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1、 1燃烧学导论：概念与应用第八章习题燃烧学导论：概念与应用第八章习题 参考答案参考答案* 本章习题：1， 3， 4，5，6，7，8 8.1 Given: the outward propagation of a spherical laminar flame into an infinite medium of unburned gas Find: an expression for the radial velocity of the flame front, v Assumptions: spherical symmetry Approach: use conservation of ma
2、ss and definition of flame speed Flame front 2344 3uLffinfbfbmS Awhere Armr =Conservation of mass: 22d d dd dt4u4h s,dbuLbinf fffL fubmSSm t rrrt rvt =Comments: since ub,the flame front moves into the unburned gas at a speed greater than the flame speed. This is due to the outward expansion of the b
3、urned gases. And there is also a flame sheet assumption. 8.2 GIVEN: the thermodynamics employed in chapter 8 SHOW: that )() 1(ubPCTTch+= for an adiabatic flame ASSUMPTIONS: cP of species are constant and equal APPROACH: apply conservation of energy across flame and )(refPo fTTchh+= For adiabatic fla
4、me:RPHH= XfPhhh+=+) 1( substituting )(refPo fTTchh+= XrefuPo fFrefuPo fPrefbPo fTTchTTchTTch+=+)()()()(1( * 该答案来自于清华大学热能系姚强姚强教授。 2rearranging ,Pr,(1)()()(1)()ooo ffXf FPurefPurefbr fpehhhc TTc TTc TT+=+ (1)()(1)()CurefbrpefphTTCC TT=+ (1)()CbPuhTT C=+ 8.3 GIVEN: methane combustion FIND: the laminar
5、flame using the global single-step kinetics of chapter 5 and compare these results with those of propane in example 8-2. ASSUMPTIONS: simplified theory for laminar flame speed developed in chapter 8 applies, =1, no fuel or oxidizer in products. APPROACH: use expression for flame speed developed in c
6、hapter 8 and find “ fm KREuA24358= Note: for methane units of A are 1/s, so no conversion needed to work in 3mkmole. For other fuels, this may not be true. See problem 5-14. 3 . 13 . 034)()(exp()(22OFOFua MWY MWY TREAmkmole dtCHd where 3/1938. 0 173863.27315. 8325.101mkg TMWRPmixu= and kmolekgMWYMWi
7、i mix/63.27)85.28945. 0 043.16055. 0()(11=+=33 . 13 . 0840874. 0)043.16)1938. 0(0275. 0()043.16)1938. 0(0275. 0)(173824358exp()103 . 1 ( mskmol dtCHd =34/402. 1)0874. 0(043.16“ mskgdtCHdMWmFf=& thermal diffusivity evaluated at : KTTTub1259)2002217(21)(21=+=+=smTcTkP/100 .58)1183(161. 11066.79 )()(26
8、3 = scmsmmSuf L/5/05. 0161. 1402. 1) 111.17)(1058(2“ ) 1(22/162/1=+=+= &COMMENTS: Note that this does not agree with experimental data, primarily due to the slow kinetics of methane combustion. The slow kinetics coupled with the simplicity of our model result in incorrect modeling of methane combust
9、ion. Consequently, you would not want to use this in situations where accuracy is important. 8.4 Solution: Determine the flame speed from example 8.2 SL=38.9cm/s and then use mass conservation across the flame to find vb=293cm/s 48.5 Solution: The flame front assumes an angle with the flow direction
10、 such that the velocity profile normal to the flame front equals the flame speed. sinLuSV=, where Vu is the unburned gas velocity. For the angle to be constant, Vu must be Constant, assuming SL is constant. Therefore, the velocity at the tube exit must have a uniform, or tophat profile. Comments: No
11、te that Vu must be greater than SL to have a conical flame shape. If Vu=SL, the flame would be flat. And if VuSL, the flame could not be maintained at the tube exit and would be flashback if the tube diameter were greater than the quenching diameter. 8.6 GIVEN: ve=75cm/s of C3H8-air mixture, scmSHCL
12、/35 83,= FIND: Flame cone angle ASSUMPTIONS: Uniform velocity profile at nozzle exit SOLUTION: With reference to Fig.8.3b, We write Vu=Ve and Vu,n=SL So, =sin-1(SL/ve)= sin-1(35/75)= sin-1(0.466) Finally, =27.8 COMMENTS: The fundamental concept in determining the flame cone angle is that the velocit
13、y of the unburned mixture entering normal to the flame must equal the laminar flame speed. 8.7 Approach: Using the following schematic, determine the slope dz/dr and integrate to determine the flame shape. 22 LavS=, dtandLza rS= =, Thus, 22d dLLvSz rS=, 5222222 0()( )1/LLLLvRSvSz rdzdrdrSSr=where SL
14、, V0 and R are constants. Note that this cannot be integrated directly but dz/dr can be expanded into a series and then integrated, neglecting higher order terms. 8.8 GIVEN: Equations 8-20 and 8-22 2/1“ ) 1(2ufmSL&+= 8-20 175. 0PTTu 8-27 FIND: the pressure and temperature dependencies of laminar fla
15、me speed. ASSUMPTIONS: ideal gas, global single-step kinetics, Tb used to estimate “ mf& APPROACH: Substitute 8-27 into 8-20 and determine the temperature and pressure dependences of uf/ “ m& recognizing that the molar concentrations are functions of T and P. )exp(“ m2 uuufjibuAFFOFTREAMW dtFdMW=&and the concentration iii MWY)(= where bRTP= and uuRTP= )()()()(exp(“ m22ufjOOjbiFFibbuA Fu MWYRTP MWY RTP TREAMWPRT=&so, )exp()exp(“ m11ufbuAnn bun bnbuA uTREPTTTPTREPT &where n=i+j substituting this and 8-27 into 8-20 2/112175. 0)exp(
