Combustion Presentation _audio

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1) Combustion Chemistry

CHAPTER 3- Material balance for

SIMPLE reactive system

Combustion –

the rapid reaction of a fuel with oxygen.*Note: Refer Felder pp.142.

When a fuel is burned, carbon in the fuel reacts to form either CO2 orCO, hydrogen forms H2O, and sulfur form SO2.

Partial or incomplete combustion – a combustion reaction in whichCO is formed from a hydrocarbon.

COMBUSTION REACTIONS





Example:

C + O2 → CO2 Complete combustion of carbon

C3 H8 + 5O2 → 3CO2 + 4H2 O Complete combustion of propane

C3 H8 + 3.5O2 → 3CO + 4H2 O Incomplete combustion of propane


Composition on a wet basis –

mole fraction of a gas that containswater.

Composition on a dry basis (Orsat Analysis)– mole fraction of thesame gas that without the water.

Stack gas or flue gas –

product gas that leaves a combustion furnace.



2) Theoretical and Excess Air



Theoretical Oxygen –

the moles (batch) or molar flow rate(continuous) of O2 needed for complete combustion of all the fuel fedto the reactor, assuming that all carbon in the fuel is oxidized to CO2 and all hydrogen is oxidized to H2O.

Theoretical Air – the quantity of air that contains theoretical oxygen.

Percent Excess Air:

*Note: Refer Felder pp.145.


100%

airof moles

airof molesairof moles AirExcess%

ltheoretica

ltheoretica fed



Example (Felder pp.143)

A stack gas contains 60.0 mole% N2 , 15.0% CO2 , 10.0% O2 , and the balanceH2O. Calculate the molar composition of the gas on a dry basis.



Basis of calculation.. gaswetmol100

Comp.

Mol (wet

basis)

O2 10

CO2 15

N2 60

H2O 15

Total 100 11.8%%O

100

85

10%O

2

2

Mol (Dry

basis)

% Mol

(Dry basis)

10 11.8

15 17.7

60 70.5

85 1.00




An Orsat analysis (a technique for stack analysis) yields the following dry basis composition: 65.0 mole% N2 , 14.0% CO2 , 11% CO and 10.0% O2.

A humidity measurement shows that the mole fraction of H2O in the stackgas is 0.0700. Calculate the stack gas component on a wet basis.



Basis of calculation.. basisdrymol100





Comp.Mol (Dry

basis)

O2 10

CO2 14

N2 65CO 11

H2O 0

Total 100

Mol (Wetbasis)

% Mol(Dry basis)

10

14

6511

x 0.0700

100 + x 1.00

mol7.527x 70.93x

70.070x-x x0.07x7

x100

x0.070

n

n y

total

OH

OH 2

2





Comp.Mol (Dry

basis)

O2 10

CO2

14

N2 65

CO 11

H2O 0

Total 100

0930. 107.527

10

y 2O

Mol (Wetbasis)

Mol fraction(Wet basis)

10 0.093

14 0.130

65 0.605

11 0.102

7.527 0.070

107.527 1.00




100 mol/h of butane (C4H10) and 5000 mol/h of air are fed into a combustionreactor. Calculate the percentage of excess air.



Find the theoretical air needed

OH 5CO4O2

13 H C 222104

22104

2104

Omol650Omol

2

13100 H Cmol100

Omol2

13 H Cmol1





mol3095

0.21

mol650n

mol650nn0.21

mol650nl,theoreticaO

airltheoretica

Oairltheoretica

O2

2

2

1.6%airexcessof Percentage

100%3095

35 airexcessof Percentage

100%n

nn airexcessof Percentage

airltheoretica

airltheoreticaairexcess

6

095000




Ethane is burned with 50% excess air. The percentage conversion of theethane is 90%; of the ethane burned, 25% react to form CO and the balancereact 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.



OH 3CO2O2

5 H C

OH 3CO2O2

7 H C

2262

22262





Analyze the information….

