mae 5310: combustion fundamentals

1. 1 MAE 5310: COMBUSTION FUNDAMENTALS Coupled Thermodynamic and Chemical Systems: Well-Stirred Reactor (WSR) Theory October 8, 2009 Mechanical and Aerospace Engineering Department Florida Institute of Technology D. R. Kirk

2. 2

3. 3 APPLICATION OF CONSERVATION LAWS

4. 4 APPLICATION OF CONSERVATION LAWS

5. 5 WSR SUMMARY Solving for temperature and species mass fraction is similar to calculation of adiabatic flame temperature (Glassman, Chapter 1) The difference is that now the product composition is constrained by chemical kinetics rather than by chemical equilibrium WSR (or PSR) is assumed to be operating at steady-state, so there is no time dependence Compared with the constant pressure and constant volume reactor models considered previously The equations describing the WSR are a set of coupled (T and species concentration) nonlinear algebraic equations Compared with constant pressure and constant volume reactor models which were governed by a set of coupled linear, 1st order ODEs Net production rate term, although it appears to have a time derivative above it, depends only on the mass fraction (or concentration) and temperature, not time Solve this system of equations using Newton method for solution of nonlinear equations Common to define a mean residence time, tres, for gases in WSR

6. 6 EXAMPLE 1: WSR MODELING Develop a WSR model using same simplified chemistry and thermodynamic used in previous example Equal constant cp’s, MW’s, one-step global kinetics for C2H6 Use model to develop blowout characteristics of a spherical reactor with premixed reactants (C2H6 and Air) entering at 298 K. Diameter of reactor is 80 mm. Plot f at blowout as a function of mass flow rate for f = 1.0 and assume that reactor is adiabatic

7. 7 EXAMPLE 1: RESULTS AND COMMENTS Decreasing conversion of fuel to products as mass flow rate is increased to blowout condition Decreased temperature as flow rate is increased to blowout condition Mass flow rate for blowout is about 0.193 kg/s Ratio of blowout temperature to adiabatic flame temperature is 1738 / 2381 = 0.73 Repeat calculations at various equivalence ratios generates the blowout characteristic curve Reactor is more easily blown out as the fuel-air mixture becomes leaner Shape of blowout curve is similar to experimental for gas turbine engine combustors

8. 8 EXAMPLE 2: GAS TURBINE COMBUSTOR CHALLENGES Based on material limits of turbine (Tt4), combustors must operate below stoichiometric values For most relevant hydrocarbon fuels, ys ~ 0.06 (based on mass) Comparison of actual fuel-to-air and stoichiometric ratio is called equivalence ratio Equivalence ratio = f = y/ystoich For most modern aircraft f ~ 0.3 Summary If f = 1: Stoichiometric If f > 1: Fuel Rich If f < 1: Fuel Lean

9. 9 EXAMPLE 2: WHY IS THIS RELEVANT? Most mixtures will NOT burn so far away from stoichiometric Often called Flammability Limit Highly pressure dependent Increased pressure, increased flammability limit Requirements for combustion, roughly f > 0.8 Gas turbine can NOT operate at (or even near) stoichiometric levels Temperatures (adiabatic flame temperatures) associated with stoichiometric combustion are way too hot for turbine Fixed Tt4 implies roughly f < 0.5 What do we do? Burn (keep combustion going) near f=1 with some of ingested air Then mix very hot gases with remaining air to lower temperature for turbine

10. 10 SOLUTION: BURNING REGIONS

11. 11 COMBUSTOR ZONES: MORE DETAILS Primary Zone Anchors Flame Provides sufficient time, mixing, temperature for “complete” oxidation of fuel Equivalence ratio near f=1 Intermediate (Secondary Zone) Low altitude operation (higher pressures in combustor) Recover dissociation losses (primarily CO ? CO2) and Soot Oxidation Complete burning of anything left over from primary due to poor mixing High altitude operation (lower pressures in combustor) Low pressure implies slower rate of reaction in primary zone Serves basically as an extension of primary zone (increased tres) L/D ~ 0.7 Dilution Zone (critical to durability of turbine) Mix in air to lower temperature to acceptable value for turbine Tailor temperature profile (low at root and tip, high in middle) Uses about 20-40% of total ingested core mass flow L/D ~ 1.5-1.8

12. 12 EXAMPLE 2: GAS TURBINE ENGINE COMBUSTOR Consider the primary combustion zone of a gas turbine as a well-stirred reactor with volume of 900 cm3. Kerosene (C12H24) and stoichiometric air at 298 K flow into the reactor, which is operating at 10 atm and 2,000 K The following assumptions may be employed to simplify the problem Neglect dissociation and assume that the system is operating adiabatically LHV of fuel is 42,500 KJ/kg Use one-step global kinetics, which is of the following form Ea is 30,000 cal/mol = 125,600 J/mol Concentrations in units of mol/cm3 Find fractional amount of fuel burned, h Find fuel flow rate Find residence time inside reactor, tres

13. 13 EXAMPLE 2: FURTHER COMMENTS Consider again the WSR model for the gas turbine combustor primary zone, however now treat temperature T as a variable. At low T, fuel mass flow rate and h are low At high T, h is close to unity but fuel mass flow rate is low because the concentration [F] is low ([F]=cFP/RT), which reduces reaction rate In the limit of h=1, T=Tflame and the fuel mass flow rate approaches zero For a given fuel flow rate two temperature solutions are possible with two different heat outputs are possible

14. 14 EXAMPLE #3: HOW CHEMKIN WORKS Detailed mechanism for H2 combustion Reactor is adiabatic, operates at 1 atm, f=1.0, and V=67.4 cm3 For residence time, tres, between equilibrium and blow-out limits, plot T, cH2O, cH2, cOH, cO2, cO, and cNO vs tres.

15. 15 EXAMPLE #3: HOW CHEMKIN WORKS