Combustion: Flame Theory and Heat Produced. Arthur Anconetani Oscar Castillo Everett Henderson

Transcription

1 Combustion: Flame Theory and Heat Produced Arthur Anconetani Oscar Castillo Everett Henderson

2 What is a Flame?! Reaction Zone! Thermo/Chemical characteristics

3 Types of Flame! Premixed! Diffusion! Both can be Laminar or Turbulent

4 Premixed! Mixed before Combustion! Characteristics! Reacts Rapidly! Constant Pressure! Propagates as Thin Zone! Ex: Spark Engine Temp Intensity

5 Diffusion! Mixed during Combustion! Characteristics! Reaction occurs at Fuel/Air interface! Controlled by the Mixing of the Reactants! Ex:Diesel Engines

6 Laminar! Premixed! Simplest flame type! Ex: Bunsen burner! Diffusion! Ex: Candle

7 Turbulent! Premixed! Faster heat release than laminar! Ex: Indirect fuel injection engines! Diffusion! Ex: Direct fuel injection engines

8 Chemical Energy! The energy inside fuel can be considered potential energy! Combustion unleashes that potential energy! How do we calculate the amount of energy released?

9 Basic Chemistry! Hydrocarbon fuels! Air! Nitrogen (79%)! Oxygen (21%)! 1 mol O 2 :3.76 mol N 2! Common Products: H 2 O, CO 2, N 2

10 Basic Chemistry-Moles! Amount of mass of an element or compound that contains Avogadro s number of atoms or molecules! Avogadro s number = E 23! For example one mole of Hydrogen contains 6.022E 23 Hydrogen atoms.! Molar mass is the amount of mass in one mole of a substance.

11 Basic Chemistry Molar Mass Mass is conserved in chemical equations 1 kmol H 2 + ½ kmol O 2 1 kmolh 2 O 1kmol H kg kmol H kmol O 32kg + 2 kmol O 1kmol H 2 O 18.02kg 2 kmol H 2 O 18.02kg = kg

12 Balancing an Equation Original Chemical Equation: C 4 H 10 + a(o N 2 ) bco 2 + ch dn 2 Write equations for each element, solve: C: 4 = b H: 10 = 2c c = 5 O: 2a = 2b + c a = 6.5 N: 2(3.76)a = 2d d = Final Balanced Equation: C 4 H (O N 2 ) 4CO 2 + 5H N 2

13 Focusing on the Problem! We have! Basics of Flame Theory! Balanced Equations! What s Missing?

14 Enthalpy! Definition h = u + Pv! Reference State! 25 C! 1 atm! Δh = h(t,p) h(t,p) ref

15 Enthalpy! Enthalpy of Formation! Energy exchanged during compound formation! N 2, O 2, & H 2 have h form =0! Total Enthalpy h=h form + Δh Total Enthalpy h of formation h at T, P h h at reference T, P

16 More Enthalpy! Enthalpy of Combustion H R =H P-ideal Q loss! Higher and Lower Heating Values! Liquid H 2 O! Vapor H 2 0 H P-loss H combustion H P-Ref

17 Energy Equation Q-W = ΔU Q= ΔU+W= ΔU+ PΔV Q = ΔH Q = H p H r

18 Combustion Chamber! Burned and Unburned regions! Flame propagation! Constant Pressure

19 Heat Loss Example A mixture of 1kmol of gaseous methane and air, originally at reference state, burns completely in a combustion chamber, at constant pressure. Determine the amount of heat the chamber loses if the Product temperature measured after combustion is 890K. Balanced ( ) CH 4 + O N 2 CO 2 + H 2O + N 2 ( ) CH O N 2 CO 2 + 2H 2O N 2 Q H p H r Q ( nh ) P ( nh ) R

20 Heat Loss Example Q H p H r Q ( nh ) P ( nh ) R ( nh ) P ( nh ) R 1 ( h CO2 ) 1h CH4 2h ( H2O ) ( ) h N2 ( ) 2 ( h O2 ) ( ) h N2 h h form + h + ( ( )) h h form ht ( ) h T ref A lot of terms Lets look at two of them.

21 Heat Loss Example-CO 2 H CO2 ( ) h form + ( ht ( p ) ht ( ref )) 1 h CO2 h f h at 298 K h at 890 K (kj/kmol) Carbon Dioxide -393, CO 2 Water Vapor -241, H 2 O Oxygen O Nitrogen N Methane Ch Octane C 8 H H CO ( ) H CO kj

22 Heat Loss Example-O 2 H O2 ( ) 2 h form + ( ht ( R ) ht ( ref )) 2 h O2 h f h at 298 K h at 890 K (kj/kmol) Carbon Dioxide -393, CO 2 Water Vapor -241, H 2 O Oxygen O Nitrogen N Methane Ch Octane C 8 H H O2 2[ 0 + ( ) ] H O2 0

23 Heat Loss Example The remaining terms are evaluated, using the above techniques. ( ) ( h CH4 + 2h O h N2 ) Q h CO2 + 2h H2O h N2 Q [ ( ) + 2( ) ( 17899) ] [ ( ) ( 0) ] Q kj kj of heat was lost to the surroundings.

24 Departures From Ideal! Combustion not always complete! Insufficient Mixing! Insufficient Air! May Lead to Knocking

25 Adiabatic Flame Temperature! Adiabatic Conditions! Limiting Value of Flame Temperature! Iterative Process

26 AFT Example This problem has the same set of assumptions as the last problem. The only difference is that now we are assuming adiabatic flame conditions CH 4 + 2(O N 2 ) CO 2 + 2H 2 O N 2 Q = H P -H R H P = H R h CO2 2 ( h H2O ) ( ) h N2 h CH4 ( ) 2 ( h O2 ) ( ) h N2

27 AFT Example evaluate the products: ( ( ) ( )) H CO2 h formco2 + h CO2 T P h CO2 T ref ( ( ) ( )) H H2O 2h formh2o + h H2O T P h H2O T ref ( ( ) ( )) H N h formn2 + h N2 T P h N2 T ref None of these enthalpy terms can be fully evaluated since T p is unknown

28 Keeping it Real! Efficiency! How far does AFT fall from actual?! Factors influencing! Dissociation! Chamber not really adiabatic

29 Conclusion! Premixed and Diffusion! Laminar and Turbulent! Finding Qin! Balancing Chemical Equation! Energy Balance Equation! Finding Adiabatic Flame Temperature! Gives Limit of Product Temperature! Dissociation, other factors decrease temperature