Department of Chemical Engineering Faculty of Engineering & Engineering Technology Abubakar Tafawa Balewa University, Bauchi CHE411: Chemical Reaction Engineering I Homework Submission Date: 23rd November, 2022 Problem 1: Water containing a short-lived radioactive species flows continuously through a well-mixed holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed stream. This is not bad, but we'd like to lower it still more. One of our office secretaries suggests that we insert a baffle down the middle of the tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this would help? If not, tell why; if so, calculate the expected activity of the exit stream compared to the entering stream. Solution Before insertion of baffle CAo CA1 = 1/7 CAo π1, π1 After insertion of baffle CAo CA2 = ? CAo π1 /2, π1 /2 π1 /2, π1 /2 For the single mixed reactor, the performance equation is π1 = ππ1 = πΆπ΄0 − πΆπ΄1 ππΆπ΄1 πΆπ΄0 −1 =7−1=6 πΆπ΄1 After insertion of baffle, we have two well mixed tanks in series with the following performance equation For tank 1 π1 πΆπ΄0 − πΆπ΄1 = 2 ππΆπ΄1 πΆπ΄1 1 = πΆπ΄0 1 + ππ1 2 π1 πΆπ΄1 − πΆπ΄2 = 2 ππΆπ΄2 πΆπ΄2 1 = πΆπ΄1 1 + ππ1 2 For tank 2 Therefore πΆπ΄2 πΆπ΄2 πΆπ΄1 1 1 1 1 1 )( )=( )( )= = =( ππ ππ πΆπ΄0 πΆπ΄1 πΆπ΄0 1+3 1+3 16 1 + 21 1 + 21 Hence, for the two well mixed tanks in series the expected activity of the exit stream is 1/16 of the feed stream. Problem 2: Aqueous feed containing reactant A (CA0 = 2 mol/liter) enters a plug flow reactor (10 liter) which has a provision for recycling a portion of the flowing stream. The reaction kinetics and stoichiometry are and we wish to get 96% conversion. Should we use the recycle stream? If so, at what value should we set the recycle flow rate so as to obtain the highest production rate, and what volumetric feed rate can we process to this conversion in the reactor? Solution Given π΄ → π ππ΄ = 0, πΆπ΄0 = 2 −ππ΄ = 1πΆπ΄ πΆπ πππ , πππ‘ππ πππ πππ‘ππ. πππ π = 10 πππ‘ππ, ππ΄π = 0.96 Recall that the performance equation for recycle reactor is ππ΄π π 1 = (π + 1) ∫ π π ππ π΄π πΉπ΄0 (−ππ΄ ) π΄ ππ΄1 = π +1 The optimum recycle ratio that minimizes the reactor volume or space time (or maximizes the production rate) is given by ππ΄π 1 | −ππ΄ π π΄π Which can also be written as = 1 πππ΄ π΄π (−ππ΄ ) ∫π (ππ΄π −ππ΄π ) 1 | −ππ΄ π π΄π π 1 (ππ΄π − ππ΄π ) = ∫π π΄π (−π ) πππ΄ π΄π (1) π΄ We know that πΆπ΄ = πΆπ΄0 (1 − ππ΄ ) πΆπ = πΆπ΄0 ππ΄ 2 −ππ΄ = ππΆπ΄ πΆπ = ππΆπ΄0 (1 − ππ΄ )πΆπ΄0 ππ΄ = ππΆπ΄0 ππ΄ (1 − ππ΄ ) (2) Substituting eq. 2 into the left-hand side of eq. 1 gives 1 1 | (ππ΄π − ππ΄π ) = 2 | (ππ΄π − ππ΄π ) −ππ΄ π ππΆπ΄0 ππ΄ (1 − ππ΄ ) π π΄π π΄π But ππ΄π = π ππ΄π π +1 Therefore 1 | −ππ΄ π π΄π (ππ΄π − ππ΄π ) = π ππ΄π ) π +1 π ππ΄π π ππ΄π (ππ΄π − 2 ππΆπ΄0 π +1 (1− π +1 = ) ππ΄π (π +1) 2 ππΆπ΄0 π ππ΄π (1+π −π ππ΄π ) (3) Now let us know evaluate the right-hand side of eq. 1 ππ΄π ππ΄π 1 1 πππ΄ = ∫ πππ΄ 2 ππ΄π (−ππ΄ ) ππ΄π ππΆπ΄0 ππ΄ (1 − ππ΄ ) ππ΄π 1 1 = ∫ ππ 2 ππΆπ΄0 ππ΄π ππ΄ (1 − ππ΄ ) π΄ From Table of integrals, we get 1 1 π₯ ∫ = ππ ( ) π₯(ππ₯ + π) π ππ₯ + π Thus, ππ΄π ππ΄π 1 1 1 ππ΄ )| πππ΄ = 2 ππ ( 2 ∫ 1 − ππ΄ π ππΆπ΄0 ππΆπ΄0 ππ΄π ππ΄ (1 − ππ΄ ) ∫ π΄π ππ΄π 1 1 ππ΄π = 2 ππ (1 − π ) − 2 ππ (1 − π ) ππΆπ΄0 ππΆπ΄0 π΄π π΄π Substituting ππ΄π = π ππ΄π π +1 gives π ππ΄π ππ΄π ππ΄π 1 1 1 1 π +1 ) πππ΄ = ππ ( )− ππ ( 2 ∫ 2 2 π ππ΄π 1 − ππ΄π ππΆπ΄0 ππ΄π ππ΄ (1 − ππ΄ ) ππΆπ΄0 ππΆπ΄0 1−π +1 = Thus 1 + π (1 − ππ΄π ) 1 ) 2 ππ ( ππΆπ΄0 π (1 − ππ΄π ) ππ΄π ∫π π΄π 1 (−ππ΄ ) πππ΄ = 1 2 ππΆπ΄0 ππ ( 1+π (1−ππ΄π ) π (1−ππ΄π ) ) (4) By substituting eq. 3 and eq. 4 in eq.1, we get ππ΄π (π + 1) 1 + π (1 − ππ΄π ) 1 = ) 2 2 ππ ( ππΆπ΄0 π ππ΄π (1 + π − π ππ΄π ) ππΆπ΄0 π (1 − ππ΄π ) Or 1 + π (1 − ππ΄π ) π +1 = ππ ( ) π (1 + π − π ππ΄π ) π (1 − ππ΄π ) Substituting ππ΄π = 0.96 gives π +1 1 + 0.04π ) = ππ ( π (1 + 0.04π ) 0.04π Solving by trial-and-error method gives π = 0.28 The production rate is then calculated from the following equation for recycle reactor ππ΄π π 1 = (π + 1) ∫ π π πππ΄ π΄π (−π ) πΉπ΄0 π΄ ππ΄1 = π +1 (π + 1) 1 + π (1 − ππ΄π ) π = ππ ( ) 2 πΉπ΄0 ππΆπ΄0 π (1 − ππ΄π ) 2 πππΆπ΄0 1 πΉπ΄0 = (π + 1) 1 + π (1 − ππ΄π ) ππ ( ) π (1 − ππ΄π ) Substituting the known values gives the production rate as πΉπ΄0 = 10(1)(2)2 (0.28 + 1) 1 = 6.9 πππ/πππ 1 + 0.28(1 − 0.96) ) ππ ( 0.28(1 − 0.96) And the volumetric feed rate is calculated as π= πΉπ΄0 6.9 = = 3.45 πππ‘ππ/πππ πΆπ΄0 2 Thus, for the given reactor, by setting the recycle ratio at 0.28 we can achieve the highest production rate of 6.9 mol/min and the volumetric feed rate we can process to get 96% conversion in the reactor is 3.45 liter/min.
