CN2116-HW7-Solution (XJP_ 2011)

8/11/2019 CN2116-HW7-Solution (XJP_ 2011)

1/12

CN2116 Homework Set #7 (due 14-03-2011) CN2116-XJP-2011-HW7

1

1. The F curve is shown below for a real reactor

What is the mean residence time?

Solution:

min5.2)4)(5.0()2)(5.0(2

1)(

)(

)()(

1

00

0

=+===

=

=

tdFdtttEt

dttEdF

dttEtF

m

t

2. The following data were obtained from a tracer test to a reactor.

t (s) 0 5 10 15 20 25 30 35

C(t) (mg/dm

3

) 0 0 0 5 10 5 0 0

(a) Plot C(t), E(t) and F(t) curves.

(b) Evaluate the mean residence time and the variance.

Solution:

(a)Plot C(t)

Plot E(t)


2/12


2

100

)(

)(

)()(

100)10)(2030(2

1)10)(1020(

2

1)(Area

0

30

tC

dttC

tCtE

dm

smgdttC

==

=+==

t(s) 0 5 10 15 20 25 30

C(t)(mg/dm3) 0 0 0 5 10 5 0

E(t) (s1) 0 0 0 0.05 0.1 0.05 0

0)(,30For

)30(01.0)()(,3020For

)10(01.0)()(,2010For

0)(,10For

2

1

=

==

==

=

tEt

ttEtEt

ttEtEt

tEt

Plot F(t)

5.33.0005.0)30(01.05.0)(5.0)(,3020For

5.01.0005.0)10(01.0)(0)(,2010For

)()(

2

202022

2

101011

0

+=+=+=

+==+=

=

ttdttdttEtFt

ttdttdttEtFt

dttEtF

tt

tt

t

(b) sdtttdtttdtttEdtttEdtttEtm

20)30)(01.0()10)(01.0()()()(30

20

20

10

30

20 2

20

10 1

0=+=+==

2230

20 2

220

10 1

22

0

2

0

22 6.16)()()()()( stdttEtdttEttdttEtdttEttmmm

=+===


3/12


3

3. A pulse input (N0= 1 mol, v = 4 liters/min) to a vessel gives the results shown in thefollowing figure.

(a)Check the material balance with the tracer curve to see whether the results areconsistent.

(b)If the result is consistent, determine the volume of the vessel V, the mean residence

time tm, and sketch the E(t) curve.

Solution

(a)Check to see if the results are consistent

By material balance:lit

mol

lit

mol

v

NArea

min25.0

min/4

10 ===

From the graph:lit

mol

lit

molArea

min25.0min)5)(05.0(

==

So the results are consistent.

(b) 1

0

min2.025.0

05.0

25.0

)(

)(

)()(

====

tC

dttC

tCtE

lit10min)/lit4min)(5.2(

min5.2)2.0()(0

5

0

=====

===

vtVv

Vt

dttdtttEt

mm

m


4/12


4

4. The first-order reaction, BA , with 1min8.0 =k is carried out in a real reactor with thefollowing RTD function

For 20 t then 22 )()( = ttE min-1(hemi circle)

For 2>t then 0)( =tE

(a)What is the mean residence time?(b)What is the variance?(c)What is the conversion predicted by the segregation model?(d)What is the conversion predicted by the maximum mixedness model?

Solution

(a)Mean residence time

By definition

=0

1)( dttE . The area of the semicircle representing the E(t) is given by

12

2

==

A and

2

= . For constant volumetric flow min8.02

===

mt .

(b)Variance

==

2

0

222

0

22 )()()()( dtttdttEt

Using polymath:

159.02 =Polymath results

POLYMATH Report

Ordinary Differential Equations 29-Nov-2010

Calculated values of DEQ variables

Variable Initial value Minimal value Maximal value Final value

1 E 0 0 0.7980869 0.0166534

2 E2 0 0 0.7980869 0.0166534

3 sigma 0 0 0.1593161 0.1593161

4 t 0 0 1.596 1.596

5 t1 1.596174 1.596174 1.596174 1.596174

6 tau 0.7980869 0.7980869 0.7980869 0.7980869

Differential equations

1 d(sigma)/d(t) = (t-tau)^2*E


5/12


5

Explicit equations

1 tau = (2/3.14)^0.5

2 t1 = 2*tau

3 E2 = (t*(2*tau-t))^(1/2)

4 E = if (t


6/12


6

8 Xbar 0 0 0.4447565 0.4447565


1 d(Xbar)/d(t) = X*E

Explicit equations1 tau = (2/3.14)^0.5

2 t1 = 2*tau

3 E2 = (t*(2*tau-t))^(1/2)

4 E = if (t


7/12


7




1 E 0.0166534 0 0.7980866 0

2 E1 0.0166534 0 0.7980866 03 F 1. -0.0005053 1. -0.0005053

4 k 0.8 0.8 0.8 0.8

5 lam 1.596 0 1.596 0

6 tau 0.7980869 0.7980869 0.7980869 0.7980869

7 x 0 0 0.4445289 0.4445289

8 z 0 0 1.596 1.596


1 d(x)/d(z) = -(-k*(1-x)+E/(1-F)*x)

2 d(F)/d(z) = -E

Explicit equations

1 k = .8

2 lam = 1.596-z

3 tau = (2/3.14)^0.5

4 E1 = (tau^2-(lam-tau)^2)^0.5

5 E = if (lam


8/12


8

[Discussion Topics]

1. A liquid macrofluid reacts according to RA as it flows through a vessel. Find theconversion of A for the following flow patterns and kinetics.

