Unformatted text preview: ChE 222: Entropy Balance for Thermodynamics Problems We start with the second law of thermodynamics, in equation form: (1) Modifying this equation to be in differential form, rather than integral form (above) produces: (2) Now, recall from lecture we defined the following equation for the change in entropy of the surroundings (dSsurr/dt): (3) Recall here that the subscript "out" refers to streams leaving the system (not the surroundings) and that the subscript "in" refers to streams entering the system (not the surroundings). Inserting equation (3) into equation (2) yields our overall entropy balance for a given system: Now, let's apply this to problem 3.17 in Koretsky. Steam at 8 MPa and 500C flows through a throttling device, where it exits at 100 kPa. Determine the entropy change for this process. System: The throttling valve Stream: 1 8 MPa 500C Throttling Valve Stream: 2 100 kPa T2 = ?C (4) Mole balance: Here dNsys/dt = 0 because the system is at steadystate, so the mole balance becomes which results in Now, we write an energy balance: Here, dUsys/dt = 0 because the system is at steadystate, = 0 because the system boundaries don't deform and there's no shaft (mechanical) work, and = 0 because the throttling process takes place too fast for there to be any heat transfer to/from the surroundings. This reduces the energy balance to: which can be rewritten as or directly as This result highlights that the process is isenthalpic, meaning the enthalpy is constant between the inlet and outlet streams for the system. From the steam tables (via the NIST website, http://webbook.nist.gov/chemistry/fluid) we find that at T1 = 500C and P1 = 8 MPa, = 3399.5 kJ/kg and = 6.7266 J/g/K. This means that By iteratively searching for an enthalpy of 3399.5 kJ/kg at a pressure of 100 kPa = 0.1 MPa, we find the outlet temperature of T2 = 458C and = 8.7177 J/g/K. To find the change in entropy for this process, we write an entropy balance for the system: Here dSsys/dt = 0 because the system is at steadystate, and = 0 because the throttling process takes place too fast for there to be any heat transfer to/from the surroundings. This reduces the entropy balance to: Or, using the mole balance result ( for this process, = 8.7177 6.7266 = 1.9911 J/g/K. This is the change in entropy for the process. As a sanity check, this is saying that the entropy of the system increases during this process, which is allowed by the second law of thermodynamics. Also, this makes sense because the throttling process happens quickly, and quick processes are usually irreversible, and irreversible processes generate entropy. ): ...
View
Full Document
 Spring '08
 Benjamin
 Thermodynamics, Entropy, Heat, Entropy balance

Click to edit the document details