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1 MINISTRY OF THE RUSSIAN FEDERATION FOR CIVIL DEFENSE, EMERGENCIES AND RELIEF OF THE CONSEQUENCES OF NATURAL DISASTERS Academy of the State Fire Service A.S. Androsov, E.P. Saleev EXAMPLES AND TASKS on the course THEORY OF COMBUSTION AND EXPLOSION Moscow 5
2 MINISTRY OF THE RUSSIAN FEDERATION FOR CIVIL DEFENSE, EMERGENCY SITUATIONS AND RELIEF OF THE CONSEQUENCES OF NATURAL DISASTERS Academy of the State Fire Service A.S. Androsov, E.P. Saleev EXAMPLES AND TASKS on the course THEORY OF COMBUSTION AND EXPLOSION Tutorial Approved by the Ministry of the Russian Federation for Civil Defense, Emergencies and Disaster Relief as a teaching aid for higher educational institutions of the Ministry of Emergency Situations of Russia Moscow 5
3 UDC BBK A ISB N Androsov A.S., Saleev E.P. Examples and tasks for the course. Theory of combustion and explosion. Tutorial. - M.: Arfltvbz GPS EMERCOM of Russia, p. Reviewers: Department of General and Special Chemistry of the Academy of the State Fire Service of the Ministry of Emergency Situations of Russia, Department of Fire Engineering and Services of the Academy of the State Fire Service of the Ministry of Emergency Situations of Russia. Examples and tasks for the course Theory of Combustion and Explosion are compiled on the basis of many years of experience in teaching the discipline at the Academy of the State Fire Service of the Ministry of Emergency Situations of Russia in such a way that they can serve as a guide for course design. In order to ensure methodological unity with the theoretical part of the course, at the beginning of each chapter, examples of problem solving are given, as well as basic calculation formulas. The appendix contains tables of values most commonly used in solving problems in this course. Designed for cadets, listeners and adjuncts of educational institutions of the Ministry of Emergency Situations of Russia of a fire-technical profile. Chapters 1, 3 are written by Ph.D. tech. Sciences Associate Professor Androsov A.S., Chapter 4 Cand. tech. sciences senior researcher Saleev E.P. ISB N Academy of State Fire Service EMERCOM of Russia, 5
4 Contents Page Chapter 1. Material and heat balances of combustion processes Calculation of the amount of air required for combustion of substances Calculation of the volume and composition of combustion products Calculation of the heat of combustion of substances Calculation of combustion and explosion temperatures Chapter. Concentration limits of flame propagation (ignition) Chapter 3. Temperature indicators of fire hazard Calculation of temperature limits of flame propagation (ignition) Calculation of flash and ignition temperatures Calculation of standard auto-ignition temperature. 61 Chapter 4. Parameters of the explosion of vapor-gas mixtures Calculation of the maximum pressure of the explosion Calculation of the TNT equivalent of the explosion and the safe distance according to the action of air shock waves .. 63 Appendix
5 Chapter 1. Material and heat balances of combustion processes The theoretical basis for calculating material and heat balances are the fundamental laws of conservation of matter and energy Calculation of the amount of air required for combustion of substances Calculation formulas For practical calculations, it is assumed that air consists of 1% oxygen and nitrogen. Thus, the volume ratio of nitrogen and oxygen in the air will be: φ φ Ο 79 3.76, (1.1) 1 Ν where φ Ν, φ Ο, respectively, the volume (% vol.) content of nitrogen and oxygen in the oxidizing environment. Therefore, for 1 m 3 (kmol) of oxygen in the air there are 3.76 m 3 (kmol) of nitrogen. The mass ratio of nitrogen and oxygen in air is 3.3% O and 76.7% N. It can be determined based on the expression: φ φ NOMMNO ,9, (1.) 1 3 where M,MON are the molecular masses of oxygen and nitrogen, respectively . For convenience of calculations, combustible substances are divided into three types (Table 1.1): individual chemical compounds (methane, acetic acid, etc.), substances of complex composition (wood, peat, shale, oil, etc.), a mixture of gases (generator gas, etc.). 5
6 Table 1.1 Type of combustible substance Calculation formulas Dimension Individual substance B (1.3, a) n no + n kmol m 3 ; N kmol m 3 Substance of complex composition Mixture of gases Г (no + nn) В ngm g (1.3, b) CSO В,69 + H (1.4) φg O i φo В in (1.5) 1 m 3 kg m 3 kg m 3 kmol; m 3 kmol Here B is the theoretical amount of air; n G, no, nn the amount of fuel, oxygen and nitrogen obtained from the equation of the chemical reaction of combustion, kmol; M G is the molecular weight of the fuel; the volume of 1 kmole of gas under normal conditions (.4 m 3); C, H, S, O mass content of the corresponding elements in the fuel composition, %; ϕ Г i concentration of the i-th combustible component, % vol.; ϕ O oxygen concentration in the combustible gas, % vol.; no i the amount of oxygen required for the oxidation of one kmole of the i-th combustible component, kmol. To determine the volume of air during combustion under conditions other than normal, use the equation of state for ideal gases P T P T 1 1, (1.6) where P is normal pressure, Pa; T normal temperature, K; air volume under normal conditions; P 1, 1, T 1 - respectively pressure, volume and air temperature, characterizing the given combustion conditions. Practical amount of air The volume of air that actually enters the combustion zone. The ratio of the practical volume of air to the theoretical one is called the coefficient of excess air α: 1 V α. (1.7) The difference between the practical and theoretical air volumes is called excess air B: B B B B. (1.8) 6
7 Equations (1.7) and (1.8) imply that В В (α-1). (1.9) If the oxygen content in the combustion products is known, then the excess air coefficient is determined by the formula φ α 1+ O В (1 φ) O, (1.1) theoretical volume of combustion products. For substances in which the volume of combustion products is equal to the volume of air consumed (for example, carbon), formula (1.1) is simplified: 1 α. (1.11) 1 In the case of the formation of products of incomplete combustion (CO, H, CH 4, etc.), formula (1.11) takes the form φ O 1 α 1 φ, (1.11, a) + CH 4 where φ O, φ co, φ CH, φ 4 H content of the corresponding substances in combustion products, % vol. If the oxygen content in the oxidizing environment differs from its content in the air, then formula (1.1) can be written as: α 1+ and, accordingly, the formula (1.11) φ O (φ φ) OOOO (1.1) φo α, (1.13) φ φ where φ O is the initial oxygen content in the oxidizing environment, % vol.; theoretical volume of the oxidizing medium. Often in fire engineering calculations it is required to determine the mass of air that went to combustion, where ρ is the air density, kg / m 3. Obviously, m in in ρ in, (1.14) 7
8 ρ φ М + φ М PT N N O O В. (1.15) PT 1 After substituting constant values into formula (1.15), we obtain 3 P ρв 3, 47 1, (1.16) T where P is atmospheric pressure, Pa; T air temperature, K. Examples Example 1. Determine the theoretical mass and volume of air required for the combustion of 1 m 3 of methane under normal conditions. Solution. A combustible substance is an individual chemical compound, therefore, formula (1.3, a) must be used to calculate the volume of air. We write the equation for the chemical reaction of CH 4 combustion in air CH 4 + O + 3.76 N CO + HO + 3.76 N. From the equation we find n O; n 3.76 7.5; N n 1 CH, then 4 + 7.5 V 9.5 m 3 / m 3 or kmol / kmol. 1 Using formula (1.14), taking into account equation (1.15), we calculate the mass of air,79 8 +,1 3 m B 9.5 9.5 1.8 1, kg / m 3.,4 Determine the theoretical volume of air required for the combustion of 1 kg of benzene. Decision. A combustible individual chemical compound, therefore, to calculate according to formula (1.3, b), we write the equation for the chemical reaction of combustion C 6 H 6 + 7.5 O + 7.5 3.76 N 6 CO + 3 HO + 7.5 3.76 N, find n 1; n75; n 7.5 3.76 8,. O, N Molecular weight of benzene M The volume of 1 kmole of gas under normal conditions is 4 m 3 (7.5 +,) 8.4 V 1.3 m 3 /kg
9 Example 3. Determine the volume and mass of air required for combustion of 1 kg of organic mass of the composition: C 6%, H 5%, O 5%, N 5%, W 5% (humidity), if the excess coefficient air α.5; air temperature 35 K, pressure 995 Pa. Decision. Since the combustible substance is of complex composition, the theoretical amount of air under normal conditions is determined by the formula (1.4) 6 5 V, 9m 3 /kg. 3 8 From formula (1.7) we calculate the practical amount of air under normal conditions α, 5 5, m 3 /kg. In B, We find the amount of air that went to the combustion of a substance under given combustion conditions. Using formula (1.6), we obtain m 14, V (RT) 16.8 m 3 / kg.8 1, (RT) V ρ V 18.9 V kg / kg. PRI me R 4. Determine the volume of air required for combustion of 5 m 3 mixture of gases, consisting of% CH 4; 4%CH; 1% CO; 5% N and 5% O if the excess air ratio is 1.8. Solution. A combustible mixture of gases, therefore, to calculate the volume of air that went to combustion, we use formula (1.5). To determine the stoichiometric coefficients for oxygen no i, we write the equation for the reactions of combustion of combustible components in oxygen CH 4 + O CO + HO, CH +.5O CO + HO, CO +.5O CO, +, 5 4 +, then B 5, 7m 3 / m 3. 1 For combustion of 5 m 3 of a gas mixture, the required theoretical volume of air will be B 5 5, 7 8, 5 m 3. The practical amount of air: 18, 3, m 3. B, 9
10 Example 5. Determine the coefficient of excess air during combustion acetic acid if 3 m 3 of air was supplied for combustion of 1 kg. Solution. To determine the excess air coefficient using formula (1.7), it is necessary to calculate its theoretical amount. The molecular weight of acetic acid is 6. CH 3 COOH + O + 3.76 N CO + HO + 3.76N; (+,) 3 76.4 V 3.6 m 3 /kg. 1 6 Then the coefficient of excess air according to the formula (1.7) is equal to 3, α, 8. 3, 6 Combustion proceeded with a lack of air. PRI mme R 6. Determine the volume of air used for the oxidation of 1 m 3 of ammonia, if the oxygen content in the combustion products was 18%. Decision. We determine the theoretical amount of air required for the combustion of 1 m 3 of ammonia: then NH 3 +.75O +.75 3.76N.5N + 1.5HO +.75 3.76N, 75+, 75 3, 76 V 3, 6 m 3 /m 3. 1 To determine the coefficient of excess air according to formula (1.1), it is necessary to calculate the theoretical amount of combustion products 1 m 3 ammonia (1., formula 1.14) 1.5 +,5 +, 75 3.76 4.8 m 3 /m 3. 1 Excess air coefficient 18 4.8 α 1+ 9,. 3, 6 1 (18) The volume of air involved in the combustion process of 1m 3 ammonia, we determine from the formula (1.7) 9 3, m 3 /m 3. B, 1
