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Therefore the process is impossible. 20 Steam 20 bar = 2 MPa and 300° C H$ = 30235 . 6 kJ kg Final pressure = 1 bar. For reference saturation conditions are P = 01 . 46 S$ L = 13026 . 36 V V H$ = 26755 . 3594 U$ = 25061 . : =M 1 1 2 2 2 1 dt From Steam Tables T = 250° C ⇒ H2 = 30235 . 2158 kJ/kg K P = 01 . : S$ = 81245 . kJ kg K dS & $ + M& S$ + S& = 0 = MS 1 2 2 gen dt S&gen . : M& 1S$1 + M 2 2 2 1 2 ⇒ Two-phase mixture. Solve for fraction of liquid using entropy balance. 3594) + (1 − x )⋅ 13026 .

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I 42315 =F = 089805 . H 47315 M . 9628 Given the approximations, the two results are in quite good agreement. For what follows, the energy balance result will be used. Therefore, the mass of water finally present (per m3) is × M L (initial) = 15506 M L (final) = 0897 . kg L L o $ occupying V = M ( final) × V 150 C = 15506 m3 . 001091 = 01692 . 8308 m3 = = 2115 . 1692. 0135 of the mass in the tank. 18 kg. 3% has been withdrawn. 96 kg being withdrawn. This is in excellent agreement with the more rigorous finite difference calculations done above.

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