AAE 439

Ch5 –1

5. COMBUSTION AND THERMOCHEMISTRY

AAE 439

Ch5 –2

Overview

 Definition & mathematical determination of chemical equilibrium,

 Definition/determination of adiabatic flame temperature,

 Prediction of composition and temperature of combusted gases as a function of initial temperature,

 Prediction of amounts of fuel & oxidizer,

 Thermochemical changes during expansion process in nozzle.

 Performance Parameters:

CF= 2γ 2

γ −12

γ +1

⎛⎝⎜

⎞⎠⎟

γ +1γ −1

1−p

e

p0

⎛

⎝⎜⎞

⎠⎟

γ −1γ

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥+

pe− p

a

p0

⋅ ε

c* =

RT0

γγ +1

2

⎡

⎣⎢

⎤

⎦⎥

γ +1γ −1

Performance depends on: T, MW, p0, pe, pa, γ

AAE 439

Ch5 –3

Overview

 Important Concepts & Elements of Analysis

 Conversion of Chemical Energy to Heat

 Simple Treatment of Properties of Gases

 Balancing Chemical Reactions - Stoichiometry

 Adiabatic Flame Temperature

 Chemical Equilibrium and Gibbs Free Energy

 Nozzle Expansion Effects

 Thermochemical Calculations

AAE 439

Ch5 –4

5.1 THERMODYNAMICS OF GAS MIXTURES

AAE 439

Ch5 –5

Perfect Gas

 Perfect Gas Law relates pressure, temperature and density for a perfect gas/mixture of gases :

 Universal Gas Constant:

 Gas Constant:

 Calorically Perfect Gas:  Internal Energy

 Enthalpy

 Specific Heat Relationships:

 Definition of “Mole”: A mole represents the amount of gas, which contains Avogadro’s number of gas

molecules: 6.02·1023 molecules/mol.

pV = nℜT = mRT ⇔ p v = RT

ℜ = 8.314

Jmol ⋅K

R = ℜ

M

du = cvdT u

2− u

1= c

v(T

2−T

1)

dh = cp

dT h2− h

1= c

p(T

2−T

1)

c

p− c

v= R γ =

cp

cv

AAE 439

Ch5 –6

Gibbs-Dalton Law

 Properties of a mixture is determined by the properties of constituents according to Gibbs–Dalton Law:

 The pressure of a mixture of gases is equal to the sum of the pressure of each constituent when each occupies alone the volume of the mixture at the temperature of the mixture.

 The internal energy and the entropy of a mixture are equal, respectively, to the sums of the internal energies and the entropies of its constituents when each occupies alone the volume of the mixture at the temperature of the mixture.

 Temperature

 Pressure

 Volume

 Energy

 Entropy

 Enthalpy

Tmix= T

1= T

2=…= T

N

p

mix= p

1+ p

2+ p

3…+ p

N= p

ii=1

N

∑

Vmix= m

mixv

mix= m

1v

1= m

2v

2=…= m

Nv

N

E

mix= m

mixe

mix= m

1e

1+ m

2e

2+…+ m

Ne

N= m

ie

ii=1

N

∑

Smix= m

mixs

mix= m

1s1+ m

2s2+…+ m

Ns

Nsmix

= Smix

nmix

Hmix= m

mixh

mix= m

1h

1+ m

2h

2+…+ m

Nh

Nh

mix= H

mixn

mix

VContainer

TContainer

pContainer

“Bar” denotes Property with respect to Molar Quantity

AAE 439

Ch5 –7

Mixture of Gases

 Composition of a gas mixture is expressed by either the constituent mass fractions or mole fractions.

