VTU-NPTEL-NMEICT Project

Transcription

1 VTU-NPTEL-NMEICT Project Progress Report The Project on Development of Remaining Three Quadrants to NPTEL Phase-I under grant in aid NMEICT, MHRD, New Delhi SME Name : Course Name: Type of the Course Module Subject Matter Expert Details Dr.A.R.ANWAR KHAN Prof & H.O.D Dept of Mechanical Engineering Applied Thermodynamics web II DEPARTMENT OF MECHANICAL ENGINEERING, GHOUSIA COLLEGE OF ENGINEERING, RAMANARA Page 1 of 29

2 Sl. No. 1. CONTENTS DISCRETION Quadrant a. Animations. b. Videos. c. Illustrations. a. Wikis. b. Open Contents Quadrant Quadrant -4 a. Problems. b. Assignments c. Self Assigned Q & A. d. Test your Skills. Page 2 of 29

3 MODULE-II COMBUSTION THERMODYNAMICS QUADRANT-2 Animations 1) 2) 3) 4) 5) arge_resonance_energy_t_page_11.html 6) 7) Videos 1) 2) 3) sonance_energy_t_page_11.html 4) 5) Page 3 of 29

4 Illustrations Introduction All conventional fossil fuels, whether, solid, liquid or gaseous, contain basically carbon and hydrogen which invariably react with the oxygen in the air forming carbon dioxide, carbon monoxide or water vapour. The heat energy released as a result of combustion can be utilized for heating purposes or for generation of high pressure steam in a boiler or as power from an engine or a gas turbine. The solid fuels are burned in beds or in pulversied from suspended in the air stream. The liquid fuels are burned either by vaporising and mixing with air before ignition, when they behave like a gaseous fuel. The gaseous fuels are either burned in burners when the fuel and air are premixed or the fuel and air flow separately in to a burner or a furnace and simultaneously mix together as combustion proceeds. The Kg-mole or gram-mole is widely used in combustion calculations as a unit of weight. The molecular weight of any substance in kg represents one kilogram mole or 1K mole. 1Kmol of hydrogen has a mass of 2.016Kg and 1Kmol of carbon has a mass of 12Kg. Consider a reaction CH 4 + 2O 2 CO 2 + 2H 2 O ( ) ( ) kg of methane reacts with 64Kg of oxygen to form 44kg of carbon dioxide and kg of water. We can also simply state that 1Kmol of methane reacts with 2Kmol of oxygen to form 1Kmol of carbon dioxide and 2K mol of water, this has advantage of permitting easy conversion between the mass and volumetric quantities for the gaseous fuel and the product of combustion. If the gases are considered ideal then according to Avogadro hypothesis, all gases contain the same number of molecules per unit volume. It implies that 1K mole of any gaseous substance occupies the volume of 22.4m 3 at NTP i.e., 1.013bar and 273K. CH CO 2 + 2H 2 O 1 volume of methane reacts with 2 volume of oxygen to form one volume of CO 2 and two Page 4 of 29

5 volumes of H 2 O. Therefore in any reactions, the mass in confirmed but the no. of mol or volumes may not be considered. (i) C +O 2 CO 2 On mass basis (12) + (32) (44) 1 Mol (C) + 1 Mol (O 2 ) 1 Mol (CO 2 ) (ii) 2C +O 2 2CO On mass basis (24) + (32) (56) Volume basis 2 Mol (C) + 1 Mol (O 2 ) 2Mol (CO) (iii) 2C +O 2 2CO 2 On mass basis (56) + (32) (88) Combustion Stoichometry 2 Mol (CO) + 1 Mol (O 2 ) 2 Mol (CO 2 ) A balanced chemical equation for complete Combustion of the reactions with no excess air in the product is known as a stiochiometric equation. A stiochiometric mixture of the reactants is one in which the molar proportions of the reactants are exactly as given by the stiochiometric coefficients, so that no excess of any constituent is present. In general a chemical reaction may be written as aa + bb - cc + dd Where the reactants A and B react to form the products C and D. The small letters a, b, c and d are known as the stiochiometric coefficients. For the combustion of any fuel the most common oxidizer is air which is a mixture of 21% O 2 and 79% N 2 (on volume basis). One mol of oxygen is accompanied by 79/21 (3.76) mol of Nitrogen. The Chemical equation for the stiochiometric combustion of carbon with air is written as C + O N 2 CO N 2 The minimum amount of air required for the complete combustion of a fuel is known as Page 5 of 29

