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COMBUSTION THERMODYNAMICS

COMBUSTION THERMODYNAMICS. Introduction: A fuel is a substance which releases energy in the form of heat while undergoing combustion. The fuels are classified as chemical fuels and nuclear fuels.

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COMBUSTION THERMODYNAMICS

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  1. COMBUSTION THERMODYNAMICS Introduction: A fuel is a substance which releases energy in the form of heat while undergoing combustion. The fuels are classified as chemical fuels and nuclear fuels. A chemical fuel is a substance which releases energy in the form of heat while undergoing combustion. The main combustible elements in any chemical fuel are carbon and hydrogen. Some fuel contains small quantities of sulphur which is also combustible. However presence of sulphur in fuel is considered undesirable. Fuels are further classified as below;

  2. Combustion or burning is the chemical process in which the inflammable matter in a substance combines with oxygen at a temperature above the spontaneous ignition temperature of that substance and results in the evolution of heat and light. The combustion process involves the oxidation of constituents in the fuel, that are capable of being oxidized and it can be represented by a chemical equation. These equations indicate the required amount of oxygen combined with required amount of fuel. In a chemical reaction the terms, reactants and the products are frequently used. Reactants comprise of initial constituents which start the reaction while products comprise of final constituents which are formed by the chemical reaction.

  3. The smallest particle which can take part in a chemical change is called an atom. It is rare to find elements to exist naturally as single atom. Some elements have atoms which exist in pairs, each pair forming a molecule, and the atoms of each molecule are held together by stronger inter-atomic forces. The symbols and molecular weight of some important elements, compounds and gases are given in table below:

  4. Generally combustion requires reaction with an oxidant or supporter of combustion. Oxygen is the main oxidant, but the halogens (chlorine and fluorine), hydrogen peroxide and also nitric acid, may act as oxidants as in rocket propulsion. Air is the commonest oxidant because it is cheap and readily available. The following analysis of air is used in combustion calculations:

  5. Composition of fuels: An accurate chemical analysis by mass of the different elements in the fuel is called the ultimate analysis. The elements usually included being carbon, hydrogen, nitrogen, oxygen and sulphur. The ultimate analysis is normally used for solid and liquid fuels and is also known as gravimetric analysis. Another analysis of solid fuels like coal is called proximate analysis, gives the percentage by mass of moisture, volatile matter, combustible solid (fixed carbon) and ash. Gaseous fuels are usually mixtures of several gases like hydrogen, methane, oxygen, carbon dioxide, carbon monoxide, nitrogen and other hydrocarbons. The composition of gaseous fuels is normally given by percent volume known as volumetric analysis.

  6. Combustion equations: In a combustion chamber proportionate masses of air and fuel enter where the chemical reaction takes place, and then the combustion products pass to the exhaust. By the conservation of mass the mass flow remains constant (i.e. total mass of products = total mass of reactants), but the reactants are chemically different from the products, and the products leave at a higher temperature. The total number of atoms of each element concerned in the combustion remains constant, but the atoms are rearranged into groups having different chemical properties. This information is expressed in the chemical equation which shows (i) the reactants and the products of combustion, (ii) the relative quantities of the reactants and products. The two sides of the equation must be consistent, each having the same number of atoms of each element involved. Some important combustion equations are given below:

  7. 1. Combustion of hydrogen: 2H2 + O2 → 2H2O This equation tells us that: 2 volumes of Hydrogen + 1 volume of Oxygen → 2 volumes of H2O This shows that there is volumetric contraction on combustion. The H2O may be liquid or a vapor depending on whether the product has been cooled sufficiently to cause condensation. The proportions by mass are obtained by using atomic weights as follows: 2H2 + O2 → 2H2O 2(2 X 1) + 2 X 16 → 2(2X1 + 16) 4kg of H2 + 32kg of O2 → 36kg of H2O 1kg of H2 + 8kg of O2 → 9kg of H2O

