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Appello 08 06 2023 english

Full exam for Fundamentals of Chemical Processes in the Energy Engineering degree programme at Politecnico di Milano. The document covers: 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 𝑁2 + 3 2 𝐻2 ↔ 𝑁𝐻3 The inlet stream (S1), consisting

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Full exam for Fundamentals of Chemical Processes in the Energy Engineering degree programme at Politecnico di Milano. The document covers: 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 𝑁2 + 3 2 𝐻2 ↔ 𝑁𝐻3 The inlet stream (S1), consisting

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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 𝑁2 + 3 2 𝐻2 ↔ 𝑁𝐻3 The inlet stream (S1), consisting of 60% H2 and 40% N2 (molar basis), enters at T1=200 Β°C. Stream (S1) is preheated in the heat exchanger (HE) by the effluents (Stre am (S3)) from re actor R1 up to T2= 300 Β°C, before entering in reactor R1, which operates adiabatically and isobarically at P=250 atm. Assuming that: β€’ Equilibrium conditions are reached at the reactor outlet β€’ Ideal gas behavior β€’ 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 T4 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 π›₯𝐺𝑅 0 = βˆ’54300.8 + 116.54 βˆ™ 𝑇 [J/mol] and T in [K], where 600 [K] < T < 1500 [K] Reference: ideal gas, 1 atm. π›₯𝐻𝑅 0(298𝐾) = βˆ’45720 [J/mol] 𝐢̃𝑝,𝑖 = π‘Žπ‘– + 𝑏𝑖 βˆ™ 𝑇 + 𝑏𝑖 βˆ™ 𝑇2 + 𝑏𝑖 βˆ™ 𝑇3 [J/mol] and T in [K] Species a b C d TC [K] PC [bar] Ο‰ 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 NH3 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 (zi) reported in the following table along with Antoine equation coefficients Ai, Bi e Ci. Specie z [-] Ai Bi Ci MTBE 0.50 5.896 708.69 179.9 MeOH 0.20 8.0724 1574.99…

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