6 theoretical air. However in practice it is difficult to achieve complete combustion with theoretical air. Therefore fuel requires some excess air for different application and may vary from 5% ~ 20% and in gas turbine it may go up to 400% of theoretical quantity. Theoretical air required for complete combustion. If the fuel composition is known, the requirement of oxygen or air can be calculated either by mass balance or by mole method. Consider a equation C + O 2 CO 2 (12) + (32) (44) Or 1 + 8/3 11/3 Or 1 Kg C + 8/3 Kg O 2 11/3 kg CO 2 Similarly 2 C + O 2 2CO 2(12) + (32) 2(28) Or 1 Kg C + 32/24 Kg O 2 56/24 kg CO Or 1 Kg C + 4/3 Kg O 2 7/3 kg CO Similarly 1 Kg CO + 4/7 Kg O 2 11/7 kg CO 2 On molal basis 1K mol C + 1K mol O 2 1K mol CO 2 1K mol C + ½ K mol O 2 1K mol CO 1K mol C + ½ K mol O 2 1K mol CO 2 Conversion of Gravimetric analysis to volumetric basis and vice versa If the composition of fuel is given on gravimetric (or weight) basis it can be converted to volumetric (or mole) basis as follows. Divide the weight of each constituents of the mixture by its molecular weight. This will give the relative volume (or mole) of each constituents. Add all the relative volumes of the constituents then, Indivisual (relative) volume of the constituents X 100 Total volume of all the constituents Page 6 of 29

7 Will give the %age by volume of each constituents in the fuel. If the volumetric composition of a fuel is given, it can be converted to gravimetric (or weight) basis as follows. Multiply the indivisual volume of each constituent by its molecular weight. This will give relative weight of each constituent. Add all the relative weights of the constituents then Indivisual weight of the constituents X 100 Total(reative)weights of the constituents Will give the %age by weight of each constituent in the fuel. Dew point of products: The product of combustion containing water vapour are known as wet products. The water vapour present in combustion product is cooled down to a point of condensation the vapour turn in to liquid and volume will be reduced. Knowing the partial pressure exerted by the water before condensing, it is possible to find the saturation temp. corresponding to partial pressure from the steam tables. Enthalpy of reaction Enthalpy of a reaction is defined as the difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction. For combustion process, the enthalpy of a reaction is usually referred to as the enthalpy of combustion it is obviously a very useful property for analyzing the combustion processes of fuels. However there are so many different fuels and fuel mixtures that is not practical to list enthalpy of combustion values for all possible cases. Besides, the enthalpy of combustion is not of much use when the combustion is incomplete. Therefore a more practical approach would be have a more fundamentally property to represent the chemical energy of an element or compound at some reference state. This property is the enthalpy of formation which can be viewed as the enthalpy of a substance at a specified state due to its chemical composition. To establish a starting point it is assigned the enthalpy of formation for all stable elements such as O 2, N 2, H 2 and C a value of zero at standard reference state of 25 o C and 1 atm. For all stable compounds. In a chemical reaction bonds are broken in the reactants and new bonds formed in the products. Energy is required to break bonds and energy is released when bonds are formed. The energy associated with a chemical reaction depends on the number and type of bonds broken and/or Page 7 of 29

8 formed. Every chemical species has a certain amount of "heat content," or enthalpy, H, which cannot be measured. However, differences in enthalpy can be measured. The net energy change for a reaction performed at constant pressure is the enthalpy change for the reaction. This enthalpy change, H, has units kj/mol and is defined: [C + H (fuel)] + [O 2 + N 2 (Air)] -> (Combustion Process) -> [CO 2 + H 2 O + N 2 (Heat)] where C = Carbon, H = Hydrogen, O = Oxygen, N = Nitrogen (1) H = H(products) H(reactants) If energy is given off during a reaction, such as in the burning of a fuel, the products have less heat content than the reactants and H will have a negative value; the reaction is said to be exothermic. If energy is consumed during a reaction, H will have a positive value; the reaction is said to be endothermic. The enthalpy change for a chemical change is independent of the method or path by which the change is carried out as long as the initial and final substances are brought to the same temperature. This observation, known as HESS'S LAW, has important practical utility. Thermochemical equations may be treated as algebraic equations: they may be written in the reverse direction with a change in the sign of H even though the reverse reaction may not actually occur; they may be added and subtracted algebraically; the equation and associated Page 8 of 29

9 H value may be multiplied or divided by factors. Hess's Law allows the calculation of enthalpy changes that would be difficult or impossible to determine directly, i.e. by experiment. The enthalpy change for the reaction: (2) 2C (s) + O2 (g) 2CO (g) cannot be determined directly because carbon dioxide will also form. However, H can be measured for: (3) C (s) + O2 (g) CO2 (g) H = kj (4) 2CO (g) + O2 (g) 2CO2 (g) H = kj Multiplying equation (3) by 2 gives equation (5), and reversing equation (4) gives equation (6): (5) 2C (s) + 2O2 (g)2co2 (g) H = kj (6) 2CO2 (g)2co (g) + O2 (g) H = kj Adding equations (5) and (6) gives the desired information: (2) 2C (s) + O2 (g) 2CO (g) H = kj For a reaction in which a compound is formed from the elements, the enthalpy change is called the heat of formation, H f o, for the compound. The superscript "o" indicates standard conditions of one atmosphere pressure. Equation (2) and (3) are such reactions. Some others: (7) S (s) + O2 (g)so2 (g) H = kj (8) Mg (s) + Cl2 (g)mgcl2 (s) H = kj In reactions (2), (3), (7), and (8) H for the reaction is H f o for the compound. For the reaction: Page 9 of 29

