Combustion and Fuels

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Combustion and Fuels

• Usu. Hydrocarbons or Hydrogen in Air are used
• Other fuels include CO, NH3, CS2, H2S,
• If you use an oxidizer other than air, there is a
  host of new fuels that can be used (rocket fuels)
• Why hydrocarbons?
• Many are liquids - high density, easy to
  transport
• Lots of it although in the wrong places, e.g.,
  Middle East, Saudi Arabia, Iran, Iraq
• Relatively non-toxic fuel combustion products
• Relatively low explosion hazard
• Why Air? (Air 0.21O2 0.79N2)
• Its there and FREE! (although much spent to
  reduce air pollution)

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Types of Hydrocarbons
Alkanes - CnH2n2
H
H
H
H
H
H
H - C - H
H - C - C - H
H - C - C - C - H
H
H
H
H
H
H
Methane
Propane
Ethane
Napthenes - Alkanes w/ Rings
H
H
C
Cyclopropane (not very stable -lots of strain in
C-C bonds)
H - C - C - H
H
H
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Types of Hydrocarbons
Olefins- Contains one or more CC
H
H
H
H
H
H
H
H
H
C C
C C - C
H
C C - C C
H
H
H
H
H
H
Ethylene
Propene
Butadiene
Acetylenes - Contains one or more C?C bond
H - C ? C - H Acetylene (very reactive)!
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Types of Hydrocarbons
Aromatics - contains one or more six-sided ring
structures
H
C
Benzene
H
H
C C
C C
H
H
C
H
Alcohols - contains one or more -OH groups
H
H
H
H - C -O H Methanol
H - C - C - O H Ethanol
H
H
H
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More Fuels
Hydrogen H - H Carbon Monoxide
C O Ammonia H - N
H
H
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Stoichiometry
Balancing chemical reactions when products are
known (assumed)
Example 1 CH4 ? (O2 3.76 N2) --gt ? CO2 ?
H2O ? N2
from Dr. Pearlmans crystal ball
Balance to get 1 CH4 2 (O2 3.76 N2) --gt 1
CO2 2 H2O 2 (3.76) N2
Fuel - Air Ratio (FAR) is defined as
ni moles of i Mi molecular weight of species i
where
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FAR
In our example, FAR
1 mole CH4 x 16g/mol CH4
(2 mole O2 x 32g/mol O2 2 x 3.76 mol N2 x
28g/mol N2)
FAR 0.058
AFR Air to fuel ratio 1/ FAR 17.2
Also, Xf mass fuel/ mass total mass fuel/
(mass air mass fuel)
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Equivalence Ratio

• f lt 1 Excess air (lean mixture)
• f gt 1 Excess fuel (rich mixture)

If 1 mol CH4 burns with 1 mol air, then
Reactants 1 CH4 1 (O2 3.76 N2)
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Equivalence Ratio
What is f for methane-air where the mole fraction
of air and products is unknown?
1 CH4 ? (O2 3.76 N2) --gt ? CO ? CO2 ? H2O
? N2

• Cant Solve! - too many unknowns
• 5 unknowns
• 4 equations (1 for each atom)

Need to consider chemical equilibrium to solve!
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Thermodynamic Calculations of Combustion Processes
Motivation Cannot always determine the amount
of various products based on stoich.
Considerations - many possible products to choose
from. Chemical Thermodynamics to the rescue
Lets us determine which products are obtained if
we wait long enough (as short as 1ms to forever).
Quiz What chemical reaction yields the
highest known flame temperature? H2 -O2 at f
1 (T3079 K at 1atm) NO!!
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Thermodynamic calculations of combustion processes
The HIGHEST KNOWN temperature is associated
with C4N2 - O3 at f 1 (T5516 K at 1atm)
Ozone
Carbon Sub-Nitride
C4N2 O3 -gt 4 CO N2
WHY ?
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Why?
H2-O2 has more heat release per unit mass and per
mole of reactants, but the Products are NOT just
H2O
H2 O2 -gt H2O
Rather,
H2 O2 -gt 0.58H2O 0.05O2 0.15H2 0.08H
0.11OH 0.33O
Water dissociates into smaller molecules and 1.
Produces more mol of products to soak up more
heat, lowering the flame temp. 2. Energy
is partly used to break the water into these
other things, thus reducing the flame temp.
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Why?

• If we prevent water from dissociating, flame temp
  increases to
• 4998K. Still lower.
• Why?
• Water is triatomic molecule w/ many degrees of
  freedom
• (vibrate rotate). Energy is stored in these
  d.o.f.
• How does carbon sub-nitride do it?
• CO and N2 have fewer d.o.f., so each one soaks
  up less energy
• CO and N2 are very stable even at 5500K (dont
  dissociate)

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Goal of Combustion Thermodynamics
Given Initial State (P, T, Yi) and an assumed
Process (Constant P, Constant V, Constant S, ),
find the final state of the mixture.
Three common processes in engine apps 1.
Compression from low P/ high Vol. To high P/ low
Vol. (usu. Pressure or volume is known and
the composition is frozen (doesnt
change). This process is usually assumed
Isentropic (constant entropy)
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Goal of Combustion Thermodynamics

• Combustion - Usu. assumed constant pressure or
  constant volume.
• Composition must change to provide heat
  release.
• Expansion - from high P/low V to low P/ high V.
• The volume ratio or the final pressure ratio is
• usu. given. Again, assumed isentropic.

