Mini Project Reaction Engineering GROUP 9 (Stage 2) (1)

Mini Project Reaction Engineering (CKB 20104) Semester January 2017 Group Number: ____9___

Student Name *According to the work task distributed

Section

Student ID

1)SHARIFAH NASUHA BINTI S. MAHADI

LO1

55213115264

2)MUHAMMAD AFNAN BIN SYIHABUDDIN

LO1

55213115409

3)MUHAMMAD HAFIZAN HAKIMIN BIN MAHADZIR

LO1

55213115277

4)MUHAMMAD HILMI BIN JAMALUDIN

LO1

55213115219

Calculation Basis: 5.5

  

Type of Reactor: Continuous Stirred Tank Reactor (CSTR)

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Table of Contents STAGE 2

1.0 MASS BALANCE ...................................................... ............................................................................. .............................................. .............................. ....... 3 2.0 STOCHIOMETRIC TABLE........................................... ................................................................. ............................................ ........................... ..... 6 3.0 REACTION KINETICS ...................................................... ............................................................................ ........................................... ..................... 16

4.0 REACTOR DESIGN AND SIZING .......................................... ................................................................. .................................... ............. 21

.................................................................. ............................................. ............................................. ............................ ..... 32 REFFERENCES ...........................................

APPENDICES ........................................... ................................................................. ............................................ ............................................. ................................ ......... 33















1.0

Mass Balance 1.1 Calculate an overall mass balance.

₃

₂ ₅

CH OOC H

A

+

ṁAo = 121.3803

₃

+ NaOH

B

C

 

₂ ₅

CH COONa +

C H OH

+

D

CSTR

ṁA = 36.4159 ṁBo = 275.4993

 

 

ṁB = 236.9302

 

   ṁD = 44.4238  ṁC = 79.1058

OVERALL MASS BALANCE Mass flow rate total in MAo +

MBo

   396.8796 

+

=

MCo

+

 

Mass flow rate total out MDo

=

 

MA

+

 

MB

+

 

MC +

121.3803 + 275.4993 = 36.4159 + 236.9302 +79.1058 + 44.4238 = 396.8757

 

MD

 















1.2

Start with balanced stoichiometric equation.

SPECIES

SYMBOL

CH₃COOC₂H₅ NaOH CH₃COONa C₂H₅OH Total

A B C D

1.3

INITIAL (

  )

FAo = 1.3776 FBo = 6.888 FTo = 8.2656 Table 1.0 : Stoichiometric table

REMAINING (

  )

F A = 0.4133 FB = 5.9237 FC = 0.9643 FD = 0.9643 FT = 8.2656

Present your calculations in a MASS BALANCE TABLE with the IN and OUT numbers as shown below. The units must be presented in quantities of g/s.

Table 2.0 : Mass Balance table.

 )

TYPE OF SPECIES

IN (

CH₃COOC₂H₅

ṁAo = FAo x mw(NaOH) =1.3776 x 88.11 = 121.3803 ṁBo = FBo x mw (CH₃COOC₂H₅) = 6.888 x 39.997 = 275.4993

NaOH

C₂COONa

-

C₂H₅OH

-

Total

ṁTo= ṁAo + ṁBo = 121.3803 + 275.4993 = 396.8796

 

OUT ( )

ṁA = FA x mw(NaOH) = 0.4133 x 88.11 = 36.4159 ṁB = FB x mw (CH ₃COOC₂H₅) = 5.9237 x 39.997 = 236.9302 ṁC = FC x mw(CH₃COONa) = 0.9643 x 82.0344 = 79.1058 ṁD = FD x mw(C₂H₅OH) = 0.9643 x 46.06844 = 44.4238 ṁT = ṁA + ṁB + ṁC + ṁD = 36.4159+ 236.9302+ 79.1058 + 44.4238 = 396.8757















1.4

Based on the given production rate, determine the change of each species mass flow rate if the production rate is increased from 0 to 100%. Plot a Figure that charted the changes for each species mass flow rate versus production rate increment (0, 20, 40, 60, 80, and 100%).