11 Example 7. Determine the volume of the oxidizing medium, consisting of 6% O and 4% N, necessary for the combustion of 1 kg of isopropyl alcohol, if its temperature is 95 K, pressure 6, kPa. Solution. Since the oxidizing medium differs in composition from air, we determine by formula (1.1) the volume ratio of oxygen and nitrogen to be 4:6.67. Isopropyl alcohol combustion reaction equation C 3 H 7 OH + 4.5O + 4.5.67N 3CO + 4HO + 4.5.67N. The theoretical volume of the oxidizing medium under normal conditions is calculated by the formula (1.3, b). The molecular weight of the fuel is 6: (4.5 + 4.5.67) 1 6.4 os.8 m 3 /kg. The volume of the oxidizing medium under given combustion conditions is determined from the formula (1.6) () 4.9 OS RT.35 6.73 m 3 /kg. Example 8. Determine the mass of dinitrotoluene, C 7 H 6 (NO), burned in a sealed volume of 1 m 3, if the oxygen content in the combustion products was 1%. Solution. Since the combustion products contain oxygen, the combustion proceeded in excess air. The excess coefficient is determined by formula (1.1). C 7 H 6 (NO) + 6.5O + 6.5 3.76N 7CO + 3HO + N + 6.5 3.76N. Molecular mass of fuel 18. Theoretical volume of air (6.5 + 6.5 3.76), 4 V 38, m 3 /kg Theoretical volume of combustion products (formula 1.14) (, 5,) 3 76.4 4.4 m 3 /kg, 4 α 1+, 1 (1) Practical volume of air used for combustion, 55 38.9 7 m 3 /kg. AT 11
12 Then the mass of the burnt dinitrotoluene m g is determined from the ratio P 1 m G 1.3 kg. 9.7 V Control tasks 1. Determine the mass and volume (theoretical) of air required for combustion of 1 kg of methyl, ethyl, propyl and amyl alcohols. Construct a graph of the dependence of the volume of air on the molecular weight of alcohol .. Determine the theoretical volume of air required for the combustion of 1 m 3 of methane, ethane, propane, butane and pentane. Construct a graph of the dependence of the volume of air on the position of the substance in the homologous series (carbon content in the molecule of the substance). 3. Determine the theoretical mass of air used for combustion of 1 kg of methane, methyl alcohol, formic aldehyde, formic acid. Explain the reason for the influence of the composition of a substance on the volume of air required for their combustion. 4. Determine the volume and mass of air that went into combustion of 1 kg of wood of the composition: C 47%, H 8%, O 4%, W 5%, if the excess air coefficient is 8; pressure 9 GPa, temperature 85 K. 5. How much air, kg, was supplied to the combustion of 1 kg of carbon if the oxygen content in the combustion products was 17%? 6. How much air, kg, is required to be supplied for combustion m 3 of generator gas composition: CO 9%, H 14%, CH 4 3%, CO - 6.5%, N - 45%, O -.5%, if the coefficient excess air is equal to, 5? 7. Determine the amount of burned toluene, kg, in a room with a volume of 4 m 3 if after a fire in the absence of gas exchange it is established that the oxygen content has decreased to 17%. 8. How much chlorine, m 3, was supplied to the combustion of 3 m 3 of hydrogen, if the excess oxidizer in the combustion products was 8 m 3? 9. Determine the excess air in the combustion products of the gas mixture of composition: CO 15%, C 4 H 1 45% O 3%, N 1%, if the excess air coefficient is 1.9. 1. How much oxidizing medium, m 3, consisting of 5% oxygen and 5% nitrogen, is necessary for the combustion of 8 kg of ethyl acetate, if the excess coefficient is 1; temperature 65 K, pressure 85 GPa. 11. Determine the coefficient of excess of the oxidizing medium, consisting of 7% oxygen and 3% nitrogen, if during the combustion of sulfur the content is 1
13 oxygen dropped to 55%. Determine the amount of burned sulfur (kg) if the volume of the room is 18 m 3 How much anthracite (assume that the carbon content is 1%) burned out in a room with a volume of 15 m 3 if the combustion ceased when oxygen decreased to 13%. Gas exchange is ignored. 13. Calculate the mass and volumetric air flow required for the combustion of a gas fountain with a flow rate of 3 million m 3 / day, consisting of CH 4 8%, CO 1%, HS 5%, O 5% at an air temperature of 7 ° C and pressure 15 kPa. Homework Calculate the volume and mass of the oxidizing medium required for the combustion of the i-th combustible substance (Table 1.). Version number Combustible substance Chemical formula Quantity of fuel Composition of the oxidizing medium 1 Methyl alcohol CH 3 OH kg Air Aniline 3 Mixture of gases 4 Nitrobenzene 5 Compound substance C 6 H 7 N CO 45% N 15% C 4 H 8 1% O 3% 5 kg O 7% N 3% kg Air C 65 % O % H 5 % S 1 % T 8 K R 98 Pa α.5 g Air Normal α 1.4 6 Ethylene CH 4 5 m 3 O 5 % N 75 % 7 Sulfur O 6 % S kg N 4 % 8 Complex substance C 9 % H 3 % N 5 % O % 1 kg Air Normal α.5 T 35 K P1 Pa α 1.8 T 3 K P 95 Pa α 1.5 13
14 Variant number 9 Mixture of gases Combustible substance Chemical formula CH 4 15% C 3 H 8 7% O 1% H 5% oxidizing Conditions of combustion of a combustible medium 5 m 3 Air Normal α 1.9 1 Aluminum Al 15 kg O 4% N 58% Normal α.8 11 Alloy Mg% Al 8% 8 kg Air T 65 K P 9 Pa α 1.5 1 Formic acid CHO 5 kg Air Normal α 1, 13 Dimethyl ether (CH 3) O 1 kg Air 14 Mixture of gases 15 Complex substance 16 Glycerol HS 5% SO 15% CO 15% H 3% O 15% C 8% H 8 % W 1% T 8 K P116 Pa α 4, 15 m 3 Air Normal α 1.4.7 kg Air C 3 H 8 O 3 1 kg Air 17 Acetylene CH 15 l Cl 18% N 8% 18 Gas mixture 19 Ethyl acetic acid ester Methylethyl ketone 1 Chlorobenzene Nitrotoluene CH 4 3% O 8% N 15% H 47% T 6 K P11 Pa α 1.4 T 35 K P113 Pa α 1.9 Normal α 1.8 3 m 3 Air Normal α 3, C 4 H 8 O 5 kg Air T 7 K R 85 Pa α 1.5 C 4 H 8 O 5 kg Air Normal α.5 T 35 K C 6 H 5 Cl 7 kg Air P 1 Pa α.8 C 7 H 7 NO 1 kg O 5% N 7 5% T 8 K R 98 Pa α 1.4 14
15 Variant number 3 Mixture of gases Combustible substance Chemical formula NH 3 5 % C 4 H 1 5 % C 4 H 8 15 % CO 3 % O 5 % Amount of fuel l End of table 1. Composition of the oxidizing medium Combustion conditions Air 4 Butyl alcohol C 4 H 1 O 4 kg Air 5 Dibromohexane C 6 H 1 Br 3 kg 6 Compound substance 7 Mixture of gases C 7% S 5% H 5% O% C 3 H 8 1% CO 79% H 5% O 5% N 1% O 65% N 35% 15 kg Air Normal α 1.8 T 65 K P1 Pa α 1.8 T 8 K P 98 Pa α 1.7 T 85 K R 1 Pa α.8 1 m 3 Air Normal α 3.5 1.. Calculation of the volume and composition of combustion products In order to simplify the calculation, all combustible substances are divided into three types: individual, complex, mixtures of combustible gases (Table 1.3) Type of combustible substance Individual substance Substance of complex composition n Calculation formulas (1.17) ng ni (1.18) n M Г C CO 1.86 1 (1.19) HWHO 11, + 1.4 1 1 (1.) S SO.7 1 ( 1.1) 1 7C+ 1 HO +,63S +,8 N (1.) N 1 8 T a b l e 1.3 Dimension m 3 ; kmol m 3 kmol m 3 kg m 3; kmol kg kg 15
16 Type of combustible substance Mixture of gases i Calculation formulas i 1 ni φ n ii + φngi. 1 ng (1.3) Dimension m 3 ; kmol m 3 kmol Here is the theoretical volume of combustion products; n НГi the amount of the i-th combustion product in the reaction equation, kmol; n G amount of fuel, kmol; volume of 1 kmole of gas; M is the molecular weight of the fuel; NGi is the volume of the i-th reaction product; C, H, S, O, N, W content of the corresponding elements (carbon, hydrogen, sulfur, oxygen, and nitrogen) and moisture in the combustible substance, wt %; ϕ Гi content of the i-th combustible component in the gas mixture, % vol.; ϕ NGi content of the i-th non-combustible component in the composition of the gas mixture, % vol. The practical (total) volume of combustion products consists of the theoretical volume of combustion products and excess air or + (α 1) + Δ (1.4) V B. (1.5) Composition of combustion products, i.e. the content of the i-th component is determined by the formula φ i 1, (1.6) i i where ϕ i is the content of the i-th component in the combustion products, % vol.; i volume of the i-th component, m 3, kmol; Σ i total volume of combustion products, m 3, kmol. When burning in excess air, the combustion products contain oxygen and nitrogen O.1ΔВ; (1.7) N + 79, (1.8) N, B where N is the theoretical volume of nitrogen in combustion products, m 3, kmol, 79. (1.9) N B 16
17 Examples Example 1. What amount of combustion products will be released during the combustion of 1 m 3 of acetylene in air if the combustion temperature was 145 K. Solution. Combustible individual chemical compound (formula 1.17). Let us write the equation for the chemical reaction of combustion CH +.5O +.5 3.76N CO + HO +.5 3.76N The volume of combustion products under normal conditions + 1+.5 3.76 1.4 m / m 3. 1 The volume of combustion products at 145 K 1.4 145 (RT) 65.9 m3/m Ex. Determine the volume of combustion products during the combustion of 1 kg of phenol, if the combustion temperature is 1 K, the pressure is 95 Pa, the excess air coefficient is 1.5. Solution. Combustible individual chemical compound (formula 1.18). Let us write the equation for the chemical reaction of combustion C 6 H 5 OH + 7O + 7 3.76N 6CO + 3HO + 7 3.76N. Molecular weight of fuel 98. Theoretical volume of combustion products under normal conditions () 3.76.4 8.1 m 3 /kg Practical air volume under normal conditions (1.5) (.76) (1.5 1) 8.1 +, 4 11.9 m 3 /kg The volume of combustion products under specified conditions 11, (RT) 55.9 m 3 /kg Example 3. Determine the volume of combustion products during the combustion of 1 kg of organic mass of the composition: C 55%, O 13%, H 5%, S 7%, N 3%, W 17%, if the combustion temperature is 117 K, the excess air coefficient is 1.3. 17
18 D olution. Combustible substance of complex composition (formulas). Theoretical composition of combustion products under normal conditions 55 CO 1.86 1. m 3 /kg; H O 11, + 1.4.6 +.8 m 3 / kg; SO, 7.5 m 3 / kg; N,63+.8 3 4.7 m 3 /kg. 1 8 Total theoretical volume of combustion products under normal conditions 1+.8 +.5 + 4.7 6.55 m 3 /kg. The practical volume of combustion products under normal conditions, 55+, (1.3 1) 6.55+ 1.8 8, 35 m 3 /kg. 3 8 The practical volume of combustion products at a combustion temperature of 8, (RT) 35, 8 m 3 /kg. 73 Example 4. Calculate the volume of combustion products during the combustion of 1 m 3 of a gas mixture consisting of C 3 H 6 7%, C 3 H 8 1%, CO 5%, 15%, if the combustion temperature is 13 K, excess air coefficient, 8. Ambient temperature 98 K. Solution. Fuel - mixture of gases (formula 1.3) C 3 H 6 + 4.5O + 4.5 3.76N 3CO + 3HO + 4.5 3.76N, C 3 H 8 + 5O + 5 3.76N 3CO + 4HO + 5 3.76N. The volume of combustion products is determined by the formula (1.3) 1 CO (), 45 m 3 / m 3; 1 1 H O (), 4 m 3 / m 3. 1 Since the gas mixture contains oxygen, it will oxidize part of the combustible components, therefore, the air consumption will decrease (formula 1.5). eighteen
