Advanced Thermal and Fluid Engineering

Running head: ADVANCED THERMAL AND FLUID ENGINEERING
ADVANCED THERMAL AND FLUID ENGINEERING
Name of the Student
Name of the University
Author Note

ADVANCED THERMAL AND FLUID ENGINEERING1
Part one: Simple initial value problem
In this particular task the equation of motion of a cylinder through which water flows in a
variable rate is modelled and then the vibration of the cylinder is obtained by numeric
solution by FDM approach. The energy extracted by the electricity is obtained from the
power generated though the vibration of the cylinder. This is modelled by a damper with a
certain damping constant c and a spring with spring constant k as shown in the figure below.
Last digit of student id is j = 2
Given, Um= 1 + 2/10 = 1.2 m/s, U0 = 0.1*1.2 = 0.12 m/s, D = 10 + 2 = 12 cm = 12*10^(-2)
m, Mass of cylinder m = 50 kgs, Water density ρ = 1024 kg/m^3, Stiffness of spring K = 200
N/m, Damping coefficient c = 10 N*m/sec, md = 1 and CA = CD = 1.
The KC number is given by,
KC = Um*T/D
Hence, for KC = 2 => T = 2*D/Um = 0.2 sec.
The entire time of 0.2 sec is divided in small steps.
Flow velocity V r =u−V

ADVANCED THERMAL AND FLUID ENGINEERING2
Now, initially, V r (0) = V(0) = U0 and V r (end) = U0
Now, the Morrison equation is given by,
Fwater=C A∗md∗( d V r
dt )+ ( ½ )∗ρ∗CD∗A P∗|V r |∗V r
Now, value of Fwater will be => mass of water*g = 1024*(2*pi*D/2)*L*9.8 = 5990 N.
The Morrison equation by finite difference approximation is given as,
CA∗md∗Vr ( t +dt )−Vr ( t )
dt + ( ½ )∗ρ∗CD∗AP∗|V r ( t )|∗V r ( t )=0
 Vr ( t +dt )= (− ( ½ )∗ρ∗CD∗A P∗|V r ( t )|∗V r ( t ) ) dt
CA∗md
+Vr ( t )
From V r (t +dt ) Obtained from the above iteration equation, V(t) can be easily obtained for
each points in the time range [0,T].
The discrete numerical integration relation that gives approximate power P is given by,
P = 1
N ∑
n=1
N
c∗(V ¿¿ n)2 ¿
N = total number by which T is divided.
MATLAB code for Vibration speed:
function Power = question1(KC,c)
j = 2;
%%% defining parameters of the system
% amplitude of oscillatory flow in m/s
Um = 1 + (j/10);

ADVANCED THERMAL AND FLUID ENGINEERING3
% steady state flow in m/s
U0 = 0.1*Um;
% diameter in m
D = (10 + j)*1e-2;
% length of the cylinder
L = 1;
% mass of cylinder
m = 50;
% density of the water in Kg/m^3
rho = 1024;
% spring stiffness N/m
K = 200;
% the surface of the cylinder is the projected area
Ap = 2*pi*(D/2)*L;
% added mass coefficient
md = 1;
CA = 1.8; CD = 1.8; T = KC*D/Um;
Fwater = 1024*(2*pi*D/2)*L*9.8;