Modelling and Simulation of Marine Systems Questions 2022

Electrical Engg.
Modelling and Simulation of Marine Systems / JEE506
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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