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Analyse and Model Engineering System

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Added on  2023/01/19

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This document discusses various topics related to Analyse and Model Engineering System, including finding the determinant of a matrix, determining current flow in a DC circuit, solving equations using different methods, and analyzing voltage and pressure. It also includes examples of solving differential equations and using Laplace Transform.

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Analyse and Model Engineering
System

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TASK 2.0
2.1 To find the determinant of 3x3 matrix -
1 2 3
0 -4 1
0 3 -1
Solution: Determinant of 3 x 3 matrix can be calculated by using formula -
A = a b c
d e f
g h i
then, determinant of A can be determined by using -
A = a (ei – fh) – b (di – fg) + c (dh - eg)
Let determinant of given matrix is given by A
i.e.
A = 1 2 3
0 -4 1
0 3 -1
then, determinant of A = [ 1 ( -4 x -1 – 1 x3) - 2 ( 0 x -1 – 1 x 0 ) + 3 (0 x 3 – 1 x 0) ]
= [ 1 (4 – 3) – 2 ( 0 – 0) + 3 ( 0 – 0 )]
= 1
1
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2.2 Three closed loops of a D.C. Circuit is given by -
2 x + 3 y – 4 z = 26
x – 5 y – 3 z = -87
-7 x + 2 y + 6 z = 12
To determine current flow, by Kirchoff's laws
Solutions: Kirchoff's Laws – this law states that at any node or junction in an electrical circuit,
algebraic sum of entire flow of current meet at zero in a network of conductors. Therefore, using
this law, current flow in given closed loops of DC, can be determined by Using Loop Current
Analysis to find unknown variables -
2 x + 3 y – 4 z = 26 ...(i)
x – 5 y – 3 z = -87 ...(ii)
-7 x + 2 y + 6 z = 12 ...(iii)
Using substitution method,
From equation (i),
x = 26 – 3 y + 4 z ...(iv)
2
Put this value of x into equation (ii),
26 – 3 y + 4 z - 5 y – 3 z = -87
2
26 – 3 y + 4 z – 10 y – 6 z = -174
- 13 y – 2 z = -200
13 y = 200 – 2z
y = 200 – 2z
13
Put this value of y in equation (iv) then,
x = 26 – 3 (200 – 2z ) + 4 z
13
2
x = 338 – 600 – 3 z + 52 z
26
2

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x = 49 z – 262
26
Now, put value of x and y in equation (iii)
then,
-7 (49 z – 262) + 2 (200 - 2z) + 6 z = 12
26 13
- 343 z + 1834 + 800 – 8z + 156 z = 312
-195 z = -2322
z = 11.90
so, x = 49 (11.90) – 262 = 12.35 and
26
y = 200 – 2z = 200 – 2 x 11.90 = 13.55
13 13
So, x = 12.35, y = 13.55 and z = 11.90
2.22 Find value of x, y, z by using Guassian elimination method
Linear vector equation, in matrix form -
2 3 -4 x = 26
A = 1 -5 3 y = -87
-7 2 6 z = 12
Now, using Guassian Elimination method,
2 3 -4 26
1 -5 -3 -87
-7 2 6 12
Use R1 → ½ R1
1 3/2 -2 13
1 -5 -3 -87
-7 2 6 12
3
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Use R2 → R2 - R1
1 3/2 -2 13
0 -13/2 -1 -100
-7 2 6 12
Use R3 → R3 + 7 R1
1 3/2 -2 13
0 -13/2 -1 -100
0 25/2 -8 103
So, solving this,
we get, x = 10
y = 14
z = 9
2.23 Matrix solution using inverse matrix
Let A X = B
2 3 -4 x 26
A = 1 -5 3 X = y B = -87
-7 2 6 z 12
Using inverse matrix -
so, X = B A-1
Now, A-1 = 1 AT
det.A
So, A-1 = 1 -36 27 -33
-21 -26 -16 -25
-11 -10 -13
A-1 = 1 -36 27 -33
-21 -26 -16 -25
-11 -10 -13
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So,
x 1 -36 27 -33 26
y = -21 -26 -16 -25 -87
z -11 -10 -13 12
x = 10
y = 14
z = 9
5

