Ch E 416 – Assignment 8 Solutions

---

(1) Batch Distillation

 

 

(2) Multicomponent Distillation (shortcut method)

 

2.1          Minimum number of trays using the Fenske equation

 

From HYSYS ,

where   LK = light key (n-butane), and

            HK = heavy key (i-Pentane)

Recoveries of LK and HK components in distillate are 98% and 2%, respectively

 

 

 

 

2.1          Minimum reflux ratio using the Underwood method

 

Relative volatilities (with respect to isopentane) and feed flow rates of the components

 

component

propane

isobutane

n-butane

isopentane

n-pentane

Flow  rate in feed (kmol/hr)

0.2

0.2

0.2

0.2

0.2

Relative volatility

11.744

4.155

2.796

1.000

0.713

 

With n-butane as the light key and isopentane the heavy key component, the assumption of no distribution of components lighter than the light key and heavier than the heavy key component means that the distillate flow component rates and compositions are as follows (recovery of LK 98%, of heavy key 2% in distillate)

 

 

component

propane

isobutane

n-butane

isopentane

n-pentane

Flow  rate in distillate (kmol/hr)

 

0.2

 

0.2

 

0.98(0.2)

 

0.02(0.2)

 

0.0

Mole fractions

1/3

1/3

0.98/3

0.02/3

0.0

 

Solve the equation

 

For the value of  satisfying 1.000<< 2.796

 

Use of fzero of Matlab yields:

=1.5766

 

We use  to calculate minimum reflux ratio from the equation:

 

 

 

From which, Rmin = 0.6597

 

2.3 Minimum reflux ratio if isobutane is the light key; heavy key unchanged

 

Assuming that lighter than light key and heavier than heavy key do not distribute, all component flow rates in the distillate are known except that of n-butane.

 

From the equations,

 

,

 

There are three equations and three unknowns (Rmin, n-butane flow rate in distillate, and D). Simultaneous solution yields:

 

Rmin = 0.5944

 

2.4 Design using HYSYS

 

You should get:

Minimum number of trays = 5.7

Actual number of trays     = 17.4

Optimal Feed stage          = 8.3

 

See A8.hsc

 

(3) Multicomponent Distillation (Rigorous method)

3.1

 

Inputs:

     Number of trays = 18, with total condenser and default reboiler

     Distillate rate     = 0.6052

     Reflux ratio        = 0.65

 

You should get:

 

 

Rigorous

Shortcut case

 

Flow rate [kmol/h]

0.6052

0.3948

0.6052

0.3948

mole frac

 

 

 

 

 

Distillate

Bottoms

Distillate

Bottoms

C3

3.30E-01

1.80E-08

3.30E-01

5.19E-06

iC4

3.29E-01

1.52E-03

3.29E-01

2.09E-03

nC4

2.98E-01

5.02E-02

3.17E-01

2.00E-02

iC5

3.69E-02

4.50E-01

2.00E-02

4.76E-01

nC5

5.39E-03

4.98E-01

2.96E-03

5.02E-01

 

 

 

        CMO is valid for this mixture of similar hydrocarbons

 

3.2

 

Switch specs of distillate rate with recovery spec of the LK in distillate

 

You should get a new distillate rate of 0.629 kmol/h

 

3.3

 

In HYSYS, for example, using a distillate rate = 0.45, a reflux ratio = 0.4 and number of trays = 50 with feed on the 25th tray, a pinch condition of class 1 results. See figure below:

 

 

 

 

3.4

 

In HYSYS, for example, using a distillate rate = 0.75, a reflux ratio = 0.3 and number of trays = 50 with feed on the 25th tray, a pinch condition of class 2 results. See figure below.

 

 

 

 

Posted November 30, 2006

Return to top