1

MAE 5310: COMBUSTION FUNDAMENTALS

Coupled Thermodynamic and Chemical Systems:Well-Stirred Reactor (WSR) Theory

October 29, 2012

Mechanical and Aerospace Engineering DepartmentFlorida Institute of Technology

D. R. Kirk

2

WELL-STIRRED REACTOR THEORY OVERVIEW

• Well-Stirred Reactor (WSR) or Perfectly-Stirred Reactor (PSR) is an ideal reactor in which perfect mixing is achieved inside the control volume

• Extremely useful construct to study flame stabilization, NOx formation, etc.

3

APPLICATION OF CONSERVATION LAWS

0,,

,,,

outiiniiii

ii

iii

outiiniicvi

YYmVMW

Ymm

MWm

mmmVdt

dm

Rate at which mass of i accumulateswithin control volume

Rate at which mass of i isgenerated within control volume

Mass flow of i into control volume

Mass flow of i out of control volume

Relationship between mass generation rate of aspecies related to the net production rate

4

APPLICATION OF CONSERVATION LAWS

N

iiniini

N

iiouti

inout

N

jjj

iii

outicvii

ThYThYmQ

hhmQ

MWX

MWXY

TXfTXf

1,

1,

1

,,

Outlet mass fraction, Yi,out is equalto the mass fraction within the reactor

Conversion of molar concentration into mass fraction(see slide 2)

So far, N equations with N+1 unknowns, need to close setApplication of steady-flow energy equation

Energy equation in termsof individual species

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, res, for gases in WSR

RT

PMWm

V

mix

res

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 at blowout as a function of mass flow rate for ≤ 1.0 and assume that reactor is adiabatic

0

01

023.0

023.0

,,

Pr

65.11.075.1

,

65.11.075.1

,

inPinFFFf

OxF

OxFGOxinOx

OxFGFinF

TTcYYh

YYY

YYRT

PMWVk

F

AYYm

YYRT

PMWVkYYm

• Set of 4 coupled nonlinear algebraic equations with unknowns, YF, YOx, YPr, and T

– Treat mass flow rate and volume as known parameters

• To determine reactor blowout characteristic, solve nonlinear algebraic equations on previous slide for a sufficiently small value of mass flow rate that allows combustion at given equivalence ratio

• Increase mass flow rate until failure to achieve a solution or until solution yields input values

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

EXAMPLE 2: GAS TURBINE COMBUSTOR CHALLENGES

• Based on material limits of turbine (Tt4), combustors must operate below stoichiometric values

– For most relevant hydrocarbon fuels, s~ 0.06 (based on mass)

• Comparison of actual fuel-to-air and stoichiometric ratio is called equivalence ratio

– Equivalence ratio = = stoich

– For most modern aircraft ~ 0.3

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 > 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 < 0.5

• What do we do?– Burn (keep combustion going) near =1 with

some of ingested air– Then mix very hot gases with remaining air to

lower temperature for turbine

10

SOLUTION: BURNING REGIONS

Air

Com

pres

sor

Tur

bine

~ 1.0T>2000 K

~0.3

PrimaryZone

11

COMBUSTOR ZONES: MORE DETAILS

1. Primary Zone

– Anchors Flame

– Provides sufficient time, mixing, temperature for “complete” oxidation of fuel

– Equivalence ratio near =1

2. 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 res)

– L/D ~ 0.7

3. 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

EXAMPLE 2: GAS TURBINE ENGINE COMBUSTOR

• Consider 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, • Find fuel flow rate

• Find residence time inside reactor, res

5.125.011 exp105 oxfuela

fuel XXRT

Ex

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 are low

– At high T, is close to unity but fuel mass flow rate is low because the concentration [F] is low ([F]=FP/RT), which reduces reaction rate

– In the limit of =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

=1, kerosene-air mixtureV=900 cm3

P=10 atm

LHVmQ f

14

EXAMPLE #3: HOW CHEMKIN WORKS• Detailed mechanism for H2 combustion

• Reactor is adiabatic, operates at 1 atm, =1.0, and V=67.4 cm3

• For residence time, res, between equilibrium and blow-out limits, plot T, H2O, H2, OH, O2, O, and NO vs res.

15

EXAMPLE #3: HOW CHEMKIN WORKS

Input quantities in CHEMKIN:Chemical mechanismReactant stream constituentsEquivalence ratioInlet temperature and pressureReactor volumeres (1 ms ~ essentially equilibrated conditions)

Tflame and H2O concentration drop as res becomes shorter

H2 and O2 concentrations rise

Behavior of OH and O radicals is more complicated

NO concentration falls rapidly as res falls below 10-2 s