  8. 2. Combustion of carbon: (i) Complete combustion of carbon to carbon-dioxide C + O2 → CO2 1 volume Carbon + 1 volume Oxygen → 1 volume CO2 And 1kg C + 8/3 kg O2 → 11/3 kg CO2 (ii) The incomplete combustion of carbon. The incomplete combustion of carbon occurs when there is an insufficient supply of oxygen to burn the carbon completely to carbon-dioxide. 2C + O2 → 2CO 1kg C + 4/3 kg O2 → 7/3 kg CO If a further supply of oxygen is available then the combustion can continue to completion. 2CO + O2 → 2CO2 1kg CO + 4/7 kg O2 → 11/7 kg CO2

  9. 3. Combustion of sulphur to sulphur-dioxide: S + O2 → SO2 1kg S + 1kg O2 → 2kg SO2 4. Combustion of methane: CH4 + 2O2 → CO2 + 2H2O 1kg CH4 + 4kg O2 → 11/4 kg CO2 + 9/4 kg H2O

  10. Theoretical Air and Excess Air: The minimum amount of air that supplies sufficient oxygen for the complete combustion of all the carbon, hydrogen, and any other elements in the fuel that may oxidize is called the theoretical air. When complete combustion is achieved with theoretical air, the products contain no oxygen. Complete combustion is not attained in practice unless more than the amount of theoretically required air is used. The need of excess air is due to the difficulty of obtaining intimate contact between the air and the fuel. It is also partly due to the need to complete the combustion within the combustion space. The excess air is the amount of air supplied over and above theoretical air.

  11. The complete combustion of methane with minimum amount of theoretical air and 150% theoretical air respectively is written as:   Mass of excess air supplied may be determined by the mass of unused oxygen found in the flue gases. We know that in order to supply one kg of oxygen, we need 100/23 kg of air. Therefore, mass of air supplied = (100/23) X mass of excess oxygen.

  12. Stoichiometric Air-Fuel (A/F) Ratio: Stoichiometric (or chemically correct) mixture of air and fuel is one that contains just sufficient oxygen for complete combustion of fuel. A weak mixture is one which has excess air. A rich mixture is one which has a deficiency of air. The percentage of excess air is given as under: An alternate method of expressing the same is in terms of mixture strength;

  13. For a stoichiometric mixture, the mixture strength is 100%. A weak mixture is one whose mixture strength is less than 100%, while a rich mixture has mixture strength of more than 100%.

  14. Theoretical or stoichiometric or minimum air required for complete combustion: We know that fuel mainly consists of constituents like carbon, hydrogen and sulphur and we also know that for combustion; Carbon requires 2.67 times its own mass of oxygen Hydrogen requires 8 times its own mass of oxygen Sulphur require its own mass of oxygen Now consider 1kg of fuel whose ultimate analysis gives: Mass of carbon = C kg Mass of hydrogen = H2 kg Mass of sulphur = S kg Then oxygen required to burn C kg of carbon = 2.67C kg H2 kg of hydrogen = 8 H2 kg S kg of sulphur = S kg

  15. Therefore total oxygen required for complete combustion of 1 kg of fuel = (2.67C + 8H2 + S) kg As fuel already contains O2 kg of oxygen which we assume can be used for combustion. Then total oxygen required for complete combustion of 1 kg of fuel is, = (2.67C + 8H2 + S – O2) kg Since air contains 23% of oxygen on mass basis, i.e. 1 kg of oxygen is associated with 100/23 = 4.35kg of air. Therefore, Minimum or theoretical air required for complete combustion of 1kg of fuel, = 100/23 (2.67C + 8H2 + S – O2) kg

  16. Theoretical or stoichiometric or minimum air required for complete combustion: Consider 1m3 of gaseous fuel whose ultimate volume analysis is: Volume of carbon monoxide = CO m3 Volume of hydrogen = H2 m3 Volume of methane = CH4 m3 Volume of ethylene = C2H4 m3 Then oxygen required to burn CO m3 of carbon = 0.5 CO m3 H2 m3 of hydrogen = 0.5 H2 m3 CH4 m3 of sulphur = 2CH4 m3 C2H4 m3 of hydrogen = 3C2H4 m3 Therefore total oxygen required for complete combustion of 1 m3 of fuel = (0.5CO + 0.5H2 + 2CH4 + 3C2H4) m3

