Balloons and Pressure
Date: Summer 2012
If I have an elastic balloon (or bladder) that contains 300 psi gage pressure of a compressible gas and a subject the outside surface of that balloon to 25 psi gage pressure of a non-compressible liquid, what happens to the pressure inside the balloon (or bladder), assuming the temperature remains constant? Or another way to put the question would be, does it take more that 300 psi pressure of the liquid to affect the pressure inside the bladder?
Hi Rick -- the pressure on the inside of the balloon must equal the
pressure exerted on it. The balloon exists in equilibrium -- it
expands or contracts until the pressures balance out. So, *any* change
in the external pressure would have some effect on the balloon (even
if very small). If you were to take the balloon to the bottom of a
swimming pool, it would shrink a little (because the pressure at the
bottom of the pool is higher). The pressure exerted on it is not just
the external fluid pressure, though; it is the sum of the pressure
from the external fluid plus the force due to the elasticity of the
balloon. So, the answer depends on the nature of the balloon. If the
balloon were very "stiff", then the pressure in the balloon would
increase slightly, but less than the 25psi difference exerted
externally. If the balloon were very stretchy, then the internal
pressure change would be closer to the 25psi. The "incompressible"
designation you placed on the external fluid is irrelevant because you
have specified the pressure, not the system volume.
Hope this helps,
This is an application of the Ideal Gas Law,
(Pressure)(Volume)=(molar number of gas)(Ideal Gas Constant)(Temperature)
You can assume that the gas volume within the balloon follows this law. Since the volume of the gas is constant, and since you stated that the temperature is constant, then the right hand side will stay constant for the volume regardless of whether it compresses or not. Call this constant K. The measured pressure (without the additional liquid) is 300 psi; so given the volume of the balloon, you can calculate what the constant will be.
Now, the tricky part is that the balloon is stretched and pulling in on the gas. It is stretched to the point where the inward force at its surface exactly balances the measured pressure. If this inward force changes, as from an externally-applied force from the incompressible liquid (25 psi), then the balloon will change its shape to where again the inward force from inner surface of the balloon exactly matches the outward force of the enclosed gas. The balloon will not have to apply as much force as before, and it will simply shrink as a result. The volume of the balloon will thus decrease, but the gas inside will still satisfy the equation PV=K. Thus, the gas pressure must increase. The difficulty is determining by how much. If we assume that the pressure created by the balloon depends on its stretched volume we can write
Where Pb is the balloon pressure as a function (F) of its volume (V). This pressure must exactly balance the pressure of the enclosed volume, along with that of the applied pressure (25 psi):
P=F(V) + 25 (1)
With the ideal gas law, we have
Thus, we have two equations (1, 2) and two unknowns (P,V). Unfortunately, at this point we need to know F(V), i.e., how much force/area (or pressure) the balloon creates at different volumes. The equation will depend on the particular balloon material and its elasticity. If you are interested in pursuing the solution further, you can apply a generalized version of Hooke’s law (treating the balloon material like a stretched spring). Unfortunately, it is not a trivial problem.
One further element to consider is the effects of the atmospheric pressure of the background. Including this pressure will affect the final pressure (because F(V) is nonlinear), but it doesn’t change the fundamental reasoning.
Kyle Bunch, PhD, PE
The volume of the balloon would expand until the pressure inside the balloon (300 psi) (P1) matched the pressure outside of the balloon (25 psi) (P2).
If the P2 is increased to 325 psi, the volume of the balloon would decrease until P1 was 325 psi.
From the following URL:
P1 = 300 psi
P2 = 25 psi
For Isothermal process,
Known Ratio of P2/P1
V2 = V1/(P2/P1) = V1/(25/300) = V1/(1/6)
V2 = 6 V1
The balloon will expand its volume by six times to reach 25 psi.
First thing one must do to answer this is to delete all information that is
irrelevant to the question, and restate it in a simpler way. In this case,
the only thing that matters regarding the outside of the bladder is the
pressure. What fluid (compressible or not, liquid or gas) is irrelevant!
You start with a very heavy but flexible bladder that can withstand high
pressure, then fill it with a gas (obviously a "compressible gas" since
all gases are compressible) to a pressure of 300 psig.
At this point we can summarize that there is 300 psig inside the
bladder and 0 psig outside.
Now, we increase the pressure that acts on the outside of the bladder
to 25 psig.
Quite clearly, the pressure inside must increase, since the increased
outside pressure tends to squeeze the bladder and its contents,
resulting in a smaller bladder size. Squeezing the bladder and its
contents to a smaller volume, quite clearly compresses its contents
and thus increases the pressure inside the bladder.
There are some issues you have not addressed. Or put another way, it does not matter that the liquid is non-compressible if it is in a container that has moveable walls. In that case the gas @ 300 psi would push against the non-compressible liquid until the gas pressure and the liquid pressure attained equal pressures (also assuming no solubility of the gas in the liquid). If by non-compressible you mean that the liquid is immovable, because it is in a rigid solid container, then there is no difference whether the liquid is present or not present, the gas would expand to fill the volume contained in the rigid container. If the non-compressible liquid is in the rigid container, then the volume available to the gas is the volume of the rigid container minus the volume of the non-compressible liquid.
Click here to return to the Engineering Archives
Update: November 2011