10 (9) 2S (s) + 3O2 (g) 2SO3 (g) H = kj The heat of reaction is associated with the formation of two moles of SO3. But heat of formation is per mole of compound, so Hf o for SO3 is half of 790.4, or kj. Extensive listings of heats of formation are available in handbooks. With these values of H o f, you can calculate virtually any heat of reaction. The heat of a reaction is the sum of Hf o values for the products minus the sum of Hf o values for the reactants. Expressed as a formula: o o (10) Hrxn = Hf products- H f reactants Internal Energy of Combustion: It is defined as the difference between the internal energy of the products and the internal energy of the reactants when complete combustion occurs at a given temperature and pressure. U c = U p U R = p n e (h f + h pv ) - R n i (h f + h pv ) Combustion efficiency It is defined as the ratio of ideal fuel-air to the actual fuel-air ratio comb = (F/A) ideal (F/A) actl Page 10 of 29

11 QUADRANT-3 Wikis rreversible_reactions Open Contents: Applied Thermodynamics by R. K. Rajput Applied Thermodynamics for Engineering Technologists by Eastop Applied Thermodynamics by B. K. Venkanna B. V. S Basic and Applied Thermodynamics by Nag Applied Thermodynamics by D. S. Kumar A textbook of applied thermodynamics, steam and thermal... by S. K. Kulshrestha Applied thermodynamics by Anthony Edward John Hayes Page 11 of 29

12 Problems: QUADRANT-4 1) A coal sample gave the following analysis by weight, carbon 85 %, hydrogen 6%, OXYGEN 6%, The Remainder Being Incombustible. Determine minimum weight of air required per kg of coal for chemically correct composition. Solution: Element wt. (kg) O 2 required (kg) C= *8/3=2.27 H 2 = *8=0.48 O 2 = TOTAL O 2 =2.75 Weight of O 2 to be supplied = wt of O 2 needed wt of O 2 already present in fuel Weight of air needed = 2.69*(100/23) =11.70kg = =2.69kg 2) The percentage composition of a sample of liquid fuel by weight is C=84.8%, H 2 =15.2%. Calculate. i) the weight of air needed for the combustion of 1kgof fuel. ii) The volumetric composition of the products of combustion if 155 excess air is supplied. Solution: Element wt. (kg) O 2 required (kg) Dry products (kg) C= *8/3=2.261 (0.848*11)/3 =3.109(CO 2 ) H 2 = *8=1.216 Total O 2 =3.477 Minimum air needed for combustion Excess air supplied Weight of oxygen in excess air =3.477*(100/23) = 15.11kg = (15.11*15)/100 = 2.266kg = 2.266*(23/100) =0.521kg Total air supplied = minimum air + excess air = = kg Weight of N 2 in flue gases = *(77/100) = 13.38kg. Page 12 of 29

13 Volumetric composition of the product of combustion Name of the gas Weight (x) Molecular weight (y) CO 2 O 2 N Proportional volume(z)=x/y z= Percentage volume =z/ z* % 2.89% 84.60% 3) A single cylinder was supplied with a gas having the following % volumetric analysis CO=5, CO 2 =10, H 2 =50, CH4=25, N 2 =10. The percentage volumetric analysis of dry gases was CO 2 =8, O 2 =6 & N 2 =86.determine the air-fuel ratio by volume. Solution: Combustion equations are : 2H 2 + O 2 = 2H 2 O 1 vol + 1/2 vol = 1 vol 2CO + O 2 = 2CO 2 1 vol + 1/2 vol = 1 vol CH 4 + 2O 2 = CO 2 + 2H 2 O 1 vol + 2 vol = 1 vol + 2 vol. Gas Vol (m 3 ) O 2 needed (m 3 ) Products (m 3 ) CO 2 N 2 CO CO H CH N Total Volume of air required = /21 = 3.69 m 3 Page 13 of 29

14 Volume of nitrogen in the air = /100 = 2.92 m 3 Dry combustion products of 1 m 3 of gases (V) contain 0.4 m 3 of CO m 3 of N 2 (as given in the table) m 3 of N 2 (From air supplied for complete combustion) = 3.42 m 3 Excess air supplied = (O 2 *V) / (21-O 2 ) = (6.0*3.42) / (21-6) =20.52/15= 1.37 m 3 Total quantity of air supplied = = 5.06 m 3. Air fuel ratio = Volume of air = 5.06 = 5 Volume of fuel 1 4) A sample of fuel has the following percentage composition : Carbon = 86 per cent ; Hydrogen = 8 per cent ; Sulphur = 3 per cent ; Oxygen = 2 per cent ; Ash = 1 per cent. For an air-fuel ratio of 12 : 1, calculate : (i) Mixture strength as a percentage rich or weak. (ii) Volumetric analysis of the dry products of combustion. Solution. Element, wt. (kg) C = 0.86 H 2 = 0.08 S = 0.03 O 2 = 0.02 Weight of oxygen to be supplied per kg of fuel = = 294 kg. O 2 reqd. (kg) 0.86*(8/3) = *8 = *(1/1) = 0.03 Total O 2 = 2.96 Weight of minimum air required for complete combustion =(294 *100)/23= kg Hence correct fuel-air ratio = (1/1278): 1 But actual ratio is (1/12): 1. (i) Mixture strength = (1278/12) 100 = 106.5% This shows that mixture is 6.5% rich. (Ans.) Deficient amount of air = = 0.78 kg Page 14 of 29

