Chemical Engineering GATE 2016 Solution

8/18/2019 Chemical Engineering GATE 2016 Solution

1/43

Chemical Engineering

GATE 2016 Solution

The Gate CoachBest GATE, IES, PSU Coaching since 1997


2/43

Chemical Engineering GATE 2016 Solution

Visit us at www.thegatecoach.com| New Batches are starting for GATE 2017 from Feb 14th & March 5th

General Aptitude

Q. 1 – Q. 5 carry one mark each.

Q.1 The volume of a sphere of diameter 1 unit is ________ than the volume of a cube of side 1 unit.

(A)

least (B) less (C) lesser (D) low

Ans: (B) less

Q.2 The unruly crowd demanded that the accused be _____________ without trial.

(A) hanged (B) hanging (C) hankering (D) hung

Ans: (A) hanged

Q.3 Choose the statement(s) where the underlined word is used correctly:

(i) A prone is a dried plum.

(ii) He was lying prone on the floor.

(iii)

People who eat a lot of fat are prone to heart disease.

(A) (i) and (iii) only (B) (iii) only (C) (i) and (ii) only (D) (ii) and (iii) only

Ans: (D) (ii) and (iii) only

Q.4 Fact: If it rains, then the field is wet.

Read the following statements:

(i) It rains

(ii) The field is not wet

(iii) The field is wet

(iv) It did not rain

Which one of the options given below is NOT logically possible, based on the given fact?

(A) If (iii), then (iv). (B) If (i), then (iii).

(C) If (i), then (ii). (D) If (ii), then (iv).

Ans: (C) if (i) then (ii)


3/43



Q.5 A window is made up of a square portion and an equilateral triangle portion above it. The base of

the triangular portion coincides with the upper side of the square. If the perimeter of the window

is 6 m, the area of the window in m2 is ___________.

(A) 1.43 (B) 2.06 (C) 2.68 (D) 2.88

Ans: (B) 2.06

Solution:

2 2

6

5

3 3 6 6 6 6 3 36 362.066

4 4 5 5 5 5 4 25 25

side

total area a a

Q.

6 –

Q. 10 carry two marks each.

Q.6 Students taking an exam are divided into two groups, P and Q such that each group has the same

number of students. The performance of each of the students in a test was evaluated out of 200

marks. It was observed that the mean of group P was 105, while that of group Q was 85. The

standard deviation of group P was 25, while that of group Q was 5. Assuming that the marks were

distributed on a normal distribution, which of the following statements will have the highest

probability of being TRUE?

(A) No student in group Q scored less marks than any student in group P.

(B)

No student in group P scored less marks than any student in group Q.

(C) Most students of group Q scored marks in a narrower range than students in group P.

(D) The median of the marks of group P is 100.

Ans: (C) Most students of group Q scored marks in a narrower range than students in group P.


4/43



Q.7

A smart city integrates all modes of transport, uses clean energy and promotes sustainable use of

resources. It also uses technology to ensure safety and security of the city, something which critics

argue, will lead to a surveillance state.

Which of the following can be logically inferred from the above paragraph?

(i)

All smart cities encourage the formation of surveillance states.

(ii) Surveillance is an integral part of a smart city.

(iii) Sustainability and surveillance go hand in hand in a smart city.

(iv)

There is a perception that smart cities promote surveillance.

(A) (i) and (iv) only (B) (ii) and (iii) only

(C) (iv) only (D) (i) only

Ans: (B) (ii) and (iii) only

Q.8 Find the missing sequence in the letter series.

B, FH, LNP, _ _ _ _.

(A) SUWY (B) TUVW (C) TVXZ (D) TWXZ

Ans: (C) TVXZ

Q.9 The binary operation □ is defined as a □ b = ab+(a+b), where a and b are any two real numbers.

The value of the identity element of this operation, defined as the number x such that a □ x = a,

for any a, is .

(A) 0 (B) 1 (C) 2 (D) 10

Ans: (A) 0


5/43



Q.10 Which of the following curves represents the function,

Here, x represents the abscissa and represents the ordinate.

