Combustion Thermodynamicsy

Dr. Rui ChenDr. Rui ChenDepartment of Aeronautical and Automotive Engineering

Loughborough University

1. INTRODUCTION1. INTRODUCTION

Fuels Gasoline Kerosene Diesel Heavy fuel

C 85.5 86.3 86.3 86.1• Gasoline: assume iso octane (C H ) C 85.5 86.3 86.3 86.1

H 14.4 13.6 12.8 11.8

S 0.1 0.1 <0.9 2.1

• Gasoline: assume iso-octane (C8H18)• Diesel: assume deodecane (C12H26)

AirO2 N2

% by volume 21.0 79.0

% by mass 23.3 76.7

• Molar mass of air = 0.21 x 32 + 0.79 x 28 = 28.8 (kg/kgmol)

Fuel and Air MixtureS /f• Stoichiometric air/fuel ratio

• Rich mixture• Weak (or lean) mixture

ratio air/fuel tricStoichiome=φ

ratio air/fuel Actual=φ

φλ 1

tii /f lt iSt i hiratio air/fuel Actual

==

Fuel/air equivalent ratio φ:

Air/fuel equivalent ratio λ:

2

φratioair/fueltricStoichiome

2. COMBUSTION2. COMBUSTION

O2HO2H 222 →+ 22 COOC →+ 2COO2C 2 →+

Basic Combustion Chemical Equations

22222 N2179O2HN

2179O2H ⎟

⎠⎞

⎜⎝⎛+→⎟

⎠⎞

⎜⎝⎛++

Stoichiometric Combustion of Hydrocarbon Fuels

22222mn cNObHaCON2179OHC ++→⎥⎦

⎤⎢⎣⎡ ++ x

Sto c o et c Co bust o o yd oca bo ue s

Weak Combustion of Hydrocarbon FuelsWeak Combustion of Hydrocarbon Fuels

222222mn dOcNObHaCON2179O

4mnλHC +++→⎥⎦

⎤⎢⎣⎡ +⎟

⎠⎞

⎜⎝⎛ ++

Rich Combustion of Hydrocarbon Fuels

eCOcNObHCOaN2179O

4mnλHC 22222mn +++′→⎥⎦

⎤⎢⎣⎡ +⎟

⎠⎞

⎜⎝⎛ ++

3

Assuming complete combustion of the hydrogen

22222mn cNObHaCON79OHC ++→⎥⎦⎤

⎢⎣⎡ ++ x

Stoichiometric Combustion of Hydrocarbon Fuels

22222mn 21 ⎥⎦⎢⎣

Atom balance: naan =⇒= :C2/ 2 :H mbbm =⇒=

4/ 2x2 :O mnxba +=⇒+=

⎟⎠⎞

⎜⎝⎛ +=⇒=×

42179 22

2179 :N mnccx

The stoichiometric combustion equationq

22222mn N421

79OH2

CON2179O

4HC ⎟

⎠⎞

⎜⎝⎛ +++→⎥⎦

⎤⎢⎣⎡ +⎟

⎠⎞

⎜⎝⎛ ++

mnmnmn

The stoichiometric air/fuel ratioThe stoichiometric air/fuel ratio

4mn

NN

AFRfuel

airmole stoich, +=== x

287932 ⎞⎜⎛ ×+

4

mn12

2821

32AFR

mmAFR mole stoich,

f

amass stoich, +×

⎠⎜⎝

×+==

Example:Octane C8H18, and λ = 1.2 (fuel rich combustion)Assume a ratio of CO to H2 in the products: nCOAssume a ratio of CO to H2 in the products:

The rich combustion equation

2nHnCOR =

Atom balance:

833.02.1

11===

φλ222222mn fHeCOcNObHCOaN

2179O

4mnλHC ++++′→⎥⎦

⎤⎢⎣⎡ +⎟

⎠⎞

⎜⎝⎛ ++

1818mm ⎞⎛⎞⎛( ) ( ) 3.854

184

18810.83324m

4mn12λa =−⎟

⎠⎞

⎜⎝⎛ +×−×=−⎟

⎠⎞

⎜⎝⎛ +−=′

92

182mb ===

11.404

1882179833.0

4mn

2179c =⎟

⎠⎞

⎜⎝⎛ +××=⎟

⎠⎞

⎜⎝⎛ += λ

( ) ( ) 1754188833012mnλ12e ⎞⎜⎛ +××⎞

⎜⎛ +

5

( ) ( ) 175.44

8833.0124

nλ12e =⎠

⎜⎝

+×−×=⎠

⎜⎝

+−=

3. FIRST LAW APPLIED TO COMBUSTION3. FIRST LAW APPLIED TO COMBUSTION

Non-flow System

UUUWQ ∆=−=− 12

∫∆2

dTCUwhere

∫=∆1

dTCU V

In the case of combustion, we have to include chemical energy in the internal energy

CS UUU ∆+∆=∆Where

= change in sensible internal energySU∆ g gy= change in chemical internal energy= change in (total) internal energy

S

CU∆U∆

6

Complete Combustion at Constant Volume

Figure 3.3 Internal energy vs. temperature of combustion

Internal Energy of Combustion at Τ0 and is given the symbol ∆U0. i.e.

