化工原理习题答案英文

Problems and Solutions Distillation

1、 A continuous fractionating column is used to separate 4000kg/h of a mixture of 30

percent CS 2and 70 percent CCl 4. Bottom product contain 5 percent CS 2at least, and the rate of recovery of CS 2in the overhead product is 88% by weight,required. Calculate (a) the moles flow of overhead product per hour .(b) the mole fractions of CS 2and CCl 4in the overhead product, respectively

Solution: Form overall material balance

1F D W F D W Fx Dx Wx =+'''=+ (）

（2）

Known by the justice of the problem

0.88D F Dx Fx ''= （3）

Take the place of 3 types and enter 2 types

0.880.122880/0.05400028801220/0.880.8840000.30.943

1120

F F w w F

D Fx Fx Wx kg h

x D F W kg h Fx x D '''=+'⨯4000⨯0.3

=='

=-=-='⨯⨯'===F

0.12Fx W=

The unit converts ：

0.943/76

0.970.943/760.057/154

D x ==+（mole fraction ）

0.97760.0315478.3/112014.3/78.3

m M kg kmol

D Kmol h =⨯+⨯=∴==

2、 A liquid containing 40 mole percent methanol and 60 mole percent water

is to be separated in a continuous fractional column at 1 atm pressure .Calculate the value of q under the three following conditions (a) the feeding is liquid at 40 C (b) the feeding is saturated liquid. The equilibrium data for methanol-water liquid at 1 atm pressure are given in the attached table. If the column is fed with 100koml/h.The molar fractions of methanol in overhead product and bottom product are 0.95 and 0.04,respectively.A reflux ratio at the top of column is 2.5.Calculate (a) the mole flow of overhead product per hour (b) the mole flow of liquid in rectifying column (c) the mole flow of vapor in stripping column .Assume that the constant molar flow applies to this system . Solution: Form overall material balance

f D w F D W

Fx Dx Wx =+=+

Solve the eqution we can have ：

()100(0.40.04)

39.6/0.950.0410039.660.4/F w D w F x x D kmol h

x x W kmol h

--=

==--=-=

And 2.539.699/39.699138.6/L RD kmol h

V D L kmol h ==⨯==+=+=

These upper values are fixed under the three feed conditions 。

(1)V V q F '=+-

The feed at 40℃，q=1.07

138.6(1.071)100145.6/V kmol h '∴=+-⨯= Feed saturated liquid ，q=1

138.6/V V kmol h '==

3、The feeding at dew point

is fed to a continuous fractionating column ,and

the equilibrium relations are given by equations: y=0.723x + 0.263 ( in rectifying column); y=1.25x – 0.0187 (in stripping column) .Calculate(a) the compositions of feeding, overhead product and bottom product , respectively(b) the reflux ratio

Solution: The slope of the operating line in rectifying column is

0.7231

R

R =+

We can get ：R=2.61

The intercept of the rectifying line on y axis is

0.26310.263(2.611)0.95

D

D X R X =+∴=+= The intersection of the stripping line and the diagonal is Y=X=Xw

so Xw=0.0748

Form the intersection of these two operating lines, gives: 0.7230.263 1.250.01870.535

0.7230.53350.2630.65

X X X Y +=-==⨯+=

Because the feed at dew point ,so the feed line is horizon ,and the composition of feed is 0.65F

X Y ==