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

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

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, 20225.6

SOLUTION

−𝑟𝐴 = (0.04𝑚𝑖𝑛−1)𝐶𝐴 − (0.01𝑚𝑖𝑛−1)𝐶𝑅

 {𝑘1 = 0.04𝑚𝑖𝑛−1

𝑘2 = 0.01𝑚𝑖𝑛−1

𝑉 = 2𝑚3 = 2000𝐿

𝜐0 = 100 𝐿 𝑚𝑖𝑛⁄

𝐶𝐴0 = 100 𝑚𝑚𝑜𝑙 𝐿⁄

𝐾 = 𝑘1

𝑘2 =

𝑋𝐴𝑒

1 − 𝑋𝐴𝑒 =>

0.04 0.01 =

𝑋𝐴𝑒

1 − 𝑋𝐴𝑒 => 𝑋𝐴𝑒 = 0.8 = 80%

−𝑟𝐴 = 𝑘1𝐶𝐴0(1 − 𝑋𝐴) − 𝑘2𝐶𝐴0𝑋𝐴

= 𝑘1𝐶𝐴0(1 − 𝑋𝐴) − 𝑘1 (1− 𝑋𝐴𝑒)

𝑋 𝐴𝑒 𝐶𝐴0𝑋𝐴

= 𝑘1𝐶𝐴0[(1 − 𝑋𝐴) − 1− 𝑋𝐴𝑒

𝑋 𝐴𝑒 𝑋𝐴]

= 𝑘1𝐶𝐴0[(1 − 𝑋𝐴) − 1−0.8

0.8 𝑋𝐴]

= 𝑘1𝐶𝐴0 (1 − 1.25𝑋𝐴)

𝜏 = 𝑉

𝜐0 = 𝐶𝐴0∫

𝑑𝑋𝐴 (−𝑟𝐴)

𝑋 𝐴

0

 𝑉

𝜐 0 = 𝐶𝐴0∫ 𝑑𝑋𝐴

𝑘 1 𝐶 𝐴0 (1−1.25𝑋 𝐴 )

𝑋 𝐴

0

 2000

0.04 ×100 × (1−1.25𝑋 𝐴 )

𝑋𝐴 0

 𝑋𝐴 = 0.506 = 50.6%

SOLUTION

Half-life time: t1/2= 5.2 days

Mean residence time t= 30 days

k = 0.693

𝑡1/2 =0.693

5.2 = 0.1333𝑑−1

For Mixed Flow Reator:

𝐶𝐴

𝐶𝐴0 =

1

1 + 𝑘𝑡 =

1

1 + 0.1333 × 30= 0.2

Fraction of activity = 1 - 𝐶𝐴

𝐶𝐴0 = 1 – 0.2 = 0.8

 % of activity = 80%

5.8

SOLUTION

-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 :

XAe = 𝐾𝐶−𝐶𝑅0/𝐶𝐴0

𝐾𝐶+1

𝐾𝑐 =𝑘1

𝑘 2 = 4

 XAe = 0.8 = 80%

Actual conversion

𝑉

𝐹 𝐴0 = 𝑋𝐴

−𝑟 𝐴 = 2000

100 = 𝑋𝐴

0.04𝐶 𝐴 − 0.01𝐶 𝑅

CA = CA0 (1-XA)

CR = CR0 + CA0 XA

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:

𝐴𝑒𝑛𝑧𝑦𝑚𝑒→ 𝑅, −𝑟𝐴 = 0.1𝐶𝐴

1 + 0.5𝐶𝐴

𝑚𝑜𝑙 𝑙𝑖𝑡𝑒𝑟 𝑚𝑖𝑛

SOLUTION

We already have {

𝑣0 = 25 ( 𝑙

min)

𝐶𝐴0 = 2(𝑚𝑜𝑙

𝑙 )

𝑋𝐴 = 95%

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

From

−𝑟𝐴 = 0.1𝐶𝐴

1+0.5𝐶 𝐴

𝐶𝐴 = 𝐶𝐴0(1 − 𝑋𝐴) = 2 x (1-0.95)=0.1 (mol/l)

𝑣0 = − ∫

𝑑𝐶𝐴

−𝑟𝐴

𝐶 𝐴𝑓

𝐶𝐴0

= − ∫ 1 + 0.5𝐶𝐴

0.1𝐶𝐴

0.1

2

𝑑𝐶𝐴 ≈ 39.457

→ 𝑉 = 39.457 × 25 ( 𝑙

min) = 986.433 (𝑙𝑖𝑡𝑒𝑟)

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), −𝑟𝐴 = (10 𝑚𝑖𝑛−1) 𝐶𝐴 Find the expected conversion in a 22-liter reactor

Solution

A → 2.5 (products) ⇒ Irreversible reaction

−𝑟𝐴 = (10 𝑚𝑖𝑛−1) 𝐶𝐴 ⇒ 𝑘 = 10 (𝑚𝑖𝑛−1) (FIRST ORDER)

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

𝜀𝐴 =2.5 − 1

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

We use equation (5.21):

⟺ 𝑘 × 𝑉

𝜐0 = −(1 + 𝜀𝐴) ln(1 − 𝑋𝐴) − 𝜀𝐴𝑋𝐴

⟺ 10 ( 1

𝑚𝑖𝑛) × 22 (𝐿) ×

1

100 (𝑚𝑜𝑙𝑚𝑖𝑛)

× 2 (𝑚𝑜𝑙

𝐿 ) = −(1 + 1.5) ln(1 − 𝑋𝐴) − 1.5 × 𝑋𝐴

⇒ 𝑋𝐴 = 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)

0.8+0.2 = 0 pA0 = 0.8P0 = 0.8 x 1 = 0.8 (atm) at t0 = 20 (s)

 CA0 = 𝑝𝐴0

𝑅𝑇 = 0.8 (𝑎𝑡𝑚) 0.082 (𝐿.𝑎𝑡𝑚𝑚𝑜𝑙.𝐾) × (100+273)(𝐾) = 0.026 (𝑚𝑜𝑙

𝐿 ) Calculate τ:

𝑉

𝐹𝐴0 = 𝜏

𝐶𝐴0

 τ = tf – t0 = 𝑉𝐶𝐴0

𝐹𝐴0 = 208 (𝐿) × 0.026 (

𝑚𝑜𝑙

100 ×36001 (𝑚𝑜𝑙𝑠 ) = 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 (𝑚𝑜𝑙

𝐿 ) For mixed flow reactor with εA = 0:

-rA = 𝐶𝐴0− 𝐶𝐴

𝜏 = 0.026 − 0.0041 (

𝑚𝑜𝑙

195 (𝑠) = 1.12 x 10-4 (𝑚𝑜𝑙

𝐿.𝑠)

τ = 𝐶𝐴0 𝑋𝐴

−𝑟 𝐴 => XA = 𝜏×(−𝑟𝐴)

𝐶 𝐴0 = 195(𝑠) × (1.12×10

−4 )(𝑚𝑜𝑙𝐿.𝑠) 0.026 (𝑚𝑜𝑙𝐿 ) = 0.84

 XA = 84%