Test_2_Sp04 - TEST II THERMO I 3/31/04 OPEN BOOK AND NOTES...

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TEST II THERMO I 3/31/04 OPEN BOOK AND NOTES 1.) (33%) Find the entropy change s 2 - s 1 from an initial state of P 1 = 101.32 kPa, T 1 = 25°C to a final state of P 2 = 400 kPa, T 2 = 500°C for: a) air (Gas constant R = 0.28700 kJ/(kg•K)) s 2 - s 1 = s o (T 2 ) - s o (T 1 ) - R ln (P 2 /P 1 ) From Table 5s for air, s o (T 2 ) = s o (500°C) = 7.6852 kJ/(kg•K) s o (T 1 ) = s o( 25°C) = 6.6999 kJ/(kg•K) s 2 - s 1 = 7.6852 - 6.6999 - 0.28700 ln (400/101.32) = 7.6852 - 6.6999 - 0.28700 x 1.3732 = 0.5912 kJ/(kg•K) b) H 2 0 (do not treat as an ideal gas.) Using the steam tables, Tables 10s or 11s show that state 1 is subcooled , state 2 is superheated . Does not act as ideal gas!! From Table 12s, s 2 = s(P=400 kPa, T = 500°C) = 8.1914 kJ/(kg•K) Entropy for a subcooled liquid is independent of pressure, so we can choose s 1 = s L (T 1 ), and from Table 10s, s 1 = s l (25°C) = 0.3670 kJ/(kg•K) Thus, s 2 - s 1 = 8.1914 - 0.3670 = 7.8244 kJ/(kg•K) (Note the large entropy change compared with air. ... this is because of the
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This test prep was uploaded on 03/19/2008 for the course ME 326 taught by Professor Schmidt during the Spring '07 term at University of Texas.

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Test_2_Sp04 - TEST II THERMO I 3/31/04 OPEN BOOK AND NOTES...

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