SolHW8ME2320 - v = 0.09533 m 3/kg(error = 3.7 b P R = 0.159 and T R = 1.12 therefore from compressibility chart Z ≈ 0.96 v = Zv ideal

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3 – 91 3 – 92 Given: Superheated water vapor @ 10 MPa and 150 K Find: v based on a) ideal gas eqn. b) compressibility chart; c) Tables Solution: R = 0.4615 KJ/(kg K); T cr = 647.1 K; P cr = 22.06 MPa; a)Ideal gas: v = RT/P = (0.4615)(723)/(3500)
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Unformatted text preview: v = 0.09533 m 3 /kg (error = 3.7 %) b) P R = 0.159 and T R = 1.12, therefore from compressibility chart Z ≈ 0.96. v = Zv ideal = (0.96)(0.09533) = 0.0916 m 3 /kg (error = 0.4%) c) Tables: v = 0.09196 m 3 /kg...
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This note was uploaded on 04/07/2008 for the course ME 232 taught by Professor Monefort during the Spring '08 term at Western Michigan.

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SolHW8ME2320 - v = 0.09533 m 3/kg(error = 3.7 b P R = 0.159 and T R = 1.12 therefore from compressibility chart Z ≈ 0.96 v = Zv ideal

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