15 Amount of air saved by burning 1 kg of C to CO instead of CO 2 = Oxygen saved 100/23 = * =5.8 kg Hence 0.78/58 = kg of carbon burns to CO and as such = kg of carbon burns to CO 2. CO formed = /3= kg CO 2 formed = /3= kg N 2 supplied = = 9.24 kg SO 2 formed = = 0.06 kg. ii) The percentage composition of dry flue gases is given as below: Dry products Weight (x) Molecular weight (y) CO CO 2 N 2 SO Proportional volume(z)=x/y Σ z = Percentage volume =z/ z* % 15.03% 81.97% 0.22% 5) A fuel (C 10 H 22 ) is burnt using an air-fuel ratio of 13: 1 by weight. Determine the complete volumetric analysis of the products of combustion, assuming that the whole Amount of hydrogen burns to form water vapour and there is neither any free oxygen nor any free carbon. The carbon burns to CO 2 and CO.Air contains 77% of nitrogen and 23% of oxygen by weight. Solution: Combustion equation is 2C 10 H O 2 = 20CO H 2 O = or = Air required for complete combustion= (992 *100)/ (284/ 23) = 15.2 kg/kg of fuel Air actually supplied = 13 kg/kg of fuel Deficiency of air = = 2.2 kg/kg of fuel Also 1 kg of C requires(4/3)*(100/23)= 5.8 kg of less air to burn to CO instead of CO 2. Page 15 of 29

16 Hence 22/58 = kg C is burnt to CO; And = kg of C is burnt to CO 2. Weight of CO 2 formed = /3= kg Weight of CO formed = /3= kg Weight of H 2 O formed = (22/142) 9 = kg Weight of N 2 from air = = kg. Products of combustion CO 2 CO H 2 O N Weight (x) Molecular weight (y) Proportional volume(z)=x/y Σ z = Percentage volume =z/ z* % 6.254% % % 6) The following results were obtained in a trial on a boiler fitted with economizer : CO 2 CO O 2 N 2 Analysis of gas entering the economizer Analysis of gas leaving the economizer (i) Determine the air leakage into the economizer if the carbon content of the fuel is 80 percent. (ii) (ii) Determine the reduction in temperature of the gas due to air leakage if atmospheric temperature is 20 C and flue gas temperature is 410 C. Ash collected from ash pan is 15 per cent by weight of the fuel fired.take: C p for air = kj/kg K and c p for flue gas = 1.05 kj/kg K. Solution: (i) Air supplied = Air supplied on the basis of conditions at entry to the economizer = (80.3 *80)/33(8.3+0) = kg Air applied on the basis of conditions at exit = (80.6*80)/33(7.9+0) = kg Air leakage = = 1.28 kg of air per kg of fuel. (Ans) For each kg of fuel burnt, the ash collected is 15% i.e., 0.15 kg. Weight of fuel passing up the chimney = = 0.85 kg Page 16 of 29

17 Total weight of products = Weight of air supplied per kg of fuel+ Weight of fuel passing through chimney per kg = = 24.3 kg Heat in flue gases per kg of coal = Weight of flue gases Specific heat Temperature rise above 0 C = (410 0) = kj of fuel Heat in leakage air = Weight of leakage air Specific heat Temperature rise of air above 0 C = (20 0) = kj. We can still consider, in the mixture, the gas and the air as separate and having their own specific heats, but sharing a common temperature t. For heat balance : ( ) t = t = t = C Fall in temperature as a result of the air leakage into the economizer = = 18.7 C. 7) The chemical formula for alcohol is C 2 H 6 O. Calculate the stoichiometric air/fuel ratio by mass and the percentage composition of the products of combustion per kg of C 2 H 6 O. Solution:. Chemical equation for complete combustion of given fuel can be written as follows : C 2 H 6 O + 3O 2 2CO 2 + 3H 2O (1 46) + (3 32) = (2 44) + (3 18) For complete combustion of 1 kg of C 2 H 6 O, oxygen required = (3 32)/46= kg of oxygen = (100/23) = kg of air A : F ratio for complete combustion = : 1. Ans. Page 17 of 29

