LAMINAR PREMIXED LAMINAR PREMIXED FLAMESFLAMESOVERVIEWOVERVIEWApplications: Applications: Heating appliancesHeating appliancesBunsen burnersBunsen burnersBurner for glass product manufacturingBurner for glass product manufacturingImportance of studying laminar premixed flames:Importance of studying laminar premixed flames:Some burners use this type of flames as shown by Some burners use this type of flames as shown by examples aboveexamples abovePrerequisite to the study of turbulent premixed Prerequisite to the study of turbulent premixed flames. Both have the same physical processes flames. Both have the same physical processes and many turbulent flame theories are based on and many turbulent flame theories are based on underlying laminar flame structure.underlying laminar flame structure.PHYSICAL DESCRIPTIONPHYSICAL DESCRIPTIONPhysical characteristicsPhysical characteristicsFigure 8.2 shows typical flame temperature profile, Figure 8.2 shows typical flame temperature profile, mole fraction of reactants,mole fraction of reactants,R R, and volumetric heat , and volumetric heat release, .release, .Velocity of reactants entering the flame, Velocity of reactants entering the flame, u u = flame = flame propagation velocity, Spropagation velocity, SL LProducts heated Products heated product density (product density (b b) < reactant ) < reactant density (density (u u). Continuity requires that burned gas ). Continuity requires that burned gas velicity, velicity, b b >= unburned gas vel., >= unburned gas vel., u uu u u u A = A = b b b b AA(8.1) (8.1) Q&For a typical hydrocarbon-air flame at PFor a typical hydrocarbon-air flame at Patm atm, , u u//b b 77 considerable acceleration of the considerable acceleration of the gas flow across the flame (gas flow across the flame (bb toto u u).).Laminar premixed flame structureA flame consists of A flame consists of 2 zones2 zones::Preheat zonePreheat zone, where little heat is released, where little heat is releasedReaction zoneReaction zone, where the bulk of chemical energy , where the bulk of chemical energy is releasedis releasedReaction zoneReaction zone consists of 2 regions:consists of 2 regions:Thin regionThin region (less than a millimeter)(less than a millimeter), where , where reactions are very fastreactions are very fast Wide regionWide region ((several millimeters)several millimeters), where , where reactions reactions are sloware slowInIn thin region (fast reaction zone), destruction of the fueldestruction of the fuel molecules and creation of many intermediate species occur.molecules and creation of many intermediate species occur. This region is dominated by bimolecular reactions toThis region is dominated by bimolecular reactions to produce CO produce CO.. Wide zone Wide zone (slow reaction zone) (slow reaction zone) is dominated by radicalis dominated by radical recombination reactions andrecombination reactions and final burnout of CO final burnout of CO viavia CO +CO + OHOH CO CO2 2 +H +HFlame coloursinFlame coloursin fast-reaction zone fast-reaction zone: :IfIf air > air > stoichiometric proportions,stoichiometric proportions, excited CH radicals excited CH radicals resultresult inin blue radiation blue radiation.. IfIf air < air < stoichiometric proportions, the zone appearsstoichiometric proportions, the zone appears blue- blue-green green as a result of radiation fromas a result of radiation from excited C excited C2 2. .In In both flame regionsboth flame regions, , OH radicalsOH radicals contribute to contribute to the visible radiationthe visible radiationIf the flame is If the flame is fuel-rich (much less air)fuel-rich (much less air), , sootsoot will will form, with its consequent blackbody continuum form, with its consequent blackbody continuum radiation. Although soot radiation has its radiation. Although soot radiation has its maximum maximum intensityintensity in the infraredin the infrared (recall Wiens law for (recall Wiens law for blackbody radiation), the spectral sensitivity of the blackbody radiation), the spectral sensitivity of the human eye causes us to see a human eye causes us to see a bright yellowbright yellow (near (near white) to dull white) to dull orange emissionorange emission, depending on the , depending on the flame temperatureflame temperatureFigure 1. Spectrum of flame coloursTypical Laboratory Premixed FlamesTypical Laboratory Premixed FlamesThe typical Bunsen-burner flame is a dual flame: The typical Bunsen-burner flame is a dual flame: a a fuel rich premixed inner flamefuel rich premixed inner flame surrounded by surrounded by a a diffusion flamediffusion flame. Figure 8.3 illustrates a Bunsen . Figure 8.3 illustrates a Bunsen burner.burner.The diffusion flame results when The diffusion flame results when CO and OHCO and OH from from the rich inner flame encounter the ambient air. the rich inner flame encounter the ambient air. The shape of the flame is determined by the The shape of the flame is determined by the combined effects of the velocity profile and heat combined effects of the velocity profile and heat losses to