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Department of Mechanical Engineering ME 322 – Mechanical Engineering Thermodynamics Lecture 36 Combustion Reactions

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Combustion Processes 2 Why do mechanical engineers have to know about combustion? Consider a combustion chamber in a gas turbine cycle, Air from compressor Air to turbine Our current modelWhat really happens Air from compressor Combustion products to the turbine Fuel input How is this analyzed??

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Combustion Fuels –Stored chemical energy Combustion Reaction –Transforms the chemical energy stored in the fuel to thermal energy (heat) Goals of this section of the course –Understand combustion chemistry –Use combustion chemistry to determine the heat released during a combustion process Heat of reaction 3

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Combustion The combustion of a fuel requires oxygen, Fuel In the most general sense, a fossil fuel makeup is, 4 The Greek letters signify the atomic composition of the fuel. For example...

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Combustion Oxidant The 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. 5

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Combustion Products (for fuels with no sulfur content) Complete Combustion: CO 2, H 2 O, and N 2 Incomplete Combustion: CO 2, H 2 O, N 2, CO, NO x Combustion with Excess Oxygen: CO 2, H 2 O, N 2, and O 2 NOTE: Fuels containing sulfur have the potential of introducing sulfuric acid into the product stream. 6

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Combustion Terminology Theoretical or Stoichiometric Air –The amount of air required for complete combustion of the fuel Determined by balancing the combustion reaction Excess or Percent Theoretical Air –The amount of air actually used in the combustion process relative to the stoichiometric value Can cause incomplete combustion or excess oxygen 7

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Combustion Terminology 8 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 Ratios Equivalence Ratio

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Equivalence Ratio and Products Stoichiometric ( =1) –CO 2, H 2 O, N 2 Lean ( < 1 with T < 1800 R) –CO 2, H 2 O, N 2, O 2 Rich ( 1 with T < 1800 R) –CO 2, H 2 O, N 2, O 2, CO, H 2 Rich ( 1 with T > 1800 R) –CO 2, H 2 O, N 2, O 2, CO, H 2, H, O, OH, N, C(s), NO 2, CH 4 9 ME 322 Advanced courses ME 422 & 433

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Stoichiometric (Complete) Combustion 5 equations 5 unknowns ( 0 through 4 ) Atomic Balances Stoichiometric Combustion of a General Fuel in Air 10

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Lean Combustion PTA = Percent Theoretical Air expressed as a decimal 6 equations 6 unknowns ( 0 through 5 ) Requires a stoichiometric balance first (to get 0 ) Lean Combustion of a General Fuel in Excess Air 11

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Example – Octane Combustion Given: Gasoline (modeled as octane - C 8 H 18 ) burns completely in 150% theoretical air (or 50% excess air). Find: (a)the A/F ratios (mass and molar) (b)the equivalence ratio (c)the dew point of the products of combustion at assuming that the products are at 1 atm 12

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Example – Octane Combustion 13 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!

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Example – Octane Combustion 1. Balance the stoichiometric reaction to get 0 2. Balance the reaction with 150% theoretical air 14 Solution strategy... 3. Calculate the required (A/F) ratios, the equivalence ratio, and the dew point temperature of the products

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Example – Octane Combustion Stoichiometric Reaction 15

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Example – Octane Combustion Combustion in 150% theoretical air 16

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Example – Octane Combustion The molar (A/F) ratio can now be found... 17

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Example – Octane Combustion The mass-based (A/F) ratio can be found knowing the molecular masses of the air and the fuel, 18 The molecular mass of the air is, The molecular mass of the fuel is,

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Example – Octane Combustion Now, the mass-based (A/F) can be found... 19 Once the (A/F) ratios are determined, the equivalence ratio can be found,

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Example – Octane Combustion The dew point of the products is the temperature where the water vapor condenses, T dp = T sat at P w (partial pressure of the water vapor) 20

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Example – Problem 15.42 21 Given: Combustion exhaust with 9.1% CO 2, 8.9% CO, 82% N 2, and no O 2 Find: a)fuel model C n H m b)mass percent of carbon and hydrogen in fuel c)molar air/fuel ratio and percent theoretical air (PTA) d)dew point temperature at.106 MPa

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Example – Problem 15.42 22 STEP 1: Write balance equation using ORSAT data C n H m + a(O 2 + 3.76 N 2 ) 9.1 CO 2 + 8.9 CO + bH 2 O + 82 N 2 STEP 2: Solve for unknowns and write fuel model C n H m n = ? m = ? a=? b=?

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Example – Problem 15.42 23 STEP 3: Compute molar mass of fuel & mass composition Mfuel = 18lbmol C /lbmol fuel *(12lbm/lbmol C ) + 33lbmol H /lbmol fuel *(1lbm/lbmol H ) = 249 lbm/lbmol fuel C: 18*(12)/249 87% H: 33*(1)/249 13%

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Example – Problem 15.42 24 STEP 4: Calculate molar air/fuel ratio 21.8 * (1 + 3.76) /1 = 103.8 moles air / mole fuel STEP 5: Write equation for stoichiometric combustion C 18 H 33 + 26.25(O 2 + 3.76 N 2 ) 18 CO 2 + 16.5 H 2 O + 98.7 N2 STEP 6: Find theoretical air %TA = (21.8 / 26.75) * 100 = 83%

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Example – Problem 15.42 25 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) = 116.5 Pw =.106 * (16.5/116.5) =.015 Mpa Tsat (.015 MPa) = 54 C

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