18 Also 46 kg of fuel produces products of combustion = = 142 kg 1kg of fuel produces(142/46)= kg of products of combustion (i.e CO and H 2O) Hence CO 2 produced by fuel = 88/ = or 191.3%. (Ans) H 2 O produced by fuel =54/ = or 117.4%. (Ans) 8) Calculate the amount of theoretical air required for the combustion of 1 kg of acetylene (C 2 H 2 ) to CO 2 and H 2O. Solution. For combustion of acetylene (C 2 H 2 ) the stoichiometric equation is written as C 2 H 2 + x O 2 a CO 2 + b H 2O (i) Balancing the carbon atoms on both sides of the combustion eqn. (i), we get 2C = a C i.e. a = 2 Now balancing hydrogen atoms on both sides, we get 2H = 2bH b = 1 Thus, eqn. (i) becomes C 2 H 2 + x O 2 2 CO H 2O Now, balancing oxygen atoms in the above equation 2x = = 5 i.e., x = 2.5 Hence, the final combustion eqn. (i) is C 2 H O 2 2 CO H 2O...(ii) Thus, for combustion of C 2 H 2 in air, we get C 2 H O (79/21) N 2 2CO 2 + H 2O + 2.5(79/21)N 2 On a mass basis, this becomes ( ) C 2H (2 16) O (79/21) (2 14) N 2 2 ( ) CO 2 + ( ) H 2O + 2.5(79/21)(2 14) N 2 26 kg C 2 H kg O N 2 88 kg CO kg H 2 O kg N 2 or 1 kg C 2 H kg O kg N kg CO kg H 2O kg N 2 i.e., Amount of air = = kg of air per kg of C 2H 2 Hence amount of theoretical air required for combustion of 1 kg acetylene = kg. Page 18 of 29

19 9) One kg of octane (C 8 H 18) is burned with 200% theoretical air. Assuming complete combustion determine : (i) Air-fuel ratio (ii) Dew point of the products at a total pressure 100 kpa. Solution: The equation for the combustion of C 8 H 18 with theoretical air is C 8 H O (79/21) N 2 8CO 2 + 9H 2O (79/21) N 2 For 200% theoretical air the combustion equation would be C 8 H 18 + (2) (12.5) O 2 + (2) (12.5) (79/21) N 2 8CO 2 + 9H 2O + (1) (12.5) O 2 + (2) (12.5) (79/21) N 2 Mass of fuel = (1) ( ) = 114 kg/mole Mass of air = (2) (12.5) (1+(79/21))*28.97 = kg/mole of fuel (i) Air-fuel ratio: Air-fuel ratio, A/F = Mass of air /Mass of fuel=3448.8/114= i.e., A/F = (Ans) (ii) Dew point of the products, tdp: Total number of moles of products = (2) (12.5) (79/21)= moles/mole fuel Mole fraction of H 2O = 9/123.5 = Partial pressure of H 2O = = 7.28 kpa The saturation temperature corresponding to this pressure is 39.7 C which is also the dew- point temperature. Hence tdp = 39.7 C. 10) One kg of ethane (C 2 H 6) is burned with 90% of theoretical air. Assuming complete combustion of hydrogen in the fuel determine the volumetric analysis of the dry products of combustion. Solution: The complete combustion equation for C 2 H 6 is written as: C 2 H O 2 2CO 2 + 3H 2 O The combustion equation for C 2 H 6 for 90% theoretical air is written as: Page 19 of 29

20 C 2 H 6 + (0.9) (3.5) O 2 + (0.9) (3.5) (79/21) N 2 a CO 2 + b CO + 3H 2O + (0.9) (3.5) (79/21) N 2 By balancing carbon atoms on both the sides, we get 2 = a + b...(i) By balancing oxygen atoms on both the sides, we get (0.9) (3.5) (2) = 2a + b (ii) Substituting the value of b (= 2 a) from eqn. (i) in eqn. (ii), we get (0.9) (3.5) (2) = 2a + 2 a = a + 5 a = 1.3 b = 2 a = = 0.7 Thus the combustion equation becomes: C 2 H 6 + (0.9) (3.5) O 2 + (0.9) (3.5) (79/21) N 2 1.3CO CO + 3H 2O + (0.9) (3.5) (79/21) N 2 Total number of moles of dry products of combustion = (0.9) (3.5) (79/21) = = moles/mole of fuel Volumetric analysis of dry products of combustion is as follows : CO 2 = 13/ = 9.38% (Ans) CO = 07/ = 5.05%. (Ans) N 2 =11.85/ = 85.56% (Ans) 11) Methane (CH 4 ) is burned with atmospheric air. The analysis of the products on a dry basis is as follows : CO 2 = 10.00%, O 2 = 2.37%, CO = 0.53%, N 2 = 87.10%. (i) Determine the combustion equation ; (ii) Calculate the air-fuel ratio ; (iii) Percent theoretical air. Solution: (i) Combustion equation : From the analysis of the products, the following equation can be written, keeping in mind that this analysis is on a dry basis. Page 20 of 29