(A)

(B)

(C)

(D) Ans: (C)


6/43



Chemical Engineering

Q. 1 – Q. 25 carry one mark each.

Ans: (C) Gauss - Seidel

Ans: (A)

Solution:

sinat L e bt

=2 2

( )

b

s a b

[By first shifting property]

Ans: (D)

Solution:

2 25r z a b and 1 1

4tan tan

3

b

a


7/43



Ans: (B)

Solution:

Stream P is splitting into Q & R, hence their composition will be same

Overall mass balance:

P = Q + R

Q + R = 100 (1)

Doing component mass balance:

Ethanol in P = Ethanol in Q + Ethanol in R

P X 0.3 = Q X 0.3 + R X 0.3

P = Q + R (2)

So it is obvious, we cannot use component balance over splitter, so least

parameter required is flow rate of any of 2 exit streams

So, i.e. = 1

Ans: – 58 kJ / mol


8/43



Solution:

For infinitely dilute solution of component (1)

1 20 & 1 x x

So, 2 21 2 60 (1) 100 0 (1)h x x x

1 58h

Ans: (C) 34.8

Solution: We know that

2

11 4

2

f V P

g gd

2

1 11

0.2

2222

2 2 2

0.221 1 1 21

1

2 0.2

2 2

2 0.2

1 1

1.8

22

1

4. .

2

.

.

34.8

f V P

d

V d m V

P f V

P f V V d m V

P V

P V

V P P Kpa

V


9/43



Q.7 In a cyclone separator used for separation of solid particles from a dust laden gas, the separation

factor is defined as the ratio of the centrifugal force to the gravitational force acting on the particle. S r

denotes the separation factor at a location (near the wall) that is at a radial distance r from the centre of

the cyclone. Which one of the following statements is INCORRECT?

(A) S r depends on mass of the particle

(B) S r depends on the acceleration due to gravity

(C) S r depends on tangential velocity of the particle

(D) S r depends on the radial location ( r ) of the particle

Ans: (A) Sr depends on mass of the particle

Q.8 A vertical cylindrical vessel has a layer of kerosene (of density 800 kg/m3) over a layer of water (of

density 1000 kg/m3). L-shaped glass tubes are connected to the column 30 cm apart. The interface

between the two layers lies between the two points at which the L-tubes are connected. The levels (in

cm) to which the liquids rise in the respective tubes are shown in the figure below.

The distance ( x in cm, rounded off to the first decimal place) of the interface from the point at

which the lower L-tube is connected is _______

Ans: 10 cm

Solution:

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42


10/43



Where

1

1

. . [20 (30 )] . .

800 9.81 [20 (30 )] 1000 9.81

800 9.81[50 ] 9810

392400 7848 9810

392400 1962

k water P g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 9.81 x 42

= 412 020 Pa.

Putting P1 and P2 in equation (1)

412020 = 392400+1962 x

1962 x = 19620

X = 10 cm

Q.9 A composite wall is made of four different materials of construction in the fashion shown below.

The resistance (in K/W) of each of the sections of the wall is indicated in the diagram.

3

0.25

1

0.7 Direction of heat flow Direction of heat flow


11/43



The overall resistance (in K/W, rounded off to the first decimal place) of the composite wall, in the

direction of heat flow, is _______

Ans: (A) 3.9

Solution: Overall resistance = 3 + 0.2 + 0.7 = 3.9 K/W

Q.10 Steam at 100oC is condensing on a vertical steel plate. The condensate flow is laminar. The average

Nusselt numbers are Nu1 and Nu2, when the plate temperatures are 10oC and 55oC, respectively. Assume

the physical properties of the fluid and steel to remain constant within the temperature range of interest.

Using Nusselt equations for film-type condensation, what is the value of the ratio Nu2 / Nu1 ?

(A) 0.5 (B) 0.84 (C) 1.19 (D) 1.41

Ans: (C) 1.19


1 1

3 34 4

1

4

( ) ( )0.943 0.943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

g h k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

4901.19

45

u

u

N

N

Q.11 A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K the vapour

pressures of pure benzene and pure toluene are 92 kPa

and 35 kPa, respectively. The mixture follows Raoult’s

law. The equilibrium vapour phase mole fraction


12/43



(rounded off to the second decimal place) of benzene in contact with this liquid mixture at 350 K is

_______

Ans: 0.396

Solution:

From Raoult’s law

. v i i i

p x p

& we’ i i

i

p pyi

p p

Mol fraction of benzene in vapor phase is given by

0.2 920.396

0.2 92 0.8 35

V

B B B

B V T

B B T T i

x p py

x P x p p

Q.12 Match the dimensionless numbers in Group-1 with the ratios in Group-2.

