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CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING …

CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING CALCULATIONSL ecture 11 combustion ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os LECTURE 11. combustion ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE2A combustion ProcessLECTURE 10. Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING .

CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING CALCULATIONS Lecture 11 Combustion Processes Prof. Manolito E Bambase Jr. Department of Chemical Engineering.

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Transcription of CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING …

1 CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING CALCULATIONSL ecture 11 combustion ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os LECTURE 11. combustion ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE2A combustion ProcessLECTURE 10. Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING .

2 University of the Philippines Los Ba os SLIDE3 CHEMICAL ReactiionsAssociated with combustion ProcessesC + O2========>CO2C + >CO2H + >H2OS + O2========>SO2 LECTURE 10. Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE4 Terms Associated with combustion ProcessesOrsatAnalysisRefers to the type of gas analysis which eliminates water as a component (dry-free basis). If water is included in the report, it is termed wet-basis AirThe amount of air required for complete combustion of C, H, and S.

3 It does not depend on how much material is actually but what can be AirThe amount of air in excess of that required for complete combustion . The % excess air is the same as % excess 10. Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE5 Example 11-1. Theoretical and Stoichiometric AirIn a given process, 100 kmol of carbon is burned in a furnace. It has been found that 20% of the carbon undergoes incomplete combustion resulting to CO production.

4 The rest of the carbon undergoes complete combustion . Determine the amount of air required (in kmol) if 50% excess O2must be Reactions:C + O2========>CO2C + >COLECTURE 10. Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE6 Example 11-1. Theoretical and Stoichiometric AirCalculate for theoretical O2needed:Assume that all the carbon is burned completely to kmol C (1/1) = 100 kmol O2It is not correct to do the following:C CO2:100 kmol C ( )(1/1) = 80 kmol O2C CO:100 kmol C ( )( ) = 10 kmol O2 LECTURE 10.

5 Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE7 Example 11-1. Theoretical and Stoichiometric AirTotal O2required stoichiometricallybased on the actual process:Stoichiometric O2= (80 + 10) kmol = 90 kmolTheoretical O2is based not on what is stoichiometricallyneeded according to what is actually Air = (100 kmol)(1 ) = kmolAnd the actual air supplied:Actual Air = kmol ( ) = kmolLECTURE 10.

6 Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE8 Example 11-2. combustion of Propane (C3H8)Fuels for motor vehicles other than gasoline are being eyed because they generate lower levels of pollutants than does propane (C3H8) has been suggested as a source of economic power for vehicles. Suppose that in a test, 20 kg of C3H8is burned with 400 kg of air to produce 44 kg of CO2and 12 kg of the percent excess 10.

7 Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE9 Example 11-2. combustion of Propane (C3H8)Write the overall combustion reaction for the fuel assuming it is burned completely:C3H8+ 5O2========>3CO2+ 4H2 OFor 20 kg of C3H8, the theoretical O2required is:38238238381kmol C H5 O20 kg C H= kmol kg C H1C H LECTURE 10. Solving Material Balance Problems Involving Reactive ProcessesProf.

8 Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE10 Example 11-2. combustion of Propane (C3H8)The actual O2supplied is221kmol air1air400 kg air= kmol O29 kg The percent excess air (or O2) kmol O kmol O%excess air = 100=28% kmol OLECTURE 10. Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE11 Example 11-3.

9 combustion of Methane (CH4)Generation of methane-rich biogas is a way to avoid high waste-disposal costs, and burning it can meet up to 60% of the operating costs for such waste-to-energy the complete combustion of kg of methane (CH4) in biogas with 300 kg of air. Determine the % excess of air, and the total moles and composition of the flue 10. Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE12 Example 11-3.

10 combustion of Methane (CH4)Degrees of Freedom Analysis: Atomic BalanceUnit: Reactorunknowns (P,x1,x2,x3,x4)+5independentatomic specie(s)independent nonreactivemolecularspecie(s)other equations:Degreesof freedom0 LECTURE 10. Solving Material Balance Problems Involving Reactive ProcessesProf. Manolito E Bambase Jr. Department of CHEMICAL ENGINEERING . University of the Philippines Los Ba os SLIDE13 Example 11-3. combustion of Methane (CH4)Write the atomic species balances (mole basis):(1)C:16 kg CH4(1/16)(1) = Px1(2) H:16 kg CH4(1/16)(4) = Px4(3) O:300 kg Air (1/29)( )(2) = 2Px2+ 2Px1+ Px4(4) N:300 kg Air (1/29)( )(2) = 2Px3(5) x:x1+ x2+ x3+ x4= 1 LECTURE 10.


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