7-207.39 (cont’d) (c)We may analyze the overall process in several ways, each of which leads to the same result. Let us first note that the net rate of heat input to the system is ±±±QQQ=−=−=−evaporatorcondenserBtu min20002500500and the compressor work Wcrepresents the total work done on the system. The system is closed (no mass flow in or out). Consider a time interval Δtminbg. Since the system is at steady state, the changes ΔU, ΔEkand ΔEpover this time interval all equal zero. The total heat input is ±QtΔ, the work input is ±WtcΔ, and (Eq. 8.3-4) yields ±±...Qt W tW QccΔΔ−=⇒==−××=−−05001341 109 486 1011834Btu 1 minhpmin60 sBtu shp 7.40Basis: Given feed rates ±.n1080(mol / h)0.2 C HCHC, 1.1 atm38410o±±nn3 84 10o0(mol C H / h)(mol C H/ h)227 C±Q(kJ / h)±n20o(mol / h)0.40 C H.60 C H25 C, 1.1 atm0Molar flow rates of feed streams:
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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.