Chemical Engineering Computation
with MATLAB® 2nd Edition by
Yeong Koo Yeo
Complete Chapter Solutions Manual
are included (Ch 1 to 11)
** Immediate Download
** Swift Response
** All Chapters included
** MATLAB Programs
** Lecture Slides
,Table of Contents are given below
1. Introduction to MATLAB®
2. Numerical Methods with MATLAB®
3. Physical Properties
4. Thermodynamics
5. Fluid Mechanics
6. Chemical Reaction Engineering
7. Mass Transfer
8. Heat Transfer
9. Process Control
10. Optimization
11. Computational Intelligence
,Chapter 2 Numerical Methods with MATLAB
Linear Systems
2.1 In the photosynthesis reaction, water reacts with carbon dioxide to give glucose and oxygen. This
reaction can be expressed as
𝑥1 𝐶𝑂2 + 𝑥2 𝐻2 𝑂 → 𝑥3 𝑂2 + 𝑥4 𝐶6 𝐻12 𝑂6
Determine the values of coefficients 𝑥1 , 𝑥2 , 𝑥3 , and 𝑥4 to balance the equation. Is it possible to
determine these values? If not, under what conditions can the solutions be found?
2.1(Solution)
Carbon balance: 𝑥1 = 6𝑥4 , oxygen balance: 2𝑥1 + 𝑥2 = 2𝑥3 + 6𝑥4 , hydrogen balance: 2𝑥2 = 12𝑥4 .
Rearrangement of these equations gives
𝑥1
1 0 0 −6 𝑥 0
2
[2 1 −2 −6 ] [𝑥 ] = [0]
3
0 2 0 −12 𝑥 0
4
We can use the backslash operator to get the solution:
>> x = A\b
x=
0
0
0
0
The given equations can be rewritten as
𝑥1 − 6𝑥4 = 0, 𝑥2 − 6𝑥4 = 0, 𝑥3 − 6𝑥4 = 0 ⇒ 𝑥1 = 6𝑥4 , 𝑥2 = 6𝑥4 , 𝑥3 = 6𝑥4
Thus if we set 𝑥4 = 1, we have 𝑥1 = 𝑥2 = 𝑥3 = 6.
2.2 Four reactors are connected by pipes where directions of flow are depicted by means of arrows as
shown in Figure P2.218. The flow rate of the key component is given by the volumetric flow rate
𝑄 (𝑙𝑖𝑡𝑒𝑟/𝑠𝑒𝑐) multiplied by the concentration 𝐶 (𝑔/𝑙𝑖𝑡𝑒𝑟) of the component. The incoming flow rate is
assumed to be equal to the outgoing rate. Using the flow rates given below, calculate the concentration at
each reactor:
𝑄13 = 75 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 , 𝑄24 = 20 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 , 𝑄33 = 60 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 ,
𝑄21 = 25 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 , 𝑄32 = 45 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 , 𝑄43 = 30 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐
FIGURE P2.2
2.2(Solution)
Material balance for each reactor can be expressed as follows:
Reactor 1: 350 + 𝑄21 𝐶2 = 𝑄13 𝐶1 ⇒ 350 + 25𝐶2 = 75𝐶1 ⇒ 75𝐶1 − 25𝐶2 = 350
Reactor 2: 𝑄32 𝐶3 = 𝑄21 𝐶2 + 𝑄24 𝐶2 ⇒ 45𝐶3 = 25𝐶2 + 20𝐶2 ⇒ 45𝐶3 − 45𝐶2 = 0
, Reactor 3:
𝑄13 𝐶1 + 𝑄43 𝐶4 = 𝑄32 𝐶3 + 𝑄33 𝐶3 ⇒ 75𝐶1 + 30𝐶4 = 45𝐶3 + 60𝐶3 ⇒ 75𝐶1 + 30𝐶4 − 105𝐶3 = 0
Reactor 4: 150 + 𝑄24 𝐶2 = 𝑄43 𝐶4 ⇒ 150 + 20𝐶2 = 30𝐶4 ⇒ 30𝐶4 − 20𝐶2 = 150
These equations can be rearranged as
75𝐶1 − 25𝐶2 = 350, −45𝐶2 + 45𝐶3 = 0, 75𝐶1 − 105𝐶3 + 30𝐶4 = 0, −20𝐶2 + 30𝐶4 = 150
The following commands produce desired outputs:
>> A = [75 -25 0 0;0 -45 45 0;75 0 -105 30;0 -20 0 30]; b = [350 0 0 150]'; C = A\b
C=
7.4444
8.3333
8.3333
10.5556
2.3 Paraxylene, styrene, toluene and benzene are to be separated with the array of distillation columns
shown in Figure P2.3.19 Determine the molar flow rates (𝑘𝑔𝑚𝑜𝑙/𝑚𝑖𝑛) of 𝐷1 , 𝐷2 , 𝐵1 , and 𝐵2 .
