化学反应工程--第四版--第一章作业

O4H2COO3OH2CHO2H2HCHOOOH2CH2223223进入反应器的原料中，甲醇：空气：水蒸气＝2：4：1.3（摩尔比），反应后甲醇转化率达72％，甲醛的收率为69.2％。计算（1）反应的选择率；（2）反应器出口的气体组成。1-1银催化剂上进行甲醇氧化为甲醛的反应反应器出口组成如下：CH3OH：27.397×(1-72%)=7.671molCH2O：27.397×69.2%=18.959molCO2：27.397×72%-18.959=0.767molH2O：18.959+0.767×2+17.808=38.01molO2：11.468-18.959/2-1.5×0.767=0.838molN2：42.756mol组成物质的量ni0(mol)CH3OH27.397H2O17.808O211.468N242.756注：空气中O2占20.93%N2占78.03%解：（1）反应选择率S=Y/x=69.2%/72%=96.11%（2）设进入反应器的原料量为100mol，反应器进料组成如下：O4H2COO3OH2CHO2H2HCHOOOH2CH2223223组成物质的量ni(mol)摩尔分数yi(%)CH3OH7.6717.02HCHO18.95917.35CO20.7670.702H2O38.30135.04O20.8380.767N242.75639.12小计109.29211-2甲醇合成过程中含有组分CO、CO2、H2、CH3OH、H2O、N2和CH4。已知反应器进口各组分的摩尔分数yi0,i代表各组分。取CO和CO2加氢反应为关键反应，分别导出:（1）出口气体中关键组分摩尔分数为yCO2和yCO时；（2）出口气体中关键组分摩尔分数为yCH3OH和yCO2时,其余各组分摩尔分数的计算表达式。OHCOHCOOHOHCH3HCOOHCH2HCO222232232(1)(2)(3)解：取CO和CO2加氢反应为关键反应，反应方程式为：其中(2)＝(1)＋(3)，故可取(1)、(2)或(1)、(3)为关键反应式。进口出口组分摩尔分率摩尔流量摩尔流量摩尔分率COCO2H2CH3OHH2ON2CH4小计11(1)出口气体中关键组分摩尔分数为yCO2和yCO，取1、2反应。0COy02COy02Hy02OHy02Ny04CHyCOy2COy2HyOHy22Ny4CHyOHCHy300COTyN002COTyN002HTyN002OHTyN002NTyN004CHTyN0TNCOTyN2COTyN)(3)(2222000000COTCOTCOTCOTHTyNyNyNyNyN002NTyN004CHTyN)()(223000000COTCOTCOTCOTOHCHTyNyNyNyNyN2220000COTCOTOHTyNyNyN)(22200000COTCOTCOTCOTTTyNyNyNyNNN03OHCHy003OHCHTyN00022221221COCOCOCOTTyyyyNN)(22200000COTCOTCOTCOTTTyNyNyNyNNN由可得00000000000000000000004224222222222222233222222221221221221221))(221(221)()(32)32(221221CHCOCOCOCOCHNCOCOCOCONCOCOCOCOOHCOCOOHCOCOCOCOCOCOOHCHOHCHCOCOCOCOHCOCOCOCOHyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy进口出口组分摩尔分率摩尔流量摩尔流量摩尔分率COCO2H2CH3OHH2ON2CH4小计11(2)出口气体中关键组分摩尔分数为yCH3OH和yCO20COy02COy02Hy02OHy02Ny04CHyCOy2COy2HyOHy22Ny4CHyOHCHy300COTyN002COTyN002HTyN002OHTyN002NTyN004CHTyN0TN2233000000)(COTCOTOHCHTOHCHTCOTyNyNyNyNyN2COTyN)()(222332000000COTCOTOHCHTOHCHTHTyNyNyNyNyN002NTyN004CHTyNOHCHTyN32220000COTCOTOHTyNyNyNOHCHTOHCHTTTyNyNNN332200003OHCHy003OHCHTyN000000000000003344332223322223233323332322212121212121)()(212122121)2(OHCHOHCHCHCHOHCHOHCHNNCOOHCHOHCHCOOHOHCOOHCHCOOHCHCOOHCHOHCHCOCOOHCHOHCHOHCHCOOHCHHHyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyOHCHTOHCHTTTyNyNNN332200000332121OHCHOHCHTTyyNN得OHOHCH3HCOOHCH2HCO232232232210COOHCHTTyyNN得)21()21()21)(())(21(2)21)((3443222322223232332220000000OHCHCHCHOHCHNNCOOHCHCOOHOHCOOHCHCOCOOHCHCOCOOHCHOHCHCOHHyyyyyyyyyyyyyyyyyyyyyyy23220COTOHCHTTTyNyNNN由1-3确定下列反应物系的独立反应数。(1)[CO、CO2、H2、CH3OH、H2O、CH4、CH3OCH3、C4H9OH、N2][H、C、O、N](2)[CO2、MgO、CaO、MgCO3、CaCO3][Ca、Mg、C、O]解(1)：写出原子矩阵NOCHrccNOCHNOHNOH29433423222943342322HCOCHCHCHOHOHCHCOCOH131HCOCHCHCHOHOHCHHCOCOji2000000000110112100421011100532120012/1,2000000000110110210421010110106424200][NOCHccNOCHrrrNOCHrrOHNNOHNOHOHCHHCOCHCHCHOHCOCOH94HCOCHCHCHOHOHCHCOCOH432HCOCHCHCHOHOHCHCOCOH233943342222294334232229433423220000010000311101001732100102532100011000000000311101000732110100532120012/1200000000031110100042101110053212001由此可知，矩阵[βji]的秩为4，而反应组分数为9，因此独立反应数等于9－4＝5。若选定H2O、CH4、CH3OCH3、C4H9OH、CH3OH为关键组分,5个独立反应的各计量系数向量可表示为：TTTTT1,0,0,0,0,,,,0,1,0,0,0,,,,0,0,1,0,0,,,0,0,0,1,0,,,,0,0,0,0,1,,,,453525155443424144433323133423222122413121111010000010000010000010000010000010000311101001732100102532100014544434241353433323125242322211514131211OHCHHCOCHCHCHOHCOCOH3943342222NOCHOHN可得出可能的反应方程式为：010000010000010000010000010000010000311101001732100102532100014544434241353433323125242322211514131211OHCHHCOCHCHCHOHCOCOH3943342222NOCHOHNOHCHCOHOHHCCOCOHOCHCHCOCOHCHCOCOHOHCOCOH32942233224222223753322OCrrrrrrMgCaOCMgCa00000101000101011001)2(,33112110010101010100][123413CaCOMgCOCaOMgO2COji33解(2)：写出原子矩阵TT1,0,,,0,1,,,32221213121111100100000101000101011001323122211211CaCOMgCOCaOMgO2CO33可得出可能的反应方程式为：-1-1001CO2+MgO=MgCO3-10-101CO2+CaO=CaCO3由此可知，矩阵[βji]的秩为3，而反应组分数为5，因此独立反应数等于5－3＝2。若选定MgCO3和CaCO3为关键组分,两个独立反应的各计量系数向量可表示为：;;AAVAWARdNdNkfkfdVdW证明(1)：AARbbRbWARbRiiiAAppRpWAVvRvipvivRbWpWvdNdNVWdVdWkfdWdVVSSSdNdNWdVdWkfdWSSdVSSSSkkkS或者1-5gCgPgfgwyRTkRTkRTkRTkkP2AACACCAyAggdNPyPkCkkykydWRTRT－,fp又，理想气体，接近于所以有AwAfACAACACAPAggdNkfkfdWkdNPkCkPkPdWRTRT证明(2)：－－