ME 475/675 Introduction to

CombustionLecture 8

Announcements

• HW 3 Ch 2 (57), Due Monday• Modify MathCAD program from lecture notes with

Dissociation• Last time: CO2 as a function of temperature and pressure: • ; three unknowns: , , • ;

• For more complete analysis, add additional reactions• For add ; one more unknown: • One more constraint: ;

• + ,+ • …

Equilibrium Products of Combustion•Combine Chemical Equilibrium (2nd law) & Adiabatic Flame

Temperature (1st law) • For Example: Propane and air combustion • Ideal

• Four products for a range of air/fuel ratios: •Now consider seven more possible dissociation products:

•What happens as air/fuel (equivalence) ratio changes

Flame temperature and major mole-fractions vs • Equivalence Ratio • At , O2, CO, H2 all present due to

dissociation. Not present in “ideal” combustion• at .15• at .05

• and decrease for • For decreases faster• For decreases faster

Fuel RichGet CO, H2

Fuel LeanO2

%

Tad[K]

“Old”

“New”

Minor-Specie Mole Fractions

• NO, OH, H, O• < 4000 ppm = 0.4%• Peak near

ppm

1%

Simple Product Calculation method

• No minor species

•

• Assume and are known

• What is a good assumption for lean or stoichiometric mixtures ?• • c = e = 0 (no CO or H2), but now include • 3 unknowns (b, d, f), 3 atom balances (C, H, O)

Atomic Balance for Lean combustion • • C: so • H: so • O: so

• • if

• Mole Fractions• ; ; ; • http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/index.htm

Comparison

• Total number of modes decreases as increases• Does not include CO or H2 at • Not bad for a simple model for

0.6 0.7 0.8 0.925

30

35

40

4541.667

25.8

Ntot Phi( )

10.6 Phi

0.6 0.7 0.8 0.90

0.05

0.1

0.15

0.20.155

0

Xco2 Phi( )

Xh2o Phi( )

Xo2 Phi( )

Xn2 Phi( )

10

10.6 Phi

CO2

H2O

O2

N2/10

1000 1500 2000 2500 3000 35000.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

T [K]

Kp

For Rich combustion

• ; no (or fuel)

• 4 unknows: b, c, d and e• 3 Atom balances: C, H, O• Need one more constraint

• Consider “Water-Gas Shift Reaction” equilibrium

• Not dependent on P since number of moles of products and reactants are the same• ;

• See plot from data on page 51• KP = 0.22 to 0.1635 for T = 2000 to 3500 K

Atomic Balances

• C: • (in terms of b and “knowns”)

• O:

• H:

• Plug into equilibrium constraint• • ; b is the only unknown!• Expand and collect terms to get

Solution

• Since , use “-” root

• Mole fraction can be calculated from b• http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/index.htm

Comparison

• Total number of moles continues to decrease as increases• Does not include O2 at • Not bad for a simple model for • More accurate models may be developed by including more equilibrium reactions and constants

using computer programs• Computer Programs Provided by the Book

• Appendix F, pp. 713-4, for Complex Reactions

0.6 0.7 0.8 0.925

30

35

40

4541.667

25.8

Ntot Phi( )

10.6 Phi

Φ≤1

1 1.1 1.2 1.320

22

24

2625.8

20.429

Ntot Phi( )

1.41 Phi

1 1.2 1.4 1.6 1.80

0.05

0.1

0.15

Xco2 Phi( )

Xco Phi( )

Xh2o Phi( )

Xh2 Phi( )

Xn2 Phi( )

10

Phi

CO2

H2O

H2

N2/10

CO

Air Preheaters

• Preheating the air using exhaust or flue gas increases the flame temperature • Recuperators uses heat transfer across a wall• Regenerators use a moving ceramic or metal matrix

Exhaust or Flue Gas Recirculation

• Inserting exhaust gas into the reactants reduces flame temperature, which can reduce pollution (oxides of nitrogen, NO, NO2)