000 RP UUU −=∆

7

Figure 3.2 Constant volumeb i

Change in internal energy of real processes

To evaluate UP2-UR1, we need to consider both the chemical and sensible energies. Assume the combustion process occurs in three stages:

• change in the reactants from T1 to T0; (this will involve sensible energy only)

• constant volume combustion at T0;constant volume combustion at T0; (this will involve chemical energy only)

• change in the products from T0 to T2. (this will involve sensible energy only)

Figure 3.4 Change in internal energy of real combustion process

( ) ( )1000212 RRPPRP UUUUUUUQ −+∆+−=−=

8

ExampleExampleA stoichiometric mixture of C3H8 (Propane), initially at 15oC, is burnt in a constant volume vessel. Heat is then removed from the vessel until the final temperature of the products is 1500oC.

Determine the heat transfer per kg of fuel. Take ∆U0=-48,800 kJ/kg

The mean cv of each constituent in kJ/kg-K is as follows: 15 to 25oC 25 to 1500oC C3H8O2 N

1.660 0.657 0 743

- - 0 859N2

CO2 H2O

0.743 - -

0.8590.911 1.848

Solution:

Constant volume: W=0

From first law: ( ) ( )1000212 RRPPRP UUUUUUUQ −+∆+−=−=From first law:

Chemical equation:

( ) ( )1000212 RRPPRPQ( ) ( )∑∑ −+∆+−=

RiVi

PiVi TTCmUTTCmQ 10,002,

2222283 N21795O4H3CON

2179O5HC ⎟

⎠⎞

⎜⎝⎛++→⎥⎦

⎤⎢⎣⎡ ++

Molecule mass:C3H8: 44 kg/kmoleO2: 32 kg/kmoleN2: 28 kg/kmole

2121 ⎠⎝⎦⎣

9

gCO2: 44 kg/kmoleH2O: 18 kg/kmole

Solution (Continue):

Assuming 1 kg of fuel:

Products ( )

[ ] ( )kJ/kg8023649

44/251500859.0288.18848.1184911.0443

02,

=−×××+××+××=

−∑P

iVi TTCm

Reactants

kJ/kg80.23649=

( )

[ ] ( ) 44/1525743.0288.18657.0325660.144

10,

−×××+××+×=

−∑R

iVi TTCm

Internal energy of combustion

kJ/kg 38.129=

kJ/kg 488000 −=∆U

Therefore, heat transfer is fuel of kJ/kg 25020.82−=Q

10

Enthalpy of combustion: - For steady flow or constant pressure combustion define the th l f b ti ∆H i th ∆U

Complete Combustion at Constant Pressure

enthalpy of combustion, ∆H0 in the same way as ∆U0.

( ) ( )1000212 RRPPRP HHHHHHH −+∆+−=−Relationship between ∆H0 and ∆U0 : - From the definition of enthalpy, h=u+pv, we have that

( )RP nnTRUH −+∆=∆ 0000

Fuel Calorific Value: 00CV UH ∆−≅∆−≅

Enthalpy of Formation and Reaction EnthalpyEnthalpy of Formation and Reaction EnthalpyEnthalpy of formation, HF: - The formation of a chemical compound involves either the release or the absorption of energy. This energy is characterised by the term: Enthalpy of Formation.

Reaction enthalpy, H: - The enthalpy of a substance at a temperature other than the reference temperature of 25oC is given by

Relationship between HF and ∆H0 :

( )16.298−+= TCHFH P

∑∑ −=∆ HFnHFnH

11

Relationship between HF and ∆H0 : ∑∑ −=∆R

iiP

ii HFnHFnH 0

4. ADIABATIC FLAME TEMPERATURE4. ADIABATIC FLAME TEMPERATURE

The adiabatic flame temperature, Tf, is the reaction temperature for the case of no work and heat transfer

transfer.

Consider a flow process:

The first law is: Q-W=HP2-HR1

but, Q=W=0, therefore, HP2-HR1=0

This can be shown on a H-T diagram.

Using enthalpy of combustion:

( ) ( ) 010002 =−+∆+ RRPP HHHHH( ) ( ) 010002 =−+∆+− ∑∑

RPii

PPii TTcmHTTcm

or, using reaction enthalpy:

where T T

( )[ ] ( )[ ] 00102 =−+−−+ ∑∑R

PiiiP

Piii TTCHFnTTCHFn

12

where, T2=Tf

Example:Example:Question:Gaseous Methane (CH4) and air both enter a combustion chamber at 10oC. If burning occurs at constant pressure, consider a stoichiometric

i t d i th d t i l l t th di b ti fl

HF at 25oC KJ/kgmol

CP KJ/kgmol-K

CH4 Air

-748730

35.129.0

( ) 222224 NO2HCO0.79N0.21OCH yx ++→++

mixture and, using the data given, calculate the adiabatic flame temperature.

