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Process Calculations Lecture notes in Chemical engineering, Lecture notes of Process Control

SASTRA University Process Control

Process Calculations Lecture notes in Chemical engineering

Typology: Lecture notes

2021/2022

Available from 08/30/2023

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Exercise:

5. 100 moles of methane are fed to the reactor at 311K, air is used 50% excess and is

supplied at 373K. If the percent conversion is 60, calculate the heat that must be

removed for the product stream to emerge at 478K. Data:

Cpmfor air (373 – 298K) = 29.2908 kJ/kmol.K Compt Cpm(311 -298K)

J/mol.K

Cpm(478 – 298K)

J/mol.K

CH4

HCHO

H2O

36.044

“

40.193

29.2866

30.0821

41.2902

34.2392

ΔHR= -283.094 kJ/mol

Partial preview of the text

Download Process Calculations Lecture notes in Chemical engineering and more Lecture notes Process Control in PDF only on Docsity!

Exercise:

5. 100 moles of methane are fed to the reactor at 311K, air is used 50% excess and is

supplied at 373K. If the percent conversion is 60, calculate the heat that must be

removed for the product stream to emerge at 478K. Data:

Cpm for air (373 – 298K) = 29.2908 kJ/kmol.K Compt^ Cpm (311 -298K)

J/mol.K Cpm (478 – 298K) J/mol.K CH 4 N 2 O 2 HCHO H 2 O

ΔHR = -283.094 kJ/mol

Contd……

5. 100 moles of methane are fed to the reactor at 311K, air is used 50% excess and is

supplied at 373K. If the percent conversion is 60, calculate the heat that must be

removed for the product stream to emerge at 478K. Data:

Cpm for air (373 – 298K) = 29.2908 kJ/kmol.K

Compt Cpm (311 -298K) J/mol.K Cpm (478 – 298K) J/mol.K CH 4 N 2 O 2 HCHO H 2 O

Ans: -14034 kJ

ΔHR = -283.094 kJ/mol

Hess’s law:

Calculate the heat of formation of ZnSO 4 from its elements using Hess’s law. Zn + S → ZnS ΔH1 = - 184.23 kJ/mol 2ZnS + 3O 2 → 2ZnO+2SO 2 ΔH2 = - 929.5 kJ/mol 2 SO 2 + O 2 → 2SO 3 ΔH3 = -196.8 kJ/mol ZnO + SO 3 → ZnSO 4 ΔH4 = -230.3 kJ/mol Ans: -977.7 kJ/mol

Hess’s law:

Calculate the heat of formation of Chloroform gas from its elements using Hess’s law. C+O 2 →CO 2 ΔH1 = - 393.51 kJ/mol H 2 + ½ O 2 →H 2 O ΔH2 = - 285.83 kJ/mol ½ H 2 + ½ Cl 2 →HCl ΔH3 = -167.57 kJ/mol CHCl 3 + ½ O 2 + H 2 O → CO 2 + 3 HCl ΔHc = -2230.91 kJ/mol Ans: -100.44 kJ/mol

Contd……

9. SO 2 gas is oxidized in 100% excess air with 70% conversion to SO 3. The gases enter

the converter at 400 oC and leave at 450 oC. Datum temperature is 0oC.

Cpm (cal/mol.C): SO 3 -15.5, SO 2 -11, O 2 -7.5 and N 2 -7.1.

ΔHR = -23490 cal/mol

Compute the kcal absorbed in heat exchanger of the converter per kmol of SO 2 sent.

Ans: -13944.55 kcal

Process Calculations Lecture notes in Chemical engineering, Lecture notes of Process Control

Related documents

Partial preview of the text

Download Process Calculations Lecture notes in Chemical engineering and more Lecture notes Process Control in PDF only on Docsity!

Exercise:

5. 100 moles of methane are fed to the reactor at 311K, air is used 50% excess and is

supplied at 373K. If the percent conversion is 60, calculate the heat that must be

removed for the product stream to emerge at 478K. Data:

Cpm for air (373 – 298K) = 29.2908 kJ/kmol.K Compt^ Cpm (311 -298K)

ΔHR = -283.094 kJ/mol

Contd……

5. 100 moles of methane are fed to the reactor at 311K, air is used 50% excess and is

supplied at 373K. If the percent conversion is 60, calculate the heat that must be

removed for the product stream to emerge at 478K. Data:

Cpm for air (373 – 298K) = 29.2908 kJ/kmol.K

ΔHR = -283.094 kJ/mol

Hess’s law:

Hess’s law:

Contd……

9. SO 2 gas is oxidized in 100% excess air with 70% conversion to SO 3. The gases enter

the converter at 400 oC and leave at 450 oC. Datum temperature is 0oC.

Cpm (cal/mol.C): SO 3 -15.5, SO 2 -11, O 2 -7.5 and N 2 -7.1.

ΔHR = -23490 cal/mol

Compute the kcal absorbed in heat exchanger of the converter per kmol of SO 2 sent.