Energy Engineering - Fundamentals of Chemical Processes

Full exam

Fundamentals of Chemical Processes Prof. Gianpiero Groppi January 17 2022 Problem 1 (18 points) S R: CH 4 + H 2O ↔ 3 H 2 + CO WGS: H 2O + CO ↔ CO 2 + H 2 Syngas is produced in a reforming reactor (R1) where Steam Reforming (SR) and Water Gas Shift (WGS) reactions simultaneously occur. Feed stream (1) contains methane and water with a 1:3.5 molar ratio at T 1 = 500 K. The stream (1) is preheated in the heat exchanger HE up to T 2 = 800 K, using the exhaust stream (5) from the burner as hot fluid, and fed to the Steam Reformer (R1) operating at P=30 atm. Equilibrium conditions are reached at the R1 outlet with a C selectivity to CO ������� ������������������ = ������ ̇3, ������������ ������̇1, ������������ 4 −������̇3, ������������ 4 � equal to 60 %. The steam reformer is heated by the burner, which provides the required heat duty Q. The burner is fed at 298 K by stream (4), which consists of methane and dry air (21% O 2 - 79% N 2) with 20% air excess. The burner secures complete methane combustion. Outlet stream (6) from the heat exchanger is at T 6 = 650 K. Assuming: • Ideal gas behavior of all the streams • Ideal behavior with no heat losses of the heat exchanger • Isobaric reactors Calculate: 1. Temperature T 3 and molar composition of outlet steam (3) from reactor R1 T3 H2 CO CO 2 CH 4 H2O 2. The heat Q entering in R1 per mole of feed methane to the steam reformer Q/n CH4,(1) 3. Exhaust burner temperature T 5 and molar CH 4 feed flow rate to the burner per mole of CH 4 feed to the reformer T5 nCH4,(4) / n CH4,(1) Thermodynamic data: ∆������������,������������0 (������)=53717 −60 .25 ∙������ ������������������������������������� � T in K, 600 K < T < 1500 K ������������������,������������������0 =−8514 +7.71 ⋅������ ������������������������������������� � T in K, 600 K < T < 1500 K Reference: ideal gas at: 1 atm, 298 K ������������������,������������0 (298 ������)=+49270 ������������������������������������� � Steam Reforming ������������������,������������������0 (298 ������)=−9830 ������������������������������������� � WGS ������������������������������������ ,������������4 0 (298 ������)=−191591 ������������������� ������������������ � Species a b c d ������̃������ ,������(������)= H2 6.483 2.215 x 10 -3 -3.298 x 10 -6 1.826 x 10 -9 CO 7.373 -3.070 x 10 -3 6.662 x 10 -6 -3.037 x 10 -9 CO 2 4.728 17.540 x 10 -3 -13.380 x 10 -6 4.097 x 10 -9 CH 4 4.598 12.450 x 10 -3 2.860 x 10 -6 -2.709 x 10 -9 H2O 7.701 0.4595 x 10 -3 2.521 x 10 -6 -0.859 x 10 -9 O2 6.715 -0.879 x 10 -3 4.171 x 10 -6 -2.544 x 10 -9 N2 7.440 -3.240 x 10 -3 6.400 x 10 -6 -2.790 x 10 -9 Problem 2 (12 points) Consider a feed mixture F consisting of n-hexane, n-octane and n-decane, with composition (molar fractions - z i) reported in the following table along with the Antoine equation coefficients: A i, B i e C i: Antoine Equation: P VAP [bar], T [K] log10�P������������������ (T)�=Ai− Bi T+Ci Specie z [ -] Ai Bi Ci n-C6H14 0.3 4.00266 1171.53 -48.784 n-C8H18 0.5 4.04867 1355.126 -63.633 n-C 10H22 0.2 4.07857 1501.268 -78.67 The feed stream F, which is at 8 bar and 450 K, is laminated isoenthalpically before entering in the flash drum D 1 at 5 bar. D 1 is heated with a duty of 20 kJ per feed mole. The vapor stream (V 1) from D 1 is further isoenthalpically laminated and fed to a second flash drum (D 2) at 3 bar. Flash (D 2) allows to recover in the vapor stream V 2 50% of n-hexane in the feed stream F. Assuming an ideal gas behaviour of the vapor phases and an ideal mixture behavior with negligible Poynting factor of the liquid phases, calculate: 1. Physical state of feed stream F Stato 2. Vaporization ratio α 1 and temperature T1 of flash D 1 α1 T1 3. Vaporization ratio α 2 and temperature T2 of flash D 2 α2 T2 Data: Specific heat at constant pressure are constant with temperature for all the species in the liquid and the vapor phases. Evaporation enthalpies are provided at normal boiling temperature (Teb at 1 atm). Species C P liq [J/mol/K] C P vap [J/mol/K] ΔH VAP [J/mol] n-C 6H14 195.6 143.3 31573 n-C 8H18 255.7 175.9 41495 n-C 10H22 313.3 233.1 51392 Note: • The exam is closed book. It is forbidden to use personal notes, course slides and textbooks. • Report the procedures, the explanations and the results in the paper (also the intermediate results). • Print the Excel spreadsheet: each printed page must contain your name and surname. • Both the paper and the Excel prints must be submitted for the evaluation. • No correction will be given in case only the Excel prints are submitted. • No correction will be given in case the procedure is not reported.