Table 3.0 : Mass balance table with the changes of each species mass flow rate. Production Rate Increment (%) 0

ṁAo

ṁBo

ṁA

ṁB

ṁC

ṁD

ṁTo

ṁT

(g/s)

(g/s)

(g/s)

(g/s)

(g/s)

(g/s)

(g/s)

(g/s)

121.3821

275.5033

36.41463

236.933

79.10814

44.42514

396.8854

396.8809

20

145.6582

330.6032 3 30.6032

43.69745

284.3187

94.93009

53.31035

476.2614

476.2566

40

169.9351

385.7051

50.98053

331.7063

110.752

62.19556

555.6402

555.6345

60

194.2112

440.8049 4 40.8049

58.26335

379.0924

126.5732

71.08032

635.0161

635.0092

80

218.4872

495.9048

65.54617

426.478

142.3951

79.96553

714.3921

714.3849

100

242.7642

551.0067 5 51.0067

72.82926

473.8657

158.2339

88.85074

793.7709

793.7795

600 500 S / G E 400 T A R W300 O L F S 200 S A M 100 0 0

20

40

60

80

100

120

PRODUCTION RATE INCREMENT % ṁAo

ṁBo

ṁA

ṁB

ṁC

ṁD

Figure 1.0 : The changes for each species mass flow rate versus production production rate increment.















2.0

STOCHIOMETRIC TABLE

2.1 Develop a general stoichiometric table for the r eaction system using only symbols to represent the variables. [1]

A

+

B

CH3COOC2H5 + species CH3COOC2H5

NaOH

symbol A

→

C

+

D

→ CH3COONa + C2H5OH initial(mol/s) FA0

change -FA0X - FA0X

NaOH

B

FB0 = FA0.ѲB

CH3COONa

C

FC0 = FA0.ѲC

C2H5OH

D

FD 0 = FA0.ѲD

   + FA0X  +

  FA0X

FT0 = FA0 + FA0.ѲB + FA0.ѲC + FA0.ѲD Table 1 : stochiometric tabble symbol

remaining(mol/s) FA = FA0-FA0X

   FC = FA0.ѲC + FA0X 

FB = FA0.ѲB - FA0X

FD = FA0.ѲD +

  FA0X

FT = FA+ FB+ FC+ FD

















2.2 Subsequently develop another stoichiometric table and substitute in the real NUMBERS based on own calculations using the basis given in the assigned task . [1] Production rate (sodium acetate) : 5.5 Million lbm/year Convert ibm to gram (1 lbm = 453.592 37 Gram)

= 5.5 Million

 .       ℎ x    x   x  ℎ x  

= 79.108 g/s

Given conversion X : 70%

Molar mass Sodium Acetate : 82.0343 g/mol Fc

= 79.108 g/s x = 0.9643

Fc = Fc0 +

 

   X

 =  =

. . 

= 1.3776 

  . 

















EtAc initial conc. , CB0: 0.1 M NaOH initial conc., CA0 : 0.5 M ѲB =

 . M  = . M

SPECIES

: ѲB = 5

Temperature : 340 K NaOH conversion X :70%

SYMBOL INITIAL (mol/s)

CHANGE (mol/s)

REMAINING (mol/s) FA = FA0-FA0X = 1.3776-0.96432 1.3776-0.96432 = 0.41328

NaOH

A

FA0 = 1.3776

-FA0X - 1.3776(0.7) = - 0.96432

CH3COOC2H5

B

FB0 = FA0.ѲB =(1.3776)(5) = 6.888

- FA0X

   - (1.3776)(0.7)  = - 0.96432

CH3COONa

C

FC0 = FA0.ѲC =0

   + (1.3776)(0.7)  + FA0X

= + 0.96432 C2H5OH

D

FD0 = FA0.ѲD =0

   + (1.3776)(0.7)  + FA0X

= + 0.96432 TOTAL

FT O = 1.3776 + 6.888+ 0 + 0 = 8.2656 mol/s Table 2 : stochiometric table with real numbers and value

 

FB = FA0.ѲB - FA0X = 6.888 - 0.96432 = 5.92368 FC = FA0.ѲC +

  FA0X

= 0 + 0.96432 = 0.96432 FD = FA0.ѲD +

  FA0X

= 0 + 0.96432 = 0.96432

FT = 0.41328 +5.92368 + 0.96432 + 0.96432 = 8.2656 mol/s

















2.3 Based on the given production rate, determine the change of each species mass flowrate/mass if one of the raw material mat erial capacity is increased from 0 to 100%. Develop the stoichiometric tables for each scenario (0, 20, 40, 60, 80, and 100%). 1 For 0% INCREASE MASS FLOWRATE : Molar mass Sodium Acetate : 82.0343 g/mol Production rate = 79.108 g/s

Fc

= 79.108 g/s x

Fc = Fc0 +

   = 0.9643 .  