FIGURE P2.3
2.3(Solution)
Material balance for each component is given by:
Xylene: 0.07𝐷1 + 0.18𝐵1 + 0.15𝐷2 + 0.24𝐵2 = 0.15 × 80 = 12
Styrene : 0.04𝐷1 + 0.24𝐵1 + 0.1𝐷2 + 0.65𝐵2 = 0.25 × 80 = 20
Toluene : 0.54𝐷1 + 0.42𝐵1 + 0.54𝐷2 + 0.1𝐵2 = 0.4 × 80 = 32
Benzene : 0.35𝐷1 + 0.16𝐵1 + 0.21𝐷2 + 0.01𝐵2 = 0.2 × 80 = 16
These equations can be rearranged as 𝐴𝑥 = 𝑏, which can be solved by using the backslash operator:
with MATLAB® 2nd Edition by
Yeong Koo Yeo
Complete Chapter Solutions Manual
are included (Ch 1 to 11)
** Immediate Download
** Swift Response
** All Chapters included
** MATLAB Programs
** Lecture Slides
,Table of Contents are given below
1. Introduction to MATLAB®
2. Numerical Methods with MATLAB®
3. Physical Properties
4. Thermodynamics
5. Fluid Mechanics
6. Chemical Reaction Engineering
7. Mass Transfer
8. Heat Transfer
9. Process Control
10. Optimization
11. Computational Intelligence
,Chapter 2 Numerical Methods with MATLAB
Linear Systems
2.1 In the photosynthesis reaction, water reacts with carbon dioxide to give glucose and oxygen. This
reaction can be expressed as
𝑥1 𝐶𝑂2 + 𝑥2 𝐻2 𝑂 → 𝑥3 𝑂2 + 𝑥4 𝐶6 𝐻12 𝑂6
Determine the values of coefficients 𝑥1 , 𝑥2 , 𝑥3 , and 𝑥4 to balance the equation. Is it possible to
determine these values? If not, under what conditions can the solutions be found?
2.1(Solution)
Carbon balance: 𝑥1 = 6𝑥4 , oxygen balance: 2𝑥1 + 𝑥2 = 2𝑥3 + 6𝑥4 , hydrogen balance: 2𝑥2 = 12𝑥4 .
Rearrangement of these equations gives
𝑥1
1 0 0 −6 𝑥 0
2
[2 1 −2 −6 ] [𝑥 ] = [0]
3
0 2 0 −12 𝑥 0
4
We can use the backslash operator to get the solution:
>> x = A\b
x=
0
0
0
0
The given equations can be rewritten as
𝑥1 − 6𝑥4 = 0, 𝑥2 − 6𝑥4 = 0, 𝑥3 − 6𝑥4 = 0 ⇒ 𝑥1 = 6𝑥4 , 𝑥2 = 6𝑥4 , 𝑥3 = 6𝑥4
Thus if we set 𝑥4 = 1, we have 𝑥1 = 𝑥2 = 𝑥3 = 6.
2.2 Four reactors are connected by pipes where directions of flow are depicted by means of arrows as
shown in Figure P2.218. The flow rate of the key component is given by the volumetric flow rate
𝑄 (𝑙𝑖𝑡𝑒𝑟/𝑠𝑒𝑐) multiplied by the concentration 𝐶 (𝑔/𝑙𝑖𝑡𝑒𝑟) of the component. The incoming flow rate is
assumed to be equal to the outgoing rate. Using the flow rates given below, calculate the concentration at
each reactor:
𝑄13 = 75 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 , 𝑄24 = 20 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 , 𝑄33 = 60 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 ,
𝑄21 = 25 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 , 𝑄32 = 45 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐 , 𝑄43 = 30 𝑙𝑖𝑡𝑒𝑟 ⁄𝑠𝑒𝑐
FIGURE P2.2
2.2(Solution)
Material balance for each reactor can be expressed as follows:
Reactor 1: 350 + 𝑄21 𝐶2 = 𝑄13 𝐶1 ⇒ 350 + 25𝐶2 = 75𝐶1 ⇒ 75𝐶1 − 25𝐶2 = 350
Reactor 2: 𝑄32 𝐶3 = 𝑄21 𝐶2 + 𝑄24 𝐶2 ⇒ 45𝐶3 = 25𝐶2 + 20𝐶2 ⇒ 45𝐶3 − 45𝐶2 = 0
, Reactor 3:
𝑄13 𝐶1 + 𝑄43 𝐶4 = 𝑄32 𝐶3 + 𝑄33 𝐶3 ⇒ 75𝐶1 + 30𝐶4 = 45𝐶3 + 60𝐶3 ⇒ 75𝐶1 + 30𝐶4 − 105𝐶3 = 0
Reactor 4: 150 + 𝑄24 𝐶2 = 𝑄43 𝐶4 ⇒ 150 + 20𝐶2 = 30𝐶4 ⇒ 30𝐶4 − 20𝐶2 = 150
These equations can be rearranged as
75𝐶1 − 25𝐶2 = 350, −45𝐶2 + 45𝐶3 = 0, 75𝐶1 − 105𝐶3 + 30𝐶4 = 0, −20𝐶2 + 30𝐶4 = 150
The following commands produce desired outputs:
>> A = [75 -25 0 0;0 -45 45 0;75 0 -105 30;0 -20 0 30]; b = [350 0 0 150]'; C = A\b
C=
7.4444
8.3333
8.3333
10.5556
2.3 Paraxylene, styrene, toluene and benzene are to be separated with the array of distillation columns
shown in Figure P2.3.19 Determine the molar flow rates (𝑘𝑔𝑚𝑜𝑙/𝑚𝑖𝑛) of 𝐷1 , 𝐷2 , 𝐵1 , and 𝐵2 .
FIGURE P2.3
2.3(Solution)
Material balance for each component is given by:
Xylene: 0.07𝐷1 + 0.18𝐵1 + 0.15𝐷2 + 0.24𝐵2 = 0.15 × 80 = 12
Styrene : 0.04𝐷1 + 0.24𝐵1 + 0.1𝐷2 + 0.65𝐵2 = 0.25 × 80 = 20
Toluene : 0.54𝐷1 + 0.42𝐵1 + 0.54𝐷2 + 0.1𝐵2 = 0.4 × 80 = 32
Benzene : 0.35𝐷1 + 0.16𝐵1 + 0.21𝐷2 + 0.01𝐵2 = 0.2 × 80 = 16
These equations can be rearranged as 𝐴𝑥 = 𝑏, which can be solved by using the backslash operator:
