How Does External Pressure Affect Internal Pressure Variation in a Sphere?

[Jenq]

TL;DR Summary

pressure variations of a sphere subjected to different pressures

Hi

A sphere with radius r is buried in an elastic ideal medium at pressure P1.
Inside the sphere I use energy E to create a variation of pressure of dP.
What variation dPx I would measure if the sphere was buried in a medium at pressure P2 using the same energy E?
Is it possible to solve this problem?

Thank you

Jen

 

[jbriggs444]

This "elastic medium". How does it respond to displacement? i.e. how does it respond when a spherical volume within begins to expand? How elastic is it?

For instance, is it a perfect fluid which responds by moving away with no increase in pressure at all?

 

[Lnewqban]

How flexible are the walls of the sphere?
If perfectly rigid, how much would the internal pressure be affected by changes in the external pressure?

 

[Jenq]

This "elastic medium". How does it respond to displacement? i.e. how does it respond when a spherical volume within begins to expand? How elastic is it?

For instance, is it a perfect fluid which responds by moving away with no increase in pressure at all?

Yes, I am looking for a solution in the simplest conditions, e.g. perfect elastic medium that doesn't increase in pressure if the sphere expands.

 

[Jenq]

How flexible are the walls of the sphere?
If perfectly rigid, how much would the internal pressure be affected by changes in the external pressure?

I don't know the answer, for a perfectly rigid wall is it possible to find a relationship between the 2 pressures in steady conditions?

 

[jbriggs444]

Yes, I am looking for a solution in the simplest conditions, e.g. perfect elastic medium that doesn't increase in pressure if the sphere expands.

You asked for a result as a change in pressure in a medium that deflects with no change in pressure. That's an easy answer: Zero.

If you wanted to know about deflection instead, that's easy as well. $\Delta E = P \Delta V$. Solve for delta V.

Edit: If you want to work on deriving that formula above, we can help with that. A good starting point would be a piston arrangement rather than an expanding sphere.

 

[Lnewqban]

I don't know the answer, for a perfectly rigid wall is it possible to find a relationship between the 2 pressures in steady conditions?

I believe that the original problem should have clarified that important detail.

Please, see:
https://en.m.wikipedia.org/wiki/Ideal_gas_law

A rigid wall or container would not allow any change in internal volume; therefore, if internal temperature does not change (because there is no thermal energy transfer through those walls), internal pressure must remain the same, no matter how much the medium pressure changes.

The opposite extreme would be walls that can be expanded much without offering much resistance or elastic force.
In that case, the internal pressure value would be very much close to the external one.
Think of a soap bubble, for example; its internal pressure must be equal to the atmospheric pressure surrounding it.

A rubber balloon would be an intermediate situation between those two extremes.

 

[Jenq]

You asked for a result as a change in pressure in a medium that deflects with no change in pressure. That's an easy answer: Zero.

If you wanted to know about deflection instead, that's easy as well. $\Delta E = P \Delta V$. Solve for delta V.

Edit: If you want to work on deriving that formula above, we can help with that. A good starting point would be a piston arrangement rather than an expanding sphere.

Ah sorry I misunderstood, I meant the pressure exerted by the external environment on the sphere does not change if the pressure / volume inside the sphere changes.

The equation is useful to me: so we can say that $P_1 \Delta V_1 = \Delta E = P_2 \Delta V_2$ where $P_1$ and $P_2$ are the external pressures, right? If so, yes I would like to read a reference on how to derive that equation.
Thanks

 

[Jenq]

I believe that the original problem should have clarified that important detail.

Please, see:
https://en.m.wikipedia.org/wiki/Ideal_gas_law

A rigid wall or container would not allow any change in internal volume; therefore, if internal temperature does not change (because there is no thermal energy transfer through those walls), internal pressure must remain the same, no matter how much the medium pressure changes.

The opposite extreme would be walls that can be expanded much without offering much resistance or elastic force.
In that case, the internal pressure value would be very much close to the external one.
Think of a soap bubble, for example; its internal pressure must be equal to the atmospheric pressure surrounding it.

A rubber balloon would be an intermediate situation between those two extremes.

I see, then I think the case should fall into an intermediate situation. Otherwise in a pure elastic case the sphere should inflate till the internal pressure balances the external one $\Delta P_x = \Delta P + P_2 - P_1$ ?

Thank you for the useful link.

 

[jbriggs444]

Ah sorry I misunderstood, I meant the pressure exerted by the external environment on the sphere does not change if the pressure / volume inside the sphere changes.

The equation is useful to me: so we can say that $P_1 \Delta V_1 = \Delta E = P_2 \Delta V_2$ where $P_1$ and $P_2$ are the external pressures, right? If so, yes I would like to read a reference on how to derive that equation.

Instead of a sphere, assume that we have a cylinder under a pool of water. One end of the cylinder is closed. The other end is open to the water. There is a piston mounted in the cylinder. The piston is free to move within the cylinder. It makes a tight seal with the sides. Friction is negligible.

We remove the water behind the piston so that only vacuum remains. Obviously, the pool water is pushing on the piston. So we have to brace the piston, holding it stationary against the water pressure.

Now we use some energy, pushing the piston away from our brace on the vacuum side and out toward the open pool.

Let us call the pool pressure at the depth of the cylinder "P".
Let us say that the piston has cross-sectional area "A".
Let us say that we have used energy "E" to move the piston.

We ask: "much volume does the piston cover when we apply this energy E"?We can start with how much force there is on the piston. Force is equal to pressure times area.$$F=PA$$How much distance ($\Delta x$) has the piston covered? Work is equal to force times distance. So$$E=F\Delta x$$Solving for $\Delta x$ and substituting $PA$ for $F$, we get:$$\Delta x = \frac{E}{PA}$$Multiply through by PA and we get:$$PA\Delta x = E$$How much volume ($\Delta V$) has the piston swept out? Well that is area times displacement -- $A \Delta x$. So we get $$P \Delta V = E$$QED