Problems CRE

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


3/22

-rA = -dCA/dt = k CACB

i.e., -rA = k (CAo - CAoXA) (CBo - CBoXB)


4/22

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


5/22

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


6/22

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


7/22

24.75(1 - 2XA,1 + XA,12) = XA,1

24.75 - 50.5 XA,1 + 24.75 XA,12 = 0

XA,1 = 0.82


8/22

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


9/22

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


10/22

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%.


11/22

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


12/22

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


13/22

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.


15/22


16/22

For the second o rder equat ion, equn.1 can be written ast = CAoXA/ kCA

2


17/22

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)


18/22


19/22

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


20/22

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?


21/22

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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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.

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