《化学反应工程》(第四版)第三章课后习题答案详解

习题3-1解答习题3-1解答解：根据CSTR基本方程AxAAArdxct00)(，ccAB时当1/00220)1()(AAAxkcr)1(1)1()1(1)1(00200201AAAxAAAxAAAxxkcxxdkcxkcdxtAA，xA时当5.0)(30.5)5.01(5.0307.0615.011ht，xA时当9.0)(70.47)9.01(9.0307.0615.011ht，xA时当99.0)(70.524)99.01(99.0307.0615.011ht，ccAB时当5/00)5)(1()(20AAAAxxkcr)1(55ln41])5()1([41)5)(1(10000002AAAxAAxAAAxAAAAxxkcxdxxdxkcxxdxkctAAA，xA时当5.0)(78.0)5.01(55.05ln307.0615.0412ht，xA时当9.0)(79.2)9.01(59.05ln307.0615.0412ht，xA时当99.0)(81.5)99.01(599.05ln307.0615.0412ht分析：等当量配料，随转化率提高，反应时间迅速增长；若采用过量组分配料，随转化率提高，反应时间增长放慢。习题3-2解答26.5ABAACCkCrLmolCCBA/02.000;DCB：A已知)1(0AfAAfxkCx：t根据);/(min6.5molLk(min)64.169)95.01(02.06.595.02133t：mVR时和当(反应时间与反应体积无关）习题3-3解答（1）PFR)1(1lnAfpxk（2）CSTR)1(AfcAfcxxk解：（1）（2）操作条件不变，等体积反应器，则303.2)9.01(1ln)1(1ln)1(AfpAfcAfcxxxk697.0303.3303.2Afcx习题3-4解答平推流;11ln111Axk22221AAxxk全混流;21RTEekk/0122/0/01211ln/112AAARTERTExxxekekkk]6.011ln/7.017.0ln[100068.83314.8142312T;9347.0100068.83314.8142312TKT4412习题3-5解答;/5.10.3210LmolCALmolCB/0.10.2210;000RPCC;8.0Bfx;533.05.18.00.100ABfBAfCxCx;7.0)533.01(5.1AC;2.0)8.01(1BC;8.08.00.10BfBRPxCCC;)(00fAAfARrxCVVAfARPBARxCCCCCVV00)7.18(min/004.0533.05.1)8.08.07.12.07.08(1.00LVmin/2004.05.0200201mLVVV习题3-6解答：已知;2AAkCrmin)/(4.17molmlk;/75.0mlgmin;/14.70LV;/14.70LmolCA解：（1）串联两个体积0.25m3的CSTR试求下列方案的转化率。min351014.725.0)(32212021100AAAAAAARkCxxCkCxCVV35.41014.7104.173535)1()1(3602212211AAAAAAkCxxxxx;035.47.935.4211AAxx62.01Ax35.4)1(62.0222AAxx035.47.997.4222AAxx80.02Ax（2）一个0.25m3的CSTR，后接一个0.25m3的PFRmin351014.725.032202110021AAxxAAAAAARkCdxCkCxCVV;35.4)1()1(2122211AAxxAAAAxdxxx62.01Ax;35.4)1(262.022AxAAxdx;35.462.011112Ax;98.6112Ax86.098.6112Ax（3）一个0.25m3的PFR，后接一个0.25m3的CSTR351014.725.0)(321120021001AAAAxAAARkCxxCkCdxCVVA35.4)1()()1(22120211AAAxAAxxxxdxA;35.41)1(11Ax81.035.5111Ax;35.4)1()81.0(222AAxx035.47.916.5222AAxx88.02Ax（4）两个0.25m3的PFR串联min351014.725.0322002100211AAAxxAAAxAAARkCdxCkCdxCVV35.4)1()1(21121021AAAxxAAxAAxdxxdx35.41111111121AAAxxx;35.41111Ax;35.5111Ax;35.435.5112Ax90.07.9112Ax;81.01Ax习题3-7讲解3-7已知：AP;PAACkCr);/(1032skmolmksmV/002.030;/230mkmolCA00PC，xA时问98.0下列各种情况下的反应体积。