Feed: 50% excess air

Percentage conversion of C2H6: 90.0% @ 0.90 – 25% form CO, balanceform CO2



Basis of calculation: 100 mol of C2H6 fed

Process flow chart:

Process

unit

100 mol C2H6 n1 mol C2H6

n2 mol O2

n3 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

n0 mol air

0.79 mol N2/mol

0.21 mol O2/mol





Write system equation andoutline a solution procedure

Additional information – fractionalconversion C2H 6 = 0.90

621

162

62

62

H Cmol10n

mol100.1100nH Cunreactedof Moles

mol900.90100reactedH C Moles

mol100 fedH C Moles

0.90 f



Processunit


n2 mol O2

n3 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

n0 mol air

0.79 mol N2/mol

0.21 mol O2/mol

Find the amount of air fed into the process unit:

Calculate the theoretical O2 needed

Calculate the theoretical air needed

From % excess air, calculate the amount of excess air fed to the process unit





Processunit


n2 mol O2

n3 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

n0 mol air

0.79 mol N2/mol

0.21 mol O2/mol

mol16670.21

Omol350n

Omol350n0.21

2airltheoretica

2airltheoretica

Calculate the theoretical O2 needed

22262

262

OlTheoreticaOmol350Omol2

7

100 H Cmol100

Omol2

7 H Cmol1

:combustioncompleteFor






Processunit


n2 mol O2

n3 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

n0 mol air

0.79 mol N2/mol

0.21 mol O2/mol

Calculate the amount of excess air fed to the process unit

2500moln

0.501667

1667n

0.50n

nn

100%n

nnair%excess

0

0

airltheoretica

airltheoretica0

airltheoretica

airltheoretica0

N 2 in = N 2 out = 0.79 x 2500 = 1975 mol N 2

n3 = 1975 mol N 2




SIMPLE reactive systemProcess

unit

100 mol C2H6 10 mol C2H6

n2 mol O2

1975 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

2500 mol air

0.79 mol N2/mol

0.21 mol O2/molGiven 75% ethane reacted form CO2

COmol135n

formedCOmol135267.5 H Cmol67.5

COmol2 H Cmol1

:1reactionFrom

mol67.5mol900.75CO formtoreactH C Amount

25

262

262

262

Given 25% ethane reacted form CO

COmol45n

formedCOmol45222.5 Cmol22.5

COmol2 Cmol1

:2reactionFrommol22.5mol900.25CO formtoreactC

4

2

2

2

6

6

6

H

H

H Amount





Process

unit


n2

mol O2

1975 mol N2

45 mol CO

135 mol CO2

n6 mol H2O

50% excess air

2500 mol air

0.79 mol N2/mol

0.21 mol O2/mol

Atom C: 0 unk.

Atom H: 1 unk. (n6)

Atom O: 2 unk. (n2, n6)

Analyze the atomic balance for:

mol2702mol540 n 60-6002n 2n60600

OH mol1

H mol2OH moln

H Cmol1

H mol6H Cmol10

H Cmol1

H mol6H Cmol100

666

2

26

62

62

62

62

Solve the atomic balance on H …





Process

unit


n2 mol O2

1975 mol N2

45 mol CO

135 mol CO2

n6 mol H2O

50% excess air 2500 mol air

0.79 mol N2/mol

0.21 mol O2/mol

mol232.52

mol465 n 4652n 452n1050

OH mol1

Omol1OH mol

molCO1

Omol2COmol1

COmol1

Omol1COmol45

Omol1

Omol2Omoln

Omol1

Omol2Omol25000.21

222

2

2

2

2

2

2

2

2

270270

27035

2

Solve the atomic balance on O…



Component Mole Composition (y)

O2 232.5 0.097

CO 45 0.019

CO2 135 0.056

C2H6 10 0.004

N2 1975 0.824

Total 2397.5 100%



Process

unit


232.5 mol O2

1975 mol N2

45 mol CO

135 mol CO2

270 mol H2O

50% excess air

2500 mol air

0.79 mol N2/mol

0.21 mol O2/mol

Moles ratio of water todry stack gas:

gasstackdrymolOH mol0.113

gasstackdrymol2397.5

OH mol270

2

2




A hydrocarbon gas is burned with air. The dry basis product gas compositionis 1.5 mol%CO, 6.0 mol% CO2 , 8.2 % O2 , and 84.3% N2. There is no atomicoxygen in the fuel. Calculate the ratio of hydrogen to carbon in the fuel gasand speculate on what the fuel might be. Then calculate the percent excess airfed to the reactor.



Analyze the information….

