Water mass for pressurised vessel

Data: $V_0=0.5\ \text{m}^3$, test pressure $\Delta p=3000\ \text{bar}=300\ \text{MPa}$, $K=2000$ MPa, the empty vessel expands by 0.6 %.

Step 1: volume of the vessel at 3000 bar (expanded by 0.6 %):

$$V_v=0.5\times1.006=0.503\ \text{m}^3$$

Step 2: compression of water. From $K=-\dfrac{\Delta p}{\Delta V/V}$:

$$\frac{\Delta V}{V}=\frac{\Delta p}{K}=\frac{300}{2000}=0.15$$

So the water occupies 85 % of its original (atmospheric) volume at 3000 bar. The water must fill the vessel (0.503 m³) at that pressure:

$$V_w(1-0.15)=0.503\ \Rightarrow\ V_w=\frac{0.503}{0.85}=0.5918\ \text{m}^3$$

Step 3: mass of water (density of water at atmospheric pressure 1000 kg/m³):

$$m=\rho V_w=1000\times0.5918=591.8\ \text{kg}$$

Answer: about 592 kg of water must be pumped in (about 92 kg more than the 500 kg that would fill the vessel at atmospheric pressure).