40 Combustion Calculations weights and the molecul: ons soni ea eaael engineering practoca sme ofthe lements and the compounds ‘Substance Chemical formulae Atomic weight Molecular weight | Carbon 2 12 12 Sulphur = al cu Hydrogen : : 2 oxygen 02 16 32 Nitrogen Ne u 28 Carbon dioxide CO2 = 44 carbon monoxide co = 28 Water vapour 120 ~ 18 Sulphur dioxide S02 - 64 ‘Methane CH - 16 Ethylene CoH = 28 aie = = 29 COMPLETE COMBUSTION Combustion is a chemical reaction between a fuel and an oxidant which proceeds above some minimum temperature called the spontaneous ignition temperature to which the reac- tants must be heated. _ Incombustion of gases & vapours, the proportions of fuel & oxidant must be between the limits of inflammability, and vary with the particular fuel, oxidant, diluent, direction of flame Propagation, pressure, temperature, presence of catalyst, etc. In the combustion of all fuels, it ‘desirable to know the stoichiometric proportions i.e. the amount of oxidant which on comple- {@of combustion with a given amount of fuel would lead to the products like carbondioxide wa Water (HO), sulphur dioxide (SO,) and nitrogen (N,). This idealized concept is the basis combustion calculations. yin’ Baiority of hydrocarbon fuels have as active ingredients the elements —carbo ae i ding of the basic che ), oxygen (0), nitrogen (N) and sulphur (S). An understan‘ mee ence pserved in fuel in the the fuel in n (C), ical tusfonan2 Which these elements are involved is fundamental to practic form of» 7°St Solid liquid and gaseous fuels. Inert ingredients are often 0 Toisture and ash, the latter being the inorganic residue after combustion of 403404 FUELS, FURNACES AND py air or oxygen to constant mass at a temperature of around 800°C. Gaseous fuels may, water vapour, carbon dioxide, or nitrogen, which are diluents, which generally ing. (ain volume of combustion products and detract from the maximum temperature attainapj,° th Generally combustion requires reaction with an oxidant or supporter of combustion, ¢, is the main oxidant, but the halogens (chlorine and fluorine), hydrogen peroxide and ais) acid, may act as oxidants as in rocket propulsion. ite Air is the commonest oxidant because it is cheap and readily available. A typical yoy percentage composition of dry air is : trie Np = 78.09, Op = 20.95, Ar = 0.93, CO, Kr = 0,0001, NO = 0.000 05, Hy = 0.000 05, 0: x This composition varies with altitude and slightly by int the following analysis are used in combustion calculations : 0.03, Ne = 0.0018, He = 0.005, cH, 000 04, Xe = 0.000 008. dustrial pollution. By convents, 4 = 0.002 Air components by volume % by weight % On 21.00 23, Ne 79.00 7 Total 100.00 100 Assuming the fixed proportion of oxygen to ‘nitrogen’, which includes all the inerts, the stoichiometric oxygen is related to the stoichiometric air requirement. Combustion calculations are based on a number of assumptions : (a) Avogadro's hypothesis : Equal volumes of gases measured under the same conditions of temperature and pressure contains equal numbers of molecules. (6) The ideal gas law generally applies, since most combustion reactions involve high temperature and a total pressure close to one atmosphere. (i.e. PV = nRT) (c) Conservation of chemical elements : The total mass of C, H, 0, N, S before reaction is the same as that after reaction. (d) Dalton’s law of partial pressures applies to mixture of gases. Each component at the same temperature existing on its own in a vessel would exert a partial pressure p,, thus ie, P=Ep; (e) Conservation of energy, for a given mass of fuel, oxidant and diluent. ie. Total energy of reactants = Total energy of products. _,_ Enthalpy tables normally use 0°C as the datum temperature for zero enthalpy. Hence se=- sible heats can be calculated from this datum. Heats of chemical reaction, determined by calorimeter are generally referred to a datum temperature of 25°C and the preheats of reactants and products are corrected to this temperature in calorimetry. (A Most combustion work is at constant pressure (in practice near one atmosphere) $0” stant pressure conditions may be assumed by convention unless stipulated otherwise _{¥) Theoretical air is defined as the minimum quantity of air per unit mass of fuel required which is sufficient on complete combustion to give CO», HyO, SO, and Ny as products. Theoret cal air is a fixed quantity for a given fuel and is caleulated by the stoichiometry/chemical 2° tion of the various combustible constituents of the fuel. Similarly, theoretical flue gas ree! eee