Dry basis: 0.015 mol CO/mol, 0.06 mol CO2/mol,0.082 mol O2/mol, 0.843 mol N2/mol



Basis of calculation: 100 mol of dry basis product

Process flow chart:

Process

unit

n1 mol C

n2 mol H1.5 mol CO

8.2 mol O2

84.3 mol N2

6.0 mol CO2

n4 mol H2O

n3 mol air

0.79 mol N2/mol

0.21 mol O2/mol



N 2 in = N 2 out = 84.3 mol N 2

Calculate the air fed to the unit

mol106.700.79

mol84.3

n

mol84.3n0.79

3

3

C R 3 r b c or





Processunit

n1 mol C

n2 mol H1.5 mol CO

8.2 mol O2

84.3 mol N2

6.0 mol CO2

n4 mol H2O

106.70 mol air

0.79 mol N2/mol

0.21 mol O2/mol

Atom C: 1 unk. (n1)

Atom H: 2 unk. (n2, n4)

Atom O: 1 unk. (n4)

Analyze the atomic balance for:

mol7.5n 61.5n

COmol1

Cmol1COmol6

COmol1

Cmol1COmol1.5Cmoln

11

2

21

Solve the atomic balance on C…

CHAPTER 3 Material balance for





Processunit

n1 mol C

n2 mol H1.5 mol CO

8.2 mol O2

84.3 mol N2

6.0 mol CO2

n4 mol H2O

106.70 mol air

0.79 mol N2/mol

0.21 mol O2/mol

mol14.9n n1.51216.444.814

OH mol1

Omol1OH moln

COmol1

Omol1COmol1.5

COmol1

Omol2

COmol6Omol1

Omol2

Omol8.2Omol1

Omol2

Omol106.700.21

44

2

24

22

22

22

Solve the atomic balance on O…

Solve the atomic balance on H …

mol9.8n

H mol1

H mol2H mol1H moln

2

2

22

2

94

OO.






Calculate the H/C in the fuel

)(CH methane CH/molmol4CH/molmol3.977.5

29.8

n

n4

C

H

Percentage of excess air

2actionReOH O2

1 2H

1Reaction CO OC

22

22

mol14.95n

H 2mol

Omol0.5H mol29.8

C1mol

Omol1Cmol7.5n

nnn

ltheoreticaO

22ltheoreticaO

2ltheoreticaO1ltheoreticaOltheoreticaO

2

2

222






airmol71.190.21

Omol14.95n

Omol14.95n0.21

2airltheoretica

2airltheoretica


Calculate the amount of excess air fed to the process unit

49.9%100%71.19

71.19106.7air%excess

100%n

nnair%excess

airltheoretica

airltheoretica0




Exercise

A researcher burned n-Pentane (C5H12) with excess air in a continuouscombustion chamber.

The analysis on the product gas and report shows that product gas contains0.304 mole% pentane, 5.9 mole% oxygen, 10.2 mole% carbon dioxide and the balance nitrogen on a dry basis. Based on 100 mol of dry product gas,

i) Draw and label a flowchart for this process

(2 marks)

ii) Calculate the mol of n-Pentane and air fed in the combustion chamber,the percentage of excess air and the fractional conversion of n-Pentanein this process.

(8 marks)



O H COO H C 222125

658




Exercise

Ethylene (C2H4) has been commercially used for production of ethanol(C2H5OH) by hydration process:

However, some of the product is converted to diethyl ether ((C2H5)2O) in theside reaction:

The feed to the reactor contains ethylene (C2H4), steam (H2O) and inert gas(G). A sample of the reactor effluent gas is analyzed and found to contain 43.3mole% ethylene, 2.5 mole% ethanol, 0.14% ether, 9.3% inert gas andthe balance water. Based on 100 mol of effluent gas,

i) Draw and label a flowchart for this process (3 marks)

ii) Calculate the molar composition of the reactor feed, the percentageconversion of ethylene, the fractional yield of ethanol and theselectivity of ethanol production relative to diethyl ether production.

(9 marks)



OH H C O H H C 52242

O H O H C OH H C 2252522

CHAPTER 4 ENERGY balance for



A common practice is to arbitrarily designate a reference state for asubstance at which U or H is declared to equal zero, and then tabulateU and/or H for the substance relative to the reference state. *Note: ReferFelder pp. 339 and 359

In Chapter 7 (Felder), U and H are state properties of a species; theirvalues depend only on the state of the species – primarily on itstemperature and state of aggregation (solid, liquid or gas) and, to a

lesser extent, on its pressure (and for mixtures of some species, on itsmole fraction in the mixture).

When a species passes from one state to another, both U and H forthe process are independent of the path taken from the first state tosecond one Hypothetical Process Path *Note: Refer Felder pp 360

CHAPTER 4- ENERGY balance for

non reactive system

Reference State

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