化工原理试题答案

1、姓名班级化工原理（吸收）试题答案学号成绩液膜0.07045. y=mx6. NA=ky(y-y i)7. (1)解吸；0.894(2)Na=K y(y-y i) '吸收。1.B选择题：2. C.C.3.D4. C5. B6. B7. A8. A三、判断题:1. X 2.V5. X 6. X3.V7"4. X8 V、填空题:1.0.03920.0007050.0007062. 液膜阻力气膜阻力H/kG3. 气膜、4. 0.0658四、问答题:1. 如图2.pp = C/H ,=Ex,E称为亨利系数,y = mxH称为溶解度系数，m称为相平衡常数。它们只适合于稀溶液的情五、况。计

2、算题：=(yi-y2)/(y i/m)=m n1 * 答*(L/G) min=(y1-y2”X1L/G=1.2(L/G) min=(y1-y2)/x1X1=y1/ ( 1.2m) =0.08/(1.2 X 2.5)=0.0267S=m/(L/G)=m/(1.2 X mX n )=1/(1.2 n )=1/(1.2 098)=0.8503y1/y2=y 1/(y1(1- n )=1/(1- n )=1/0.02=50Nog=1/(1-S)L(1-S)y1/y2+S=14.2h=H ogX Nog=0.6 X 14.2=8.50 m>6 m ,所以不合用。2 * 答*-6-4-2-1Ky=P

3、kG=100X 3.84X 10 =3.84 X 10 kmol.m .s-4-2-2 -1Kx=CkL=55.6 X 1.83X 10 =1.02X 10 kmol.m .s-21/Ky=1/ky+m/kx=1/(3.84 X 10)+m/(1.02 X 10 )m=E/P ,E=plxm=p/Px=6.7/(100 X 0.05)=1.341/Ky=2604+131.4=2735.4-4-2-1Ky=3.656X 10 kmol.m .s ( y)。3 *答* y1=0.01y2=y1(1- n )=0.01(1-0.9)=0.001L/G=(y 1-y2)/x1,X1=(y 1-Y 2)

4、/(L/G)=(0.01-0.001)/1.5=0.0062 HOG=G/Kya=100/(0.785 X 0.5 X 3600)/0.1=1.415 m Nog=1/(1-S )L(1-S)Y 1/Y2+SS=m/(L/G)=1/1.5=0.667Nog=4.16h=HOGX Nog=1.415 X 4.16=5.88 m。4.答案m=E/P E=mP=0.25 X 850/760=0.28(大气压). 是稀溶液-3-1 H = p s/M s.E =1000/(0.28 X 18)=198.4(kmol.m .atm ) 根据1/K G = 1/ k G +1/(H k l)即1/K G

5、=1/1.25 + 1/(198.4 X 0.85)1 -2 -1 K Gk g =1.25 kmol.m .h .atm 因此，是气膜控制过程。131已知：D=0.8m , Ls =1200kg.h ,G=1000Nm .hy1=0.013$ =0.995x2=0y =0.75x-3-1K ya = 0.055kmol.m .s .atmLs =1200/(3600 18)=0.0185 kmol/sG =1000/(22.4 3600)=0.0124kmol/s为低浓度吸收- (L/G)=0.0185/0.0124=1.49X1 =(G/L)(y 1 -y2)+x 2=(0.013-0.000065)/1.49=0.00868y2=y1(1- $ )=0.013(1-0.995)=0.000065/Z = Hogx Nog 而 Nog=(y1-y2)/ Ym* yi=yi-yi=0.013-0.75 X 0.00868=0.006490* y2=y2-y2=0.000065-0=0.000065 ym=( 屮- y2)/ln( y2)=(0.006490-0.000065)/In (0.006490/0.000065)=0.001396 N og=(0.013-0.000065)/0.0013