Ans: (D) P – II, Q – III, R – II

Q.13 For what value of Lewis number, the wet-bulb temperature and adiabatic saturation temperature

are nearly equal?

(A) 0.33 (B) 0.5 (C) 1 (D) 2

Ans: (C) 1


13/43



Ans: 2

Solution: Upper limit of reaction of reaction rate will be at

0

0 0

0

0

0

10 (10)2

1 5 15

A A

A A

A

A

A

A

C C

C C r

C C

C

Q.15 The variations of the concentrations (C A, C R and C S) for three species (A, R and S) with time, in an

isothermal homogeneous batch reactor are shown in the figure below.

Select the reaction scheme that correctly represents the above plot. The numbers in the reaction

schemes shown below, represent the first order rate constants in unit of s ‒ 1.


14/43



Ans: (C)

Q.16 Hydrogen iodide decomposes through the reaction 2HI ⇋ H2 + I2. The value of the universal gas

constant R is 8.314 J mol ‒ 1K ‒ 1. The activation energy for the forward reaction is 184000 J mol ‒ 1. The ratio

(rounded off to the first decimal place) of the forward reaction rate at 600 K to that at 550 K is _______

Ans: 28.5

Solution:

From Arrhenius law,

2

1 1 2

1 1E k ln

k R T T

1 2550 & 600where T K T K

So 2

1

184000 1 1

8.314 550 600

k ln

k

2 128.5 Ans k k


15/43



Ans: (D) P – III, Q – I, R – II

Q.18 What is the order of response exhibited by a U-tube manometer?

(A) Zero order (B) First order

(C) Second order (D) Third order

Ans: (C) Second Order

Q.19 A system exhibits inverse response for a unit step change in the input. Which one of the following

statement must necessarily be satisfied?

(A) The transfer function of the system has at least one negative pole

(B) The transfer function of the system has at least one positive pole

(C) The transfer function of the system has at least one negative zero

(D) The transfer function of the system has at least one positive zero

Ans: (D) The transfer function of the system has at least one positive zero.

Q.20 Two design options for a distillation system are being compared based on the total annual cost.

Information available is as follows:

Option P Option Q

Installed cost of the system (Rs in lakhs) 150 120

Cost of cooling water for condenser (Rs in lakhs/year) 6 8


16/43



Cost of steam for reboiler (Rs in lakhs/year) 16 20

The annual fixed charge amounts to 12% of the installed cost. Based on the above information, what is

the total annual cost (Rs in lakhs /year) of the better option?

(A) 40 (B) 42.4 (C) 92 (D) 128

Ans: (A) 40

Q.21 Standard pipes of different schedule numbers and standard tubes of different BWG numbers are

available in the market. For a pipe / tube of a given nominal diameter, which one of the following

statements is TRUE?

(A) Wall thickness increases with increase in both the schedule number and the BWG number

(B) Wall thickness increases with increase in the schedule number and decreases with increase

in the BWG number

(C) Wall thickness decreases with increase in both the schedule number and the BWG number

(D) Neither the schedule number, nor the BWG number has any relation to wall thickness

Ans: (B)

Q.22 Terms used in engineering economics have standard definitions and interpretations. Which one of

the following statements is INCORRECT?

(A) The profitability measure ‘return on investment’ does not consider the time value of money

(B) A cost index is an index value for a given time showing the cost at that time relative to a

certain base time

(C) The ‘six-tenths factor rule’ is used to estimate the cost of an equipment from the cost of a

similar equipment with a different capacity

(D) Payback period is calculated based on the payback time for the sum of the fixed and the

working capital investment

Ans: (D)

Q.23 India has no elemental sulphur deposits that can be economically exploited. In India, which one of

the following industries produces elemental sulphur as a by-product?

(A) Coal carbonisation plants (B) Petroleum refineries

(C) Paper and pulp industries (D) Iron and steel making plants


17/43



Ans: (B)

Q.24 Two paper pulp plants P and Q use the same quality of bamboo as a raw material. The chemicals

used in their digester are as follows:

Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes

Which one of the following statements is CORRECT?

(A) Plant P and Plant Q both use the Sulfite process

(B) Plant P and Plant Q both use the Kraft process

(C) Plant P uses Sulfite process

(D) Plant P uses Kraft process

Ans: (D)

Q.25 Match the industrial processes in Group-1, with the catalyst materials in Group-2.

