Generalities
The problem has spherical symmetry, so it makes sense to use spherical coordinates ($r$, $\theta$, $\phi$). We can divide the vessel into differential elements like the one shown in this post.
Deformation and strain
Only radial deformations are allowed by the spherical symmetry, so let's parametrize the deformation by
$$r \rightarrow r + u(r)$$
Assuming small deformations:
$$dr \rightarrow dr(1 + u'(r))$$
$$r\,d\theta \rightarrow (r + u(r))\,d\theta$$
$$r\cos\theta\,d\phi \rightarrow (r + u(r))\cos\theta\,d\phi$$
The associated strains:
$$\epsilon_{rr} = \frac{dr(1 + u'(r)) - dr}{dr} = u'(r)$$
$$\epsilon_{\theta\theta} = \frac{(r + u(r))\,d\theta - r\,d\theta}{r\,d\theta} = \frac{u(r)}{r}$$
$$\epsilon_{\phi\phi} = \frac{(r + u(r))\cos\theta\,d\phi - r\cos\theta\,d\phi}{r\cos\theta\,d\phi} = \frac{u(r)}{r}$$
Symmetry
By spherical symmetry we have:
$$\epsilon_{\theta\theta} = \epsilon_{\phi\phi} = \epsilon_{tt}$$
$$\sigma_{r\theta} = \sigma_{r\phi} = \sigma_{\theta\phi} = 0$$
$$\sigma_{\theta\theta} = \sigma_{\phi\phi} = \sigma_{tt}$$
$$\frac{\partial\,\sigma_{rr}}{\partial\,\theta} = \frac{\partial\,\sigma_{\theta\theta}}{\partial\,\theta} = \frac{\partial\,\sigma_{\phi\phi}}{\partial\,\theta} = \frac{\partial\,\sigma_{rr}}{\partial\,\phi} = \frac{\partial\,\sigma_{\theta\theta}}{\partial\,\phi} = \frac{\partial\,\sigma_{\phi\phi}}{\partial\,\phi} = 0$$
Equilibrium condition
The equilibrium condition can be expressed in spherical coordinates as
$$2\sigma_{rr} + r\frac{\partial\,\sigma_{rr}}{\partial\,r}-\sigma_{\theta\theta}-\sigma_{\phi\phi} = 0$$
or, using the symmetry conditions,
$$2\sigma_{rr}(r) + r\sigma_{rr}'(r) - 2\sigma_{tt} = 0$$
Hooke's law
Applying Hooke's law for isotropic materials we get
$$\epsilon_{rr} = \frac{1}{E}(\sigma_{rr} - 2 \nu \sigma_{tt})$$
$$\epsilon_{tt} = \frac{1}{E}(\sigma_{tt} - \nu (\sigma_{rr} + \sigma_{tt}))$$
where $E$ is the elastic modulus and $\nu$ is the Poisson's ratio.
Math
Combining the previous results we get the following equations:
$$u'(r) = \frac{1}{E}(\sigma_{rr} - 2 \nu \sigma_{tt})$$
$$\frac{u(r)}{r} = \frac{1}{E}(\sigma_{tt} - \nu (\sigma_{rr} + \sigma_{tt}))$$
$$2\sigma_{rr}(r) + r\sigma_{rr}'(r) - 2\sigma_{tt} = 0$$
From there we can get the following differential equation for the deformation (by a tedious but quite straightforward path):
$$2 u'(r) - 2 \frac{u(r)}{r} + r\,u''(r) = 0$$
From there, it's easy to get
$$u(r) = A\,r + \frac{B}{r^2}$$
and, as a consequence,
$$\sigma_{tt} = E \frac{A}{1 - 2\nu} + E\frac{B}{1 + \nu}\frac{1}{r^3}$$
$$\sigma_{rr} = E \frac{A}{1 - 2\nu} - 2 E\frac{B}{1 + \nu}\frac{1}{r^3}$$
where $A$ and $B$ are integration constants determined to be consistent with the boundary conditions.
Solving the original problem
The original problem has zero external pressure, internal pressure $P$, interior radius $R_i$ and exterior radius $R_o$. For continuity, we must have
$$\sigma_{rr}(R_i) = -P$$
$$\sigma_{rr}(R_o) = 0$$
Getting the constants from these boundary conditions:
$$A = \frac{P(1 - 2\nu)}{E}\frac{R_i^3}{R_o^3 - R_i^3}$$
$$B = \frac{P(1 + \nu)}{2 E}\frac{R_o^3 R_i^3}{R_o^3 - R_i^3}$$
Getting the stresses:
$$\sigma_{rr} = P\frac{R_i^3}{R_o^3 - R_i^3}\left(1 - \frac{R_o^3}{r^3}\right)$$
$$\sigma_{tt} = P\frac{R_i^3}{R_o^3 - R_i^3}\left(1 + \frac{1}{2}\frac{R_o^3}{r^3}\right)$$
The pressurizer is at a higher temperature than the reactor core by design and it does indeed contain both water and steam. Note that the Wikipedia article on PWRs says it is “partially filled with water”. The rest of its volume is occupied by steam. Boiling in the reactor core is undesirable in a PWR, so by having the pressurizer operate at a higher temperature than the core, the pressure in the whole primary coolant loop is maintained at the value where it boils at the temperature in the pressurizer. Since the pressurizer is now the hottest part of the primary coolant system, boiling in the core (and the rest of the primary coolant loop) is then avoided.
The proportion of water/steam content of the pressurizer can then adjust to coolant volume changes (e.g. caused by temperature changes in the other parts of the primary coolant system) by boiling or condensing within the limits of the pressurizer’s volume. If the pressure of the primary coolant system drops, part of the water inside the pressurizer flows out of it, equalizing the pressure and part of it evaporates, slightly lowering the temperature inside the pressurizer. The other way around applies for pressure increases.
This design is used because it achieves a very soft response of the pressure in reaction to changes in water volume and it does this to a certain extent passively and without moving mechanical parts. Consider the following cases of what will happen if a small extra volume of water enters quickly:
- If the gas volume is filled with an ideal gas and is thermally isolated, pressure and temperature rise (adiabatic compression).
- If the temperature can equalize, the pressure increase will be less (isothermal compression).
- If the gas is water vapor instead of an ideal gas, the isothermally compressed steam will be supersaturated, so a part of it will condense to water. This has higher density than steam, so the pressure increase will be even lower.
- Without gas in the system, any volume change would have to be absorbed by the compressibility of the water and the elasticity of the piping, both of which are very stiff.
The isothermal case is conceptually similar to an expansion tank. With condensation/evaporation, the response is softer and there is no need for a flexible membrane or movable piston to separate the gas from the liquid. With active temperature control, the pressure changes can be reduced further, but the soft response of the passive behaviour means that a longer reaction time of the active control can be tolerated and the pressure will still be within limits.
, but these are set up for high pressures (hundreds of psi) and small vessels (a meter or two at most). I'm looking at relatively low pressures — one atmosphere or less — but a radius of hundreds, maybe thousands of meters.
Best Answer
If you take a one meter long section of your cylinder and imagine cutting it in half, the force separating the halves is pressure*diameter*1m This is resisted by 2*wall thickness*1m of wall. So the stress in the wall is the ratio of these:$$\text{stress}=\frac {\text{pressure*diameter}}{2* \text{wall}}\\ \text{wall}=\frac {\text{pressure*diameter}}{2* \text{stress}}$$ Note that the wall thickness scales linearly with the diameter.