Reactoon Eng Crswrk 2

Ibrahim Aliyu AhmedID: 200480447 Module leader: Dr Xiaojun Lai

PEME 3311: Reaction Engineering


Question 1.

We wish to explore various reactor setups for the transformation of A into R. The

feed contains 99% A, 1% R; the desired product is to consist of 10% A, 90% R.

The transformation takes place by means of the elementary reaction

A + R → R + R

with rate constant k = 1 litre/(mol . min). The concentration of active materials is

CA0 + CR0 = CA+ CB = C0 =1

throughout.

What reactor holding time will yield a product in which CR = 0.9 mol/litre

(a) In a plug flow reactor,

(b) In a mixed flow reactor,

(c) In a minimum-size setup with recycle?

(d) In a minimum-size setup of combined 2-reactor system without recycle?

Solution

a) The system is at constant density since it does not change F total.

−r A=k C ACR

CR=1−CA

−rA= (1 )C A (1−C A )=CA (1−C A)

Since,

C A0=0.99 (1 )=0.99mol/L

Then, we can write;

τp=∫0.1

0.99 dCA

−r A

=∫0.1

0.99 dCA

C A(1+C A)

Ibrahim Aliyu Ahmed 200480447 2


τp=−11

ln1−CA

C A∫0.1

0.99

¿−ln1−0.99

0.99+ln

1−0.10.1

=6.79min=407.4 s

b) As for the mix flow reactor, the time is;

τm=CAo X A

−rA=C Ao−C A

−r A

=CAo−C A

CA (1−C A)= 0.99−0.1

0.1(1−0.1)=9.89min=593.4 s

c) A graph is drawn to show the variation of –rA with the CA

Table 1: showing data for CA –rA used to plot the graph.

CA 0.99 0.8 0.6 0.4 0.2 0.1(-rA) 0.009 0.16 0.24 0.24 0.16 0.09

0 0.2 0.4 0.6 0.8 1 1.20

0.05

0.1

0.15

0.2

0.25

0.3

-rA vs CA

CA

-rA

Figure 1: showing the plotted graph of –rA against CA

After analysis of the graph above, it was found out that the flow rate of CA is at the maximum point. So the value can be seen below.

d (−rA)dCA

=¿C A (− 1) + (1 − C A) (1) = −C A + 1 − C A = −2C A + 1 = 0

Therefore,

CA = 0.5 mol/L



Fig:

τm= 0.99−0.50.5(1−0.5)

=1.96minutes

τm=∫0.1

0.5 dCA

C A (1−C A )=−1

1ln

1−C A

CA∫0.1

0.5

¿−ln1−0.5

0.5+ ln

1−0.10.1

=2.197minutes

τTotal=1.96+2.197=4.15minutes

d)

Figure 2: showing a minimum-size set up of a combined reactor (mixed flow and

plug flow reactor)

In a minimum –size setup of combined reactor system without recycle. The optimum set up would be a mixed flow reactor up to the point having maximum rate followed by a plug flow reactor. Therefore we obtain a point of maximum rate:

−τ A=K C ACR=1∗CA∗(1−C A)

−r A=C A (1−C A )=CA−C A2

At maximum rate dr A

dC A

=0

0=1−2CA

C A=0.5



But X A=CAo−CA

CAo

=0.99−0.50.99

=0.495

Figure 3: diagram of a minimum set up

τ m=CAo xA

−r A

¿ 0.99−0.495

0.5−0.52=1.9602

C p=CAo ∫0.495

0.9 dxA

−r A

C AO×areaunder curve=0.99×2.694687=2.6677

So, the total holding time = τ m+τ p=1.9602+2.6677=4.63minutes

Question 2

We plan to replace the present MFR with one having double the volume. For the

same aqueous feed (10 mol/L) and same feed rate find the new conversion. The

reaction kinetics is;

A → R, -rA = k CA1.5



And present conversion is 70%.

Solution

Figure 4: showing the exiting reactor

So, we can write;

Vvo

=C Ao X A

KCA o

1.5 (1−XA)1.5=0.7

KCAo (0.3 )1.5

0.5 =kVCAo

0.5

vo=4.26

Ibrahim Aliyu Ahmed 200480447

Vo

Vo

XA = 0.7

CAo = 10 mol/L

Vo

Vo

XA ’

CAo = 10 mol/L

6


Figure 5: showing the doubled volume reactor

Then,

2Vvo

=C AoX ' A

kC Ao1.5(1−X 'A)

1.5=2kVCAo

0.5

vo=2 (4.26 )=8.52

M=X ' A

(1−X 'A)1.5=8.52

As for XA , it was found by making some approximations. This value was found out to be;

XA’=0.794

The table below shows the data used to calculate XA .

Table 2: showing the data for XA’ and M

XA’ 0.8 0.75 0.77 0.79M 8.94 6 6.98 8.21

0.74 0.75 0.76 0.77 0.78 0.79 0.8 0.810123456789

10

graph of M vs XA

XA

M

Figure 6: a graph of plotted value of M against XA’.

