Calculation of Potential Energy Stored in Compressed Water.

Consider a cylinder capped on one end with an end cap and on the other end with a piston (see Fig. 3). The cylinder is filled with a mass, $M$, of water which is pressurized by applying a force, $F$, to the piston. The force that must be applied is $F=pA$ where $A$ is the area of the piston and $p$ is the water pressure. The work done in compressing the water is $E = \int F dh$,

  $$E = A \int_{h_o}^{h(p)} p~dh$$ (9)

where $h(p)$ and $h_o$ are the piston positions at pressure $p$ and at atmospheric pressure, respectively. This quantity of energy is stored in the water as potential energy and represents the maximum that might hypothetically be converted to kinetic energy during vessel failure.

Figure 3: Schematic used for calculation of the potential energy stored in the compressed water.

The fact that the mass of water is fixed can be used to determine the relationship between the piston position and the water pressure. Note that a fixed mass implies that

  $$\rho\left(p\right) V\left(p\right) = \rho_o V_o$$ (10)

where $\rho\left(p\right), V\left(p\right)$ are the water density and volume of the water at pressure, $p$. The quantities $V_o, \rho_o$ are measured at atmospheric pressure. The dependence of water density on pressure can be expressed as

  $$\rho\left(p\right) = \rho_o \left(1 + \gamma p \right)$$ (11)

where $\gamma \equiv \frac{1}{\rho}\frac{\partial \rho}{\partial p}$ is the compressibility of water. Table 3 demonstrates that the compressibility is nearly constant over the range of pressures relevant to the PTV.

Table 3: The compressibility of water at a temperature of 20$^\circ$C. The compressibility varies by less than 20% over the range of pressures and temperatures that are applicable to the PTV.

Pressure (psi)

Compressibility $(psi)^{-1}$
0	$3.17 \cdot 10^{-6}$
2500	$3.03 \cdot 10^{-6}$
5000	$2.90 \cdot 10^{-6}$
7500	$2.78 \cdot 10^{-6}$
10000	$2.67 \cdot 10^{-6}$

The density can also be expressed as $\rho =\frac{M}{A h}$ where $A$ is the cross-sectional-area of the vessel. Substituting this into (11) yields the following equivalent relationships between the pressure and the piston position,

  $$p = \gamma^{-1}\left(\frac{h_o}{h} - 1 \right)~~,~~~ h = \frac{h_o}{1+\gamma p}~~.$$ (12)

Substituting (12) into (9) yields

  $$E(p) = \frac{A}{\gamma} \int_{h_o}^{\frac{h_o}{1+\gamma p}} \left(\frac{h_o}{h} - 1\right)~dh~.$$ (13)

Evaluating this integral gives

  $$E(p) = \frac{V_o}{\gamma}\left[ \frac{\gamma p}{1 + \gamma p} - \ln\left ( 1 + \gamma p \right) \right]$$ (14)

Equation (14) is exact but unnecessarily complicated. Note that even at a maximum pressure of 10 kpsi then $\gamma p \doteq 0.03$ and therefore $\gamma p \ll 1$. Computing the Taylor series of (14) and neglecting third and higher order terms yields a greatly simplified expression with an error of approximately only 3%:

  $$E(p) = \frac{\gamma}{2} V_o p^2~.$$ (15)