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

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

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 8 +5 𝑂 2 +3.76 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

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

Minor-Specie Mole Fractions: NO, OH, H, O 1% 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

Simple Product Calculation method 𝐶 𝑥 𝐻 𝑦 +𝑎 𝑂 2 +3.76 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 2 +3.76 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)

Atomic Balance for Lean combustion Φ≤1 𝐶 𝑥 𝐻 𝑦 +𝑎 𝑂 2 +3.76 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 4 1 Φ −1 = 𝑥+ 𝑦 4 1−Φ Φ Check: if Φ=1, then 𝑓=0 𝑁 𝑇𝑜𝑡 =𝑏+𝑑+𝑓+3.76𝑎=𝑥+ 𝑦 2 + 𝑥+ 𝑦 4 1−Φ Φ +3.76 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 4 1−Φ Φ 𝑁 𝑇𝑜𝑡 ; 𝜒 𝑁 2 =3.76 𝑥+ 𝑦 4 Φ 𝑁 𝑇𝑜𝑡 MathCAD Solution http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/index.htm

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

For Rich combustion Φ>1 𝐶 𝑥 𝐻 𝑦 +𝑎 𝑂 2 +3.76 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 1 𝑃 𝐻 2 𝑃 𝑜 1 𝑃 𝐶𝑂 𝑃 𝑜 1 𝑃 𝐻 2 𝑂 𝑃 𝑜 1 = 𝜒 𝐶 𝑂 2 1 𝜒 𝐻 2 1 𝜒 𝐶𝑂 1 𝜒 𝐻 2 𝑂 1 𝑃 𝑃 𝑜 0 = 𝑏 𝑁 𝑇𝑜𝑡 1 𝑒 𝑁 𝑇𝑜𝑡 1 𝑐 𝑁 𝑇𝑜𝑡 1 𝑑 𝑁 𝑇𝑜𝑡 1 = 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 0.1635 for T = 2000 to 3500 K

Atomic Balances 𝐶 𝑥 𝐻 𝑦 +𝑎 𝑂 2 +3.76 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

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 2 2 −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 http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/index.htm For Homework, re-derive these equations with pure oxygen (no nitrogen)

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. 713-4, for Complex Reactions

More Complex 𝐶 𝑂 2 Dissociation Before: 𝐶 𝑂 2 ↔𝐶𝑂+ 1 2 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+ 1 2 𝑂 2 +__𝑂 add 𝑂 2 ↔2𝑂; one more unknown: 𝜒 𝑂 Need one more constraint: 𝐾 𝑃, 𝑂 2 = 𝑃 𝑂 𝑃 𝑜 2 𝑃 𝑂 2 𝑃 𝑜 1 = 𝜒 𝑂 2 𝜒 𝐶 𝑂 2 1 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 1 𝜒 𝑂 2 1 2 𝜒 𝐶 𝑂 2 1 𝑃 𝑃 𝑜 1 2 ; 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 + 1 2 𝑔 𝑓, 𝑂 2 𝑜 −1 𝑔 𝑓,𝐶 𝑂 2 𝑜 𝜒 𝐶 𝑂 2 + 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 …

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: www.mhhe.com/turns3e 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

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

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

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