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Consider a half sphere of radius $R$, partially filled with an incompressible liquid with a density $\rho$, up to a height $h.$ I'd like to find the pressure field inside the liquid.

For $| x | \leqslant x_l$, it's easy: indeed, we know that $P(x,h)=P_0$, and moreover $\frac{\textrm{d}P}{\textrm{d}z} = -\rho g$, so

$$ P(x,z) = P_0 + \rho g (h - z)$$

Also, when $\theta_l \ll 1$, $x_l \approx R$, so $P$ should not vary too much between $x_l$ and $R$: for any $x$, $$ P(x,z) \approx P_0 + \rho g (h-z)$$

However, in general cases, I really don't know how we could find the pressure field for $|x| > x_l$. The equation $\frac{\textrm{d}P}{\textrm{d}z} = -\rho g$ is still true, yet I can't achieve to find the pressure next to side of the half sphere: how could we find it ?

Edit:

JezuzStarusst's answer is valid away from the side of the sphere, since there's no horizontal force acting on the fluid. However, near the side, surface tension appears. How could we take it into account ?

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    $\begingroup$ Without being sure, shouldn't the pressure be the same everywhere at a given height? Otherwise there would be a flow of particles in one direction or the other. $\endgroup$
    – JezuzStardust
    Oct 24, 2016 at 20:07
  • $\begingroup$ @JezuzStardust Yes, you're right! Thanks, you can post it as an answer so that I can close this topic. However, this works only when there is no horizontal force acting on the fluid, but it is acted on by surface tension near the side of the sphere: how would the pression be modified if we take this into account ? $\endgroup$
    – Spirine
    Oct 24, 2016 at 20:11

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I believe that the pressure must be the same at a given height $z$, since otherwise there would be a flow in one direction or the other.

At $|x|> x_l$ the height of the fluid is less, but there is a compensating normal force from the containing vessel.

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  • $\begingroup$ What about if we take surface tension into account? $\endgroup$
    – Spirine
    Oct 24, 2016 at 20:18
  • $\begingroup$ I am not sure how to deal with the surface tension. Sorry. Hopefully someone else can answer that. $\endgroup$
    – JezuzStardust
    Oct 24, 2016 at 20:19

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