# Calculate work done by a time-dependent pressure to a spherical hole

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• miraboreasu
In summary: Which can be simplified to: $$\rho\frac{du}{dt}=\frac{\partial}{\partial r}+\frac{2\sigma_{rr}-\sigma_{\theta \theta}-\sigma_{\phi...}u}{r}$$In summary,
miraboreasu
Hello,

Suppose I have a spherical hole in a elastic infinite space. I apply a time-dependent pressure to the inner surface of the spherical hole.
I know p = f(t).

If I only consider this as an elastic problem, no failure happened, how can I calculate the work done by p during the time from 0 to t0?

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miraboreasu said:
Hello,

Suppose I have a spherical hole in a elastic infinite space. I apply a time-dependent pressure to the inner surface of the spherical hole.
I know p = f(t).

If I only consider this as an elastic problem, no failure happened, how can I calculate the work done by p during the time from 0 to t0?
You could if you knew how the volume of the hole was changing over time. But all we know about the material around the hole is that it is "elastic" and "infinite". That is not enough to tell us how it deforms under pressure.

russ_watters
miraboreasu said:
I apply a time-dependent pressure to the inner surface of the spherical hole.
If the material is elastic, you need only consider the initial and final pressures. You can ignore the path taken between the end points, since work done may be recovered without loss due to plastic deformation or failure.

Do you know Young's modulus, E, for the material ?

The pressure on the spherical boundary causes an elastic deflection outwards. You will need to integrate the compression of the material from the sphere's radius to infinity. There are probably some simple approximations that can be made.

As measuring any internal parameter inside that elastic infinite space, perhaps the pressure and volume of the injected fluid could be more feasibly measured at the external source.
Pressure–volume work: 1 J = 1 Pascal x 1 cubic meter

Of course, the above would not be possible if the pressure increases due to an internal exothermic chemical reaction.

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This is the spherical equivalent of applying a time varying force to the end of a semi-infinite rod. In this case, only radial displacements are present, but stresses are developed in all 3. principal directions (radial, latitudinal, and longitudinal). Still, this is a 1D problem in the radial direction. The local stresses are governed by the differential form Hooke's law in 3D. The stress-equilibrium equation in spherical coordinates should be employed, including the internal terms. You will end up with a wave equation in the radial direction. I recommend solving this first for a step change in pressure at the spherical surface. When that solution is in hand, you can represent the pressure as a series of step changes with time, and use superposition to get the time dependent solution (convolution).

Baluncore
Chestermiller said:
This is the spherical equivalent of applying a time varying force to the end of a semi-infinite rod. In this case, only radial displacements are present, but stresses are developed in all 3. principal directions (radial, latitudinal, and longitudinal). Still, this is a 1D problem in the radial direction. The local stresses are governed by the differential form Hooke's law in 3D. The stress-equilibrium equation in spherical coordinates should be employed, including the internal terms. You will end up with a wave equation in the radial direction. I recommend solving this first for a step change in pressure at the spherical surface. When that solution is in hand, you can represent the pressure as a series of step changes with time, and use superposition to get the time dependent solution (convolution).
Thanks so much, why only radial displacements are present?

miraboreasu said:
Thanks so much, why only radial displacements are present?
It is an obvious consequence of symmetry. If the setup is spherically symmetric then everything that follows must be spherically symmetric.

Radial displacements can be spherically symmetric. Tangential displacements cannot be -- which tangential direction would be chosen? If you cannot pick a direction, the only possible deterministic prediction is zero.

The radial displacement is ##u(t,r)##. The nonzero components of the strain tensor are $$\epsilon_{rr}= \frac{\partial u}{\partial r}$$and$$\epsilon_{\theta \theta}=\epsilon_{\phi \phi}=\frac{u}{r}$$
The non-zero components of the stress tensor are $$\sigma_{rr}=\lambda(\epsilon_{rr}+\epsilon_{\theta \theta}+\epsilon_{\phi \phi})+2\mu \epsilon_{rr}=\lambda\left(\frac{\partial u}{\partial r}+2\frac{u}{r}\right)+2\mu\frac{\partial u}{\partial r}$$
$$\sigma_{\theta \theta}=\lambda(\epsilon_{rr}+\epsilon_{\theta \theta}+\epsilon_{\phi \phi})+2\mu \epsilon_{\theta \theta}=\lambda\left(\frac{\partial u}{\partial r}+2\frac{u}{r}\right)+2\mu\frac{u}{r}$$
$$\sigma_{\phi \phi}=\lambda(\epsilon_{rr}+\epsilon_{\theta \theta}+\epsilon_{\phi \phi})+2\mu \epsilon_{\phi \phi}=\lambda\left(\frac{\partial u}{\partial r}+2\frac{u}{r}\right)+2\mu\frac{u}{r}$$where ##\lambda## and ##\mu## are the Lame' elastic constants.
The differential force balance in the radial direction for this symmetric situation reduces to: $$\rho\frac{\partial^2 u}{\partial t^2}=\frac{\partial \sigma_{rr}}{\partial r}+\frac{2\sigma_{rr}-\sigma_{\theta \theta}-\sigma_{\phi \phi}}{r}$$

