# Calculate the work done by pressure rupturing a spherical containment

• A
• miraboreasu
In summary, the conversation discusses a simulation in which a spherical object is subjected to a time-dependent pressure on the inner surface, causing it to break into smaller triangles. The goal is to calculate the energy brought by the pressure to the system and the conversation explores different methods to do so. The simulation data includes nodal coordinates, velocities, and displacements for three vertices of one triangle, and the ultimate goal is to determine the total energy brought into the system by the pressure. The question also mentions knowing the energy needed to break the sphere into parts, which is denoted as E.

#### miraboreasu

I am post-processing a simulation.

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 in the normal direction all the time. After t1=0.1s, the spherical is broken, and each little triangle is disconnected.
Assuming during t1, p is constantly 1 (10*0.1). My ultimate goal is to calculate the energy brought by p to this system

My idea is to use p*area*displacement for 1 triangle, then do the same thing for all other triangles

Here is what I have from the simulation (for one triangle).
Nodal coordinates (vector in x,y,z) for three vertices of the triangle
p1: 2.48309 2.51276 2.45388
p2: 2.4875 2.50415 2.45103
p3: 2.47773 2.50283 2.45452
Nodal velocities (vector in x,y,z) for three vertices of the triangle
v1: -11.352 4.68846 -58.9501
v2: -10.2788 -1.54017 -60.6666
v3: 12.043 6.94501 -34.1632
Nodal displacements (vector in x,y,z) for three vertices of the triangle
d1: -0.00023 0.000131 -0.00071
d2: -0.00025 6.02E-05 -0.00066
d3: -0.00027 0.000148 -0.00066

I write the following MATLAB code to compute the area of this triangle from nodal coordinates
'''
p1=[2.48309 2.51276 2.45388];
p2=[2.4875 2.50415 2.45103];
p3=[2.47773 2.50283 2.45452];

edge12=p2-p1;
edge13=p3-p1;

area = 0.5*norm(cross(edge12,edge13),2)
'''

Then I use p*area to get the force, but I don't know how to get the right (or approximated) displacement, since force should be in the same direction as displacement, what I have is the Nodal displacements shown above.

Any idea to figure out the energy brought by p to this system during t1? It will be great if the answer is explained in detail

That should equal the amount of energy that is initially needed to increase the internal pressure of the sphere from zero (respect to atmospheric pressure) to the rupture pressure.

• berkeman
Lnewqban said:
That should equal the amount of energy that is initially needed to increase the internal pressure of the sphere from zero (respect to atmospheric pressure) to the rupture pressure.
Can you please be more specific? Assume I know the energy needed to tear the sphere into parts, which is E.
Besides, the data I provide is all I have. What should I use to calculate it?

Welcome to PF.

miraboreasu said:
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 in the normal direction all the time. After t1=0.1s, the spherical is broken, and each little triangle is disconnected.
Assuming during t1, p is constantly 1 (10*0.1). My ultimate goal is to calculate the energy brought by p to this system
Could you please say more about what you are trying to simulate/calculate? If you are wanting to calculate how much energy you need to use to compress a gas to rupture a spherical enclosure, the post-rupture dynamics seem irrelvant to me. You would need to simulate the strength and deformation of the enclosure based on the material properties of the enclosure material, and the properties of the gas that is going to cause the rupture.

The more background that you can give us about your project and simulation, the better we can help you. berkeman said:
Welcome to PF.

Could you please say more about what you are trying to simulate/calculate? If you are wanting to calculate how much energy you need to use to compress a gas to rupture a spherical enclosure, the post-rupture dynamics seem irrelvant to me. You would need to simulate the strength and deformation of the enclosure based on the material properties of the enclosure material, and the properties of the gas that is going to cause the rupture.

The more background that you can give us about your project and simulation, the better we can help you. Hey berkeman,

I am post-process my simulation. My simulation is to add a dynamic loading to a sphere surface. I know the energy needed to tear the sphere into parts.

I know how much pressure is added, since it is added as a boundary condition. Under this pressure, the sphere is broken into triangle as it was meshed, and I have these data from the output of the simulation
Nodal coordinates (vector in x,y,z) for three vertices of the triangle
p1: 2.48309 2.51276 2.45388
p2: 2.4875 2.50415 2.45103
p3: 2.47773 2.50283 2.45452
Nodal velocities (vector in x,y,z) for three vertices of the triangle
v1: -11.352 4.68846 -58.9501
v2: -10.2788 -1.54017 -60.6666
v3: 12.043 6.94501 -34.1632
Nodal displacements (vector in x,y,z) for three vertices of the triangle
d1: -0.00023 0.000131 -0.00071
d2: -0.00025 6.02E-05 -0.00066
d3: -0.00027 0.000148 -0.00066

I want to know the total energy brought into this system by my boundary condition pressure p

miraboreasu said:
Can you please be more specific? Assume I know the energy needed to tear the sphere into parts, which is E.
Besides, the data I provide is all I have. What should I use to calculate it?
Energy in = energy out

It is like a compressed spring: elastic energy is accumulated, and then, released.
Nothing else will separate the little triangles.
Once detached, they should move radially out at same speed than liberated air.

The energy of the expanding air should dissipate with the square of the radial distance from the center of the sphere.
...Or so I believe. miraboreasu said:
I am post-process my simulation. My simulation is to add a dynamic loading to a sphere surface. I know the energy needed to tear the sphere into parts.
How do you know the value of that energy? What were your calculations/simulations to tell you that? What is the material and its thickness? What are its properties that are germaine to its rupturing?

berkeman said:
How do you know the value of that energy? What were your calculations/simulations to tell you that? What is the material and its thickness? What are its properties that are germaine to its rupturing?
The simulation outputs this energy. It is considered an elastic material.

## 1. How is the work done by pressure calculated?

The work done by pressure in rupturing a spherical containment is calculated by multiplying the pressure exerted on the surface of the containment by the change in volume of the containment. This can be represented by the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume.

## 2. What is the unit of measurement for work done by pressure?

The unit of measurement for work done by pressure is joules (J). This is the standard unit of measurement for work in the International System of Units (SI).

## 3. How does the size of the containment affect the work done by pressure?

The size of the containment does not directly affect the work done by pressure, as long as the pressure and change in volume remain constant. However, a larger containment may require more pressure to rupture, resulting in a higher amount of work done.

## 4. Can the work done by pressure be negative?

Yes, the work done by pressure can be negative if the pressure decreases and the volume of the containment increases. This means that the containment is expanding and doing work on its surroundings.

## 5. What other factors may affect the work done by pressure in rupturing a spherical containment?

The work done by pressure may also be affected by the material and structural strength of the containment, as well as the presence of any external forces or obstacles. Additionally, the temperature and composition of the substance inside the containment may also play a role in determining the work done by pressure.