛 + 𝜐𝐴𝜉Species Moles Balances:

For most problems it is easier to apply mole balances, but for some

problems, such as problems with complex or unknown reaction

equations, element balances are preferred.

Element Material Balances, EMB: Number of atoms enter the

reaction EQUAL number of atoms leave the reaction

38

Chapter 5: Element Material Balances, EMB

𝑛1,H2O

𝑛3=100 mol

𝑦3,H2CO3=0.05

𝑦3,H2O= 0.95

𝑛3,H2CO3

𝑛3,H2OAbsorber

𝑛2,CO2

1

2

3

Example 5: Carbon dioxide is absorbed in water in the process shown

below. The reaction isCO2 + H2O H2CO3

Apply the element balance to find the unknowns in the flow chart?

39

Chapter 5: Element Material Balances

Example 5.11: HydrocrackingResearchers in the field oil industry study the hydrocracking of pure components,

such as octane (C8H18) to understand the behavior of cracking reactions.

In one such experiment for the hydrocracking of octane (C8H18) , the cracked

products had the following composition in mole percent: 19.5% C3H8, 59.4% C4H10,

and 21.1% C5H12.

You are asked to determine the molar ratio of hydrogen consumed to octane reacted

for this process.

Solution:

40

Chapter 5: Element Material Balances

𝑛1,C8H18

𝑛3𝑛3,C3H8𝑛3,C4H10𝑛3,C5H12𝑦3,C3H8=0.195

𝑦3,C4H10 =0.594

𝑦3,C5H12 =0.211

Lab

Reactor

𝑛2,H2

1

2

3

Species Product Moles

percentage

C3H8 19.5%

C4H10 59.4%

C5H12 21.1%

41

Chapter 5: Material Balances Involving Combustion

Wet basis: all the gases resulting from a combustion process including the

water vapor, known as Flue or stack gas.

Dry basis: all the gases resulting from a combustion process not9includingthe water vapor.

Complete combustion: the complete reaction of the hydrocarbon fuel

producing CO2, SO2, and H2O.

Partial combustion: the combustion of the fuel producing at least some

CO.

Theoretical air (or theoretical oxygen): the minimum amount of air (or

oxygen) required to be brought into the process for complete combustion.

Sometimes this quantity is called the required air (or oxygen).

CH4+ Air CO2 (g)+ 2H2O (g) + Energy

42


% Excess air =excess airrequired air

∙ 100

= excess O2 0.21

required O2 0.21∙ 100 =

excess O2required O2

∙ 100

% Excess air =O2 in(enter the process)−O2 required

O2 required∙ 100

O2 excess = O2,in (enter the process) – O2 required

Excess air (or excess oxygen): excess air (or oxygen) is the amount of air

(or oxygen) in excess of that required for complete combustion.

The calculated amount of excess air does not depend on how much material is

actually burned but what is possible to be burned. Even if only partial

combustion takes place.

43


Example 5.12: Excess Air

Fuels other than gasoline are being eyed for motor vehicles because they

generate lower levels of pollutants than does gasoline. Compressed propane

is one such proposed fuel.

Suppose that in a test 20 kg of C3H8 is burned with 400 kg of air to

produce 44 kg of CO2 and 12 kg of CO.

What was the percent excess air?

Solution:C3H8+ 5O2 3CO2 + 4H2O (g)

44


Example 5.13: A Fuel Cell to Generate Electricity From Methane

Fuel cell is an open system into which fuel and air are fed, and the outcome

are electricity and waste products. Figure bellow is a sketch of a fuel cell in

which a continuous flow of methane (CH4) and air (O2 plus N2) produce

electricity plus CO2 and H2O.

Special membranes and catalysts are needed to promote the reaction of

CH4.

Based on the data given in Flow chart, you are asked to calculate the

composition of the products in stream 3.

Solution:

45


𝑚1,CH4=16.0 kg

𝑛1,CH4

𝑚2,𝑎𝑖𝑟=300 kg

𝑛2,𝑎𝑖𝑟𝑛2,O2𝑛2,N2𝑦2,O2=0.21

𝑦2,N2 =0.79

Lab

Reactor

𝑛3𝑛3,CO2𝑛3,N2𝑛3,O2𝑛3,H2O

1

3

2Air H2O CO2 CH4 MW(g/mol) 29 18 44.0 16

46


Example 6: Combustion of EthaneEthane is burned with 50% excess air. The percentage conversion of the

ethane is 90%.

