生物反应工程第二章作业

生物反应工程第二章2-4.下面是Monod在其发表论文时提供的另一组有关在乳糖溶液中进行菌种分批培养时得到的实验数据：试根据上述数据，按Monod方程确定其动力学参数。序号t/hCs/(g/L)Cx/(g/L)Cx平均Cx1/Cs103平均Rx平均Cx/Rx10.5215815.822.8719.36.13.461541.20.3812422.829.26.4268.16.842111.5437530.3211429.237.88.633.58.26.8751.40.379437.848.510.743.1510.28.918921.50.362548.559.611.154.054030.833331.60.371959.666.56.963.0552.18.648653.70.38266.567.81.367.155003.19.62846平均Rx=Cx/t平均Cx/Rx=Ks/max(1/Cs)+1/max作图可得：max=0.97(h-1) Ks=36(g/h)2-5.在有氧的条件下，杆菌以甲醇为底物进行生长，在进行分批培养时得到数据如下所示：时间/hCx/(g/L)Cs/(g/L)（Cx）00.29.23-1.6094420.2119.21-1.555940.3059.07-1.1874480.988.03-0.0202101.776.80.57098123.24.61.145.60.921.166.150.0771.186.201.取线性相关的点：时间/hCx/(g/L)Cs/(g/L)（Cx）80.988.03-0.0202101.776.80.57098123.24.61.145.60.921.处理数据得：序号ttCs平均CsCxCx平均Cx1/Cs*103平均Rx平均Cx/Rx181028.036.87.4150.981.770.791.375134.861770.3953.2101226.84.65.71.773.21.432.485175.43860.7153.3121424.60.922.763.25.62.44.4362.318841.23.作图得: (1)max=0.299(h-1) Ks=0.269(g/h) (2)Yx/s=(Cxt-Cx0)/(Cs0-Cst)=0.65 (3)td=ln2/max=0.693/0.299=2.318(h)2-6.在一连续操作的机械搅拌实验反应器中用乳糖培养大肠杆菌，该反应器体积为1L，加入乳糖的初试浓度为Cs0=160mg/L。当采用不同加料速度时，得到一下结果。F/(L/h)Cs/(mg/L)Cx/(mg/L)0.2415.60.410150.8401211006试求大肠杆菌生长的动力学参数。解：由FCx=VRrx=rx/Cx=F/VR 定义F/VR=D称为稀释度，1/D=Ks/max*1/Cs+1/max1/Cs=max/Ks*(1/D)-1/Ks经过数据处理得FCs0CsCx1/D1/Cs0.2160415.650.250.41601015