Ch E 416 - Assignment 6 Solutions

[Question 1]

[Question 2]

Note: If you subtracted the last equilibrium stage to obtain the number of trays needed inside the column, you answer is correct. If you have considered the number of trays as the total number of equilibrium stages including the reboiler, your answer will be marked as correct. We apologize for the inconveniences this could have caused. In any case, there is only a difference in the stripping section.

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 (x_i , y_i) of top operating line and the q-line.

Enriching operating line:

q – line:

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

2. Start at

3. Calculate from

4. Calculate from

5. Calculate from

6. Calculate from

7. Calculate from

8. Calculate from

9. Calculate from

10. Calculate from

11. Calculate from

12. Calculate from

13. Calculate from

< , then switch to the Stripping operating line

;

14. Calculate from

15. Calculate from

16. Calculate from

17. Calculate from

18. Calculate from

19. Calculate from

20. Calculate from

21. 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

2. Calculate from

3. Calculate from

4. Calculate from

5. Calculate from

6. Calculate from

7. Calculate from

8. Calculate from

9. Calculate from

10. 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:

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

2. Now find R_min 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 (x_i , y_i) of top operating line and the q-line.

Enriching operating line:

q – line:

Enriching section:

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

;

3. Compute coefficients of quadratic.

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

4. Estimate Ne,

Striping section:

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

6. Compute coefficients of quadratic

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

7. 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.

[Question 1]

[Question 2]

Posted November 03, 2006

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