Chapter 13
Reacting Mixtures and Combustion
Photo courtesy of www.freefoto.com
Fundamentalsproductsfuel oxidizer reactants products
Chemical Equations:
2 2 2
11H O 1H O
2
2 2 2
11 kmol H kmol O 1 kmol H O
2
2 2 22 kg H 16 kg O 18 kg H O
Modeling Air:
• 79% Nitrogen, 21% Oxygen• Only the Oxygen reacts: Nitrogen considered inert• Mair = 28.97 kg/kmol or lb/lbmol (Tables A-1)
Air Fuel Ratio:
mass of air moles of air
mass of fuel moles of fuel air air
fuel fuel
M MAF AF
M M
1a
2b
1*b2*a
2*b2*1
OH*bO*aH*1 222
Fundamentals
Chemical Equations:
Common fuels modeled as simple hydrocarbons:• Natural Gas Methane (CH4)• Gasoline Octane (C8H18)• Diesel Dodecane (C12H26)
Theoretical Air: The minimum amount of air that provides the necessary oxygen for complete combustion (i.e. For one mole of octane the theoretical air is 59.5 moles)
2*d2*0.79*a N)
1*c2*b2*0.21*a O)
2*c18*1 H)
1*b8*1 C)
47.0d
59.5a
9c
8b
22222188 N*d OH*c CO*b)0.21N(0.21O*aHC*1 Stoichiometric Coefficients (Four equations and four unknowns)
Fundamentals
2*d2*0.79*59.5*1.5 N)
2*e1*92*82*0.21*59.5*1.5 O)
25.6e
70.5d
222222188 O*eN*d OH*9 CO*8)0.79N(0.21O*59.5*5.1HC*1
Stoichiometric Coefficients (Two equations and two unknowns)
Percent Excess Air:The percent of air supplied that is in excess of the theoretical air
Example: Combustion of Octane with 50% excess air (or 150% theoretical)
222222188 O*25.6N*70.5 OH*9 CO*8)0.79N(0.21O*25.89HC*1
QuizOne kg/min of methane is burned in a combustor with 25% excess air. The temperature and pressure of the air and fuel are 25oC and 101kPa respectively. The design velocity for each intake is 15 m/s. Determine the diameter of the air intake line in meters.
QuizOne kg/min of methane is burned in a combustor with 25% excess air. The temperature and pressure of the air and fuel are 25oC and 101kPa respectively. The design velocity for each intake is 15 m/s. Determine the diameter of the air intake line in meters.
CH4+a(0.21O2+0.79N2) bCO2+cH2O+dN2
C) 1*1 = b*1 b=1H) 1*4 = c*2 c=2
O) a*2*0.21 = b*2+c*1 a=9.524
Theoretical air = 9.524 kmol(air)/kmol(fuel)
Theoretical Air
QuizOne kg/min of methane is burned in a combustor with 25% excess air. The temperature and pressure of the air and fuel are 25oC and 101kPa respectively. The design velocity for each intake is 15 m/s. Determine the diameter of the air intake line in meters.
Actual Mass flow rate of air
kg/min 21.5 21.5*kg/min 1.0 AF*mm
5.2104.16
97.28*9.11
M
M*AFAF
fuelair
fuel
air
Actual Air = Theoretical * (1+%excess) = 9.524*1.25
= 11.9 kmol(air)/kmol(fuel)
QuizOne kg/min of methane is burned in a combustor with 25% excess air. The temperature and pressure of the air and fuel are 25oC and 101kPa respectively. The design velocity for each intake is 15 m/s. Determine the diameter of the air intake line in meters.
