(TIỂU LUẬN) CHEMICAL REACTION ENGINEERING (CH3347) CLASS CC01—GROUP 2 SEMESTER 212

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY - VNU
OFFICE FOR INTERNATIONAL STUDY PROGRAM
FACULTY OF CHEMICAL ENGINEERING

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CHEMICAL REACTION ENGINEERING (CH3347)
CLASS CC01—GROUP 2 --- SEMESTER 212
INSTRUCTOR: Ph.D. CHÂU NGỌC ĐỖ UYÊN
Student’s Name

Student’s ID

LÊ PHẠM NHÃ THY

1953016

LÊ NGỌC THẢO

1952123

ĐẬU NGUYỄN ANH THƯ

1952476

VŨ THỊ PHI YẾN

1952542

TRƯƠNG HẠNH THANH TUYỀN

1915806

LÝ THANH THUÝ VY

1852885

Ho Chi Minh city, 2022


SOLUTION
1−
( 0.01 −1
− = (0.04 )  −

 1 = 0.04 
 {−1
 2 = 0.01 
−1
 = 2  3 = 2000


)

 0 = 100 ⁄


0

= 100 ⁄

 =



1



2

=


0.04

=>
=
1 − 
0.01
1− 



− =   (1 −  ) −  



1 0



2 0 
 (1−  )

1
= 1  0 (1 − 
 ) −





 

0 

1− 


= 1  0 [(1 − 
 ) −   ]



1−0.8

= 1  0 [(1 − 
 ) − 0.8  ]


= 1  0 (1 − 1.25 
)




=

∫
0
 =
(− )
0
0

=>  = 0.8 =
80%






0

= 

2000
100

0



∫0







 1  0 (1−1.25  )

= 100 ∫
0





0.04×100× (1−1.25  )

  = 0.506 =
50.6%


SOLUTION
Half-life time: t1/2= 5.2 days
Mean residence time t= 30 days
0.693

k=

 1/2

=

0.693
5.2

= 0.1333 
−1

For Mixed Flow Reator:

1
1
=
=
= 0.2


1 +  1 + 0.1333 ×
30
0



Fraction of activity = 1 - 



0

 % of activity = 80%

= 1 – 0.2 = 0.8


SOLUTION
-r = 0.004 C – 0.01 C
A

A

R

V = 2m3 = 2000 liter
FA = 100 l/min ,
CA0 = 100 mmol/l = 0.1 mol/l
At equation :
X

=

 −

 /
0 0

 +1

Ae


1

 =  = 4
2

 XAe = 0.8 = 80%
Actual conversion

 0

=


−

=

2000
100

=


0.04  − 0 .01 

CA = CA0 (1-XA)
CR = CR0 + CA0 X
A

Substituting all values, we get XA = 0.072 = 7.2 %


5.9. A specific enzyme acts as catalyst in the fermentation of reactant A. At a given
enzyme concentration in the aqueous feed stream (25 liter/min) find the volume of plug
flow reactor needed for 95% conversion of reactant A (CA0 = 2 mol/liter). The kinetics of
the fermentation at this enzyme concentration is given by:


→

,

0.1 


− = 1
+ 0.5  . 


SOLUTION



 0 = 25 min
( )

We already have { 

0



= 2(  )

 = 95%

For the volume of plug flow reactor, we have this equation with  = 0

From
− =

0.1 

1+0.5 

 =  0 (1 −  ) = 2 x (1-0.95)=0.1 (mol/l)




→  =−∫
0



 0

0.1


=−∫
−
2

1 + 0.5 
  ≈ 39.457
0.1 



→  = 39.457 × 25 ( ) = 986.433(
)
min


5.10. A gaseous feed of pure A (2 mol/liter, 100 mol/min) decomposes to give a variety
of products in a plug flow reactor. The kinetics of the conversion is represented by
A → 2.5 (products), − = (10 −1 )

Find the expected conversion in a 22-liter reactor.
Solution
A → 2.5 (products) ⇒ Irreversible reaction
(
−1 ) (FIRST
− = (10 −1 )  ⇒  = 10
ORDER)
For this stoichiometry, A → 2.5 (products), the expansion factor is
 =

2.5 − 1
= 1.5
1

We use the equation (5.8) in Handbook Chemical Reaction Engineering by Octave
Levenspiel:

We use equation (5.21):

⟺ ×


= −(1 +  ) ln(1 −  ) − 
 0




 

1
1

) × 22() ×
⟺10 (
× 2 ( ) = −(1 + 1.5) ln(1 −  ) − 1.5 ×

 

100 ( )

⇒  = 0.8997


Example 3.9

The feed contains 20% mole inerts:
0.8A -> 0.4R + 0.4S
0.2 inerts -> 0.2 inerts
=> εA =

(0.8+0.2)−(0.8+0.2)

=0

0.8+0.2

pA0 = 0.8P0 = 0.8 x 1 = 0.8 (atm) at t0 = 20 (s)
 C

=
A0



0



0.8 ()

= 0.082 (.) × (100+273)() = 0.026 (
.




)

Calculate τ:



0

=


 0

 τ = tf – t0 =



0

 0



=

208() × 0.026 ( )
1

100×
( )
3600

= 194.69 (s) ≈ 195 (s)



 tf = τ + t0 = 195 + 20 = 215 (s)
According to the data table, using interpolation, at t = 215 (s), pA = 0.125 (atm)
 CA = 0.0041 (




)

For mixed flow reactor with εA = 0:


-rA =
τ=

 0−  

 0
−





=

0.026 − 0 .0041 ( )

=> XA =

 XA = 84%.

195 ()
×(− )


 0

= 1.12 x 10-4 (


.

)

−4 
195() × (1.12×10 )(
. )
=
=

0.026 ( )

0.84