   X

EtAc initial conc. , CB0: 0.1 M NaOH initial conc., CA0 : 0.5 M

 =  = 1.3776   ѲB =

 . M  = . M

SPECIES

: ѲB = 5

Temperature : 340 K NaOH conversion X :70%


CHANGE (mol/s)


NaOH

A

FA0 = 1.3776

-FA0X - 1.3776(0.7) = - 0.96432

CH3COOC2H5

B

FB0 = FA0.ѲB =(1.3776)(5) = 6.888

- FA0X

   - (1.3776)(0.7)  = - 0.96432

CH3COONa

C

FC0 = FA0.ѲC =0

  + (1.3776)(0.7)  + FA0X

= + 0.96432 C2H5OH

D

FD0 = FA0.ѲD =0

 

+ FA0X

 

FB = FA0.ѲB - FA0X = 6.888 - 0.96432 = 5.92368 FC = FA0.ѲC +

  FA0X

= 0 + 0.96432 = 0.96432 FD = FA0.ѲD +

  FA0X

















For 20% INCREASE MASS FLOWRATE: From Task 2 table, the value of FA is 0.41328 mol/s Find mass of Raw material A (Ethyl acetate)

Molacular weight of Ethyl Acetate : 88.11g/mol MA = FA

2

 × molar mass of A = 0.41328  × 88.11   = 36.4141 g/s

New mass flowrate after 20% increment = 43.69 g/s

Fa = mass flowrate Find new FA0

SPECIES

    = 43.69 × ×      .  FA0 =

 (−)

= 0.4958 mol/s

= 1.6526 mol/s


CHANGE (mol/s)


NaOH

A

FA0 = 1.6526

-FA0X - 1.6526(0.7) = - 1.15682

CH3COOC2H5

B

FB0 = FA0.ѲB =(1.6526)(5) = 8.263

- FA0X

   - (1.6526)(0.7)  = - 1.15682

CH3COONa

C

FC0 = FA0.ѲC =0

   + (1.6526)(0.7)  + FA0X

= + 1.15682 C2H5OH

D

FD0 = FA0.ѲD =0

   + (1.6526)(0.7)  + FA0X

= + 1.15682

 

FB = FA0.ѲB - FA0X = 8.263- 1.15682 = 7.1061 FC = FA0.ѲC +

  FA0X

= 0 + 1.15682 = 1.15682 FD = FA0.ѲD +

  FA0X

= 0 + 1.15682 = 1.15682


















Molacular weight of Ethyl Acetate : 88.11g/mol

× molar mass of A   × = 0.41328 88.11  

MA = FA

= 36.4141 g/s New mass flowrate after 40% increment = 50.9797 g/s

Fa = mass flowrate = 50.9797

 ×    

   ×  . 

Find new FA0

SPECIES

FA0 =

 (−)

= 0.5785 mol/s = 1.9283 mol/s


CHANGE (mol/s)


NaOH

A

FA0 = 1.9283

-FA0X - 1.9283(0.7) = - 1.34981

CH3COOC2H5

B

FB0 = FA0.ѲB =(1.9283)(5) = 9.6415

- FA0X

  - (1.9283)(0.7)  = - 1.34981

CH3COONa

C

FC0 = FA0.ѲC



+ FA0X

 

FB = FA0.ѲB - FA0X = 9.6415- 1.34981 = 8.29169 FC = FA0.ѲC +



FA0X





















× molar mass of A   × = 0.41328 88.11  

MA = FA


Fa = mass flowrate = 58.2625 Find new FA0

SPECIES NaOH

 ×    

   ×  .  FA0 =

 (−)

= 0.6612 mol/s = 2.204 mol/s

SYMBOL INITIAL (mol/s) A

FA0 = 2.204

CHANGE (mol/s)

REMAINING (mol/s)

-FA0X

FA = FA0-FA0X



















× molar mass of A   × = 0.41328 88.11  

MA = FA

= 36.4141 g/s New mass flowrate after 80% increment = 65.545g/s


SPECIES

 ×    

   ×  .  FA0 =

= 0.7438 mol/s

 (−)

= 2.479 mol/s


CHANGE (mol/s)

REMAINING (mol/s)



















× molar mass of A   × = 0.41328 88.11  

MA = FA



SPECIES NaOH

 ×    

   ×  .  FA0 =

 (−)

= 0.8265 mol/s = 2.755 mol/s

SYMBOL INITIAL (mol/s) A

FA0 = 2.755

CHANGE (mol/s)

REMAINING (mol/s)

-FA0X

FA = FA0-FA0X

















Determine the change of each species mass flowrate if one of the raw material capacity increase.

