Assignment 1: Modeling and Simulation of Marine Systems (JEE506)

This document presents a comprehensive solution to an assignment on modeling and simulation of marine systems, specifically for the JEE506 course. The assignment focuses on two key tasks: modeling and simulation using MATLAB. The modeling aspect involves deriving differential equations for a Maxon brushed DC motor and a fluid storage level system. This includes developing transfer functions and state-space models for both systems. The simulation task requires the use of MATLAB to simulate the derived equations, generate numerical solutions, and implement state-space models. The solution includes MATLAB M-files, simulation results, and plots. The document provides detailed mathematical derivations, transfer function calculations, state-space model representations, and block diagrams for both the DC motor and the fluid storage level system, demonstrating a complete understanding of the modeling and simulation process.

Electrical Engg.
Modelling and Simulation of Marine Systems / JEE506
Student Name :
Student ID :

Question 1 )
Maxon Brushed DC motor
The part 10 is chosen which has following parameters :
If the motor is on no load condition, w = constant , θ = 0 ,
K i0 = b w0
. i0 = no load current = 7.76 mA = 0.00776 A
. w0 = no load speed = 1420 rpm = 148.627 rad / s
. k = Torque constant = 321 mNm/A = 0.321 Nm/A
. b = k i0 / w0 = 0.321 x 0.00776 / 148.627 = 0.00001676 Nms / rad
1) Mathematical equations for the motor system :
Relation between input voltage , armature current , shaft speed and angular
displacement :
Voltage value :
V = L di / dt + Ri + k θ’ (1)
Applying KVL for electrical elements and k θ’ for angular motion.
Balancing the torque :
. k i = J θ’’ + b θ’ (2)
The sum of all the torque values is equal to 0.
Angular speed = w = θ’

2) Transfer function
Finding the value of ‘i’ from equation 2
. i = 1 / k ( Jθ’’ + bθ’ )
Substituting the value of ‘i’ in equation 1.
V = L/k (Jθ’’’ + bθ’’) + R / k ( J θ’’ + b θ’ ) + kθ’ (3)
1) Shaft speed and Input voltage
Since w = θ’
Substituting the value above in equation 3
V = L/k ( Jw’’ + bw’ ) + R / k ( J w’ + b w ) + kw
Taking the Laplace Transform of the above equation with zero initial conditions :
V (s) = L/k ( s2J W ( s ) + bsW ( s ) ) + R / k ( Js W ( s ) + b W ( s ) ) + k W ( s )
W(s) / V(s) = 1/ [ ( J L / k ) s2 + ( b L + J R / k ) s + ( b R / k + k ) ]
2) Shaft angle and input voltage
V = L/k (Jθ’’’ + bθ’’) + R / k ( J θ’’ + b θ’) + kθ’
Taking the Laplace Transform of the above equation with zero initial conditions :
V(s) = L/k (Js3θ(s) + bs2θ(s))+R/k(J s2 θ(s) + b sθ(s)) + ksθ(s)
V(s)/ θ(s)= L/k (Js3 + bs2)+R/k(J s2 + b s) + ks
V(s)/ θ(s)= L/k Js3 + Lb/ks2+RJ/k s2 + Rb/k s + ks
. θ(s) / V(s)= 1 / [ L/k Js3 + Lb/ks2+RJ/k s2 + Rb/k s + ks ]
3) State Space Model
Let x1 = θ, x2 = θ’ , x3 = i, D = b

V = L x3’ + R x3 + k x2
Balancing the torque :
. k x3 = J x2’ + D x2
Let u = V is input then
Matrix form gives :
. x1’ = 0 x1 + x2 + 0 x3 + 0 u
. x2’ = 0 x1 –D/J x2 –k/J x3 + 0 u
. x3’ = 0 x1 –k/L x2 –R/L x3 + 1/L u
Let output = θ, output equation is θ = x1 .
4)
Block Diagram
Voltage value :
V = L di / dt + Ri + k θ’
Balancing the torque :
. k i = J θ’’ + b θ’
Angular speed = w = θ’
V = L di / dt + Ri + k w
Taking Laplace Transform :
V(s) = L s I(s) + R I (s) + k w
I (s) / V(s) – kw = 1 / Ls + R
T = k i
Taking Laplace Transform :
T (s) / I (s) = k

. k i = J w’ + b w
Taking Laplace Transform :
W(s)/ T(s) = 1/Js + b , w = 0
. θ (s) / W(s) = 1/s
MATLAB Plot :

Problem 2)
Fluid Storage Level System
Relation ( Equation ) between the inlet flow rate and the liquid level in the tank :
. qo = Outlet flow rate
. qo = kv x h
Here, kv = drain value constant = 0.0015 m2 / s
. h = liquid level in the tank ( in m )
Considering the mass balance across the liquid storage tank :

. ρ [ qo ( t ) – qi ( t ) ] = - ρ dV / dt
The storage tank is cylindrical in shape.
The pressure cannot go above the value of 250 kPa.
Internal diameter of the storage tank = d = 18 cm
Internal radius of the storage tank = r = d / 2 = 18 / 2 cm = 9 cm = 0.09 m
Volume of the storage tank = V = ∏ r x r x h = ∏ x 0.09 x 0.09 x h = 0.025434 h
Differentiating :.
. dV / dt = 0.025434 dh / dt
. qi ( t ) = d V ( t ) / dt + qo ( t )
. qi ( t ) = 0.025434 dh / dt + kv x h
2)
Transfer function = H ( s ) / Q ( s )
Taking the Laplace transform of the equation obtained in part 1 and assuming initial
conditions as 0 :
Q ( s ) = 0.025434 s H ( s ) + kv x H ( s )
H ( s ) / Q ( s ) = 1 / ( 0.025434 s + kv ) = 1 / ( 0.025434 s + 0.0015 )
3)
State Space Model :
. qi ( t ) = 0.025434 dh / dt + kv x h

Putting x1 = h , u = q
The state equation can be given by
. u = 0.025434 (x1)’ + 0.0015 x1
. 0.025434 (x1)’ = u - 0.0015 x1
. (x1)’ = – 0.059 x1 + 39.317 u
If the output is taken as ‘h’, then output equation can be given as :
Y = x1 = h
4)
Block diagram:
Q ( s ) ---- [1 / ( 0.025434 s + 0.0015 ) ] ---- H ( s )
MATLAB PLOTS :

Qi = Q0 when dh/dt = dV/dt =0, its at t=0 s.