How does the pressure change with time?

Attached is the question. I believe this is how I would set up the equation. At timestep 0 = xdot = 0

This is the initial code that I have but I need to get pressure vs time graph.

% Define System parameters

Force = 100; % input force 100 N

m = 50; % mass 50 kg

c = 1000; % friction constant 1000 Ns/m

A = .02; % 0.02 m2

dt = .001; % fixed time step of 1 millisecond

ts = 0; % simulation start time

tf = 2; % simulation end time

t = ts:dt:tf; % simulation time vector [0 .001 .002 ... 50]

x = zeros(size(t)); % displacement vector [0 0 0 ...]

xddot = ones(size(t));

xdot = zeros(size(t)); % Velocity vector [0 0 0 ...]

f = Force*ones(size(t)); % force vector [10 10 10 ...]

xddot(1) = 1/m*(f(1)); % initial value for acceleration

for i = 1:length(t)-1

x(i+1) = xdot(i)*dt + x(i);

xdot(i+1) = xddot(i)*dt + xdot(i);

xddot(i+1) = 1/m*(f(i+1) - c*xdot(i));

end

%Dis plot

subplot(311)

plot(t,x)

% Velocity plot

subplot(312)

plot(t,xdot)

% Acceleration plot

subplot(313)

plot(t,xddot)

Answer 1

[Sam Chak]

Hi @sanjeet mansoor

I'm unfamiliar with your Hydraulic system. If the system is correctly described by the following equation

then the simulation probably looks like this:

tspan  = linspace(0, 2, 2001);
x0     = [0.1 0];       % initial displacement
[t, x] = ode45(@odefcn, tspan, x0);
plot(t, x(:, 1)), grid on, xlabel('t')

function dxdt = odefcn(t, x)
    dxdt  = zeros(2, 1);
    Load  = 100;    % Force due to Load 100 N
    m     = 50;     % Mass 50 kg
    c     = 1000;   % Friction constant 1000 Ns/m
    A     = 0.02;   % Area 0.02 m2
    omega = 6;      % adjust this parameter (increase for faster response)
    P1    = (- 2*m*omega*x(2) - m*(omega^2)*x(1) + c*x(2))/A;           % Pressure if Load is unknown
    P2    = (- 2*m*omega*x(2) - m*(omega^2)*x(1) - Load + c*x(2))/A;    % Pressure if Load is known
    dxdt(1) = x(2);
    dxdt(2) = (A*P2 + Load - c*x(2))/m;     % Decide either P1 or P2
end

[sanjeet mansoor]

sorry, the pressure starts at 10000 and ends at 5000 Pa. I tried changing x0, but im not sure what that is.

[Sam Chak]

@sanjeet mansoor, no worries. But I don't see anywhere in the equations provided by you, can enter the initial pressure 10 kPa and the final pressure 5 kPa. Unless this is a Boundary-Value Problem (BVP).

If you merely want to plot the Time vs. Pressure graph, then you should provide the formula/equation for the Pressure as a function of Time. That's Simple and Direct.

If you want to fix Pa, Pa, and make the pressure to decrease linearly over time, that is also possible.

Anyhow, you must describe your problem in clarity. If you don't understand a particular part of the problem, then also need to describe the specific part you don't understand in clarity.