1. ME 475/675 Introduction to Combustion
Lecture 2
Ideal stoichiometric hydrocarbon combustion, Mixture molecular weight, Air/Fuel mass ratio

2. Ideal Stoichiometric Hydrocarbon Combustion
air
CxHy + a(O2+3.76N2)  (x)CO2 + (y/2) H2O a N2
a = number of oxygen molecules per fuel molecule, 𝑁 𝑂 2 𝑁 𝐹𝑢𝑒𝑙
Number of air molecules per fuel molecule is a(1+3.76)
If a = aST = x + y/4, then the reaction is Stoichiometric
No O2 or Fuel in products
This mixture produces nearly the hottest flame temperature
If a < x + y/4, then reaction is fuel-rich (oxygen-lean, “fuel” left over)
If a > x + y/4, then reaction is fuel-lean (oxygen-rich, O2 left over)
Equivalence Ratio (of fuel)
Φ= 𝑎 𝑆𝑡 𝑎 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝑁 𝑂 2 𝑁 𝐹𝑢𝑒𝑙 𝑆𝑡 𝑁 𝑂 2 𝑁 𝐹𝑢𝑒𝑙 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝑁 𝐹𝑢𝑒𝑙,𝐴𝑐𝑡𝑢𝑎𝑙 𝑁 𝐹𝑢𝑒𝑙,𝑆𝑡 𝑁 𝑂 2 ,𝑆𝑡 𝑁 𝑂 2 ,𝐴𝑐𝑡𝑢𝑎𝑙
Φ=1→ Stiochiometric
Φ>1→ Fuel Rich
Φ<1→ Fuel Lean

3. Air to fuel mass ratio [kg air/kg fuel] of reactants
𝐴 𝐹 = 𝑚 𝐴𝑖𝑟 𝑚 𝐹𝑢𝑒𝑙 = 𝑁 𝐴𝑖𝑟 𝑀𝑊 𝐴𝑖𝑟 𝑁 𝐹𝑢𝑒𝑙 𝑀𝑊 𝐹𝑢𝑒𝑙 = 𝑁 𝑂 2 𝑁 𝐹𝑢𝑒𝑙 𝑁 𝐴𝑖𝑟 𝑁 𝑂 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 =𝑎 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙
For a stoichiometric mixture
𝐴 𝐹 𝑆𝑡 =4.76 𝑎 𝑆𝑡 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 ; 𝑤ℎ𝑒𝑟𝑒 𝑎 𝑆𝑡 =𝑥+ 𝑦 4
Need to find molecular weights

4. Molecular Weight of a Pure Substancex
Only one type of molecule:
AxByCz…
Molecular Weight
MW = x(AWA) + y(AWB) + z(AWC) + …
AWi = atomic weights
Inside front cover of book
Examples
𝑀𝑊 𝑂 2 = 2( AW 𝑂 ) = 2( ) = 𝑘𝑔 𝑘𝑚𝑜𝑙
𝑀𝑊 𝐻 2 𝑂 = 2( AW 𝐻 ) + ( AW 𝑂 ) = 2( ) + ( ) = 𝑘𝑔 𝑘𝑚𝑜𝑙
𝑀𝑊 𝑃𝑟𝑜𝑝𝑎𝑛𝑒 = 𝑀𝑊 𝐶 3 𝐻 8 = = 𝑘𝑔 𝑘𝑚𝑜𝑙
See page 701 for fuels

