Written Assignment 7 Solutions_updated Nov 3.pdf

Note the process is not externally reversible since

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reversible. (Note: The process is not externally reversible, since temperature is transferred across a finite temperature difference). In this case the following approach may be applied to find the entropy change in the system : 2 1 2 1 2 1 gen S T Q S (1) Because the process is internally reversible, so in the system S gen = 0. The equation (1) can be simplified to: T Q S 2 1 2 1 (2) In equation (2), T denotes the temperature of the system, since we are just considering the entropy change in the system and not the total entropy generated throughout the process (in which case we would use the temperature of the surroundings). (See example 7-21 in the textbook for more information). ΔS 1-2 = m·(s 2 - s 1 ) (3) The relevant properties can be found on table A-4. s 2 = s g at 120 o C = 7.1292 kJ/kg·K s 1 = s f at 120 o C + x 1 ·s fg at 120 o C = 1.5279 + 0.65 × 5.6013 kJ/kg·K = 5.168745 kJ/kg·K Put the values of m , s 2 and s 1 into the equation (3): ΔS 1-2 = 8kg × (7.1292 5.168745)kJ/kg·K = 15.68364 kJ/K
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