CO2 H2O (steam) N2

-393522-241827

0

53.343.534.2Solution:

Stoichiometric mixture:

O balance:N balance:

Hence, we have

R ti th l

524.92242.0 =⇒+= xx524.7258.1 =⇒= yyx

( ) 222224 N524.7O2HCO0.79N0.21O524.9CH ++→++

( )298+= TCHFHReaction enthalpy:

Reactants:Methane:Air:Total:

( )298−+= TCHFH P

( ) 753402982831.35748734

−=−×+−=CHnH( )[ ] 4143298283290524.9 −=−×+×=airnH

fuelofkJ/kgmol79483414375340 −=−−=∑ HTotal:

Products:CO2:H2O:

fuelof kJ/kgmol79483414375340∑R

H

( )298−=∆ ff TTfCO TnH ∆×+−= 3.53393522

2

[ ] ffOH TTnH ∆+−=∆+−×= 0.874836545.4324182722

13

N2Total:

[ ] ffOH 2 [ ] ffN TTnH ∆+=∆+×= 3.25702.340524.72

fuel of kJ/kgmol 6.397877176 fP

TH ∆+−=∑

Solution (Continued):( )

First law: :

But W=0, and (for adiabatic reaction) Q=0, yield

∑∑ −=−RP

HHWQ

∑∑ ∑∑ =RP

HH

794836.397877176 −=∆+− fT

C2006o=∆ fT f

K23042982006 =+=fT

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5. DISSOCIATION5. DISSOCIATION Equilibrium Constant, KPThe proportions in which different gases will be present in a mixture can be predicted using

E ilib i C t t K C id th tian Equilibrium Constant KP. Consider the reaction

D of kgmols dC of kgmols cB of kgmols bA of kgmols a ⋅+⋅⇔⋅+⋅

The equilibrium constant, KP, at any pressure p is given by ( ) ( )( ) ( )ba

dD

cC

P pppp

K =where p is the partial pressure of constituent x ( ) ( )BA ppwhere, px is the partial pressure of constituent x.

It should be noted that KP has the units (atmospheres)y

where the value of y depends upon the relative number of kgmols and is given by ( ) ( )badcy +−+=

Note also that the numerical value of KP for a particular reaction will depend on the form of the chemical equation. For example,

22 COO1CO ⇔+ 2COO2CO ⇔+22 COO2

CO ⇔+22 2COO2CO ⇔+

will have different KP values.

KP can be shown to be a function of temperature only and a typical graph of K versus temperature is

15

Figure 3.5 KP versus temperature and a typical graph of KP versus temperature is shown in Figure.

Example:Example:Question:For a soichiometric mixture of heptane (C7H16) and air, calculate the exhaust product composition at 2600K and 10 atmosphere assuming that the products consist of CO2, CO, H2O, H2, O2 and N2 only.

Solution:First, write down the chemical equation

have 5 unknowns, get 3 equations from atom balance( )

2222222167 N21

1179fHeOdCOObHaCON2179O11HC ×

+++++→⎟⎠⎞

⎜⎝⎛ ++

C: (1)O: (2)H: (3)

We can get a further 2 equations from KP relationships

7=+ da11222 ×=+++ edba

1622 =+ fb

g q p

22 COO21CO ⇔+ ( )

2/1

2/1CO, 2

−

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=t

P nP

edaK

OHO21H 222 ⇔+ ( )

2/1

2/1OH, 2

−

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=t

P nP

efbK

where, (4)

and p = pressure of products

211179 ×

+++++= fedbant

16

Solution (Continued):

At 2600K we can obtain from tables ( ) ( ) 5577162191log == KKAt 2600K, we can obtain from tables

Hence,(5)

( ) ( ) 5577.16 ,219.1log22 CO,CO,10 == PP KK

( ) ( ) 954.104 ,021.2log OH,OH,10 22== PP KK

2/12/1

2/1 360.525577.16 −⋅=⎟⎟⎠

⎞⎜⎜⎝

⎛=

⋅ tt

nnP

eda

(6)

These 6 simultaneous equations could now be solved for the 6 unknowns probably using an iterative

⎠⎝ t

2/12/1

2/1 894.331954.104 −⋅=⎟⎟⎠

⎞⎜⎜⎝

⎛=

⋅ tt

nnP

efb

q p y gtechnique. For example, could solve as follows:

a. Estimate value for a b. Calculate d from equation (1)c Combine equations (4) and (5) eliminating nt and e to obtain ratio b/fc. Combine equations (4) and (5) eliminating nt and e to obtain ratio b/fd. Substitute into equation (3) to obtain f and be. Determine e from equation (2)f. Calculate nt and determine a from equation (4)g. Compare with assumed value for a (step a.) and iterate until obtain good agreement. This iterative

17

method is ideally suited for solution by computer. For many applications, more dissociation species will need to be considered and a more efficient iterative method would be employed.