Question 3

Reactant A in a liquid either isomerizes or dimerizes

A → R (desired), rR = k1 CA



A + A → S (unwanted), rS = k2 CA2

I. write (R/A) and [R/(R+S)]

II. If in batch reactor, CA0 =1 mol/L. When reaction is completed (i.e.

CA =0), CS = 0.18mol/L in resultant mixture, what is the ratio of rate

constants – k1/k2?

III. Find CRf when XA=80% in PFR

IV. Find CRf when XA=80% in MFR

Solution

(I)

φ ( RA )= r R

−rA=

k1C A

k 1C A+2k2CA2

φ ( RR+S )= rR

r R+rS=

k1C A

k 1CA+kCA2

(II) C s=CR=C Ao−CA−C s=1−0−0.18=0.82

CRmax=k1

2k2(1+

2k2

k1

C Ao)

table 2: showing the values for K and CR

K=k1/k2 5 4

CR calculated by (1) 0.84 0.81



4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 50.795

0.80.805

0.810.815

0.820.825

0.830.835

0.840.845

CR vs K

K = k1/k2

CR

Figure 7: showing graph of CR against K

The product of distribution for a reactor of flow in piston is similar to the idea discontinuous reactor.

Hence, the following values were found to be;

K=4.32 ,k 1

k 2

=4.32∧k 1=4.32k2

(III)CRf=? x A=80 %(0.8) For PFR

c Rf=−1∆C A

∫C Ao

C Rf K 1C A

K1C A+2 K2CA2dC A

C Ao−1mol/Litre

CRf=1 (1−0.8 )=0.2mol /Litre

CRf=−1∆C A

∫1

0.21

1+2 K2

K1

C A

dC A

CRf=−1∆C A [ K1

2K2

ln( K 1

2K 2

+C A)]∫10.2

❑

¿− 1∆CA

[ 2.16 ln 2.36−2.16 ln 3.16 ]=0.788mollitre



(IV) CRf=? x A=80%(0.8) For MFR

CRf= yield∗(CAo−CA)

¿( K1C A

K1CA+2K 2C A2 )(C Ao−C A)

¿ [ 1

1+2 K2

K1

C A ]∆C A

¿ [ K1

2 K2

K1

2 K2

+CA ]∆C A

¿ [ 2.162.16+0.2 ] [ 1−0.2 ]=0.7322

mollitre

Question 4

Chemical X, a powdered solid, is slowly and continuously fed for half an hour

into a well-stirred vat of water. The solid quickly dissolves and hydrolyses to Y,

which then slowly decomposes to Z as follows:

Y → Z, -rY = kCY, k = 1.5hr-1

The volume of liquid in the vat stays close to 3 m3 throughout this operation, and

if no reaction of Y to Z occurred, the concentration of Y in the vat would be 100

mol/m3 at the end of the half-hour addition of X.



(a) What is the maximum concentration of Y in the vat and at what time is this

maximum reached?

(b) What is the concentration of product Z in the vat after 1 hour?

Solution 4

The material balances for this kind of reaction can written as;

Addition = reaction + accumulation

There are two types of processes in this case:

1) Between 0 to 15mins Addition, reaction and accumulation

2) After 15mins Reaction and accumulation

For the first process,

We have the rate of addition of Y below;

Molesaddedmin

=100

mol∗3m3

m3

15min=20

molmin

The initial concentration of CY is 0 until the rate of reaction starts to increase exceeding 20mol/min. This may not take place within the 15min of addition. The maximum concentration of Y is at the period of 15minutes and over. Therefore, the maximum rate of reaction for the first 15 minutes can be estimated.

Maximum possible reaction rate15h

100mol

m3

1h60min

=2.5molmin

Due to the reaction, the speed of the reaction during the initial 15 minutes will be

decreased because the value of CY which is 100 mol/m3 will never be reached.

Since the addition rate = 20 mol/min, CY then increases at the point of the

addition.

So,

Addition=reaction+Accumulation

20molmin

=1.5h

1h60min

CY (3m3 )+ ddt

(3CY )



dCY

dt=0.025CY=

203

CY=e−∫ P (t )dt

dt+constant

Where; P (t) = 0.025 y Q (t) =( 203 )

CY=e−0.025 t {800

3e0.025 t+ constant}

When t = 0 CY = 0 where the constant = - 800

3

CY=800

3e−0.025 t {e0.025t−1}=800

3{1−e−0.025t (15)}=83.39

mol

m3

the maximum value of CY is 83.39mol/m3 .

moles of Y reacted = 100-83.39 = 16.61 mol/m3

Another method to solve question 4:

For this solution we have;

X Y Z

n1 = 0 n2 = 1, k2 = 1.5hr-1

C10 = 1—mol/m3

Where 100mole/m3

1 /2hr=200

molm3hr

Therefore,

K=k 2CXo

k1

=(1.5 )(100)

200=0.75

(a) Y max can be obtain at 30 minutes



CYmax

CXo

=1k

(1−e−k )= 10.75

(1−e−0.75 )=0.7035

C y max=0.7035∗100=70.35mol

m3

(b) after the period of 30 minutes, it was observed that X has disappeared while Y and Z present

CY

CXo

= 1K

(e+k−k2 t−e−k2 t )= 10.75

(e−0.75−1.5−e−1.5 )=0.3323

Therefore;

CZ

CXo

=1−0.3323=0.6677

C z=0.6677∗100=66.77mol

m3

Question 5

The reversible 1st order gas reaction

A R

is to be carried out in a mixed flow reactor. For operations at 300 K the volume

of reactor required is 100 litres for 60% of conversion of A.

