Lost work exchanger

Sufficient vapor is compressed to provide the necessary 65 M BTU/ hr. This requires about 2000 hp, or 5.0 M BTU/hr, of work. The lost work of the two condensers has been eliminated but two heat exchangers are needed to maintain heat balance. Some throttling must still be done. 

A simplified flow diagram for this segment of the process is shown in Figure 4, and thermodynamic analysis in Table III. The refrigerated exchangers and cold box represent about 30% of the lost work of the process. However, the tower itself has a very high percentage of the lost work in the system. Thus the details of the tower heat and material balance were examined in search of ways to improve its efficiency. 

Two streams of steam are brought into thermal contact as shown in Figure 6- b. The cooler stream enters at 150 and exits at 300 °C and has a flow rate of 1 kg/s. The hot stream enters at 500 °C and exits at 400 °C. All streams are at constant pressure of 1 bar. Determine the unknown flow rate, the amount of heat exchanged between the streams, the rate of entropy generation, and the lost work. 

This represents 8.83/45.05 = 0.196, or 19.6% of the total lost work for the cycle. This lost work results because of a frictional pressure drop of 2 psi through the heat exchanger and the rather large temperature driving force for heat transfer. 

This table shows clearly that the reactor and cooler are, by far, the largest contributors to the inefficiency of the process. Some reduction in lost work can be achieved by replacing the partial condenser with two or three heat exchangers operating with coolants at different temperature levels. But what can be done with the reactor Would it be better to operate it at a lower or higher temperature Should a larger excess of hydrogen be used Clearly, there is room for considerable improvement in the reactor operation. See Exercise 9 23. 

The value of AT) sometimes called the approach temperature, is a key design variable for a heat exchanger network. It has an impact on lost work associated with heat transfer. The net lost work Exioss for a heat flow of q between high temperature T and low temperature T2 is... 

In the example above, the electron transfer was direct, that is, the electrons were exchanged directly from the zinc metal to the cupric ions. But such a direct electron transfer doesn t allow for any useful work to be done by the electrons. Therefore, in order to use these electrons, indirect electron transfer must be done. The two half-reactions are physically separated and connected by a wire. The electrons that are lost in the oxidation half-reaction are allowed to flow through the wire to get to the reduction half-reaction. While those electrons... 

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