HW 4 Solution 2015

# HW 4 Solution 2015 - CHE4473 Kinetics Spring 2015 Homework...

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Unformatted text preview: CHE4473 Kinetics Spring 2015 Homework N°4: Design of Ideal Isothermal Reactors (Variable Density) Due Tuesday FEB 3rd 2015 In all these problems you should notice that when the reaction stoichiometry shows a variation in the number of moles and the reaction is in the gas phase concentration is a function of conversion _____________________________________________________________________________________________________ 1. A homogeneous gas-­‐phase reaction A è3R has a reported rate expression at 215°C -­‐rA = 10-­‐2 CA½ [mol/Liter sec] Find the space-­‐time (VR/v0) needed to reach 80% conversion (XA) of a 50% A-­‐50% inert feed to a plug flow reactor (PFR) operating at 215°C and 5 atm (CAo = 0.0625 mol/Liter). __________________________________________________________________________________________________ 2. The homogeneous gas-­‐phase decomposition of phosphine 4 PH3(g) è P4(g) + 6 H2 (g) proceeds at 649°C with the first-­‐order rate -­‐r(PH3) = (10/h) CPH3 a) What size of plug flow reactor (PFR) operating at 649°C and 460 kPa can produce 80% conversion of a feed consisting of 40 mol/h of pure phosphine? b) What size of plug flow reactor (PFR) operating at 649°C and 11.4 atm is needed for 75% conversion of 10 mol/hr of phosphine in a 2/3 phosphine-­‐1/3 inert feed? __________________________________________________________________________________________________ 3. A gaseous feed of pure A (1 mol/Liter) enters a CSTR mixed flow reactor (2 Liters) and reacts as follows: 2A è R -­‐rA = 0.05 CA2 [mol/Liter. Sec] Find what feed rate (v0) (liter/min) will give an outlet concentration CA = 0.5 Mol/Liter. __________________________________________________________________________________________________ 4. Gaseous reactant A decomposes as follows: A è 3R -­‐rA = (0.6 min-­‐1) CA Find the conversion of A (XA) in a 50% A-­‐50% inert feed (v0 = 180 liter/min, CAo = 300 mmol/Liter) to a 1 m3 CSTR mixed flow reactor. __________________________________________________________________________________________________ 5. A 1 Liter/sec flow of a 20% ozone-­‐80% air mixture passes through a plug flow reactor (PFR) at 1.5 atm and 93°C. Under these conditions ozone (O3) decomposes by the following homogeneous reaction 203 è 302 , -­‐rozone = k Cozone2 k = 0.05 [Liter/mol . sec] What reactor size is needed for 50% decomposition of ozone? __________________________________________________________________________________________________ 6. For the first-­‐order reaction A è B on a CSTR, start from the CSTR design equation to derive the expression kτ = X/(1-­‐X) – What assumptions are made? Obtain an equivalent equation for the second-­‐order reaction 2A è 5B (gas phase). ChE 4473 Homework 4 Solution Spring 2015 1. For this stoichiometry and with 50% inerts, two volumes of feed gas would give four volumes of completely converted gas; thus: tau = V/v0 4−2 ! = = 1 2 pfr design eq: Using the plug flow design equation: V = Fa0* dX / (-ra) = !! !!" ! !!!" ! = !! −! ! ! ! ! !! 1 − ! 1 + ! ! ! ! = ! ! !! !.! ! 1 + ! 