5. Mixtures containing n components
𝑁 𝑖 = number of moles of species 𝑖
𝑖=1, 2,..𝑛
Total number of moles in system
𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝑖=1 𝑛 𝑁 𝑖
Mole Fraction of species i
𝜒 𝑖 = 𝑁 𝑖 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝑁 𝑖 𝑖=1 𝑛 𝑁 𝑖
Mass of species 𝑖= 𝑚 𝑖 = 𝑁 𝑖 𝑀𝑊 𝑖
Total Mass
𝑚 𝑇𝑜𝑡𝑎𝑙 = 𝑖=1 𝑛 𝑚 𝑖
Mass Fraction of species i
𝑌 𝑖 = 𝑚 𝑖 𝑚 𝑇𝑜𝑡𝑎𝑙 = 𝑚 𝑖 𝑖=1 𝑛 𝑚 𝑖
Useful facts:
𝑖=1 𝑛 𝜒 𝑖 = 𝑖=1 𝑛 𝑌 𝑖 =1
but 𝜒 𝑖 ≠ 𝑌 𝑖
Mixture Molar Weight: 𝑀𝑊 𝑀𝑖𝑥 = 𝑚 𝑇𝑜𝑡𝑎𝑙 𝑁 𝑇𝑜𝑡𝑎𝑙
𝑀𝑊 𝑀𝑖𝑥 = 𝑚 𝑖 𝑁 𝑖 = 𝑁 𝑖 𝑀𝑊 𝑖 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝜒 𝑖 𝑀𝑊 𝑖 (weighted average)
𝑀𝑊 𝑀𝑖𝑥 = 𝑚 𝑖 𝑁 𝑖 = 𝑚 𝑇𝑜𝑡𝑎𝑙 𝑚 𝑖 / 𝑀𝑊 𝑖 = 𝑌 𝑖 / 𝑀𝑊 𝑖
Example
𝑀𝑊 𝐴𝑖𝑟 = 𝜒 𝑖 𝑀𝑊 𝑖 =0.21 𝑀𝑊 𝑂 𝑀𝑊 𝑁 2
=0.21∗ 2∗ ∗ 2∗
=0.21∗ ∗ =28.85 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒
Remember and/or write inside front cover of your book
Relationship between 𝜒 𝑖 and 𝑌 𝑖
𝑌 𝑖 = 𝑚 𝑖 𝑚 𝑇𝑜𝑡𝑎𝑙 = 𝑁 𝑖 𝑀𝑊 𝑖 𝑁 𝑇𝑜𝑡𝑎𝑙 𝑀𝑊 𝑀𝑖𝑥 = 𝜒 𝑖 𝑀𝑊 𝑖 𝑀𝑊 𝑀𝑖𝑥
𝜒 𝑖 = 𝑌 𝑖 𝑀𝑊 𝑀𝑖𝑥 𝑀𝑊 𝑖

6. Stoichiometric Air/Fuel Mass Ratio
For Hydrocarbon fuel CxHy
𝐴 𝐹 𝑆𝑡 = 𝑎 𝑆𝑡 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙
𝑀𝑊 𝐴𝑖𝑟 = 𝜒 𝑖 𝑀𝑊 𝑖 = 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒
𝑀𝑊 𝐹𝑢𝑒𝑙 = 𝑀𝑊 𝐶 𝑥 𝐻 𝑦 =𝑥 𝑦( )
aSt = x + y/4
𝐴 𝐹 𝑆𝑡 = 𝑥+ 𝑦 4 (4.76)28.85 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 𝑥 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 +𝑦( 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 ) 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 𝑥 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 𝑥 = 𝑦 𝑥 𝑦 𝑥
For 𝐶 𝑥 𝐻 𝑦 , 10< 𝐴 𝐹 𝑆𝑡 <35
Constraints on y/x later (see page 291 for some 𝐴 𝐹 𝑆𝑡 =𝜈)

7. Equivalence Ratio Φ
Φ= 𝑎 𝑆𝑡 𝑎 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝑎 𝑆𝑡 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 𝑎 𝐴𝑐𝑡𝑢𝑎𝑙 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 = 𝐴 𝐹 𝑆𝑡 𝐴 𝐹 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝐹 𝐴𝑐𝑡𝑢𝑎𝑙 𝐹 𝑆𝑡 𝐴 𝑆𝑡 𝐴 𝐴𝑐𝑡𝑢𝑎𝑙
Φ=1→ Stiochiometric
Φ>1→ Fuel Rich
Φ<1→ Fuel Lean
𝑎= 𝑎 𝑆𝑡 Φ = 𝑥+ 𝑦 4 Φ ; 𝐴 𝐹 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝐴 𝐹 𝑆𝑡 Φ = 𝑥+ 𝑦 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 Φ
CxHy + a(O2+3.76N2)
% Stoichiometric Air (%SA)= 100% Φ = 𝐹 𝑆𝑡 𝐹 𝐴𝑐𝑡𝑢𝑎𝑙 𝐴 𝐴𝑐𝑡𝑢𝑎𝑙 𝐴 𝑆𝑡 ∗100%
% Excess Oxygen (%EO) = (%SA)-100%

