2Combustion products to the turbine Combustion ProcessesWhy do mechanical engineers have to know about combustion? Consider a combustion chamber in a gas turbine cycle,Our current modelWhat really happensFuel inputAir from compressorAir to turbineAir from compressorCombustion products to the turbineHow is this analyzed??
3Combustion Fuels Combustion Reaction Stored chemical energyCombustion ReactionTransforms the chemical energy stored in the fuel to thermal energy (heat)Goals of this section of the courseUnderstand combustion chemistryUse combustion chemistry to determine the heat released during a combustion processHeat of reaction
4Combustion The combustion of a fuel requires oxygen, Fuel In the most general sense, a fossil fuel makeup is,The Greek letters signify the atomic composition of the fuel.For example ...
5CombustionOxidantThe oxidant must contain oxygen. The most abundant ‘free’ source is atmospheric air. By molar percent, atmospheric air is considered to be ...For every mole of oxygen involved in a combustion reaction, there are 79/21 = 3.76 moles of nitrogen.
6Combustion Products (for fuels with no sulfur content) Complete Combustion: CO2, H2O, and N2Incomplete Combustion: CO2, H2O, N2, CO, NOxCombustion with Excess Oxygen: CO2, H2O, N2, and O2NOTE: Fuels containing sulfur have the potential of introducing sulfuric acid into the product stream.
7Combustion Terminology Theoretical or Stoichiometric AirThe amount of air required for complete combustion of the fuelDetermined by balancing the combustion reactionExcess or Percent Theoretical AirThe amount of air actually used in the combustion process relative to the stoichiometric valueCan cause incomplete combustion or excess oxygen
8Combustion Terminology In many combustion processes, one of the parameters we are interested in is how much air (or oxygen) is required per unit quantity (moles or mass) of fuel.Air-Fuel and Fuel-Air RatiosEquivalence Ratio
9Equivalence Ratio and Products Stoichiometric (F=1)CO2, H2O, N2Lean (F < 1 with T < 1800 R)CO2, H2O, N2, O2Rich (F > 1 with T < 1800 R)CO2, H2O, N2, O2, CO, H2Rich (F > 1 with T > 1800 R)CO2, H2O, N2, O2, CO, H2, H, O, OH, N, C(s), NO2, CH4ME 322Advanced courses ME 422 & 433
10Stoichiometric (Complete) Combustion Stoichiometric Combustion of a General Fuel in AirAtomic Balances5 equations5 unknowns(n0 through n4)
11Lean Combustion Lean Combustion of a General Fuel in Excess Air PTA = Percent Theoretical Airexpressed as a decimal6 equations6 unknowns(x0 through x5)Requires a stoichiometricbalance first (to get n0)
12Example – Octane Combustion Given: Gasoline (modeled as octane - C8H18) burns completely in 150% theoretical air (or 50% excess air).Find:the A/F ratios (mass and molar)the equivalence ratiothe dew point of the products of combustion at assuming that the products are at 1 atm
13Example – Octane Combustion In order to calculate the air-fuel ratios and the equivalence ratio, we need to know how much air is used in the combustion reaction. This is determined by balancing the combustion reaction.In order to determine the dew point of the products, we need to know the molar composition of the products. This is also determined by balancing the combustion reaction.Everything depends on the correct balance of the combustion reaction!
14Example – Octane Combustion Solution strategy ...1. Balance the stoichiometric reaction to get n02. Balance the reaction with 150% theoretical air3. Calculate the required (A/F) ratios, the equivalence ratio, and the dew point temperature of the products
16Example – Octane Combustion Combustion in 150% theoretical air
17Example – Octane Combustion The molar (A/F) ratio can now be found ...
18Example – Octane Combustion The mass-based (A/F) ratio can be found knowing the molecular masses of the air and the fuel,The molecular mass of the air is,The molecular mass of the fuel is,
19Example – Octane Combustion Now, the mass-based (A/F) can be found ...Once the (A/F) ratios are determined, the equivalence ratio can be found,
20Example – Octane Combustion The dew point of the products is the temperature where the water vapor condenses,Tdp = Tsat at Pw (partial pressure of the water vapor)
21Example – Problem 15.42Given: Combustion exhaust with 9.1% CO2, 8.9% CO,82% N2, and no O2Find:fuel model CnHmmass percent of carbon and hydrogen in fuelmolar air/fuel ratio and percent theoretical air (PTA)dew point temperature at .106 MPa
22Example – Problem 15.42STEP 1: Write balance equation using ORSAT dataCnHm + a(O N2) 9.1 CO CO + bH2O + 82 N2STEP 2: Solve for unknowns and write fuel model CnHmn = ? m = ? a=? b=?
23Example – Problem 15.42STEP 3: Compute molar mass of fuel & mass compositionMfuel = 18lbmolC/lbmolfuel*(12lbm/lbmolC) lbmolH/lbmolfuel*(1lbm/lbmolH) = 249 lbm/lbmolfuelC: 18*(12)/249 87% H: 33*(1)/249 13%
24Example – Problem 15.42 STEP 4: Calculate molar air/fuel ratio 21.8 * ( ) /1 = moles air / mole fuelSTEP 5: Write equation for stoichiometric combustionC18H (O N2) 18 CO H2O N2STEP 6: Find theoretical air %TA = (21.8 / 26.75) * 100 = 83%
25Example – Problem 15.42 STEP 7: Find dew point of combustion products # moles of H2O (in original equation) = 16.5# moles of other combustion products = 100 # moles of all products (in original equation) = 116.5Pw = .106 * (16.5/116.5) = .015 Mpa Tsat (.015 MPa) = 54 C