A hydrocarbon stream in a petroleum refinery is to be separated at 1,500 kPa into two products
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A hydrocarbon stream in a petroleum refinery is to be separated at 1,500 kPa into two products under the conditions shown below. Using the data given, compute the minimum work of separation, Wmin , in kJ/h for To = 298.15K.
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kmol/h Feed Product 1 Component Ethane 30 30 200 Propane n-Butane 192 370 4 n-Pentane 350 50 n-Hexane Feed Product 1 Product 2 Phase condition Liquid Vapor Liquid Temperature, K Enthalpy, kJ/kmol Entropy, kJ/kmol-K 364 394 313 19,480 36.64 25,040 33.13 25,640 54.84 Product Product Product Feed 3 Liquid 304 Phase condition Liquid 305 Liquid 299 Liquid 314 Temperature, F Enthalpy. 29.750 29,290 29,550 28,320 Btu/lbmol 12.47 Entropy, 15.32 13.60 14.68 Btu/lbmol-°R
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From Eq 4 Table 21 For the feed stream in n 30 200 370 350 50 10...View the full answer
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