Assignment__6 - Assignment #6 PROBLEM 6.1 PROBLEM...

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Assignment #6 PROBLEM 6.1 PROBLEM STATEMENT: A system consisting of n=3 particles can exist in any of the three system states, A, B, or C shown below. Each system state has four energy levels, e i =0, 1, 2, and 3, and the probability p i of a level is n i /n (n i being the number of particles occupying level i.) (a) Show that the total energy is the same for all three system states. (b) By observation (no calculations), which system state appears to have the least random and which the most random distribution of particles in energy levels? (c) Calculate the entropy, , for each system state. If the system were to transition spontaneously from one state to another, what order would they proceed in and which state would be the equilibrium state? For simplicity, you may take the value of k=1. DIAGRAM DEFINING SYSTEM AND PROCESS: GIVEN: A system of 3 particles with 4 energy levels. FIND: Total energy and entropy for each system state. ASSUME: k=1
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GOVERNING RELATIONS: Total energy, and entropy, QUANTITATIVE SOLUTION: a). The total energy at each system state is given by . At State A: At State B: At State C: The total energy is the same for all three system states. b). State A has the least random distribution of particles in energy levels as the three particles are at the same level. State B has the most random distribution because all the particles are placed at different levels c). The entropy at each system state is given by . At State A: At State B: At State C: Now, system transition will occur from a high entropy to a lower entropy. Thus, the order to reach the equilibrium state would be State B à C à A. DISCUSSION OF RESULTS: This problem demonstrates that the Boltzmann relation produces numerical results for S which match our intuitive understanding of entropy. The least random distribution of particles (all in the same state) results in the lowest value for S, while the most random distribution (all particles in different states) results in the highest value. PROBLEM 6.3
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PROBLEM STATEMENT: What is the change in specific entropy of water (Btu/ lbm- ° R) when it is cooled isobarically from 200 psia and 500 ° F to a final specific volume of 2 ft 3 /lbm, and what is the work (Btu/lbm) for the process? DIAGRAM DEFINING SYSTEM AND PROCESS: P 1 = 200 psia T 1 = 500°F P 2 = 200 psia v 2 = 2 ft 3 /lbm GIVEN: H 2 O, isobaric cooling State 1: P 1 = 200 psia , T 1 = 500°F = 960°R State 2: P 2 = 200 psia , v 2 = 2 ft 3 /lbm FIND: and GOVERNING RELATIONS: Work, for isobaric process QUANTITATIVE SOLUTION: From Table 12e, the specific entropy and volume at State 1 are From Table 11e at O=200 psia it can be seen that State 2 exists as a liquid-vapor mixture since The quality at State 2 can thus be calculated Thus, the specific entropy at State 2 is The change in specific entropy is Work for the process is
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DISCUSSION OF RESULTS: Note that the negative change in entropy is possible since there was cooling of the H 2 O during the process. PROBLEM 6.6
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This note was uploaded on 09/01/2010 for the course ME 320 taught by Professor Dr.kinne during the Spring '07 term at University of Texas.

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Assignment__6 - Assignment #6 PROBLEM 6.1 PROBLEM...

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