19 In this case, it is more convenient to determine the theoretical volume of nitrogen by the formula (1.9) 4, N,79 13, m 3 /m 3. 1 Theoretical volume of combustion products, 45 +.4 + 13, 18.5 m 3 /m 3. Practical the volume of combustion products (formulas 1.4, 1.5) 4.5 + (.8 1) 4.5 m 3 / m 3. 1 The volume of combustion products at a temperature of 13 K 4.5 13 (RT) 183.4 m 3 / m P Example 5. Determine the composition of the combustion products of methyl ethyl ketone. Solution. In this formulation of the problem, it is more rational to determine directly from the combustion equation the volume of products in kmol released during the combustion of 1 kmole of fuel CH 3 COC H 5 + 5.5O + 5.5 3.76N 4CO + 4HO + 5.5 3.76N, CO 4 kmol; 4 H O kmol; N, 7 kmole; 7 and 8, pray. Using formula (1.6), we find the composition of combustion products 4 1 ϕ H 14 O ϕco %, ϕ N 1 () 7%. 8.7 Example 6. Determine the volume and composition (% vol.) of the combustion products of 1 kg of mineral oil composition: C 85%, H 15%, if the combustion temperature is 145 K, the excess air coefficient is 1.9. Decision. Using the formulas (), we determine the volume of combustion products 85 CO 1.86 1.6 m 3 /kg; 1 15 H O 11, 1.7 m 3 /kg; 1 1 N () 9.1 m 3 /kg. 1 Theoretical volume of combustion products under normal conditions 19
20 1.6 + 1.7 + 9.1 1.4 m 3 /kg. The practical volume of combustion products under normal conditions (formula 1.5) 85 1.4 +, (1.9 1) 1.4 + 1.5, 9 m 3 /kg. 3 The volume of combustion products at a temperature of 145 K.9 145 (RT) 11.7 m 3 /kg. 73 Obviously, the composition of combustion products does not depend on the combustion temperature; therefore, it is reasonable to determine it under normal conditions. According to formulas (1.6, 1.8) 1.6 1.1 1.5 1 ϕ CO 7.1%; ϕ 9.4 O %;.9.9 (9.1 +.79 1.5) 1 1.7 1 ϕ N 76, %; ϕ 7, 3 HO,9 %.,9 Example 7. Determine the amount of burned acetone, kg, if the volume of carbon dioxide released, reduced to normal conditions, was 5 m 3. Solution. Let us write the equation for the combustion reaction of acetone in air CH 3 COCH 3 + 4O + 4 3.76N 3CO + 3HO + 4 3.76N. It follows from the equation that during combustion, 3.4 m 3 of carbon dioxide is released from 58 kg (molecular weight of acetone). Then, for the formation of 5 m 3 of carbon dioxide, m G of fuel 5 58 m G 43, kg must react. 3.4 Example 8. Determine the amount of burnt organic mass of the composition: C 58%, O %, H 8%, N %, W 1% in a room with a volume of 35 m 3, if the content of carbon dioxide was 5%. Decision. Let's determine the volume of released carbon dioxide Vyd 35.5 17.5 m3. .1 m 3 / kg. one
21 Let's determine the amount of the burned substance 17.5 m Г 15.9 kg. 1.1 Example 9. Determine the time when the content of carbon dioxide in a room with a volume of 48 m 3 as a result of burning wood (C 45%, H 5%, O 4%, W 8%) was 8%, if the specific mass burnout rate of wood is 8 kg / (m s), and the combustion surface is 38 m. When solving, gas exchange with the environment is not taken into account, dilution as a result of the release of combustion products is neglected. Solution. Since dilution by combustion products is not taken into account, we determine the volume of carbon dioxide released as a result of combustion, corresponding to 8% of its content in the atmosphere 8 48 CO 38.4 m 3 1 from expression (1.19) we determine how much should burn combustible material to release a given volume of carbon dioxide 38.4 m G 46 kg. 1,86,45 The burning time is determined based on the ratio m τ, υ Г F m where τ is the burning time; m G mass of burnt wood, kg; υ m mass burnout rate of wood, kg/(m s); F burning surface, m; 46 τ 151 s.5 min.,8 38 Control tasks 1. Determine the volume and composition (% vol.) of combustion products of 1 m 3 ethylene, propylene, butylene, if the combustion temperature is 18 K, the pressure is 98 Pa. Construct a graph of the dependence of the volume of combustion products and the content of individual components on the molecular weight of the fuel .. Determine the volume of combustion products and the content of water vapor and oxygen during combustion of 1 kg of hexane, heptane, octane, decane, if the combustion temperature is 13 K, the pressure is GPa, the excess coefficient is 1
22 combustion air 1.8. Construct a graph of the dependence of the volume of combustion products and oxygen content on the molecular weight of the fuel. 3. Determine the volume and composition of the combustion products of 1 kg of wood with the composition C 49%, H 6%, O 44%, N 1%, if the combustion temperature is 15 K, the excess air coefficient is 1.6. 4. How many combustion products, reduced to normal conditions, are formed as a result of the combustion of 5 m 3 of a gas mixture of the composition H 45%, C 4 H 1%, CO 5%, NH 3 15%, O 15%, if combustion proceeded at an excess coefficient air, equal to 3,? 5. Determine how much crude oil of the composition: C 85%, H 1%, S 5% burned out in a volume of 5 m 3, if the content of sulfur dioxide was 5 m 3. Calculate at what oxygen content the combustion ceased. 6. After what time will the CO content in a room with a volume of 3 m 3 as a result of the combustion of hexanol from a surface of 8 m be 7%? Mass burnout rate of hexane, 6 kg/(m s). 7. Determine the content of SO (% vol.) in a volume of 1 m 3 per.5 m and 4 minutes of burning oil of the composition: C 8%, H 8%, S 1%, if its burnout rate from an area of 5 m was 4 kg /(m s). Construct a graph of the dependence of the content of sulfur dioxide on the burning time. 8. Determine the volume of combustion products released for 5 minutes after ignition of the gas mixture of the composition: C H 3%, H%, O 15%, HS 18%, CO 15% and the content of carbon dioxide, if the excess air coefficient is 1.5, combustion temperature 13 K. Gas consumption 5 m 3 / s, gas temperature 95 K. Homework Calculate the volume of products formed, m 3, and their nitrogen content (% vol.) during combustion of the i-th substance (Table 1.4). Table 1.4 Variant number Combustible substance Chemical formula Quantity of fuel Composition of the oxidizing medium 1 Diethyl alcohol (CH 5) O 1 kg Air Acetic acid CH 4 O 5 kg "3 Alloy Mg% Al 8% 1 kg" Combustion conditions T g 15 K P114 Pa α.5 T g 1 K P 98 Pa α.6 T g 8 K P 95 Pa α 1.6
23 To be continued 1.4 Option number 4 Mixture of gases Combustible substance Chemical formula CH 4% C 3 H 8 65% O 15% Quantity of fuel Composition of the oxidizing medium 1 m 3 "5 Octyl alcohol C 8 H 18 O 1 kg" 6 Compound substance 7 Mixture of gases 8 Aniline C 9% H 5% O 5% NH 3 1% C 4 H 1 8% N 7% O 3% 1 kg "1 m 3" C 6 H 7 N 1 kg "9 Diethyl ether (CH 5) O 5 kg "1 Mixture of gases 11 Nitrobenzene 1 Compound substance 13 Mix of gases CO 7% C 3 H 8 5% O 5% C 6 H 5 NO C 7% H 6% O 14% W 1% CH 4 6% CO 3% H 1% 14 Dimethyl ether (CH 3) O 1 kg 15 Glycerin 16 Complex substance C 3 H 8 O 3 C 8% H 1% O 8% 1 m 3 O 4% N 58% kg Air 1 kg "1 m 3 "1 kg O 3% N 7% O 7% N 73% 1 kg Air Combustion conditions T g 148 K P113 Pa α.4 T g 13 K P1 Pa α.5 T g 13 K P 97 Pa α 1.6 T g 16 K P113 Pa α 1, T g 155 K P 94 Pa α 1.7 T g 16 K P113 Pa α 1.4 T g 14 K P113 Pa α.5 T g 18 K P 87 Pa α 1.8 T g 13 K R 97 Pa α 1.3 T g 15 K R113 Pa α 1, T g 18 K R 87 Pa α 1.8 T g 16 K R113 Pa α.1 T g 135 K R 99 Pa α 1.8 3
24 To be continued 1.4 Variant number 17 Mixture of gases Combustible substance Chemical formula CH 6 6 % C 3 H 8 3 % H 5 % O 5 % Amount of fuel Composition of the oxidizing medium 1 m 3 - "- 18 Methyl ethyl ketone C 4 H 8 O 1 kg - "- 19 Complex substance Nitrotoluene 1 Mixture of gases C 6% H 7% O - 1% W 1% 4 kg - "- C 7 H 7 NO kg - "- NH 3 4% C 3 H 8 4% H 1% O 1% Dibromohexane C 6 H 1 Br 1 kg 1 m 3 - "- O 5% N 5% 3 Dinitrobenzene C 6 H 4 (NO) 1 kg Air 4 Carbon disulfide CS kg - "- 5 Dichlorobenzene C 6 H 4 Cl 5 kg - "- 6 Formic acid 7 Ethyl acetate C 7% S 5% H 5% O% 1 kg O 8% N% C 4 H 8 O 1 kg Air Combustion conditions T g 165 K P113 Pa α.6 T g 148 K R 91 Pa α 1.7 T g 11 K R113 Pa α 1.4 T g 134 K R1 Pa α.6 T g 18 K R113 Pa α 1.7 T g 14 K R 9 Pa α.3 T g 165 K R 81 Pa α 1.1 T 17 K R 97 Pa α 1.6 T 13 K R 99 Pa α 1.4 T 6 K R 98 Pa α.5 T g 15 K R1 Pa α 1.5 4
25 1.3. Calculation of the calorific value of substances Calculation formulas When calculating the heat balance in a fire, as a rule, the net calorific value is determined. The amount of heat released during the combustion of a unit mass (volume) of fuel in the gaseous state of water Q Q Q, B H where Q is the highest calorific value; Q n lower calorific value; Q is the heat of evaporation of water formed during the combustion of a substance. c Type of fuel Substances Individual substances Substances of complex composition (Mendeleev's formula) Mixture of gases 9(OS) 5.1(9H + W) (1.31) 1 Q H QHφgi (1.3) 1 kJ/kg kJ/mol; kJ/m 3 where H i, H j, respectively, the heat of formation of one kmole of the i-th final product of combustion and the j-th starting material; n i, n j, respectively, the number of kmoles of the i-th reaction product and the j-th starting material in the combustion reaction equation; C, H, S, W, respectively, content, % wt. carbon, hydrogen, sulfur and moisture in the composition of the substance; O sum of oxygen and nitrogen, % wt; Q H i net calorific value of the i-th combustible component of the gas mixture, kJ/kmol; ϕ gi content of the i-th combustible component in the gas mixture, % vol. The calculation of the calorific value of gas-air mixtures is carried out according to the formula CM 1 Q H Q Hφ G, (1.33) 1 CM where Q N is the calorific value of the gas-air mixture, kJ / m 3, kJ / kmol; Q N is the lower calorific value of a combustible substance, kJ / m 3, kJ / kmol; ϕ g fuel concentration in a mixture with an oxidizer, % vol. 5