the tube wall. losses to the tube wall. Secondary diffusion flame results when CO and H products from rich inner flame encounter ambient air Fuel-rich pre-mixed inner flame For the flame to remain For the flame to remain stationarystationary, , SSL L = normal component of = normal component of u u = = u u sinsin (8.2)(8.2). Figure . Figure 8.3b illustrates vector diagram.8.3b illustrates vector diagram.Example 8.1Example 8.1. A premixed laminar flame is . A premixed laminar flame is stabilized in a one-dimensional gas flow where the stabilized in a one-dimensional gas flow where the vertical velocityvertical velocity of the unburned mixture, of the unburned mixture, u u, varies , varies linearly with the horizontal coordinate, x, as shown linearly with the horizontal coordinate, x, as shown in the lower half of Fig. 8.6. Determine the flame in the lower half of Fig. 8.6. Determine the flame shape and the distribution of the local angle of the shape and the distribution of the local angle of the flame surface from vertical. Assume the flame surface from vertical. Assume the flame flame speed Sspeed SL L is independent of position and equal to is independent of position and equal to 0.4m/s (constant)0.4m/s (constant), a nominal value for a , a nominal value for a stoichiometric methane-air flame. stoichiometric methane-air flame. SolutionSolutionFrom Fig. 8.7, we see that the local angle, From Fig. 8.7, we see that the local angle, , which , which the flame sheet makes with a vertical plane is (Eqn. the flame sheet makes with a vertical plane is (Eqn. 8.2)8.2) = arc sin (S= arc sin (SL L//u u), where, from Fig. 8.6,), where, from Fig. 8.6,u u (mm/s) = 800 + (1200 800)/20 x (mm) (mm/s) = 800 + (1200 800)/20 x (mm) (known).(known).u u (mm/s) = 800 + 20x.(mm/s) = 800 + 20x.So,So, = arc sin (400/(800 + 20x (mm))= arc sin (400/(800 + 20x (mm))and has values ranging from 30and has values ranging from 30o o at x = 0 to19.5at x = 0 to19.5o o at x at x = 20 mm, as shown in the top part of Fig. 8.6.= 20 mm, as shown in the top part of Fig. 8.6.To calculate the flame position, we first obtain an To calculate the flame position, we first obtain an expression for the local slope of the flame sheet expression for the local slope of the flame sheet (dz/dx) in the x-z plane, and then integrate this (dz/dx) in the x-z plane, and then integrate this expression with respect to x find z(x). From Fig. expression with respect to x find z(x). From Fig. 8.7, we see that8.7, we see that,which, for,which, for u u=A + Bx,=A + Bx, becomes becomes Integrating the above with A/SIntegrating the above with A/SL L = 2 and B/S= 2 and B/SL L ==0.05 yields0.05 yields-10 ln[(x-10 ln[(x2 2+80x+1200)+80x+1200)1/2 1/2+(x+40)]+(x+40)]-20-203+10 ln(203+10 ln(203+40)3+40)The flame position z(x) is plotted in upper half of The flame position z(x) is plotted in upper half of Fig. 8.6.Fig. 8.6.( )/ 111 11tanu LLx Sdzdx S _ ,/ 1111Ldz ABxdx S 1 _+ 1 1 , ]x. 1 111dz xz(x) dx (x x ) 11 1111 1dx 11 _ _ + + + , ,SIMPLIFIED ANALYSISSIMPLIFIED ANALYSISTurns (2000) proposes simplified laminar flameTurns (2000) proposes simplified laminar flamespeed and thickness on one-dimensional flame.speed and thickness on one-dimensional flame.Assumptions used:Assumptions used:One-dimensional, constant-areaOne-dimensional, constant-area, steady flow. One-, steady flow. One-dimensional flat flame is shown in Figure 8.5. dimensional flat flame is shown in Figure 8.5. Kinetic and potential energies, viscous shear work, Kinetic and potential energies, viscous shear work, and thermal radiation are all neglected.and thermal radiation are all neglected.The small pressure difference across the flame is The small pressure difference across the flame is neglected; thus, neglected; thus, pressure is constantpressure is constant..The diffusion of heat and mass are governed by The diffusion of heat and mass are governed by Fourier's and Fick's lawsFourier's and Fick's laws respectively (respectively (laminar laminar flowflow). ). Binary diffusion is assumed.Binary diffusion is assumed.The Lewis number, Le, which expresses the The Lewis number, Le, which expresses the ratio of thermal diffusivity, ratio of thermal diffusivity, , to mass diffusivity, , to mass diffusivity, D, i.e.,D, i.e.,is unityis unity, , pkLeD C D u pkCThe The Cp mixture Cp mixture f( f(temperature, composition).temperature, composition). This is equivalent to assuming that individual This is equivalent to assuming that individual species specific heats are all equal and species specific heats are all equal and constant.constant.Fuel and oxidizer form products in a single-step Fuel and oxidizer form products in a single-step exothermic reaction. Reaction isexothermic reaction. Reaction is1 kg fuel + 1 kg fuel + kg oxidiser kg oxidiser (( + 1)kg products+ 1)kg productsThe oxidizer is present in stoichiometric or The oxidizer is present in stoichiometric or excess proportions; thus excess proportions; thus fuel is completely fuel is completely consumed at the flame. consumed at the flame. For this simplified system, For this simplified system, SSL L and and found arefound are(8.20)(8.20)and and or or (8.21)(8.21)where is where is volumetric mass ratevolumetric mass rate of fuel and of fuel and is is thermal diffusivity. Temperature profile is assumed thermal diffusivity. Temperature profile is assumed linear from Tlinear from Tu u to Tto Tb b over the small distance, as over the small distance, as shown in Fig. 8.9.shown in Fig. 8.9.