The integral of vdP is essential for calculating work in flow machines, particularly in the context of a reversible compressor undergoing a polytropic process. This approach differs from the traditional PdV expression used in cylinder-piston systems. The work done by the compressor can be derived from the first law of thermodynamics, specifically for steady-state conditions, where the differential work is expressed as dW_s = vdP. This formulation is crucial for understanding energy balances in flow systems.
PREREQUISITESEngineering students, mechanical engineers, and professionals working with flow machines and thermodynamic systems will benefit from this discussion, particularly those focusing on energy efficiency and work calculations in compressors.
Chestermiller said:Is your system a flow system, such that fluid is flowing into the system and out of the system through inlets and outlets to a control volume? An example would be a nozzle.
Thank you very much, now I understand the concept, I'm working on how to do the balance in such a way that I can demonstrate it in this case.robphy said:
For a differential section of the compressor operating at steady state, the open system version of the first law of thermodynamics tells us the $$dW_s=dh=Tds-vdP$$ where h is the enthalpy per unit mass of gas, s is the entropy per unit mass, v is the volume per unit mass, and ##W_s## is the shaft work. For adiabatic reversible operation, ds = 0, so $$dW_s=vdP$$Emmanuel S said:Indeed. In fact, the problem was with a reversible compressor; the process was polytropic, and after performing the mass and energy balance, when calculating the compressor's specific work, the professor established w = ∫ vdP
And he said that for flow machines, the work took that expression, not PdV.