26 The specific rate (intensity) of heat release during combustion is q Q H υ M, (1.34) where q is the specific intensity of heat release kW/m; υ m mass burnout rate, kg/(m s). The rate of heat release during combustion, the heat of fire is equal to q Q H υ M F, (1.35) where q n is the intensity of heat release, kW; F burning area, m. Examples Example 1. Determine the net calorific value of acetic acid, if the heat of its formation is 485.6 kJ / mol. Solution. To calculate using formula (1.3), we write the equation for the combustion of acetic acid in oxygen CH 3 COOH + O CO + HO; (396.9 + 4.1485.6) 79.6 3 Q kJ/mol 79.6 1 kJ/kmol. Н To calculate the amount of heat released during the combustion of 1 kg of fuel, it is necessary to divide the obtained value by its molecular weight (64) 3 79.6 1 Q Н 1384 kJ/kg. 64 EXAMPLE Calculate the net calorific value of the organic mass of the composition: C 6%, H - 8%, O 8%, S %. D e x e n i e. According to the formula of D.I. Mendeleev (1.31) () 5, Q N 339.9 8 kJ/kg. Example 3. Determine the net calorific value of a gas mixture consisting of CH 4 4%, C 4 H 1%, O 15%, HS 5%, NH 3 1%, CO 1%. Solution. For each combustible component of the mixture, using formula (1.3), we find the heat of combustion (Table 1.6). 6
27 Heat Equation for the reaction of fuel formation, 1-3 kJ/kmol CH 4 + O CO + HO 75 C 4 H 1 + 6.5O 4CO + 5HO 13.4 Q N Heat of combustion, 1-3 kJ/kmol T a b f a,9 + 4, 75 86.3 Q 4 396, 13.5 H 666.1 HS + 1.5O HO + SO 1.1 Q 4, + 97.5 1.1 338.6 H NH 3 +.75O 1.5HO +.5N 46.1 Q 1.5 4, 46.1 317, N kJ/kmol. To determine the calorific value of 1 m 3 of a gas mixture, it is necessary to divide the obtained value by the volume occupied by 1 kmole of gas under standard conditions (4.4 m 3): 3 178.5 1 Q H 5776 kJ / m 3. 4.4 P p and measure 4. Calculate the calorific value of 1 m 3 of a stoichiometric hexane-air mixture. Solution. We find the stoichiometric composition of the combustible mixture according to the combustion reaction equation C 6 H .5O + 9.5 3.76N 6CO + 7HO + 9.5 3.76N. The entire volume of the reacted components (1 + 9.5 + 9.5 3.76) is taken as 1%, and the amount of fuel (1 kmol) will correspond to the stoichiometric concentration 1 1 φ G, %. 1+ 9.5 + 9.5 3.76 The heat of combustion of 1 m 3 of hexane is determined by the formula (1.3) Q 6 396, 167, 399.6 kJ / mol, N 399.6 H 1 4.4 3 3 Q 1 16, kJ/m 3. 7
28 The volume of one kmole of gas under standard conditions is 4.4 m 3. The heat of combustion of 1 m 3 of a stoichiometric hexane-air mixture is determined by the formula (1.33) 3 16, 1, Q 355 kJ / m 3. 1 EXAMPLE 5. Determine the intensity of heat release on an organic mass fire (composition in the example), if the burnout rate is 15 kg/(m s) and the fire area is 15 m2. Solution. According to formula (1.35): 3 q 646 .5 kW 59.5 MW. P 1 Control tasks 1. Determine the net calorific value of 1 m 3 of ethane, propane, butane, pentane and hexane. Construct the dependence of Q n on the molecular weight of the fuel. Heat of formation of combustible substances: ethane 88.4 kJ/mol, propane 19.4 kJ/mol, butane 3.4 kJ/mol, pentane 184.4 kJ/mol, hexane 11, kJ/mol.. Calculate the heat of combustion of 1 m 3 acetylene-air mixture at the lower and upper concentration limits of ignition, as well as at stoichiometric concentration. The concentration ignition limits (CPV) of acetylene are equal to -81,%. N o t e. Plot the net calorific value as a function of fuel concentration in the air. When calculating the calorific value of the mixture at the VKVV, it must be taken into account that only part of the fuel can be completely oxidized in air, the rest of the fuel will not enter into a combustion reaction due to the lack of an oxidizing agent. 3. Determine the net calorific value of 1 kg of wood composition C 49%, H 8%, O 43%. What is the specific intensity of heat release in a fire if the mass burnout rate is 1 kg/(m s)? 4. For the condition of the previous problem, determine the change in the calorific value and specific intensity of heat release at a moisture content in wood (over 1%) in the amount of 3, 5, 1 and 15%. The burn-out rate of wet wood will decrease to .9,8.6 and.5 kg/(m s) respectively. Construct a graph of Q n and q as a function of the moisture content in a combustible material. Note. To solve the problem, it is necessary to recalculate the composition of wood taking into account moisture in such a way that the content of all components is 1%. eight
29 5. Determine the intensity of heat release, kW, during the combustion of a gas mixture of composition: CO 15%, C 4 H 8 4%, O%, H 14%, CO 11%, if the expiration rate is 8 m 3 / s Calculation of the combustion temperature and explosion Combustion temperature temperature of combustion products in the chemical reaction zone. This is the maximum temperature of the flame zone. The combustion and explosion temperatures are determined from the heat balance equation Q H n C i 1 p (v) i (TG T) In this case, the adiabatic combustion temperature and the actual combustion temperature. (1.36) * QН TG T +, (1.37) C T Г T + pi Q C pi, (1.38) * where T Г and Т Г are the adiabatic and actual combustion temperatures, respectively; T - initial temperature; i volume of the i-th product of combustion; Q N is the lowest heat of combustion of a substance; Q is the heat used to heat the combustion products; C i is the heat capacity of the i-th combustion product at a constant volume. In this case, Q Q Н (1 - η), (1.39) where η is the proportion of heat losses as a result of energy radiation, chemical and mechanical underburning. The calculation of the combustion temperature using formula (1.37) or (1.38) can only be carried out by the method of successive approximations, since the heat capacity of gases depends on the combustion temperature (Table 1.7) 9
30 Determined parameters p/n 1 Volume and composition of combustion products Lower calorific value or amount of heat used to heat combustion products (in the presence of heat losses) 3 Average value of the enthalpy of combustion products 4 Combustion temperature T 1 by average enthalpy using Table 1a or 1b , focusing on nitrogen (the highest content in combustion products) 5 Heat content of combustion products with temperature T 1 (Table 1a, 1b appendix) 6 If Q< Q Н (), то T >T 1 (in i (1.) kmol / kmol, m 3 / kg Q or Q Н (1.3) kJ / kmol, kJ / kg Note Table 1.7 QН () HCP (1.4) i 1 Q Hi i (1.41) H i - enthalpy of the i-th product of combustion; i - / volume of the i-th product of combustion if Q > QН (), then T< T 1) 7 Q по формуле (1.41) 8 Расчет проводим до получения неравенства Q < QН () < Q 9 Температура горения (Н())(1) T Q Q T T T Г 1 + (1.4) Q Q Температура взрыва, протекающего в изохорно-адиабатическом режиме (при постоянном объеме) рассчитывается по уравнению теплового баланса (1.36) по методике, приведенной в табл Отличие заключается в том, что при расчетах вместо средней энтальпии продуктов горения и их теплосодержания (пп. 3-7) используется значение внутренней энергии газов (табл. приложения). Внутренняя энергия газов U C v T, где С v теплоемкость при постоянном объеме, кдж/(моль К), кдж/(м 3 К). Действительная температура горения на пожаре для большинства газообразных, жидких и твердых веществ изменяется в достаточно узких пределах (13-18 К). В связи с этим расчетная оценка действительной температуры горения может быть значительно упрощена, если теплоемкость продуктов горения выбирать при температуре 15 К: 3
31 Qн TG T +, (1.43) * С * where C Pi is the heat capacity of the i-th combustion product at 15 K (Table 1.8). Substance kJ/(m 3 K) Pi i Heat capacity ,4 31, Air 1.44 3.6 1-3 Examples Example 1. Determine the adiabatic temperature of the combustion of ethyl alcohol in air. Decision. The calculation is carried out according to the scheme given in Table Since the combustible individual substance, to determine the volume and composition of combustion products, we write the equation for the chemical reaction of combustion CH 5 OH + 3O + 3 3.76N CO + 3HO + 3 3.76N. Therefore, combustion products consist of: CO mol, HO 3 mol, N 11.8 mol, 16.8 mol. From Table 3 of the application we find the heat of formation of fuel - 78, kJ / mol Q H 396, - 78, 14, kJ / mol. 3. Average enthalpy of combustion products 14, H cf 76.3 kJ/mol. 16.8 4. Since H cf is expressed in kJ / mol, according to Table. 1a of the application, we select, focusing on nitrogen, the first approximate combustion temperature T 1 1 o C. 5. Calculate the heat content of combustion products at 1 o C using the formula (1.41) Q 114.7 + 93.4 11.8 133.7 kJ / mol . 31
32 6. Compare Q H and Q, since Q > QH, choose the combustion temperature equal to o C. 7. Calculate the heat content of combustion products at o C: Q 18.6 + 88.1.8 11.8 135 kJ/mol. 8. Since Q< Q < Q, определим температуру горения по формуле (1.4) Н (14, 135)(1) T + 1 о С. 133,7 135 Г П р и м е р. Определить адиабатическую температуру горения органической массы, состоящей из С 6 %, Н 7 %, О 5 %, W 8 %. Р е ш е н и е. 1. Так как горючее представляет собой сложное вещество, состав продуктов горения рассчитываем по формулам () 6 CO 1,86 1,1 м /кг; 1,4, 88 H O 11, 1 + м 3 /кг; N ,1 м 3 /кг. 1 8 Общий объем продуктов горения равен 7, 1 м3 /кг.. Определим низшую теплоту cгорания вещества по формуле Д.И. Менделеева (1.31) Q Н 339,9 5-5,1() 3958,4 кдж/кг. 3. Определим среднюю энтальпию продуктов горения 3958,4 H CP 3417,7 кдж/м 3. 7,1 4. Так как величина энтальпии рассчитана в кдж/м 3, первую приближенную температуру выбираем по табл. 1б приложения. Ориентируясь на азот, принимаем Т 1 1 о С. 5. Рассчитываем теплосодержание продуктов горения при 1 о С по формуле (1.41) Q 5118, 1,1,9 5,1 5144,5 кдж/кг 6. Из сравнения Q Н и Q Q Н >Q choose the second approximate temperature equal to 19 o C. 7. Calculate the heat content of combustion products at 19 C 3
33 Q 4579.7 1.5, 5.1 498.8 kJ/kg. 8. Since Q< QН < Q, определим температуру горения (3958,4 498,8)(1 19) T Г о С,8 П р и м е р 3. Рассчитать действительную температуру горения фенола (H обр 4, кдж/моль), если потери тепла излучением составили 5 % от Q н, а коэффициент избытка воздуха при горении,. Р е ш е н и е. 1. Определим состав продуктов горения: C 6 H 5 OH + 7O + 7 3,76N 6CO + 3H O + 7 3,76N, 6 моль; 3 моля; 6, 3 моля, CO H O (,76)(, 1) 39, 98 N В моля, 75, 3 моля.. Определим низшую теплоту сгорания фенола (формула 1.3): Q Н 7 396, - 1 4, 35,7 кдж/моль, так как по условию задачи 5 % тепла теряется, определим количество тепла, пошедшее на нагрев продуктов горения (теплосодержание продуктов горения при температуре горения) (формула 1.39) Q 35,7(1 -,5) 65,5 кдж/моль. По формуле (1.43) определим действительную температуру горения 65,5 Т Г К. 3 1 (5,81 6,3+ 3,6 39,98) П р и м е р 4. Рассчитать температуру взрыва метановоздушной смеси стехиометрического состава. Р е ш е н и е. Расчет проводим по схеме, представленной в табл Объем и состав продуктов горения СН 4 + О + 3,76N СО + Н О + 3,76N. Продукты горения: CO 1 кмоль/кмоль, H O моль/моль, N 3,76 7,5 кмоль/кмоль.. Низшая теплота сгорания: Q Н 1 396,6 + 4, кдж/моль. 3. Средняя внутренняя энергия продуктов горения QН 86 U ср 76,8 кдж/моль. 1,5 33
