Thermodynamics filled in class notes_Part_7

Thermodynamics filled in class notes_Part_7 - 2.1. CARNOT...

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Unformatted text preview: 2.1. CARNOT 21 T ( K ) P ( kPa ) v ( m 3 kg ) 1 400 1 . 148 10 2 1 . 000 2 1200 5 . 36886 10 3 . 06415 3 1200 1 . 78955 10 3 . 19245 4 400 3 . 82667 10 2 3 . 000 Table 2.1: State properties for Carnot cycle. one gets k = R + c v c v , (2.33) kc v = R + c v , (2.34) c v ( k 1) = R, (2.35) c v = R k 1 = . 287 kJ kg K 1 . 4 1 = 0 . 7175 kJ kg K . (2.36) Recall the first law: u 2 u 1 = 1 q 2 1 w 2 . (2.37) Recall also the caloric equation of state for a CPIG: u 2 u 1 = c v ( T 2 T 1 ) . (2.38) Now process 1 2 is isentropic, so it is also adiabatic, hence 1 q 2 = 0, so one has u 2 u 1 = 1 q 2 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright =0 1 w 2 , (2.39) c v ( T 2 T 1 ) = 1 w 2 , (2.40) parenleftbigg . 7175 kJ kg K parenrightbigg (1200 K 400 K ) = 1 w 2 , (2.41) 1 w 2 = 5 . 7400 10 2 kJ kg . (2.42) The work is negative as work is being done on the system in the compression process....
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

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Thermodynamics filled in class notes_Part_7 - 2.1. CARNOT...

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