# Work calculation from applied pressure

4 views (last 30 days)
Miraboreasu on 24 Sep 2022
Commented: William Rose on 30 Sep 2022
A spherical is meshed by many little triangles.A time-dependent pressure (p=10*t) is equally applied to the inner surface of a spherical. After t1=0.1s, the spherical broken and each little triangle is disconnected.
The data I have is the nodes locations (m) of each little triangle like
point1 2.48309 2.51276 2.45388
point2 2.4875 2.50415 2.45103
point3 2.47773 2.50283 2.45452
and velocity (m/s) of each nodes
v1 -11.352 4.68846 -58.9501
v2 -10.2788 -1.54017 -60.6666
v3 12.043 6.94501 -34.1632
How can I calculate the work has been done by pressure p
I think need to ignore the work done by the interation of each triangle

William Rose on 24 Sep 2022
You ask "How can I calculate the work has been done by pressure p".
Do you mean the work done from time t=0 to time t=0.1 seconds, when the sphere breaks apart?
The pressure-volume work from time 0 to 0.1 is
W=integral from t=0 to 0.1 of {P(t)dV(t)}
where P(t) is the pressure at time t, (and you told us that P=10*t), and V(t) is the volume at time t.
If the triangle has mass, then the pressure may also do mechanical work, accelerating the triangle. But you did not tell us the triangle's mass or its acceleration. If the triangle is made of elastic material, then the pressure may do work stretching the triangle. But you did not tell us the triangle's spring constant.
The computation of pressure-volume work requires knowing the total volume of the sphere at each instant. We cannot determine this from knowing just how the corners of one triangle are moving. The overall movement may be a combination of inflation/deflation plus translation plus rotation. We do not even know on which side of the triangle the pressure is applied. I assume the pressure on the "other" side is zero.
William Rose on 26 Sep 2022
I am hesitant to give advice, since I do not have a clear picture in my mind of what is going on.
• You said initially that this triangle is part of a sphere that is being inflated by a pressure gradient. Then the corner velocities (which you gave in your initial post) should be normal to the surface. But they're not. Why not?
• You said initially that this triangle is part of a sphere that is being inflated by a pressure gradient. Then the area should be increasing. But it's decreasing. Why?
• The displacements which you specified in your most recent comment are not proportional to the velocities which you specified in your initial posting. One would expect the displacements to be the velocities times a time increment. Why is this not the case?
If the velocities are after the burst, then the movement is not due to the pressure, at least not solely, and certainly not in any simple way.
Pressure times change in volume equals work done. You are correct that you can also compute the work as force times displacement. Specifically, the work is the inner product of the force vector times the displacement vector. The force vector is the pressure (scalar) times the area (scalar) times a unit vector normal to the surface. You use a vector normal to the surface, because pressure pushes in all directions, and creates a force normal to the surface against which it pushes.
William Rose on 30 Sep 2022
Thank you for accepting this answer.

### Categories

Find more on Lighting, Transparency, and Shading in Help Center and File Exchange

R2022b

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!