  17. As fuel already contains O2 m3 of oxygen which we assume can be used for combustion. Then total oxygen required for complete combustion of 1 kg of fuel is, = (0.5CO + 0.5H2 + 2CH4 + 3C2H4 – O2) m3 Since air contains 21% of oxygen on volume basis, i.e. 1 m3 of oxygen is associated with 100/21 = 3.76m3 of air. Therefore, Minimum or theoretical air required for complete combustion of 1kg of fuel, = 100/21 (0.5CO + 0.5H2 + 2CH4 + 3C2H4 – O2) kg

  18. Air-fuel Ratio from analysis of products of combustion: By making analysis of the combustion products, the air-fuel ratio can be calculated by the following methods. 1. When fuel composition is known i)Carbon Balance Method- Quite accurate when combustion takes place with excess air and when no free (solid) carbon is present in the products. ii) Hydrogen Balance Method- Suitable when solid carbon is suspected to be present in the products. iii)Carbon-Hydrogen Balance Method- Used when there is some uncertainty about percentage of N2 present. 2. When fuel composition is unknown Only Carbon-Hydrogen Balance Method can be employed.

  19. Analysis of exhaust (flue) gas: The combustion products are mainly gaseous. When a sample is taken for analysis it is usually cooled down to a temperature which is below the saturation temperature of the steam present. The steam content is therefore not included in the analysis, which is quoted as the analysis of the dry products. Since the products are gaseous, it is usual to quote the analysis by volume. An analysis which includes the steam in the exhaust is called a wet analysis. Practical analysis of combustion products: The most common means of analysis of the combustion products is the Orsat analysis using Orsat apparatus. It gives an analysis of dry products of combustion.

  20. A burette- used to measure the volume of the exhaust gas sample inside the apparatus at any time. • A gas cleaner- to filter out solid particles in the exhaust gas sample. • A leveling bottle partly filled with water and connected to the bottom of the burette by tubing. • Three absorption pipettes- each containing different absorbent chemical solutions used to absorb different constituents of the gas sample. The pipettes are interconnected by means of a manifold fitted with cocks with S1, S2, S3 and contain different chemicals to absorb CO2, CO and O2. Each pipette is also fitted with a number of small glass tubes which provide a greater amount of surface. These tubes are wetted by absorbing agents and are exposed to the gas under analysis.

  21. The pipette 1, 2, 3 contain the following chemicals: Pipette 1: caustic soda (KOH) to absorb CO2 Pipette 2: alkaline solution of pyrogallic acid to absorb O2 Pipette 3: acid solution of cuprous chloride to absorb CO Further-more the apparatus has a leveling bottle and a three way cock to connect the apparatus either to gases or atmosphere. The procedure is as follows: 100cc of gas whose analysis is to be made is drawn into the bottle by lowering the leveling bottle. The stop cock S1 is then opened and the whole flue gas is forced to pipette 1. The gas remains in this pipette for some time and most of the CO2 is absorbed. The leveling bottle is then lowered to allow the chemical to come its original level.

  22. The volume of gas thus absorbed is read on the scale of the measuring bottle. The flue gas is then forced through pipette 1 for a number of times to ensure that the whole of the CO2 is absorbed. Further the remaining flue gas is then forced to pipette 2 which absorb whole of O2. The reading on the measuring bottle will be the sum of volume of CO2 and O2. The oxygen content then is found by subtraction. Finally, as before the sample of gas forced through the pipette 3 to absorb CO completely. The amount of N2 in the sample can be determined by subtracting from total volume of the gas the sum of CO2,CO and O2 contents.

  23. Enthalpy of formation (ΔHf): A combustion reaction is a particular kind of reaction in which the products are formed from reactants with the release or absorption of energy as heat is transferred to and from the surroundings. In some substances like hydrocarbon fuels the heat of reaction or combustion may be calculated on the basis of known values of the enthalpy of formation, ΔHfof the constituent of the reactants and products at the temperature T0 (reference temperature). The enthalpy of formation(ΔHf) is the increase in enthalpy when a compound is formed from its constituent elements in their natural form and in a standard state. The standard state is 250C, and 1 atmospheric pressure. Therefore the enthalpy of combustion at T0 or the constant pressure heat of combustion at T0, may be given as:

  24. Enthalpy of reaction: The enthalpy of reaction hR, is defined as the difference between the enthalpy of products at a specified state and the enthalpy of the reactants at the same state for a complete reaction. For combustion processes, the enthalpy of reaction is usually referred as the enthalpy of combustion hC, which represents the amount of heat released during a steady flow combustion process when 1 k-mol (or 1 kg) of fuel is burned completely at a specified temperature and pressure (Fig. below).

