ME300Lecture08

ME300Lecture08 - Lecture 8: Continue Exergy Analysis for...

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Lecture 8: Continue Exergy Analysis for Closed Systems Example 8-1 iven: nal state: igid tank Given: final state: p 2 = 1.5 bar team Rigid tank Steam m = 1 kg p = 1 bar CM nvironment @ W 1 T 1 = 100 o C Environment @ T o =20 o C, p o =1 bar Find: » Work done, W = ? hange in exergy E? 8.1 » Change in exergy, » Irreversibility, » Second law effectiveness, ε =? d
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Continue Example 8.1 Assumptions: » Closed system » Adiabatic, Q = 0 » Δ KE = Δ PE = 0 Solution: diagram T =1.5 bar » T-s diagram p 2 1.5 bar p 1 =1 bar 1 T sat (p=1bar)=99.63 o C 8.2 s
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Continue Example 8.1 8.3
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Continue Example 8.1 8.4
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Continue Example 8.1 8.5
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Continue Example 8.1 8.6
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Open Systems Exergy Analysis Definition of Useful Work for Open Systems: . . m ut CV W Surroundings @ T o ,p o . m out B A . in . Q o Reservoir T Assumptions: » Surroundings are at fixed T o and p o Q j @ T j 8.7 » System interacts only with surroundings and one or more heat reservoirs
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Continue Open Systems Exergy Analysis 1 st law, energy balance: d 22 CV o j in out ji n o u t in out dE vv Q Q W m hg z m z dt 2 2        2 nd law, entropy balance: j CV o in in out out CV Q dS Q ms m s t T T Solve entropy balance equation for Q o and then substitute energy balance equation: no u t oj dt in energy balance equation: CV CV o o j in o in out o out o CV n o u t j dE dS T TQm T s m T s T dt dt T   8.8 j in out n o u t in out Q W m h gz m h gz 
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Continue Definition of Useful Work for Open Systems
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ME300Lecture08 - Lecture 8: Continue Exergy Analysis for...

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