obtained by the complete Combustion of fuel using theoretical amount of air. Bae ion is general en as 1 i iqui i fuels or flue pos eee taken 100g of solid or liquid fuel and 100 Kg moles of & (h) Excess air is the practical amount of a ied i oce' ly to ensure that, under the conditions of eden an am process is likely to go to completion. Normally 10, 15-20, 20-25 and 50-10% excess #! supplied with gaseous, liquid, pulverised fuel and solid lumpy fuels, respectively. <3 usual ustio®, Ss LATION’ 405 actual air ~ theor theoretical air _ actual O, ~ theoretical O, r theoretical 0, actual air used , factol = theoretical air + At 40% excess air, the value of excess air factor is 14 ai pxcess IS f corrosion, the c for the purposes 0! 7 convention is that sulphur burns to sulphu @ Bxcebractice 1-3% of sulphur in the gases may form SO,, part of which se eabeestaa e ios wustioP process and is completed in the upper atmosphere) ca in the fuel is assumed to take part in the combustion re: Thus explosives i sows in the oceie ae ustion reaction. Thus explosives con ti i cept for a very small amount of nitrogen which forms NO and NO,, nitrogen (N) in teh pat onventionally assumed to form nitrogen gas (Np) on combustion the fuel » Basic complete combustion reactions are ; @ c+ 02 = €0, imole 1 mole 1 mole (molar or volume basis) inkg 32kg 44 kg (mass basis) | 1 H, + 5% a HO 1 imele mole 1 mole (molar or volume basis) akg 16 kg 18 kg (mass basis) S + O = 80, 1mole imole = 1 mole (molar or volume basis) 32kg - S2kg 64 kg (mass basis) ‘These reactions obey the laws of algebra, as each reaction can be multiplied through by any cmstant factor, and reactions can subsequently be added and subtracted. Reactions are always shown in a positive sense. Some common combustion reactions are : Combustibles Chemical Reactions carbon monoxide CO + 3 = COz e . wydrogen sulphide HS + 1302 = Hx0 + S02 methane CH, + 202 = CO2 + 2H20 unsaturated hydrocarbon | C25Hs5 + 1502 = 10CO2 + 10H20 Propane C3Hg + 502 = 3CO2 + 4H20 butane 1 CaHio + $5 02 = 4002 + 5H20 thane Cols + 802 = 2COz + 2H20 Pentane CsHiz + 802 = 5CO2 + 6H20 acetylene 1 Cally + 25 02 = 2002 + H20 Ethylene C2Hy + 302 = 2CO2 + 2H20 | benzene | Celle + 74.02 = 6CO2 + 3120 Octane Celio + 124029 8CO + 9HL0FUELS. FURNACES AND ep 406 ; 7 im) Analyses of solid and liquid fuels are normally eee ae mass basis, whit os (m) Analyses of solid a : te gas analyses are generally ¢_.°*% els are normally analysed on a volume b ste ga : ; porta fuels are Pa al is because condensation takes place from the sample. Moisture ig ¢ volume on the d s becaus le mined separately, if it is measured at all (n) Ultimate COs is defined as the em stely burnt with stoichiometric ai ee ea s in the solid and liquid fuels are expresied as elements in eit Yere, while for gaseous fuels, they are given in volume percent or mols Percent. Usual yore (on dry basis) for fuel gas & flue gas are used. SO, & CO, a ported tog Orsat analysis as SO) is also absorbed in KOH in Orsat appara 4 ong with COs. Im case solid & liguid fuels and some gaseous fuels (having Nz < 3%), contribution to the N, conten: respective flue gas is negligible and hence can be neglected. Combustion air is usually assun to be dry having average molecular weight of 29. Nm° refers to volume in m? at N.7p 0°C & 760 mm Hg) % CO, in waste gas (dry basis) when the ie INCOMPLETE COMBUSTION The formation of black smoke, soot, tar, partial decomposition products and unburnt fi are all symptoms of incomplete combustion. CO, Hy or CH, analysed in the waste gas indicate incomplete combustion. The process may be carried out deliberately as in gasification and sometimes for reducing atmos- pheres in the heat treatment of metals. More often it is due to insufficient air for complete combustion or bad mixing of fuel and air. FLUE GAS COMPOSITION-ULTIMATE CO, Fig. 40.1 indicates the approximate variation in. = composition (dry) of the waste gas for a given fuel for Defsiem Verlaan! ee s - 5 Air Air \ir a varying amount of air assuming complete reaction. : of the oxygen available. be 40.1, Combustion air Vs flue gas components Equations for Stoichiometri 1. (a) Carbon burning to CO, (Ai C+0, = CO, Inserting the values of 12 units by weight +2 ic or chemically correct mixtures inalysis by weight) the molecular weights, x 16 units by wei, If we consider the unit of wei eh = 44 units by weight= AL CULATIONS: pnccaut srt Fi burning to carbon mono: Ge : (ay care Monoxide (Anal yee hy wn 26-40, = 2¢0 werahey gx 12k (C) 2% 18K OH = 2.12 +4) ga ka of carbon * 32 kg of oxygen = 56 kg pare ‘o7 »,4 7 oN Momeriede " 1 ka (C) + 3 kB (Oa) = 5 ke (COY g to carbon monoxide (Analysis by 2C + 0, = 2CO Neglecting the volume of solid carbon, we find that volume of O, combines with carbon to give 2 volumes of CO js an increase in volume here. g.