  25. It is expressed as hR = hC = Hprod - Hreact Which is -393.520 kJ/k-mol for carbon at the standard reference state. The enthalpy of combustion of a particular fuel is different at different temperatures and pressures

  26. Internal energy of reaction: When the combustion process is carried out at constant volume, the non-flow energy equation, Q = (U2 – U1) + W, can be applied to giveQ = (UPo – URo) Where, W = 0 for constant volume combustion U1 = URo U2 = UPo The internal energy change is independent of the path between the two states and depends only on the initial and final values and is given by the quantity Q. The heat transferred during this constant volume process is called the internal energy of combustion at T0 (or constant volume heat of combustion), and is denoted by . Thus is a negative quantity since the internal energy of the reactants includes the potential chemical energy and heat is transferred from the system.

  27. Combustion efficiency: In evaluating the performance of an actual combustion process a number of different parameters can be defined depending on the nature of the process and the system considered. The combustion efficiency in a gas turbine can be defined as Where, (F/A) = Fuel-Air ratio for adiabatic and complete combustion and in which the products would attain the adiabatic flame temperature.

  28. In case of steam generator or boiler In case of an internal combustion engine

  29. Adiabatic flame temperature: In a given combustion process that takes place adiabatically and with no work or changes in kinetic or potential energy involved, the temperature of the products is referred to as adiabatic flame temperature. With the assumption of no work and no changes in K.E or P.E, this is the maximum temperature that can be achieved for the given reactants because any heat transfer from the reacting substances and any incomplete would tend to lower the temperature of the products.

  30. The following points to be noted: • The maximum temperature achieved through adiabatic complete combustion varies with the type of reaction and percent of theoretical air supplied. An increase in the Air-Fuel ratio will affect a decrease in the maximum temperature. • For a given fuel and given pressure and temperature of the reactants, the maximum adiabatic flame temperature that can be achieved is with a stoichiometric mixture. • The adiabatic flame temperature can be controlled by the amount of excess air that is used.

  31. Heating values of the fuel: In case of a constant pressure process (steady flow process), the negative of the enthalpy of combustion is called heating value at constant pressure. This represents the heat evolved during the combustion at constant pressure. Similarly, the negative of the internal energy of combustion is called the heating value at constant volume and it represents the heat released during the constant volume combustion process.

  32. In general, the heating value also called calorific value of a fuel is the quantity of heat energy released by the combustion of the fuel. For solid and liquid fuels it is expressed in kJ/kg of fuel, while for gaseous fuels the unit used is kJ/m3 of fuel at standard temperature and pressure. Heating values of fuel can be experimentally determined using calorimeters. E.g., Bomb calorimeter, Boyce calorimeter and Junkers gas calorimeter.

  33. The Higher Heating Value (HHV) or Higher Calorific Value (HCV) is the heat released when H2O in the products of combustion is in the liquid state. The Lower Heating Value (LHV) or Lower Calorific Value (LCV) is the heat released in the reaction when H2O in the products is in the vapor state. Therefore, LCV = HCV – mH2O.hfg Where, mH2O is the mass of water formed during the combustion hfg is the latent heat of vaporization.

  34. Example 1 Calculate the minimum volume of air required to burn one Kg of coal having the following composition by weight C = 72.4%, H2 5.3%, N2 = 1.81, O2 = 8.5%, moisture 7.2% S = 0.9% and ash 3.9% On weight basis: Taking 1kg coal as basis weight of oxygen required to burn 1kg of coal  C + O2 → CO2 0.724 x 32/12 = 1.93 kg 0.53x 16/2 = 0.424 kg 0.009x32/32 = 0.009 kg Total O2 = 2.363 kg per kg of coal