21 x CH 4 + y O 2 + z N CO CO O 2 + a H 2O N 2 To determine all the unknown co-efficients let us find balance for each of the elements. Nitrogen balance : z = 87.1 Since all the nitrogen comes from the air, z/y=79/21; y = 87.1/ (79/21)= Carbon balance : x = = Hydrogen balance : a = 2x = = Oxygen balance. All the unknown co-efficients have been solved for, and in this case the oxygen balance provides a check on the accuracy. Thus, y can also be determined by an oxygen balance y = / (21.06/2) = Substituting these values for x, y, z and a, we have, CH O N CO CO O H 2 O N 2 Dividing both sides by 10.53, we get the combustion equation per mole of fuel, CH O N CO CO + 2H 2O O N 2 (ii) Air-fuel ratio A/F : The air-fuel ratio on a mole basis is = moles air/mole fuel The air-fuel ratio on a mass basis is found by introducing the molecular weights A/F = ( )/ (12+ (1* 4))= kg air/kg fuel. (Ans) The theoretical air-fuel ratio is found by writing the combustion equation for theoretical air, CH 4 + 2O (79/21) N 2 CO 2 + 2H 2 O + (2) (79/21) N 2 A/F theor. = = kg air/kg fuel. (Ans) (iii) Percent theoretical air: Per cent theoretical air = 18.96/ = 110%. 12) The following analysis relate to coal gas : H 2 = 50.4%, CO = 17 %, CH 4 = 20 %, C 4 H 8 = 2 %, O 2 = 0.4 %, N 2 = 6.2 %, CO 2 = 4 %. (i) Calculate the stoichiometric A/F ratio. (ii) Find also the wet and dry analyses of the products of combustion if the actual mixture Is 30 per cent weak? Page 21 of 29

22 Solution: The example is solved by a tabular method ; a specimen calculation is given below : For CH 4 : CH 4 + 2O 2 CO 2 + 2H 2 O i.e., 1 mole CH moles O 2 1 mole CO moles H 2 O There are 0.2 moles of CH 4 per mole of the coal gas, hence 0.2 moles CH moles O moles H 2 O O 2 required for the CH 4 in the coal gas = 0.4 moles per mole of coal gas. The oxygen in the fuel (0.004 moles) is included in column 4 as a negative quantity. Product Moles/mole Combustion equation O 2 Products H 2 O fuel moles/mole fuel CO H 2O CO CH 4 C 4 H 8 O 2 N 2 CO H 2 + O 2 2H 2O 2CO + O 2 2CO 2 CH 4 + 2O 2 CO 2 +2H 2O C 4 H 8 + 6O 2 4CO 2 +4H 2O (i) Stoichiometric A/F ratio : Air required = 0.853/0.21 = 4.06 moles/mole of fuel (where air is assumed to contain 21% O 2 by volume) Stoichiometric A/F ratio = 4.06/1 by volume. (Ans) (ii) Wet and dry analyses of the products of combustion if the actual mixture is 30% weak : Actual A/F ratio with 30% weak mixture = (30/100) 4.06 = = 5.278/1 Associated N 2 = = 4.17 moles/mole fuel Excess oxygen = = moles Total moles of N 2 in products = = moles/mole fuel. Analysis by volume of wet and dry products: Product Moles/mole fuel % by vol. (dry) % by vol. (wet) CO Page 22 of 29

23 H 2O O 2 N Total wet = H 2O = Total dry = Frequently asked Questions. 1) Write a short note on excess air. 2) What do you mean by stoichiometric air-fuel (A/F) ratio? 3) What is enthalpy of formation (ΔHf) =? 4 What is the difference between higher heating value (HHV) and lower heating value (LHV) of the fuel? 5) What is adiabatic flame temperature? 6) The following is the analysis of a supply of coal gas : H2 = 49.4 per cent ; CO = 18 per cent ; CH4 = 20 per cent ; C4H8 = 2 per cent ; O2 = 0.4 per cent ; N2 = 6.2 per cent ; CO2 = 4 per cent. (i) Calculate the stoichiometric A/F ratio. (ii) Find also the wet and dry analyses of the products of combustion if the actual mixture is 20 per cent weak. 7) ) Methane (CH 4 ) is burned with atmospheric air. The analysis of the products on a dry basis is as follows : CO 2 = 10.00%, O 2 = 2.37%, CO = 0.53%, N 2 = 87.10%. (i) Determine the combustion equation ; (ii) Calculate the air-fuel ratio ; (iii) Percent theoretical air. 8) A fuel (C 10 H 22 ) is burnt using an air-fuel ratio of 13: 1 by weight. Determine the complete volumetric analysis of the products of combustion, assuming that the whole Amount of hydrogen burns to form water vapour and there is neither any free oxygen nor any free carbon. The carbon burns to CO 2 and CO.Air contains 77% of nitrogen and 23% of oxygen by weight. Page 23 of 29