These analyses are needed to determine the work
input/ output, efficiency, heat release, etc.
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Assumptions

• Assumptions that we will use
• Ideal Gases
• Adiabatic
• Kinetic and Potential Energy Negligible
• Combustion Process is Constant P or Constant V
• Compression/ Expansion processes reversible
• -gt isentropic (constant entropy)

Lets first consider the combustion process,
then the compression and expansion processes
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1st Law of Thermodynamics
For a control mass, dE ?Q ?W where E
energy of system, Q heat transfer to system,
W work transferred to the system E U PE
KE, so dE dU d(PE) d(KE) ?Q ?W For
d(PE) d(KE) 0, ?Q 0 (adiabatic) ?W -
PdV for simple compressible substance Get, dU
PdV 0 -gt dH 0 for constant pressure process
(recall, H UPV) --gt Hreactants H products
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1st Law of Thermodynamics

• Also, recall hH/M (M mass constant for a
  fixed control mass)
• Since, Hreactants Hproducts
• Hreactants/Mreactants Hproducts /Mproducts
• Thus, hreactants hproducts
• For constant volume process (dV0)
• dU 0 or U H - VP constant
• or h - vP constant (vV/M, h H/M)
• Thus, (h - vP)reactants (h - vP)products for
  constant V process

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Total Enthalpy
Enthalpy/mole of species i
Total enthalpy of a mixture
Moles of species i

• Thermal contribution
• cp,i (T-298) if cp is constant
• a f(T) only for ideal gases

• Chemical contribution
• per mol
• o standard state
• f formation
• i species i

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For Constant Pressure Process
hreactants hproducts
and recall hH/M and
same sum for products
(Energy Equation)
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For Constant Volume Process
Know PVnRT
PvnRT(n/M)RT RT
Recall, h-vP constant
same sum for products
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For Constant Volume Process
In constant volume process, usu. we want to know
the final pressure
Recall, RRo/M, therefore Rf Ro/Mf and Ri
Ro/Mi
where MfMitotal mass of reactants (products)
doesnt change ? Rf Ri
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Heating Value of a Fuel
For a const. P process we said
(Chemical enthalpy change per unit mass)
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LHS
is total thermal enthalpy of mixture
Avg Cp per unit mass
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RHS
Fuel Mass Total Mass
-QR Heating Value of the Fuel (per unit mass of
fuel)
fuel mass/(fuelair mass) FAR/(1FAR)
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Example of QR Calculation
Combustion of iso-octane-air 1C8H18
25/2(O23.76N2) -gt 8CO2 (g) 9H2O (g)
25/2(3.76)N2
1mol(-59.74kcal/mol)25/2(00)
8(-94.1)9(-57.8)25/2(0)

1mol (0.114 kg/mol)
Table 4.4
10640 kcal/kg fuel 4.5 x 107 J/kg fuel (most
HCs 4-5x107)
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LHS RHS
If we assume
Not bad if mostly N2
Good approximation to energy equation to get
Tproducts Tflame
Note For constant volume combustion everything
else same except replace by
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Example
3H2 O2 -gt 2H2O 1H2 Pi 1atm,
Treactants400K Question What is Tproducts if
Pf1atm (constant pressure process)?
Reactants H2 cp6.9 cal/mol/K at 298K
O2 cp7.0 cal/mol/K at 298K Products H2O
cp8.03 cal/mol/K at 298K
3mol(6.9cal/mol/K)(400-298)K0)1mol(7.0)(400-298
)0
hreactants
3mol(0.002kg/mol)1mol(0.032kg/mol)
7.4 x 104 cal/kg
2mol(8.03cal/mol/K)(T-298)-52800)1mol(6.9)(T-298
)0
hproducts
2mol(0.018kg/mol)1mol(0.002kg/mol)
7.4 x 104 cal/kg
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Example (cont.)
Set hreactants hproducts Solve for T (5459K)
Too high!! Why? Because cp?const.
Alternative, use cp?const, look up (h-h298) in
tables A.5 A.12
3mol(707cal/mol0)1mol(723cal/mol0)
7.4 x 104 cal/kg
hreactants
3mol(0.002kg/mol)1mol(0.032kg/mol)
2mol(h-h298)H2O-57800cal/mol1mol((h-h298)H20)
hproducts
2mol(0.018kg/mol)1mol(0.002kg/mol)
Need to find T such that hreactants hproducts
Guess, then reinterate. At T4014K,
(h-h298)H2O44064 cal/mol (h-h298)H230455
cal/mol Evaluate hproducts 7.8x104 cal/kg,
getting close, reiterate!