解：（1）单个PFRAxAAARrdxCVV000根据AAxAAAAxAAAAAAxxdxkCxCxkCdxC000000)1(1)1(sxxkCxdxkxdxkcVVAAAxAAxAAARAA2.389201.0892.32])1([ln22)1(298.05.005.05.000原积分可改写为的对称函数是关于被积函数;5.0)1(1)(:AAAxxxxf30778.02.389002.0mVVR（2）单个CSTRAAfARrxCVV00根据)(2500)98.01(201.01)]1()[1(00000sxCCxkCxCVVAfAAAfAAfAR)(5002.02500250030mVVR（3）两个CSTR串联21201100)(AAAAAAARrxxCrxCVV根据1m2m2020120101010)1()()1(AAAAAAAAAAAAAxCxkCxxCxCxkCxC;)1()()1(122121AAAAAxxxxx85.01Ax98.02Ax)85.01(201.01)1(11001AARxkCVV)(67.0002.033.33333.333301mVVR)(34.167.02231mVVRR98.0)98.01(201.0)85.098.0()1()(2201202AAAAARxxkCxxVV);(63.33102sVVR)(66.063.331002.032mVR习题3-8讲解3-8已知：RAACkCkr21;32*Ax;31Afx?21Afx使如何调节加料速率，AR解：当反应达到平衡转化率时021RAACkCkr23/213/21)1()1(***0*0021AAAAAAAARxxxCxCCCCkk;21313231212101kkkkVVR2121021212121kkkkVVRAfAfAfAfAAfAAfARAAfAfAAfARxkxkxxCkxCkxCCkCkxCrxCVV210201021000)1()1()(3114121/21/2212121210102kkkkkkkkVV010231VV加料速率减小到原来的1/3倍，可使转化率达到0.5。习题3-9讲解3-9已知：AL2AMAACkr11)(异构反应212)(AACkr二聚反应进料浓度为CA0。试证明在PFR中，L最大收率为012)1ln(ACkk，式中证明：AfAPxSrrYmaxmaxmax)(SdCCkkkdCCkCkCkdCrrCACCAACCAAAACCALLfAAfAAfAfA0002112211对于PFR：AALfAfACCACCCCCsdCSAfA000。，CCLA值最大时显然当0;00AfACCACCsdCSAfA00002110000000maxmax11AAACAAACAAAAfAAfACAAfdCCkkkCsdCCCCCCCsdCxSY00AAfAAfCCCx)1ln(111ln1ln1)(012102102121021max0AACAAACkkkCkkCkkCkkdCkkYA证毕。习题3-10解答已知：AB+1LMB+2LBACCkr11BLCCkr2230/1.0mkmolCAB大量过量，tcAcMt10.0550.038t20.010.042?)1(12kk?:)2(max,LCBSTR?)3(Ax【分析与解答】结果为：解：（1）由于是间歇反应器，所以;111ABACkCCkr液相反应，前后体积不变；由此可确定L浓度。B组分大大过量，近似为一级反应；由此可简化速率方程。LBLCkCCkr222tcAcLcMt10.0550.0070.038t20.010.0480.0422121)()(12LLAAccLLccAArdcrdctt2121048.0007.0201.0055.0112925.1705.1007.0048.0ln101.0055.0ln1kkkkckdcckdcttLLAA129.1705.1925.11212kkkk?:)2(max,LCBSTR122)(:210max,kkkALkkcC根据0346.0346.01.0)129.11(1.0)/1(1129.1129.11//120max,1212kkkkALkkcC?)3(Ax;00AAAAcccx;10tkAAecc;/ln1212kkkktopt390.01.0)1129.1129.1lnexp(1.0)1//ln(0)/ln(0121212121kkkkAkkkkkAAececc61.01.0039.01.000AAAAcccx习题3-11解答解：22)1(1211AAANMLLALcccrrrrrrs，c，Af时只有当由此可以看出0有最大值Lc对于CSTR：;)1(102AfALfAfccccsS20)1(AfAfALfcccc020max,)01(0AALccc对于PFR：由于PFR中的选择率沿不同管截面而发生变化，其总选择率应为瞬时选择率的积分：22)1(1211AAANMLLALcccrrrrrrsAfALAfAccAAAfAccAcccccdccccsdcSAAfAfA002000)1(1021111)1(10AAfccAALccdcccAAf，c，Af时只有当由此可以看出0有最大值Lc000max,111011AAALcccc证毕。习题3-12解答在平推流反应器中进行基元反应ALMNk1k2k3L为目的产物，已知，求L的最大收率及最优接触时间。312kkk解：AfALxSccY0m