34 4. According to the table. application, we take the first approximate explosion temperature (for nitrogen) T 1 7 o C. 5. Calculate the internal energy of combustion products at T 1: U 1 pgi U i 1 18.9 + 1.4 + 7.5 7.86, kJ / mol. 6. Comparison of the value of Q H and U 1 shows that T 1 is too high. 7. Choose T 5 o C. U 1 118.3 + 94.3 + 7.5 64.3 789, kJ / mol. 8. Since U 1 > Q H > U T vzr, (7 5) 54 o C. 86, 789, Control tasks 1. Determine how the adiabatic combustion temperature changes in the homologous series of saturated hydrocarbons (for example, methane, propane, pentane and heptane). Construct a graph of the dependence of the combustion temperature on the molecular weight of the combustible substance. Determine how the adiabatic combustion temperature of wood composition changes: C 49%, H 8%, O 43%, if the moisture content (over 1%) is 5, 15%. Construct a graph of the dependence of the combustion temperature on the moisture content of the fuel. Note. When solving the problem, it is necessary to recalculate the composition of wood so that the amount of all components (including water) is 1%. 3. Determine how the adiabatic temperature of benzene combustion will change in air and in an oxidizing environment containing 5, 3, and 4% oxygen. Construct a graph of the dependence of the combustion temperature on the oxygen content. 4. Calculate the actual combustion temperature of a gas mixture consisting of 45% H, 3% C 3 H 8, 15% O, 1% N, if the heat loss was 3% of Q H, and the excess air coefficient during combustion is 1.8 . 5. Determine the amount of burned anthracite (C 1%) in a room with a volume of 18 m 3, if the average volume temperature increased from 35 to 65 K. 98.1% air) if the heat loss by radiation is % of the net calorific value. 34
35 7. Determine how the combustion temperature of acetylene will change when it is diluted with nitrogen in an amount of 1.3%, if the heat loss by radiation is 5% of the net calorific value, the excess air coefficient is 1. Plot temperature versus nitrogen content in acetylene. 8. Determine the burning time of toluene, at which the temperature in a room with a volume of 4 m 3 will increase from 95 to 375 K, if its burnout rate is 15 kg / (m s), and the fire area is 5 m. When calculating, neglect the increment in the volume of combustion products over consumed air. Homework Calculate the combustion temperature of the i-th substance (Table 1.9). Option number Combustible substance Chemical formula Composition of the oxidizing medium 1 Mixture of gases CO 4%, C 3 H 8 5%, CO 1% Air T a b l e 1.9 Combustion conditions α 1.4 η.5 Substance C 8%, H 5%, α 1.6 complex composition S 6%, W 9% - "- η.3 3 Propionic acid C 3 H 6 OO 5%, N 75% α 1.3 η.4 4 Glycerin C 3 H 8 O 3 Air α 1, η,35 5 Acetic butyl ether C 6 H 1 O - “- α 1.4 η,15 6 Ethylbenzene C 8 H 1 - “- α 1.5 η, 7 Substance of complex composition C 8%, H 8%, O 5%, W 5% - “- α 1, η,35 8 Gas mixture CO 6%, H 4% - “- α 1.8 η.4 9 Ammonia NH 3 - “- α 1, η , 1 Hexane C 6 H 14 - "- α 1.4 η,15 11 Nitroethane CH 5 NO - "- α 1.5 η, 1 Hexyl alcohol C 6 H 14 O Air α, η,1 35
36 Variant number End of table 1.9 Combustible substance Chemical formula Composition of the oxidizing medium Combustion conditions C 75%, H 8%, - “- α 1, complex composition C 1%, W 5% η,4 14 Substance 15 Mixture of gases CH 4 7%, NH 3%, O 1% 16 Formic acid 17 Substance of complex composition 18 Substance of complex composition CH OO 5%, N 75% C 56%, H 14%, O %, W 1% C 78%, H 1%, O 1% 19 Mixture of gases CO 75%, CH 4 5% Gas mixture C 3 H 8 7%, C 4 H 1%, O 1% C 85%, H 1%, O 5% 1 Substance of complex composition Gas mixture CH 6 75%, CH 4 %, O 5% 3 Substance of complex composition C 7%, H 16%, O 14% 4 Mixture of gases CO 5%, CH 4 3%, CO % 5 Substance of complex composition C 77%, H 13%, N 4%, O 6% - "- α 1.8 η, α, η.3 Air α 1, η.4 - "- α 1.6 η.15 - "- α 1.9 η, - "- α 1.8 η, - "- α 1.4 η,3 - "- α 1.7 η, - "- α 1, η,35 - "- α 1.9 η,15 - "- α 1, η,45 6 Ethylene CH 4 O 3% N 7% α 1.5 η.4 7 Amyl alcohol C 5 H 1 O Air α, η.15 36
Chapter 37 Flame propagation (ignition) concentration limits The lower (upper) flame propagation concentration limit is the minimum (maximum) concentration of fuel in the oxidizer that can ignite from a high-energy source with subsequent combustion spread to the entire mixture. Calculation formulas The lower concentration limit of ignition ϕ H is determined by the limiting calorific value. It has been established that 1 m 3 of various gas-air mixtures at the LCVV emits during combustion a constant average amount of heat of 183 kJ, called the limiting heat of combustion. Therefore, φ Q 1 PR N, (.1) QН if we take the average value of Q PR equal to 183 kJ / m 3, then ϕ Н will be equal to φ Н Q where Q Н is the lower calorific value of the combustible substance, kJ / m 3. Lower and the upper CPV can be determined by the approximation formula H ϕ () 1 H B, (.) an + b where n is the stoichiometric coefficient at oxygen in the chemical reaction equation; a and b are empirical constants, the values of which are given in Table..1. T a b l e.1 Concentration limits Flammability values a b Lower 8.684 4.679 Upper n 7.5 1.55.56 n > 7.5.768 6.554 37
38 The concentration limits of ignition of vapors of liquid and solid substances can be calculated if the temperature limits are known φ Н (В) pн(В) 1, (.3) p ) ignition limit, Pa; p - ambient pressure, Pa. Saturated vapor pressure can be determined from the Antoine equation or from Table. 4 appendices B lg P A, (.4) С + t where A, B, C are Antoine constants (table 1 of the appendix); t - temperature, C (temperature limits). To calculate the concentration limits of ignition of mixtures of combustible gases, the Le Chatelier rule is used where φ P 1 n () CM 1 φ H (V), (.5) μ i φ N (V) i V lower (upper) CPV of a mixture of gases,% about.; ϕ н(в)i - Н(В) Р lower (upper) limit of ignition of the i-th combustible gas %, vol.; µ i - molar fraction of the i-th combustible gas in the mixture. It should be kept in mind that Σµ i 1, i.e., the concentration of combustible components of the gas mixture is taken as 1%. If the concentration limits of ignition are known at a temperature T 1, then at a temperature T they are calculated by the formulas (.6), (.7) φ, lower concentration ignition limit, respectively, at temperatures Т and Т 1 ; φ VG and 1 φ VG upper concentration limit of ignition, respectively, at temperatures T 1 and T; T G combustion temperature of the mixture. 38
39 Approximately, when determining the LEL of TG, 155 K is taken, while determining the VKVL, 11 K. When the gas-air mixture is diluted with inert gases (N, CO, H O vapors, etc.), the ignition region narrows: the upper limit decreases, and the lower one increases. The concentration of an inert gas (phlegmatizer), at which the lower and upper limits of ignition are closed, is called the minimum phlegmatizing concentration ϕ f. The oxygen content in such a system is called the minimum explosive oxygen content MVSK O) the oxygen content below the MVSK is called safe of the specified parameters is carried out according to the formulas О without Calculation 1 ; (.8) 1 φf; (.9) 4.844 φ,φ 4, (.1) 1 O where ΔH f is the standard heat of formation of the fuel, J/mol; hi, h" i, h depending on the type of chemical element in the fuel molecule and the type of phlegmatizer (Table 11 of the Appendix), mi is the number of atoms of the i-th element (structural group) in the fuel molecule Calculation of these parameters can be carried out using another physically more transparent method by solving the heat balance equation ( 1.36) under the following two conditions: - at the phlegmatization point, the combustible mixture has a limiting combustion temperature of 15 K; - the mixture is stoichiometric when carbon is oxidized to CO, hydrogen to HO. The heat balance equation (1.36) in the case of dilution with a neutral gas is represented as : Q Н (Т) Г Т С ni + Срф nф Рi, (.11) * where Т Г is the limiting combustion temperature 15 K; C Pi, C Rf, respectively, the heat capacity of the i-th combustion product and neutral gas (phlegmatizer), kJ / (mol K); n i the number of moles of the i-th combustion product of the stoichiometric mixture, mol/mol; n f is the number of moles of neutral gas at the point of the phlegmatizer, mol/mol. 39
40 From (.11) n Q (T G T) СРi С (Т Т) N f RF Г ni (.1) Taking the volume of all components of the gas-air mixture as 1%, the concentration (% vol.) of each of them ni φi 1 (.13) n + n + n + n Г ON f Examples Example 1. Determine the lower concentration limit of ignition of butane in air from the limiting heat of combustion. Decision. For the calculation according to the formula (.1) in Table. 3 applications we find the lowest calorific value of a substance 88.3 kJ / mol. This value must be converted to another dimension kJ / m 3: 88, kJ / m 3., 4 Using the formula (.1), we determine the LEQV φ H 1.4%. 18.7 13 According to the table. 4 applications, we find that the experimental value of ϕ Н is 1.9%. The relative calculation error, therefore, amounted to 1.9 1.4 H 1 5%. 1.9 EXAMPLE Determine the concentration limits of ignition of ethylene in air. D elution. Calculation of CPV is carried out according to the approximation formula. We determine the value of the stoichiometric coefficient for oxygen Thus, n 3, then C H 4 + 3O CO + HO. 1 φ H 3.5 8.679%; 1 φ 18, 1.55 3.56 3 V +%. Let us determine the relative calculation error. According to the table 4 applications experimental limit values are 3, 3,: 4
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Saratov 2010
Compiled by: R.P. Volkov, lecturer at FGOU SPO "SGPPC named after Yu. A. Gagarin"
Internal reviewer: O.G. Stegalkina - lecturer
FGOU SPO "SPPK named after Yu.A. Gagarin"
Guidelines for solving problems and performing independent work on the course "Physical and chemical foundations for the development and cessation of combustion in a fire" for students of all forms of education in the specialty "Fire Safety".
