Lecture 11 Combustion Processes

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1 CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING CALCULATIONS Lecture 11 Combustion Processes

2 LECTURE 11. Combustion Processes A Combustion Process 2

3 Chemical Reactiions Associated with Combustion Processes C + O 2 ========> CO 2 C + 0.5O 2 ========> CO 2H + 0.5O 2 ========> H 2 O S + O 2 ========> SO 2 3

4 Terms Associated with Combustion Processes Orsat Analysis Refers 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 analysis. Theoretical Air The amount of air required for complete combustion of C, H, and S. It does not depend on how much material is actually but what can be burned. Excess Air The amount of air in excess of that required for complete combustion. The % excess air is the same as % excess O 2. 4

5 Example Theoretical and Stoichiometric Air In 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. The rest of the carbon undergoes complete combustion. Determine the amount of air required (in kmol) if 50% excess O 2 must be satisfied. Relevant Reactions: C + O 2 ========> CO 2 C + 0.5O 2 ========> CO 5

6 Example Theoretical and Stoichiometric Air Calculate for theoretical O 2 needed: Assume that all the carbon is burned completely to CO kmol C (1/1) = 100 kmol O 2 It is not correct to do the following: C CO 2 : 100 kmol C (0.80)(1/1) = 80 kmol O 2 C CO: 100 kmol C (0.20)(0.5/1) = 10 kmol O 2 6

7 Example Theoretical and Stoichiometric Air Total O 2 required stoichiometrically based on the actual process: Stoichiometric O 2 = ( ) kmol = 90 kmol Theoretical O 2 is based not on what is stoichiometrically needed according to what is actually burned. Theoretical Air = (100 kmol)(1/0.21) = kmol And the actual air supplied: Actual Air = kmol (1.5) = kmol 7

8 Example Combustion of Propane (C 3 H 8 ) Fuels for motor vehicles other than gasoline are being eyed because they generate lower levels of pollutants than does gasoline. Compressed propane (C 3 H 8 ) has been suggested as a source of economic power for vehicles. Suppose that in a test, 20 kg of C 3 H 8 is burned with 400 kg of air to produce 44 kg of CO 2 and 12 kg of CO. Calculate the percent excess air. 8

9 Example Combustion of Propane (C 3 H 8 ) Write the overall combustion reaction for the fuel assuming it is burned completely: C 3 H 8 + 5O 2 ========> 3CO 2 + 4H 2 O For 20 kg of C 3 H 8, the theoretical O 2 required is: 20kg C H 1kmol C H 5O =2.27kmol O kgC3H 8 1C3H 8 9

10 Example Combustion of Propane (C 3 H 8 ) The actual O 2 supplied is 1kmol air 1air 400kg air =2.90kmol O 29kg air 0.21O 2 2 The percent excess air (or O 2 ) is 2.90kmol O -2.27kmol O 2.27kmol O 2 2 %excess air = 100=28% 2 10

11 Example Combustion of Methane (CH 4 ) 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 plants. Consider the complete combustion of 16.0 kg of methane (CH 4 ) in biogas with 300 kg of air. Determine the % excess of air, and the total moles and composition of the flue gas. 11

12 Example Combustion of Methane (CH 4 ) Degrees of Freedom Analysis: Atomic Balance Unit: Reactor unknowns (P,x 1,x 2,x 3,x 4 ) +5 independent atomic specie(s) independent nonreactive molecular specie(s) other equations: Degrees of freedom 0 12

13 Example Combustion of Methane (CH 4 ) Write the atomic species balances (mole basis): (1) C: 16 kg CH 4 (1/16)(1) = Px 1 (2) H: 16 kg CH 4 (1/16)(4) = Px 4 (3) O: 300 kg Air (1/29)(0.21)(2) = 2Px 2 + 2Px 1 + Px 4 (4) N: 300 kg Air (1/29)(0.79)(2) = 2Px 3 (5) x: x 1 + x 2 + x 3 + x 4 = 1 13

14 Example Combustion of Methane (CH 4 ) Simplifying the equations (1) C: 1 = Px 1 (2) H: 4 = Px 4 (3) O: 4.34 = 2Px 2 + 2Px 1 + Px 4 (4) N: = 2Px 3 (5) x: x 1 + x 2 + x 3 + x 4 = 1 14

15 Example Combustion of Methane (CH 4 ) If composition of flue gas is expressed in terms of actual number of moles (n s) instead of mole fractions (x s) C: 1 = n 1 H: 4 = n 4 O: 4.34 = 2n 2 + 2n 1 + n 4 N: = 2n 3 n: n 1 + n 2 + n 3 + n 4 = P 15