Solution

(a)Using the segregation model, dttEtXXbatch

)()(0

=

For a batch reactor we have

0A

A

C

r

dt

dX =

From the reaction kinetics, we have

zero order reaction, minmol/lit3 = Ar and mol/lit60 =AC

Solving for X(t), we have

2minfor t1,X

min2for,5.05.0)(

>=

== ttXttX

From the flow pattern (a), we have

3t0for,min3

1)( 1 = tE

Therefore, the mean conversion for a zero order reaction is

3

2)

3

1()1()

3

1()5.0(

3

2

2

0=+= dtdttX

(b)For a batch reactor we have

0A

A

C

r

dt

dX

=

From the reaction kinetics, we have

zero order reaction, minmol/lit03.0 = Ar and mol/lit1.00=AC

Solving for X(t), we have:

min33.3for t1,X

min33.3for,3.03.0)(

>=

== ttXttX


9/12


9

From the flow pattern (b), we know nothing leaves the reactor before 4 min, so

everything has reacted. 1=X

2. A step tracer input was used on a real reactor with the following results:

For 10t min, then CT= 0

For 3010 t min, then CT= 10 g/dm3

For 30>t min, then CT= 40 g/dm

3

The second-order reaction BA with min/moldm1.0 3 =k is to be carried out in the realreactor with an entering concentration of A of 1.25 mol/dm

3at a volumetric flow rate of 10

dm3/min. Here kis given at 325 K.

(a)What is the mean residence time tm?(b)What is the variance?(c)What conversion do you expect from an ideal PFR, an ideal CSTR, and a real reactor

with the same tm?

(d)What is the conversion predicted by the Segregation model and the Maximum Mixednessmodel?

Solution

(a)The cumulative distribution function F(t) is given:

The real reactor can be modeled as two parallel PFRs:

The relative 30minmin,10),(43)(

41)( 2121 ==+= tttE

Mean residence time

=+==1

0min25)75.0)(20()1)(10(tdFtm

or


10/12


10

min254

3

4

1)(

4

3)(

4

1)( 21

0 021

=+=

+==

dttttdtttEtm

(b)Variance:

2222

2

2

1

0 021

222

min75)2530(4

3)2510(

4

1)(

4

3)(

4

1

)(4

3)(

4

1)()()(

=+=+=

+==

mm

mm

tt

dtttttdttEtt

(c)Conversion X from an ideal PFR, an ideal CSTR, and a real reactor.

For a PFR, second-order, liquid phase, irreversible reaction with min/moldm1.0 3 =k ,

min25= and 30 mol/dm25.1=AC

758.01 0

0 =+

=A

A

Ck

CkX

For a CSTR, second-order, liquid phase, irreversible reaction with min/moldm1.0 3 =k ,

min25= and 30 mol/dm25.1=AC

572.0)1(

02 ==

XCk

X

XA

For two parallel PFRs, min101= and min302= , 001 4/1 AA FF = and 002 4/3 AA FF = ,

second-order, liquid phase, irreversible reaction with min/moldm1.0 3 =k , min25=

and 30 mol/dm25.1=AC

3

01

0101 /556.0)

11( dmmol

Ck

CkCC

A

AAA =

+=

3

02

0202 /263.0)

11( dmmol

Ck

CkCC

A

AAA =

+=

731.04

3

4

1

00

201000

=

=A

AAA

Cv

CvCvCv

X

(d)Conversion predicted by the Segregation model.

731.014

3

14

1

)(4

3)(

4

1

1)()(

20

20

10

10

021

0

0

0

=+

++

=

+

+==

A

A

A

A

A

A

kC

kC

kC

kC

dttttkC

tkCdttEtXX


11/12


11

Conversion predicted by the Maximum Mixedness model

XF

EXkC

d

dX

XkCkCr

XF

E

C

r

d

dX

A

AAA

A

A

)(1

)()1(

)1(

)(1

)(

2

0

220

2

0

+=

==

+=

We need to change the variable such the integration proceeds forward:

XzTF

zTEXkC

dz

dXA

)(1

)()1( 20

=

Solved by Polymath:

706.0=X

POLYMATH Report POLYMATH Report




1 ca 1.25 0.3593529 1.25 0.3672986

2 cao 1.25 1.25 1.25 1.25

3 E 0 0 1.25 0

4 E1 1.25 1.25 1.25 1.25

5 E2 1.25 1.25 1.25 1.25

6 E3 0 0 0 0

7 EF 0 0 2.427988 0

8 F 0.9999 -0.0001081 0.9999 -0.0001081

9 k 0.1 0.1 0.1 0.1

10 lam 40. 0 40. 0

11 ra -0.15625 -0.15625 -0.0129134 -0.0134908

12 t1 10. 10. 10. 10.

13 t2 30. 30. 30. 30.

14 x 0 0 0.7125177 0.7061611

15 z 0 0 40. 40.


1 d(x)/d(z) = -(ra/cao+E/(1-F)*x)

2 d(F)/d(z) = -E


12/12


12

Explicit equations

1 cao = 1.25

2 k = .1

3 lam = 40-z

4 ca = cao*(1-x)5 t1 = 10

6 t2 = 30

7 E3 = 0

8 ra = -k*ca^2

9 E2 = 0.75/(t2*2*(1-0.99))

10 E1 = 0.25/(t1*2*(1-0.99))

11E = if ((lam>=0.99*t1)and(lam=0.99*t2)and(lam

CN2116-HW7-Solution (XJP_ 2011)

Documents

Transcript of CN2116-HW7-Solution (XJP_ 2011)