Group-1 Group-2

P Ethylene polymerisation I Nickel

Q Petroleum feedstock cracking II Vanadium pentoxide

R Oxidation of SO2 to SO3 III Zeolite

S Hydrogenation of oil IV Aluminium triethyl with titanium chloride

promoter

(A) P-IV, Q-III, R-II, S-I (B) P-I, Q-IV, R-III, S-II

(C) P-I, Q-II, R-III, S-IV (D) P-II, Q-III, R-IV, S-I

Ans: (A) P – IV, Q – III, R – II, S – I


18/43


19/43



Ans: 2.55

Ans: 2.65

Solution:

f (x) = x3 + 5, Interval [1, 4]

By applying mean value theorem

1 ( ) ( )( ) f b f a

f cb a

X=2.65

Ans: (A) 94.67

Solution:


20/43



Put h = 1, a = 0, b = 4

y0 = 3, yn = 55, y1 = 10, y2 = 21, y3 = 36,

Ans: (A) 1802

Solution: At steady state energy balance equation can be written as:

Energy IN – Energy OUT = heat solution + energy removed

Energy removed = heat of solution + energy OUT – energy IN

We have:

Heat of solution = 2 4

2 4

650 42600

kJ kg H SO kJ x

kg H SO h h

Energy out =2.8

10 (40 25) 420kJ kJ

kg solution K

kg solution K h

Energy In =2

2

2.86 (25 10) 378

kJ kJkg H O K

kg H O K h


21/43



So, Energy removed

= 2600 + 378 + 420

= 1802kJ

h

Energy removed = 1802kJ

h


22/43




23/43


24/43



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RT v b

P

RT b b T

P

Rb T b

P

v Rb

T P

h v v T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400 /

T T

f i f i T T

dh b dP

h h b P P

J mol

Ans: - 1.65


25/43



Ans: - (C) 2.4

Solution:

3 3

1

2

sec

120

240

36 0.01m mhr

P kPa

P kPa

Q

Required head to develop 2 1 2 1P P

y y g

=3(240 120) 10

12

9810

x

= 12.232+12

= 24.232 m

Power required = 4Q

= 9810 X 0.01 X 24.232

= 2.377 kW


26/43



Ans: (C) 2.52 rpm

Solution:

We have,

Power No =3 5

i

P

P N D

For constant power no.

3 5

3 5 ( )

i

i

Pconst

P N D

P N D I

Volume of tank =

2

4 V D H

( )V

D H given

3 ( )V

V D II

From equation (I )and (II)

3 5

3

i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 3

1

i i

v V

N D N DP

V D D

If all linear dimensions have to be doubled


27/43



2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 3

2 1

2/3

2 1 1

2/3

2

( 4 )

14 2.52

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans: 2.15 mm

Solution: -

We know that , for Re( 0.1)N

terminal setting velocity under stokes law regime is given by ,

2

2

2

( )

18

10 (1180 1000)

18 0.1

1000

P p f

t

f

p

t

t p

d gv

d x x v

x

v d

In Stoke’s law max value of particle Reynolds number is 0.1 and it is given by

Re 0.1 p f

p

f

d v

N

Putting values,


28/43



2 3(1000 ) 100.1

0.1

p pd x d x

3

3 3

8

1 1010

10 1

2.15

p

p

mmd m x mm

m

d mm

Ans: 49 m3/ h

Ans: 0.125 m / s.


2

/ 2 f

U

3

3 2

.

1 10 . 20.05

0.05 10 1000 1

0.125 / sec.

u

y

u

u m


29/43



Ans: (B) 84.2


13

0.8 1/3

0.8

0.023 Re Pr

0.023 Pr

Nu

G d

mG A

Since, mass flow rate in the tubes is doubled, Nu and there by hi gets affected

0.81

3

0.8

20.23 Pr

2 i

i

G d Nu

hNu

Nu h

We know that,

0

0

1 1 1 1,

i

i i

h hU h h h


30/43



i.e.i

U h

and 0.82i

U h

0

S i

S P

T T UAn

T T mC

0 2

S i

S p

T T U An

T T m C

Dividing,0.8

0

135 20

2135 90

2135 20

135

n

nT

0

0.9382 1.1487115

135n

T

0

0

0

1150.8167

135

84.2

nT

T C


31/43



Ans: (A) 26

Ans: 0.25

Solution: - Given that 1 20.1 , 0.2r m r m

11 120, 1F F

We know that, A1 F12 = A2 F21 hence1

21

2

AF

A

22 2

1 121 22 2

2 2

0.14 1 0.254 40.2

r r F r r


32/43



Ans: 1

Solution: Minimum reflux ratio is given by

R min =

1

1 1

.9 71

.7 .5

D X Y

Y x

Ans: (C) 10.5

Solution:

