# Identification of the Wear Margin of a Pipeline-Machine Subsystem
## 4 The Thermodynamic Model of an Open Pipeline-Machine Subsystem
### Identification of a Hydrophore Filling Subsystem  
A hydrophore is a tight pressure vessel of volume \(V_{h}\), which, before filling, contains in its lower part a liquid of volume \(V_{1}\) and gas (mostly air) of pressure \(p_{1h}\) above the liquid. After filling, the volume occupied by the liquid increases to the value of \(V_{2}\) and gas pressure increases to \(p_{2h}\) (Figure 7). Knowing the filling time, we can define the value of what we call herein vessel filling capacity, \(\dot{V}_{zh}\), of a hydrophore as  
\[\dot{V}_{zh}\ =\ \frac{V_{2}-V_{1}}{\Delta t_{N}} \tag{54}\]  
where \(V_{2}-V_{1}\) represents the loss of compressed air volume or the increase in water volume and \(\Delta t_{N}\) represents filling time. For cost-effectiveness, the hydrophore should be as small as possible but have such functional volume that the actual work cycle of the device is not shorter than the limit value corresponding to the allowable frequency of pump switching [45]. The gas being filled into a hydrophore becomes compressed, so its volume decreases from \(V_{2}-V_{1}\) to \(V_{2}-V_{2}\). Depending on the conditions, compression can be isothermal, polytropic or adiabatic. The volume \(V_{2}-V_{2}\) of gas compressed to the same pressure \(p_{2}\) depends on the type of compression. The work required for compression is called the external compression work \(L_{1-2}\). The value of exponent \(n\) of the compression polytropic curve for a specific system can be determined experimentally by using known thermodynamic relations [46] between the quantities of the polytropic process of perfect gases. \[\frac{T_{2}}{T_{1}}\ =\ \left(\frac{p_{2h}}{p_{1h}}\right)^{\frac{n-1}{n}}\ =\ \left(\frac{V_{h}-V_{1}}{V_{h}-V_{2}}\right)^{n-1} \tag{55}\]  
After finding the logarithms of last two terms on both sides,  
\[n\ =\ \frac{\log\left(\frac{p_{2h}}{p_{1h}}\right)}{\log\left(\frac{V_{h}-V_{ 1}}{V_{h}-V_{2}}\right)} \tag{56}\]  
When the calculated exponent of the polytropic curve assumes a value of \(n>1\), external compression work can be determined from the following equation [46]:  
\[L_{1-2}\ =\ \frac{p_{1h}\cdot(V_{h}-V_{1})-p_{2h}\cdot(V_{h}-V_{2})}{n-1} \tag{57}\]  
When \(n\approx 1\) we can assume that the compression is isothermal. External compression work of isothermal process [46] is given as follows:  
\[L_{1-2}\ =\ p_{1h}\cdot V_{1}\cdot\ln\frac{(V_{h}-V_{2})}{(V_{h}-V_{1})}\ =\ p_{2h}\cdot(V_{h}-V_{2})\cdot\ln\frac{p_{1h}}{p_{2h}} \tag{58}\]  
Figure 7: (**a**) Liquid volumes and gas pressures in the hydrophore before and after filling the hydrophore. (**b**) Air pressure in the hydrophore, where \(V_{1}\) is the initial volume of the liquid, \(V_{2}\) is the final volume of the liquid, \(V_{h}\) is the total volume of the hydrophore, \(p_{1h}\) is the initial gas pressure, \(p_{2h}\) is the final gas pressure and \(L_{1-2}\) is the work performed during gas compression (hatched area). In the hydrophore, we can assume that the energy required to compress gas is equal to external work \(L_{1-2}\) of the pump and that the liquid of volume \((V_{2}-V_{1})\) obtains energy equal to external work \(L_{1-2}\). The external work of the pump, neglecting the height above reference plane, is equal to the pressure energy \(p_{c}\) of the liquid with volume \((V_{2}-V_{1})\) The pressure \(p_{c}\) is a mean pressure acting on the liquid during the filling of the hydrophore vessel. \[L_{1-2}\ =\ m\frac{p_{c}}{\rho}\ =\ (V_{2}-V_{1})\cdot p_{c} \tag{59}\]  
\[p_{c}\ =\ \frac{L_{1-2}}{(V_{2}-V_{1})} \tag{60}\]  
Efficiency \(\eta_{eln}\) of hydrophore filling can be determined as the quotient of the compression work and the electric energy required to perform the compression work:  
\[\eta_{eln}\ =\ \frac{L_{1-2}}{E_{el}} \tag{61}\]  
Pressure \(p_{1}\) in the hydrophore can be changed by supplying compressed air to the hydrophore. An increase in pressure \(p_{1}\) at the same volume change \(\Delta V\) will result in greater compression work and increased pressure energy.