4310 Combustion Introduction Lecture 14

8/8/2019 4310 Combustion Introduction Lecture 14

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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


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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.


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APPLICATION OF CONSERVATION LAWS

( ) 0,,

,,

,

=+

=

=

+=

outiiniiii

ii

iii

outiinii

cvi

YYmVMW

Ymm

MWm

mmmVdt

dm

Rate at which mass of i accumulates

within control volume

Rate at which mass of i is

generated within control volume

Mass flow of i into control volume

Mass flow of i out of control volume

Relationship between mass generation rate of a

species related to the net production rate


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APPLICATION OF CONSERVATION LAWS

[ ]( ) [ ]( )

[ ]

[ ]

( )

( ) ( )

=

=

=

==

==

=

N

i

iniini

N

i

iouti

inout

N

j

jj

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 set

Application of steady-flow energy equation

Energy equation in terms

of individual species


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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

PMW

m

V

mix

res

=

=


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EXAMPLE 1: WSR MODELING

Develop a WSR model using same simplified chemistry and thermodynamic used in previous example

Equal constant cps, MWs, one-step global kinetics for C2H

6

Use model to develop blowout characteristics of a spherical reactor with premixed reactants (C2H

6and

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

65.11.0

75.1

,

65.11.0

75.1

,

=+

=

=

=

inPinFFFf

OxF

OxFGOxinOx

OxFGFinF

TTcYYh

YYY

YYRT

PMWVk

F

AYYm

YY

RT

PMWVkYYm

Set of 4 coupled nonlinear algebraic equations with unknowns, YF, Y

Ox, Y

Pr, 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


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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


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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

Summary

If = 1: Stoichiometric

If > 1: Fuel Rich

If < 1: Fuel Lean


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EXAMPLE 2: WHY IS THIS RELEVANT?

Most mixtures will NOT burn so far away fromstoichiometric

Often called Flammability Limit

Highly pressure dependent

Increased pressure, increased flammabilitylimit

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 areway too hot for turbine

Fixed Tt4 implies roughly < 0.5

What do we do?

Burn (keep combustion going) near =1 withsome of ingested air

Then mix very hot gases with remaining air tolower temperature for turbine


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SOLUTION: BURNING REGIONS

Air

Comp

ressor

Turb

ine

~ 1.0

T>2000 K

~0.3

Primary

Zone


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COMBUSTOR ZONES: MORE DETAILS

1. Primary Zone

Anchors Flame

Provides sufficient time, mixing, temperature for complete oxidation of fuel Equivalence ratio near =1

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 res)

L/D ~ 0.7

1. 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


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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 (C12H

24) 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

Eais 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 oxfuel

afuel XX

RTEx =


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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 mixture

V=900 cm3

P=10 atm

LHVmQ f =


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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 .


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EXAMPLE #3: HOW CHEMKIN WORKS

Input quantities in CHEMKIN:

Chemical mechanismReactant stream constituents

Equivalence ratio

Inlet temperature and pressure

Reactor volume

res (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

4310 Combustion Introduction Lecture 14

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