Elementary Principles 264 - 7.38 (cont'd) Energy balance: H...

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7-19 7.38 (cont’d) ΔΔ Δ ± , ± , ± , ± ± ± ² ± ² ± ² ± ² ± ² ± ² ± (. ) ( ± )( . ) ±± EEQ W ll vv l l pk s Hm H m H m H mH mH mH m m mm vl in in in in Energy balance: kJ / kg = =⇒ + = ⇒= + = + + 0 00 3149 417 5 1 26754 There is no value of ± m l between 0 and 1 that would satisfy this equation. (For any value in this range, the right-hand side would be between 417.5 and 2675.4). The two-phase assumption is therefore incorrect; the effluent must be pure vapor. (c) Energy balance = = ± ² ± ² ² () H T out out in in in out out out 3149 kJ / kg = bar, T 337 C Table B.7 1 1 D (This answer is only approximate, since Δ ± E k is not zero in this process). 7.39 Basis: 40 lb min circulation m (a) Expansion valve R = Refrigerant 12 40 lb m R( l )/min H ² = 27.8 Btu/lb m 93.3 psig, 86°F 40 1 77 8 9 6 lb min lb R lb lb R( / lb Btu / lb m / / ) ² ., ² . xv xl HH v v == Energy balance: neglect out in Δ ± , ± , ± , ± ² ± ² EWQ E H
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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