final_sum06_s

final_sum06_s - TDEC202 Energy II(Solutions to Final Exam...

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Unformatted text preview: TDEC202 - Energy II (Solutions to Final Exam) Summer Term, 2006 M. Carchidi –––––––––––––––––––––––––––––––––––– Solution to Problem 1 a.) The entropy change per mass of the surroundings (which is f xed at 25 ◦ C) is ∆ s surroundings = q T surroundings = +20 25 + 273 = +0 . 0671141 kJ/(kg · K). b.) We start with a saturated liquid of R-134a at P 1 = 900 kPa. From Table A-12, we see that h 1 = h f @900 kPa = 101 . 61 kJ/kg. and s 1 = s f @900 kPa = 0 . 37377 kJ/(kg · K). From energy F ow we must have h 1 = q + h 2 , so that h 2 = h 1 − q = 101 . 61 − 20 = 81 . 61 kJ/kg which is a saturated mixture, since 38 . 43 kJ/kg = h f @200 kPa < h 2 < h g @200 kPa = 244 . 46 kJ/kg. The quality of the refrigerant when it leaves the throttle is x 2 = h 2 − h f @200 kPa h g @200 kPa − h f @200 kPa = 81 . 61 − 38 . 43 244 . 46 − 38 . 43 = 0 . 210 , which means that s 2 = s f @200 kPa + x 2 ( s g @200 kPa − s f @200 kPa ) = 0 . 15457 + (0 . 210)(0 . 93773 − . 15457) = 0 . 319034 kJ/(kg · K). The entropy change per mass of the refrigerant is then ∆ s refrigerant = s 2 − s 1 = 0 . 319034 − . 37377 = − . 054736 kJ/(kg · K). c.) The entropy generated per mass is s generated = ∆ s surroundings + ∆ s refrigerant = (0 . 0671141) + ( − . 054736) = +0 . 0123...
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This note was uploaded on 04/20/2008 for the course TDEC 202 taught by Professor Michaelcarchidi during the Spring '07 term at Drexel.

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final_sum06_s - TDEC202 Energy II(Solutions to Final Exam...

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