24 Assignments: 1) Write a short note on excess air. 2) What do you mean by stoichiometric air-fuel (A/F) ratio? 3) What is enthalpy of formation (ΔHf) =? 4 What is the difference between higher heating value (HHV) and lower heating value (LHV) of the fuel? 5) What is adiabatic flame temperature? 6) During a trial in a boiler, the dry flue gas analysis by volume was obtained as CO2 = 13%, CO = 0.3%,O2 = 6%, N2 = 80.7%. The coal analysis by weight was reported as C = 62.4%, H2 = 4.2%, O2 = 4.5%, moisture= 15% and ash 13.9%. Estimate : (a) Theoretical air required to burn 1 kg of coal. (b) Weight of air actually supplied per kg of coal. (c) The amount of excess air supplied per kg of coal burnt. [Ans. 8.5 kg ; 11.5 kg ; 3 kg] 7) In a boiler trial, the analysis of the coal used is as follows : C = 20%, H2 = 4.5%, O2 = 7.5%, remainderincombustible matter. The dry flue gas has the following composition by volume : CO2 = 8.5%, CO = 1.2%, N2 = 80.3%, O2 = 10%. Determine : (i) Minimum weight of air required per kg of coal. (ii) Percentage excess air. [Ans. (i) 3.56 kg, (ii) 63.2%] 8) Calculate the stoichiometric air-fuel ratio for the combustion of a sample of dry anthracite of the following composition by mass : C = 90 per cent ; H2 = 3 per cent ; N2 = 1 per cent ; Sulphur = 0.5 per cent ; ash = 3 per cent. If 20 per cent excess air is supplied determine : (i) Air-fuel ratio (ii) Wet analysis of the products of combustion by volume. [Ans /1 (i) 13.5/1 ; (ii) CO2 = 16.3%, H2O = 0.03%, SO2 = 3.51%, N2 = 80.3%] 9) The percentage composition of a fuel by weight is as follows : Carbon = 89.3 per cent ; Hydrogen = 5 per cent ; Oxygen = 4.2 per cent ; Nitrogen = 1.5 per cent and the remainder ash. Determine the stoichiometric air-fuel ratio by mass. If 30 per cent excess air is supplied, find the percentage composition of dry flue gases by volume. [Ans ; CO2, = 14.3%, O2 = 4.9%, N2 = 80.8%] Page 24 of 29

25 Self Answered Question & Answer 1) Find the stoichiometric air-fuel ratio for the combustion of ethyl alcohol (C2H6O), in a petrol engine. Calculate the air-fuel ratios for the extreme mixture strengths of 90% and 120%. Determine also the wet and dry analyses by volume of the exhaust gas for each mixture strength. [Ans. 8.96/1 ; 9.95/1 ; 7.47/1, Wet analysis : CO2 = 11.2%, H2O = 16.8%, O2 = 1.85%, N2 = 70.2%, Dry analysis : CO2 = 13.45%, O2 = 2.22%, N2 = 84.4% Wet analysis : CO2 = 6.94%, CO = 6.94%, H2 = 20.8%, N2 = 65.3% Dry analysis : CO2 = 8.7%, CO = 8.7%, N2 = 82.5%] 2) The following is the analysis of a supply of coal gas : H2 = 49.4 per cent ; CO = 18 per cent ; CH4 = 20 per cent ; C4H8 = 2 per cent ; O2 = 0.4 per cent ; N2 = 6.2 per cent ; CO2 = 4 per cent. (i) Calculate the stoichiometric A/F ratio. (ii) Find also the wet and dry analyses of the products of combustion if the actual mixture is 20 per cent weak. [Ans. (i) 4.06/1 by volume ; (ii) Wet analysis : CO2 = 9.0%, H2O = 17.5%, O2 = 3.08%, N2 = 70.4%. Dry analysis : CO2 = 10.9%, O2 = 3.72%, N2 = 85.4%] 3) Calculate the stoichiometric air-fuel ratio for the combustion of a sample of dry anthracite of the following composition by mass : C = 90 per cent ; H2 = 3 per cent ; N2 = 1 per cent ; Sulphur = 0.5 per cent ; ash = 3 per cent. If 20 per cent excess air is supplied determine : (i) Air-fuel ratio (ii) Wet analysis of the products of combustion by volume. [Ans /1 (i) 13.5/1 ; (ii) CO2 = 16.3%, H2O = 0.03%, SO2 = 3.51%, N2 = 80.3%] 4) The percentage composition of a sample of fuel was found to be C = 85%. H2 = 9%, S = 3%, O2 = 1.5%, Ash = 1.5%. For an air-fuel ratio of 12 : 1. Calculate : (i) The mixture strength as a percentage rich or weak. (ii) The volumetric analysis of the dry products of combustion. [Ans. 8.83% rich ; CO2 = 12.16%, CO = 2.54%, N2 = 84.83%, SO2 = 0.47%] 5) A fuel has the following composition by weight : Carbon = 86% ; hydrogen = 11.75% and oxygen = 2.25%. Calculate the theoretical air supply per kg of fuel, and the weight of products of combustion per kg of fuel. Page 25 of 29