1. What are the equilibrium conversion rates at 300k and 400K

respectively?

2. What is the activation energy for k2?

3. What should be the volume of the reactor for the same feed rate and

conversation but with operations at 400 K?



Data:

k1 = 103 exp (-2416/T)

Hr = -8000 J/mol at 300K

Cp = CpR – CpA =0

K = 10 at 300 K

Feed consists of pure A

Total pressure is constant

Solution 5

For this solution, it must be noted that V0 varies with changes in temperature and also, HΔ r is constant due to Cp = 0.Δ

So,

m=V m

vo=CAo X A

−rA=

C AoX A

k 1C Ao (1−X A ) – ¿¿

vo=Vm(k1 (1−X A)−k2 X A

X A)

k 1( 130k )=0.318min−1

k 2(300k)=k1

K=0.318

10=0.0318min−1

vo=100[0.318 (1−0.6 )−0.0318 (0.6 )]

0.6=18.02

Lmin



K400K=K300K exp [−∆H r

R ( 1400

− 1300 )]=10 exp [−8000

8314 ( 1400

− 1300 )]=4.485

k 1(400K)=2.382min−1

k 2(400K)=2.3824.485

=0.531min−1

vo(400K )=vo(300K )( 400300 )=18.02( 400

300 )=24.03L

min

V=vo( X A

k1 (1−X A )−k2 X A)=24.03 ⌊ 0.6

2382 (1−0.6 )−0.531(0.6)⌋=22.73 L

Alternative solution for question 5

1. The reaction was assumed to be an elementary reaction. That is; n=1 for forward and backward reaction.

Since H is independent of temperature, then:

ln ( KK 300

)=∆H r

R ( 1T

− 1300 )

K=K 300 exp[−∆ H r

R ( 1T

− 1300 ) ]

Substitute for K300, Hr gives us;

K=10 exp [ 8000R ( 1

T− 1

300 )]K=e2.3×e

8000R ( 1

T− 1

300 )=e

(8000R ( 1

T− 1

300 )+2.3)

=e8000RT

−0.90744 (1)

but at equilibrium



K=C R

C A

=CAo X Ae

CAo(1−X Ae)=

X Ae

1−X Ae

So , X Ae=K

K+1 (2)

Therefore we can evaluate the equilibrium conversion at any temperature by empting the value of the T into equation 1 to find K then solving for XAe using equation 2.

At 300K,

K=e8000

( 8.314×300 )−0.90744=9.974

At 400K,

K=e8000

( 8.314×400 )−0.90744=4.473

At 300K

X Ae=9.974

9.974+1=0.9089

At 400K,

X Ae=4.473

4.473+1=0.8173

2. Activation energy for K2

At 300K,

K1=103e(−2416

T )

K=e( 8000RT

−0.90744)

Therefore,



K=K1

K= 103 e

(−2416T )

e( 8000RT

−0.9074)=103 e

(−2416T

−962.23T

+0.9074)=103 e

( (−3378.2324 )T

+0.9074)

Therefore, the activation energy for K2 is:

K2 = 3378.2324

3. The volume of the reactor for the same feed rate and conversation but with

operations at 400 K ?

The performance equation for a MIR is;

τ=C Ao X A

−γA

= Vvo

V=reactor volume

vo= volumetric flowrate

at initial conditions: T= 300K, XA=0.6

As evaluated in the first question, at 300K XA= 0.9089

So, at 300K

Vvo

=CAo X A

−γ A

=CAo X A

K1CA−K2CR

Since the feed consist of pure, then;

C AoX A

K1C A−K 2CR

= 1

K1(1−X A

X Ae

)

Vvo

= 1

K1( 1−X A

X Ae)

At 300k

K1=103e−2416

300 =0.31804



VV 0

= 1

0.31804 (1−0.6

0.9089)= 1

0.31804 (0.339801)=9.2516

Since V = 100 Liters

V O=100

9.2516=10.80895

At 400k

VV O

= 1

K 1(1−X A

X Ae

)

k 1=103 e−2416

400 =2.381559

¿ X Ae at 400k also∈(5.1 )=0.8173

V=V O

K1(1−X A

X Ae

)= 10.80895

2.381559(1− 0.60.8173

)=17.07 liters .


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Reactoon Eng Crswrk 2