1
$\begingroup$

If I were to fully fill a sphere with water without any air trapped inside, how to find the pressure of the water at, say, the center of the sphere?

$\endgroup$
  • 3
    $\begingroup$ Hello and welcome to Engineering.SE! Your question could use some improvement. What are your own thoughts on that problem? Elaborate a bit more. $\endgroup$ – idkfa Sep 14 '15 at 10:30
  • $\begingroup$ Pressure is proportional to depth, regardless of the container shape. $\endgroup$ – Chris Mueller Sep 14 '15 at 13:23
7
$\begingroup$

$$p = p_0 + \rho g \frac{d_{sphere}}{2}$$

$p_0$ would be 1 Bar I assume, you are not specific under what conditions you fill the sphere and I assume the spheres border are made from a solid material.

The rest is just that the pressure is not dependent on the shape of the container. You just calculate the height of the water column for you desired point, which for the center would be $\frac{d_{sphere}}{2}$.

$\endgroup$
4
$\begingroup$

The pressure at the center would be the pressure at the top plus the extra due to the depth of one radius. The difference in pressure due to height is the same inside the sphere as water anywhere else in the same gravity (using the simplifying assumption that water is incompressible, which is quite valid for most human-scale circumstances on earth).

The difficulty you have is deciding what the pressure is at the top of the sphere. If there is a little hole at the top so that the pressure is atmospheric, then it's simple. Otherwise, you have to know how much the sphere is pressurizing the water, which you haven't told us anything about.

$\endgroup$
  • $\begingroup$ I am obligated by my education to recite my fluids professor's mantra: "There is no such thing as an incompressible fluid, only an incompressible flow." The compressibility certainly is negligible, but it does exist. $\endgroup$ – Trevor Archibald Sep 14 '15 at 20:14
  • $\begingroup$ @Trevor: Agreed. I'm not saying water is incompressible, only that treating it as such is a valid simplification in many ordinary circumstances. $\endgroup$ – Olin Lathrop Sep 14 '15 at 22:40
  • $\begingroup$ Interestingly enough! I was revising Bernoulli's equation when this question suddenly popped out in my mind. I was thinking: "Hmm...If the gauge pressure at a open water surface level is 0, then what is the gauge pressure of water in a solid sphere?" I tried to use Bernoulli's equation to get the answer but to no avail as I can't find the pressure at top the sphere. Thanks for the responses. Let me elaborate further the question. If the solid sphere is completely inelastic (meaning it can't change its dimension), and the radius is 1. Then what is the pressure at the center again? Thank you! $\endgroup$ – Steven Sep 15 '15 at 11:25
  • $\begingroup$ @Steven: It's still whatever the pressure at the top is plus a 1 radius head. The way you've defined the problem though, it will be hard to determine what the pressure at the top is. It's not clear whether the sphere is trying to compress the water or not. $\endgroup$ – Olin Lathrop Sep 15 '15 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.