8. Example
For extra credit, this problem may be clearly reworked and turned in at the beginning of the next class period.
Problem 2.11, Page 91: In a propane-fueled truck, 3 percent (by volume) oxygen is measure in the exhaust stream of the running engine. Assuming “complete” combustion without dissociation, determine the air-fuel ratio (mass) supplied to the engine.
Also find
Equivalence Ratio: Φ
% Stoichiometric air: %SA
% Excess Oxygen: %EA
ID: Are reactants Fuel Rich, Fuel Lean, or Stoichiometric?
Work on the board

9. Thermodynamic Systems (reactors)
Closed systems
Rigid tanks, piston/cylinders
1 = Initial state; 2 = Final state
Mass: 𝑚 1 = 𝑚 2 =𝑚
Chemical composition inside can change
But atoms are conserved
1st Law
1 𝑄 2 − 1 𝑊 2 =Δ𝐸=𝑚( 𝑒 2 − 𝑒 1 )
𝑒=𝑢+ 𝑣 𝑔𝑧
1 𝑄 2 − 1 𝑊 2 =𝑚 𝑢 2 − 𝑢 𝑣 − 𝑣 𝑔 𝑧 2 − 𝑧 1
How to find internal energy 𝑢 for mixtures, and change 𝑢 2 − 𝑢 1 when composition changes due to reactions (not covered in Thermodynamics I)
1 𝑊 2
1 𝑄 2
m, E

10. Open Systems (control volume)
Steady State, Steady Flow (SSSF)
Fixed volume, no moving boundaries
Properties are constant and uniform
Inside CV (Dm = DE = 0), and
At ports (ℎ=𝑢+𝑃𝑣)
One inlet and one outlet: 𝑚 𝑖 = 𝑚 𝑜 = 𝑚
Atoms are conserved
Energy: 𝑄 𝐶𝑉 − 𝑊 𝐶𝑉 = 𝑚 ℎ 𝑜 − ℎ 𝑖 + 𝑣 𝑜 2 2 − 𝑣 𝑖 𝑔 𝑧 𝑜 − 𝑧 𝑖
Composition and temperature of inlet and outlet may not be the same due to reaction
Need to find ℎ 𝑜 − ℎ 𝑖 (not covered in Thermodynamics I)
Dm=DE=0
Inlet i
Outlet o
𝑄 𝐶𝑉
𝑚 𝑖 𝑒+𝑃𝑣 𝑖
𝑚 0 𝑒+𝑃𝑣 𝑜
𝑊 𝐶𝑉

11. Combustion Thermochemistry
We use thermodynamics to evaluate the internal energy, enthalpy and entropy of a systems at different states
The difference in energy and enthalpy between the products and reactants is used with the first law to predict
The heat of combustion (energy released during combustion, due to changes in chemical bonds) if the products are assumed to be at the same temperature as the reactants
The adiabatic flame temperature (product temperature assuming all reaction heat release stays in the system). Determined by assuming the products have the same energy as the reactants
The steady state product composition for a reaction may be determined from entropy considerations
Need to find
The internal energy, enthalpy and entropy of mixtures of gases, and
How to account for effects of chemical bonds
These are not covered in ME 311 Thermodynamics I, but we we’ll cover them now

12. Ideal Gas Equation of State
𝑃𝑉=𝑁 𝑅 𝑈 𝑇
Universal Gas Constant
𝑅 𝑈 =8.315 𝑘𝐽 𝑘𝑚𝑜𝑙𝑒 𝐾 =8315 𝐽 𝑘𝑚𝑜𝑙𝑒 𝐾
Inside book front cover
kJ = kN*m= kPa*m3
𝑃𝑉= 𝑁∗𝑀𝑊 (𝑅 𝑈 /𝑀𝑊)𝑇=𝑚𝑅𝑇
Specific Gas Constant
R = 𝑅 𝑈 /𝑀𝑊
MW = Molecular Weight of that gas
𝑃𝑣=𝑅𝑇;𝑣= 𝑉 𝑚 = 1 𝜌
𝑃=𝜌𝑅𝑇
Number of molecules
N*NAV
Avogadro's Number, 𝑁 𝐴𝑉
6.022∗ 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑘𝑚𝑜𝑙𝑒
6.022∗ 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑚𝑜𝑙𝑒
Number of molecules in 12 kg of C12

