Ch E 416 - Assignment 6 Solutions

Question 1

(1a) McCabe-Thiele

Equilibrium stages required, N=10

Number of trays = 10 – 1 = 9

The MATLAB file McCabeT.m was used to generate the above graph. It just adds graphics to the Lewis program. Study it, if you are curious.

(1b) Lewis method

1. First find the intersection point (xi , yi) of top operating line and the q-line.

Enriching operating line:

q – line:

(which agrees visually with the result from McCabe-Thiele above)

1. Start at

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

< , then switch to the Stripping operating line

;

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

< , then this is the last step

Number of equilibrium stages required, 10

Number of trays = 9

(1c)Minimum number of trays by Lewis method

1. Start at

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

1. Calculate  from

< , then this is the last step

Number of minimum equilibrium stages required, 5

Minimum number of trays = 4

(1d) Minimum reflux by analytical method.

First find the intersection (xi,yi) of q-line and the equilibrium line. In MATLAB:

1.

S=solve('y=3.8*x/(1+(3.8-1)*x)','y=x*0.55/(0.55-1)-0.45/(0.55-1)');

S.x

ans =

-.95524796812001114944356021381365

.30589731876936179879420956446299

S.y

ans =

2.1675252943689025159865735946611

.62612549928189113480707719898968

So the intersection point is

1. Now find Rmin from

R=solve('.6261255=.3058973*R/(R+1)+0.95/(R+1)')

R = 1.0113865674540843061291916202258

(1e) Smoker by hand

Given,

a=3.8, R=1.2, xd=0.95, xb=0.05, zf=0.45, q=0.55.

1. First find the intersection point (xi , yi) of top operating line and the q-line.

Enriching operating line:

q – line:

Enriching section:

1. Next find (K1,K2) for enriching section from,

;

c(1) = m*(alpha-1) = 1.5273;

c(2) = m + b*(alpha-1) –alpha = -2.0455;

c(3) = b = 0.4318 ;

K1=0.2626, K2=1.0767

1. Estimate Ne,

Striping section:

1. Next find (K1,K2) for stripping section from,

c(1) = m*(a -1) = 5.7474;

c(2) = m + b*(a -1) -a=-1.8947;

c(3) = b = -0.0526;

K1 = -0.0258, K2 = 0.3554

1. Estimate Ns,

Number of equilibrium stages required, 10

Number of trays = 9

(1f) Case studies

>> [Ne,Ns]=smoker(alpha,1.0,xd,xb,zf,q)

Ne = 8.6076 + 4.0196i

Ns = 8.6960 + 4.4205i

« The reflux ratio is less than the minimum required for infinite number of stages: Rmin = 1.0113865, Therefore, it is a Pinch condition »

>> [Ne,Ns]=smoker(alpha,1.5,xd,xb,zf,q)

Ne =  4.0341

Ns =  3.7760

« Ntot = 8 »

>> [Ne,Ns]=smoker(alpha,2.0,xd,xb,zf,q)

Ne =  3.4305

Ns =  3.1518

« Ntot = 7 »

>> [Ne,Ns]=smoker(alpha,4.0,xd,xb,zf,q)

Ne =  2.8135

Ns =  2.5213

« Ntot = 6 »

>> [Ne,Ns]=smoker(alpha,10.0,xd,xb,zf,q)

Ne =  2.5266

Ns =  2.2293

« Ntot = 5 »

NOTE: The number of stages required decreases with increasing reflux. This will decrease the capital cost, but increase the operating cost.

Effect of purity change on number of stages.

Base case with R=1.2

>> [Ne,Ns]=smoker(alpha,R,0.975,xb,zf,q)

Ne =  6.8594

Ns =  5.6056

« Ntot = 13 »

>> [Ne,Ns]=smoker(alpha,R,0.99,xb,zf,q)

Ne =  9.0446

Ns =  6.3871

« Ntot = 16 »

>> [Ne,Ns]=smoker(alpha,R,0.999,xb,zf,q)

Ne = 13.2003

Ns =  7.2140

« Ntot = 22 »

NOTE: Increasing the purity increases the number of stages.

Question 2

(2.a) The MATLAB function is ass6_2a.m

Initial guess for R=1.2.

Ne=5,Alpha=3.8, xd=0.98,.z=0.45,q=0.55.

R = fsolve('ass6_2a',1.2,opt,5,3.8,.98,.45,.55)

Norm of      First-order   Trust-region

Iteration  Func-count     f(x)          step         optimality    radius

0          2         5.85129                          35.5               1

1          4         1.08851        0.16498           5.29               1

2          6       0.0938249        0.20573          0.788               1

3          8      0.00166356        0.11901          0.079               1

4         10     8.31396e-07      0.0210457        0.00169               1

5         12     2.24974e-13    0.000492229       8.78e-07               1

Optimization terminated: first-order optimality is less than options.TolFun.

R =  1.7113 ç This reflux ratio will fit exactly 5 stages in the top section.

(2. b) The MATLAB function is ass6_2b.m

Note: Whatever the method used to get to the answer will be considered as correct. The goal here is not to grade you on how well you program, but instead to illustrate the influence of the changes made on the process through the operational variables.

Option 1: R=1.7113. Initial guess for xb=0.02

Number of stages is 6.

xb = fsolve('ass6_2b',0.02,opt,6,3.8,.98,1.7113,.45,.55)

Norm of      First-order   Trust-region

Iteration  Func-count     f(x)          step         optimality    radius

0          2         1.35432                          75.3               1

1          3         1.35432      0.0179969           75.3               1

2          5        0.698565     0.00449923           69.1          0.0045

3          7        0.249169      0.0101146            116          0.0112

4          9      0.00608688     0.00215219             13          0.0112

5         11     5.19756e-06    0.000468013          0.358          0.0112

6         13     4.07173e-12    1.45073e-05       0.000317          0.0112

7         15     6.40752e-26    1.28631e-08       3.97e-11          0.0112

Optimization terminated: first-order optimality is less than options.TolFun.

xb =  0.0080 ç Bottoms product purity corresponding to exactly 6 stages in bottom section.

Note: If you considered 6 trays in the bottom section, the number of stages should be 7

Option 2: R=1.7113. Initial guess for xb=0.02

Number of stages is 7.

Norm of      First-order   Trust-region

Iteration  Func-count     f(x)          step         optimality    radius

0          2         4.68182                           140               1

1          3         4.68182      0.0334614            140               1

2          5         2.16088     0.00836536            160         0.00837

3          6         2.16088      0.0134711            160          0.0209

4          8         1.07747     0.00336777            158         0.00337

5          9         1.07747     0.00681549            158         0.00842

6         11        0.559632     0.00170387            143          0.0017

7         13        0.146797     0.00391718            179         0.00426

8         15      0.00238924    0.000818338           17.5         0.00426

9         17     8.59217e-07    0.000136327           0.32         0.00426

10         19     1.15439e-13    2.68585e-06       0.000117         0.00426

11         21     4.20384e-25    9.85204e-10       2.24e-10         0.00426

Optimization terminated: first-order optimality is less than options.TolFun.

xb =  0.0036 ç Bottoms product purity corresponding to exactly 7 stages in bottom section. The composition will be less than the other case. You have considered an additional separation stage.

Posted November 03, 2006