Written Assignment 7 Solutions_updated Nov 3.pdf

# State 3 saturated mixture at 4581 o c and 10 kpa to

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State 3: saturated mixture at 45.81 o C and 10 kPa. To determine u 3 , the quality (x 3 ) must first be determined from s 3 . Since the process from state 2 to state 3 is reversible and adiabatic, s 3 = s 2 = s = 7.1292 kJ/kg∙K. The relevant properties can be found on table A-5. 8640460825 . 0 4996 . 7 6492 . 0 1292 . 7 3 3 fg f s s s x u 3 = u f at 10 kPa + x 3 ·u fg at 10 kPa = 191.79 + 0.8640460825 × 2245.4 kJ/kg = 2131.919074 kJ/kg The values of m, u 3 and u 2 can then be substituted into equation (8). kJ kg kJ kg W 847412 . 3175 / ) 9 . 2528 919074 . 2131 ( 8 3 2 W 2-3 = 3175.85 kJ (d) The total entropy generated can be expressed as: S gen_1-3 = S gen1-2 + S gen2-3

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CHE 243: Engineering Thermodynamics 5 The process from state 2 to 3 is reversible and adiabatic, therefore S gen2-3 = 0. S gen_1-3 = S gen1-2 = ΔS 1-2 – Q 1-2 / T surr (9) Q 1-2 was determined above to be: 6166.075835 kJ. T surr = 200 o C = 473.15 K ΔS 1-2 = m(s 2 -s 1 ) = 8kg ( 7.1292 5.168745 ) kJ/kg·K = 15.68364 kJ/K The values of ΔS 1-2 , Q 1-2 and T surr can then be substituted into equation (9). S gen_1-3 = 15.68364kJ/K 6166.075835kJ / 473.15K = 2.651671628 kJ/K
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