(a) Carbon monoxide burning to carbon dioxide (Analysis by weigh 2CO + 0, =2C0, right) 2x 28 kg + 32 kg =2x 44 kg ide+4 ry ___ kg of carbon monoxide + 7 kg of oxygen = "7 kx of carbon dioxide volume ipycarbon burnin n monoxide burning to carbon dioxi $ we 2CO + 0, = 2C0, RTT 2 vol CO + 1 vol. O = 2 vol. CO, There is molecular contraction in volume. 4.(a) Sulphur burning to sulphur dioxide (Analysis by weight) S+0,=S0, 32 kg + 32 kg = 64 kg i Lkg of S + 1 kg of 02 > 2 kg of SO, (6) Sulphur burning to sulphur dioxide (Analysis by volume) ' S +0, =S0, Negligible volume of S +1 vol. of O, = 1 vol. of SO, ‘There is no molecular contraction or expansion in volume. 5, (a) Methane, complete combustion (Analysis by weight) CH, + 20, = CO, + 2H,0 16 kg + 64 kg = 44 kg + 36 kg 1 eg of CH, + 4g of Op = 2 eg of C0, + § Ws of HO (b) Ca: (b) Methane, complete combustion (Analysis by volume) CH, + 202 = CO2 + 2H,0 1 vol. + 2 vol. = 1 vol. +2 vol. 1 vol. of CH, + 2 vol. of O2 = 1 vol. of COa +? vol. of H here is no molecular expansion or contraction in volume. nalysis by weight) 8.(a) Ethylene, complete combustion (A\ CH, + 802 = 2CO; + 2H,0 28 kg +96 kg = 88 kg + 96 ke " Ls 0 Dg of CH +24 keg of On» 2 Ha of On 7 HEE 0 (steam)FUFLS. FURNACES AND REF 4p 408 Me ) (b) Ethylene, complete combustion (Analysis by volume CH, + 30, = 2CO, + 2H,0 1 volume +3 volume = 2 volume + 2 volume 1 volume of Call, +3 volume of 0, = 2 volume of COp + 2 volume of 1,0 (steam, There is no molecular contraction or expansion in volume. Je, Kilomole and Mole Fraction “ Mole ots subetance ie the mass of the substance numerically equal to the molecular ws ie of the substance. Thus mole can be expressed as Weight of the substance in kj Mole = Xfoiecular weight of the substance because the equations of che ol lly useful in the case of gaseous fuels a reactions are also molecular equations, Avogadro's law states that under the same conditinn of pressure and temperature, a fixed volume of any gas contains the same number of molecule: In other words, the molecular volume of all the gases is the same under the same conditions of pressure and temperature. Thus molecular volume for any gas can be expressed as pV =MRT 7 where, M-= No. of moles of gas and, R = characteristic gas constant 8487 a i 2 5 3 where, 7 is expressed in °K, p is expressed in kgf/m? and Vis expressed in m?. It ean also be expressed a8 Vg = 298487 Vinal = where, _p is expressed in kgf/cm” Kilomole or kg.mole Molecular weight of a substance is a number proportional to the mass of the molecule. Thus init mass is expressed in any system of units, the molecule will be correspondingly ealled ‘Therefore, 1 kilomole of oxygen will mean 32 kg of oxygen. 1 gram mol. of oxygen will mean 32 grams of oxygen. Similarly 1 kg mole of any gas will occupy a volume of 22.4 Nm’. Example 1. The volumetric composition of a gaseous fuel is the following : Hz = 50% ; CH, = 20% ; C2Hy = 2% ; CO, = 5% ; CO = 16% ; Ny = 7%. Determine the molecular weight, and the dei the mixture is 1.03 kgflem?. Determine also weight Solution. The universal gas constant ‘MR’ for any gas is equal to 848. Therefore, we have Constituent Percentage compositic an eee by volume" | Molecular fraction | Molecular weight. M #1 : ae 05 2 a a 0. 16 ae a : 28 co, : 0.02 0.05 44 co 16 fi 0.16 28 2 1 0.07 28 Therefore, molecular wei ight of the mixture : Monsters ® ; misture O5%2 402% 16+ 00228 +005 444 016% 28 0.07 28 <184412 FUELS, FURNACES AND REFRACTOR volume of Op in air. Usually CO, percentage varies from 10 to 12% by volume in coal fired p, with natural draught and 10 to 15% by volume in boilers with artificial draught. These figures are the result of dilution of the product of combustion by excess air Example 6. The analysis of coal used in a boiler trial is as follows : 82% carbon, go, hydrogen, 4% oxygen, 2% moisture and 8% ash. Determine the theoretical minimum air required for complete combustion Of 1 kg of coal If the actual air supplied is 18 kg per kg of coal the hydrogen is completely burned and go, of carbon is burned to CO>, the remainder to CO, determine the volumetric analysis of the dry products of combustion. . Solution. For complete combustion, all the carbon in the fuel must burn to CO,, therefore, 0. Minimum air required = 11.6C + 34.8 (# - ge 35 S = 11.6 x 0.82 + 34.8 (0 06 ~ ot ier, low ower = 0.5 + 1.912 = 11.412 kg per kg of fuel. But actually only 80% of the available carbon is burnt to CO2. Therefore, CO, actually produced = 0.8 0.82 44 = 2.403 keg of CO, ‘Therefore, oxygen actually required by the portion of carbon burning to CO, =0.8x0.82x 2 = 1.749 kg of Op Similarly 20% of carbon is burnt to CO, It produces CO =0.2x0.82x a 0.383 kg of CO The oxygen actually required for 20% carbon burning to CO =0.8 x 0.82 xis = 0.219 kg of Op Also oxygen required by hydrogen = 0.06 x 8 = 0.48 kg of On and, water produced .06 x 9 = 0.54 kg of HzO Air actually supplied is 18 kg. It contains Op 23 =18x 23, “To Therefore, free O. in the flue gas = 4.14 (from air supplied) + 0.04 (from fuel) ~ 1.749 (for CO.) ~ 0.219 (for CO) - 0.48 (for H,0) = 1.732 kg per kg of coal and, nitrogen in the flue gas= 18 x 0.77 = 13.87 kg per kg of coal. 4.14 kg of Op 1g of constituent ‘Molecular i ke (o Proportional % volume of Cinstituent gas | 80% PEF He of coal weight: weight constituent gos A B C=A/B D= $ ee COs 2.403 cor 44 0.0546 Sade e 0.383 28 0.0137 a 2 1.732 32 0.0541 oon Ne . 13.87 28 0.4953, 80.2% Total 0.6177 100CALCULATIONS 4pusTION om xample 7 The analysis of the coal in boiler trial was C = 81%, Hy = 4.5%, 0, = Bator incombustible me Orsat analysis of the dry flue gas was CO, = 8.3%, CO = 1.4%, O2 = 10%, Np = 80.3% (by uifferenc- Determine © (a) the weight of air supplied per kg of coal (p) the percentage of excess air. solution. % Volumetric | Molecular | Proportional | Analysis by | Carbon per kg | Weight of carbon unstituent | COMP. weight | weight | weight | of constituent |per hg of dry flus gas consti cos A B C=AxB E F=DxE 12 0.0335 COs 83 44 365.2 0.1228 Zz co 14 28 39.2 0.0132 = eee 2 10.0 32 320.2 0.1076 - - N 80.3 28 2249 0.756 = = 4 Total 100 2913.4 0.03916 ‘Therefore, for 0.81 kg of carbon in 1 kg of coal the flue gas will be 2 aT g = 20.68 kg of dry flue gas per kg of coal. Water formed .045 x 9 = 0.405 kg per kg of coal. Incombustibles 1-(0.81 x 0.045 + 0.08) = 0.065 kg/kg of coal. Therefore, air supplied per kg of coal = 20.68 x 0.405 ~ (1 - 0.065) = 21.085 - 0.935 = 20.15 kg per kg of coal. Minimum air required per kg of coal 0. = 1160-348 H ~G}-sassen 6x 0.81 +os( 045 - 0.08 ) oa) = 9.4 + 1.2175 = 10.6175 kg air per kg of coal Therefore, percentage excess of air _ 20.15 ~ 10.6175 _ 9.5325 _ = Toe17s —* 100-45. erg % 100 = 89.8%, Example 8. (a) Calculate the amount of air required for theoretically complete combustion °f 100 kg of coal of the following composition, C = 82%, H = 6%, Oo = 4%, ash = 8%, moisture = 2%. Solution. Minimum 0, required for combustion of earbon = 82 x 2 = 218.7 kg Minimum O, required for combustion of Hy = 6x 2 = 48 kg ©, available with the coal = 4 kg414 FUELS. FURNACES AND REFRACToRipg 18.7 + 48 - 4 = 262.7 kg Hence, Net 0, required for combustion of coal . 100 . Air required for combustion of coal = 262.7 “93 = 1141 ke, Example 8. (b) Calculate amount of air required for theoretically complete combustion of 100 Nm of blast furnace gas of the following composition (by volume %) CO = 17, CO = 22.1, Hy = 4.9, No= 558, O, = 0.2. Solution. 0, required for combustion of CO = 1. 4 I Oy» required for combustion of Hy = 1/2 x 4.9 m® = 2.45 m O» available with the gas =0.2m* 3 Net 0» required for complete combustion = 11.05 + 2.45 - 0.2 = 13.3 m’ 100 Air required for complete combustion of gas = 13.3 x “97° /2.x 22.1 m3 = 11.05 m? = 63.4 m*. Air-Fuel Ratio ‘Air-fuel ratio can be easily found out either by volume for gaseou: solid/liquid fuels in the following manner : In case of example ‘8 a’ above, air/fuel ratio (by weight) 141 _ og 7 TAL keke In case of example ‘8 b’ above, air/fuel ratio (by volume) 63. 3/Nm? T00 = 0,634 Nm*/Nm’ Nm* means volume of gas in m’ at NTP (i.e. 0°C and 760 mm Hg). Example 8. (c) For calculation of volume of products of combustion, consider 100 Nm* of same (B.F. gas) fuel as mentioned in example &(b). is fuels or by weight for Solution. Constituen Ozygen required for Products of combustion, Nm? ts | Volume % | complete combustion, Nm* CO H20 N. fe C02 7 7 = = 2 02 (0.2) - i co 22.1 11.05 22.1 - - He 49 2.45 ~ 49 Ne 55.8 i - 55.8 Total 100 13.3 39.1 49 55.8 sn eesiee trogen ofthe fuel, nitrogen of the air required for combustion will also be preseat in the products of com vustion volume of air required for combustion, as calculated in example Volume of N, in the air required for combustion = 63.4 x 29. = 50.1 Nm? 100 : Now, we have following products of combustion (for 100 Nm‘ of B.F. gas)LATIONS rcatc 415 a [ Consttents Volume, Nm® “% by volume | C02 ae 26.1 i H20 49 31 N (55.8 + 50.1) = 105.9 108 Total 149.9 100 of products of combustion and the fuel gas (by volume) patio 7 100 Nm’ of B.F. gas of the above mentioned composition will require 63.4 m’ of air Beil complete combustion and will produce 149.9 m° of wet flue gases on combus- for 5 qd ot volume and composition of dry flue gas, we may neglect H,0 from the flue gases sth Constituents Volume, Nm* % by volume C02 39.1 27 Ne 105.9 73 Total 145.0 100 Iuess Air for Combustion of Fuel : Complete combustion is not attained in practice unless more than the amount of air teretically required is used. The need of excess air is due to the difficulty of obtaining intimate cna between the air and the fuel. It is also partly due to the need to complete the combustion wihin the combustion space. 7 For maximum efficiency, it is essential that correct percentage of excess air should be used. _itmuch ar is used fuel will be wasted because the extra air will have to be heated and ‘ition heat will thus pass out of the system as sensible heat in the chimney gases. If too mma used combustion will not be complete and the chimney gases will carry away unused etal heat in the form of unburnt combustible gases, such as carbon monoxide (CO), : na (H,) and methane (CH,). 's the coefficient of excess air is defined as = ‘Actual air used for combustion _ Re per = Theoretically required Air for complete combustion oe Ute tee of excess air = (E - 1) x 100 ee Air on Products of Combustion : “bremabmng 22d composition of products of combustion can be calculated, wh Aaa ae as shown in the following example. ‘We fing & (4) Assuming 10% excess air for combustion of B.F. gas as C ‘efficient of excess air, ‘E en using excess in example 8(b) and tion. Total — : Volume Volume of air required = 1.1 x 63.4 = 69.7 Nm : Bithie = 69.7 -63.4=6.3Nm° ha "Ne of products of combustion a = volume of products of combustion with theoretical peiceee pememiairi 100) 08 eaeFUELS, FURNA( CES Ay AND ay TR, 416 7 1 69 = 1.32 Nm? Qp in excess air = 700 498 in? =6.3- = Amount (Nm) per 100 m a Constituents [Or of fuel wos _|__ "7 *elume 39.1 —_| 3.15 | (105.9 + 4.98) = 110.88 n 1.32 0.85 Total 156.2 im | is by Weight : £ Volumetric Analysis to Analysi i is known it can be converted to a ir the volumetric analysis of any gas is know! eee snus walume of each constituent by its own molecular weight, This vill multnyng ght ofthe constituents, then by additing up these weights and dvd Pr the total, the analysis by weight is obtained. ea (e) A typical example is shown below by considering the BLP. gas of thy son Conversion of Example 8. composition as in example 8 (b). Solution. Constituents % Vol. A ata re io CO: V7 4 748.0 25400 On 02 32 64 0.218 co 22.1 28 618.0 21.000 i “s 7 98 0.332 os 558 28 1560.0 53.05 aa 190 2942.2 100% Weight of Flue Gas per kg of Fuel Burned : The i i aca weight of Ove gan will be more than that of the fuel on account of the air supplied. anes with the weight of carbon a te het ey comparing the weight of carton 9 combustion. e fuel, as there is no loss of carbon during the prow In the case of a steam boil " pit. This should be subtracted from theese ome ua rmed earbon inthe cinders in the el. hydrogen int naa Caving the fire gate will contain steam, due to the combination oUt analysed is, therefore dry peer tis steam will condense asthe gases coo) The ue sont and donot contain the lor il Be seen thatthe fue gases contain theese airsupple Weight of carbon in the eae a BY combustion Neither of these condition Wi Hence, by carbon balance, : Weight of flue gases per kg of fuel burned = —Weight of carbon in 1 kg fuel Weight of carbon in I ke et ategasons x 0 considering the same B.F. gas in Example ‘Be! and ‘Be' the wei 25.4 coe 12) 21 arbor in CO2 100 "44 © 28” 109 = 9.0694 + 0.09 = (Since COs = 25.4% and CO = 9 1 the weight of carbon in the flue gases ee i i iciames joo Now mple Be into Weight analysis as below neceasar jo? 3 0 v1 1594 b ¥ to convert th Volume % Molecular weight 27 “4 73 28 100 Proport 37.9 12 _ earbon in 1 kg of flue gases = 99 x 43 = 0.1035 kg to wail of flue gases per kg of fuel burned = coweight ae sds, ratio of flue gases to fuel (by weight) = 1.54 wpotetYatue of a Mixture of Gases alo’ Me value of a mixture of gases can be found out, if we know the calorific values o Calor constituents. ai lorie value of some individual combustible substances are given below in Keal/Nm° Nace —jon monowide (CO) = 3040 | Hydrogen (Ha) = 2590 ehane (CH) = 8560 Unsaturated hydrocarbons CmHn = 17000 keal/Nin? Trample 8.