Energy Engineering - Fundamentals of Chemical Processes

Full exam

Fundamentals of Chemicals Processes Prof. Gianpiero Groppi June 8th 2023 EXERCISES: Problem 1 (18 points) Ammonia is produced by the following synthesis reaction according to the process scheme reported in the figure. 1 2 ������2+3 2 ������2↔ ������������ 3 The inlet stream (S1), consisting of 60% H 2 and 40% N 2 (molar basis), enters at T1=200 °C. Stream (S1) is preheated in the heat exchanger (HE) by the effluents (Stream (S3)) from reactor R1 up to T2= 300 °C, before entering in reactor R1, which operates adiabatically and isobarically at P=250 atm. Assuming that: • Equilibrium conditions are reached at the reactor outlet • Ideal gas behavior • Ideal behavior of HE with negligible heat losses Calculate: 1. Temperature T 3 and molar composition of outlet stream (S3) from reactor R1 2. Temperature T 4 of stream (S4) 3. Hydrogen conversion Repeat calculations considering the volumetric behavior of an ideal mixture of real gases using the RKS equation of state. Thermodynamic data ������������ ������0= −54300.8+116.54∙������ [J/mol] and T in [K], where 600 [K] < T < 1500 [K] Reference: ideal gas, 1 atm. ������������ ������0( 298������) =−45720 [J/mol] ������̃ ������ ,������ =������ ������+������ ������∙������+������ ������∙������ 2+������ ������∙������ 3 [J/mol] and T in [K] Species a b C d T C [K] PC [bar] ω H2 2.7140E+01 9.2740E -03 -1.3180E -05 7.6450E -09 33.0 12.9 -0.216 N2 3.1150E+01 -1.3570E -02 2.6800E -05 -1.1680E -08 126.2 33.9 0.039 NH 3 2.7310E+01 2.3830E -02 1.7070E -05 -1.1850E -08 405.5 113.5 0.250 Nome e cognome: …………………………………………………….…… Numero di matricola: ………………………………………………..……….. Problem 2 (12 points) Consider a mixture of methyl-tert-butyl-ether (MTBE), methanol (MeOH) and isobutane (IB) with molar composition (z i) reported in the following table along with Antoine equation coefficients A i, B i e C i. Specie z [ -] Ai Bi Ci MTBE 0.50 5.896 708.69 179.9 MeOH 0.20 8.0724 1574.99 238.87 IB 0.30 6.91048 946.35 246.68 Feed stream F at P f=10 atm and T f=105 °C is splitted in F 1 and F 2. Stream F 1 is isoenthalpically expanded and fed to a flash drum D 1 operated at P 1 = 4 atm and heated by Q 1= 11 kJ per mole of F 1. The vapor stream V 1 from flash D1 and the stream F 2 are isoentalpically expanded and fed to the second flash D 2 operated at P 2 = 1.5 atm and T 2 = 50°C. The flash is designed to achieve a vaporization ratio α 2 = V 2/(V 1+F 2) =0.55. Assuming an ideal gas behavior of the vapor phase and an ideal mixture behavior with negligible Poynting factor of the liquid phase, Calculate: 1. The physical state of stream F 2. The vaporization ratio α1 of flash D 1 3. The split ratio F 2/F1 4. The heat duty (Q2) of flash D 2 per mole of total feed F Data: Specific heats at constant pressure of all the species do not change with temperature in the liquid and the vapor phases. Evaporation enthalpies are provided at normal boiling temperature (T Boil at 1 atm). Species CP liq [J/mol/K] CP vap [J/mol/K] ΔH VAP (T Boil , 1 atm ) [kJ/mol] MTBE 192 196 30.47 MeOH 80 60 35.14 IB 132 129 22.21 Antoine Equation: P SAT [atm], T [°C] PSAT ( T) =10 �Ai− B i T+C i �/760 Parametri RKS ������= 0.42748 ⋅������⋅������������������������� ⋅������������� ������������ 2 ������=0. 08664⋅������ ������������������ ⋅������������ ������������ ������=�1+������⋅�1− ������� ������� �2 ������=0.48+1.574⋅������−0.176⋅������ 2 ������= ������⋅������ ������������������������� ⋅�������2 ������=������ ⋅������ ������ ������������������ ⋅������ Coefficiente di fugacità (Z radice dell’equazione cubica associata) ln������( ������,������) =������−1−������ ������ ⋅ln������� +������ ������ �−ln( ������−������) Risoluzione dell’equazione cubica associata all’equazione di stato ������ 3+������⋅������ 2+������⋅������+������=0 ������= −1 ������= ������−������−������2 ������=−������⋅������ ������= ������−������2 3 ������=2������ 3 27 −������ ⋅������ 3 +������ ������=������ 2 4 +������ 3 27 Se D>0, esiste 1 soluzione reale. ������=�−������ 2 + √������ � 1 3 + �−������ 2 − √������ � 1 3 −������ 3 Se D = 0, si hanno 3 soluzioni reali di cui 2 coincidenti. ������ 1=−2⋅�−������ 2 � 1 3 −������ 3 ������ 2=������ 3=�−������ 2 � 1 3 −������ 3 Se D< 0, esistono 3 soluzioni reali distinte. ������1= 2⋅������13������������������ ������� 3�−������ 3 ������2=2⋅������ 1 3�������������������2������ +������ 3 �−������ 3 ������3= 2⋅������13������������������ �4������+������ 3 �−������ 3 ������= �−������3 27 ������������������( ������) =−������ 2������