Diameter of air intake
m16.0m02.0*4A4
D
m02.0s60
min1*
m/s15*kg/m18.1
kg/min5.21
V)(
mA
AV)(m
kg/m18.11kPa
1kJ/m*ol28.97kg/km
298K*K-ol8.314kJ/km
101kPaM
TR
P
2
23
air
air
airair
33
airair
Enthalpy: Reacting SystemsTabular enthalpies inadequate due to arbitrary reference datums
Standard Reference State (Stable Elements): Tref = 298.15 K, pref = 1 atm
First Law:2 2 2 2 2 22 2
CO C O CO CO C C O OCO C OcvQ m h m h m h n h n h n h
2 0 (standard reference state)C Oh h
2
2
2 for CO at 1 atmcvCO f
CO
Qh h
n
)(formation ofenthalpy ofhh
Enthalpy: Reacting Systems
First Law:
, , ,f ref ref fh T p h h T p h T p h h
kJ/kmol387939
kJ/kmol9364kJ/kmol19945kJ/kmol393520
)K298()K550(kJ/kmol393520
2
2
222
CO
cv
COcv
refCOCOCOcv
n
Q
nQ
ThThnQ
0000
2222
222222
OCCOCO
ofCOcv
OO
ofOCC
ofCCOCO
ofCOcv
nnhhnQ
hhnhhnhhnQ
2222 OOCCCOCOcv hnhnhnQ
Standard Reference State
Tref = 25 oCPref = 1 atm
Energy Balances: Steady State
cvcvP R f fe i
e iP RF F
Q Wh h n h h n h h
n n
2 2 2 2 23.76 3.762 4 4 4
cvcvCO H O N F O N
F F
Q W b b b bah h a h h a h a h
n n
2 2 23.762 4
P CO H O Nb b
h ah h a h
2 23.764 4
R F O Nb b
h h a h a h
Energy Balances: Steady State
2 2 2 2 2
4 2 2 2 2 21*CH 2(O 3.76N ) CO 2H O 2*3.76N
2 2*3.76 2 2*3.76cvCO H O N F O N
F
cvP R
F
f ref
Qh h h h h h
n
Qh h
n
h h h T h T
Energy Balances: Steady State
22 2
22 2
4
-
-
-
2 7.52
169300 15829 4028 2 104040 13494 4258 7.52 11410 3730
289353 Btu/lbmol(fuel)
32210 Btu/lbmol(fuel)
28935
f fP ref ref refNCO H O
P NCO H O
P
fRCH
cvP R
F
h h h T h T h h T h T h T h T
h
h
h h
Qh h
n
- -3 32210 257143 Btu/lbmol(fuel)
Closed System Energy Balance
P R P RP R P R
Q W U U nu nu n h RT n h RT
f fP RP R
Q W n h h RT n h h RT
Fuel Enthalpies
RP e ie iP R
h n h n h Enthalpy of Combustion
For Example: A Control Volume at Steady State
cvcvf f f fe i e i e ie ie i e iP R P R P RF F
Q Wn h h n h h n h n h n h n h
n n
LHVRPh
LHV (Lower Heating Value): The enthalpy of combustion when the reactants and products are at the standard reference state and the water formed by combustion is a gas
HHV (Higher Heating Value): …water formed by combustion is a liquid
Adiabatic Flame Temperature
When no power produced, and combustion carried out adiabatically, Tp reaches a theoretical maximum.
When using tables, requires iteration to determine!
f fe ie iP R
n h h n h h
Adiabatic Flame Temperature
4
22 2
22 2
-
0
32210 Btu/lbmol(fuel)
2 7.52
169300 4028 2 104040 4258 7.52 3730
?
cvP R
F
fRCH
f fP ref ref refNCO H O
PNCO H O
P R
Qh h
n
h h
h h h T h T h h T h T h T h T
h h T h T h T
h h
T
Fuel Cells
Solid Oxide Fuel Cell Proton Exchange Membrane Fuel Cell
Third Law of Thermodynamics
The absolute entropy of a pure-crystalline substance at the absolute zero of temperature is
zero. Clip art courtesy of MS Office 2000
, ln ii ii
ref
y ps T p s T R
p Ideal Gas
, , , ,ref refs T p s T p s T p s T p
Entropy BalancesControl Volumes at Steady State, Reacting System
2 2 2 2 2
/0 3.76 3.76
4 4 2 4cvjj
F O N CO H O N
j F F
Q T b b b bs a s a s as s a s
n n
Closed, Reacting System
P Rb
QS S
T
1
P R bF F
Qns ns
n T n
Chemical ExergyTotal Exergy = Thermo-mechanical Exergy + Chemical Exergy
Thermo-mechanical Exergy found in Chapter 7
For a Hydrocarbon: CaHb
2
2 2 2
2 2
/ 4ech
0 0 0( ) / 2e ee , ln
4 2
a b
O
f O CO H O g a b
CO H O
yb bg a g ag g T p RT
y y
Chemical ExergyFor Carbon Monoxide: CO
2
2 2
2
1/ 2ech
0 0 0 e
1e , ln
2O
CO O CO
CO
yg g g T p RT
y
For Water: H2O 2 2
2
ch
0 0 0( ) ( ) e
1e , lnH O l H O g
H O
g g T p RTy
For N2, O2, CO2 ch
0 e
1e lnRT
y
For mixture of gas phases of Ideal Gases at T0, p0
ch
0 ee ln i
ii i
yRT y
y
Exergy Summary
2
ch0 0 0 0 0e e
2
Vu u p v v T s s gz
2
chf 0 0 0e e
2
Vh h T s s gz