CHEMICAL and BIOMOLECULAR ENGINEERING 140 Exam 1 Friday, September 28, 2018 Closed Book Name: Section: Total score: /100 Problem 1: /30 Problem 2: /35 Problem 3: /35
1. (30 points) Short answer questions: 3 points per question i) For a reactor with chemical reaction A ---> B, under what conditions does the reaction rate equal the time rate of change of concentration of B, r B = dc B /dt? Batch reactor (test tube) ii) For a reversible reaction A <--> B, what defines the condition of chemical equilibrium? Equal forward and reverse rates iii) For a non-reacting system with one species flowing in at a rate different from the rate of that species flowing out, can the system be at steady state? Why or why not? No, total mass is changing with time in the system (Accumulation) iv) What is the most common reason to use recycle in a chemically reacting system? Use more of the reactants to increase overall conversion v) Why is purge used in NH 3 manufacture? To avoid buildup of inert argon (or oxygen!) vi) List 2 possible results obtained from a material balance problem solution that would be physically impossible. Negative flow rate, negative concentration, mole/mass fraction > 1, fractions don t add to 1... vii) How many system boundaries can be drawn for the reactor with bypass shown below? 4: overall, mixer, reactor, splitter
viii) Write the general balance equation of some quantity X in words. Rate flow X in - rate flow X out + rate generation X - rate consumption X = rate accumulation of X ix) What terms are zero in the balance equation of a batch system? Flow in, flow out x) What terms are zero in the balance equation of a steady state system? Accumulation
2. (35 points) Lorena makes a polymer (P) in benzene (B), at a concentration of 1g polymer to 10 g benzene. She then adds 2 g hexane (H) per g benzene, which causes the polymer to precipitate. The mixture goes to a filtration unit, where the precipitate stream contains 90 wt% polymer and 10 wt% hexane. The precipitate stream accounts for 50% of the polymer fed to the filtration unit, while the other 50% remains with the benzene and the rest of the hexane (the filtrate), and is thrown away. a. (10 points) How much benzene, hexane, and polymer must be fed to this precipitation and filtration process to end up with 10 g precipitate? Given that the precipitate is 90 wt% polymer, 10 g precipitate will contain 9 g polymer. Another 9 g polymer is thrown away for a total of 18 g polymer input. Since there is ten times the amount of benzene as polymer, 180 g benzene must be input. Given that there is twice as much hexane as benzene, 360 g hexane must be input.
Sad about losing so much polymer, she decides to recycle the benzene / hexane / polymer in a continuous process. She purges 10% of the recycle stream, and mixes the rest with the input polymer / benzene stream. She adds enough hexane to the precipitator to make sure that the hexane:benzene ratio in the precipitator is 2:1. The filter has the same characteristics as above (it filters out 50% of the polymer fed to it, and the precipitate stream is 90 wt% polymer, 10 wt% hexane). b. (20 points) What flow rates of benzene, hexane, and polymer must now be fed to the overall process to end up with 10 g / min of precipitate? In 10 g/min precipitate, there are 9 g/min P. since the filter loses 50% of polymer fed, 9g polymer ends up in the bottom feed, and 18g must be fed. 10% of the 9g lost is purged (0.9 g/min) and the other 90% is recycled (8.1 g/min). A balance on the precipitator shows that in order to get 18 total g/min P, 9.9 g/min P must be input and added to the recycle stream. Since the ratio of polymer to benzene is 1:10, 99 g/min B must be input. An overall mass balance on benzene shows that the amount of benzene purged must equal the amount input (99 g/min). Since this only accounts for 10% of the benzene coming out of the filter, the other 90% is in the recycle stream (891 g/min). Added to the input, there is a total of 990 g/min benzene put into the precipitator. Given that twice as much hexane as benzene is in the precipitator, a total of 1980 g/min hexane is in the precipitator. Of that amount, only 1 g/min leaves the system via the precipitate, so 1979 g/min are in the filtrate. Since 10% of the filtrate is purged, 197.9 g/min H are purged. An overall mass balance shows that the amount input must be 197.9 + the 1g in the precipitate = 198.9 g/min H fed.
c. (5 points) Comment (briefly!) on a possible benefit and downside of adding this recycle stream. Possible benefits include: Higher percentage of P recovered Less waste Lower amounts of hexane and benzene required Possible downsides include: Higher flow rates require larger reactors Expensive to build this new thing
3. (35 points) a. Write a balanced reaction for the synthesis of ammonia (NH 3 ) from N 2 and H 2 (3 points) N2 + 3 H2 2 NH3 b. 5 mol/min N 2 and 12 mol/min H 2 are fed to a reactor running this reaction. What is the limiting reactant? (3 points) What is the percent excess of the other reactant? (3 points) If all the N2 is used, 3*5 = 15 mol/min H2 would be required. Since there is not enough H2 for this, H2 is the limiting reactant. To use 12 mol/min H2, 12/3 = 4 mol/min N2 are required. Therefore the percent excess of N2 is (5-4)/4 = 25% excess. c. At 100% conversion, how many mol/min of NH 3 would be produced? (3 points) If 4 moles of N2 are used, 4*2 = 8 mol/min NH3 are produced. d. 6 mol/min of NH 3 are produced. What is the conversion of N 2? (3 points) What is the conversion of H 2? (3 points) What is the extent of reaction? (3 points) If 6 moles of NH3 are produced, 6/2 = 3 mol/min of N2 are used. Therefore, the conversion of N2 is 3/5 = 60% conversion. Of the 12 mol/min H2 available, 6*(3/2) = 9 mol/min are used. Therefore, the conversion of H2 is 9/12 = 75% conversion. The equation for extent of reaction, ni = nio - vi*extent, applies to all species involved. Using NH3, 6 = 0-2*extent. Therefore extent = 3 mol/min. e. In real life, N 2 gas is often in the form of air, with O 2 also present. Given that air is 21 mol% O 2, 79 mol% N 2, how much O 2 is fed in this process? (2 points) Given that 5 mol/min N2 are fed, 5 * (0.21/0.79) = 1.33 mol/min O2 are fed. f. The reaction N 2 + 2O 2 -> 2NO 2 can also occur as an undesired side reaction. If 6 mol/min of NH 3 is still produced, what is the maximum amount of NO 2 that could be produced? (3 points) If 6 mol/min NH3 are produced, 3 mol/min N2 are used and 2 mol/min N2 are remaining. However, only 1.33 mol/min O2 are available, so O2 is the limiting reactant. If the reaction has 100% conversion, 1.33*(2/2) = 1.33 mol/min NO2 are produced. g. If the amount of NO 2 you found in (f) is produced, what is the yield of NH 3? (3 points) What is the yield of NO 2? (3 points) What is the selectivity for NH 3? (3 points)
Since the maximum possible amount of NH3 made is 8 mol/min (from part c), the yield is 6/8 = 75% yield. In the NO2 reaction, 1.33 moles are produced which is equal to the maximum possible number of moles, so the yield is 1.33/1.33 = 100% yield. Selectivity is defined as the moles of desired product (in this case, NH3) divided by the moles of byproduct (NO2) produced. Given that 6 mol/min NH3 and 1.33 mol/min NO2 are produced, selectivity = 6/1.33 = 4.5.