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If we substitute the equations for the stresses into the differential force balance of the previous post, we obtain $$\rho\frac{\partial ^2u}{\partial t^2}=(\lambda +2\mu)\frac{\partial}{\partial r}\left(\frac{1}{r^2}\frac{\partial(r^2u)}{\partial r}\right)$$or, equivalently, $$\rho\frac{\partial ^2\xi}{\partial t^2}=(\lambda +2\mu)r^2\frac{\partial}{\partial r}\left(\frac{1}{r^2}\frac{\partial\xi}{\partial r}\right)$$where ##\xi=r^2u##. At r = R, the boundary condition is $$\sigma_{rr}=\lambda\left(\frac{\partial u}{\partial r}+2\frac{u}{r}\right)+2\mu\frac{\partial u}{\partial r}=-P$$or, equivalently, $$(\lambda+2\mu)\frac{1}{r^2}\frac{\partial (r^2u)}{\partial r}-4\mu\frac{u}{r}=-P$$or$$(\lambda+2\mu)\frac{\partial \xi}{\partial r}-4\mu\frac{\xi}{R}=-PR^2$$
The boundary condition at large r is ##\xi=0##.

Initially the displacement is zero at all locations $$\xi(0,r)=0$$The initial velocity is also zero at all locations, except at the very boundary r = R (involving zero mass in the limit), where it is finite. From the solution for a sudden force applied to the end of a rod (at short times, the compression wave is very close to the boundary in our system, so that spherical curvature can be neglected at short times), we find that, for our case, $$\left(\frac{du}{dt}\right)_{t=0,r=R}=\frac{P}{\sqrt{\rho(\lambda+2\mu)}}$$or, equivalently, $$\left(\frac{d\xi}{dt}\right)_{t=0,r=R}=\frac{PR^2}{\sqrt{\rho(\lambda+2\mu)}}$$
This completes the formulation of the linearly elastic model equations for a suddenly pressurized spherical cavity.

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hutchphd
It just occurred to me that there is a steady state (i.e., time-independent) solution to the equations I presented in the previous few posts that applies in the limit of vary gradually applied pressure (over time). This is analogous to applying a gradually varying compressive force on the two ends of an elastic rod. The steady state solution for the case of a pressurized elastic cavity is: $$u=\frac{PR^3}{4\mu r^2}$$and, at the cavity boundary $$u(R)=\frac{PR}{4\mu}$$Based on this, the work done by the pressure per unit area of cavity surface is $$w=\int{Pdu}=\frac{P^2R}{8\mu}$$Since the surface area of the cavity is ##4\pi R^2##, the total work done by the pressure is $$W=\frac{\pi P^2R^3}{2\mu}=\frac{(1+\nu)}{E}\pi P^2R^3$$where ##\nu## is the Poisson ratio and E is the Young's modulus.

miraboreasu and hutchphd
Chestermiller said:
It just occurred to me that there is a steady state (i.e., time-independent) solution to the equations I presented in the previous few posts that applies in the limit of vary gradually applied pressure (over time). This is analogous to applying a gradually varying compressive force on the two ends of an elastic rod. The steady state solution for the case of a pressurized elastic cavity is: $$u=\frac{PR^3}{4\mu r^2}$$and, at the cavity boundary $$u(R)=\frac{PR}{4\mu}$$Based on this, the work done by the pressure per unit area of cavity surface is $$w=\int{Pdu}=\frac{P^2R}{8\mu}$$Since the surface area of the cavity is ##4\pi R^2##, the total work done by the pressure is $$W=\frac{\pi P^2R^3}{2\mu}=\frac{(1+\nu)}{E}\pi P^2R^3$$where ##\nu## is the Poisson ratio and E is the Young's modulus.
Thank you so much, I have an idea about using the final equation for W. I can discretize the $f(t)$ with timesteps, then for each timestep, I have $r$ which is R+u

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miraboreasu said:
Thank you so much, I have an idea about using the final equation for W. I can discretize the $f(t)$ with timesteps, then for each timestep, I have $r$ which is R+u
That would not be the correct thing to do. To this degree of approximation, W is the cumulative work up to time t, and P in the equation is already the pressure at the instant t. Also, using R+u is useless, because the equations are already linearized, and that would introduce a second-order term that is not valid.