( )/ 111 1FLumS 1 + 1 ]&( )11uFm 1 1+ 1 ]&1LS Fm&FACTORS INFLUENCING FLAME SPEED FACTORS INFLUENCING FLAME SPEED ((SSL L) AND FLAME THICKNESS () AND FLAME THICKNESS ())1. Temperature (T1. Temperature (Tu u and Tand Tb b))Temperature dependencies of STemperature dependencies of SL L and and can be can be inferred from Eqns 8.20 and 8.21. Explicit inferred from Eqns 8.20 and 8.21. Explicit dependencies is proposed by Turns as followsdependencies is proposed by Turns as follows (8.27)(8.27)where where is thermal diffusivity, Tis thermal diffusivity, Tu u is unburned gas is unburned gas temperature, temperature, , T, Tb b is burned is burned gas temperature.gas temperature.. 111 1( )( )uu pk TT T PCT( ). 11b uT T T +

(8.28)(8.28)where the exponent n is the overall reaction order, where the exponent n is the overall reaction order, RRu u = universal gas constant (J/kmol-K), E= universal gas constant (J/kmol-K), EA A = = activation energy (J/kmol)activation energy (J/kmol)Combining above scalings yields and applying Eqs Combining above scalings yields and applying Eqs 8.20 and 8.218.20 and 8.21SSL L (8.29)(8.29) (8.30)(8.30)/F um & [ ]1. exp( /( )nn nub u A u bTF T P T E R TP . / ( )/ 1111 1 11exp1n nAu bu bET TT PRT _ ,. / / 1111 1 1exp1n nAbu bET T PRT _ ,For hydrocarbons, For hydrocarbons, n n 2 and E2 and EA A 1.67.101.67.108 8 J/kmol J/kmol (40 kcal/gmol). Eqn 8.29 predicts (40 kcal/gmol). Eqn 8.29 predicts SSL L to increase by to increase by factor of 3.64factor of 3.64 when Twhen Tu u is increased from 300 to is increased from 300 to 600K. Table 8.1 shows comparisons of S600K. Table 8.1 shows comparisons of SL L and and The empirical SThe empirical SL L correlation of Andrews and correlation of Andrews and Bradley [19] for stoichiometric methane-air flames,Bradley [19] for stoichiometric methane-air flames,SSL L (cm/s) = 10 + 3.71.10-4[T(cm/s) = 10 + 3.71.10-4[Tu u(K)](K)]2 2(8.31)(8.31)which is shown in Fig. 8.13, along with data from which is shown in Fig. 8.13, along with data from several experimenters. several experimenters. Using Eqn. 8.31, an increase in TUsing Eqn. 8.31, an increase in Tu u from 300 K to from 300 K to 600 K results in 600 K results in SSL L increasing by a factor of 3.3increasing by a factor of 3.3, , which compares quite favourably with our estimate which compares quite favourably with our estimate of 3.64 (Table 8.1).of 3.64 (Table 8.1).Table 8.1Table 8.1 Estimate of effects of TEstimate of effects of Tu u and Tand Tb b on Son SL L and and using Eq 8.29 and 8.30using Eq 8.29 and 8.30Case A: referenceCase A: referenceCase C: TCase C: Tb b changes due to heat transfer or changes due to heat transfer or changing equivalent ratio, either lean or rich.changing equivalent ratio, either lean or rich.Case B: TCase B: Tu u changes due to preheating fuelchanges due to preheating fuelCaseCaseA (ref)A (ref)BBCCTTu u (K)(K)300300600600300300TTb b (K)(K)2,0002,0002,3002,3001,7001,700SSL L/S/SL,A L,A113.643.640.460.46//A A110.650.651.951.95Pressure (P)Pressure (P)From Eq. 8.29, if, again, n From Eq. 8.29, if, again, n 2, S2, SL L f (P).f (P).Experimental measurements generally show a Experimental measurements generally show a negative dependence of pressure. Andrews and negative dependence of pressure. Andrews and Bradley [19] found thatBradley [19] found thatSSL L (cm/s) = 43[P (atm)](cm/s) = 43[P (atm)]-0.5 -0.5(8.32)(8.32)fits their data for P > 5 atm for methane-air flames fits their data for P > 5 atm for methane-air flames (Fig. 8.14). (Fig. 8.14). Equivalent Ratio (Equivalent Ratio ())Except for very rich mixtures, the primary effect of Except for very rich mixtures, the primary effect of on Son SL L for similar fuels is for similar fuels is a result of how this a result of how this parameter affects flame temperaturesparameter affects flame temperatures; thus, we ; thus, we would expect S would expect S L,max L,max at a slightly rich mixture and at a slightly rich mixture and fall off on either side as shown in Fig. 8.15 for fall off on either side as shown in Fig. 8.15 for behaviour of methane. behaviour of methane. Flame thickness (Flame thickness () shows the inverse trend, ) shows the inverse trend, having a minimum near stoichiometrichaving a minimum near stoichiometric (Fig. 8.16). (Fig. 8.16). Fuel TypeFuel TypeFig. 