Thermodynamics Quiz 7

Thermodynamics Quiz 7 - Δ E = 1 Q 2 − 1 W 2 The work is...

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650:351 Thermodynamics Doyle Knight Department of Mechanical and Aerospace Engineering Rutgers - The State University of New Jersey New Brunswick, NJ USA Tel: 732 445 4464 · Email: [email protected] OFce hours: Tuesday and ±riday, 9:00 am to 11:30 am and by appointment

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Quiz No. 7 A piston-cylinder confguration has 3 kg oF ammonia at 50 kPa and -20 C. It is then heated to 60 C at constant pressure through the bottom oF the cylinder From external hot gas at 250 C. ±ind the heat transFer to the ammonia and total entropy generation. Solution ±rom the specifed pressure and temperature For State 1, the remaining state variables can be Found From Table B.2.2, s 1 = 6 . 3187 kJ/kg-K ,u 1 = 1312 kJ/kg ,v 1 = 2 . 4463 m 3 /kg Similarly, the remaining state variables can be obtained For State 2, s 2 = 6 . 9038 kJ/kg-K ,u 2 = 1443 . 0 kJ/kg ,v 2 = 3 . 2421 m 3 /kg The ±irst Law oF Thermodynamics applied to the ammonia in the piston is
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Unformatted text preview: Δ E = 1 Q 2 − 1 W 2 The work is 1 W 2 = i pdv = p i dv = p ( V 2 − V 1 ) = pm ( v 2 − v 1 ) = (50)(3)(3 . 2421 − 2 . 4463) = 119 . 4 kJ 650:351 Thermodynamics Page 1 Quiz No. 7 since the pressure is constant. The change in energy in the ammonia is Δ E = m ( u 2 − u 1 ) = (3)(1443 . − 1312 . 3) = 392 . 1 kJ Thus, 1 Q 2 = Δ E + 1 W 2 = 119 . 4 + 392 . 1 = 511 . 5 kJ The total entropy generation is Δ S net = Δ S cv + Δ S surr where Δ S cv = m ( s 2 − s 1 ) = (3)(6 . 9038 − 6 . 3187) = 1 . 755 kJ/K and Δ S surr = − Q cv T o + s m e s e − s m i s i = − 511 . 5 (250 + 273 . 15) = − . 978 kJ/K Thus, Δ S net = 1 . 755 − . 978 = 0 . 777 kJ/K 650:351 Thermodynamics Page 2...
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This test prep was uploaded on 04/07/2008 for the course MECHANICAL 351 taught by Professor Knight during the Spring '08 term at Rutgers.

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Thermodynamics Quiz 7 - Δ E = 1 Q 2 − 1 W 2 The work is...

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