Spherical vessel 1.9 m, volume increase 400 cm3

Data: $d = 1900$ mm, $t = 6$ mm, $\delta V = 400\ \text{cm}^3 = 4\times10^5\ \text{mm}^3$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.25$.

For a thin sphere, $\sigma = \dfrac{pd}{4t}$ in all directions, the linear strain is $\varepsilon = \dfrac{\sigma}{E}(1-\nu)$ and

$$\frac{\delta V}{V} = 3\varepsilon = \frac{3pd}{4tE}(1-\nu)$$

Calculation

$$V = \frac{\pi d^3}{6} = \frac{\pi(1900)^3}{6} = 3.5914\times10^{9}\ \text{mm}^3$$ $$\frac{4\times10^5}{3.5914\times10^{9}} = 1.1138\times10^{-4} = \frac{3p(1900)(0.75)}{4(6)(2\times10^5)} = 8.9063\times10^{-4}\,p$$ $$p = 0.1251\ \text{N/mm}^2$$

Answer: $p \approx 0.125\ \text{N/mm}^2$.