13. Partial Pressure 𝑃 𝑖 and volume fraction 𝑉 𝑖 𝑉 of a specie in a mixture at pressure 𝑃 and volume V
Each specie acts as if it was the only component at the given V and T
Specie 𝑖: 𝑃 𝑖 𝑉= 𝑁 𝑖 𝑅 𝑢 𝑇
Mixture: 𝑃 𝑉=𝑁 𝑅 𝑢 𝑇
Ratio: 𝑃 𝑖 𝑃 = 𝑁 𝑖 𝑁 = 𝜒 𝑖
𝑃 𝑖 = 𝜒 𝑖 𝑃
𝑃 𝑖 = 𝜒 𝑖 𝑃 =𝑃 𝜒 𝑖 =𝑃
Volume Fraction
Each specie acts as if it was the only component at the given P and T
Specie 𝑖: 𝑃 𝑉 𝑖 = 𝑁 𝑖 𝑅 𝑢 𝑇
Mixture: 𝑃 𝑉=𝑁 𝑅 𝑢 𝑇
Ratio: 𝑉 𝑖 𝑉 = 𝑁 𝑖 𝑁 = 𝜒 𝑖

14. End 2015

15. Extensive and Intensive System Properties
Intensive Properties
Independent of system size
Examples
Per unit mass (lower case)
v = V/m [m3/kg]
u = U/m [kJ/kg]
h = H/m [kJ/kg]
Denoted using lower-case letters
Exceptions
Temperature T [°C, K]
Pressure P [Pa]
Molar Basis (use bar )
V = vm = N 𝑣
U = um = N 𝑢
H = hm = N ℎ
N number of moles in the system
Useful because chemical equations deal with the number of moles, not mass
Extensive thermodynamic properties depend on System Size (extent)
Examples
Volume V [m3]
Internal Energy E [kJ]
Enthalpy H = E + PV [kJ]
Test: cut system in half
Denoted with CAPITAL letters

16. Calorific Equations of State for a pure substance
𝑢=𝑢 𝑇,𝑣 =𝑢(𝑇)≠𝑓𝑛(𝑣)
ℎ=ℎ 𝑇,𝑃 =ℎ(𝑇)≠𝑓𝑛(𝑃)
For ideal gases
Differentials (small changes)
𝑑𝑢= 𝜕𝑢 𝜕𝑇 𝑣 𝑑𝑇+ 𝜕𝑢 𝜕𝑣 𝑇 𝑑𝑣
For ideal gas
𝜕𝑢 𝜕𝑣 𝑇 = 0; 𝜕𝑢 𝜕𝑇 𝑣 = 𝑐 𝑣 𝑇
𝑑𝑢= 𝑐 𝑣 𝑇 𝑑𝑇
𝑑ℎ= 𝜕ℎ 𝜕𝑇 𝑃 𝑑𝑇+ 𝜕ℎ 𝜕𝑃 𝑇 𝑑𝑃
𝜕ℎ 𝜕𝑃 𝑇 = 0; 𝜕ℎ 𝜕𝑇 𝑃 = 𝑐 𝑃 𝑇
𝑑ℎ= 𝑐 𝑃 𝑇 𝑑𝑇
Specific Heat [kJ/kg C]
Energy input to increase temperature of one kg of a substance by 1°C at constant volume or pressure
How are 𝑐 𝑣 𝑇 and 𝑐 𝑃 𝑇 measured?
Calculate 𝑐 𝑝 𝑜𝑟 𝑣 = 𝑄 𝑚Δ𝑇 𝑝 𝑜𝑟 𝑣 𝑘𝐽 𝑘𝑔𝐾
𝑐 𝑝 = 𝑐 𝑝 ∗𝑀𝑊; 𝑐 𝑣 = 𝑐 𝑣 ∗𝑀𝑊 𝑘𝐽 𝑘𝑚𝑜𝑙 𝐾
m, T Q w
P = wg/A = constant
𝑐 𝑝
m, T Q
𝑐 𝑣
V = constant

