Thermodynamics SECOND LAW OF THERMODYNAMICSFigure 22 Real Process Cycle Compared to Carnot CycleSolution:h= 1 - (T_{c} / T_{h})h= 1 - (60 + 460)/(340 + 460)= 1 - 520/800= 35% as compared to 18.0% actual efficiency.An open system analysis was performed using the First Law of Thermodynamics in the previouschapter. The second law problems are treated in much the same manner; that is, an isolated,closed, or open system is used in the analysis depending upon the types of energy that cross theboundary. As with the first law, the open system analysis using the second law equations is themore general case, with the closed and isolated systems being "special" cases of the open system.The solution to second law problems is very similar to the approach used in the first law analysis.Figure 23 illustrates the control volume from the viewpoint of the second law. In this diagram,the fluid moves through the control volume from section in to section out while work is deliveredexternal to the control volume. We assume that the boundary of the control volume is at someenvironmental temperature and that all of the heat transfer (Q) occurs at this boundary. We havealready noted that entropy is a property, so it may be transported with the flow of the fluid intoand out of the control volume, just like enthalpy or internal energy. The entropy flow into thecontrol volume resulting from mass transport is, therefore,_{in}s_{in}, and the entropy flow out of themcontrol volume is_{out}s_{out}, assuming that the properties are uniform atmRev. 0 Page 75 HT-01