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3.1
a) Equation of voltage is given by -
vc = 120(1 – e -t/CR ),
Here, R = 47kΩ and C = 15 uF
at t = 0 sec,
vc = 120(1 – e0 )
vc = 0 V
Now, t = 1 sec,
vc = 120(1 – e -1/15x47 )
vc = 120 (1 – e-1/705 )
= 120 (1 – 0.9985)
= 120 x 0.0015 = 0.18V
At time t = 2 sec,
vc = 120(1 – e -2/15x47 )
vc = 120 (1 – e-2/705 )
= 120 (1 – 0.9976)
= 120 x 0.0024 = 0.288V
At time t = 3 sec,
vc = 120(1 – e -2/15x47 )
vc = 120 (1 – e-2/705 )
= 120 (1 – 0.9976)
= 120 x 0.0024 = 0.288V
Therefore, at different values of t, voltage can be determined with fixed coulomb charges as
shown in following graph –
Time Value of
voltage
0 sec 0
1 sec 0.18 V
2 sec 0.288 V
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b) Given atmospheric pressure p, altitude
Altitude (h) Pressure (p)
500 73.9
1500 68.42
3000 61.6
5000 53.56
8000 43.41
Where,
p = a. ekh
then, ekh = p/a
or, kh = log p –log a
or, k = 1/h (log p –log a)
at h = 500, p = 73.9
k = 1/500 (log 73.9 – log a)
= 1/500 (1.86 – log a) ….(i)
Similarly, at h = 1500, p = 68.42
k = 1/1500 (log 68.42 – log a)
= 1/1500 (1.83 – log a) …(ii)
Similarly, on solving both equations,
1/500 (1.86 – log a) = 1/1500 (1.83 – log a)
1.86 – log a = 1/3 (1.83 – log a)
Log a = 1.375
Or, a = e1.375 = 3.95
Therefore, k will be = 0.001
7
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3.2
Given, y = 2.5 (ex – e-x) + x -25
Using bisection method
Take interval [1,2]
Then, iteration can be find in following way –
A b y(a) y(b) c = a+b/2 y(c)
1 2 -18.15 -4.875 1.5 -12.85
1 1.5 -18.15 -12.85 1.25 -15.725
1.25 1.5 -15.725 -12.85 1.375 -14.375
1.375 1.5 -14.375 -12.85 1.4375 -13.6125
Using Newton’s method
xn+1 = xn – f(xn) / f’(xn)
So, at n = 1 and x = 1
y’ = 2.5 (ex + e-x)
so, x2 = 1 + (-18.15)/7.65
= 1 – 2.3725 = -1.3725
X3 = -1.3725 + (-14.375)/8
= -1.3725 -1.79 = -3.16
8

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3.3
30 x 10-3
Given, q = i.dt
0
Using, trapezoidal rule,
Time (ms) Current A
0 0
5 4.8
10 9.1
15 12.7
20 8.8
25 3.5
30 0
9
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3.4
Mathematical model -
dp = 3(1+t) – p
dt
it can be written as -
p + dp/dt = 3(1+t)
which is in the form first order linear differential equation, as dy/dx + Py = Q
so, it can solve as -
P (I.F) = Q x I.F dt
here, Integrating Factor (I.F.) = e∫P.dt
in context with given equation, here, P =1, Q = 3(1+t) and I.F. = e∫1.dt = et
so, p. et = 3(1+t) et .dt
p . et = 3[ et. dt + t.et .dt ]
p . et = 3.et + 3 [ t. et - d(t)/dt et .dt ]
p . et = 3 et + 3 (t -1) et
p . et = 3 et + 3t – 3 et
p = 3 t + e1-t
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4.21
Differential equation of a LCR circuit is given by
L d2 i/dt2 + R di/dt + 1/C I = 0
on simplifying,
put d i/dt = s
then, equation will transform -
s2 + R/L s + 1/C = 0
s = -R ± √s2 – 4L/C
2L
here, L = 0.1 H, R = 5Ω and C = 20 uF
then, s = -5 ± √52 – 4x 0.1/20
2 x 0.1
s = -5 ± 4.9 = -0.5 and -49.5
0.2
4.20
d2 y/dx2 + 6 dy/dx + 13 y = 0
Let dy/dx = D
then, equation will reduce to
D2 + 6D + 13 = 0
on solving, D = - 3 + 2i < 0
then,
y (x) = e-3x [ C1 cos (2x) + C2 Sin (2x)]
here, C1 and C2 are arbitrary constant,
11

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4.30
Now, Using Laplace Transform of the function,
y (x) = e-3x [ C1 cos (2x) + C2 Sin (2x)]
we get,
y(0) = 1
y(x) - y'(0) = 0
12
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