17. Molar Specific Heat Dependence on Temperature
Monatomic molecules: Nearly independent of temperature
Only translational kinetic energy
Multi-Atomic molecules: Increase with temperature and number of molecules
Also possess rotational and vibrational kinetic energy

18. Internal Energy and Enthalpy
Once cp(T) and cv(T) are known, internal energy change can be calculated by integration
𝑢 𝑇 = 𝑢 𝑟𝑒𝑓 + 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑣 𝑇 𝑑𝑇
ℎ 𝑇 = ℎ 𝑟𝑒𝑓 + 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑃 𝑇 𝑑𝑇
Appendix A (pp , bookmark)
𝑐 𝑝 𝑇 :𝑡𝑎𝑏𝑙𝑒𝑠 𝑎𝑛𝑑 𝑐𝑢𝑟𝑣𝑒 𝑓𝑖𝑡𝑠
Note 𝑐 𝑣 = 𝑐 𝑝 − 𝑅 𝑢
𝑐 𝑝 = 𝑐 𝑝 /𝑀𝑊

19. Mixture Properties
Use these relations to calculate mixture enthalpy and internal energies (per mass or mole) as functions of the properties of the individual components and their mass or molar fractions.
u and h depend on temperature, but not pressure
Individual gas properties are on pp as functions of gas and T
Enthalpy
𝐻 𝑚𝑖𝑥 = 𝑚 𝑖 ℎ 𝑖 = 𝑚 𝑇𝑜𝑡𝑎𝑙 ℎ 𝑚𝑖𝑥
𝒉 𝒎𝒊𝒙 (𝑻)= 𝑚 𝑖 ℎ 𝑖 𝑚 𝑇𝑜𝑡𝑎𝑙 = 𝒀 𝒊 𝒉 𝒊 (𝑻)
𝐻 𝑚𝑖𝑥 = 𝑁 𝑖 ℎ 𝑖 = 𝑁 𝑇𝑜𝑡𝑎𝑙 ℎ 𝑚𝑖𝑥
𝒉 𝒎𝒊𝒙 (𝑻)= 𝑁 𝑖 ℎ 𝑖 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝝌 𝒊 𝒉 𝒊 (𝑻)
Internal Energy
𝒖 𝒎𝒊𝒙 (𝑻)= 𝒀 𝒊 𝒖 𝒊 (𝑻)
𝒖 𝒎𝒊𝒙 𝑻 = 𝝌 𝒊 𝒖 𝒊 𝑻

20. Standardized Enthalpy and Enthalpy of Formation
Needed for chemically-reacting systems because energy is required to form and break chemical bonds
Not considered in Thermodynamics I
Needed to find 𝑢 2 − 𝑢 1 and ℎ 𝑜 − ℎ 𝑖
ℎ 𝑖 𝑇 = ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 +Δ ℎ 𝑠,𝑖 (𝑇)
Standard Enthalpy at Temperature T =
Enthalpy of formation at standard reference state: Tref and P° +
Sensible enthalpy change in going from Tref to T = 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑃 𝑇 𝑑𝑇
Appendices A and B pp

21. Normally-Occurring Elemental Compounds
For example:
O2, N2, C, He, H2
ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 = 0
𝑇 𝑟𝑒𝑓 =298K
Use these as bases to tabulate the energy for form of more complex compounds
Example:
At 298K (1 mole) O ,390 kJ  (2 mole) O
To form 2O atoms from one O2 molecule requires 498,390 kJ/kmol of energy input to break O-O bond
ℎ 𝑓,𝑂 𝑜 𝑇 𝑟𝑒𝑓 = 498,390 kJ 2 𝑘𝑚𝑜𝑙 𝑂 =249,195 𝑘𝐽 𝑘𝑚𝑜𝑙 𝑂
ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 for other compounds are in Appendices A and B