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Unformatted text preview: Spring 2006 1 ME 201 Thermodynamics Homework #5 Solutions 1. (5 pts) Calculate the entropy change for N 2 as it goes from 250 K and 1000 kPa to 1300 K and 60 kPa. Solution: Substance Type: Ideal Gas (N 2 ) Problem Type: Process State 1 State2 T 1 = 250 K T 2 = 1300 K P 1 = 1000 kPa P 2 = 60 kPa Using our air tables the entropy change will be given by Δ s =   R ln P P 2 1 2 1 φ φ ⋅ & ¡ ¢ £ ¤ ¥ From the N 2 tables we have φ 2 = 236.831/ 28.013 = 8.454 kJ / kg K ⋅ φ 1 = 186.370 / 28.013 = 6.653 kJ / kg K ⋅ The gas constant for N 2 is R = 0.2968 kJ/kg ⋅ K, so that upon substituting we get Δ s = 8.454  6.653  0.2968)ln 60 1000 = 2.636 kJ / kg K ( & ¡ ¢ £ ¤ ¥ ⋅ . ME 201 Thermodynamics Spring 2006 2 2. (20 pts) For the two processes given below, determine the final temperature, pressure, specific volume, and the changes in internal energy, enthalpy, and entropy. a. Air at 8400 R and 66 psia goes isothermally to 1.7 psia. Solution: Substance Type: Ideal Gas (Air) Problem Type: Process (Isothermal) State 1 State2 T 1 = 8400 R T 2 = 8400 R P 1 = 66 psia P 2 = 1.7 psia Bold values are calculated We note that to fix state 2 we need another property, but since the process is...
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This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.
 Spring '06
 SOMERTON

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