Numerical Method for Simple Initial Value Problem

Contents
Part one: Simple initial value problem............................................................................................2
Introduction:.................................................................................................................................2
Equation of motion.......................................................................................................................2
Numerical method........................................................................................................................5
MATLAB program for solving the problem................................................................................6
Results and Discussions...............................................................................................................7
Discussions.............................................................................................................................11
Discussions:............................................................................................................................16
Conclusion.................................................................................................................................17
Part 2: Hydrodynamic force on a square cylinder.........................................................................18
Introduction................................................................................................................................18
Results........................................................................................................................................19
Discussions:...............................................................................................................................21
Conclusion.................................................................................................................................22

SIMPLE INITIAL VALUE PROBLEM
Introduction
In this section, an energy extraction device is discussed which consists of cylinder elastically
mounted on an electricity generator from which the extraction of power takes place by use of a
damper of dumping coefficient as shown in the figure below. The direction of flow of the fluid is
normal to the axis of the cylinder. The fluid that flows past the cylinder undergo some
modifications to consist of oscillatory flow and steady flow.
The prediction of various parameters like damping coefficient (c) to displacement and Keulegan-
carpenter (KC), vibrational velocity and the resultant power extracted is done using dynamic
simulation that uses MATLAB. A serious analysis of the hydrodynamic forces that acts upon the
cylinder are required to understand the forces that act on any water vessel.
The Equation of Motion
Below is the procedure for deriving an equation of motion.
Before the start, draw a free body diagram for the cylinder.
From the diagram, it can clearly be seen that there are three forces that acts on the cylinder.

The three forces include:
Fdamping= damping force = - cV (negative sign because the force is against the motion of the
cylinder)
Fwater= force from the water.
Fspring=spring force = - KX
FTotal = Fwater + Fdamping + Fspring
By using Morrison equation, you can calculate Fwater using the formula below
FWATER=C A md
d V r
dt + 1
2 ρC D AP |V r|V r
For
 CA is the inertia coefficient.
 Md mass of displaced fluid
 CD is the drag coefficient
 Ap is the area of the project equals to the DL of the cylinder
 Vr is the water velocity relative to the cylinder and equals to Vr= u-V
 u is the incoming velocity which is equal to u=U0+Umsin(ὠt)
 v is the velocity of the cylinder
Applying second Newton law
∑ F=ma=m dv
dt
FTotal= m dV
dt

Fwater + Fdamping + Fspring = m dV
dt
m dV
dt =C A
md
d V r
dt + 1
2 ρ CD A P|V r|V r−cV −KX
m dV
dt =C A
md ( du
dt −dV
dt )+ 1
2 ρCD AP |u V |(u−V )−cV −KX
m dV
dt +C A md
dV
dt =C A md
du
dt + 1
2 ρCD AP |u V |(u−V )−cV −KX
dV
dt = 1
m+C A md (C A md
du
dt + 1
2 ρ CD AP|uV |(u−V )−cV −KX )
∴ The equation of motion of the cylinder is
d2 X
d t2 = 1
m+C A md (CA md
du
dt + 1
2 ρC D AP |u dX
dt |(u− dX
dt )−C dX
dt −KX )
NUMERICAL METHOD
The numerical solution is found by using fourth-order Runge Kutta method

It is important to convert the second order differential equation into two first order equations as
below
dX
dt =V
dV
dt =f ¿V, X, t)
Where f ( V , X , t ) = 1
m+C A md ( CA md
du
dt + 1
2 ρ CD AP |u V |(u−V )−CV −Kx )
The fourth order- Kutta method is
X ( n+1 ) ∆ t= Xn ∆ t + ∆ t
6 (k X 1+2 k X 2+ 2 kX 3 +k X 4 )
V ( n+1 ) ∆ t =V n ∆ t + ∆ t
6 (k V 1+ 2k V 2+ 2 kV 3 +kV 4 )
where

PROGRAMMING OF THE MATLAB FOR SOLVING THE PROBLEM
% A MATLAB program for solving the problem of the vibration of a
% circular cylinder in a water flow
% The code below is using the Fourth Runge-Kutta method
clear all;
pi = acos(-1); % pi = 3.14159
Um = 1.4; % Fluid Velocity amplitude
Uo=0.14; % Uniform flow velocity
T =1.2; % oscillatory flow period KC=10
D = 0.14; % cylinder diameter in meter
L = 1; % cylinder length
m = 50; % cylinder mass
rho = 1024; % fluid density
K = 200; % spring stiffness
c =100; % damping constant
CA = 1; % added mass coefficient
CD = 1.8; % drag coefficient
omega = 2*pi/T; % angular frequency of the flow
md = rho*pi*(D*D)/4*L; % added mass coefficient
Ap = D*L; % projected area
dt = T/40; % time step
net = (T/dt*5)*12 ; % %total step to be calculated
X(1:ndt+1) = 0; % save displacement X from step 0 to step ndt