1. ME 475/675 Introduction to Combustion
Lecture 8
Dissociation, Problem X2, Equilibrium Produces of Combustion, Simple models for lean and rich combustion

2. Announcements HW 3: X2, Ch 2 (57), Due Monday
Modify MathCAD program from lecture notes with 𝜒 𝑁 2 =0
ABET assessment visitor during last 20 minutes of Monday’s lecture
Dr. Sriram Somasundaram

3. Equilibrium Products of Hydrocarbon Combustion
Combine Chemical Equilibrium (2nd law) & Adiabatic Flame Temperature (1st law)
For Example: Propane and air combustion
Ideal
𝐶 3 𝐻 𝑂 𝑁 2 →3𝐶 𝑂 2 +4 𝐻 2 𝑂+18.8 𝑁 2 +(0) 𝑂 2
Four products for a range of air/fuel ratios: 𝐶 𝑂 2 , 𝐻 2 𝑂, 𝑁 2 , 𝑂 2
Now consider seven more possible dissociation products:
𝐶𝑂, 𝐻 2 , 𝐻, 𝑂𝐻, 𝑂, 𝑁𝑂, 𝑁
What happens as air/fuel (equivalence) ratio changes
Φ= 𝐴/𝐹 𝑆𝑡𝑜𝑖𝑐ℎ𝑖𝑜 𝐴/𝐹 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝐹 𝐴 𝐴𝑐𝑡𝑢𝑎𝑙 𝐹 𝐴 𝑆𝑡𝑜𝑖𝑐ℎ𝑖𝑜

4. Data: Flame temp and major mole-fractions vs Φ
Equivalence Ratio Φ= 𝐹 𝐴 𝐴𝑐𝑡𝑢𝑎𝑙 𝐹 𝐴 𝑆𝑡𝑜𝑖𝑐ℎ𝑖𝑜
At Φ=1
O2, CO, H2 all present due to dissociation.
Not present in “ideal” combustion
𝜒 𝐻 2 𝑂 𝑀𝑎𝑥 at Φ=1.15
𝑇 𝐴𝑑 𝑀𝑎𝑥 at Φ=1.05
𝑇 𝐴𝑑 − 𝑇 𝑅𝑒𝑓 = 𝐻 𝐶 𝑁 𝑇𝑜𝑡 𝑐 𝑝,𝑚𝑖𝑥
𝐻 𝐶 and 𝑁 𝑇𝑜𝑡 𝑐 𝑝,𝑚𝑖𝑥 decrease for Φ>1
For Φ<1.05: 𝑁 𝑇𝑜𝑡 𝑐 𝑝,𝑚𝑖𝑥 decreases faster
For Φ>1.05: 𝐻 𝐶 decreases faster
For Φ<1 need to include O2
For Φ>1 need to include CO and H2
Tad[K]
𝜒 𝑖 %
“Old”
“New”
Fuel Lean O2
Fuel Rich
Get CO, H2

5. Minor-Specie Mole Fractions: NO, OH, H, O1%
Not considered in Ideal combustion
For all
𝜒 𝑖 < 4000 ppm = 0.4%
Peak near Φ=1
𝜒 𝑂𝐻 >10 𝜒 𝑂
𝜒 𝑁(𝑛𝑜𝑡 𝑠ℎ𝑜𝑤𝑛) ≪ 𝜒 𝑂
𝜒 𝑖
ppm

