New! Hire Essay Assignment Writer Online and Get Flat 20% Discount!!

ENGG952 Engineering Computing

Published : 23-Sep,2021  |  Views : 10

Question:

Write the finite-difference equation of the differential equation. The report must clearly provide the detailed derivation of the technique. Programing to obtain the solution for time duration 0 ? t ?1 . (For explicit finite difference method, you can use EXCEL or matlab. For implicit Crank-Nicholson method, you must use the matlab).Please demenstrate that appropriate are needed to solve u until steadystate solution is reached.Plot the nondimensional temperature versus nondimensional length for a few typical values of nondimensional times, which can demonstrate the evloution of the tempeature at different time.

Mass-Spring-Damper system for an Ordinary Differential Equation Free vibrations in a mechanical system are caused by initial conditions of the parameters: displacement, velocity, or acceleration where no external force is allowed to interact. The mechanical system in the free vibration will oscillate with its natural frequency and eventually settle down to zero as a result of damping effects.

Dividing through by mass, m,The natural frequency of the system and the damping ratio are replaced as shown in the following equation,The solution to the differential equation.

The transient solution of the system is given as,The constants A and B are obtained when the initial conditions are inserted. The final solution is obtained as,From the initial conditions,

The Runge-Kutta Heun Method considering a certain range or period of time

tspan=[0:0.1:50];

y0=[0.02;0];

[t,y]=ode45('unforced2',tspan,y0);

plot(t,y(:,1));

grid on

xlabel('time')

ylabel('Displacement')

title('Critically-damped Modelling of MSD system (C=40)')

%% order 647236

tspan=[0:0.1:50];

y0=[0.02;0];

[t,y]=ode45('unforced1',tspan,y0);

plot(t,y(:,1));

grid on

xlabel('time')

ylabel('Displacement')

title('Under-damped Modelling of MSD system (C=5)')

hold on;

%% order 647236

function yp=unforced2(t,y)

c=40;

m=20;

k=20;

yp= [y(2); (-((c/m)*y(2))-((k/m)*y(1)))];

%% order 647236

function yp=unforced1(t,y)

c=5;

m=20;

k=20;

yp= [y(2); (-((c/m)*y(2))-((k/m)*y(1)))];

The graphical illustration of the behavior of the mass-spring-damper for an under-damped and critically-damped system are shown in the same axes. The critically damped reduces to zero faster than the under-damped. The damping coefficient determines the rate at which the signal reduces its amplitude or magnitude as it travels. heat conduction in an insulated rod. The non-dimensional form is expressed.

The explicit finite difference method and the implicit crank-Nicholson method are used to solve this equation. The explicit method of parabolic PDE is simple to program though governed by a criterion for the solution to converge. The steps used are restricted by the criterion. There are implicit methods employed as well as they are more stable. There are two implicit method namely the simple implicit method and the Crank-Nicholson method which are simple with low accuracy and complex with high accuracy respectively. The explicit schemes are conditionally stable for the parabolic partial differential equations. The case study is on heat conduction equation over time variable with one spatial dimension.

• Implicit method

The elliptic parabolic PDEs can be solved using the finite Difference Method. The dimension of time in this case is open and the initial condition is set at t=0.

Step 1: the temperature for the next time step is calculated and indexed as shown below,

For the ith node: -

Space domain at time l: Time domain at node i:

The time derivative, unfortunately, is less accurate than the spatial derivative. The central difference in terms of x, indexed by I,To ensure stability the time step is limited by a space interval such that,

• Crank-Nicholson method

Clear all; % clear memory of variables and figures

alfa = 0.835;

x0 = 0;                             % starting x

xm = 10;                            % ending x

t0 = 0;                             % starting t

tn = 0.5;                            % ending t

m = 5;                              % number of interval in x direction

n = 300;                            % number of interval in t direction

x = linspace(x0,xm,m+1);            % x coordinates of the nodes

t = linspace(t0,tn,n+1);            % t coordinates of the nodes

del_x = (xm - x0)/m;                % delta x

del_t = (tn - t0)/n;                % delta t

hh = del_x*del_x;                   % constant

r = hh/del_t;                       % constant

p = 2*r/alfa;                       % constant

%%%%%   Setting the initial conditions    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%f = inline('sin(x)');              % initial condition expression

T(1,1:m+1) = 0;                     % set T at t = 0

%%%%%   Setting the boundary conditions   %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

T(2:n+1,1) = 100.0;                 % T at x = x0

T(2:n+1,m+1) = 50.0;                % T at x = xm

%%%%% Setting the matrix form and solve them at each time step

A_left(1:m-1,1:m-1) = 0;

A_right(1:m-1,1:m-1) = 0;

for i = 1:m-1

A_left(i,i) = -2 - p;

A_right(i,i) = 2 - p;

if i < m-1;

A_left(i,i+1) = 1;

A_right(i,i+1) = -1;

end

if i > 1;

A_left(i,i-1) = 1;

A_right(i,i-1) = -1;

for j = 1:n

b = A_right*T(j,2:m)';

b(1) = b(1) - T(j,1) - T(j+1,1);

b(m-1) = b(m-1) - T(j,m+1) - T(j+1,m+1);

T(j+1,2:m) = A_left\b;

%this figure is only suitalbe for dt=0.1sec only.

figure(1)

x=x0:del_x:xm

plot(x,T(11,:),x,T(31,:), x,T(51,:), x,T(101,:),x,T(201,:),x,T(301,:))

legend('t=0.0s','t=0.1s','t=0.2s','t=0.3s','t=0.4s','t=0.5s');

xlabel('x-direction (Non-dimensional x)'); ylabel('temperature (non dimensional temperature)')

To calculate the temperatures for the first step seeks to move from the known to the unknown,

 time step time step Node 1 Node 2 Node 3 Node 4 Node 5 Node 6 time step time step x=0 x=2 x=4 x=6 x=8 x=10 time step 0 t=0.0 100 0 0 0 0 50 time step 1 t=0.1 100 2.0875 0 0 1.04375 50 time step 2 t=0.2 100 4.087847 0.043577 0.021788 2.043923 50 time step 3 t=0.3 100 6.005589 0.127546 0.064455 3.002794 50 time step 4 t=0.4 100 7.845018 0.248933 0.12711 3.922523 50 time step 5 t=0.5 100 9.610185 0.404958 0.208882 4.805161 50

Plot for the non-dimensional temperature versus the non-dimensional length for a few typical values of non-dimensional times. They are used to demonstrate the evolution of the temperature at different times as shown in the table above. The time differs per time step.

Get An Awesome Price Quote For Your Paper â€“ Absolutely FREE!

Our Amazing Features

No missing deadline risk

No matter how close the deadline is, you will find quick solutions for your urgent assignments.

100% Plagiarism-free content

All assessments are written by experts based on research and credible sources. It also quality-approved by editors and proofreaders.

500+ subject matter experts

Our team consists of writers and PhD scholars with profound knowledge in their subject of study and deliver A+ quality solution.

Covers all subjects

We offer academic help services for a wide array of subjects.

Pocket-friendly rate

We care about our students and guarantee the best price in the market to help them avail top academic services that fit any budget.

Getting started with MyEssayAssignmentHelp is FREE

15,000+ happy customers and counting!

Rated 4.7/5 based on
1491 reviews
;