The guidelines consider examples of solving typical problems in the section “Fundamentals of combustion processes. Material and heat balance of combustion processes" discipline "Physical and chemical bases of development and cessation of combustion in a fire"; given options for tasks for independent problem solving.
Printed in the printing house of the FGOU SPO “SGPPC named after Yu.A. Gagarin"
INTRODUCTION
Guidelines for solving problems and performing independent work on the section “Fundamentals of combustion processes. Material and heat balance of combustion processes" of the discipline "Physical and chemical bases of development and cessation of combustion in a fire" are designed to train fire safety engineers within the framework of the work program of the discipline "Physical and chemical bases of development and cessation of combustion in a fire" in the specialty 280104.
Guidelines for solving problems are drawn up in full accordance with the State Educational Standard of Higher Professional Education, taking into account the specifics professional activity GPS employees. Tasks are designed to consolidate the theoretical course and methods of practical calculations in this section of the discipline. Guidelines will help students master the material of the discipline being studied, which is necessary for the successful work of a fire safety engineer in any field of his activity.
The guidelines include: brief theoretical provisions, general provisions for calculating the material and heat balance of combustion processes of gaseous and condensed substances, the nature of the flame glow, combustion temperature, as well as a large number of examples of solving typical problems and reference information necessary for solving problems.
The structure and content of guidelines for solving problems provide for the possibility of self-working out by trainees of the material for each section of the discipline.
Starting to study the course, it is necessary to imagine that the basis of all phenomena occurring in a fire is the combustion process. Knowledge of the essence of this phenomenon, the laws of combustion, the mechanisms and methods for its termination are necessary for the successful work of a fire safety engineer in any field of his activity.
1. Write structural formulas, draw up equations for the reactions of combustion of combustible substances in air and calculate stoichiometric coefficients.
1.1. amylbenzene, abietic acid, allylamine;
1.2. amyl diphenyl, adipic acid, allyl isothiocyanate;
1.3. amylene, acrylic acid, alnapht;
1.4. amylnaphthalene, allyl acetate, altax;
1.5. amyltoluene, allylidene diacetate, amylamine;
1.6. anthracene, allyl caproate, amyl nitrate;
1.7. acenaphthene, allyl alcohol, amyl nitrite;
1.8. acetylene, amyl acetate, amyl sulfide;
1.9. benzene, amyl butyrate, amyl trichlorosilane;
1.10. butylbenzene, amylxylyl ether, amylchloronaphthalene;
1.11. butylcyclohexane, amillaurate, aminalon;
1.12. butylcyclopetane, amyl methyl ketone, amino azo dye;
1.13. hexadecane, amyloleate, aminocaproic acid;
1.14. hexane, amyl salicylate, aminopelargonic acid;
1.15. hexylcyclopentane, amyl stearate, aminocyclohexane;
1.16. heptadecane, amylphenyl methyl ether, ampicillin;
1.17. heptane, amnlphenyl ether, anginine;
1.18. decane, amyl formate, aniline;
1.19. diamylbenzene, anisole, antrimid;
1.20. diamylnaphthalene, acetal, atophane;
1.21. divinylacetylene, acetaldehyde, aceclidine;
1.22. dihydrocyclopentadiene, acetylacetone, acetanilide;
1.23. diisobutylene, acetysalicylic acid, acetyl chloride;
1.24. diisopropylbenzene, acetyltributylcitrate, acetoacetanilide;
1.25. dimethylenecyclobutane, acetomethoxane, acetonitrile;
1.26. ditolylmethane, acetone, acetoxime;
1.27. diphenyl, acetonylacetone, acetoethylamide;
1.28. diphenylmethane, acetopropyl alcohol, benzamide;
1.29. diethylcyclohexane, acetoacetic ether, benzyldiethylamine;
1.30. dodecane, acetophenone, benzylthiol;
1.31. isobutylbenzene, benzaldehyde, benzyl chloride;
1.32. isobutylcyclohexane, benzantrone, benzyl cyanide;
1.33. isooctane, benzhydrol, benzimidazole;
1.34. isopentane, benzyl acetate, sodium benzoate;
1.35. isoprene, benzyl benzoate, benzoyl chloride;
1.36. isopropenylbenzene, benzyl salicylate, benzoxazolone;
1.37. isopropylacetylene, benzyl cellosolve, benzene sulfazide;
1.38. methylcyclohexane, benzyl ethyl ether, benzenesulfamide;
1.39. methylcyclopentane, benzylsuccinic acid, benzenesulfonic acid;
1.40. octyltoluene, methoxybutyl acetate, benzonitrile.
2. Write structural formulas and determine during the combustion of which combustible substance the greater number of moles of combustion products will be released?
2.1. benzophenone and benzophenone tetracarboxylic acid;
2.2. borneol and butanal;
2.3. butanoic acid and butyl acetate;
2.4. butylacetylricinoleate and butylacetoacetate
2.5. butylbenzyl sebacate and butyl benzoate;
2.6. butyl butyrate and butyl vinyl ether;
2.7. butyl glycol and butyl glycol acetate;
2.8. butylglycide ether and butyldiethyl adipate;
2.9. butylisovalerate and butylcapronate;
2.10. butyl carbitol and butyl lactate;
2.11. butyl laurate and butyl methacrylate;
2.12. butyl methyl ketone and butyl oleate;
2.13. butylpropionate and butylricinooleate;
2.14. butyl stearate and butylphenyl ether;
2.15. butyl formate and butylethyl acetaldehyde;
2.16. butyl ethyl ketone and butyl ethyl ether;
2.17. valeric acid and valeric aldehyde;
2.18. vanillin and vetiveryl acetate;
2.19. vetiver alcohol and vetinyl acetate;
2.20. vetinone and vinylallyl ether;
2.21. vinyl acetate and vinyl butyrate;
2.22. vinyl isobutyl ether and vinyl isooctyl ether;
2.23. vinyl isopropyl ether and vinyl crotonate;
2.24. vinyl methyl ketone and vinyl oxyethyl methacrylate;
2.25. vinyl octadecyl ether and vinyl propionate;
2.26. vinyl trimethyl nonyl ether and vinyl ethyl ether;
2.27. vinyl ethyl ether and tartaric acid;
2.28. vitamin A (acetate) and vitamin C;
2.29. gallic acid and hexanal;
2.30. hexanoic acid and hexyl acetate;
2.31. hexyl butyrate and hexyl diethyl hexahydrophthalate;
2.32. hexyl methacrylate and hexyl methyl ketone;
2.33. hexyl alcohol and hexyl propionate;
2.34. hexyl formate and hexyl cellosolve;
2.35. heliotropin and heptadecyl alcohol;
2.36. heptanal and heptylacetate;
2.37. heptyl butyrate and heptyl diphenyl ketone;
2.38. heptyl isobutyl ketone and heptyl methyl ketone;
2.39. heptyl alcohol and heptyl propionate;
2.40. heptyl formate and hydroquinone.
SOLUTION.
1. We compose the equations for the reactions of combustion of combustible gases of the mixture in air:
C 2 H 2 + 2.5 (O 2 + 3.76 N 2) \u003d 2 CO 2 + H 2 O + 2.5*3.76N2 ,
C 3 H 8 + 5 (O 2 + 3.76 N 2) \u003d 3 CO 2 + 4 H 2 O + 5*3.76N2.
2. Calculate the theoretical volumes of air and combustion products during complete combustion of 1 m 3 of the gas mixture (formulas 8 and 15):
3. Calculate the actual volumes of air and combustion products, taking into account a 40% excess of air (α = 1.4).
4. Since the volume of the combustible mixture was 10 m 3, the actual volumes of air and combustion products will be 176.7 and 192.9 m 3, respectively.
ANSWER: The combustion of 10 m 3 of a complex gas mixture requires 176.7 m 3 of air, while 192.9 m 3 of combustion products are formed.
EXAMPLE:Determine the volumes of air and combustion products during the combustion of 2 kg of a combustible substance having an elemental composition: C = 50%; H = 10%; N = 10%; ash = 12%; moisture = 18%. Assume that the air and combustion products are under normal conditions.
SOLUTION:
1. To solve the problem, we use formulas (9) and (16).
When burning 2 kg of combustible substance, 14.34 and 16.14 m 3 of air and combustion products are formed, respectively.
ANSWER: When burning 2 kg of combustible substance, 14.34 m 3 of air is consumed and 16.14 m 3 of combustion products are formed.
TASKS FOR INDEPENDENT SOLUTION
1. Determine the volume of air required for the combustion of 50 m 3 of acetylene at α=1, 7.
2. Determine the volumes of air, combustion products and the percentage of combustion products in 2 m 3 of ethane. Take the temperature of the combustion products to be 1200 K, pressure 101.3 kPa, excess air α=1.2.
3. Determine the volume of air required for the combustion of 15 m 3 of butane at a temperature of 10С and a pressure of 750 mm Hg. Art., if combustion proceeds with an excess air coefficient equal to 1.4 (α=1.4).
6. Calculate the amount of amylbenzene that can burn in a closed room with a volume of 200 m 3 if combustion stops at a residual oxygen content of 12%. The initial temperature in the room is 24 ° C, the pressure is 98 kPa.
7. Determine how much butyl acetate can burn in a room with a volume of 200 m 3 if its combustion stops at an oxygen content of the air equal to 13.8% (normal conditions).
8. Determine the volumes of combustion products and air during the combustion of 7 kg of hexane. The combustion process proceeded at a temperature of 33°C and a pressure of 730 mm. rt. Art. The temperature of the combustion products is assumed to be 1300 K.
9. Determine the volumes of combustion products and air during the combustion of 11 kg of acetone. The combustion process proceeded at a temperature of 30 about C and a pressure of 720 mm Hg. Art. The temperature of the combustion products is assumed to be 1300 K.
10. Determine the volume of combustion products and air during the combustion of 17 kg of toluene. The combustion process proceeded at a temperature of 30 about C and a pressure of 745 mm Hg. Art. The temperature of the combustion products is assumed to be 1100 K.
11. Calculate the volume of air and the volume of combustion products during complete combustion of 6 kg of cellulose, consisting of 80% carbon, 13% hydrogen and 7% oxygen, if combustion occurs at a temperature of 25 ° C and a pressure of 95 kPa. The excess air coefficient is 1.4.
12. Determine the volume of air required for the combustion of 6 kg of diethyl ether at a temperature of 15 ° C and a pressure of 750 mm Hg. Art. The excess air coefficient was 1.3.
13. Determine how much benzene burned in a closed room with a volume of 180 m 3, if it is known that its combustion stopped when the oxygen content in the air was 14.6%. The temperature before the fire was 19 o C and pressure 100 kPa.
15. Determine the coefficient of excess air if 212 m 3 of air is consumed for the combustion of 8 kg of ethyl acetate at a temperature of 25 ° C and a pressure of 760 mm Hg. Art.
16. Calculate the excess air coefficient and the percentage of carbon dioxide in the combustion products if 70 m 3 of air is consumed for the complete combustion of 4 kg of ethyl propyl ether (C 5 H 12 O) at a temperature of 22 ° C and a pressure of 92 kPa.