The ethane burned: 25% reacts to form CO and 75 % reacts to form CO2?

Calculate the molar composition of the stack gas on a dry and wet basis and

the mole ratio of water to dry stack gas?

50% 𝑒𝑥𝑐𝑒𝑠𝑠 𝑎𝑖𝑟𝑛2,𝑎𝑖𝑟𝑦2,O2=0.21

𝑦2,N2 =0.79

Combustion

Unit

𝑛3,𝐶2𝐻6𝑛3,CO2𝑛3,CO𝑛3,N2𝑛3,O2𝑛3,H2O

1 3

2

𝑛1,𝐶2𝐻6=100 mol

C2H6+ 7 2O2 2CO2 + 3H2O

C2H6+ 5 2O2 2CO + 3H2O

47


50% 𝑒𝑥𝑐𝑒𝑠𝑠 𝑎𝑖𝑟𝑛2,𝑎𝑖𝑟𝑦2,O2=0.21

𝑦2,N2 =0.79

Combustion

Unit

𝑛3,𝐶2𝐻6𝑛3,CO2𝑛3,CO𝑛3,N2𝑛3,O2𝑛3,H2O

1 3

2

𝑛1,𝐶2𝐻6=100 mol

C2H6+ 7 2O2 2CO2 + 3H2O

C2H6+ 5 2O2 2CO + 3H2O

excess air = 50%

fC2H6 =90%

25% C2H6 reacts to form CO

75 % C2H6 reacts to form CO2

48


Example 7: Combustion of a Hydrocarbon Fuel of Unknown Composition

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

composition is 1.5 mole% CO, 6.0% C02,8.2% 02, 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?

Calculate the percent excess air fed to the reactor?

𝑛2,𝑎𝑖𝑟𝑦2,O2=0.21

𝑦2,N2 =0.79

Combustion

Unit

𝑛3=100 mol (dry gas )

𝑛3,CO2𝑛3,CO𝑛3,O2𝑛3,N2

𝑛4,H2O

1 3

2

𝑛1,𝐶𝑛1,𝐻

ElementsInput

Moles %

CO 1.5%

CO2 6.0%

O2 8.2%

N2 84.3%

Output composition

on a dry basis

49


𝑛2,𝑎𝑖𝑟𝑦2,O2=0.21

𝑦2,N2 =0.79

Combustion

Unit

𝑛3=100 mol (dry gas )

𝑛3,CO2 = 1.5 mol

𝑛3,CO = 6.0 mole

𝑛3,O2 = 8.20 mol

𝑛3,N2 =84.3 mol

𝑛4,H2O

1 3

2

𝑛1,𝐶𝑛1,𝐻

Output composition on a dry basis

50


Example 5.14: Combustion of CoalA local utility burns, the moisture in the input fuel was 3.90%.

The air on the average contained 0.0048 kg H2O/kg dry air.

The refuse showed 14.0% unburned coal, with the remainder being ash.

You are asked to check the consistency of the data before they are stored in

a database. is the consistency satisfactory?

What was the average percent excess air used?

Elements Input Moles %

C 83.05%

H 4.45 %

O 3.36%

N 1.08%

S 0.70%

Ash 7.36%

Input composition on a dry basis

ElementsInput

Moles %

CO2+SO2 15.4%

CO 0

O2 4%

N2 80.6%

Output composition

on a dry basis

51


𝑚𝑡𝑜𝑡

𝑚2,𝑎𝑖𝑟=300 kg

𝑛2,𝑎𝑖𝑟𝑦2,O2=0.21

𝑦2,N2 =0.79

Lab

Reactor

𝑛3𝑛3,CO2𝑛3,N2𝑛3,O2𝑛3,H2O

1 3

2

𝑚1=100 kg

𝑚1,𝐶=83.05 kg

𝑚1,𝐻=4.45 kg

𝑚1,O=3.36 kg

𝑚1,N= 1.08 kg

𝑚1,S=0.70 kg

𝑚1,Ash =7.36 kg

chapter 5: extent of reactionsfalhuthali.kau.edu.sa/getfile.aspx?id=238734&lng=ar&fn... ·...

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