  35. But 0.085kg O2 is available in coal, therefore O2 required = 2.363 – 0.085 = 2.278kg per Kg of coal. Air contains 23% of oxygen by weight. Therefore the weight of the air supplied is   2.278x 100/23 = 9.9 kg per kg of coal Density of air required at NTP P v = mRT P = m/v RT = ρRT,  ρ = Molecular weight Volume = P/RT = 1.013x105/287x273 = 1.29 kg/m3 Therefore volume of air required = 9.9(kg)/1.29(kg) = 7.67 m3

  36. On mole basis Consider 100kg of coal  C = 72.4/12= 6.03K mol, O2 = 8.5/32 = 0.265K mol H2 = 5.3/2 = 2.65K mol, H2O = 7.2/18 = 0.4K mol  N2 = 1.8/28 = 0.064K mol, S = 0.9/32 = 0.028K mol  1K mol C + 1K mol O2 → 1Km CO2  Therefore 6.03 K mol of carbon requires 6.03 K mol of oxygen  1 K mol H2 + ½ K mol O2→ 1K mol H2O H2 - 2.65 x ½ = 1.325K mol,S -0.028x1 = 0.028 Total O2 required 6.03 + 1.325 + 0.028 = 7.383  The oxygen present in coal 0.265K mol  Net O2 required = 7.383 – 0.265 = 7.118K mol  Air required  7.118x100/21= 33.89K mol / 100kg of coal = 0.3389K mol / 1kg coal  Volume of air supplied 0.3389K mol/kg X 22.4m3= 7.59m3/kg of coal

  37. Example 2 Calculate the volumetric analysis of the flue gases when coal burns with 20% excess air from the previous calculation the actual air required 33.89K mol/100kg coal. Therefore the actual air is 33.89 x 120/100=40.67K mol/ 100 kg coal The amount of N2 associated with this  40.67 x 79/100 = 32.13K mol  The amount of O2 present 40.67 x 21/100 = 8.54K mol  The actual amount of O2 required was 7.118K mol excess O2 will appear in exhaust gas = 8.54 – 7.118 = 1.422K mol.

  38. Therefore: CO2 = 6.03K mol SO2 = 0.028K mol N2 = 32.13K mol (air) + 0.064 (fuel) = 32.194K mol O2 = 1.422K mol os excess oxygen. Therefore the Total volume = (6.03 + 0.028 + 32.194 + 1.422) = 39.674K mol   The volumetric composition of the gas CO2= (6.03/39.674) x 100 = 15.12% SO2= (0.028/39.674) x 100 = 0.07% N2 = (32.13/39.674) x 100 = 81.15% O2= (1.422/39.674) x 100 = 3.58%

  39. 3.The composition of dry flue gases obtained by burning a liquid fuel containing only hydrogen and carbon is CO2 10.7%, O2 5.1%, N2 84.2%. Calculate the composition of fuel by weight and excess air used. Solution: consider 100K mol of dry flue gases. They will contain 10.7K mol of O2 (from CO2) + 5.1K mole of (as max. oxygen) = 15.8K mol Using nitrogen balance the actual air used 84.2 x 100/79 = 106.58K mol of dry flue gases and oxygen in the air supplied 106.58 x 21/100 = 22.38K mol. Therefore the amount of O2 present in the water produced by the combustion of H2 is 22.38 – 15.8 = 6.58K mol O2.

  40. We know that 1 K mole of H2 combines with ½ K mol O2 to produce water. Therefore the amount of hydrogen present is 6.58x2 = 13.16K mol/100K mol of dry flue gases, and the carbon present is 12X10.7 = 128.4kg/100K mol of dry flue gas. Therefore the composition of fuel (by weight) is 128.4kgC and 26.32Kg H2 on the %age basis. C = (128.4/(128.4+26.32) x 100 = 82.99% H = (26.32/(128.4+26.32) x 100 = 17.01% Excess air supplied The amount of O2 required to burn 10.7K mol C is 10.7K mol and to burn 13.16K mol H2 is 13.16 X ½ = 6.58 Total O2 required = 10.7 + 6.58 = 17.28K mol/100K mol of dry flue gases %age of excess air = (22.38 – 17.28)/(17.28) x 100 = 29.5%

