Analyse and Model Engineering System

This document provides solutions and explanations for various topics related to Analyse and Model Engineering System. It includes determinants of a matrix, current equations obtained from closed loop, inverse matrix method, voltage of a capacitor, value of constant a and k, differential equation solution, determining charge stored, and formulating a differential equation for a first order RC circuit.

Analyse and Model Engineering
System

TASK 2.0
2.1 Determinant of matrix
1 2 3
0 -4 1
0 3 -1
Solution
Determinant of given matrix = 1*[(-4)*(-1) – (1)*(3)] – 2*[(0)*(-1) – (1)*(0)] + 3*[(0)*(3) – (-
4)*(0)]
Determinant = 1
2.21 Current equations obtained from closed loop
2x + 3y – 4z = 26
x – 5y -3z = -87
-7x +2y + 6z = 12
Solution
2 3 -4 | 26
1 -5 -3 | -87
-7 2 6 | 12
Using Gaussian elimination form:
R1 < --->R3
R2 ---> R2 + (1/7)*R1
R3 ---> R3 + (2/7)*R1
R3 ---> R3 + (25/33)*R2
-7 2 6 | 12
0 (-33/7) (-15/7) | -85.28
0 0 (-43/11) | -35.18
1

Again converting this into equation form:
-7x +2y +6z = 12
(-33/7) y + (-15/7) z = -85.28
(-43/11) z = -35.18
By third equation we have z = 8.99
On putting this value in second equation we get y = 14
Similarly by substituting values of y and z, we have x = 10
x = 10
y = 14
z = 8.99
2.22 Inverse matrix method
The matrix form of equations is:
2 3 -4 x 26
1 -5 -3 y = -87
-7 2 6 z 12
Solution
The inverse of coefficient matrix A is determined as follows:
2 3 -4 | 1 0 0
1 -5 -3 | 0 1 0
-7 2 6 | 0 0 1
On reducing the matrix to echelon form we have the inverse matrix as:
(-8/43) (-26/129) (-29/129)
(5/43) (-16/129) (2/129)
(-11/43) (-25/129) (-13/129)
From the given equation we have
AA¯X = A¯B
2

x = (-8/43) (-26/129) (-29/129) 26
y = (5/43) (-16/129) (2/129) * -87
z = (-11/43) (-25/129) (-13/129) 12
x = 10
y = 14
z = (-280/43)
TASK 3
3.1 A. Voltage of capacitor
Solution
Vc = 120 [1 – e^(t/RC)]
t = 2 seconds
R = 47 k ohm
C = 15 micro Farad
Vc = 120 [1 – e^(2/47k*15u)]
= 120 * 112.96
Vc = 13.5 kv
3

3.1 B. Value of constant a and k
Solution
Altitude (h) Pressure (P)
500 73.9
1500 68.42
3000 61.6
5000 53.56
8000 43.41
P = a e^(kh)
Taking log both the sides
log p = log [a e^(kh) ] = log [a] + log [e^(kh)]
log p = log [a] + kh
The above equation is of form y =mx + c
When it is plotted on graph it is observed that a straight line is passed through points and it
verifies the equation p = a e^(kh)
The gradient of line = k = [log 73.1 – log 43.41] / [500-8000] = [-7 * 10^(-5) ]
Similarly
log y = kx + log a
when x = 0 then log y = log a => y = a
at x = 0, vertical intercept = 76 so a = 76
Thus a = 76 and k = [-7 * 10^(-5) ]
3.2 Differential equation solution
y = 2.5 (e^x – e^-x) + x – 25
Solution
i) Using bisection method
f (x) = 2.5 (e^x – e^-x) + x – 25
4

f (1) = 2.5 (e^1 – e^-1) + 1 – 25 = -18.12 which is negative
f (3) = 2.5 (e^3 – e^-3) + 3 – 25 = 28.08 which is positive
Thus one root of y lies in between 1 and 3.
x1 = (1+3)/ 2 = 2
f(x1) = f(2) = 2.5 (e^2 – e^-2) + 2 – 25 = -4.86 which is negative
Thus the next root will lie between 2 and 3
x2 = (2+3) /2 = 2.5
f(x2) = f(2.5) = 2.5 (e^2.5 – e^-2.5) + 2.5 – 25 = 7.7 which is positive
Hence the root will lie between 2 and 2.5
x3 = (2+2.5) / 2 = 2.25
Since we have to take value up to only two decimal point the approximate value is x = 2.25
ii) Using newton's method
xn+1 = xn – f(xn) / f’(xn)
here, f(xn) = y = 2.5 (e^x – e^-x) + x – 25
so, differentiating it with respect to x, we get -
f’(xn) = y' = 2.5 (ex + e-x)
put n = 1 and x = 1
so, x2 = 1 + (-18.15)/7.65
= 1 – 2.3725 = -1.3725
X3 = -1.3725 + (-14.375)/8
= -1.3725 -1.79 = -3.16
5

