Work calculation from applied pressure

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

Answer 1

[William Rose]

@Miraboreasu,

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.

[Miraboreasu]

Suppose the triangle is made of elastic material, and Young's modolus is E.

For the volume I know the volume when the sphere is intact, but I only have the locations of the nodes of each triangle when the sphere breaks.

[William Rose]

@Miraboreasu,

If you give the Young's modulus then the thickness is also required. (This is why I requested the spring constant, rather that E_Y, in my previous answer.) Even if you know the triangle's thickness, it is still complicated, because the three corners all have differennt relative velocities. Therefore each corner is moving in a different direciton. Therefore the triangle's area and shape are changing.

After the sphere has burst, the pressure shoudl be the same on both sides of the triangle, so the pressure work will be zero after bursting.

[William Rose]

@Miraboreasu, The attached script computes the rate of change of area of the triangle, and plots the triangle and velcity vectors at the corners. The area is decreasing with time. If the corners continutue on their presnt trajectories, the area will decrease for a short time and then will start to increase, as the corners pass by one another and continue moving. In the figure below, corners 1,2,3 are colored R,G,B respectively.

Script console output:

>> showTriangle

d(dist21)/dt=6.27~=6.27

d(dist32)/dt=-13.08~=-13.03

d(dist13)/dt=-11.67~=-11.63

Area(t0)=4.8445e-05, Area(t1)=4.8396e-05, dArea/dt~=-4.84e-02

dt for approximation=1.00e-06

Mean Velocity=-3.2,3.4,-51.3

Good luck.

[Miraboreasu]

I found the displacement (m) of each node

d1 0.000131475 -0.000706995 0.000754736

d2 6.02E-05 -0.000662299 0.000711364

d3 0.000147876 -0.000661116 0.000729507

My idea is thatI calculated the areas of each triangle from the edge vectors

I think I can use the pressure times area to get the force, then times displacement.

But I have the displacement for three nodes, how to make them the same direction as my force?

[William Rose]

@Miraboreasu,

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]

@Miraboreasu,

Thank you for accepting this answer.