8.17 shows SFig. 8.17 shows SL L for Cfor C1 1-C-C6 6 paraffins (single paraffins (single bonds), olefins (double bonds), and acetylenes bonds), olefins (double bonds), and acetylenes (triple bonds). Also shown is H(triple bonds). Also shown is H2 2. S. SL L of Cof C3 3HH8 8 is used is used as a reference. as a reference. Roughly speaking the CRoughly speaking the C3 3-C-C6 6 hydrocarbons all hydrocarbons all follow the same trend as a function of flame follow the same trend as a function of flame temperature. Ctemperature. C2 2HH4 4 and Cand C2 2HH2 2 S SL L > the C> the C3 3-C-C6 6 group, group, while CHwhile CH4 4SSL L lies somewhat below. lies somewhat below. HH2 2's S's SL,max L,max is many times > that of Cis many times > that of C3 3HH8 8. Several . Several factors combine to give Hfactors combine to give H2 2 its high flame speed: its high flame speed: i.i.the thermal diffusivity (the thermal diffusivity ()of pure H)of pure H2 2 is many is many times > the hydrocarbon fuels; times > the hydrocarbon fuels; ii.ii.the mass diffusivity (the mass diffusivity (DD) of H) of H2 2 likewise is much > likewise is much > the hydrocarbons; the hydrocarbons; iii.iii.the reaction the reaction kinetics for Hkinetics for H2 2 are very rapidare very rapid since since the the relatively slow CO relatively slow CO COCO2 2 step that is a major step that is a major factor in hydrocarbon combustion is absent. factor in hydrocarbon combustion is absent. Law [20] presents a Law [20] presents a compilation of laminar compilation of laminar flame-speed data for flame-speed data for various pure fuels and various pure fuels and mixtures shown in mixtures shown in Table 8.2.Table 8.2.Table 8.2Table 8.2 SSL L for for various pure fuels various pure fuels burning in air for burning in air for = = 1.0 and at 1 atm1.0 and at 1 atmFuelFuelSSL L (cm/s)(cm/s)CHCH4 44040CC2 2HH2 2136136CC2 2HH4 46767CC2 2HH6 64343CC3 3HH88 4444HH2 2210210FLAME SPEED CORRELATIONS FLAME SPEED CORRELATIONS FOR SELECTED FUELSFOR SELECTED FUELSMetghalchi and Keck [11] experimentally Metghalchi and Keck [11] experimentally determined Sdetermined SL L for various fuel-air mixtures over a for various fuel-air mixtures over a range of temperatures and pressures typical of range of temperatures and pressures typical of conditions associated with reciprocating internal conditions associated with reciprocating internal combustion engines and gas turbine combustion engines and gas turbine combustors.combustors.Eqn 8.33 similar to Eqn. 8.29 is proposedEqn 8.33 similar to Eqn. 8.29 is proposedSSL L = S= SL,ref L,ref (1 2.1Y(1 2.1Ydil dil) (8.33)) (8.33)for Tfor Tu u >> 350 K. 350 K. , _ _

, ,uuref refT PT P The subscript ref refers to reference conditions The subscript ref refers to reference conditions defined bydefined byTTu,ref u,ref = 298 K, P= 298 K, Pref ref = 1 atm= 1 atm and and SSL,ref L,ref = B= BM M + B+ B2 2(( - - M M))22 (for reference conditions)(for reference conditions)where the constants where the constants BBM M, B, B2 2, and , and M M depend on fuel depend on fuel typetype and are given in Table 8.3. and are given in Table 8.3. Exponents of T and P, Exponents of T and P, and and are functions of are functions of , , expressed asexpressed as = 2.18 - 0.8(= 2.18 - 0.8( - 1) - 1) (for non-reference conditions)(for non-reference conditions) = -0. 16 + 0.22(= -0. 16 + 0.22( - 1) - 1) (for non-reference conditions) (for non-reference conditions)The term YThe term Ydil dil is the mass fraction of diluent present is the mass fraction of diluent present in the air-fuel mixture in Eqn. 8.33 to account for in the air-fuel mixture in Eqn. 8.33 to account for any recirculated combustion products. This is a any recirculated combustion products. This is a common technique used to control NOcommon technique used to control NOx x in many in many combustion systems combustion systems Table 8.3Table 8.3 Values for BValues for BM M, B, B2 2, and , and M M used in Eqn used in Eqn 8.33 [11]8.33 [11]Fuel M BM (cm/s) B2 (cm/s)MethanolMethanol1.111.1136.9236.92-140.51-140.51PropanePropane1.081.0834.2234.22-138.65-138.65Iso octaneIso octane1.131.1326.3226.32-84.72 -84.72 RMFD-303RMFD-3031.131.1327.5827.58-78.54-78.54Example 8.3Example 8.3Compare the laminar flame speeds of gasoline-airCompare the laminar flame speeds of gasoline-airmixtures with mixtures with = 0.8 for the following three cases:= 0.8 for the following three cases:i.i.At ref conditions of At ref conditions of T = 298 K and P = 1 atmT = 298 K and P = 1 atmii.ii.At conditions typical of a spark-ignition engine At conditions typical of a spark-ignition engine operating at wide-open throttle: operating at wide-open throttle: T = 685 K and P T = 685 K and P = 18.38 atm= 18.38 atm..iii.iii.Same as condition ii above, but with Same as condition ii above, but with 15 percent 15 percent (by mass) exhaust-gas recirculation(by mass) exhaust-gas recirculation SolutionSolutionRMFD-303RMFD-303 research fuel has a controlled research fuel has a controlled composition simulating composition simulating typical gasolinestypical gasolines. The . The flame speed at 298 K and 1 atm is given byflame speed at 298 K and 1 atm is given bySSL,ref L,ref = B= BM M + B+ B2 2(( - - M M))2 2From Table 8.3,From Table 8.3,BBM M = 27.58 cm/s, B= 27.58 cm/s, B2 2 = -78.38cm/s, = -78.38cm/s, M M = 1. 13.= 1. 