(g) Considering the B.F. gas of the composition as in example ‘8b we find Solution. Combustible constituents Volume % CLV. of gas, keal/Nm*__| co. 22.1 3040 Ho 49 2590 Hence, C.V. of B.F. gas = 0.221 x 3040 + 0.049 x 2590 = 800 kcal/Nm* Density of a Mixture of Gases : Density of a mixture of gases can also be found out from the density of its fundamental ‘ustituents. Densities of some gases are given below : Srsttuents Volume % Density, kg/Nm>__| Remarks | 00; 44 1 Kg mol of any gas at ‘ " mah 22.4 Nm*. Hence Densit by, o 32 c ay =143 °8 22. > Molecular w Ms 28 4 22.1 Bes * 49 20,0894 Ra” N 28 —=1.25 | 55.8 Biol 4FUELS, FURNACES ES AND g EPR A. 418 example (8b) Ty Hi density of B.F. a8 of composition in ¢ xamp! < lence densi 25 x 55.8 « 1.25 197 pox 1age2a.t x 125 +49 0.0894 + 55.8 125)=13y, =r 197+ Nn Gases 5 Mixture of Specific Heat (C,) of @ pane constant pressure can also be calculated a Specific heat of a mixture Of RAT cific heat can be found out either on yoj, Pha, its constituents 46¢ or Keal/Kg °C respectively. © baa, fic heat of : cate and their units are Keal/m i sure between 0°C to 500°C (on volume basig Cy.mean specific heat a OF sey nstant press gases is tabulated below —— Constituents us o 0.334 Na 0.319 co 0.321 He 0.312 Considering a flue gas of following composition (by volume%) CO2 25 On 0.85 Na 7 kcal G, of dry flue gases = is (25 0.48 + 0.85% 0.834 + 71x 0.319) = 0.38 Heat Loss in Flue Gases : Total heat loss in the flue gases is the sum of heat carried away by dry flue gases and the heat carried away by steam in the flue gases caused by the combustion of the hydrogen or due ‘to moisture in the fuel. Heat carried away by dry flue gases per M* of fuel burned = V- C, : (tz ~ ty) Heat carried away by the steam formed per m° of fuel burned = W - (Hy ~ H.1) Hence, total heat loss in flue gases = V- C, (ta — ty) + W (Ago - hy). where, _V= Volume of dry flue gases per kg of fuel C, = Mean specific heat of dry flue gases ¢, = Temperature of flue gases leaving the furnace , = Ambient temperature W = Weight of H,O formed per M? of fuel burned Hs, = Total heat of super heated steam at temperature ty H,, = Sensible heat of water at temperature ¢,. Thus, the heat loss in the flu 4 substi ine errs sae ey ghee ae cy, Anmble 9.4 blast furnace gas has the following volumetrie analysis: Hy = 9% CO” 4° 2%, COy = 6%, Oy = 3% and Nz = 56%. Determine the ultimate gravimetric ana!)* aancaL CULATIONS us ao oq, Ultimate analysis is given i nal compounds. Thus in this ¢ onstituent elements and not the con analysis will be expressed in terms Proportional weight 7 = Ax molecular weight of element Ha _O oan 9x2=18 y 24x 12 = 288 S raxi6~a04 | axes = 6x32=192 | 3xs2-96 | - = | 56x28 - 1568 | 384 26 oz | _ 1568 Gravimetric percentage | Constituent elements | Proportional weight, W Ww | composition D = $y, 100 | We 384 | c 384 = 3650 «100-1445 | 26 6 7 3285x100 - 0.98 | 672 ° $22 x 100 = 25.35 1568 7 HBB x 100 = 58.22 Total 2650 100 j Example 10. A coal with a calorific value of 7100 K cal/ Kg has a composition by weight % C= 0.78, He = 0.05, Oz = 0.08, 'S = 0.02 and N = 0.02, remainder is ash. It is burnt ina furnace woth 50% excess air, The flue gases enter the chimney at "325°C and the atmospheric temperature eC Caleulate the proportion of heat carried away by the flue gases. Assume perfect combus- tion C, for air = 0.24, Cy for dry products of combustion = 0.25. Heat carried away per kg of moisture in flue gases is 700 K cal. Solution, Since complete combustion is assumed, all th H,0 and $ to SO. ‘Therefore, for 1 kg of coal, we get 0.78 # = 286 kg CO, (C burning to CO») 0.05 x 9 = 0.45 kg HzO (Hz burning to H,0) 0.02 x 2 = 0.04 kg SO» (S burning to S02) ‘ad thus oxygen required is given by 0.78 x 2 = 9,078 kg 0» (C burning to C2) 0.05 x 8 = 0.4 kg Op (Hz burning to H,0) 0.02 x 1 = 0.02 kg Op (S burning 0 $0.) _ Total = 2.498 kg O2 eC burns to CO, and all the H. toACT ey complete combustion but oy le Ons . quired per kg coal for Therefore, 2.498 kg oxygen is Fe ne _ 9.498 ~ 0.08 = 2.418 ke per kg of coal iy Therefore, O2 and, with 50% excess ait ! 