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1 CH-303: Chemical Reaction Engineering Department of Chemical Engineering, IIT Roorkee B.Tech & IDD, Chemical (Third Year) Instructor: Dr. Prakash Biswas 1. The reaction AB C are conducted in a steady state isothermal, continuous stirred tank reactor. All reactions are of first order, with identical rate constants (k). The reactor volume (V) and volumetric flow rate (vo) are constant. If the feed consists of pure A at a concentration CAo, formulate material balance equation for A, B, C. 2. The rate of a liquid phase reaction of type, A +B products, is found to be independent of concentrations of A and B and equal to 1 kmol/m 3 .min at 300 K. Find the conversion in a mixed flow reactor having volume equal to 2 m 3 with feed concentration of A and B equal to 5 kmol/m 3 , feed rate equal to 1 m 3 /min and reactor temperature equal to 300 K. If the activation energy of the reaction is given as 83.1 kJ/mol, find the volume of an isothermal plug flow reactor for the same conversion and feed conditions as in the case of the above mentioned reactor but with reactor temperature kept at 320 K. 3. For a second-order liquid phase reaction, A products, occurring in an isothermal plug flow reactor, the conversion is 50%. What will be the conversion if the plug flow reactor is replaced by a continuous stirred tank reactor of the same volume? 4. An irreversible homogeneous liquid phase reaction AB + C is carried out in two isothermal flow reactor of 100 liters capacity each operating at 60 o C. Find the exit conversion if both the reactors are operated in series, when (a) Both the reactors are ideal plug flow reactors. (b) An ideal plug flow reactor is followed by an ideal back mix reactor. Additional data: Feed rate = 20 liters/min, feed concentration = 1 gmol/lit, rate constant = 0.5 min -1 . 5. In a homogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer? Contd... Tutorial-3 (Autumn-2012-13)

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Page 1: Tutoral 3 Solution

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CH-303: Chemical Reaction Engineering

Department of Chemical Engineering, IIT Roorkee

B.Tech & IDD, Chemical (Third Year)

Instructor: Dr. Prakash Biswas

1. The reaction A↔ B → C are conducted in a steady state isothermal, continuous stirred

tank reactor. All reactions are of first order, with identical rate constants (k). The reactor

volume (V) and volumetric flow rate (vo) are constant. If the feed consists of pure A at a

concentration CAo, formulate material balance equation for A, B, C.

2. The rate of a liquid phase reaction of type, A +B → products, is found to be independent of

concentrations of A and B and equal to 1 kmol/m3.min at 300 K. Find the conversion in a

mixed flow reactor having volume equal to 2 m3 with feed concentration of A and B equal

to 5 kmol/m3, feed rate equal to 1 m3/min and reactor temperature equal to 300 K. If the

activation energy of the reaction is given as 83.1 kJ/mol, find the volume of an isothermal

plug flow reactor for the same conversion and feed conditions as in the case of the above

mentioned reactor but with reactor temperature kept at 320 K.

3. For a second-order liquid phase reaction, A →products, occurring in an isothermal plug

flow reactor, the conversion is 50%. What will be the conversion if the plug flow reactor

is replaced by a continuous stirred tank reactor of the same volume?

4. An irreversible homogeneous liquid phase reaction A→B + C is carried out in two

isothermal flow reactor of 100 liters capacity each operating at 60oC. Find the exit

conversion if both the reactors are operated in series, when

(a) Both the reactors are ideal plug flow reactors.

(b) An ideal plug flow reactor is followed by an ideal back mix reactor.

Additional data:

Feed rate = 20 liters/min, feed concentration = 1 gmol/lit, rate constant = 0.5 min-1.

5. In a homogeneous isothermal liquid polymerization, 20% of the monomer disappears in

34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What

rate equation represents the disappearance of the monomer?

Contd...

Tutorial-3 (Autumn-2012-13)

Page 2: Tutoral 3 Solution

2

CH-303, Chem. Reac. Eng. Tutorial-3, Autumn: 2012-13, Dr.P.Biswas

6. Consider a municipal water treatment plant for a small community (Fig. 1). Waste water,

32 000 m3/day, flows through the treatment plant with a mean residence time of 8 h, air

is bubbled through the tanks, and microbes in the tank attack and break down the

organic material

(Organic waste) + O2 → CO2 + H2O

A typical entering feed has a BOD (biological oxygen demand) of 200 mg O2/liter, while

the effluent has a negligible BOD. Find the rate of reaction, or decrease in BOD in the

treatment tanks.

Fig. 1

7. The first-order reversible liquid reaction

A → R CA0 = 0.5 mol/lit, CRo = 0

takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while

equilibrium conversion is 66.7%. Find the rate equation for this reaction.

