ME300Lecture06

# ME300Lecture06 - Lecture 6 Review of 2nd Law Part II...

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Lecture 6: Review of 2 nd Law, Part II Example 6.1: Given: -rigid tank, finite time process - final temperature T f = 400 K Reservoir 500 K CM H 2 V = 2 m 3 @ 500 K T i = 320 K p i = 180 kPa Find: - Entropy change of the hydrogen, S CM ntropy production 6.1 - Entropy production,

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Continue Example 6.1 Assumptions: - Closed system ideal gas - H 2 is ideal gas - KE = PE = 0 Solution: 6.2
Continue Example 6.1 6.3

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Continue Example 6.1 6.4
Continue Example 6.1 6.5

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Continue Example 6.1 6.6
Continue Example 6.1 6.7

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Continue Example 6.1 6.8
Continue Example 6.1 6.9

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Continue Entropy Production Entropy Production in Open Systems: j CV in in out out Q dS ms m s T    ji n out j dt 6.10
Example 6.2 Given: CV . T 1 = 1000 K p 1 = 1 bar ir T 2 = 400 K p = 1 bar Q Air 2 300 K Water ind T 3 = 300 K p 3 = 1 bar T 4 = 350 K p 4 = 1 bar Find : » mass flow rate of air » total entropy production W kg m0 .

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ME300Lecture06 - Lecture 6 Review of 2nd Law Part II...

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