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Exercise 10 solution text

Divisi per argomento di Fundamentals of Chemical Processes per il corso di Energy Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

Fundamentals of Chemical ProcessesDivisi per argomento

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Divisi per argomento di Fundamentals of Chemical Processes per il corso di Energy Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 2023-2024 1 EXERCISE 10 Estimation of the composition at the outlet of a reactor for the production of NH3 Ammonia is produced by reacting H2 and N2 in a catalytic reactor. 3 H2 + N2 → 2 NH3 An inlet stream with known composition and at the temperature of 310°C is fed to a reactor operating at a pressure of 275 atm. At the exit of the reactor, the outlet gas stream is at 500°C. Determine the composition of the outlet stream assuming that the thermodynamic equilibrium is reached. Assume constant pressure and the volumetric behavior of an ideal mixture of real gases. DATA SPECIES Molar fraction Tc [K] Pc [bar] ω NH3 3.53 405.5 113.5 0.250 H2 62.6 33.0 12.9 -0.216 N2 20.87 126.2 33.9 0.039 *Ar 3.22 150.8 48.7 0.001 *CH4 9.78 190.4 46.0 0.011 *inert species GR°= -12972 +27.84 T [cal/mol] 600<T<1500K Reference: pure species as ideal gas at 1 atm Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 2023-2024 2 RKS parameters: 𝑎 = 0.42748 ⋅ 𝛼 ⋅ (𝑅𝑔𝑎𝑠 ⋅ 𝑇𝐶) 𝑃𝐶 2 𝑏 = 0.08664 ⋅ 𝑅𝑔𝑎𝑠 ⋅ 𝑇𝐶 𝑃𝐶 𝛼 = (1 + 𝑆 ⋅ (1 − √𝑇𝑅)) 2 𝑆 = 0.48 + 1.574 ⋅ 𝜔 − 0.176 ⋅ 𝜔2 𝐴 = 𝑎 ⋅ 𝑃 (𝑅𝑔𝑎𝑠 ⋅ 𝑇) 2 𝐵 = 𝑏 ⋅ 𝑃 𝑅𝑔𝑎𝑠 ⋅ 𝑇 Fugacity coefficient (Z indicates the root of the cubic equation): ln 𝜑 (𝑇, 𝑃) = 𝑍 − 1 − 𝐴 𝐵 ⋅ ln (𝑍 + 𝐵 𝑍 ) − ln(𝑍 − 𝐵) Solving procedure for the cubic equation: 𝑍3 + 𝛼 ⋅ 𝑍2 + 𝛽 ⋅ 𝑍 + 𝛾 = 0 𝛼 = −1 𝛽 = 𝐴 − 𝐵 − 𝐵2 𝛾 = −𝐴 ⋅ 𝐵 𝑝 = 𝛽 − 𝛼2 3 𝑞 = 2𝛼3 27 − 𝛼 ⋅ 𝛽 3 + 𝛾 𝐷 = 𝑞2 4 + 𝑝3 27 If D>0, only 1 real solution is found: 𝑍 = (− 𝑞 2 + √𝐷) 1 3 + (− 𝑞 2 − √𝐷) 1 3 − 𝛼 3 If D = 0, 3 real solutions are found (2 of which are identical): 𝑍1 = −2 ⋅ (− 𝑞 2) 1 3 − 𝛼 3 𝑍2 = 𝑍3 = (− 𝑞 2) 1 3 − 𝛼 3 If D< 0, 3 distinct real solutions are found: 𝑍1 = 2 ⋅ 𝑟 1 3 𝑐𝑜𝑠 (𝜃 3) − 𝛼 3 𝑍2 = 2 ⋅ 𝑟 1 3 𝑐𝑜𝑠 (2𝜋 + 𝜃 3 ) −…

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