8. For the decomposition A → R, CA0 = 1 mol/liter, in a batch reactor conversion is 75% after

1 hour, and is just complete after 2 hours. Find a rate equation to represent this kinetics.

microbes

200 mg O2

needed/liter

Mean residence

time t = 8 h Zero O2

needed

Waste water

32,000 m3/day

Waste water

treatment plant

Clean water 32,000 m3/day

Page 3: Tutoral 3 Solution

r-hY ~ r-~ ..fr'OJ1

A ~ 13 ~C"k

__ ~ := Y< ~ - k ~ ~ Y< (CA - ~)- 'YB ~ k Cf3 + k ~ - k <:;,. ~ ~ ('- ~ - Cl)

+'>C ;; ~Ce

~ &vvr~'- A '

1Jo

~ := Vo ~ oj- C -- ~) V

Vo ~ :: tJo CA + Ie (~- ~) V

~_> ~.<c (I of- k' v ) ~ _ ( k V ) tB'lJo 'lJtJ

~ ~~r- B:

tJ;,. ~o ~ 1-90 ~ 1- C - ~) V

~) lJo X c> - 1.90 ~ +- J( C 2. ~ - ~) V

rft>-r f.v.w-p ~ c,'

1J~ Ceo :;:: 190 ~ -I- C - '>C) V

~) 2.90 X 0 :::

CJ ::=

nO C _ 10 (' VVo c.. ,,~

Page 4: Tutoral 3 Solution

J1~~ b~ ~ ~~r- A ~~.- 1'v'-I~

.f€rw y~ ~ }-<' v(U)

•... '1.90

~ f- C- 'rA) Vv

~ --(--~) l'o

-'YA

v-~

£/R Ie.. -

IX 2- .- 3 kJ/~3~

J

3 ::=- o. it1..- -!::' 0

.- £/R. TI~ :::

UH. I X' 10 3 )e. g.J./ :x 3cro

33· 3 leJ/",,'3. m/h.e.

1X'5XO·Y• t'l 333. 3 ( .-s- 3· I >< 10 )e (2.. 2-. 3 I X 32.. 0

Page 5: Tutoral 3 Solution

,k~ [

f 0. -I1- n-_S""

r.-..-f-k s£4vvL ~ e. of../wv rw <?.hv.s , we ~" .

/ XAIe CA-" - ----k ~c I-XA)'\/

/

""Av/' - 3 XA, +- I = o·

3:t / :J-'-t2-

>tA.::: O·3g

Page 6: Tutoral 3 Solution

X'A I_~f_clx_A_- k ~ C/-XA-)

II

LJ 'xA I he..-/-t-tL Cv rvvC- "Y ~ / tJP1 d ~ e?U. 'j. (J/-.fry~- p~ ~ ~~. T~.

~)kV/ (J.5' x/(Jl)

20

p-)

v}.--

V2---'lYo

Page 7: Tutoral 3 Solution

XA- - 0·9/-79,I

-- ••......

'l/tl - ~

-- lA-z, CXAl- - XA,)

k ~ C'-XA2-J

i[

::)

t{J1)- ~~:La

{XA~ :=

Page 8: Tutoral 3 Solution

ck-F~I ""-,,,.

Ntf( 02- V's~f.

vo(!": ~f ~

tv .

"" rrw .VtJ =)

"?> P6 .~ ~"')V

(~ ~) (32 o·~ 5l,), 0 66 T "",,3·

s ~ 02-2XIO ---

~

T~ ~ yJ; rf 'rU'-P.'>-.

'2. ·0 X !of) ~ 0)../ ~.({) C6 7- f}y,3

~ fR. 1-S- W'-~,...3.ck;

:0- 2-. I1 x it> "I WvO V 3. s -

~.(1)~i~ ~

o-f i.•.....<'.fr'0-( c..=---l ::.{rc>...e.J-r' ~ J ='-'5 o'pf ~A/r ~ is in..e,ft~we ~~ (>v frysJ- t7-r~ y~ cY

~ ~~k-{-

~e. C;;o ~ ~...,..

C[>vvt -f'~ ~ ~ I.D~-. TJv.-.s y~/~~

~ ~(o~Co ) =- k (34 low;") ~ ~'8 Y

Page 9: Tutoral 3 Solution

k ::

-----y <=

..;.r~ & ~- 4is Y vW' b'J ,-

CL b t>dcJ., ye i7Le...f,y

/s ~·V~ I;)

(k,.,. k>2.) g

.J-, (1) :;:; O. 0 ~ "12.ra

(,- ~) ::::').-/3

KJ, T k>z. ::,

~ /4'Y~t?J Co->vv'. .ez..."',"': /'1"

C CD Y\.,M-. V1Y-t~ be- c~ 12... ( 'I Is A.J

- - --{])

h) Q. "C.~ ../f-..A1-6.,5- X 2./3--- =2

O'Sxv;

~ ety vJ>./yvLf'(,!

~e ~ _-- =~e.. kl.

Tl--..lA&

Page 10: Tutoral 3 Solution

·g() ~~ r'(1 ([) g (j!)

kJ ;:: o. 0 S :;+ .J7J

~~ ..ft-.e. ~ -l0x.-f''Yo--ss "on- {zY ~ cJ-,'s ""P?~~at 11

T~'6 "'r t'o..-h'0 S ~C2.- Je-P-

C!' -'2 Ie '1.L:::_(~[(?c)~~J

02~t)k{,~ [(t:)'~~,]

R (Lpf ~-<"';d v (J••J!..AA~- I2 =

h ::> YLcy",)'~;'-IR~')r~'d ,I •• ~

A ;r<'v&, ,

- -(j)

Page 11: Tutoral 3 Solution

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