16 Example Combustion of Methane (CH 4 ) Solving for the n s and P: n 1 = 1 kmol CO 2 n 2 = 0.17 kmol O 2 n 3 = 2 kmol H 2 O n 4 = 8.18 kmol N 2 P = kmol Solving for the mole fractions: x 1 = (1/11.35) x 2 = (0.17/11.35) x 3 = (8.18/11.35) x 4 = (2/11.35) = 0.09 kmol CO 2 /kmol P = 0.01 kmol O 2 /kmol P = 0.72 kmol N 2 /kmol P = 0.18 kmol H 2 O/kmol P 16

17 Example Combustion of Methane (CH 4 ) Solving for % excess air: Write the overall combustion reaction for the fuel assuming it is burned completely: CH 4 + 2O 2 ========> CO 2 + 2H 2 O For 16 kg of C 3 H 8, the theoretical air required is: 16kg CH 4 1kmol CH4 2O2 1Air 29 kg Air = 276kg Air 16 kg CH 4 1CH O 2 1kmol Air 17

18 Example Combustion of Methane (CH 4 ) Solving for % excess air: Overall combustion reaction for the CH 4 : CH 4 + 2O 2 ========> CO 2 + 2H 2 O For 16 kg of C 3 H 8, the theoretical air required is: 1kmolCH4 2O2 1Air 29 kg Air 16kg CH4 = 276 kg Air 16kg CH 4 1CH O 2 1kmol Air 300kg Air - 276kg Air %excessair = 100 = 8.7% 276kg Air 18

19 Example Combustion of Coal A local utility burns coal having the following composition on a dry basis: Component Percent C H 4.45 O 3.36 N 1.08 S 0.70 Ash 7.36 Total

20 Example Combustion of Coal The average Orsat analysis of the flue gas during a 24-hr test was: Component Percent CO 2 + SO CO 0.0 O N Total

21 Example Combustion of Coal Moisture in the fuel was 3.90% and the air on the average contained lbm H 2 O/lbm dry air. The refuse showed 14.0% combined elements as in the coal (i.e. C, H, O, N, S) and the remainder being ash. It may be assumed that these combined elements occur in the same proportions as they do in the coal. Estimate the amount of amount of flue-gas (dry basis), amount of water coming out of the process, and the %excess air. 21

22 Example Combustion of Coal 22

23 Example Combustion of Coal Basis: 100 lbm of coal Ash Balance: (100 lbm) = 0.86R R = 8.56 lbm Combustible elements in refuse 0.14(8.56 lbm) = 1.20 lbm Assuming the combustible elements (C, H, O, N, S) occur in the same proportions as they do in the coal, the quantities of the combustibles in R on an ash-free basis are: 23

24 Example Combustion of Coal Component mass (lbm) ash-free mass % Amt. in R (lbm) Amt. in R (lbmol) C H O N S Total

25 Example Combustion of Coal Find the lbmol of H and O due to water in coal: H: 100 lbm (3.9/96.1)(1/18)(2/1) = lbmol H O: 100 lbm (3.9/96.1)(1/18)(1/1) = lbmol O Find the mole fraction of H and O due to moisture in air: H: lbm H 2 O/lbm DA (29/18)(2/1) = O: lbm H 2 O/lbm DA (29/18)(1/1) =

26 Example Combustion of Coal Solve A, W, and P using (C+S), H, and N balances (C+S) Balance (mole basis): (83.05/12) + (0.70/32) = P(0.154) H Balance (mole basis): (4.45/1) A = 2W N Balance (mole basis): (1.08/14) + 2(0.79A) = 2P(0.806) + 2(0.001) 26

27 Example Combustion of Coal Solving the balance equations gives P = 44.5 lbmol A = 45.4 lbmol W = 2.77 lbmol Determine the theoretical air required to burn completely all the C, H, and S in the coal. C: (83.05/12)(1/1) = 6.92 lbmol O 2 H: (4.45/1)(1/4) = 1.11 lbmol O 2 S: (0.70/32)(1/1) = lbmol O 2 Total O 2 required = ( ) = lbmol O 2 27

28 Example Combustion of Coal Since there is already O present in the coal, this amount is subtracted from the theoretical requirement. O 2 in coal = (3.36/16)(1/2) = lbmol O 2 Corrected O 2 required = ( ) = lbmol O 2 Actual O 2 supplied = (0.21) = lbmol O 2 And the % excess air is calculated as: 9.524kmol O kmol O kmol O 2 2 %excessair = 100 = 19.8% 2 28

The material balance equations, after introducing the known values for the variables, are:

Specifications: 4 (3 independent) (one is independent, the sum is F in mol) The material balance equations, after introducing the known values for the variables, are: of equations: (e) (b) simultaneously,