ChE 344 Chemical Reaction Engineering Winter 1999 Exam I Part 1 (80%) Solution
(10 pts) 1) The trimerization 3A(g) A 3 (g,l) is carried out isothermally and without pressure drop in a PFR at 98 K and atm. As the concentration of A 3 increases down the reactor and A 3 begins to condense. The vapor pressure of A 3 at 98 K is 0.5 atm. If an equal molar mixture of A and inert, I, is fed to the reactor at what conversion of A will A 3 begin to condense? 1 A( g) 3 A 3( g,l ) Condensation begins at y = P V 0.5 atm = P T atm = 0.5 Species Entering Change Before Cond. After Cond. A( g) X F A = ( 1 X) F A = ( 1 X) I( g) --- F I = F I = A 3 ( g,l) --- X 3 F A 3 = X 3 F T = 3 X F A 3 = y A3,e F T y A3,e = F A 3 F T = X 3 3 X = 0.5 X 6 X = 1 4 4X = 6 X 6X = 6 X = 1 1
(30 pts) ) In order to study the photochemical decay of aqueous bromine in bright sunlight, a small quantity of liquid bromine was dissolved in water contained in a glass battery jar and placed in direct sunlight. The following data were obtained: Time (min) 10 0 30 40 50 60 ppm Br.45 1.74 1.3 0.88 0.6 0.44 a) Determine whether the reaction rate is zero-, first-, or second-order in bromine, and calculate the reaction rate constant in units of your choice. b) Assuming identical exposure conditions, calculate the required hourly rate of injection of bromine (in pounds) into a sunlit body of water 5,000 gal in volume in order to maintain a sterilizing level of bromine of 1.0 ppm. (Note: ppm parts of bromine per million parts brominated water by weight. In dilute aqueous solutions, 1 ppm=1 milligram per liter, molecular weight of Br = 80 Daltons.) 1 gal 3.785 liters Sunlight 1 lb 454 gms BR Products Rate law: r A = k C A α Experiment MB: dc A dt = r A ( batch) Experimental data provides C A vs. time, thus dc A dt can be calculated Combine: dc A dt = k C A α ln dc A dt = lnk +α lnc A See plot for y x vs. t Pick points:.45, 0.080 at random and 0.88, 0.09 α = ( ) ln ( +0.09 ) ( ) ln( 0.08) ln +0.08 ln.45 α = 0.99 1 pick 0.88, 0.09 3.54 = lnk + 0.99 ( )( 0.178) lnk = 3.4134 k = 0.033 min 1
Chart Estimate t (C A ) X y y x y/ x dy/dx 10.45 0.080 10 0.71 0.071 0 1.74 0.060 10 0.51 0.051 30 1.3 0.04 10 0.35 0.035 40 0.88 0.09 10 0.6 0.06 50 0.6 0.00 10 0.18 0.018 60 0.44 0.015 Problem, part b FA M.B.: rate in rate out + gen. = accum dn A dt = F A + r A V since the Br concentration is constant at 1 ppm, a steady state assumption is valid. so dn A dt = 0 r A V = F A, r A = kc A F A = 0.033 min ( 1 ppm ) 5,000 gal ppm gal 60 min F A = 85 miin 1 hr 3.785l 1 gal 1g 1000 mg lb 454g F A = 0.41 lb hr 3
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(0pts) 3) The irreversible elementary gas phase reaction A B + C is currently carried out in a packed bed reactor containing 100 kg of catalyst. The entering pressure is 0 atm and the exit pressure is 4 atm. Currently 50% conversion is achieved. It is proposed to add a CSTR with 00 Kg of catalyst downstream of the PBR. There is no pressure drop in the CSTR. The flow rate and temperature remain unchanged. a) What would be the overall conversion in such an arrangement? b) Is there a better way to carry out the reaction, and if so what is it? Current P0 = 0 100 KG P = 4 X = 0.5 Proposed = 0 100 KG P = 4 X = 0.5 00 KG X =? a) CSTR: r A out = MB RL ST dx PBR: r A = dw r A = kc A A C A = C A0 ( ) X out X in w Since ε = 0, P = ( 1 αw) 1 = 1 αw 1 P α = 0 = w 1 B + 1 C α = 0.0096 ( 1 X) 1+ εx ( ) 1 P T 0 T Combine: kc A0 1 4 0 100 ( 1 X) = dx dw 5
ε = y A0 δ, y A0 = 1, δ = 0 ε =0 C A = C A0 ( 1 X) P kc A0 dw = dx ( 1 X) kc A0 Integrate kc A0 ( 1 αw)dw = w α w 0 dx ( 1 X) 100 kc A0 ( 100 48) = 1 kc A0 = 1 5 For CSTR kc A0 = ( 1 X) wkc A0 1 X 00 5 ( 0.04 ) 1 X ( 13 1 X + X ) = X 0.5 13 4 13 X + 13 X = X 0.5 13 X 17 13 X + 17 6 = 0 X = 17 13 ± X =0.534 0.5 1 ( 1 X) 0 ( ) = X 0.5 ( ) = X 0.5 17 13 4.. 13 13 17 6 = b) Want higher conversion = ( X 0.5) w 1.3077 ± 1.1435 0.3077 no = 0.534 or 7.96 1) reduce pressure drop use larger pellets increase temperature larger k ) Use CSTR followed by PFR. 6
(10 pts) 4) The following reactions were found to occur while trying to make a desired product B (1) A B r A1 = k 1A C A A X A + X Y Species X and Y are both foul pollutants r A = k A C A r A3 = k 3A C A C X a) What is the instantaneous selectivity of B with respect to the foul pollutants X and Y? b) How would you carry out this reaction to maximize the formation of B? Additional Information k 1A =.5 e 10,000/T min 1, T in degrees Kelvin k A = 50 e 0,000/T min 1, T in degrees Kelvin k 3A = 100 e 5,000/T min 1, T in degrees Kelvin a) r B = r A1 r x = r A + r A3 r y = r A3 Selectivity of B with respect to x and y S BXY = r B r x + r y = r A1 r A = k 1A k A C A = 0.01 exp[ 10,000 /T] C A b) In order to maximize the formation of B, S BXY should be maximized. High C A use PFR Low Temperature 7
(10 pts) 5) The catalytic reaction A B to be carried out in a flow reaction system has the following rate law, kc r A = A ( 1+ K A C A ) where k = 1 min 1 K A = 1 dm 3 /mol The entering concentration of A is mol/dm 3. What type of reactor or combination of reactors would have the smallest volume to a) achieve 50% conversion? b) achieve 80% conversion? C A = C A0 ( 1 X) ( ) 1 = 1+ K A C A r A kc A [ ( )] ( ) = 1 + 1 X 1 X a) CSTR 8
b) CSTR followed by a PFR 9