Cylindrical shell: volume increase, 6 N/mm2 pressure

The extra water pumped in must fill two things: the increase in the volume of the shell (it expands) and the reduction in the volume of the water already inside (water is compressed).

Data: $d = 300$ mm, $L = 4000$ mm, $t = 12$ mm, $p = 6\ \text{N/mm}^2$, $E = 2.1\times10^5\ \text{N/mm}^2$, $\nu = 0.3$, $K = 2100\ \text{N/mm}^2$.

Increase in shell volume

$$\sigma_c = \frac{pd}{2t} = 75.0\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 37.50\ \text{N/mm}^2$$ $$\varepsilon_v = \frac{pd}{4tE}(5-4\nu) = \frac{6(300)}{4(12)(2.1\times10^5)}(5-1.2) = 6.7857\times10^{-4}$$ $$V = \frac{\pi}{4}(300)^2(4000) = 2.8274\times10^{8}\ \text{mm}^3, \qquad \delta V_{shell} = 1.9186\times10^{5}\ \text{mm}^3$$

Compression of the water

$$K = \frac{p}{\delta V_w/V} \Rightarrow \delta V_w = \frac{pV}{K} = \frac{6\times2.8274\times10^{8}}{2100} = 8.0784\times10^{5}\ \text{mm}^3$$

Additional water required

$$\delta V = \delta V_{shell} + \delta V_w = 1.9186\times10^{5} + 8.0784\times10^{5} = 9.9970\times10^{5}\ \text{mm}^3$$

Answer: additional volume of water $\approx 999.7\ \text{cm}^3$ (of which 191.9 cm³ is the expansion of the shell).