Selectivity is given by

β C/A =/

/ =

0.3/0.1

0.2/0.7= 10.5


33/43


34/43



Ans: (B) 1.08

Solution: -

1

2

11.08

BLM BLM

NA P

NA Y P

Ans: 0.2667 sec – 1

Solution: -

For reaction

1 2 A A Br k C k C

For MFR,1 2

( )

O O A A A A

m

A A B

C X C X

r k C k C

0(1 ) .

O A A A B A AC C X and C C X So ,

1 2(1 )

A

m

A A

X

k X k X

For case – I 1sec 8 mol m A

C

81 0.2

10 A

X


35/43



For case – II 5sec 5 mol m AC

51 0.5

10 A X

So for case – I 1 21 2

0.21 4 1 ( )(1 0.2) (0.2)

k k Ik k

For case – ii1 2

1 2

0.55 0.2 ( )

(1 0.5) (0.5)k k II

k k

On solving equation (I) and (II) 11

0.2667 seck

Ans: 8 m3


36/43



Solution: From the equation,

T = 350 + 25 XA

At XA = 0.8

T = 350 + 25 x 0.8 = 370 K

So, from curve at XA = 0.8 & T =370 K 3

10

. A

mol r

m s

For MFR,0

30. 100 0.8

8( ) 100

m A

A

m

A

V X

FA rA

FA X V m

r

Ans: (B)

Solution: Thiele modulus is given by,

LengthticCharactersratediffusion pore

ratereaction Intrinsic

So, for Large value of Thiele modulus, reaction rate will be high.

Therefore, we will prefer Scenario 2 over Scenario 1, as we don’t required that total pellet is covered by

catalyst.


37/43



Ans: 0.25

Solution:

It is obvious, there is delay so, it can be assumed to be PFR & CSTR in series so,

the delay is equal to P (space time of PFR)

So here delay is 5 min

P = 5 min

And we know that, for CSTR

/

( )

mt

m

e

E t

At t = 0 E (0) =1

m


38/43



So at t = 51

(0 5) 0.05 20 minm

m

E

So5

0.2520

p

m

Ans: 10.13 m3 / hr

Solution: -

For equal percentage value

f = f 0 × eBx

at x = 0.1, f = 2

at x = 0.2, f = 3

by using boundary conditions f 0 & B can be solved now at x = 0.5

f = 10.13 m3/hr

Ans: 2.5

Solution: Characterized equation is given by

1 + G open loop P = 0


39/43



2

1 101 1 0

15 25 100c

K s

On solving by using routh Array and putting bL = 0 we get 2.5cK

Ans: (A) 15.9

Solution: We know that,


40/43



Overshoot =2

80exp

100 1

B

A

can be calculate

Now, the value of T = 100 seconds

2

2

1d

t

td and is known, can be calculated

15.9 sec

Ans: (D) 7.8

Solution: Given that

3

3

150

10150 [10% ]

100

166.6

volume of liquid m

x x of the vapor space

x m v


41/43



2

2000 4000 (6000)4

106000

100

T

d C dh

dh

V =2

2

166.6

4

4

d h hd

2

2

2

2

3

3

166.66000 (6600)

4

4

( ) 6000 2 4 (166.6) (6600)

( ) 4

4 (166.6) (6600)3000

4 (166.6) (6600)

3000

466.3

7.75

T

T

d C d

d

d C d

d d d

d d

d

d

d m

Ans: (D) 6

Solution:

I.

Given process can be written as

C6H12 → C6H6 + 6 H2

Cyclo-hexane Benzene Hydrogen


42/43



II. proposed process can be written as

C6H12 + 12 H2O → 6CO2 + 18H2

Cyclo-hexane Steam Carbon di-oxide Hydrogen

Maximum ratio will be obtained for 100% conversion of cyclohexane

2

2

18

3

H formed in New processSo

H formed in old process = 6

For more details

Visit us at www.thegatecoach.com

Call us at (+91) – 9873452122, 9818652587

Mail us at [email protected]

Like us at www.facebook.com/TheGateCoach


43/43


Documents

Chemical Engineering GATE 2016 Solution