Question

PROBLEM 2 The elementary liquid phase irreversible reaction (A+B -> C) takes place in a 1 m² Mixed Flow Reactor with the equimolar mixture of A and B at the volumetric feed flow rate of 0.5 m3/min, the feed concentration of A equal to 1 mol/L, and the feed temperature of 300K. When the reaction takes place under isothermal conditions at 300K the conversion of A is 30%. When the reaction takes place adiabatically the exit temperature is 350K and the the conversion of Ais 40%. a. Calculate the rate of heat removal (in Joules/min) necessary to achieve the isothermal operation described above. b. Calculate conversion for NEW isothermal operation with the same feed but at 350K with the volumetric flow rate the same as before i.e. 0.5 m/min. Data for PROBLEM 2: Heat capacities of A, B, and Care 25, 35, and 60 J/(mol.K), respectively. Energy Balance Equation: Eq. (11-27) on p.508

Answer 1

<sub>Hi Given reaction is Please note that this equation can be solved with the heat balance easily. A+B-C Volume of the reactor is 1 m^3. Volumetric flow of A and B is 0.5dm^3/min. Feed concentration of A is 1 mol/L. It is given that the feed is equimolar thus we will have equal amount of A and B and since the volumetric flow rate of both A and B are same, feed concentration will be same. Specific heats of A, B and Care 25,35,60 J/mole. K respectively. Let us take 1 minute of operation Volume of the feed in the reactor = 1 dm^3. 1 dm^3 = 1 L. Moles of A = Concentration*volume = 0.5*1 -0.5 moles Moles of B =Concentration volume =0.5*1=0.5 moles We can assume the density of water to be 1 kg/L since we have dilute solution here. Moles of water = 1000/18 = 55.55 moles We can take the specific heat of water as 4.18 *18 = 75.24 J/mole.K. Please note that in the question it is given that in the case of the adiabatic conversion we have exit conversion as 40 %. mwi Let the heat of the reaction be H J/mole.

Heat released=Hmoles reacted Moles of A reacted=0.5"conversion=0.5*0.4-0.2 moles Heat released = H*0.2 moles In the adiabatic case we will have no heat loss thus heat released heat gained Heat gained = Eni Cpt. (Tf - Ti) Here ai is the number of moles of ith component. Cei is the specific heat of ith component. Tf is the final component and Ti is the initial temperature. WN If = 350 K and Ti = 300 Kn I Moles of A remaining will be 0.3 moles. Moles of B remaining = 0.3 moles Moles of C formed = 0.2 moles Heat gained = Heat gained by A + heat gained by B + heat gained by C+ heat gained by water. Heat gained = 0.3*25*(350-300)+0.3*35*(350-300) +0.2*60*(350-300)+55.55*75.24*(350-300) Muis 4209.58*50 210479.11 = 210.4 kj Thus H*0,2= 210.4 kl H= 1052.4 kJ Please note that this is the heat of the reaction that we are getting at 300 K. Let us solve both parts of the question now

Parta- In the case of the isothermal reaction we need to remove the heat that will be generated. When we have 30 % conversion. Heat released = H*0.5*0.3 kl I =1052.4*0.15 =157.85 kJ Thus we need to remove 157.85 kl of heat. Part 6 In the case of the feed at 350 K. Heat of the reaction at this new temperature can be calculated by using the formula. H =Ho+vi * Cpi* (Tf - Ti) Here vi is the stoichiometric coefficient of the reactant or product. Ho = 1052.4 kJ H= 1052.4+60-35-25)*(350-300) H = 1052.4 kJ We can see that the heat of the reaction here is independent of the temperature. Feed is at the temperature of 350 K. Please note that we need to assume the same heat removal capacity here. Amount of the heat released = Amount of the heat removed (isothermal operation) Let the conversion be X. H*0.5*X = 157.85 kJ

ve need to assume the same heat removal capacity here. Amount of the heat released = Amount of the heat removed (isothermal operation). Let the conversion bex H*0.5*X=157.85 k X = 0.3 or 30 %. Since heat of the reaction is independent of the temperature we can say that the conversion will not depend on the temperature. Hope this helps Feel free to comment in the case of any doubt. Thanks
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