ME 525: Combustion Lecture 3 - PowerPoint PPT Presentation

slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
ME 525: Combustion Lecture 3 PowerPoint Presentation
Download Presentation
ME 525: Combustion Lecture 3

play fullscreen
1 / 20
ME 525: Combustion Lecture 3
224 Views
Download Presentation
ursa
Download Presentation

ME 525: Combustion Lecture 3

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. ME 525: CombustionLecture 3 • Reactant(s) and Product(s) Mixtures • Introduction to the concept of enthalpy of formation. • Enthalpy and Heating Values • Calculation of the temperature of adiabatic combustion.

  2. Reaction Stoichiometry: Hydrocarbon (Fuel) Combustion with Air (Oxidizer) With the simplest of assumptions, the products of stoichiometriccombustion of a hydrocarbon fuel are CO2 and H2O. For simplest reaction thermochemistry calculations, it is assumedthat the diatomic nitrogen (N2) does not react. For elemental carbon and elemental hydrogen combustion:

  3. Stoichiometric Air: Hydrocarbon (Fuel) Combustion with Air (Oxidizer) • The amount of stoichiometric O2 that reacts with the hydrocarbon fuel CxHy is defined as the amount needed as per the simplest chemical equation that assumes that all the C in the fuel shows up as CO2 in the products and that all the H in the fuel shows up as H2O in the products. • Stoichiometric air is the amount of air that carries the stoichiometric O2. • Practical combustors/reactors are designed with fuel and air systems that can result in fuel leanor fuel richcombustion depending on the application. Equivalence Ratio, Fuel Air Ratio, Air Fuel Ratio, % Stoic Air, % Excess Air relate mass flows of fuel and air

  4. Examples of stoichiometric combustion F = 1: More Examples of stoichiometric combustion F = 1:

  5. Examples of Lean Combustion F < 1 Lean Combustion: a > astoic; Major reaction products are CO2 and H2O. Excess O2 is present in the air and it does not all react, so the products of lean combustion will contain O2. (There can be many minor products such as CO, H2, NO, CuHv including CxHy) Atom Balances: Example:

  6. Examples of Rich Combustion F < 1 Rich Combustion (a < astoic) For moderately rich combustion, major reaction products are CO, H2, CO2 and H2O. Atom balances alone are not sufficient to determine the composition of the products. Atom Balances: One more equation, for example the chemical equilibrium of the water-gas shift reaction, is needed to solve the problem: Kp is a function of temperature alone. Temperature depends on conservation of energy and external prescription of boundary/initial conditions.

  7. Reaction Thermo-chemistry Consider complete combustion of methane and 100% theoretical air at constant P and initially at standard conditions (T1 = Tref = 25°C = 298.15 K, P1 = Pref = 1.0 atm = 0.101325 MPa): What is the adiabatic flame temperature? Consider a steady state flow process where the reactants enter at the standard state and the products leave at the same pressure P1 = P2, but at an elevated temperature T2 = Tadiabatic. http://www.teachersdomain.org/resource/phy03.sci.phys.matter.fireworkcol/

  8. Constant Pressure, Steady Flow, Adiabatic Reactor -QCV= 0 T2 = ? p2 = 1 atm T1 = 25°C p1 = 1 atm CO2 + 2H2O + 7.52 N2 CH4 + 2O2 + 7.52 N2 1 2 Neglecting the kinetic and potential energy of the gases, we find from the open system form of the First Law:

  9. Reaction Thermochemistry For chemical reactions we need to write this equations in terms of the molar flow rate of each species using the index i, Dividing through by the molar flow rate of the fuel we obtain: Now specializing to the case of adiabatic, stoichiometric methane combustion depicted above we obtain:

  10. Reaction Thermochemistry We substitute for the ratios of molar flow rates to obtain Restricting our analysis to ideal gases we can drop the pressure dependence of the enthalpy, At first glance there seems to be a major problem with this equation. The products will be at temperature of over 2000 K so it would seem that their enthalpy must be much higher than the enthalpy of the reactants. What we do not account for in our study of thermodynamics with pure working substances is the tremendous energy tied up in chemical bonds. The enthalpy of formation allows us to account for the energy stored in chemical bonds.

  11. The Enthalpy of Formation For our discussions of chemical reactions and equilibrium, the enthalpy of a pure substance at the standard state is designated as the enthalpy of formation: and is usually listed on a molar basis. The enthalpy including formation and sensible enthalpy of a pure substance at a temperature T and pressure p is given by or for ideal gases

  12. The Enthalpy of Formation An element in its standard state (p = 1 atm or 0.1 MPa, T = 298 K) is defined to have an enthalpy of formation of zero. S. R. Turns, An Introduction to Combustion

  13. The Enthalpy of Formation For substances like H2O, the enthalpy of formation can be envisioned in terms of the following diagram: 2 1

  14. The Enthalpy of Formation The heat that must be removed from the control volume to bring the products to a final temperature of 298 K is the enthalpy of formation. Negative enthalpy of formation implies that the products are more stable (tightly bound) than the reactants. For liquid water Note that for water there may be two heats of formation. Water at p = 1 atm and T = 298 K is a liquid. It is convenient to define a hypothetical ideal gas standard state to obtain,

  15. Constant Pressure, Steady Flow, Adiabatic Reactor Returning to the adiabatic steady flow reactor where methane reacts with the stoichiometric amount of air: -QCV= 0 T2 = ? p2 = 1 atm T1 = 25°C p1 = 1 atm CO2 + 2H2O + 7.52 N2 CH4 + 2O2 + 7.52 N2 1 2 Neglecting the kinetic and potential energy of the gases, we find from the open system form of the First Law:

  16. Adiabatic Flame Temperature As discussed above, assuming ideal gases, the first law reduces to: Substituting using the enthalpy of formation and rearranging we obtain:

  17. Adiabatic Flame Temperature We need to iterate to solve this equation. Let’s take a wild guess that T2 = 2328 K and see if the equation is solved. From Appendices A and B, we find that at 298 K Now determine ZERO for reactants because reactants are at the standard state T1 = Tref. Interpolating in Table A.2, A.6, and A.7 we find for T2 = 2328 K

  18. Adiabatic Flame Temperature

  19. Chemical Equilibrium for Reactions of Ideal Gases In our previous discussion of the steady flow combustion of methane at 1 atm, we calculated an adiabatic flame temperature assuming complete combustion: Why is The two temperatures are different because the reaction does not go to completion. The reaction products of a combustion process is often one of the objectives of combustion calculations. Chemical equilibrium is another logical expectation for the state of the products.

  20. Chemical Equilibrium for Reactions of Ideal Gases In our previous discussion of the steady flow combustion of methane at 1 atm, we calculated an adiabatic flame temperature assuming complete combustion. • Dissociation of CO2 and of H2O are likely reasons for lack of complete combustion and associated lower temperature. • Need additional equations to find “a” and “b”