  41. Example 4 A blast furnace gas has the following volumetric analysis H2 CO-24%, CH4 – 2%, CO2-6%, O2-3% and N2-56%  Determine the Ultimate gravimetric analysis Given volumetric analysis, H2 – 9%, CO-24%, CH4 – 2%, CO2-6%, O2-3% and N2-56%  Solution: The volumetric analysis may be converted into mass or granite metric analysis by completing the table as follows:

  42. The volumetric analysis of flue gas components becomes CO-0.36, CH4 – 0.0173, CO2- 0.142, O2-0.0519 and N2-0.42

  43. 5.Determine the fuel gas analysis and air fuel ratio by weight when fuel oil with 84.9% carbon, 11.4% hydrogen, 3.2% sulphur, 0.4% oxygen and 0.1% ash by weight is burnt with 20% excess air, assume complete combustion.  Solution: Consider 1kg of fuel  Oxygen required / Kg of fuel  For burning of 1kg C - 0.849 x32/12 For burning of 1kg H - 0.114 x16/2 For burning of 1kg S - 0.032 x32/32 Total O2 required is 3.208 kg. Amount of O2 contained in the fuel = 0.004Kg Net O2 supplied / kg of fuel = 3.208 – 0.004 = 3.204 kg O2 Net air supplied = 3.204x100/23 = 13.93 kg/kg of fuel

  44. When 20% excess air supplied  Total air supplied = 13.93 x 1.2 = 16.716 kg/kg of fuel. N2 actually supplied = 16.716 x 77/100 = 12.871 kg/kg of fuel O2 actually supplied = 16.716 x 23/100 = 3.845 kg/kg of fuel Total free O2 in fuel gas = 3.845 – 6.204 = 0.641 kg/kg of fuel Total free N2 in fuel gas = 12.87 kg/kg of fuel  Flue gas analysis: C converted to CO2 = 0.849x44/12 = 3.113 kg CO2 H converted to H2O = 0.114x18/2 = 1.026 kg H2O S converted to SO2= 0.032x64/32 = 0.064 kg SO2

  45. Flue gas / kg of fuel: = 3.113 + 1.26 + 0.064 + 0.641 + 12.871 CO2 H2O SO2 O2 N2 = 17.715kg. Therefore: CO2 = (3.113/17.715)x100 = 17.573%  SO2 = (0.064/17.715)x100 = 0.36% O2 = (0.641/17.715)x100 = 3.618%  H2O = (1.026/17.715)x100 = 5.79%  N2 = (12.871/17.715)x100 = 72.656% Air fuel mixture ratio is = 16.716 : 1

  46. 6.A blast furnace gas has the following volumetric analysis.  H2 = 9%, CO = 24%, CH4 = 2%, CO2 = 6%, O2 = 3% and N2 = 56 %  Determine the ultimate gravimetric analysis.  Solution:  Total H2 in the blast furnace gas. % volumetric analysis = 9H2 + 2H4 Proportional mass = % volumetric analysis X mol. Mass of element = (9x2) + (2x4) = 18 + 8 = 26 kg.  Total ‘C’ in the blast furnace gas. % of volumetric analysis = 24C + 2C + 6C Proportional mass = (24+2+6) x 12 = 384 kg

  47. Total O2 in the blast furnace gas  % of volumetric analysis = 24xO + 6O2 + 3O2 Proportional mass = (24+16) x 9 (32) = 672 kg Total N2 in the blast furnace gas  % of volumetric analysis = 56 N2 Proportional mass of N2 = 56 x 28 = 1568 Kg.  Total weight of blast furnace gas: = 384kg C + 26kg H2 + 672kg O2 + 1568kg N= 2650kgs  Gravimetric %age compositon:  C = (384/2650)x100 = 14.49%   H2 = (26/2650)x100 = 0.98% O2= (672/2650)x100 = 25.36%  N2 = (1568/2650)x100 = 59.17%

  48. 7.The analysis of coal used in a boiler trial is as follows. 82% carbon, 6% hydrogen, 4% oxygen, 2% moisture and 8% ash. Determine the theoretical air required for complete combustion of 1kg of coal. If the actual air supplied is 18kg per kg of coal the hydrogen is completely burned & 80% carbon burned to CO2 ,the reminder is CO, Determine the volumetric analysis of the dry products of combustion.

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