0, actually supplied Oss This gives No actually supplied = 3.627% 93 influegas = 3-627 ~ 2,418 = 1.209 kg per kg coal. pen fue gas = 12.14 + 0.02 12.16 kg per kg of coal. i ‘ by dry products (CO2 +02) 2,86 + 0.04) x 0.25 x (375 — 15) = 45 x 700 = 315 kcal/kg of coal. 16 + 1.209) x 0.24 x (875 ~ 15) = 1115 katte op (assuming ¢, eal >= 0.24) to be supplied x 2.418 = 3.627 kg per kg of coal. 21.5 4 kg per kg of coal Total free ‘Total free nitrogen ae Heat carried away 252 keal eg of oy Heat carried away by steam Heat carried away by Op and No = of heat carried away per kg of coal 952 +315 + 1115 = 1682 kealiper kg of coal Total quantity Example 11. The flue gas from an industrial furnace have the following composition by volume : COq = 11.73%, CO = 0.2%, No = 0.09%, O2 = 6.81% and Nz = 81.17% Calculate the percentage excess air employed in the combustion, if the loss of carbon, j clinker and ash is 1% of the fuel used and the fuel has the following composition by weight: - C= 74%, Hy = 5%, Op = 5%, Nz = 1%, S = 1%, HzO = 9% and ash = 5%. Solution. Basis : 100 kg of the fuel charged to the industrial furnace Reactions : C+0)=CO, 1 H+} 0,=H,0 $+0,=80, Oxygen balance : Oxygen required for complete combustion = 2445 1 combustion = 75+ 5x 0.5 + = 7.447 kg mole Oxygen already present i 5 ly present in fuel = B= 0.157 kg mole Net oxygen demand from air = 1.447 - 0.157 = 7.29 kg mol: 447 — 0.157 = 7.29 kg mole. Carbon balance: Carbon lost in clinker and ash = 1 kg Carbon burnt = 74 - 1=73 kg = 73/12 = 6.08 kj Let x kg mols of flue gas are formed, eaten Therefore, by carbon balance, (0.1173 + 0.002) x = 6.8 *. x= 50.96 kg mole.0 cALCuLATIONS 421 qoe flue 628 Me m fuel a rom oir on as analysis, 0.96 x 0.8117 = 41.36 ke mole kg = 0.036 kg mole = 41.360 ~ 0.036 = 41.324 kg mole. = 21 - eupplied from air = 41.324 75 = 10.98 kg mole ox! - ~1.29 = — = 10.98 - 7.29 5300 go excess ai used = percentage excess oxygen used n 69 kg mole perce _Excess 3. . _ 3.69 theoretical * 1°°= 7.99 100 = 50.62, ple 12, Octane is burnt with 10% excess air. Calculate : m0] . ere fuel ratio by weight @ ir /fuel ratio by volume. wat int of dry exhaust gas formed per unit of fuel. @ _ of axygen in the exhaust gas per unit weight of fuel. @ a of water vapour in exhaust gas per unit weight of fuel. inalune of exhaust gas at I atmosphere and 260°C per unit weight of fuel. ‘The specific, gravity of octane may be taken as 0.7. Solution. Basis : 1 kg mole of octane burnt. Reaction :Cslyg+12.5 O,— 8CO, +9H,0 (@) Theoretical oxygen demand = 12.5 kg mole Oxygen supplied by 10% excess air = 12.5 x 1.1 = 13.75 kg mols = 13.75 x 32 = 440 kg Nitrogen supplied by air = 13.75 x 2 1.73 kg mole = 1448.4 kg Amount of air supplied = 13.75 +51.79 = 65.48 kg mole = 1888.4 kg Molecular weight of fuel =114 weight of air _ 188 weight of fuel” 114 (6) Specific gravity of octane =07 Density of octane = 0.7 gm/ce = 700 kg/m® ; 14 3 .. Volume of fuel = 799 7 0-163 m Assuming ideal gas behaviour, volume of air at N.T.P. = 65.48 x 22.4 = 1466.75 m® = 1466.75 Nm* Volume of air _ 1466.75 Volume of fuel” 0.163 (¢) Excess 0, = supplied O, — used Og = 13.75 — 12.50 = 1.25 kg mole = 8998.5422 ysis Dryfluegasanalysis _— ‘Amounts ‘ig mole 8.00 Constituents _— —~C02 a © 51.73 ae a8 1840. 4 esha 8 Weight of ary exhaust B25 = 1 Wee Weight of fuel ote ofp in the exhaust B85 _ 5 = Ms Weight of fuel sm ob Mole of water vapour in the exhaust gas . 90 = 9.079 (ee Weight of fuel al (f) Moles of exhaust gas (wet) = ° ae ae — ying ideal gas law, volume Applying 7 onRE., 6298 0.08206 250 +27) «5060 a an Volume of exhaust gas (wet) _ 3060:8 _9¢ 95 Weight of fuel 114 13, A producer gas with the ‘composition by volume, CO = 27.3%, CO, 36.7% is burnt with 20% excess air. If the combustion is 98% complete, caleul Example 020.6%, No= Si composition by volume of the flue gases. Solution. . Basis : 100 kg mole of producer gas burnt. Oxygen balance : Oy; required for CO combustion =27.8 x 0.5= 18.65 kg mole 0, persent in fuel =0.6 kg mole. Net 0» required = 13,65 ~ 0.60 = 13.05 kg mole 0, supplied by 20% excess air = 13.05 x 1.20 = 15.66 kg mole 0, wedi 88% conbusin of CO=27.8 «0.5% 0.98 = 13.96 kg mole excess = 15.66 ~ 13.38 = 2.28 ky Nitrogen balance : € mele a i air = 15.66 x = 58.91 kg mole 2 from producer gas = 66.7 kg mol Total Ny in flue gas =66.7 = 66.70 + 58.91 = Carbon dioxide balance : ean cae CO, fro co, from — aa roo re ustion of CO = Total CO, in flue gas a ze x 0.98 = 26.75 kg mole Carbon monoxide balance : 4+ 26.75 = 82.15 ke mole CO burnt = 26.75 kg mole CO left = 27.30 ~ 26.75 = 0.55 kg moleCULATIONS went 424 ft ysis naly An ‘eg mote_ Mol { 32.15 } ———Hete | 20.02 125.61 a Ta 22 a en 142 Total 160.59 eee a __ 100.00 - ‘A furnace is fired with a natural gas that con of h 4 s entirely of hydrocarb ple and sulphur compounds). The Orsi 'y of hydrocarbon + gases @ Orsat analysis of th See gte: in ECO, 8 of the flue gas gives 9.5% 0% 02M ye molar ratio of net hydrogen to carbon in the fuel ? ntage of excess air is being used ? 