26 [(Ans.) kg ; 4.21 kg] 6) For a hydrocarbon fuel having 84.8% carbon, 15.2% hydrogen by mass determine the stoichiometric air-fuel ratio. Also obtain the volumetric composition of combustion products with 15% excess air being supplied. [(Ans.) 15.11, 12.51% CO2, 2.89% O2, 84.60% N2] 7) During the combustion of coal having 22% carbon, 3.5% hydrogen, 6.5% oxygen and remaining ash by mass, the dry flue gas has volumetric composition of 8% CO2, 1.5% CO, 80.5% N 2 and 10% O 2. Calculate the minimum air required per kg of coal and % excess air. Consider air to have 23% and 77% of oxygen and nitrogen respectively by mass. [(Ans.) 3.48 kg air, 43.3% excess air] 8) C 10 H 22 is burnt with air fuel ratio of 13 by mass. Combustion is such that complete of hydrogen burns into water and there is no free oxygen and no free carbon. Also the carbon dioxide and carbon monoxide are present in combustion products. Determine volumetric analysis of combustion products. [(Ans.) 7.68% CO2, 6.25% CO, 15.32% H2O, 70.75% N2] 9) A coal sample has 66% C, 5.9% H2, 1% S, 19.9% O2, 1.5% N2, 5.6% ash by mass and calorific value of kj/kg. When this coal is burnt with 30% excess air the temperature of flue gases leaving is 300_C and ambient temperature is 17_C. Considering combustion to be complete and partial pressure of moisture to be 0.07 bar in flue gases and energy accompanied being 3075 kj/kg of steam, determine, (i) the air-fuel ratio by mass. (ii) the volumetric analysis of combustion products. (iii) the heat carried away by flue gases as percentage of heat released by coal. [(Ans.) 11.55, 19.37% O2, 4.25% H2O, 0.16% SO2, 4.91% O2, 71.31% N2, 17.36%] 10) Determine the net and gross calorific values per kg of mixture at constant pressure for stoichiometric mixture of air and C 6 H 6 (benzene) vapour at 25_C. Enthalpy of combustion for C 6 H 6 at 25_C is kj/kmol and the water is present in vapour phase in the combustion products. [(Ans.) 2861 kj/kg, 2981 kj/kg] Page 26 of 29

27 Test Your Skills: Choose the Correct Answer : 1. The smallest particle which can take part in a chemical change is called (a) Atom (b) Molecule (c) electron (d) compound. 2. A chemical fuel is a substance which releases... on combustion. (a) chemical energy (b) heat energy (c) sound energy (d) magnetic energy. 3. The most important solid fuel is (a) wood (b) charcoal (c) coal (d) all of the above. 4. For each mole of oxygen, number of moles of nitrogen required for complete combustion of carbon are (a) 20/21 (b) 2/21 (c) 77/21 (d) 79/ Modern practice is to use... excess air. (a) 5 to 10 per cent (b) 15 to 20 per cent (c) 20 to 25 per cent (d) 25 to 50 per cent. 6. Stoichiometric air-fuel ratio by mass for combustion of petrol is (a) 5 (b) 10 (c) 12 (d) An analysis which includes the steam in the exhaust is called (a) dry analysis (b) wet analysis (c) dry and wet analysis (d) none of the above. 8. The Orsat apparatus gives (a) volumetric analysis of the dry products of combustion (b) gravimetric analysis of the dry products of combustion (c) gravimetric analysis of products of combustion including H2O Page 27 of 29

28 (d) volumetric analysis of products of combustion including H2O. 9. In the Orsat apparatus KOH solution is used to absorb (a) carbon monoxide (b) carbon dioxide (c) oxygen (d) none of the above. 10. Enthalpy of formation is defined as enthalpy of compounds at (a) 25 C and 10 atmospheres (b) 25 C and 1 atmosphere (c) 0 C and 1 atmosphere (d) 100 C and 1 atmosphere. 11. Bomb calorimeter is used to find the calorific value of... fuels. (a) solid (b) gaseous (c) solid and gaseous (d) none of the above. 12. When the fuel is burned and the water appears in the vapour phase, the heating value of fuel is called (a) enthalpy of formation (b) lower heating value (c) higher heating value (d) none of the above. 13. Heat released in a reaction at constant pressure is called (a) entropy change (b) enthalpy of reaction (c) internal energy of reaction (d) none of the above (e) all of the above. 14. When the fuel is burned and water is released in the liquid phase, the heating value of fuel is called (a) higher heating value (b) lower heating value (c) enthalpy of formation (d) none of the above. 15. Choose the correct statement : (a) Number of atoms of each constituent are not conserved in a chemical reaction. (b) The mass of all the substances on one side of the equation may not be equal to the mass of all the substances on the other side. Page 28 of 29

29 (c) The number of atoms of each constituent are conserved in a chemical reaction. (d) The number of moles of the reactants in a chemical equation are equal to the number of moles of the products. ANSWERS 1. (a) 2. (b) 3. (c) 4. (d) 5. (d) 6. (d) 7. (b) 8. (a) 9. (b) 10. (b) 11. (a) 12. (b) 13. (b) 14. (a) 15. (c). Page 29 of 29