Fullname: Nguyen Khac Duy
ID: CTTT13210119 – Class: K4/01
Practice Exam
Problem #1
(i) Let the function f[c]: 𝑓
𝑐=𝑊−𝑄𝑐−𝑘𝑉
𝑐
So, the derivative of f[c]: 𝑓′[c]=−𝑄
𝑘𝑉
2
𝑐

Applying the Newton`s Method to the concentration 𝑐, we have the formula:

𝑐
𝑛+1=𝑔
𝑐
𝑛
𝑐
𝑛
𝑓
𝑐
𝑛
𝑓
𝑐
𝑛


𝑐
𝑛
𝑊−𝑄
𝑐
𝑛−𝑘𝑉
𝑐
𝑛−𝑄
𝑘𝑉
2
𝑐
𝑛
𝑐
𝑛
𝑊
𝑐
𝑛−𝑄
𝑐
𝑛
𝑐
𝑛−𝑘𝑉
𝑐
𝑛
𝑄
𝑐
𝑛
𝑘𝑉
2


𝑊
𝑐
𝑛
𝑘𝑉
𝑐
𝑛
2
𝑄
𝑐
𝑛
𝑘𝑉
2
2𝑊
𝑐
𝑛−𝑘𝑉
𝑐
𝑛
2𝑄
𝑐
𝑛+𝑘𝑉

Substituting in the parameter values (𝑊=1x
10
6
g/hr, 𝑄=1x
10
5
m
3/yr
25000
219
m
3/hr, 𝑘=0.25
m
1/2
g
1/2
hr𝑉=1x
10
6
m
3) gives:

𝑐
𝑛+1
2x1x
10
6
𝑐
𝑛−0.25x1x
10
6
x
𝑐
𝑛
2x
25000
219
𝑐
𝑛+0.25x1x
10
6
2x
10
6
𝑐
𝑛−25x
10
4
x
𝑐
𝑛
50000
219
𝑐
𝑛+25x
10
4

Starting with
𝑐
0=2
m
3/g and iterating we get

𝑐
1=4.36982
m
3/g

𝑐
2=4.62274
m
3/g

𝑐
3=4.62408
m
3/g

𝑐
4=4.62408
m
3/g
Thus the concentration at the specified conditions is:
𝑐
𝑠=4.62408
m
3/g
(ii) We have 2 formulas below:
(a)
𝑐
𝑛+1
𝑔
1
𝑐
𝑛
𝑊−𝑄
𝑐
𝑛
𝑘𝑉
2

(b)
𝑐
𝑛+1
𝑔
2
𝑐
𝑛
𝑊−𝑘𝑉
𝑐
𝑄

We take the derivative of two formulas:

𝑔
1
𝑐
𝑛−2𝑄
𝑘𝑉
x
𝑊−𝑄
𝑐
𝑛
𝑘𝑉
2𝑄(𝑄
𝑐
𝑛−𝑊(𝑘𝑉
2


𝑔
2
𝑐
𝑛−𝑘𝑉
𝑄
x
1
2
𝑐−𝑘𝑉
2𝑄
𝑐

Substituting in the parameter values (𝑊=1x
10
6
g/hr, 𝑄=1x
10
5
m
3/hr, 𝑘=0.25
m
1/2
g
1/2
hr𝑉=1x
10
6
m
3,
𝑐
𝑠=4.62408
m
3/g) gives:

𝑔
1
𝑐
𝑛
2x1x
10
5(1x
10
5
x4.62408−1x
10
6(0.25x1x
10
6
2=1.72029>1

𝑔
2
𝑐
𝑛−0.25x1x
10
6
2x1x
10
5
4.62408=0.58130<1
So the formula (b) is converge to the result found in (i) using the Newton’s method.

Problem #2
(i) The formula:
𝑥
𝑘+1=𝑔
𝑥
𝑘
1
8
𝑥
𝑘
3−1
Starting with
𝑥
0=1 and iterating we get

𝑥
1=−0.87500

𝑥
6=−1.22317


𝑥
2=−1.08374

𝑥
7=−1.22875


𝑥
3=−1.15911

𝑥
8=−1.23190


𝑥
4=−1.19466

𝑥
9=−1.23369


𝑥
5=−1.21313

𝑥
10=−1.23471

It is converging to the middle root
p
2.
(ii) The convergence of iteration scheme is bases on the derivative of 𝑔
𝑥
𝑘The mathematical