13.SSL,ref L,ref = 27.58 - 78.34(6.8 - 1.13)= 27.58 - 78.34(6.8 - 1.13)2 2 = = 19.05 cm/s19.05 cm/sTo find the flame speed at TTo find the flame speed at Tu u and P other than the and P other than the reference state, we employ Eqn. 8.33reference state, we employ Eqn. 8.33SSL L(T(Tu u, P) = S, P) = SL,refL,ref ,uu ref refT PT P _ _ , ,wherewhere = 2.18-0.8(= 2.18-0.8(-1) = 2.34-1) = 2.34 = -0.16+0.22(= -0.16+0.22(-1) = -1) = -- 0.2040.204Thus,Thus,SSL L(685 K, 18.38 atm) = (685 K, 18.38 atm) = 19.05 (685/298)19.05 (685/298)2.34 2.34(18.38/1)(18.38/1)-0.204 -0.204 ==73.8cm/s73.8cm/sWith dilution by exhaust-gas recirculation, the With dilution by exhaust-gas recirculation, the flame speed is reduced by factor (1-2.1 Yflame speed is reduced by factor (1-2.1 Ydil dil):):SSL L(685 K, 18.38 atm, 15%EGR) = (685 K, 18.38 atm, 15%EGR) = 73.8cm/s[1-2.1(0.15)]= 73.8cm/s[1-2.1(0.15)]= 50.6 cm/s50.6 cm/sQUENCHING, FLAMMABILITY, QUENCHING, FLAMMABILITY, AND IGNITIONAND IGNITIONPreviously Previously steady propagationsteady propagation of premixed of premixed laminar flameslaminar flamesNow Now transient processtransient process: quenching and ignition. : quenching and ignition. Attention to quenching distance, flammability Attention to quenching distance, flammability limits, and minimum ignition energies with heat limits, and minimum ignition energies with heat losses controlling the phenomena.losses controlling the phenomena.1. Quenching by a Cold Wall1. Quenching by a Cold WallFlames extinguish upon entering a sufficiently Flames extinguish upon entering a sufficiently small passageway. If the passageway is not too small passageway. If the passageway is not too small, the flame will propagate through it. small, the flame will propagate through it. The The critical diameter of a circular tubecritical diameter of a circular tube where a flame where a flame extinguishes rather than propagates, is referred to extinguishes rather than propagates, is referred to as the as the quenching distancequenching distance. . Experimental quenching distances are determined Experimental quenching distances are determined by observing whether a flame stabilised above a by observing whether a flame stabilised above a tube does or does not tube does or does not flashbackflashback for a particular for a particular tube diameter when the tube diameter when the reactant flow is rapidly reactant flow is rapidly shut off. shut off. Quenching distances are also determined using Quenching distances are also determined using high-aspect-ratio rectangular-slothigh-aspect-ratio rectangular-slot burners. In this burners. In this case, the quenching distance between the long case, the quenching distance between the long sides, i.e., the slit width. sides, i.e., the slit width. Tube-based quenching distances are somewhat Tube-based quenching distances are somewhat larger (larger (20-50 percent) than slit-based ones [21]20-50 percent) than slit-based ones [21]Ignition and Quenching CriteriaIgnition and Quenching CriteriaWilliams [22] provides 2 rules-of-thumb governingWilliams [22] provides 2 rules-of-thumb governingignitionignition and and flame extinctionflame extinction. . Criterion 1 -Criterion 1 -IgnitionIgnition will only occur if will only occur if enough enough energyenergy is added to heat a slab is added to heat a slab thickness steadily thickness steadily propagating laminar flamepropagating laminar flame to the to the adiabatic flame adiabatic flame temperaturetemperature..Criterion 2 -The rate of Criterion 2 -The rate of liberation of heat by liberation of heat by chemical reactionschemical reactions inside the slab must inside the slab must approximately balance the rate of heat lossapproximately balance the rate of heat loss from from the slab by thermal conduction. This is applicable the slab by thermal conduction. This is applicable to the problem of to the problem of flame quenchingflame quenching by a cold wall.by a cold wall.Simplified Quenching Analysis.Simplified Quenching Analysis. ConsiderConsider a flame that has just entered a slot a flame that has just entered a slot formed by two plane-parallel plates as shown in formed by two plane-parallel plates as shown in Fig. 8.18. Applying Williams second criterion: Fig. 8.18. Applying Williams second criterion: heat produced by reaction = heat conductionheat produced by reaction = heat conduction to to the walls, i.e.,the walls, i.e.,(8.34)(8.34)is volumetric heat release rateis volumetric heat release rate(8.35)(8.35)where where is volumetric mass rate of fuel, is volumetric mass rate of fuel, is heat of combustion is heat of combustion Thickness of the slab of gas analysed = Thickness of the slab of gas analysed = . . , cond totQ VQ& &F cQ m h &&Fm &ch Q&FindFind quenching distance, d.quenching distance, d.SolutionSolution(8.36)(8.36)A = 2A = 2L, where L is slot width (L, where L is slot width ( paper) and 2 paper) and 2 accounts for contact on both sidesaccounts for contact on both sides (left and (left and right).right).is difficult to approximate. A is difficult to approximate. A reasonable lower bound ofreasonable lower bound of ==(8.37)(8.37)where b = 2, assuming a linear distribution of T where b = 2, assuming a linear distribution of T from the centerline plane at Tfrom the centerline plane at Tb b to the wall at Tto the wall at Tw w. . In general b > 2.In general b > 2.Quenching occurs from Quenching occurs from TTb b to Tto Tw w..dTdxcond in gas walldTQ kAdx &dTdx( )/b wT Td bCombining Eqns 8.35-8.37, Combining Eqns 8.35-8.37,(8.38a) (8.38a)or or(8.38b) (8.38b)Assuming T Assuming Tw w = T = Tu u, using Eqn 8.20 (about, using Eqn 8.20 (about S SL L), and relating ), and relating

