HW1 solution

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Georgia Institute Of Technology *

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3225

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Chemistry

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Feb 20, 2024

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Homework Set 1 Solution CHBE 3225 Fall 202 3 1) Using all the information given in the problems, we have the following: where C3 = propane, C4 = butane, C5 = pentane, C6 = hexane The species that are not present as stated in the problem are crossed out and in red. Looking at the process as a whole (1 feed and 4 outlet streams): Feed flow rates are known. In the outlet streams, number of unknowns = U = 2 + 3 + 3 + 4 = 12. stream (1) (2) (3) (4) Counting the number of equations: 4 equations of overall mass balance, one for each species. 4 equations from the recovery information. For each species, 95% = one of the outlet flow / feed flow. 1 equation using the information that pentane comprises 98% of stream (4) = 9 equations total. Since we have 12 unknowns and 9 equations, we lack two more equations or pieces of information. At this point, we do need to make an additional assumption to go forward. Although there are a few that are reasonable, the easier to justify are that there is no C3 in stream (3) and that there is no C3 in stream (4) . There is no C4 in stream (3), so it’s reasonable to assume that the lighter C3 wo uld also not be present there. A new diagram reflecting the absence of C3 in the streams: Column 1 Column 2 Column 3 feed (f) C3 C4 C5 C6 (5) C3 C4 C5 C6 (1) C3 (95% rec) C4 C5 C6 (2) C3 C4 (95% rec) C5 C6 (3) C3 C4 C5 (95% rec) C6 (4) C3 C4 C5 C6 (95% rec) (6) C3 C4 C5 C6 Column 1 Column 2 Column 3 feed (f) C3 C4 C5 C6 (5) C3 C4 C5 C6 (1) C3 (95% rec) C4 (2) C3 C4 (95% rec) C5 (3) C5 (95% rec) C6 (4) C4 C5 C6 (95% rec) (6) C4 C5 C6

Homework Set 1 Solution CHBE 3225 Fall 202 3 So we have added 2 equations (C3 in 3 = 0 and C3 in 4 = 0), giving us 12 unknowns and 11 equations. Or equivalently, we have removed those two unknowns, giving us 10 unknowns and 9 equations. Either way, we need to have one more equation or remove one more unknown. One reasonable assumption here is that C4 is also not present in stream 4. If C4 is not present in the distillate, it is likely also not present in the bottoms. We will see below that this does not quite work, but let’s go with it for now. Let’s use the notation n x (i) to denote the flow rate of species x in stream ( i ). Recovery informations: n C3 (1) = 95% × n C3 (f) = 0.95 × 45 kmol/hr = 42.75 kmol/hr n C4 (2) = 95% × n C4 (f) = 0.95 × 225 kmol/hr = 213.75 kmol/hr n C5 (3) = 95% × n C5 (f) = 0.95 × 135 kmol/hr = 128.25 kmol/hr n C6 (4) = 95% × n C6 (f) = 0.95 × 500 kmol/hr = 475 kmol/hr Overall balance on C3: n C3 (2) = n C3 (f) – n C3 (1) = (45 - 42.75) kmol/hr = 2.25 kmol/hr Overall balance on C6: n C6 (3) = n C6 (f) – n C6 (4) = (500 - 475) kmol/hr = 25 kmol/hr C6 is 98% of stream (4): n (4) = n C6 (4) /0.98 = 475 kmol/hr / 0.98 = 484.7 kmol/hr Then the remainder of stream (4) is C5 (because of our assumption that C4 is not present in that stream) n C5 (4) = (484.7 – 475) = 9.7 kmol/hr But we can see now that the total amount of C5 coming out of column 3 is : n C5 (3) + n C5 (4) = 128.25 + 9.7 kmol/hr = 137.95 kmol/hr, which is more than the feed of 135 kmol/hr! There is something not consistent here. We must have C4 present in stream (4), or perhaps the recovery values given for C5 or C6 are not correct. If C4 is present in stream (4), the bottoms of a column, it seems strange that C4 is not present in stream (3), the top of that column, because C4 is the lightest component in the column. There is in fact no single correct way to make this work. We need to accept some contradiction somewhere, or claim that the problem must have been stated incorrectly. Here we will work with the following: (definitely not the only way to solve the problem) Let’s accept that C4 is somehow present in stream (4), despite it not appearing in stream (3). We can further assume that C5 is not present in stream (2) and the remainder of C5 is present in stream (4). n C5 (2) = 0. n C5 (4) = 5% × 135 kmol/hr = 6.75. Then to make n (4) to be 484.7 kmol/hr, we get: n C4 (4) = n (4) – n C5 (4) – n C6 (4) = (484.7 – 475 – 6.75) kmol/hr = 2.9 kmol/hr. The rest of the streams can be found by performing balances around the columns, e.g., For each component: n (5) = n (f) – n (1) For each component: n (6) = n (3) + n (4) All these results are tabulated below. flow rates in kmol/hr Feed 1 2 3 4 5 6 C3 45 42.75 2.25 0 0 2.25 0 C4 225 8.3 213.8 0 2.9 216.7 2.9 C5 135 0 0 128.25 6.75 135 135 C6 500 0 0 25 475 500 500 total 905 51.1 216 153.25 484.7 853.9 637.9