3.3 Determining charge stored
Given, an alternating current i
30 x 10-3
q = ∫ i. dt
0
30 x 10-3
q = i . [ t ]0
q = i . [ 30 x 10-3 - 0 ]
q = i. (30 x 10-3)
Time (ms) Current (A)
0
5
10
15
20
25
30
0
4.8
9.1
12.7
8.8
3.5
0
Charge Stored in coulombs
Using Trapezoidal Rule-
b
∫ f(x). dx = (b -a) [f(a) + f(b)]
a 2
From given table, charge at different alternating current can be calculated as -
30 x 10-3
q = ∫ i. dt
0
∆ t = 5ms
therefore,
30 x 10-3
6

q = ∫ i. dt ≈ 5/2 ( 0 + 4.8 + 9.1 + 12.7 + 8.8 + 3.5 + 0)
0
≈ 2.5 x 38.9 ≈ 97.25 coulomb
Using Simpson's Rule
b
∫ y. dx = (b -a) [(first ordinate + last ordinate) + 4 (sum of even ordinate) + 2 (sum of odd ordinates)]
a 3
therefore,
30 x 10-3
q = ∫ i. dt
0
q ≈ (5/3) [ (0+0) + 2 (4.8+12.7+3.5) + 4 (9.1 + 8.8) ]
≈ 5/3 [113.6] ≈ 189.333 coulomb
7

3.4
Mathematical model -
dp/dt = 3(1+t) – p
the above equation can be written as –
p + dp/dt = 3(1+t)
this equation is the first order linear differential equation, as
dy/dx + Py = Q
this would can be solved 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
Numerical Solution of the above differential equation in the range of 1.0, 0.2 and 2.0 at t = 1
when p =4,
at t = 1.0,
p = 3 (1.0) + e1-1.0
= 3 + e0
= 3 + 1 = 4 (which is given as the initial condition)
Now, at t = 0.2
p = 3 (0.2) + e1-0.2
= 0.6 + e0.8
= 0.6 + 2.23 = 2.83
8

p = 3 (2.0) + e1-2.0
= 6 + e-1.0
= 6 + 0.36 = 6.36
TASK 4
4.1 Formulate differential equation
A first order RC circuit consist of capacitor and resistor connected in series which can be
analyzed through first order differential equation. Since R and C are connected in series same
current i (t) flow across them.
The voltage across resistor is given as:
Vr = R i(t)
The current and voltage across the capacitor are given as :
i(t) =C dv(t) / dt where v(t) is capacitor voltage
On substituting i(t) = C dv(t) / dt into ohms law value of voltage across resistor can be
determined.
Vr (t) = RC dv(t) / dt
By applying KVL
Vtotal (t) = Vr (t) + Vc(t)
Vtotal (t) = RC dv(t) / dt + V(t)
A. Solving differential equation when t= 0 and C = 0
Solution
i(t) = C dv(t) / dt
9

The above image shows the charging and discharging of capacitor. In the first circuit C= 0 and in
the other circuit C is initially charged to V0. In both cases at t =0 switch is closed after remaining
open for long time. After closing switch the voltage equation for both charging and discharging
circuit is:
Vc (t) = RC dvc(t) / dt = Vs
Vc (t) = RC dvc(t) / dt = 0
Thus we have boundary condition vc (0) = 0 and vc (infinity) = Vs
Using above equations the solution for charging capacitor is:
Vc(t) = Vs [1 – e^ (-t/RC)]
Similarly boundary condition for discharging capacitor is vc(0)= V0 and vc (infinity) = 0. Thus
solution for discharging capacitor is:
Vc(t) = V0*[ e^ (-t/RC)]
For discharging case the current equation is given by:
Ric(t) = 1/C ∫ ic(t) dt = 0
On solving and rearranging this equation we have solution:
ic(t) = V0/R * e^ (-t/RC)
B. Current flowing through circuit
Solution
E = 20V
C = 20 micro F
R= 50 k ohm
t = 1.0 second
10

The solution for first order differential equation is given as below:
i(t) = [E/R] * e^ (-t/RC)
i (t) = [20/50k] * e^ (-1/50k20micro)
i (t) = 14.68 micro ampere
4.20 Solve differential equation
d2 y/dx2 + 6 dy/dx + 13 y = 0
Suppose dy/dx = S
then, the given differential equation will reduce to
S2 + 6S + 13 = 0
then, using quadrature formula -
x = -b ± √ (b2 – 4ac)
2a
S = -6 ± √ (62 – 4 x 1 x 13)
2 x 1
= -6 ±i4
2
on solving, S = - 3 + 2i < 0
then,
y (x) = e-3x [ C1 cos (2x) + C2 Sin (2x)]
where, C1 and C2 are arbitrary constant
11

4.21 LCR Series circuit
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
4.30 Laplace transform
Given equation –
d2 y/dx2 + 6 dy/dx + 13 y = 0
Laplace transform of second derivative equation –
Ƚ {f'' (t)} = s2 Ƚ{f'(t)} – s f(0) – f'(0)
at x = 0, y = 3 and dy/dx = 7
Ƚ {d2 y/dx2} + 6 Ƚ {dy/dx} + 13 Ƚ{y} = Ƚ{0}
s2 Y(s) – s y (0) – y'(0) + 6 [sY (s) – y (0)] + 13 Y (s) = 0
plug the initial conditions,
(s2 + 6s + 13) Y(s) – 3s – 25 = 0
Y (s) = 3s + 25
(s2 + 6s + 13)
A + B = 3s + 25
(s + 3 + √22) (s + 3 - √22) (s2 + 6s + 13)
On Solving,
A = 3.2 and B = - 0.2
then, Y (s) = 3.2 - 0.2
(s + 3 + √22) (s + 3 - √22)
12

Then, Y (s) = 3.2 e-(3+√22) – 0.2 e(-3+√22)
13