. . , Eqn 8.38b becomes , Eqn 8.38b becomesd d = 2 = 2bb / /S SL L(8.39a) (8.39a)Relating Eqn 8.21 (aboutRelating Eqn 8.21 (about ) ), Eqn 8.39a becomes , Eqn 8.39a becomesd d = 2 = 2bb Because bBecause b 2, value d is always >2, value d is always > . Values of d for fuels. Values of d for fuels are shown Table 8.4.are shown Table 8.4. ( )( )) ( ) 1/b wF cT Tm h dL k Ld b &( )11b wF ckb T Tdm h &( ) ( ) 1c p b uh c T T + Table 8.4 Flammability limits, quenching distances Table 8.4 Flammability limits, quenching distances and minimum ignition energiesand minimum ignition energiesFlammability limit Flammability limitQuenching distance, d Quenching distance, dmin minmax max Stoich-massStoich-mass air-fuelair-fuel ratio ratio ForFor =1=1AbsoluteAbsolute min, mm min, mm C C2 2H H2 20.19 0.19 13.3 13.3 2.3 2.3 - -CO CO 0.34 0.34 6.76 6.76 2.46 2.46 - - - -C C10 10H H22 220.36 0.36 3.92 3.92 15.0 15.0 2.1 2.1 - -C C2 2H H6 60.50 0.50 2.72 2.72 16.0 16.0 2.3 2.3 1.8 1.8C C2 2H H4 40.41 0.41 > 6.1 > 6.1 14.8 14.8 1.3 1.3 - -H H2 20.14 0.14 2.54 2.54 34.5 34.5 0.64 0.64 0.61 0.61CH CH4 40.46 0.46 1.64 1.64 17.2 17.2 2.5 2.5 2.0 2.0CH CH3 3OH OH 0.48 0.48 4.08 4.08 6.46 6.46 1.8 1.8 1.5 1.5C C8 8H H18 180.51 0.51 4.25 4.25 15.1 15.1 - - - -C C3 3H H8 80.51 0.51 2.83 2.83 15.6 15.6 2.0 2.0 1.8 1.8Fuel FuelMinimum ignition energy Minimum ignition energy ForFor =1 (10 =1 (10-5 -5 J)J)AbsoluteAbsolute minimum (10 minimum (10-5 -5 J)J) C C2 2H H2 23 3 - -CO CO - - - -C C10 10H H22 22- - - -C C2 2H H6 642 42 24 24C C2 2H H4 49.6 9.6 - -H H2 22.0 2.0 1.8 1.8CH CH4 433 33 29 29CH CH3 3OH OH 21.5 21.5 14 14C C8 8H H18 18- - - -C C3 3H H8 830.5 30.5 26 26Example 8.4Example 8.4. . Consider the design of a laminar-flow, adiabatic, Consider the design of a laminar-flow, adiabatic, flat-flame burner consisting of a square flat-flame burner consisting of a square arrangement of thin-walled tubes as illustrated in arrangement of thin-walled tubes as illustrated in the sketch below. the sketch below. Fuel-air mixture flows through both the tubes and Fuel-air mixture flows through both the tubes and the interstices between the tubesthe interstices between the tubes. . It is desired operate the burner with a It is desired operate the burner with a stoichiometric methane-air mixture exiting the stoichiometric methane-air mixture exiting the tubes at tubes at 300 K300 K and and 5 atm5 atm Determine the Determine the mixture mass flowrate per unit mixture mass flowrate per unit cross-sectional areacross-sectional area at the design condition.at the design condition.Estimate the Estimate the maximum tube diametermaximum tube diameter allowed so allowed so that flashback will be prevented.that flashback will be prevented.Solution SolutionTo establish a flat flame, the mean flow velocity must equalTo establish a flat flame, the mean flow velocity must equal the laminar flame at the design temperature and pressure.the laminar flame at the design temperature and pressure. From Fig. 8.14 From Fig. 8.14, ,S SL L (300K, 5atm) = 43/ (300K, 5atm) = 43/P (atm) = 43/ P (atm) = 43/5 =5 = 19.2 19.2cm/s. cm/s.The mass flux, ,The mass flux, ,is is= = == u uu u == u uS SL LAssuming an ideal-gas mixture, where Assuming an ideal-gas mixture, whereMW MWmix mix == CH4 CH4MW MWCH4 CH4 + (1 -+ (1 - CH4 CH4)MW )MWair air= 0.095(16.04) + 0.905(28.85) = 0.095(16.04) + 0.905(28.85)= 27.6 kg/kmol =5.61kg/m = 27.6 kg/kmol =5.61kg/m3 3(Stoichimetric mass ratio air/ methane = 17.2, see(Stoichimetric mass ratio air/ methane = 17.2, see Table 8.4 Table 8.4) )Thus, the mass flux is Thus, the mass flux is == u uS SL L = 5.61(0.192)== 5.61(0.192)= 1.081.08 kg/(s.m kg/(s.m2 2) )m &/ m A&m &m &We assume that if the tube diameter < the quench We assume that if the tube diameter < the quench distance (d), with some factor-of-safety applied, distance (d), with some factor-of-safety applied, the burner will operate without danger of the burner will operate without danger of flashback. flashback. Thus, we need to find the quench distance at the Thus, we need to find the quench distance at the design conditions. design conditions. Fig. 8.16Fig. 8.16 shows that dshows that dslit slit 1.7 mm. Since d1.7 mm. Since dslit slit = d= dtube tube (20-50%), use d (20-50%), use dslit slit outrightoutright (our case) with factor (our case) with factor of safety 20-50%. Data in Fig 8.16 is for slit, of safety 20-50%. Data in Fig 8.16 is for slit, design is of tube.design is of tube.Correction for 5 atm:Correction for 5 atm:Eqn. 8.39a,Eqn. 8.39a,d d /S/SL LEqn 8.27, Eqn 8.27, TT1.75 1.75/P/Pdd2 2 = = d(5atm) =1.7mm.d(5atm) =1.7mm.dddesign design 0.76 mm0.76 mmCheck whether d=0.76 mm gives laminar flow (ReCheck whether d=0.76 mm gives laminar flow (Red d < < 23002300). ). Flow is still laminarFlow is still laminar, , 1 11 11 1, , 1 1 1 1L LL LS SPd dS P S1. ( . )( . ) 1111111111111Re . 