2014 Case Day 2 Morning 9
Thread Content
Ethylene gas with a flow rate of 887 Nm3/h, a pressure of 1135 kPa, and a temperature of 40°C expands in a turbine to a pressure of 1054 kPa. Assuming an efficiency of 0.75, the enthalpy change calculated based on an adiabatic reversible expansion process is 0.0123 KJ/kg. What is the value of the isentropic expansion work (W) for this process? (A) 5.05 (B) 2.84 (C) 53885 (D) 16584.68 Efficiency needs to be taken into account when calculating the actual work done, but the actual process shouldn’t be isentropic, right? One case of isentropy is adiabatic reversible processes; in such reversible processes, the most work can be done, and the efficiency should be 100%. Therefore, the term “isentropic” should not appear in the question. What is wrong with my understanding?
I think option B is the right choice. First of all, V0=887 Nm3/h corresponds to P0=0.101 MPa； T0=273K; R=0.0083 (MPa Nm3 kmol K-1) are the values under standard conditions. The molar flow rate per hour can be calculated, and then converted into a mass flow rate. Molar flow rate: n = (P0 × V0) / (R × T0) = 39.5 kmol/h. Molecular weight: M = 28 kg/kmol. Mass flow rate: m = M × n / 3600 = 0.307 kg/s. In an adiabatic process, the increase in entropy is 0. Work done: W = 0.75; specific enthalpy: H = 9.225 × 10^-3 kJ/kg. Therefore, power: N = m × W = 2.8 W. In this problem, since V0 = 887 Nm3/h, there is no need to calculate pressure changes or anything similar. Nm3 refers to the volume of a gas at 0 degrees Celsius and 1 standard atmosphere ； N stands for Normal Condition, that is, the conditions of air are: one standard atmosphere, a temperature of 0°C, and a relative humidity of 0%.
Isentropic means that the entropy of the material entering and leaving remains unchanged; it does not imply reversibility. Do not confuse the entropy generated during a process with isentropy