Given the irreversible liquid phase reaction as following: A+ B --> 2C The react

Question

Given the irreversible liquid phase reaction as following: A+ B --> 2C The reaction is carried out non-adiabatically in a flow reactor. Equal molar of A and B enters the system at 70°C and the volumetric flow rate is 60 dm3/s with initial concentration of A = 18.5 mol/dm3. Calculate the exit reaction temperature at which the reaction is conducted when given the exit conversion as 0.65. Additional Information: Steam jacket area, A = 18 dm2 Overall heat transfer coefficient, U = 23 cal/dm2.s.K Jacket steam saturation temperature, Ta = 350 K ?Ho (273) 70,000 cal/mol

Tutorial 7.pdf-Adobe Acrobat Reader DC File Edit View Window Help Home Tools Tutorial 7.pdfx ? Sign In Chapter 7.pdf 100% 1. Given the irreversible liquid phase reaction as following Export PDF Create PDF Edit PDF Commert AB-2C The reaction is carried out non-adiabatically in a flow reactor. Equal molar of A and B enters the system at 70°C and the volumetric flow rate is 60 dm/s with initial concentration of A 18.5 mol/dm Calculate the exit reaction temperature at which the reaction is conducted when given the exit Dcombine Files ~ conversion as 0.65 Adobe Acrobat Pro DC Combine two or more files into a single PDF Steam jacket area, A-18 dm Overall heat transfer coefficient, U23 cal dms.K Jacket steam saturation temperature, Ta-350 K AHR (273)70,000 cal/mol CpA = 23cal/ mol·K CB -28cal/molK Cc -38cal/ mol Learn more Organize Pages Fill&Sign; Store and share files in the Document Cloud Learn More 11:46 PM 5/4/2018

Explanation / Answer

Solution:

Inlet Concentration of A=B= 18.5mol/dm3

Outlet Concentration of A=B=18.5(1-0.65)= 6.475mol/dm3

'A' reacted = 18.5 - 6.475 = 12.025 mol/dm3

Outlet Concentration of C = (12.025)(2)= 24.05mol/dm3

Steam Jacket is used .So Heat is added to the reactor. Q=U A dt

Heat Added = Q=(23)(18)(350-343) = 2898 cal/s

Enthalpy Balance:

Enthalpy In - Enthalpy Out + Heat Added = Heat of the reaction

Enthalpy In = (343-273)[(60)(18.5)(23) + (60)(18.5)(28)]= 3962700 cal/s

Enthalpy Out=(T-273)[(60)(6.475)(23) + (60)(6.475)(28) + (60)(24.05)(38)] =(T-273)74647.5 cal/s

Heat Added= 2898 Cal/s

Heat of the reaction= - (70000)(60)(12.025)= - 50505000 cal/s

3962700 - (T-273)74647.5 + 2898? =  - 50505000

T=1002.7 k

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