1 − ! ! ! ! We can evaluate this integral analytically, graphically, or numerically. Here, a numerical method is probably best, with Simpson’s rule serving as one viable option: !.! ! = 1 + ! 1 − ! ! ! ! = ℎℎ ℎ 1 1 + 4 1.227 + 2 1.528 + 4 2 + 1 3 0.8 = 1.331 12 With the integral evaluated, we have: . = ! . = . 2. a) Let = ! , = ! , = ! , then the reaction becomes: 4è + 6 10 −! = ℎ ! The volume of a plug flow reactor with a first order reaction and a vapor-­‐phase flow with variable density is given by: derived from pfr deq !! 1 = 1 + ! − ! ! !! 1 − ! We can evaluate the terms in this expression as follows: !! = 40 ℎ 10 = ℎ !! 460000 !! = = = 60 /! ! 8.314 922 7−4 ! = = 0.75 4 ! = 0.8 Plugging in values and evaluating the expression for volume, = . = b) !! = 10 ℎ 10 = ℎ !"#\$%,! = ! 11.4 = = 0.151 / . 08206 922 !! = !! !"#\$%,! = ! = !! = 2 0.151 = 0.100 / 3 2 6 1 ∗ + − 1 = 0.5 3 4 4 ! = 0.75 Plugging in to original equation: !! 1 = 1 + ! − ! ! !! 1 − ! 10 ℎ 1 = 1 + 0.5 ln − 0.5 ∗ 0.75 = . 10 1 − 0.75 ∗ 0.100 ℎ 3. !"# 2A è R with – ! = 0.05!! ! ! Evaluate terms: ! = solve for Xa from X = (Fa0 - Fa) / (Fa0) ! = 1−2 = −0.5 2 !! − ! 1 − 0.5 2 = = !! + ! ! 1 + −0.5 0.5 3 So, for mixed flow CSTR: −! 2 0.05 ∗ 0.5! = = = 0.0376 = . 2 !! ! 1 3 4. A è 3R occurs in a CSTR. −! = 0.6!! ! Feed is 50% A + 50% inert: !! = 0.5 ! ! = 180 !"# , = 1000 , !! = 0.3 / Start with CSTR design equation: !! !! ! = = −! !"# ! We have a gas phase reaction with a changing number of moles so we need to calculate . 3 = − 1 = 2 1 = !! = 1 Now put ! in terms of known quantities and conversion. ! !! 1 − !! 1 − ! = = = 1 + Substitute and rearrange: = 1 + ! 1 − Plug in quantities for LHS and solve for : = . 5. The reaction 2! → 3! occurs in a PFR. Let ! = ! = . −! = 0.05 ! ! We are given: !! = 0.20, ! = 1 = 1 , = 93 ° And we need to find the reactor volume at ! = 0.5. The following expression is for a 2nd order reaction occurring in a PFR with variable gas-­‐ phase density: ! + 1 ! ! = 2! 1 + ! ln 1 − ! + !! ! + !! 1 − ! Evaluate terms: !! 1.5 0.2 !! = = = 0.01 0.08206 366 11 − 10 ! = = 0.1 10 Replacing in the equation for volume: 1 0.1 + 1 ! 0.5 = 2(0.1) 1 + 0.1 ln 1 − 0.5 + 0.1 ! (0.5) + (0.01)(0.05) 1 − 0.5 = = . 6. Case 1: A è B Beginning with CSTR design equation: = −! = ! = !! 1 − !! = !! ! Sub in values: = ! Simplify and rearrange, with = ! : !! −! !"# !! ! !! 1 − ! = − We have assumed well-­‐mixed, isothermal, isobaric operation and first-­‐order kinetics with no phase change. Most importantly, we have assumed that the density of the vapor phase remains constant over the course of the reaction, as there is no change in number of moles due to the reaction. Case 2: 2A è5B Here, we will need to account for the change in gas-­‐phase density due to the reaction stoichiometry. We can no longer make the assumption that ! = !! 1 − . !! = −! !"# −! = !! 5 3 = − 1 = 2 2 Assuming the feed is 100% A: 3 = !! = 2 Now we can write the concentration accurately: ! !! 1 − !! 1 − !! 1 − !! 1 − ! = = = = = 3 ! 1 + 1 + 1 + 2 Appropriately substituting into the design equation: !! ! = ! !! 1 − 1 + Rearranging into the form of the first part of the problem: + = − ...
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• Spring '14
• LanceL.Lobban
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