17. 3 kg of acrolein burns at a temperature of 21 ° C and a pressure of 98 kPa. Calculate the volume of air that has passed into the combustion products and the percentage of water in them if combustion proceeds with an excess of air (the excess air coefficient is 1, 2).
20. Calculate the volume of a gas mixture consisting of 45% butane, 30% methane, 20% acetylene and 5% oxygen, if 80 m 3 of air is consumed for its combustion under normal conditions. The excess air coefficient is 1.6.
22. Calculate the volume of air and the volume of combustion products during complete combustion of 7 m 3 of a gas mixture consisting of 57% hydrogen, 18% carbon monoxide and 25% methane, if combustion occurs with excess air (the excess air coefficient is 1.3).
23. Calculate the volume of air and the volume of combustion products during the complete combustion of 6 kg of ceresin, consisting of 80% carbon, 15% hydrogen and 5% oxygen, if combustion occurs at a temperature of 25 ° C and a pressure of 95 kPa. The excess air coefficient is 1.5.
25. Determine the volume and composition of combustion products (in vol.%) of the gas mixture (Table 4), if combustion occurs at an excess air coefficient α (see Table 4).
Table 4
| Composition of the mixture, % | Job number | |||||||||
| 25.1 | 25.2 | 25.3 | 25.4 | 25.5 | 25.6 | 25.7 | 25.8 | 25.9 | 25.10 | |
| carbon monoxide | - | - | - | - | - | - | ||||
| Hydrogen | - | - | - | - | - | - | - | |||
| Methane | - | - | - | - | - | - | ||||
| Ethane | - | - | - | - | - | -- | ||||
| Propane | - | - | - | - | - | - | - | - | - | |
| Butane | - | - | - | - | - | - | - | - | ||
| Ethylene | - | - | - | - | - | |||||
| propene | - | - | - | - | - | - | - | - | ||
| Acetylene | - | - | - | - | - | - | - | |||
| Carbon dioxide | - | - | ||||||||
| Nitrogen | - | - | - | - | ||||||
| Oxygen | - | - | - | |||||||
| α | 1,2 | 1,3 | 1,1 | 1,2 | 1.2 | 1,2 | 1,4 | 1,1 | 1,3 |
Table 5
| job number | Substance | Elemental composition of matter | t o C | Mass of substance, kg | |||||
| C | H | O | S | W | ash | ||||
| 26.1 | Ceresin | ||||||||
| 26.2 | Coal | ||||||||
| 26.3 | Wood | ||||||||
| 26.4 | Petrol | ||||||||
| 26.5 | Oil | ||||||||
| 26.6 | fuel oil | ||||||||
| 26.7 | Kerosene | ||||||||
| 26. 8 | oil shale | ||||||||
| 26.9 | Coal | ||||||||
| 26.10 | Anthracite | 0,2 | 5,8 |
27. Determine the nature of the glow of the flame of ethylbenzene.
28. Determine the nature of the glow of the flame of acetic acid.
29. Determine the nature of the hexane flame glow.
30. Determine the nature of the glow of the flame of amyl alcohol.
31. Determine the nature of the glow of the butane flame.
32 . Determine the nature of the glow of the benzene flame.
CALCULATION OF COMBUSTION TEMPERATURE
The combustion temperature is understood as the maximum temperature to which the combustion products are heated. In engineering and fire fighting, a distinction is made between theoretical, calorimetric, adiabatic, and actual combustion temperatures.
Theoretical combustion temperature is the temperature at which the released heat of combustion of a mixture of stoichiometric composition is spent on heating and dissociation of combustion products. In practice, the dissociation of combustion products begins at temperatures above 2,000 K.
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Calorimetric combustion temperature is the temperature that is reached during the combustion of a stoichiometric combustible mixture with an initial temperature of 273 K and in the absence of losses to the environment.
Adiabatic combustion temperature is the temperature of complete combustion of mixtures of any composition in the absence of heat losses to the environment.
Actual combustion temperature is the combustion temperature achieved in a real fire. It is much lower than the theoretical, calorimetric and adiabatic, because in real conditions, up to 40% of the heat of combustion is usually lost to radiation, underburning, heating of excess air, etc.
Experimental determination of the combustion temperature for most combustible substances presents significant difficulties, especially for liquids and solid materials. However, in a number of cases, the theory makes it possible to calculate the combustion temperature of substances with sufficient accuracy for practice, based only on knowledge of their chemical formula, the composition of the initial combustible mixture and combustion products.
In the general case, the following dependence is used for calculations (approximate, since C p \u003d f (T)):
Q pg \u003d V pg * C p * T g,
where Q pg - enthalpy of combustion products;
V pg - the amount of combustion products, m 3 /kg;
C p - the average volumetric heat capacity of the mixture of combustion products in the temperature range from T 0 to T g, kJ / (m 3 * K);
T g - combustion temperature, K.
The enthalpy of combustion products is determined from heat balance equations:
Q pg \u003d Q H + Q ref - Q sweat, (24)
Q sweat = Q u + Q nedo + Q dis With , (25)
where Q use is the heat of evaporation;
Q sweat- heat loss due to radiation, underburning and dissociation combustion products.
Depending on the type of heat losses taken into account in the combustion zone (for radiation, underburning, dissociation of combustion products), one or another temperature is calculated.
During kinetic combustion of gas-vapor-air mixtures, heat losses from the combustion zone are negligibly small, therefore, for these mixtures, the actual combustion temperature is close to adiabatic, which is used in fire engineering calculations.
It is very difficult to determine the average heat capacity of a mixture of combustion products. Roughly, the enthalpy of a mixture of combustion products can be expressed as the sum of the enthalpies of its components:
Qpg =Σ (V pg) i (С р) i*T g, (26)
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where (V pg) i is the amount of the i-th component of combustion products;
C p is the average volumetric heat capacity of the ith component at T g and constant pressure
T g- combustion temperature.
When calculating the combustion temperature, the value is used Q n(lower calorific value), since at the combustion temperature water is in a gaseous state.
The values of the net calorific value of a substance (heat effect of a chemical reaction) are given in the reference literature, and can also be calculated from the corollary of Hess's law:
Q n \u003d (Σ ΔН i *n i) prod - (ΣΔН i * n i) ref, where (27)
ΔН i is the heat of formation of the i-th substance,
n i is the number of moles of the i-th substance.
According to corollary of Hess's law the thermal effect of a chemical reaction is equal to the difference between the sums of the heats of formation of the reaction products and the heats of formation of the starting materials. Recall from the course of chemistry that the heat of formation of simple substances (oxygen, nitrogen, etc.) is zero.
For example, let's calculate the heat of combustion (thermal effect) of ethane:
C 2 H 6 + 3.5*(O 2 + 3.76N 2) \u003d 2 CO 2 + 3 H 2 O + 3.76 *3.5N2.
The lower calorific value, according to the Hess consequence, is equal to:
Q n \u003d ΔH CO 2 * n CO 2 + Δ H H 2 O * n H 2 O - ΔH C 2 H 6 * n C 2 H 6 (28)
Substituting the values of the heat of formation of CO 2 , H 2 O, C 2 H 6 from reference data, determine the lower calorific value of ethane.
When a mixture of individual substances is burned, the heat of combustion of each component is first determined, and then they are summed up, taking into account the percentage of each combustible substance in the mixture:
If the fuel is a complex substance and its elemental composition is given in mass percent, then the Mendeleev formula is used to calculate the heat of combustion:
Q n c m \u003d 339.4 * C + 1257 * H-108.9 (O-N-S) -25 (9 * H + W), kJ / kg (30)
where C, H, O, N, S is the percentage of this element in the combustible substance;
W– moisture content in mass. %.
To calculate the combustion temperature, we compose the heat balance equation, assuming that the heat released as a result of combustion heats the combustion products from the initial temperature T 0 up to temperature T Mr.:
Q n (1-η) \u003d Σs rpg i * V pg i (T g -T 0)
Where η – heat loss coefficient (share of heat loss due to radiation, as well as as a result of incomplete combustion);
with RPG i – heat capacity of the i-th combustion product at constant pressure, kJ/molK;
V pg i - the volume of the i-th product of combustion, m 3.
Calculation of the volume of combustion products ( CO 2, H 2 O, SO 2, N 2) is carried out according to the following formulas:
From the heat balance equation:
The difficulty in determining the combustion temperature using this formula is that the heat capacity of a gas depends on temperature. Since gases are heated by temperature T 0 up to temperature T g, then in formula (36) it is necessary to substitute the average value of the heat capacity in this temperature range. But the combustion temperature is unknown to us and we want to find it. In this case, you can do the following. The average value of the combustion temperature of most substances in the air is approximately 1500 K. Therefore, with a small error in determining T g for calculations, we can take the average value of the heat capacity in the temperature range 273–1500 K. These values for the main combustion products are given in Table. 6.
Table 6
Average heat capacities of the main combustion products in the temperature range 273–1500 o C
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 |
The average value of the heat capacity of some gaseous substances in various temperature ranges is also given in Table. III applications.
Consider examples of solving problems for calculating the combustion temperature.
TASKS FOR INDEPENDENT SOLUTION
1. In which case, under fire conditions, during the combustion of butane, more heat will be released: with complete combustion or incomplete, proceeding by the reaction C 4 H 10 + 4.5O 2 →4CO + 5H 2 O. The answer must be confirmed by calculation using Hess's law.
2. Calculate the heat of formation of acetylene from the elements if its heat of combustion is 1411.2 kJ/mol.
3. Determine the heat of combustion of 12 kg of benzene, if the heat of its formation is 82.9 kJ/mol, the heat of formation of carbon dioxide is 396.9 kJ/mol, the heat of formation of water vapor is 242.2 kJ/mol.
4. Determine the heat of formation of pimelic acid (C 7 H 12 O 4), if the heat of combustion is 3453.5 kJ / mol.
5. Determine the heat of combustion of salicylic acid if the heat of its formation is 589.5 kJ/mol.
6. Calculate the heat of formation of methane if the combustion of 10 g of it under standard conditions releases 556.462 kJ of heat.
7. Determine the heat of combustion of benzyl alcohol (C 7 H 8 O), if the heat of its formation is 875.4 kJ / mol.
8. During the formation of octane, 208.45 kJ / mol of heat is released from the elements. Calculate its heat of combustion.
9. The heat of formation of acetone is -248.28 kJ/mol. Determine its heat of combustion and the amount of heat that is released during the combustion of 30 g of the substance.
11. Determine the calorific value of sulfophenylhydrazine (C 6 H 8 O 3 N 2 S) taking into account losses due to water evaporation. The moisture content of the substance is 20%.
12. Determine the calorific value of 4, 4 / -diaminodiphenylsulfone (C 12 H 12 O 2 N 2 S) without taking into account moisture evaporation losses.
13. Determine the heat of combustion of 4, 6-dimethylhexahydro-1, 3, 5-triazinethion-2 (C 5 H 9 N 3 S) according to the formulas of D. I. Mendeleev.
14. Determine the heat of combustion of diaminomesitylene-6-sulfonic acid (C 9 H 14 O 3 N 2 S), if the moisture content in the substance is 35%.
