HW 6 S (1) - Solutions to Homework 6 1 We can write all the...

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Solutions to Homework 6 1. We can write all the molar flowrates: F A = F A 0 (1-x) F B =3F A 0 x F I =F I 0 =F A 0 F Tot =2F A 0 (1+x) Considering ideal gases, we can write the concentration of any species as: ? ? = 𝑃 ? 𝑅𝑇 = 𝑃? ? 𝑅𝑇 = ? 𝑇 0 ? ? ? 𝑇 For A, this means: ? ? = ? 𝑇 0 (1 − ?) 2(1 + ?) The mole balance for the PFR yields, ? ? 0 ?? ?𝑉 − ?? ? = 0 ? ? 0 ?? ?𝑉 − ?? 𝑇 0 (1 − ?) 2(1 + ?) = 0 This means, ?? ?𝑉 = ?? 𝑇 0 (1 − ?) ? ? 0 2(1 + ?) As the input flor is equimolar in A and inert, we can say ? 𝑇 0 = ? ? 0 + ? ? 0 = 2? ? 0 This makes, ?? ?𝑉 = ?2? ? 0 (1 − ?) ? ? 0 2(1 + ?) = ? 𝑣 0 (1 − ?) (1 + ?) Taking v 0 =1, we can perform Euler’s method in Excel for x: ? (?) = ? (?−1) + ?? ?𝑉 ∆𝑉
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We do steps in V, until V=1 and we use Solver to make x= 0.5 at V=1 adjusting the value of k. We get k=0.881 The balance for the CSTR, ? ? 0 2 − ? ? 0 (1 − ?) − ?? ? 0 (1 − ?) (1 + ?) 𝑉 = 0 And this yields x=0.6725 which is the result we got last week with the analytical method. 2. We can write all the molar flowrates, like so: F A = F A 0 (1-x) F AA =0.5F A 0 x F H2 =0.5F A 0 x F Tot =F A 0 Now, the equilibrium constant is 𝐾 = ? ?? ? ?2 ? ? 2 = ? ?? ? 𝑇 ? ?2 ? 𝑇 ( ? ? ? 𝑇 ) 2 = ( 1 2 ? ? 0 ?) 2 [? ? 0 (1 − ?)] 2 = ? 2 4(1 − ?) 2 This yields x Eq =0.4. b) Now, we remove 90% of the H2 formed. That means that F H2 removed by membrane =0.9*F H2 F A = F A 0 (1-x) F AA =0.5F A 0 x F H2 =0.5F A 0 x - 0.9*0.5F A 0 x F Tot =F A 0 (1-0.45*x) Repeating the procedure in a), we get x Eq =0.679, which is higher (as was to be expected) than what we got previously. 3. We can write the rate expression like this: 𝑟 = ? 𝑓 [? ? ? ? ? ? 2 𝐾 ] Writing the mole balances for all the species involved,
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A. 𝑑? ? 𝑑𝑉 = −𝑟 B. 𝑑? ? 𝑑𝑉 = 𝑟 − 𝑅 ? C. 𝑑? ? 𝑑𝑉 = 2𝑟 We know R B =k c C B and that ? ? = ? 𝑇 0 ? ? ? 𝑇 = 𝑃 𝑅𝑇 ? ? ? 𝑇 With this we can perform Euler’s method in Excel: Step V F A F B F C F T C A C B C C r R B 0 0 F A 0 0 0 SUM Fi C T 0 *F A /F T ?
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