HW 3 Solution 2015 - CHE4473'Kinetics Spring'2015...

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Unformatted text preview: CHE4473'Kinetics' Spring'2015' Homework(N°3:((Design(of(Ideal(Isothermal(Reactors((Constant(Volume)((( Due'Tuesday'JAN'27th'2015' _____________________________________________________________________________________________________" 1.""We"are"operating"a"batch(reactor"to"convert"A"into"R."This"is"a"liquid"reaction,"the" stoichiometry"is"A!R,"and"the"rate"of"reaction"is"given"in"the"table"below."How"long" must"we"react"for"the"concentration"to"drop"from"CA0"="1.3"to"CAf"="0.3"mol/liter?" " '''''CA"mol/liter"G"" rA"mol/liter"min" " '''''CA"mol/liter"G"" rA"mol/liter"min" 0.1"" " " 0.1" " " " 0.7"" " " 0.10" 0.2"" " " 0.3" " " " 0.8"" " " 0.06" 0.3"" " " 0.5" " " " 1.0"" " " 0.05" 0.4"" " " 0.6" " " " 1.3"" " " 0.045" 0.5"" " " 0.5" " " " 2.0"" " " 0.042" 0.6"" " " 0.25" _____________________________________________________________________________________________________" 2.""The"aqueous"decomposition"of"A"is"studied"in"an"experimental"CSTR."The"results"in" the"table"below"are"obtained"in"steadyGstate"runs."To"obtain"75%"conversion"of"reactant" in"a"feed,"CA0"="0.8"mol/liter,"what"space"time"is"needed"in"a"plug(flow(reactor?" " """"Concentration"of"A,"mol/liter" " """"Space"Time" """In"Feed"" """" """"In"Outlet"" " """"""""""(sec)" " 0.65" 2.00" 300" 2.00" 0.92" 240" 2.00" 1.00" 250" 1.00" 0.56" 110" " " " 0.37" 1.00" 360" 0.42" 24" 0.48" 0.28" 0.48" 200" 0.20" 0.48" 560" " 3.""The"reaction"A"+"½"B"!"2"C"has"a"rate"law"–rA"="k"*"CA"*"CB"and"rate"coefficient"k"=" 0.05"liter/mol"min.""" A. The"reaction"is"conducted"in"a"wellGmixed(batch(reactor"(liquid"phase)"with" initial"concentration"CA0"="2"M;"CB0"="1"M.""Obtain"the"time"needed"to"reach"50"%," 95"%"and"99"%"conversion.""Note"that"because"the"ratio"of"initial"concentrations" of"A"and"B"match"their"stoichiometric"ratio,"CA"will"always"double"CB.,"so"this" problem"can"be"easily"solved"analytically."Why"to"reach"99"%"conversion"we"need" a"time"much"longer"than"twice"that"needed"to"reach"50"%"?"" B. Use"Polymath,"EXCEL,"or"the"software"of"your"choice"to"develop"a"numerical" solution"for"the"problem"of"part"A"and"compare"with"the"analytical"solutions"for" the"three"different"conversions.""" C. Obtain"the"numerical"solution"to"get"50"%"conversion"in"the"batch"reactor,"except" that"CA0"="1.2"M;"CB0"="2"M.""In"this"case,"the"analytical"solution"would"have"been" much"more"complicated"since"CA0" "2"CB0" ChE 4473 Homework 3 Solution Spring 2015 1. We might first think to approach this by finding a rate expression to represent the data, and then integrate the following to give time: !! ! = excel? !!! −! However, a quick plot of the data shows that a simple power law rate expression will not fit the data, as the rate begins to decrease with increasing concentration: 0.70# 0.60# rA,$mol/L$min$$ 0.50# 0.40# 0.30# 0.20# 0.10# 0.00# 0.00# 0.50# 1.00# 1.50# CA,$mol/L$$ 2.00# 2.50# With a bit more thought we see that we do not need a rate expression, but just the time to react the desired conversion. This can be done by directly solving the design graphically. ! We plot !! versus ! and integrate by finding the area under the curve between ! = 0.3 ! and !! = 1.3 : 25.0# rA,$mol/L$min$$ 20.0# 15.0# 10.0# 5.0# 0.0# 0.00# 0.50# 1.00# 1.50# 2.00# 2.50# CA,$mol/L$$ There are a number of ways to evaluate this integral numerically. I used a simple middle Riemann sum and arrived at = . . Answers should be within 10% of this but may vary. 2. We can find the rate of reaction in the CSTR using the concentrations and space time, as follows: cstr design eq !! − !,!"# −! = -r*V = Fa0 - Fa = v0(Ca0 - Ca) τ CA#feed,#mol/L CA#out,#mol/L Space#time,#s 1/6rA 6rA 0.48 0.2 560 2000 0.0005 0.48 0.28 200 1000 0.0010 1 0.37 360 571 0.0018 0.48 0.42 24 400 0.0025 1 0.56 110 250 0.0040 2 0.65 300 222 0.0045 2 0.92 240 222 0.0045 2 1 250 250 0.0040 We can now use the rate data to evaluate the following expression for the PFR: !! ! τ=− !!! −! When we plot ! versus – ! , we get a plot like this: We see that the data will not fit a simple power law rate expression, so let us evaluate the integral graphically: 2500" 1/#rA& 2000" 1500" 1000" 500" 0" 0" 0.2" 0.4" 0.6" 0.8" 1" 1.2" CA,&mol/L& !.! = τ = !.! ! = −! I evaluated this with a middle Riemann sum, other numerical techniques will generate different answers, but they should be within about 10% of this. Also note that this could have been equally well evaluated in terms of conversion with a traditional Levenspiel plot method. 3. a) dN/dt = rV N = C*V = ! ! V is a constant, −! = ! ! , ! = !! 1 − ! ! ! ! And since !! = ! !! , we can write ! = ! ! = ! !! (1 − ! ) 1 !! 1 − ! = − !! 1 − ! 1 − ! 2 !! Rearrange, separate variables, and integrate: !! ! ! 1 = 1 − ! ! 2 !! ! ! Solve for t: 2! = !! 1 − ! Plug in for various conversion: !!!.! = 20 !!!.!" = 380 !!!.!! = 1980 The reaction time increases so much because, as the reaction proceeds, the rate decreases rapidly (both A and B are depleted). Thus the final “bits” of conversion take much longer than the initial. b) Input the following differential equations into Polymath, or create an Excel spreadsheet to numerically evaluate them: ! = −! ! ! 1 = − ! ! 2 With !! = 2, !! = 1, = 0.05. The results should be approximately equivalent to those in part a, but may vary within 5% or so. The match with the analytical results depends on the step size and method with either software. c) We have the same differential equations, but with different initial values than part b. It should be very easy to modify the program used for b. !!!.! = 7.5 !!!.!" = 38 !!!.!! = 61 Again, actual results may vary within around 5%. ...
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  • Spring '14
  • LanceL.Lobban

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