《燃烧学导论：概念与应用》第四章 第二部分 习题 参考答案

  1 《燃烧学导论：概念与应用》第四章习 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 参考 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 * 本章习题： 16－18, 20 4.16 Solution: 1) 1 2 1 2 2 2[O][N [NO][N [O d[N ][O] ] ] ] ] d N [NO][Of r f rK K K Kt = − + − 1 2 1 2 2 2[O][N [NO][N [O d[N ][N] ] ] ] ] d [NO][Of r f rK K K Kt = − − + 2) 1 2 d[NO] 2 N ] d [O][fKt = 3) 1 2 2 2 [O][ ] [ [ N ] N O ] f f K K = 4) 23.7 10 st −= × 5) 91 33.164 10 m /kmol-srK = × 6) Neglecting the reverse reactions is appropriate, especially considering the reactions were calculated using the final value of 50ppm NO. At lower NO concentrations, the f/r ratios would be even greater. 7) [N]=5.487*10-11kmol/m3; xN=9.4*10-3ppm 4.17 GIVEN: the Eeldovich reaction (problem 4-16) and the following additional reaction: HNOOHN k +⎯→⎯+ 3 FIND: a) the steady-state N-atom concentration expression. b) the expression for NO formation rate, d[NO]/dt ASSUMPTIONS: equilibrium O, O2, N2, H and OH, negligible reverse reactions, APPRAOCH: 3 reactions: NNONO fk +⎯→⎯+ 12 ONOON fk +⎯→⎯+ 22 HNOOHN fk +⎯→⎯+ 3 a) 0]][[]][[]][[][ 32221 =−−= OHNkONkNOkdt Nd fff at steady-state, solving for [N]:                                                                * 该答案来自于清华大学热能系姚强教授。    2 ][][ ]][[ ][ 322 21 OHkOk NOk N ff f += b) ]][[]][[]][[][ 32221 OHNkONkNOkdt NOd fff ++= substituting for [N] as found in part a: ])[ ][][ ]][[ (])[ ][][ ]][[ (]][[][ 322 21 32 322 21 221 OHOHkOk NOk kO OHkOk NOk kNOk dt NOd ff f f ff f ff ++++= Simplifying: ]][[2])][][( ][][ 11][][[][ 21322 322 21 NOkOHkOkOHkOk NOk dt NOd fff ff f =+++= COMMENT: Note that the above expression for d[NO]/dt is the same as that derived in problem 4-16 witout the N+OH→NO+H reaction. 4.18 GIVEN: HCOOHCO K +⎯→⎯+ 21 ; )/3000exp(1017.1 35.171 TRTk u+⋅= OCOOCO K +⎯→⎯+ 22 2 ; )/000,200exp(105.2 35.1122 TRTk u−⋅= T=2000 K, P=1atm, 011.0=COx , 31068.3 −⋅=OHx , 31043.62 −⋅=Ox FIND: τchem for the 2 CO reactions and compare ASSUMPTIONS: OHCO xx >> , 2Ox SOLUTION: with the assumption that CO is the dominant species, we can use Eqn.4.75 in a straightforward way. First, we need to connect COx to [CO]: TR PxCO u CO=][ Eqn.4.40 35 /10702.6 )2000(8315 325,101011.0 mkmol−⋅== Since k1 and k2 are expressed in gmol and cm3, we connect: 38 33 33 3 5 /10702.6 100 11010702.6][ cmgmol cm m kmol gmol m kmolCO −− ⋅=⋅= Numerically evaluating k1 and k2 : sgmolcmk −⋅=⋅= /10008.4] )2000(315.8 3000exp[)2000(1017.1 31135.171   3 sgmolcmk −⋅=−⋅= /10496.1] )2000(315.8 200000exp[105.2 37122 To calculate τchem , we apply Eqn.4.75: skCO 51118111 1072.3]10008.410702.6[]][[ −−−− ⋅=⋅⋅⋅==τ skCO 997.0]10496.110702.6[]][[ 178122 =⋅⋅⋅== −−−τ COMMENT: Here we have a clear demonstration that the CO+OH reaction is much faster than CO+O2: 5 times orders-of –magnitude faster ! 如果不使用近似，利用书上公式(4.74)则有 ss 29.1;1026.4 251 =⋅= − ττ 4.20 Solution: 1 3 2 2 1 3 2 [O ][O] [ OH [H ] ] f f r r K K K K = 3 1 2 3 2 2 2 1 2 3 2 1/2 2 [O ][H ] O] [H] { ( ) } [H f f f r r r K K K K K K = 11 2 2 2 1 2 /2][H ][OH] { [O }f f r r K K K K =   4 《燃烧学导论：概念与应用》第五章习题 参考答案 本章习题：13 5.13 Solution: Based on the assumptions, the following relations can be used: 1 2 d[NO] 2 [N ] d [O]f eq eqKt = 1 1 1 383701.8 10 exp( )fK T = × − Assuming T, [O] and [N2] constant, 1 2[O[NO] [NO] 2 [ ]N (] )f f eq eq f ii K t t−=− a) 615.3NOx ppm= ; ,/ 0.294NO NO eqx x = b) 1686NOx ppm= ; ,/ 0.88NO NO eqx x = c) 新生成的 NO为 6.69ppm; 187.0/ , =eqNONO xx d) 新生成的 NO为 336.6ppm; 86.0/ , =eqNONO xx Looking at the ratio ,/NO NO eqx x , it appears that the assumption of negligible reactions was appropriate for the gas turbine combustor, particularly in the secondary combustion zone. The assumption is not valid for the gas furnace, however. The results do make sense. The Zeldovich mechanism is a strong function of temperature as shown by the fact that most of the NO produced in both the gas turbine and the furnace combustors was produced in the high temperature primary-combustion zone. Furthermore, the Zeldovich mechanism relatively slow as shown by the higher NO levels in the gas funace where residence times are much longer. 附：以下为题 5.13的具体计算过程(from ShengKai Wang) Text1: solution to 5.13 From S.K. Wang