15. Determine the lowest calorific value of wood composition: C - 41.5%; H - 6%; O - 43%; N, 2%; W– 7.5%.
16. Determine the theoretical combustion temperature of acetone using the average heat capacities.
17. Determine the theoretical combustion temperature of pentane using the average heat capacities.
18. Determine the theoretical combustion temperature of octane using average heat capacities.
19. Determine the theoretical combustion temperature of benzene using the average heat capacities.
20. Using the method of successive approximations, calculate the adiabatic temperature of the combustion of propanol.
21. Calculate the combustion temperature for a stoichiometric mixture of combustible substance with air (Table 7).
Table 7
22. Using the method of successive approximations, calculate the actual combustion temperature of a combustible substance (Table 8), if combustion occurs at an excess air coefficient α, and the proportion of heat loss by radiation is η.
Table 8
| job number | Substance name | Elemental composition of matter, wt. % | α | η | ||||||
| C | H | O | S | N | W | ash | ||||
| 22.1 | Anthracite | 0,5 | 1,0 | 21,5 | 1,1 | 0,2 | ||||
| 22.2 | Oil shale | 24,2 | 1,8 | 4,5 | 3,0 | 2,0 | 39,5 | 1,2 | 0,3 | |
| 22.3 | Kerosene | 13,7 | 0,3 | - | - | - | 1,3 | 0,4 | ||
| 22.4 | Petrol | 8,0 | 5,0 | - | 2,0 | 1,4 | 0,3 | |||
| 22.5 | Salt oil | 86,0 | 12,0 | 1,2 | 0,8 | - | - | - | 1,5 | 0,2 |
| 22.6 | fuel oil | - | - | 1,6 | 0,3 | |||||
| 22.7 | Wood | - | - | 1,7 | 0,4 | |||||
| 22.8 | Coal | - | 1,8 | 0,3 | ||||||
| 22.9 | Ceresin | - | - | - | - | 1,7 | 0,2 | |||
| 22.10 | Oil shale | 1,6 | 0,3 |
23. Determine the theoretical combustion temperature of the rubber composition: WITH = 80 %, H= 15%, S = 2%, O = 1%, N = 2 %.
24 . Determine the actual combustion temperature of the composition paper: C \u003d 55%, H \u003d 25%, N \u003d 3%, O \u003d 15%, H 2 O \u003d 2%, if heat loss due to underburning was η X=0.15, due to radiation η izl=0,20.
25. Determine the actual combustion temperature of the composition plastic: C = 70%, H = 20%, N = 5%, O = 2%, non-combustible components (fillers) amounted to 3% /, if heat loss due to underburning amounted to η X=0.20, due to radiation η izl=0.25. Excess air coefficient α = 1, 4.
APPENDIX
List of accepted designations
n is the number of moles of the substance;
β is the stoichiometric coefficient;
V to theor- theoretically necessary for combustion, m 3;
V in d- the actual (practical) volume of air that went to combustion, m 3;
V pg t- theoretical volume of combustion products, m 3;
R– gas pressure, Pa;
P 0– initial (atmospheric) pressure, Pa;
T is the temperature of the substance, K;
Q is the amount of heat, J;
Vi- the volume of the i-th gaseous substance, m 3, kmol;
α - coefficient of excess air;
m is the mass of the substance, kg;
M is the mass of one kmole of a substance, kg/kmol;
Q n– lower calorific value of substance, kJ/mol, kJ/kg;
H i- enthalpy of the i-th substance, kJ / mol, kJ / m 3;
T g– combustion temperature, K;
cpi is the heat capacity of the i-th gas at constant pressure, kJ/mol*K; kJ/m3;
η is the heat loss coefficient.
Table I
Basic physical constants of some gases
In order to simplify the calculation, all combustible substances are divided into three types: individual, complex, mixtures of combustible gases (Table 1.2.1).
Table 1.2.1
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fuel type |
Calculation formulas |
Dimension |
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individual substance |
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Substance of complex composition |
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mixture of gases |
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(1.2.2)
(1.2.3)
(1.2.4)
(1.2.5)
(1.2.7)
(1.2.8)
Here 
- theoretical volume of combustion products; 
- quantity of the i-th combustion product in the reaction equation, kmol; 
- amount of fuel, kmol; 
- volume of 1 kmole of gas; 
is the molecular weight of the fuel; 
-volume of the i-th reaction product; C, H, S, O, N - the content of the corresponding elements (carbon, hydrogen, sulfur, oxygen and nitrogen) in the combustible substance, wt%; 
-content of the i-th combustible component in the gas mixture,% vol.; 
- content ith non-combustible component in the composition of the gas mixture, % vol.
The practical (total) volume of combustion products consists of the theoretical volume of combustion products and excess air
(1.2.9)
(1.2.10)
Composition of combustion products, i.e. the content of the i-th component is determined by the formula
(1.2.11)
where 
- content I-th component in combustion products, % vol.;
- volume I- th component, m 3 , kmol;
- the total volume of combustion products, m 3, kmol.
When burning in excess air, the products of combustion contain oxygen and nitrogen.
(1.2.12)
(1.2.13)
where 
- theoretical volume of nitrogen in combustion products, m 3, kmol.
(1.2.14)
Examples
Example 1. What amount of combustion products will be released during the combustion of 1 m 3 of acetylene in air if the combustion temperature was 1450 K.
Fuel is an individual chemical compound (formula 1.2.1). We write the equation for the chemical reaction of combustion
C2H2+ 
O2+ 
N 2 \u003d 2CO 2 + H 2 O + 
N 2
The volume of combustion products under normal conditions
m 3 / m 3
The volume of combustion products at 1450 K
m 3 / m 3
Example 2. Determine the volume of combustion products during the combustion of 1 kg of phenol, if the combustion temperature is 1200 K, the pressure is 95000 Pa, the excess air coefficient is 1.5.
Fuel is an individual chemical compound (formula 1.2.2). We write the equation for the chemical reaction of combustion
C6H5OH+ 
O2+ 
N 2 \u003d 6CO 2 + 3H 2 O + 
N 2
The molecular weight of the fuel is 98.
m 3 /kg
Practical air volume under normal conditions
The volume of combustion products under given conditions
m 3 / m 3
Example 3. Determine the volume of combustion products during the combustion of 1 kg of organic mass of the composition: C-55%, O-13%, H-5%, S-7%, N-3%, W 17%, if the combustion temperature is 1170 K, excess air coefficient - 1.3.
Combustible substance of complex composition (formulas 1.2.3 - 1.2.6). Theoretical composition of combustion products under normal conditions
m 3 /kg
m 3 /kg
Total theoretical volume of combustion products under normal conditions
\u003d 1 + 0.8 + 0.05 + 4.7 \u003d 6.55 m 3 / kg
Practical volume of combustion products under normal conditions
=6,55+0,269
(1.3-1) \u003d 6.55 + 1.8 \u003d 8.35 m 3 / kg
Practical volume of combustion products at combustion temperature
=
m 3 / kg.
Example 4. Calculate the volume of combustion products during the combustion of 1 m 3 of a gas mixture consisting of C 3 H 6 -70%, C 3 H 8 -10%, CO 2 -5%, O 2 -15%, if the combustion temperature is 1300 K, excess air coefficient - 2.8. Ambient temperature 293 K.
Fuel is a mixture of gases (formula 1.2.7).
The volume of combustion products is determined by the formula (1.2.8)
m 3 / m 3
m 3 / m 3
Since the gas mixture contains oxygen, it will oxidize part of the combustible components, therefore, the air consumption will decrease (formula 1.1.5).
In this case, it is more convenient to determine the theoretical volume of nitrogen by the formula (1.2.14)
m 3 / m 3
Theoretical volume of combustion products
Practical volume of combustion products
The volume of combustion products at a temperature of 1300 K
m 3 / m 3.
Example 5. Determine the composition of the combustion products of methyl ethyl ketone.
With such a formulation of the problem, it is rational to determine directly from the combustion equation the volume of products in kmol released during the combustion of 1 kmol of fuel
kmolya; 
kmolya; 
kmolya; 
kmolya.
According to the formula (1.2.11) we find the composition of combustion products
Example 6. Determine the volume and composition of the combustion products of 1 kg of mineral oil of the composition: C-85%, H-15%, if the combustion temperature is 1450 K, the excess air coefficient is 1.9.
Solution. Using formulas (1.2.3 - 1.2.6), we determine the volume of combustion products
m 3 /kg
m 3 /kg
m 3 /kg
Theoretical volume of combustion products under normal conditions
Practical volume of combustion products under normal conditions formula (1.2.10)
The volume of combustion products at a temperature of 1450 K
m 3 /kg
Obviously, the composition of the combustion products does not depend on the combustion temperature, so it is advisable to determine it under normal conditions. By formulas (1.2.11;1.2.13)
;
;
Example 7. Determine the amount of burned acetone, kg, if the volume of carbon dioxide released, reduced to normal conditions, was 50 m 3.
We write the equation for the reaction of combustion of acetone in air
It follows from the equation that during combustion from 58 kg (molecular weight of acetone) 
m 3 carbon dioxide. Then, for the formation of 50 m 3 of carbon dioxide, Mg of the fuel must react
kg
Example 8. Determine the amount of burnt organic matter composition C-58%, O-22%, H-8%, N-2%, W-10% in a room with a volume of 350 m 3 if the carbon dioxide content was 5%.
Solution. Determine the amount of carbon dioxide released
m 3.
According to formula (1.2.6), for a substance of complex composition, we determine the volume of CO 2 released during the combustion of 1 kg of fuel,
m 3 / kg.
Determine the amount of burned material
kg.
Example 9. Determine the time when the content of carbon dioxide in a room with a volume of 480 m 3 as a result of burning wood (C-45%, H-50%, O-42%, W-8%) was 8% if the mass-specific burnout rate wood 0.008 kg / (m 2 s), and the burning surface is 38 m 2. When solving gas exchange with the environment, do not take into account dilution as a result of the release of combustion products.
Since the dilution of combustion products is not taken into account, we determine the volume of carbon dioxide released as a result of combustion, corresponding to 8% of its content in the atmosphere
m 3
From expression (1.2.3) we determine how much combustible material must burn in order to release a given volume of carbon dioxide
kg.
The burning time is determined based on the ratio
,
where 
- burning time;
Mg- mass of burnt wood, kg;
- mass burnout rate of wood, kg / (m 2 s);
F- burning surface, m 2;
min.
Assignment for independent work
Task 3: Determine the volume of combustion products during the combustion of 1 kg of a given substance, if the combustion temperature ... K, pressure ... mm Hg, = ... .
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Substance |
T p.g., K |
R, mmHg. | ||
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amylbenzene | ||||
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N-amyl alcohol | ||||
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Butyl acetate | ||||
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Butyl alcohol | ||||
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diethyl ether | ||||
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White Spirit | ||||
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ethylene glycol | ||||
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tert-amyl alcohol | ||||
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Methyl alcohol | ||||
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Amyl methyl ketone | ||||
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Butylbenzene | ||||
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Butyl vinyl ether | ||||
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Ethanol | ||||
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Butyl alcohol | ||||
Task 4: Determine the volume and composition (% vol.) of combustion products released during the combustion of 1 m 3 of combustible gas, if the combustion temperature was ... K, pressure ... mm Hg.
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Substance |
T p.g., K |
R, mmHg. |
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Acetylene | ||||
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carbon monoxide | ||||
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hydrogen sulfide | ||||
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Acetylene | ||||
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carbon monoxide | ||||
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hydrogen sulfide | ||||
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carbon monoxide | ||||
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Acetylene | ||||
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Acetylene | ||||
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carbon monoxide | ||||