034 solu kg mol of dry flue gas. gasis sat analysis, Mole of Np = 100~ (9.5+18+2.0)= 867 nygen balance | ‘i = 21 supplied by a = 86.7 x 75 = 23.05 kg mol 1 0: reported in fuel gas (dry) =9.5+ 2% 1.8 +2.0=12.4 kg mole 0, unaccounted = 23.05 - 12.4 = 10.65 kg mole = 0, reacted with Hy, 0.65 x 2= 21.3 (o)Moles of hydrogen reacted .5 + 1.8 =11.3 kg atom. Amount of carbon Moles of H Atoms of C 11.3 (b) Moles of O, required for complete combustion =Mole required for Hy + Mole required for C = 10.65 + 11,3 = 21.95 Amount of excess O = 23.05 - 21.95 = 1.1kg mole = 1.885 excess air = % excess Oy « ae x 100 = 5%. __Example 15. The exhaust gas from a hydrocarbon fuel oil fired furnace, show 10.2% COs, '% 0,and 81.9% Nz by Orsat analysis. Calculate (i) % excess air used, and (ii) kg of dry air ‘plied per kg of oil burnt in the engine. Solution. Basis : 100 kg mole of dry flue gas. Ongen balance : vin flue gas = 81.9 kg mole norte from air =819x e = 21.77 kg mole eed influegas = 10.2 (as CO,) +79 (as 0,) = 18.1 kg mole use a = 21.77 18.1 = 3.67 kg mole = 0, used for Hy. Lees in fx, 0.2 + 3.67 = 13.87 kg mole = 7.9 kg moleFUELS. FURNACES q; ND Rj FRc, x 100 = 56.96% my aa OG _ excess O2 = 73.87 4 27.34 x 2 = 14.68 kg 3.67 kg mo 22.4 ke ole (b) 0, used for Hp AG rament of pin fuel = 3097 atom Carbon in fuel Total weight of hydrocarbon fuel oil = 10.2% 12 veignt of H+ Weigh of C14 68 + 122.4 = 137.08 kg vere 0 x32 + Mole of Nox 28 Fg + 819 x 28 = 2989-8 Ke 8 _ 91.81. TOF jas analysis and air/fuel ratio by weight whey nH he open, oe, sulphur, 0.4% oxygen and 0.1% her eg re me complete combusti0n. van mt dium fuel oil but are calculated with the help of combustion, equa. Weight of ait kg dry air _ kg oil burnt Example 16. Determu h 84.9% carbon, evyurnt with 20% excess ai. Solution. Basis 100 kg of met Oxygen balance Amount of oxygen. used tions for C, S and Hy fuel oil wit 32 0, required for 0 =84.9 x 75 = 226-4 ke 0» required for S= 3.2kg 0, required for H 91.2kg Total 0, required from air = 320.8 - 0.4 = 320.4 kg Oy excess = 384.48 — 320.4 = 64.08 kg 84.48 kg, Weight of air _ 1671.65 1671.65, Weight far = = Ke Weight offuel~ 100 171% 0, present in fuel = 0.4 keg, 0p supplied = 320.4 x 1.2 1 Air supplied = 384.48 xp Flue Gas Analysis (Wet) : _ 1 Amount of Np = 1671.65 x jh; = 1287.17 kg = 45.970 kg mole oa, ee produced = "x 44 = 311.3 kg= 7.075 kg mole $0, prod =32x 4. produced =3.2x 84 =6.4kg=0.10 kg mole H,0 produced = 42-4. 18 = 20 produc $4 x 18 = 1026 kg= 5.70 kg mole o: excess = 64.08 kg = 2.003 kg mole. lue gas analysis is given below : : Constituen 2 ts Amount in kg mole Mole % = Volume % & 1075 11.63 on 0.100 0.16 a 2.003 3.29 ho 45.970 75.55 fe 5.700 937 60.848 100qn rubATIONS T10N * poiler is fired using 200 hy the o pie 172 eure and 20°C. The dry analyar pperie Pre re and 300°C is CO, — 12%, 0, 1 rT polumetric flow rate of the pas 100 kx mole of dry flue ag 0 pure saturated ‘aturated hydrocarbon gan * Of the flue gas wh eae hand N, a leaves the + Estemate ti rat formula at ost Py ane fel in Basis | got palanee # vate £48 = 85 kg mole. 21_ 44. pyair = 85X79 = 22.59 ke male , 8 CO, +O, 8 0, = 1243.0 = 1 Dp reacted with Hy = 22.59 — 15,90 5 = 7.59 x 2= 15.18 kg mole = : reel fuel (hydrocarbon) gas : porte carbon = 12kg atom An roe = 30.36 kg atom ‘amoun Atom of H _ 30.36 ‘Atom of C12 s more than 2.0, hence it is a paraffin of formula C,H, supplied 8 toa in fe BAS 5.0 kg mole 59 kg mole 15.18 x 2 = 30.36 kg mole ), repo 02 as the ratio i n___1_ _ In +2" 2.53 n=3.77=40 sore the fuel is n-butane (C,H). Volumetric flow rate : noun of fl = 200 3.45 ke mole/hr Assuming ideal gas law, volume at 20°C and 1 atmosphere is, RT _ 3.45 x 0.08206 x (273 v= Akt SAB 0.08206 2 (278 +20) 273 +20) _ 99.95 m®/hr. Example 18. A furnace is fired by a fuel gas having the volumetric composition H, = 52%, (CH= 30%, CO = 8%, Cm Hn = 3.6%, COn = 2%, Op = 0.4% and rest N», Using a certain quantity ofair in excess over stoichiometric, complete combustion of the gas is achieved, giving a dry waste osof 5m per m” of fuel burned, estimate : (a) Composition by volume of dry waste gas formed. (b) Per cent excess air used. (c) Weight of water formed per m’ of gas burned, neglecting water present in the air used For purpose of calculation, unsaturated hydrocarbon (C,,H,) can be considered as CsHs Solution. Basis 100 kg mole of fuel burnt. Reactions : CO+30, = CO, CH,+20, = CO, +2H,0 C3Hg+4.5 0. = 3CO)+3H,0 Hy +} 0, = H,0