Chestermiller said:
If we substitute the equations for the stresses into the differential force balance of the previous post, we obtain $$\rho\frac{\partial ^2u}{\partial t^2}=(\lambda +2\mu)\frac{\partial}{\partial r}\left(\frac{1}{r^2}\frac{\partial(r^2u)}{\partial r}\right)$$or, equivalently, $$\rho\frac{\partial ^2\xi}{\partial t^2}=(\lambda +2\mu)r^2\frac{\partial}{\partial r}\left(\frac{1}{r^2}\frac{\partial\xi}{\partial r}\right)$$where ##\xi=r^2u##. At r = R, the boundary condition is $$\sigma_{rr}=\lambda\left(\frac{\partial u}{\partial r}+2\frac{u}{r}\right)+2\mu\frac{\partial u}{\partial r}=-P$$or, equivalently, $$(\lambda+2\mu)\frac{1}{r^2}\frac{\partial (r^2u)}{\partial r}-4\mu\frac{u}{r}=-P$$or$$(\lambda+2\mu)\frac{\partial \xi}{\partial r}-4\mu\frac{\xi}{R}=-PR^2$$
The boundary condition at large r is ##\xi=0##.

Initially the displacement is zero at all locations $$\xi(0,r)=0$$The initial velocity is also zero at all locations, except at the very boundary r = R (involving zero mass in the limit), where it is finite. From the solution for a sudden force applied to the end of a rod (at short times, the compression wave is very close to the boundary in our system, so that spherical curvature can be neglected at short times), we find that, for our case, $$\left(\frac{du}{dt}\right)_{t=0,r=R}=\frac{P}{\sqrt{\rho(\lambda+2\mu)}}$$or, equivalently, $$\left(\frac{d\xi}{dt}\right)_{t=0,r=R}=\frac{PR^2}{\sqrt{\rho(\lambda+2\mu)}}$$
This completes the formulation of the linearly elastic model equations for a suddenly pressurized spherical cavity.
If I stick with the "dynamic version", which is $$\left(\frac{du}{dt}\right)_{t=0,r=R}=\frac{P}{\sqrt{\rho(\lambda+2\mu)}}$$
If I get your idea, to get the $$u(t,r)$$, I need to do the following?
$$\int_{R}^{\infty} 1 du=\int_{0}^{t} \frac{P(t)}{\sqrt{\rho(\lambda+2\mu)}}dt$$
and$\rho$ is density right?

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miraboreasu said:
If I stick with the "dynamic version", which is $$\left(\frac{du}{dt}\right)_{t=0,r=R}=\frac{P}{\sqrt{\rho(\lambda+2\mu)}}$$
If I get your idea, to get the $$u(t,r)$$, I need to do the following?
$$\int_{R}^{\infty} 1 du=\int_{0}^{t} \frac{P(t)}{\sqrt{\rho(\lambda+2\mu)}}dt$$
and$\rho$ is density right?
This implementation of convolution is incorrect.

Yes, ##\rho## is the density.

Chestermiller said:
This implementation of convolution is incorrect.

Yes, ##\rho## is the density.
Can you please guide me more?

miraboreasu said:
Can you please guide me more?
Do you know how to solve the full set of equations, either by expressing the displacement as a function of r times a function of time, or by solving the equations numerically?

Your implementation of superposition was incorrect because, to use it the way you tried, you need to know the velocity at r = R, not just at t = 0, but for all times subsequent to a unit step change of pressure at t = 0. If you let v*(t,R) be the velocity at r =. R for a unit step change in P at time zero, then for an arbitrary pressure variation with time, the velocity v(t,R) for arbitrary pressure variation would be $$v(t,R)=\int_0^t{v^*(t-\tau,R)P'(\tau)d\tau}$$where $$P'(\tau)=\frac{dP}{d\tau}$$