6. Simple Product Calculation method
𝐶 𝑥 𝐻 𝑦 +𝑎 𝑂 𝑁 2 →𝑏𝐶 𝑂 2 +𝑐𝐶𝑂+𝑑 𝐻 2 𝑂+𝑒 𝐻 2 + 𝑓 𝑂 2 +(3.76𝑎) 𝑁 2
Add 𝐶𝑂, 𝐻 2 𝑎𝑛𝑑 𝑂 2
Neglect “minor” species: NO, OH, H, O
𝑎= 𝑚𝑜𝑙𝑒𝑠 𝐴𝑖𝑟 𝑚𝑜𝑙𝑒𝑠 𝐹𝑢𝑒𝑙 = 𝑥+ 𝑦 4 Φ
Φ= 𝐴/𝐹 𝑆𝑡𝑜𝑖𝑐ℎ𝑖𝑜 𝐴/𝐹 𝐴𝑐𝑡𝑢𝑎𝑙
Assume 𝑥, 𝑦 and Φ are known
What is a good assumption for lean mixtures Φ≤1?
𝐶 𝑥 𝐻 𝑦 +𝑎 𝑂 𝑁 2 →𝑏𝐶 𝑂 2 +𝑑 𝐻 2 𝑂+𝑓 𝑂 2 +(3.76𝑎) 𝑁 2
c = e = 0 (no CO or H2), but now include 𝑂 2
3 unknowns (b, d, f), 3 atom balances (C, H, O)

7. Atomic Balance for Lean combustion Φ≤1
𝐶 𝑥 𝐻 𝑦 +𝑎 𝑂 𝑁 2 →𝑏𝐶 𝑂 2 +𝑑 𝐻 2 𝑂+𝑓 𝑂 2 +(3.76𝑎) 𝑁 2
C: 𝑥=𝑏 so 𝑏=𝑥
H: 𝑦=2𝑑 so 𝑑= 𝑦 2
O: 2𝑎=2𝑏+𝑑+2𝑓=2𝑥+ 𝑦 2 +2𝑓 so
𝑓=𝑎−𝑥− 𝑦 4 = 𝑥+ 𝑦 4 Φ −𝑥− 𝑦 4 = 𝑥+ 𝑦 Φ −1 = 𝑥+ 𝑦 −Φ Φ
Check: if Φ=1, then 𝑓=0
𝑁 𝑇𝑜𝑡 =𝑏+𝑑+𝑓+3.76𝑎=𝑥+ 𝑦 2 + 𝑥+ 𝑦 −Φ Φ 𝑥+ 𝑦 4 Φ
=𝑥+ 𝑦 2 + 𝑥+ 𝑦 4 Φ 1−Φ+3.76
Mole Fractions
𝜒 𝐶 𝑂 2 = 𝑥 𝑁 𝑇𝑜𝑡 ; 𝜒 𝐻 2 𝑂 = 𝑦 2 𝑁 𝑇𝑜𝑡 ; 𝜒 𝑂 2 = 𝑥+ 𝑦 −Φ Φ 𝑁 𝑇𝑜𝑡 ; 𝜒 𝑁 2 = 𝑥+ 𝑦 4 Φ 𝑁 𝑇𝑜𝑡 MathCAD Solution

8. Comparison Φ≤1 Total number of modes decreases as Φ increases
H2O O2
N2/10
Total number of modes decreases as Φ increases
Does not include CO or H2 at Φ~1
Not bad for a simple model for Φ<1
Conversely, when mixture is fuel rich, get significant CO or H2 but little O2