111. . 111111u design Ldd S / 1 11. / 1 111atm cm satm cm s2. Flammability Limits2. Flammability LimitsA flame will propagate only within a range of A flame will propagate only within a range of mixture the so-called lower and upper limits of mixture the so-called lower and upper limits of flammability. The limit is the leanest mixture (flammability. The limit is the leanest mixture ( < < 1), while the upper limit represents the richest 1), while the upper limit represents the richest mixture (mixture ( > 1). > 1). = (A/F)= (A/F)stoichstoich /(A/F)/(A/F)actualactual by mass or by mass or by moleby moleFlammability limits are frequently quoted as Flammability limits are frequently quoted as %fuel %fuel by volume in the mixtureby volume in the mixture, or as a , or as a % of the % of the stoichiometric fuel requirementstoichiometric fuel requirement, i.e., (, i.e., ( x 100%). x 100%). Table 8.4 shows flammability limits of some fuels Table 8.4 shows flammability limits of some fuels Flammability limits for a number of fuel-air Flammability limits for a number of fuel-air mixtures at atmospheric pressure is obtained from mixtures at atmospheric pressure is obtained from experiments employing "tube method". experiments employing "tube method". In this method, it is ascertained whether or not a In this method, it is ascertained whether or not a flame initiated at the bottom of a vertical tube flame initiated at the bottom of a vertical tube (approximately 50-mm diameter by 1.2-m long) (approximately 50-mm diameter by 1.2-m long) propagates the length of the tube. propagates the length of the tube. A mixture that sustains the flame is said to be A mixture that sustains the flame is said to be flammable. By adjusting the mixture strength, the flammable. By adjusting the mixture strength, the flammability limit can be ascertained.flammability limit can be ascertained.Although Although flammability limitsflammability limits are physico-chemical are physico-chemical properties of the fuel-air mixture, experimental properties of the fuel-air mixture, experimental flammability limits are related to losses from the flammability limits are related to losses from the system, in addition to the mixture properties, and, system, in addition to the mixture properties, and, hence, generally hence, generally apparatus dependentapparatus dependent [31].[31].Example 8.5. Example 8.5. A full A full CC3 3HH8 8 cylinder from a camp stove leaks its cylinder from a camp stove leaks its contents of 1.02 lb (contents of 1.02 lb (0.464 kg0.464 kg) in 12' x 14' x 8' (3.66 ) in 12' x 14' x 8' (3.66 m x 4.27 m x 2.44 m) room at 20m x 4.27 m x 2.44 m) room at 20o oC and 1 atm. C and 1 atm. After a long time fuel gas and room air are well After a long time fuel gas and room air are well mixed. Is the mixture in the room mixed. Is the mixture in the room flammableflammable??SolutionSolutionFrom Table 8.4, we see that CFrom Table 8.4, we see that C3 3HH8 8-air mixtures are -air mixtures are flammable for 0.51 < flammable for 0.51 < < 2.83. Our problem, thus, < 2.83. Our problem, thus, is to determine is to determine of the mixture filling the room. of the mixture filling the room. Partial pressure of CPartial pressure of C3 3HH8 8 by assuming ideal-gas by assuming ideal-gas behaviourbehaviour= 672.3 Pa= 672.3 PaPropane mole fraction =Propane mole fraction =F F = P= PF F/P = 672.3/101,325 = 0.00664/P = 672.3/101,325 = 0.00664and and air air = 1 - = 1 - F F = 0.99336= 0.99336The air-fuel ratio of the mixture in the room isThe air-fuel ratio of the mixture in the room is(A/F)(A/F)act act = = ( )/. ( / . )( ) 111111111111111111. ( . )( . ) 111111111F u FFroomm R MWTPV+ . 1111.( . ) 1111111111.( . ) 11111111111air airfuel fuelMWMW From the definition of From the definition of and the value of (and the value of (A/FA/F))stoich stoich from Table 8.4 (i.e. 15.6 by from Table 8.4 (i.e. 15.6 by mass ratiomass ratio), we have), we have = (= (A/FA/F))stoich stoich /(A/F)/(A/F)act act = 15.6/97.88 = 0.159= 15.6/97.88 = 0.159Since Since = 0.159 < = 0.159 < lower limit lower limit (= 0. 51), the mixture in (= 0. 