 Definition of Mass Fraction:

 Equivalent Molecular Weight:

 Perfect Gas Law

 Pressure (Gibbs-Dalton Law)

 Enthalpy

 Entropy

 where species entropy is

piV = m

iR

iT = n

iℜT

p = p

ii=1

N

∑

Mmixequiv

= 1

yi

Mi( )

i=1

N

∑=

mmix

nmix

yi=

mi

mmix

=m

i

mi

1

N

∑⇒ y

ii=1

N

∑ =1

h

mix= y

ih

ii∑

s

mix(T,p) = y

is

i(T,p

i)

i∑

s

i(T, p

i) = s

i(T, p

ref)−R ln

pi

pref

VContainer

TContainer

pContainer

AAE 439

Ch5 –8

Mixture of Gases

 Definition of Mole Fraction:

 Equivalent Molecular Weight:

 Perfect Gas Law

 Pressure (Gibbs-Dalton Law)

 Partial Pressure:

 Enthalpy

 Entropy

 where species entropy is

piV = m

iR

iT = n

iℜT

p = p

ii=1

N

∑

xi=

ni

nmix

=n

i

ni

1

N

∑⇒ x

ii=1

N

∑ =1

h

mix= x

ih

ii∑

smix

(T,p) = xisi(T,p

i)

i∑

si(T, p

i) = s

i(T, p

ref)−ℜ ln

pi

pref

VContainer

TContainer

pContainer

M

mixequiv

= xiM

ii=1

N

∑ =m

mix

nmix

pi= x

ip

AAE 439

Ch5 –9

Mixture of Gases

 Relationship between Mass and Mole Fractions:

 Other Relationships for a Gas Mixture:

 Specific Heat:

 Ratio of Specific Heat:

x

i= y

i

Mmix

Mi

c

p,mix= c

p,iy

ii=1

N

∑

γ

mix=

cp,mix

cv,mix

=c

p,mix

cp,mix

−Rmix

AAE 439

Ch5 –10

5.2 1st LAW OF THERMODYNAMICS

AAE 439

Ch5 –11

1st LTD - Fixed Mass

 First law of thermodynamics embodies the fundamental principle of conservation of energy.  Q and W are path functions and occur only at the system boundary.

 E is a state variable (property), ∆E is path independent.

m, E

Q

W

System Boundary enclosing Fixed Mass

Q − W = ΔE1→2

Heat added to system in going from state 12

Work done by system on surrounding in

going from state 12

Change in total system energy in going from state 12

Q − W = dE dt

q − w = de dt

E = m u+ 1

2v2 + g z

⎛

⎝⎜

⎞

⎠⎟

AAE 439

Ch5 –12

1st LTD - Control Volume

 Conservation of energy for a steady-state, steady-flow system.

 Assumptions:  Control Volume is fixed relative to the coordinate system.

 Eliminates any work interactions associated with a moving boundary,

 Eliminates consideration of changes in kinetic and potential energies of CV itself.

 Properties of fluid at each point within CV, or on CS, do not vary with time.

 Fluid properties are uniform over inlet and outlet flow areas.

 There is only one inlet and one exit stream.

QCV WCV

Control Surface (CS) enclosing Control Volume (CV)

Q

CV− W

CV= m e

outlet− m e

inlet+ m p

ov

o− p

iv

i( )Rate of heat

transferred across the CS, from the

surrounding to the CV.

Rate of all work done by CV,

including shaft work but excluding flow work.

Net rate of work associated with pressure

forces where fluid crosses CS, flow work.

dmCV

dt= 0

dECV

dt= 0

Rate of energy flowing out

of CV.

Rate of energy flowing into

CV.

QCV

− WCV

= m ho− h

i( ) + 12

vo2 − v

i2( ) + g z

o− z

i( )⎡

⎣⎢

⎤

⎦⎥

m e + p v( )

inlet m e + p v( )

outlet

AAE 439

Ch5 –13

TD PROCESSES in CHEM. SYSTEMS

 Chemical systems (chemical reactions) are treated as either constant-volume or constant-pressure processes.

 Energy Equation (1st Law of TD)

 Inside a rocket combustion chamber, fluid velocity (Ekin) is small and height changes of the fluid mass (Epot) is negligible. Energy contained in the fluid is governed by the internal energy of the hot combustion gas.

 Work contribution in a rocket combustion chamber results from changes in specific volume of pressure. The fluid doesn’t perform any mechanical work (Wshaft=0).