Molecular weight of ethyl acetate = 88.11 g/mol Molecular weight of sodium hydroxide h ydroxide = 39.99711 g/mol Molecular weight of sodium acetate = 82.0343 g/mol Molecular weight of ethanol = 46.06844 g/mol

RAW material A ṁA

ṁB

ṁC

ṁD

(g/s)

(g/s)

(g/s)

(g/s)

0

36.41463

236.933

79.10814

44.42514

20

43.69745

284.3187

94.93009

53.31035

40

50.98053

331.7063

110.752

62.19556

60

58.26335

379.0924

126.5732

71.08032

80

65.54617

426.478

142.3951

79.96553

100

72.82926

473.8657

158.2339

88.85074

Increment (%)

















3.0

REACTION KINETIC 3.1 Chemical equation:

CH3COOC2H5 + NaOH

CH3COONa + C2H5OH

The reaction is irreversible i rreversible reaction and non-elementary non-elementary rate law. 3 The reaction is first order with respect to NaOH and CH3COOC2H5 Overall order of the reaction is second order

3.2 Determine Rate constant, K K 1 = 25.0 L/mol.min4 T = 28˚C = 301 K Ea = 11,120 cal/mole R = 1.987 cal/mol.k

 ,    ) + ) ( ln k 2 = ln (25.0) + (  ..   ln k 2 = 5.351557879 k 2 = 0.21694 x

10





















Conversion, X = 70% = 0.70 CA =

 =  (1-X) = CA 



CA = CAO (1-X) CA = 0.1 (1-0.70) = 0.03

 

 =  (B -  x)

CB =





 



CB = CAO ( B - x) CB = 0.1(5-1(0.70)) CB = 0.43

 

kC C

O (1-X)

















3.3 Changing of temperature (Increase and decrease 10˚C) 3.3.1 Increase 10˚C T= 340 k ; New T = 350 K ln k 2 = ln k 1 +

 



 +  )

(

ln k 2 = 0.21694 x 10

 ,    +( ( ) +    ) . ..

ln k 2 = 5.82186 k 2 = 337.599

 .

-rA= kCACB

-r A= A= (337.599

   ) ( 0.06 )(0.86 .   )

















3.4 Changing of Pressure (Increase and decrease 2 MPa) In this process saponification which is the reaction between sodium hydroxide and ethyl acetate producing ester and alcohol. The pressure is ne gligible and the pressure drop does not affect liquid phase reaction. 5 3.5 Changing of conversion (Increase and decrease of 20%)

3.5.1 Increase 20% Conversion = 70% = 0.70 New conversion = 90% = 0.90

CA = CAO (1-X)

















3.5.2

Decrease 20%

Conversion = 70% = 0.70 New conversion = 50% = 0.50

CA = CAO (1-X)

CA = (0.1) (1-0.5) CA = 0.05

 

CB = CAO ( B - x)

CB = (0.1)(5-0.5) CB = 0.45 -rA= kCACB

















4.0 REACTOR DESIGN AND SIZING

4.1 Find the Volume of reactor assigned (CSTR)

 +  → +   +  →  +  Overall rate law :  =   =       

 = 0.03 0 3     = 0.43 43   K = 210.94 .

*Obtained from student 2



















4.1

Volume of reactor when the pressure is changed 

There will be no effect and changes as there ther e is no pressure involve in liquid state 6 reactant and is assumed negligible.

















4.2 Volume of reactor when the temperature is changed 4.2.1 Temperature increase by 10 10C ( at T = 350K )

ln  = ln +  ( 1  1 )    11120   1  1 ) ln  = ln210.94 . + (  340 350

















From the calculation (c. i) it shows that when temperature is added by 10 C, the volume required for the CSTR will decrease from to .

21.2629  13.2857     =    =  ln  = ln +  ( 1  1 )

















4.2.2 Temperature decrease by 10 10C ( at T = 330K )

ln  = ln +  ( 1  1 )    11120   1  1 ) ln  = ln210.94 . + (  340 330 1.987 .

















From the calculation (c. ii) it shows that when temperature is reduced by 10 C, the volume required for the CSTR will decrease from to .

21.2629  35.01     =    =  ln  = ln +  ( 1  1 )



















4.3 Volume of reactor when the conversion is changed

















From the calculation (4.3.1) it shows that when conversion is increased b y 20%, the volume

















4.3.2

Conversion decrease by 20% ( X = 0.5 )

















From the calculation (4.3.1) it shows that when conversion is reduced by 20%, the volume

















4.4

Diagram of assigned reactor





























































































































Mini Project Reaction Engineering GROUP 9 (Stage 2) (1)

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