Homework Set 1 Solution CHBE 3225 Fall 202 3 2) In this solution, we will treat the N 2 and O 2 mixture as one entity “air”, which always has N 2 : O 2 ratio of 4:1. So the mixture is simply of CO 2 and air. But if you treat N 2 and O 2 separately and list the mixture as CO 2 , N 2 , and O 2 , you should still get the same results, because the two species are always together in the same proportions. For gases at around 1 atm and 25°C (298 K) , it’s reasonable to ass ume ideal gas. We know that the minimum work of separation for an isothermal process into pure products is: W min = - n RT o (x 1 ln x 1 + x 2 ln x 2 ) where x 1 , x 2 = mol fractions of components 1 and 2 Since the problem involves separation not into pure products, we will have to take some extra steps. The approach below is slightly different from the method I went through in class in that I make an argument based on reversing separation into mixing, but really the same conceptually. First, let’s name the streams first and perform the mass balances: Feed = initial flue gas mixture, Product 1 = CO 2 -rich stream, Product 2 = air-rich stream We use as a basis 100 moles of Feed, which contains 10 vol% CO 2 . For ideal gas under constant T and P, since PV = nRT, volume fraction is the same as mol fraction. (The molar density n/V = P/RT is the same for all gases because P, R, and T are constants). Thus 100 moles of Feed contains 10 moles of CO 2 and 90 moles of air molecule. Since 90% of the CO 2 is captured, Product 1 contains 10 mol × 90% = 9 mol CO 2 . That stream has 95% purity of CO 2 . Then Product 1 contains 9 moles × (5% / 95%) air = 0.474 mol air Mass balance then gives us the composition of Product 2: 1 mol CO 2 and 89 Putting these on a table and calculating the mol fractions: CO 2 (mol) air (mol) total (mol) x CO2 x air Feed 10 90 100 0.1 0.9 Product 1 9 0.474 9.474 0.95 0.05 Product 2 1 89.526 90.526 0.011 0.989 We can write W min for each of the following steps because each involves separation into pure products: Feed → 10 mol CO 2 + 90 mol air W min,1 Product 1 → 9 mol CO 2 + 0.474 mol air W min,2 Product 2 → 1 mol CO 2 + 89.526 mol air W min,3 If we reverse the last two separations (into mixing), we get the overall process in the sum: Feed → 10 mol CO 2 + 90 mol air W min,1 9 mol CO 2 + 0.474 mol air → Product 1 – W min,2 1 mol CO 2 + 89.526 mol air → Product 2 – W min,3 + Feed → Product 1 + Product 2 W min,overall Basically, we get W min,overall = W min,1 – W min,2 – W min,3 . We use W min = - n RT o (x 1 ln x 1 + x 2 ln x 2 ) (for ideal gas, isothermal, separation into pure products) Feed: W min,1 = – (100 mol) (8,314 J/mol K)(298 K) (0.1 ln 0.1 + 0.9 ln 0.9) = 80.6 kJ Prod1: W min,2 = – (9.474 mol) (8,314 J/mol K)(298 K) (0.95 ln 0.95 + 0.05 ln 0.05) = 4.7 kJ Prod2: W min,3 = – (90.526 mol) (8,314 J/mol K)(298 K) (0.011 ln 0.011 + 0.989 ln 0.989) = 13.6 kJ W min,overall = W min,1 – W min,2 – W min,3 = 62.2 kJ for the basis of 100 moles of feed we have chosen The question asks for W min per mol of CO 2 . In our basis, we have captured 9 moles of CO 2 , so we get: W min = 62.2 kJ / (9 mol CO 2 ) = 6.9 kJ/mol CO 2 captured

