problems cre
TRANSCRIPT
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7/30/2019 Problems CRE
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7/30/2019 Problems CRE
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Two stirred tank reactors are available at a chemical plant,one of volume 100 m3 and the other of volume 30 m3. It issuggested that these tanks be used as a two stage CSTRfor carrying out an irreversible liquid phase reaction,
A + B --
Products. The two reactants only will be presentin the feed stream in equimolar amounts, CAo = CBo = 1.5gmol/lit. The volumetric feed rate will be 20 lit/min. Thereaction is first order with respect to each of the reactantsA and B i.e., second order overall. The rate constant is0.011 lit/(gmol.min). Which tank should be used as the first stage for higher overall
conversion?
With this arrangement, calculate the overall conversionobtainable under steady state conditions
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7/30/2019 Problems CRE
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-rA = -dCA/dt = k CACB
i.e., -rA = k (CAo - CAoXA) (CBo - CBoXB)
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7/30/2019 Problems CRE
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The amounts of A and B which have reacted at
any time t are equal and given by CAoXA =
CBoXB
And for the given problem CAo = CBo
Therefore, -rA,i = k CAo2(1 - XA,i)
2
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7/30/2019 Problems CRE
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Taking smaller reactor as the first reactor,
volume (V1)= 30 m3
The design equation for CSTRs in series is,
ti/CAo = (XA,i - XA,i-1) / (-rA)i
where CAo = Initial concentration of A into the
first reactor = 1.5 gmol/lit Volumetric flow rate (v) = 20 lit/min
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7/30/2019 Problems CRE
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t1 = V1/v = 30 x 1000 /20 = 1500 min
1500/ CAo = (XA,i - XA,i-1)/ k CAo2(1 - XA,i)
2
1500 = (XA,1 - 0)/ (0.011 CAo(1 - XA,1)2
) 1500 = XA,1/(0.011 x 1.5 x (1 - XA,1)
2
24.75 = XA,1/(1 - XA,1)2
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7/30/2019 Problems CRE
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24.75(1 - 2XA,1 + XA,12) = XA,1
24.75 - 50.5 XA,1 + 24.75 XA,12 = 0
XA,1 = 0.82
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7/30/2019 Problems CRE
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For the second reactor, t2 = V2/v = 100 x 1000
/20 = 5000 min
5000/ CAo
= (XA,2
- 0.82)/ k CAo
2(1 - XA,2
)2
5000 x 0.011 x 1.5 = (XA,2 - 0.82)/ (1 - XA,2)2
82.5 (1 - 2XA,2 + XA,22) = XA,2 - 0.82
83.32 - 166 XA,2 + 82.5 XA,22
= 0 XA,2 = 0.96
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7/30/2019 Problems CRE
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If the reactor is arranged with bigger reactor asthe first one:
For the first reactor:
t1 = V1/v = 1000 x 1000 /20 = 5000 min 5000/ CAo = (XA,1 - 0)/ k CAo
2(1 - XA,1)2
5000 = XA,1/(0.011 x 1.5 x (1 - XA,1)2
82.5 = XA,1/(1 - XA,1)2
82.5(1 - 2XA,1 + XA,12) = XA,1 82.5 - 166 XA,1 + 82.5 XA,1
2 = 0
XA,1 = 0.90
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7/30/2019 Problems CRE
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For the second reactor, t2 = V2/v = 30 x 1000 /20 =1500 min
1500/ CAo = (XA,2 - 0.90)/ k CAo2(1 - XA,2)
2
1500 x 0.011 x 1.5 = (XA,2 - 0.90)/ (1 - XA,2)2
24.75 (1 - 2XA,2 + XA,22) = XA,2 - 0.90 25.65 - 50.5 XA,2 + 24.75 XA,2
2 = 0
XA,2 = 0.92
From the above calculations, it is seen that thereactor with the smallest volume should be thefirst one. And overall conversion for thisconfiguration is 96%.
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7/30/2019 Problems CRE
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Problem 2
From the following data find a satisfactory
rate equation for the gas-phase
decomposition of pure A, A R + S, in a
mixed flow reactor.
t based on
inlet feed
conditions,
sec
0.423 5.1 13.5 44.0 192
XA (for CAo =
0.002
mol/lit)
0.22 0.63 0.75 0.88 0.96
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7/30/2019 Problems CRE
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Calculations:
For the mixed flow reactor, the design equation is,
t = CAoXA/-rA 1
For variable density systems,
Where, eA = fractional change in volume = (VXA = 1 - VXA = 0) / VXA = 0 = (2 - 1)/1 =1
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7/30/2019 Problems CRE
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Therefore,
2
For the f irst order equat ion, equn.1 can be written as
t = CAoXA/ kCA 3
substituting for CAo
in equation 3,t = 0.002 XA/ kCA
substituting for CA from equn.2,
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t, sec 0.423 5.1 13.5 44 192
XA 0.22 0.63 0.75 0.88 0.96
XA(1 + XA)/(1 - XA) 0.3441 2.7754 5.25 13.7867 47.04
That is, if the given data is for a first order reactiont vs. XA(1 + XA)/(1 - XA) will be a straight line, with a slope of k.
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7/30/2019 Problems CRE
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For the second o rder equat ion, equn.1 can be written ast = CAoXA/ kCA
2
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t, sec 0.423 5.1 13.5 44 192
XA 0.22 0.63 0.75 0.88 0.96
XA(1 + XA)2/(1 - XA)
2 0.5382 12.2268 36.75 215.99 2304.96
That is, if the given data is for a second order reaction
t
vs. XA(1 + XA)2
/ (1 - XA)2
will be a straight line with a slope of kCAoFrom the above data, the following graph is drawn :(graph 2)
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7/30/2019 Problems CRE
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7/30/2019 Problems CRE
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By comparing two graphs it is seen that first order kinetics is
well fitting the given data.That is, the reaction is following first order.
Slope = 0.2493 (from the graph.1)
Therefore,
k = 0.2493 sec-1
.Therefore, rate equation for the given reaction is
-rA = 0.2493 CAo (1 - XA)/(1 + XA) = 4.986 x 10-4 (1 - XA)/(1 + XA)
mol/lit.sec
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7/30/2019 Problems CRE
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Problem 3
An elementary liquid phase reaction
(irreversible first order) A R, takes place in a
PFR and the conversion is 96%. If a mixed
flow reactor of 10 times as large as the PFR ishooked up in parallel with the existing unit,
by what fraction could the production be
increased for the same 96% conversion?
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7/30/2019 Problems CRE
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Design equation for PFR: (first order reaction)
k t = -ln (1 - XA)
Design equation for CSTR: (first order
reaction)
k t = XA / (1 - XA)
For the 96% conversion in PFR,
k t = -ln (1 - 0.96)
k t = 3.219
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7/30/2019 Problems CRE
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k VPFR / vPFR = 3.219
for the same 96% conversion in CSTR,
k t = 0.96/(1 - 0.96) = 24
The CSTR to be used is of 10 times as large as the PFR.
Therefore, k VCSTR / vCSTR = 10 k VPFR / vCSTR = 24
i.e., 10 x 3.219 vPFR / vCSTR = 24
vPFR / vCSTR = 0.7456
Total volumetric rate = vPFR + vCSTR = (1 + 1/0.7456) vPFR =2.341 vPFR
That is the production will increase to 2.341 times of thepresent production rate.