Fluid expansion energy

Calculate tlie fluid expansion energy for an isotliennal expansion for a cylindriail vessel at 550°C witli initial and final pressures of 147 and 450 psi, respectively. 

Fluid expansion energy in an explosion, can be determined from the equation... 

The fluid expansion energy is calculated from the first equation for isothermal expansion ... 

To calculate expansion energy, multiply the specific expansion work by the mass of fluid released or else, if energy per unit volume is used, multiply by the volume... 

Although the specific expansion energy of propane is included in the list of fluids in Figure 6.30, Steps 3 to 5 are followed for this example. The solution for the filled vessel will be given first. 

Propane is a fluid for which speciflc expansion energy is given in Figure 6.30. Therefore, the calculation is continued with Step 5. 

Both propane and butane are fluids for which specific expansion energies are given in Figure 6.31. Therefore, calculations begin with Step 5. 

Flow through abrupt expansion Using the one-dimensional flow assumption for a single-phase incompressible fluid, the energy equation becomes... 

The blast wave produced by a sudden release of a fluid depends on many factors (AIChE, 1994), This includes the type of fluid released, energy it can produce on expansion, rate of energy release, shape of the vessel, type of rupture, and the presence of reflecting surfaces in the surroundings. Materials below their normal boiling point cannot BLEVE. 

The expansion of the reservoir fluids, which is a function of their volume and compressibility, act as a source of drive energy which can act to support primary producf/on from the reservoir. Primary production means using the natural energy stored in the reservoir as a drive mechanism for production. Secondary recovery would imply adding some energy to the reservoir by injecting fluids such as water or gas, to help to support the reservoir pressure as production takes place. 

Rare-gas clusters can be produced easily using supersonic expansion. They are attractive to study theoretically because the interaction potentials are relatively simple and dominated by the van der Waals interactions. The Lennard-Jones pair potential describes the stmctures of the rare-gas clusters well and predicts magic clusters with icosahedral stmctures [139, 140]. The first five icosahedral clusters occur at 13, 55, 147, 309 and 561 atoms and are observed in experiments of Ar, Kr and Xe clusters [1411. Small helium clusters are difficult to produce because of the extremely weak interactions between helium atoms. Due to the large zero-point energy, bulk helium is a quantum fluid and does not solidify under standard pressure. Large helium clusters, which are liquid-like, have been produced and studied by Toennies and coworkers [142]. Recent experiments have provided evidence of... 

In practice, the loss term AF is usually not deterrnined by detailed examination of the flow field. Instead, the momentum and mass balances are employed to determine the pressure and velocity changes these are substituted into the mechanical energy equation and AFis deterrnined by difference. Eor the sudden expansion of a turbulent fluid depicted in Eigure 21b, which deflvers no work to the surroundings, appHcation of equations 49, 60, and 68 yields... 

The fact that shock waves continue to steepen until dissipative mechanisms take over means that entropy is generated by the conversion of mechanical energy to heat, so the process is irreversible. By contrast, in a fluid, rarefactions do not usually involve significant energy dissipation, so they can be regarded as reversible, or isentropic, processes. There are circumstances, however, such as in materials with elastic-plastic response, in which plastic deformation during the release process dissipates energy in an irreversible fashion, and the expansion wave is therefore not isentropic. 

Energy here means the work which can be done by the fluid in expansion, ,wo ) This means that energy release during flashing must have been very rapid. 

The specific internal energy of the fluid in the expanded state U2 can be determined as follows If a thermodynamic graph is used, assume an isentropic expansion (entropy s is constant) to atmospheric pressure po- Therefore, follow the constant-entropy line from the initial state to Po- Read h- and V2 at this point, and calculate the specific internal energy U2-... 

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