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Unformatted text preview: . 8 kJ 7. m = P 1 V 1 /RT 1 , and V 2 = mRT 2 /P 2 . Since P varies linearly with V , W 12 = Z V 2 V 1 P dV = 1 2 ( P 1 + P 2 ) ( V 2V 1 ) = 37 . 5kJ and Q 12 = m ( u 2u 1 ) + W 12 = 304 . 9 kJ (tables). 8. The process is constant volume. At state 2: v 2 = v g (250 C), and m = V/v 2. v 1 = v 2 and P 1 = 100 kPa x state 1. x 1 = 0 . 029, Q 12 = 4240 . 3 kJ. 9. Same as the given info in the previous problem. 10. For water the initial state is a saturated mixture. A constanttemperature compression will therefore correspond to a constantpressure compression, until the water becomes a pure liquid phase. For air the work can be calculated from the ideal gas law. Water: W 12 =26 . 25 kJ, Q 12 =336 . 4 kJ. O 2 : W 12 = Q 12 =50 . 76 kJ....
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This note was uploaded on 09/24/2011 for the course ENG 2010 taught by Professor Staff during the Fall '11 term at Auburn University.
 Fall '11
 Staff
 Dynamics, Ode

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