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  CHEMICAL REACTION ENGINEERING CH3347 CLASS CC01—GROUP 2 -

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



CHEMICAL REACTION ENGINEERING (CH3347)

CLASS CC01—GROUP 2 - SEMESTER 212 INSTRUCTOR: Ph.D CHÂU NGỌC ĐỖ UYÊN

Student’s Name

LÊ NGỌC THẢO

VŨ THỊ PHI YẾN

Ho Chi Minh city, 20225.6

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− = (0.04 −1) − (0.01 −1)

 { 1 = 0.04 −1

2 = 0.01 −1

= 2 3 = 2000

0 =100 ⁄

0 = 100 ⁄

=

1

2

− = 1

1 0

= 1 0 [(1 − ) −1−0.0 .8 ]

=

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 =

0

100



= 0.506 = 50.6%

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Half-life time: t1/2= 5.2 days

Mean residence time t= 30 days

k =0.693=0.693= 0.1333 −1

1/2 5.2

For Mixed Flow Reator:

=

0

Fraction of activity = 1 - = 1 – 0.2 = 0.8

0

 % of activity = 80%

5.8.

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-rA = 0.004 CA – 0.01 CR

V = 2m3 = 2000 liter

FA = 100 l/min ,

CA0 = 100 mmol/l = 0.1 mol/l

At equation :

− 0 / 0

X Ae =

= 1 =4

2

 XAe = 0.8 = 80%

Actual conversion

0

CA = CA0 (1-XA)

CR = CR0 + CA0 XA

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

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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:

→

SOLUTION

0 = 25 ( min )

= 95%

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

From

= 0 (1 − ) = 2 x (1-0.95)=0.1 (mol/l)

→

0

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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

Find the expected conversion in a 22-liter reactor

Solution

A → 2.5 (products) ⇒ Irreversible reaction

− = (10 −1 ) ⇒ = 10 ( −1) (FIRST ORDER)

For this stoichiometry, A → 2.5 (products), the expansion factor is

=

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

We use equation (5.21):

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⟺10( 1)×22( )×

⇒ = 0.8997

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Example 3.9

The feed contains 20% mole inerts:

0.8A -> 0.4R + 0.4S

0.2 inerts -> 0.2 inerts

=> ε A = (0.8+0.2)−(0.8+0.2) 0.8+0.2 = 0

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

Calculate τ:

=

0 0

 τ = t f – t 0 = 0 =208 ( ) × 0.026 ( )= 194.69 (s) ≈ 195 (s)

0 100 × 36001 ( )

 tf = τ + t0 = 195 + 20 = 215 (s)

According to the data table, using interpolation, at t = 215 (s), pA = 0.125 (atm)

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-rA =

− 00.026 ( )

 XA = 84%

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