化工考研白皮书上册练习四答案

一、填空1、比表面积相等，爬流时流动阻力由固体表面积的大小决定，大，大，测定、（或预测和估算压降）2、2，0.53、少，大4、0.1hr5、6、离心力/重力，分离效率，压降7、不变，恒压降、温度浓度均匀，沟流，腾涌或节涌，散式，聚式8、0.05，26aaauuut1三、计算题1、（1）悬浮液中固体质量：100092.0280008.01121ww悬3/1054mkgkgVm2108悬kg7.168%8210831687.01000/7.168mV滤饼38313.11687.02mV滤液（2）取核算：∴可行或：取核算：∴可行23/285.05.0162.0mmKq328313.1256.0285.0mnqAV24.10n11n3321687.01745.01105.056.0mm75.1005.056.01687.02n11n3328313.19663.111256.0285.0mmqAV2、（1），（2）不变，不变2322/1667.0282635.0075.0/28025.0635.0075.0mmAVAVq滤饼滤液Kqqqe22hrs956.034455.012'sKKPP'eqqq22KhrKK677.02/956.0''5.0（3）框厚增加与q无关3、（1）23/1667.0mmq32764.3282635.01667.0mqAV3000/2.12932734.2229mkgTTpp25/1081.1msNsmqqmV/2778.02.13600/12003设沉降位于stocks区：检验假设成立mmgudptP07.61007.6)(186minsmAnquVt/10778.210102778.0)1(32101.1Re3tPpud%50')'(2minminttPPuuddmdP29.4'min（2）150℃空气设沉降位于stocks区：%9.6707.652x3/836.04232734.2229mkg25/1041.2msNsmqqmV/400.0836.03600/12003smAnquVt/1000.41010400.0)1(3mmgudptP41.81041.8)(186min21017.1Re3tPpud检验假设成立若100％除去的最小颗粒直径则mdP07.6minsmgdupPt/1008.218)(32minhkgskgAnqqVm/626/174.0)1(（3）温度↑，生产能力↓原因：4、（1）mVtqqut,,,,mqt,,233/1015.10mmnKKq295.49.075.114.3mdlAsmnqAQ/1037.83430502.0mqAV（2）获得0.0502m3滤液可得固体：相应的悬浮液质量为：相应的悬浮液体积为：相应的滤饼体积为：1000115001121悬悬浮液固体kgkgw/312.0kg82.295940502.0kg58.95312.0/82.2930856.01116/58.95m3/1116mkg30354.00502.00856.0mLAmmL16.795.4/0354.0（3），，，5.0''nnQQ2''nnLL2''nnqqsmQ/1092.5'34mmL1.10'23/01435.0'mmq