che 31. introduction to chemical engineering...

CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING CALCULATIONS

Lecture 11Combustion Processes

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños

LECTURE 11. Combustion Processes

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE

2

A Combustion Process

LECTURE 10. Solving Material Balance Problems Involving Reactive Processes


3

Chemical Reactiions Associated with Combustion Processes

C + O2 ========> CO2

C + 0.5O2 ========> CO

2H + 0.5O2 ========> H2O

S + O2 ========> SO2



4

Terms Associated with Combustion Processes

Orsat AnalysisRefers 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 AirThe 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 AirThe amount of air in excess of that required for complete combustion. The % excess air is the same as % excess O2.



5

Example 11-1. 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 O2 must be satisfied.

Relevant Reactions:

C + O2 ========> CO2C + 0.5O2 ========> CO



6


Calculate for theoretical O2 needed:

Assume that all the carbon is burned completely to CO2.

100 kmol C (1/1) = 100 kmol O2

It is not correct to do the following:

C CO2: 100 kmol C (0.80)(1/1) = 80 kmol O2

C CO: 100 kmol C (0.20)(0.5/1) = 10 kmol O2



7


Total O2 required stoichiometrically based on the actual process:

Stoichiometric O2 = (80 + 10) kmol = 90 kmol

Theoretical O2 is based not on what is stoichiometricallyneeded according to what is actually burned.

Theoretical Air = (100 kmol)(1/0.21) = 476.2 kmol

And the actual air supplied:

Actual Air = 476.2 kmol (1.5) = 714.3 kmol



8

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 gasoline.

Compressed propane (C3H8) has been suggested as a source of economic power for vehicles. 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.

Calculate the percent excess air.



9


Write the overall combustion reaction for the fuel assuming it is burned completely:

C3H8 + 5O2 ========> 3CO2 + 4H2O

For 20 kg of C3H8, the theoretical O2 required is:

3 8 23 8 2

3 8 3 8

1kmol C H 5O20kg C H =2.27kmol O44.09kgC H 1C H


Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE10


The actual O2 supplied is

22

1kmol air 1air400kgair =2.90kmol O29kgair 0.21O

The percent excess air (or O2) is

2 2

2

2.90kmol O -2.27kmol O%excessair = ×100=28%2.27kmol O



Example 11-3. 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 plants.

Consider the complete combustion of 16.0 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 gas.




Degrees of Freedom Analysis: Atomic Balance

Unit: Reactor

unknowns (P,x1,x2,x3,x4) +5

independent atomic specie(s)

independent nonreactive molecular specie(s)

other equations:

Degrees of freedom 0




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)(0.21)(2) = 2Px2 + 2Px1 + Px4

(4) N: 300 kg Air (1/29)(0.79)(2) = 2Px3

(5) x: x1 + x2 + x3 + x4 = 1




Simplifying the equations

(1) C: 1 = Px1

(2) H: 4 = Px4

(3) O: 4.34 = 2Px2 + 2Px1 + Px4

(4) N: 16.34 = 2Px3

(5) x: x1 + x2 + x3 + x4 = 1




If composition of flue gas is expressed in terms of actual number of moles (n’s) instead of mole fractions (x’s)

C: 1 = n1

H: 4 = n4

O: 4.34 = 2n2 + 2n1 + n4

N: 16.34 = 2n3

n: n1 + n2 + n3 + n4 = P




Solving for the n’s and P:

n1 = 1 kmol CO2 n2 = 0.17 kmol O2

n3 = 2 kmol H2O n4 = 8.18 kmol N2

P = 11.35 kmol

Solving for the mole fractions:

x1 = (1/11.35) = 0.09 kmol CO2/kmol Px2 = (0.17/11.35) = 0.01 kmol O2/kmol Px3 = (8.18/11.35) = 0.72 kmol N2/kmol Px4 = (2/11.35) = 0.18 kmol H2O/kmol P




Solving for % excess air:

Write the overall combustion reaction for the fuel assuming it is burned completely:

CH4 + 2O2 ========> CO2 + 2H2O

For 16 kg of C3H8, the theoretical air required is:

4 24

4 8 2

1kmolCH 2O 1Air 29kg Air16kg CH = 276kg Air16kg CH 1CH 0.21O 1kmol Air




Solving for % excess air:

Overall combustion reaction for the CH4:

CH4 + 2O2 ========> CO2 + 2H2O

For 16 kg of C3H8, the theoretical air required is:

4 2

4 8 2

1kmolCH 2O 1Air 29kg Air16kgCH4 = 276kg Air16kgCH 1CH 0.21O 1kmolAir

300kg Air - 276kg Air%excessair = ×100 = 8.7%276kg Air



Example 11-4. Combustion of Coal

A local utility burns coal having the following composition on a dry basis:

Component PercentC 83.05H 4.45O 3.36N 1.08S 0.70

Ash 7.36Total 100.00




The average Orsat analysis of the flue gas during a 24-hr test was:

Component PercentCO2 + SO2 15.4

CO 0.0O2 4.0N2 80.6

Total 100.00




Moisture in the fuel was 3.90% and the air on the average contained 0.0048 lbm H2O/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.




Basis: 100 lbm of coal

Ash Balance: 0.0736(100 lbm) = 0.86RR = 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:




Componentmass (lbm)

ash-freemass %

Amt. in R(lbm)

Amt. in R(lbmol)

C 83.05 89.65 1.076 0.0897

H 4.45 4.80 0.058 0.0537

O 3.36 3.63 0.0436 0.0027

N 1.08 1.17 0.014 0.0010

S 0.70 0.76 0.009 0.0003

Total 92.64 100.00 1.200 0.1474




Find the lbmol of H and O due to water in coal:

H: 100 lbm (3.9/96.1)(1/18)(2/1) = 0.451 lbmol H

O: 100 lbm (3.9/96.1)(1/18)(1/1) = 0.225 lbmol O

Find the mole fraction of H and O due to moisture in air:

H: 0.0048 lbm H2O/lbm DA (29/18)(2/1) = 0.0154

O: 0.0048 lbm H2O/lbm DA (29/18)(1/1) = 0.0077




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) + 0.0897 + 0.0003

H Balance (mole basis):

(4.45/1) + 0.451 + 0.0154A = 2W + 0.0537

N Balance (mole basis):

(1.08/14) + 2(0.79A) = 2P(0.806) + 2(0.001)




Solving the balance equations gives

P = 44.5 lbmolA = 45.4 lbmolW = 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 O2

H: (4.45/1)(1/4) = 1.11 lbmol O2

S: (0.70/32)(1/1) = 0.022 lbmol O2

Total O2 required = (6.92 + 1.11 + 0.022) = 8.052 lbmol O2




Since there is already O present in the coal, this amount is subtracted from the theoretical requirement.

O2 in coal = (3.36/16)(1/2) = 0.105 lbmol O2

Corrected O2 required = (8.052 – 0.105) = 7.947 lbmol O2

Actual O2 supplied = 45.35 (0.21) = 9.524 lbmol O2

And the % excess air is calculated as:

2 2

2

9.524kmol O - 7.947kmol O%excessair = ×100 = 19.8%7.947 kmol O

che 31. introduction to chemical engineering...

Documents