Calculating the leakage of water in a pressurized vessel based on initial and final pressures?

Question

Let's say you have a piping system containing liquid water under high pressure. One section of the pipe of ~70L gets isolated, and it is initially at the same pressure as when it was isolated, which is 15 MPa. We can think of this isolated portion of piping as a pressurized vessel of liquid. Assuming no change in temperature, you notice that the pressure is slowly dropping and it's currently at 7 MPa, suggesting a passing isolating valve that's at a lower pressure on the other side, or a leak in a fitting to the environment.

Would it be correct to use the difference in final and initial pressure values to calculate the initial and final densities of the water (assuming we know the temperature and it's constant), and then use the volume of isolated piping to determine the initial and final mass to calculate how much water was lost? Is this the right approach?

Answer 1

Your approach could work, but water is not very compressible, so you may need to include pipe flexibility into that. For simplicity, the pipe can be considered as thin cylindrical shell, for which axial ($\epsilon_a$) and radial ($\epsilon_r$) strain from internal pressure $P$ are ($D_m$ mean diameter, $e_n$ thickness, $E$ Young's modulus, $\nu$ Poisson's ratio):

$$\epsilon_a(P) = \frac{P\cdot D_m}{4\cdot e_n\cdot E}\cdot (1-2\nu)$$ $$\epsilon_r(P) = \frac{P\cdot D_m}{4\cdot e_n\cdot E}\cdot (2-\nu)$$

So the initial volume $V_0 = \pi\cdot r_i^2\cdot L$ increases to $V(P)$ due to the pressure:

$$V(P) = \pi\cdot \left(r_i\cdot (1+\epsilon_r(P))\right)^2\cdot L\cdot (1+\epsilon_a(P))$$

Temperature can have strong influence, so make sure it is really constant or you may have to include thermal expansions of water and pipe material.

Answer 2

Late to the party, but I think you would need an equation of state for the water (density as a function of pressure). For a gas this is the ideal gas law, but for water there is an extremely wide range of pressures corresponding to a a small range of density.

I doubt this is possible as a practical measurement.

Answer 3

Oh,it's interesting.Assuming that temperature is 20°C and the Volume 30L. If the pressure drops from 15MPa to 7MPa, the density will change from 1005.0251kg/m³ to 1001.4019kg/m³. So the mass loss of water is 108.696g. As the first two guys mentioned,this is just a very rough estimate, but the process is very simple. I'm not sure if you have done any experiments to verify this result. I will conduct a simple suppression experiment to verify it, hoping to bring some inspiration.