Pressure Field From Velocity Field

I have a velocity field in MATLAB as two 641x861 arrays (Vx and Vy) and I need to convert it to a pressure field to determine the pressure difference / gradient between two points. I've tried a handful of approaches (pressure Poisson, Euler equation, etc.) but am struggling with the fact that I already have the velocity field over time and don't need to find u* / v*, u^n+1 / v^n+1.

Right now, I'm trying to use the attached equations but the pressure field values don't seem correct. Do I need to set up a separate mesh even though I already have the velocity at specific points? I assumed that I could use the size of my Vx and Vy arrays as my mesh. I'm also unsure how to appropriately integrate the partial derivatives over a path.

Here's my current code:

imin = 2; imax = m;

jmin = 2; jmax = n;

Px = zeros(imax,jmax);

Py = zeros(imax,jmax);

dx = 1;

dy = 1;

for j = jmin:jmax

for i = imin:imax

Px(i,j) = -rho*(Vx(i,j)*((Vx(i+1,j)-Vx(i-1,j))*0.5*dx)+Vy(i,j)*((Vx(i,j+1)-Vx(i,j-1))*0.5*dy));

Py(i,j) = -rho*(Vx(i,j)*((Vy(i+1,j)-Vy(i-1,j))*0.5*dx)+Vy(i,j)*((Vy(i,j+1)-Vy(i,j-1))*0.5*dy));

end

end

Should I be going about this a different way? Thank you in advance for any feedback!

[Torsten]

How can you solve for the velocity without getting a pressure field ? The two are coupled - thus in order to solve for velocity, you automatically have to solve for pressure with the help of the continuity equation.

Further (although for dx = dy = 1, it doesn't matter)

Px(i,j) = -rho*(Vx(i,j)*(Vx(i+1,j)-Vx(i-1,j))/(2*dx)+Vy(i,j)*(Vx(i,j+1)-Vx(i,j-1))/(2*dy));
Py(i,j) = -rho*(Vx(i,j)*(Vy(i+1,j)-Vy(i-1,j))/(2*dx)+Vy(i,j)*(Vy(i,j+1)-Vy(i,j-1))/(2*dy));

instead of

Px(i,j) = -rho*(Vx(i,j)*((Vx(i+1,j)-Vx(i-1,j))*0.5*dx)+Vy(i,j)*((Vx(i,j+1)-Vx(i,j-1))*0.5*dy));
Py(i,j) = -rho*(Vx(i,j)*((Vy(i+1,j)-Vy(i-1,j))*0.5*dx)+Vy(i,j)*((Vy(i,j+1)-Vy(i,j-1))*0.5*dy));

[Mat576]

The velocity field is determined experimentally from imaging so I import it directly into MATLAB. I don’t have to solve for it, only determine a pressure field from it.

[Torsten]

In principle, the equations

Px(i,j) = -rho*(Vx(i,j)*(Vx(i+1,j)-Vx(i-1,j))/(2*dx)+Vy(i,j)*(Vx(i,j+1)-Vx(i,j-1))/(2*dy));
Py(i,j) = -rho*(Vx(i,j)*(Vy(i+1,j)-Vy(i-1,j))/(2*dx)+Vy(i,j)*(Vy(i,j+1)-Vy(i,j-1))/(2*dy));

are correct to determine dP/dx and dP/dy.

But I don't know how noisy your velocity measurements are so that you might get garbage for the pressure gradient.

[Mat576]

Thanks for your response. Do you have any insight on how to go about taking the integral of dP/dx and dP/dy over a path? I'm not sure which method is appropriate given that it's discrete data and a partial derivative.

[Torsten]

Solve the Pressure Poisson Equation (1.1.3.1) in the domain:

http://www.thevisualroom.com/poisson_for_pressure.html

You will have to know pressure on the boundaries and your velocity field must be divergence-free.