 Constant–Volume (Isochoric) Process:

 Constant–Pressure (Isobaric) Process:

E = U + Epotential

+ Ekinetic

= Q −Wshaft

−Wflow

dU = Q

dU = Q − p dV

H =U + pV

⎫⎬⎭

dH = Q

E =U ⇔ dE = dU = (δQ −δWshaft

−δWflow

)

W = − p

(ext )dV

V1

V2∫ ⇔ δWflow

= p dV

AAE 439

Ch5 –14

5.3 REACTANT AND PRODUCT MIXTURES

AAE 439

Ch5 –15

STOICHIOMETRY

 The stoichiometric quantity of oxidizer (substance A) is just that amount needed to completely burn a quantity of fuel (substance B):  An oxidizer-fuel mixture is LEAN, when there is more than a stoichiometric

quantity of oxidizer in the mixture.

 An oxidizer-fuel mixture is RICH, when there is less than a stoichiometric quantity of oxidizer in the mixture.

 Stoichiometric Chemical Reaction:  Examples:

 One mole of methane and 2 moles of oxygen form one mole of carbon dioxide and 2 mole of water.

 One mole of H2 and a half mole of O2 form one mole of H2O.

CH4+ 2O

2→CO

2+ 2H

2O

H2+ 1

2O

2→H

2O

AAE 439

Ch5 –16

STOICHIOMETRY

 Stoichiometric Oxidizer-Fuel Ratio:

 Equivalence Ratio :

 This ratio is a quantitative indicator whether a fuel-oxidizer mixture is  Lean:

 Rich:

 Stoichiometric:

 Other Parameters:  Percent Stoichiometric Oxidizer:

 Percent Excess Oxidizer:

OF

⎛

⎝⎜

⎞

⎠⎟

stoic

=m

oxidizer

mfuel

⎛

⎝⎜⎜

⎞

⎠⎟⎟

stoic

Φ =O F( )

stoic

O F= F O

F O( )stoic

% stoichiometric oxidizer = 100%

Φ

% excess oxidizer = (1−Φ)

Φ100%

OF

⎛⎝⎜

⎞⎠⎟=

noxygen

nfuel

Moxygen

Mfuel

where

AF

⎛⎝⎜

⎞⎠⎟

stoic

=m

air

mfuel

⎛

⎝⎜

⎞

⎠⎟

stoic

=n

air

nfuel

⋅M

air

Mfuel

= 4.76 ⋅a1

⋅M

air

Mfuel

Φ

! < 1

! > 1

! = 1

AAE 439

Ch5 –17

AIR (O2)/FUEL COMBUSTION

 Stoichiometric Combustion of Air and Fuel (Hydrocarbon)

 Lean Combustion of Air and Fuel

 Balancing Chemical Reaction:

 Rich Combustion of Air and Fuel

 Balancing Chemical Reaction:

C

xH

y+ a ⋅ O

2+ 3.76 N

2( ) → x ⋅CO2+ y

2⋅H

2O + 3.76a ⋅N

2

⇒ a = x + y

4

CxH

y+ a ⋅ O

2+ 3.76 N

2( ) → b ⋅CO2+ c ⋅H

2O + d ⋅O

2+ 3.76 a ⋅N

2

x & y define the hydrocarbon fuel!

CxH

y+ a ⋅ O

2+ 3.76 N

2( ) → b ⋅CO2+ c ⋅H

2O + d ⋅C

xH

y+ 3.76 a ⋅N

2

C : x = b b = x

H : y = 2c c = 12

y

O : 2a = 2b + c + 2d a = x + 14

y + d

C : x = b + x d b = x (1− d)

H : y = 2c + yd c = 12

y (1− d)

O : 2a = 2b + c a = (x + 14

y) (1− d)

AAE 439

Ch5 –18

Examples

 Example #1:  A small, low-emission, stationary gas-turbine engine operates at full load (3,950

kW) at an equivalence ratio of 0.286 with an air flowrate of 15.9 kg/s. The equivalent composition of the fuel (natural gas) is C1.16H4.32. Determine the fuel mass flow rate and the operating air-fuel ratio for the engine.

AAE 439

Ch5 –19

Examples

 Example #2:  A natural-gas-fired industrial boiler operates with an oxygen concentration of 3

mole percent in the flue gases. Determine the operating air-fuel ratio and the equivalence ratio. Treat the natural gas as methane.