管国峰版化工原理课后习题答案第七章汇总

管国峰版化工原理课后习题答案第七章汇总第六章习题1)1)苯酚(C6H50H)(A)和对甲酚(C6H4(CH3)OH)(B)的饱和蒸汽压数据为：苯酚蒸汽压OpA温度113.7114.6115.4116.3117.0对甲酚蒸汽压OpBkPakPa7.707.948.28.58.76温度t117.8118.6119.4120.0苯酚蒸汽压OpA对甲酚蒸汽压pBkPa9.069.399.7010.00kPa10.010.410.811.1911.5811.9912.4312.8513.26试按总压P=75mmHg(绝压)计算该物系的“t-x-y”数据。此物系为理想物系。解：xA?P?pB000pA?pBpxyA?AAPpB0kPa7.707.948.28.58.769.069.399.7010.00(x,y-mol分率)xA1.00.8370.6920.5580.4400.3210.2010.09520.000xB1.00.8710.7480.6240.5090.3850.2490.1220.000tOCpAOkPa113.710.0114.610.4115.410.8116.311.19117.011.58117.811.99118.612.43119.412.85120.013.262)承第1题，利用各组数据算在x=0至x=l范围内各点的相对挥发度ai,取各ai的算术平均值a,算出a对ai的最大相对误差。以平均a作为常数代入平衡方程式算出各点的“y-xi”关系，算出由此法得出各组yi值的最大相对误差。解：?i?(pOB/pOA)i,计算结果如下:113.7tOC?i1.299114.6115.4116.3117.0117.8118.6119.4120.01.3101.3171.3161.3221.3231.3241.3251.326?ni?l.318最大误差?L318?1.299?1.46%1.2992)yi?tOCxiyi?xil?(?1)xi?按1.318计，结果如下：113.7114.6115.4116.3117.0117.8118.6119.4120.01.00.8370.6920.5580.4400.3210.2010.095201.00.8710.7480.6250.5090.3840.2490.12200.384?0.385?2.60?10?30.385最大误差二3）已知乙苯（A）与苯乙烯（B）的饱和蒸汽压与温度的关系可按下式算得:式中pO的单位是mmHg,T的单位是K。问：总压为60mmHg(绝压)时，A与B的沸点各为多少C?在上述总压和65时，该物系可视为理想物系。此物系的平衡汽、液相浓度各为多少摩尔分率？解：1)令pOA?p,算得的t为A的沸点Ln60?16.0195?3279.47(/T?59.95)?TA?334.95K?61.80C令pOB?p,算得的t为B的沸点Ln60?16.019373328.57(/T?63.72)?TB?342.85K?69.70C2)p?60mmHg,t?650C?338.15KLnpOA?16.019573279.47(/338.15?59.95)?pOA?68.81mmHgLnpOB?16.019373328.57(/338.15?63.72)?pOB?48.92mmHg60?48.92xA?0.55768.81?48.9268.81?0.557?0.63960yA?4)苯(A)和甲苯(B)混合液可作为理想溶液，其各纯组分的蒸汽压计算式为式中pO的单位是mmHg,t的单位是C。试计算总压为850mmHg(绝压)下含苯25%(摩尔百分率)的该物系混合液的泡点。解：设t?104.150CLgp0A?6.906?1211(/104.157220.8)?pOA?1511mmHgLgpOB?6.95571345(/104.15?219.5)?pOB?629.9mmHgxA?850?629.9?0.251511?629.9所设正确，泡点为104.150C5)试计算总压为760mmHg(绝压)下，含苯0.37、甲苯0.63(摩尔分率)的混合蒸汽的露点。若令该二元物系降温至露点以下3,求平衡的汽、液相摩尔之比。解：1)设露点为102.250CLgP0A?6.906?1211(/102.257220.8)?P0A?1436.7mmHgLgP0B?6.95521345(/102.25?219.5)?P0B?595.3mmHg760?595.3xA?0.19571436.7?595.3即所设正确，露点为102.250C2)P?760mmHgyB?1436.7?0.1957?0.37760t?102.25?3?99,250CLgP0A?6.90671211(/99.25?220.8)?P0A?1325mmHgLgP0B?6.95571345（/99.25?219.5）?P0B?543.7mmHgxA?760?543.71325?0.2768?0.2768yA?0.4826132575437760汽相的摩尔数0.37?0.2768?0.828液相的摩尔数0.4826?0.376）有一苯（A）、甲苯（B）、空气（C）的混合气体，其中空气占2乐苯与甲苯浓度相等（均指摩尔百分数），气体压强为760mmHg（绝压）。若维持压强不变，令此三元物系降温至95C,求所得平衡汽相的组成。A、B组分均服从拉乌尔定律。已知95c时解：设原来混合气量为lkmol,汽液平衡时汽相为Ykmol,液相为Lkmol。空气苯苯甲苯0.02?yC?V760yA?1163xA（1）（2）（3）0.49?yA?V?（1?V）xA760（l?yA?yC）?475（l?xA）（4）由四个独立方程可解出xA,yA,yC,V四个未知量试差方法：设xA经（3）?yA经（4）?yC经（1）?V经（2）?xA?重设xA试差过程数据示例：设XA0,380.3780.3770.376算得的XA0.3240.3520.3630.373,yA?0.575,yC?0.0346,V?0.578kmol解得：xA?0.3767）常压下将含苯（A）60%,甲苯（B）40%（均指摩尔百分数）的混合液闪蒸（即平衡蒸锵），得平衡汽、液相，汽相摩尔数占总摩尔数的分率一一汽化率（q）为0.30。物系相对挥发度a=2.47,试求：闪蒸所得平衡汽、液相的浓度。若改用简单蒸馈，令残液浓度与闪蒸的液相浓度相同，问：储出物中苯的平均浓度为多少？提示：若原料液、平衡液、汽相中A的摩尔分率分别以xf、x、y表示，则存在如下关系：解：1)闪蒸xfqy?x?(y,x为平衡汽，液相的摩尔分率)q?lq?ly?0.70x/0.30?0.60/0.30?2.33x?2.02.47x1?(2.47?1)x解得x?0.539y?y?0.7422)简单蒸饰Ln(wl/w2)?Ln(xl/x2)?Ln(l?x2)(/l?xl)(/?1)?Ln(0.60/0.539)?2.47Ln(Cl?0.539)(/l?0.60)(/2.47?1)?0.3U?wl/w2?l.365y(平均)?xl?w(/wl?w2)2xl?x2)?0.60?(0.60?0.539)(/1.365?1)?0.7678)某二元物系，原料液浓度xf=0.42,连续精储分离得塔顶产品浓度xD=0.95。已知塔顶产品中易挥发组分回收率4=0.92,求塔底产品浓度xw。以上浓度皆指易挥发组分的摩尔分率。解：?DxD(/Fxf)即0.92?0.95D/0.42F?D/F?0.4067且W/F?l?D/F?l?0.4067?0.5933物料衡算式:Fxf?DxD?WxW?xW?0.0567即xf?(D/F)xD?(W/F)xW代入数据：0.42?0.4067?0.95?0.5933xW9)某二元混合液含易挥发组分0.35,泡点进料，经连续精福塔分离，塔顶产品浓度xD=0.96,塔底产品浓度xw=0.025(均为易挥发组分的摩尔分率)，设满足恒摩尔流假设。试计算塔顶产品的采出率D/F。若回流比R=3.2,泡点回流，写出精饰段与提饰段操作线方程。解：1)按杠杆规则D/F?(x?x)(/x?x)?(0.35?0,025)(/0.96?0.025)?0.3476感谢您的阅读，祝您生活愉快