Chemical Engineering Engineers 2017 Case Upper 8
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This post was last edited by 2005180005 on 2018-10-7 at 16:40. 7. A certain distillation system is used for the separation of a mixture of benzene and chlorobenzene. The feed rate of this mixture is 5000 kg/h, with 38% (by mass) being benzene and 62% being chlorobenzene. If the benzene content in the bottom stream exiting the reboiler is 1.0%, and the benzene content in the top product stream is 99%, with a reflux ratio of 5, what is the gas flow rate at the top of the distillation column in kg/h? (Molecular weight of benzene: 78 ； Molecular weight of chlorobenzene: 112.6) (A) 1088 (B) 9939 (C) 11327 (D) 15561. 8. Based on the conditions given in the previous question, if the feed rate of the mixture is adjusted while the reflux rate remains unchanged, the amount of liquid product at the top of the tower is 1850 kg/h. The condensation temperature is 80°C; the cooling water used in the atmospheric pressure condenser at the top of the tower serves as the coolant, with an inlet temperature of 32°C and an outlet temperature of 45°C. What is the consumption rate of cooling water in kg/h at this condensation temperature? It is known that at the liquid condensation temperature, the vaporization heat of benzene is 30.8 kJ/mol, and that of chlorobenzene is 36.6 kJ/mol; the sensible heat loss due to the cooling of the gas entering the condenser is ignored. (A) 13407 (B) 56094 (C) 67033 (D) 80440. Could xD change during the calculation for question 8? It doesn’t have to be 99%; the question isn’t clear on this point
Ignoring the change in XD, it’s impossible to solve this problem without this condition, and there’s no time during the exam to carry out such calculations. Next time I encounter a question like this, I’ll assume it remains unchanged!
This question doesn’t seem to be formulated very carefully
It’s not just a problem; it’s a serious problem. During the exam, when greeting the question-setting team, asking whether \"the reflux remains unchanged\" means that the reflux ratio R should stay the same or that the reflux rate RD for previous questions should remain unchanged
7 C F = D + W ---------(1); based on the mass of benzene, F * 0.38 = D * 0.99 + W * 0.01 --------(2). Solving these equations gives D = 1887.76 kg/h; therefore, V = (R + 1)D = 6 * 1887.76 = 11327 kg/h
This post was last edited by zhanghp30 on 2018-10-19 at 13:04; the 8D reflux ratio remains unchanged, and the purity of the material at the top of the tower stays the same. Based on 100 g of the overhead mixture, it contains 99 g of benzene; thus, the benzene content in the overhead stream is Xa = 99/78/(99/78 + 1/112.6) = 0.993. The average molecular weight of the mixture is M = 0.993*78 + 0.007*112.6 = 78.24. The volume flow rate of the vapor rising from the overhead stream is V = (R+1)*D/M = 6*1850/78.24 = 141.87 kmol/h. The average latent heat of vaporization for the mixture is r = 0.993*30.8 + 0.007*36.6 = 30.84 KJ/mol. Therefore, Qc = Vr = 141.87*30.84 = 4375.3*10^3 KJ/h. The amount of cooling water required is Wc = Qc/Cp*(t2-t1) = 4375.3*10^3/4.187/13 = 8.04*10^4 kg/h. The closest option is (D). To solve this problem, it is necessary to have a thorough understanding of the basic concepts, such as the difference between mass fractions and molar fractions, and to ensure that all units are consistent before proceeding with the calculations. Otherwise, it’s hard to get it right.