9. For Rich combustion Φ>1
𝐶 𝑥 𝐻 𝑦 +𝑎 𝑂 𝑁 2 →𝑏𝐶 𝑂 2 +𝑐𝐶𝑂+𝑑 𝐻 2 𝑂+𝑒 𝐻 2 +(3.76𝑎) 𝑁 2
No 𝑂 2 (or fuel), so 𝑓=0
4 unknows: b, c, d and e
3 Atom balances: C, H, O
Need one more constraint
Consider “Water-Gas Shift Reaction” equilibrium
𝐶𝑂+ 𝐻 2 𝑂↔𝐶 𝑂 2 + 𝐻 2
𝐾 𝑃 = 𝑃 𝐶 𝑂 2 𝑃 𝑜 𝑃 𝐻 2 𝑃 𝑜 𝑃 𝐶𝑂 𝑃 𝑜 𝑃 𝐻 2 𝑂 𝑃 𝑜 = 𝜒 𝐶 𝑂 𝜒 𝐻 𝜒 𝐶𝑂 𝜒 𝐻 2 𝑂 𝑃 𝑃 𝑜 0 = 𝑏 𝑁 𝑇𝑜𝑡 𝑒 𝑁 𝑇𝑜𝑡 𝑐 𝑁 𝑇𝑜𝑡 𝑑 𝑁 𝑇𝑜𝑡 = 𝑏𝑒 𝑐𝑑 = 𝐾 𝑃
Not dependent on P since number of moles of products and reactants are the same
Δ 𝐺 𝑇 𝑜 =1 𝑔 𝑓,𝐶 𝑂 2 𝑜 +1 𝑔 𝑓, 𝐻 2 𝑜 −1 𝑔 𝑓,𝐶𝑂 𝑜 −1 𝑔 𝑓, 𝐻 2 𝑂 𝑜 =𝑓𝑛 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 ; 𝐾 𝑃 =exp −Δ 𝐺 𝑇 𝑜 𝑅 𝑢 𝑇
See plot from data on page 51
KP = 0.22 to for T = 2000 to 3500 K

10. Atomic Balances
𝐶 𝑥 𝐻 𝑦 +𝑎 𝑂 𝑁 2 →𝑏𝐶 𝑂 2 +𝑐𝐶𝑂+𝑑 𝐻 2 𝑂+𝑒 𝐻 2 +(3.76𝑎) 𝑁 2
C: 𝑥=𝑏+𝑐
𝑐=𝑥−𝑏 (in terms of b and “knowns,” x, y, a)
O: 2𝑎=2𝑏+𝑐+𝑑=2𝑏+ 𝑥−𝑏 +𝑑=𝑏+𝑥+𝑑
𝑑=2𝑎−𝑏−𝑥 (in terms of b and “knowns”)
H: 𝑦=2𝑑+2𝑒=4𝑎−2𝑏−2𝑥+2𝑒
𝑒= 𝑦 2 −2𝑎+𝑏+𝑥 (in terms of b and “knowns”)
Plug into equilibrium constraint
𝑏𝑒= 𝐾 𝑃 𝑐𝑑
𝑏 𝑦 2 −2𝑎+𝑏+𝑥 = 𝐾 𝑃 𝑥−𝑏 2𝑎−𝑏−𝑥 ; b is the only unknown!
Expand and collect terms to get

11. Solution
𝑦 2 𝑏−2𝑎𝑏+ 𝑏 2 +𝑥𝑏 =2𝑎𝑥 𝐾 𝑃 −𝑏𝑥 𝐾 𝑃 − 𝑥 2 𝐾 𝑃 −2𝑎𝑏 𝐾 𝑃 + 𝑏 2 𝐾 𝑃 +𝑥𝑏 𝐾 𝑃
0= 𝑏 2 𝐾 𝑃 −1 +b 2𝑎 1− 𝐾 𝑃 −𝑥− 𝑦 2 + 𝐾 𝑃 𝑥 2𝑎−𝑥
𝑏= − 2𝑎 1− 𝐾 𝑃 −𝑥− 𝑦 2 ± 2𝑎 1− 𝐾 𝑃 −𝑥− 𝑦 −4 𝐾 𝑃 −1 𝐾 𝑃 𝑥 2𝑎−𝑥 2 𝐾 𝑃 −1
Since 𝐾 𝑃 −1<0, use “-” root
Mole fraction can be calculated from b
MathCAD Solution
For Homework, re-derive these equations with pure oxygen (no nitrogen)

12. Comparison Φ>1
CO2
H2O H2
N2/10 CO
Φ≤1
Total number of moles continues to decrease as Φ increases
At Φ~1 , has no not include O2, CO or H2
However, not bad for a simple model for Φ>1
More accurate models may be developed by including more equilibrium reactions and constants using computer programs
Computer Programs Provided by the Book
Appendix F, pp , for Complex Reactions