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Homework Set 1 Solution CHBE 3225 Fall 202 3 b) On Aspen, we can look up the H and S values for each of the stream given above (all at 25°C and 1 atm). For example, we can specify the feed stream as the above, and feed it to a Sep2 block where the specifications are (for one of the outlet): Flow of CO 2 = 9 mol/sec, and the mole frac of CO 2 = 0.95. Using the IDEAL model set, we find these values: H (J/mol) S (J/mol K) B (J/mol) Feed -39,500 2.600 -40,300 Product 1 -374,000 3.879 -375,200 Product 2 -4,490 0.155 -4,500 We also find flow rates to be the same as in part a: n product 1 = 9.474 mol and n product,2 = 90.526 mol. B is calculated as H – ToS where To = 25°C = 298 K. Finally, W min = (nB) out – (nB) in = (nB) Product 1 + (nB) Product 2 – (nB) Feed where n = moles of stream (see part a) = 62.2 kJ for 100 moles of feed We have the same answer for W min total as in part b). Therefore, we also have the same answers 6.9 kJ/mol CO 2 captured 3) We can set up flash calculation as we discussed in class. 3 mass balances: z I F = y i V + x i L z i , y i , x i = mol fractions in F(eed), V(apor), and L(iquid) 3 equilibrium y i P = x i P i sat (T) P i sat = vapor pressure of i 2 sum of fractions Σ y i = 1 and Σ x i = 1 1 more equation: L = V from the given ratio. (We assume that the ratio is in moles, not mass). We have a total of 9 equations. For the equilibrium, we have written Raoult’s law, which applies if assuming ideal gas and ideal solution. We can get the dependence of P i sat on temperature using Antoine’s equation, the parameters of which can be found in various sources, e.g., textbooks from your earlier courses. From Smith, Van Ness, et al.: A B C ln P sat = A – B/(T+C) methanol 16.5785 3638.27 239.500 with P in kPa, T in °C. water 16.3872 3885.70 230.170 acetone 14.3145 2756.22 228.060 Since we know the feed composition and flow rate as well as temperature and pressure, the unknowns are T, y water , y acetone , y methanol , x water , x acetone , x methanol , L and V. We thus have 9 unknowns and 9 equations. Note: We can reduce these numbers if we substitute some of the equations. For example, instead of writing a mass balance for water, we can write 100 mol/hr = L + V. Using L = V, we can find immediately that L = V = 50 mol/hr. We are then left with 7 equations with 7 unknowns. We can also write x acetone = 1 – x methanol – x water and y acetone = 1 – y methanol – y water , leaving 5 equations with 5 unknowns. Solving these equations gives us a temperature of 73.4°C, and compositions: Liquid Vapor flow rate (mol/hr) 50 50 methanol mol% 41.4% 58.6% water mol% 44.1% 15.9% acetone mol% 14.5% 25.5%