Chestermiller said:
It just occurred to me that there is a steady state (i.e., time-independent) solution to the equations I presented in the previous few posts that applies in the limit of vary gradually applied pressure (over time). This is analogous to applying a gradually varying compressive force on the two ends of an elastic rod. The steady state solution for the case of a pressurized elastic cavity is: $$u=\frac{PR^3}{4\mu r^2}$$and, at the cavity boundary $$u(R)=\frac{PR}{4\mu}$$Based on this, the work done by the pressure per unit area of cavity surface is $$w=\int{Pdu}=\frac{P^2R}{8\mu}$$Since the surface area of the cavity is ##4\pi R^2##, the total work done by the pressure is $$W=\frac{\pi P^2R^3}{2\mu}=\frac{(1+\nu)}{E}\pi P^2R^3$$where ##\nu## is the Poisson ratio and E is the Young's modulus.
$$w=\int{Pdu}=\frac{P^2R}{8\mu}$$
from
$$u=\frac{PR^3}{4\mu r^2}$$and, at the cavity boundary $$u(R)=\frac{PR}{4\mu}$$

I didn't get how to get $$du$$, I know how to get u(R)

miraboreasu said:
$$w=\int{Pdu}=\frac{P^2R}{8\mu}$$
from
$$u=\frac{PR^3}{4\mu r^2}$$and, at the cavity boundary $$u(R)=\frac{PR}{4\mu}$$

I didn't get how to get $$du$$, I know how to get u(R)
I meant to write $$w=\int{Pdu(R)}=\int{P\frac{R}{4\mu}dP}=\frac{P^2R}{8\mu}$$

Chestermiller said:
I meant to write $$w=\int{Pdu(R)}=\int{P\frac{R}{4\mu}dP}=\frac{P^2R}{8\mu}$$
I have the p(t) like this, only consider the positive part, can I still use steady state?
https://en.wikipedia.org/wiki/Blast_wave
and for this equation,
$$w=\int{Pdu(R)}=\int{P\frac{R}{4\mu}dP}=\frac{P^2R}{8\mu}$$
the integral is from where to where when du?
and the integral is from where to where when dp?

Sorry for the detail questions

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miraboreasu said:
I have the p(t) like this, only consider the positive part, can I still use steady state?
https://en.wikipedia.org/wiki/Blast_wave
This reference is for a fluid, not an elastic solid.
miraboreasu said:
and for this equation,
$$w=\int{Pdu(R)}=\int{P\frac{R}{4\mu}dP}=\frac{P^2R}{8\mu}$$
the integral is from where to where when du?
from 0 to ##PR/4\mu##
miraboreasu said:
and the integral is from where to where when dp?
from 0 to P
miraboreasu said:
Sorry for the detail questions

Chestermiller said:
This reference is for a fluid, not an elastic solid.

from 0 to ##PR/4\mu##

from 0 to P
Now I finally get it, but what value I should use from p=f(t) to for the P in the work equation

miraboreasu said:
Now I finally get it, but what value I should use from p=f(t) to for the P in the work equation
$$P=f(t)$$

Chestermiller said:
This is the spherical equivalent of applying a time varying force to the end of a semi-infinite rod. In this case, only radial displacements are present, but stresses are developed in all 3. principal directions (radial, latitudinal, and longitudinal). Still, this is a 1D problem in the radial direction. The local stresses are governed by the differential form Hooke's law in 3D. The stress-equilibrium equation in spherical coordinates should be employed, including the internal terms. You will end up with a wave equation in the radial direction. I recommend solving this first for a step change in pressure at the spherical surface. When that solution is in hand, you can represent the pressure as a series of step changes with time, and use superposition to get the time dependent solution (convolution).
If the cavity is a ellipsoid, I know the a, b, and c for it, what solution I should seek for?

miraboreasu said:
If the cavity is a ellipsoid, I know the a, b, and c for it, what solution I should seek for?
This is a much more complicated problem

## 1. How do you calculate the work done by a time-dependent pressure to a spherical hole?

The work done by a time-dependent pressure to a spherical hole can be calculated using the formula W = ∫PdV, where W is the work done, P is the pressure, and dV is the change in volume.

## 2. What is the significance of a time-dependent pressure in the calculation of work done?

A time-dependent pressure takes into account the changes in pressure over time, which can affect the work done. It allows for a more accurate calculation of the work done by the pressure on the spherical hole.

## 3. What are the units for work done in this calculation?

The units for work done are Joules (J) in the International System of Units (SI). However, other units such as ergs or foot-pounds can also be used as long as the units for pressure and volume are consistent.

## 4. Can this calculation be applied to other shapes besides a spherical hole?

Yes, this calculation can be applied to any shape as long as the pressure and volume are known. However, the formula may differ depending on the shape and the type of pressure (e.g. constant or time-dependent).

## 5. How does the work done by a time-dependent pressure to a spherical hole affect the overall system?

The work done by a time-dependent pressure to a spherical hole can affect the energy and stability of the system. It can also impact the flow of fluids or gases through the hole and the resulting changes in pressure and volume.

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