13. More Complex 𝐶 𝑂 2 Dissociation
Before: 𝐶 𝑂 2 ↔𝐶𝑂 𝑂 2 ; three unknowns: 𝜒 𝐶 𝑂 2 , 𝜒 𝐶𝑂 , 𝜒 𝑂 2
For more complete analysis, add an additional product
For 𝐶 𝑂 2 ↔𝐶𝑂 𝑂 2 +__𝑂 add 𝑂 2 ↔2𝑂; one more unknown: 𝜒 𝑂
Need one more constraint: 𝐾 𝑃, 𝑂 2 = 𝑃 𝑂 𝑃 𝑜 𝑃 𝑂 2 𝑃 𝑜 = 𝜒 𝑂 𝜒 𝐶 𝑂 𝑃 𝑃 𝑜 1
𝐾 𝑃, 𝑂 2 =exp −Δ 𝐺 𝑇, 𝑂 2 𝑜 𝑅 𝑢 𝑇 ; Δ 𝐺 𝑇, 𝑂 2 𝑜 =2 𝑔 𝑓,𝑂 𝑜 −1 𝑔 𝑓, 𝑂 2 𝑜
Already had
𝐾 𝑃,𝐶 𝑂 2 = 𝜒 𝐶𝑂 𝜒 𝑂 𝜒 𝐶 𝑂 𝑃 𝑃 𝑜 ;
𝐾 𝑃,𝐶 𝑂 2 =exp −Δ 𝐺 𝑇,𝐶 𝑂 2 𝑜 𝑅 𝑢 𝑇 ; Δ 𝐺 𝑇,𝐶 𝑂 2 𝑜 =1 𝑔 𝑓,𝐶𝑂 𝑜 𝑔 𝑓, 𝑂 2 𝑜 −1 𝑔 𝑓,𝐶 𝑂 2 𝑜
𝜒 𝐶 𝑂 𝜒 𝐶𝑂 ,+ 𝜒 𝑂 2 + 𝜒 𝑂 =1
𝑁 𝐶 𝑁 𝑂 = 𝜒 𝐶 𝑂 2 + 𝜒 𝐶𝑂 2 𝜒 𝐶 𝑂 2 + 𝜒 𝐶𝑂 +2 𝜒 𝑂 2 + 𝜒 𝑂 …

14. Computer Programs Provided by Book Publisher
Described in Appendix F (pp 113-4)
For “complex” reactions (11 product species)
Fuel: CNHMOLNK
Oxidizer: Air
Download from web:
student edition
Computer codes
Access to TPEquil, HFFlame, UVFlame
Extract All
TPEQUIL (TP Equilibrium)
Use to find
Equilibrium composition and mixture properties
Required input
Fuel CNHMOLNK
Temperature
Pressure
Equivalence ratio (with air)
to determine initial number of moles of each atom

15. HPFLAME (HP Flame) Use to find Required Input
Adiabatic flame temperature for constant pressure
Required Input
Fuel, equivalence ratio, enthalpy of reactants HR, pressure
For constant pressure: HP = HR
Find TAd
In our examples we assume ideal combustion so we knew the product composition
But this program calculates the more realistic equilibrium composition of the products from a (complex) equilibrium calculation (multiple equilibrium reactions)
But this requires TProd = TAd, which we are trying to find!
Requires program (not humans) to iterate

16. Air Preheaters
Preheating the air using exhaust or flue gas increases the flame temperature
Recuperators uses heat transfer across a wall
Regenerators use a moving ceramic or metal matrix

17. Exhaust or Flue Gas Recirculation
Inserting exhaust gas into the reactants reduces flame temperature, which can reduce pollution (oxides of nitrogen, NO, NO2)