Elementary Principles 265 - 7.39 (cont'd) (c) We may...

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7-20 7.39 (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 ±± ± QQ Q =−= = evaporator condenser Btu min 2000 2500 500 and the compressor work W c represents the total work done on the system. The system is closed (no mass flow in or out). Consider a time interval Δ t min bg . Since the system is at steady state, the changes Δ U , Δ E k and Δ E p over this time interval all equal zero. The total heat input is ± Qt Δ , the work input is ± Wt c Δ , and (Eq. 8.3-4) yields ± ± . . . Qt W t W Q cc ΔΔ −= = = −× × = 0 500 1341 10 9 486 10 118 3 4 Btu 1 min hp min 60 s Btu s hp 7.40 Basis: Given feed rates ± . n 1 08 0 (mol / h) 0.2 C H CH C, 1.1 atm 38 41 0 o ± ± n n 3 8 4 10 o 0 (mol C H / h) (mol C H / h) 227 C ± Q (kJ / h) ± n 2 0 o (mol / h) 0.40 C H .60 C H 25 C, 1.1 atm 0 Molar flow rates of feed streams:
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