ME300Lecture10

# ME300Lecture10 - Lecture 10 Example 10.1 Given R-22 T1 =...

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Lecture 10: Example 10.1 Given: Surroundings T 310 K p 1 atm -22 kg m1 . 0 . R-22 T 1 = 80 o C p = 2 MPa p 2 = p 1 Sat. liquid @ T o =310 K, p o =1 atm R22 s Q 1 ir = 47 o Find : ass ow te f ir CV Air T 3 = 37 o C p 3 = 1 atm T 4 47 C p 4 = p 3 » Mass flow rate of air » Change in stream exergy of R-22 and air » Total irreversibility 10.1 » Second law effectiveness

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Continue Example 10.1 Assumptions : T » KE = PE = 0 » adiabatic outer shell » SSSF p 1 = 2 MPa T » air is ideal gas asic quations 1 2 T sat (p 1 ) 1 Basic equations: (…) Solution: - 2 properties: s » R 22 properties: 10.2
Continue Example 10.1 10.3

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Continue Example 10.1 10.6
Review Exergy Analysis Closed System: » The exergy of a closed system is the maximum useful work that could be done as the system goes from the initial state to the dead state.      00 0 EU U p V V T S S K E P E  0 use 1 2 j 0 tot j j T WE E 1 Q T T    

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ME300Lecture10 - Lecture 10 Example 10.1 Given R-22 T1 =...

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