20200710 Past exam paper Results

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1 Chemical Processes and Technologies (Ing. Giorgia De Guido) Exam July 10th, 2020 - Exercise: Solution Exercise no. 1 Consider a liquid mixture composed of propane, n- butane, n-pentane and n -hexane, which is at a temperature of 340 K and at a pressure of 25 bar. The mass flow rate of each component in it is reported in the data section. This mixture is depressurized to 15 bar and fed to a flash drum in order to recover 95% of propane in the vapour phase. 1. At which temperature [°C] does the flash drum have to be operated? 2. Which are the molar flow rates [kmol/h] of the vapour and liquid outlet streams? 3. Which thermal power [kW] has to be supplied to/removed from the flash drum? Data Component Mass flow rate Molecular weight A B C im [kg/h] MWi [kg/kmol] propane 1808 + β 44.10 15.7260 1872.46 -25.16 n-butane 1163 58.12 15.6782 2154.90 -34.42 n-pentane 722 72.15 15.8333 2477.07 -39.94 n-hexane 2500 86.17 15.8366 2697.55 -48.78 β = 530 ( 4.5)M⋅− [kg/h] M5 = fifth digit of student number from the left (Student number/matricola = XXXXXX) The vapour pressure is calculated with the Antoine equation: ln( )ev BPA TC= − + , with T in [K] and Pev in [mmHg] Component TNBP ΔHev (@TNBP) <cPV> <cPL> Tc [K] [kcal/kmol] [kcal/(kmol K)] [kcal/(kmol K)] [K] propane 231.1 4491 15.06 28.670 369.9 n-butane 272.7 5351 21.99 34.140 425.2 n-pentane 309.2 6160 31.02 37.614 469.6 n-hexane 341.9 6896 38.86 51.700 507.4 2 Results The results are reported in the following assuming M5 = 0. Therefore: • mass flow rate of propane in the liquid feed mixture = 1673 [kg/h]. Point no. 1 The molar flow rate of each component in the liquid mixture, ,() FiFz⋅ , can be computed as: ,() i Fi i mFz MW⋅=  and: , , , 1 () () Fi Fi NC Fi i Fzz Fz = ⋅= ⋅∑  Component zF,i [-] propane 0.391234 n-butane…