51), the mixture in the room is not capable of supporting a flame.the room is not capable of supporting a flame.CommentCommentAlthough our calculations show that in the fully Although our calculations show that in the fully mixed state the mixture is not flammable, it is quite mixed state the mixture is not flammable, it is quite possible that, during the possible that, during the transient leaking processtransient leaking process, , a flammable mixture can exist somewhere within a flammable mixture can exist somewhere within the room. the room. CC3 3HH8 8 is heavier than airis heavier than air and would tend to and would tend to accumulate near the floor until it is mixed by bulk accumulate near the floor until it is mixed by bulk motion and molecular diffusion. motion and molecular diffusion. In environments employing flammable gases, In environments employing flammable gases, monitors should be located at both low and high monitors should be located at both low and high positions to detect leakage of heavy and light positions to detect leakage of heavy and light fuels, respectively.fuels, respectively.3. Ignition3. IgnitionMost of ignition uses electrical spark (pemantik Most of ignition uses electrical spark (pemantik listrik). Another means is using pilot ignition (flame listrik). Another means is using pilot ignition (flame from very low-flow fuel).from very low-flow fuel).Simplified Ignition AnalysisSimplified Ignition AnalysisConsider Williams second criterion, applied to a Consider Williams second criterion, applied to a spherical volume of gas, which represents the spherical volume of gas, which represents the incipient propagating flame created by a point incipient propagating flame created by a point spark. Using the criterion:spark. Using the criterion:Find a critical gas-volume radius, Find a critical gas-volume radius, RRcrit crit,, below which below which flame will not propagateflame will not propagateFind minimum ignition energy, Find minimum ignition energy, EEign ign,, to heat critical to heat critical gas volume from initial state to flame temperature gas volume from initial state to flame temperature ((TTu u to Tto Tb b).).Critical radius, R Critical radius, Rcrit crit, and E , and Eign ign(8.40) (8.40)( (propagation propagation) )(8.41) (8.41)where is mass flowrate/volume where is mass flowrate/volumeHeat transfer process is shown in Figure 8.20 Heat transfer process is shown in Figure 8.20(8.42) (8.42)Substitution Eqn 8.42 to 8.41 results in Substitution Eqn 8.42 to 8.41 results in. . (8.43) (8.43)R Rcrit crit is therefore determined by the flame propagation is therefore determined by the flame propagationconductionQ V Q& &1 1/ 1 1 1critF c crit critRdTm h R k Rdr &( )critb uR critT TdTdr R ( )1b ucritF ck T TRm h &If R < RIf R < Rcrit crit, it would require exothermic heat > , it would require exothermic heat > hhc cSubstituting Substituting from Eqn 8.20 into Eqn 8.43 will givefrom Eqn 8.20 into Eqn 8.43 will give(8.44)(8.44)Ignition is aimed to increase fluid from TIgnition is aimed to increase fluid from Tu u to Tto Tb b at at the onset of combustion to replace the onset of combustion to replace hhcc ((ignitionignition))(8.45)(8.45)where Ewhere Eign ign is minimum ignition energyis minimum ignition energy( )/ 1 11 critLRS( )ign crit p b uE mc T T Substitution mSubstitution mcrit crit==b b.4.4RRcrit crit3 3/3 and /3 and b b using gas using gas ideal formulae to Eqn 8.45 results inideal formulae to Eqn 8.45 results in(8.47)(8.47)where Rwhere Rb b = R= Ru u/MW/MWb b and Rand Ru u = gas constant= gas constant1, 111 _ _ _ , , ,pbignLub bcT TPRES T4. Dependencies on Pressure, Temperature4. Dependencies on Pressure, Temperatureand Compositionand CompositionUsing Eqn 8.27 and 8.29 on Eqn 8.47 Using Eqn 8.27 and 8.29 on Eqn 8.47 demonstrates effect of pressure to bedemonstrates effect of pressure to beEEign ign PP-2 -2(8.48)(8.48)(see comparison with experimental result in Fig (see comparison with experimental result in Fig 8.21)8.21)Eqn 8.47implies that in general, Eqn 8.47implies that in general, TTu u EEign ign (see (see Table 8.5). Table 8.5). EEign ign vs %fuel gives U-shaped plot (Figures 8.22 vs %fuel gives U-shaped plot (Figures 8.22 and 8.23). This figure indicates that and 8.23). This figure indicates that EEign ign is is minimum as a mixture composition is minimum as a mixture composition is stoichiometric or near it. stoichiometric or near it. If the mixture gets If the mixture gets leaner atau richer, Eleaner atau richer, Eign ign increasesincreases first gradually and then abruptly. %fuel first gradually and then abruptly. %fuel at Eat Eign ign = = to be ignited are flammability limitsto be ignited are flammability limitsFigure 8.22. Effect of %fuel on EFigure 8.22. Effect of %fuel on Eign ignFigure 8.22. Effect of %fuel on EignFigure 8.23. Effect of methane composition on EFigure 8.23. Effect of methane composition on Eign ignTable 8.5Table 8.5 Temperature influenceTemperature influenceon spark-ignition energyon spark-ignition energy Fuel Fuel Initial temp (K) Initial temp (K) E Eign ign (mJ) (mJ)n-heptanen-heptane298 298 14.5 14.5373 373 6.7 6.7444 444 3.2 3.2Iso-octane Iso-octane 298 298 27.0 27.0373 373 11.0 11.0444 444 4.8 4.8n-pentane n-pentane 243 243 45.0 45.0253 253 14.5 14.5Fuel Fuel Initial temp (K) Initial temp (K) E Eign ign (mJ) (mJ)n-heptanen-heptane298 298 7.8 7.8373 373 4.2 4.2444 444 2.3 2.3propanepropane233 233 11.7 11.7243 243 9.7 9.7253 253 8.4 8.4298 298 5.5 5.5331 331 4.2 4.2356 356 3.6 3.6373 373 3.5 3.5477 477 1.4 1.4References:References:Turns, Stephen R., An Turns, Stephen R., An Introduction to Introduction to Combustion, Concepts and ApplicationsCombustion, Concepts and Applications, 2, 2nd nd edition, McGrawHill, 2000edition, McGrawHill, 2000