GATE...GATE examination is one of the most prestigious competitive examinations conducted for Graduate engineers. Over the past few years, it has become more competitive as a number

GATESOLUTIONS

Subject-wise descriptive Solution

This book is dedicated to all

Chemical Engineers preparing for

GATE & PSUs Entrance.

2016 By Engineers Institute of India

ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may bereproduced, transmitted, stored or used in any form or by any means graphic, electronic, ormechanical & chemical, including but not limited to photocopying, recording, scanning,digitizing, taping, Web distribution, information networks, or information storage andretrieval systems.

Engineers Institute of India

28-B/7, Jia Sarai, Near IIT HauzKhas New Delhi-110016

Tel: 011-26514888

For publication information, visit www.engineersinstitute.com

ISBN: 978-93-5137-754-2

Price: Rs. 490.00

A word to the studentsGATE examination is one of the most prestigious competitiveexaminations conducted for Graduate engineers. Over the past fewyears, it has become more competitive as a number of aspirants areincreasingly becoming interested in M.Tech & government jobs due todecline in other career options.

In my opinion, GATE exam test candidates’ basics understanding ofconcepts, ability to apply numerical approach. A candidate is supposedto smartly deal with the syllabus not just mugging up concepts.

Thorough understanding with critical analysis of topics and ability to express clearly aresome of the pre-requisites to crack this exam. The questioning & examination pattern haschanged in few years, as numerical answer type questions play a major role to score a goodrank. Keeping in mind, the difficulties of an average student, we have composed this booklet.

Established in 2006 by a team of IES and GATE toppers, we at Engineers Institute of Indiahave consistently provided rigorous classes and proper guidance to engineering students overthe nation in successfully accomplishing their dreams. We believe in providing exam-oriented teaching methodology with updated study materialand test series so that our studentsstayahead in the competition. Many current and past year toppers associate with us forcontributing towards our goal of providing quality education and share their success with thefuture aspirants. Past students of EII are currently working in various departments and PSU’sand pursuing higher specializations. The basic objective of this book is to maintain thestandard and uniformity in both teaching and learning. I sincerely express my appreciationand thanks to Mr. Pradeep Kumar Singh (M.Tech IITR) and team of EII who directly orindirectly contributed towards this book.

R.K. Rajesh

DirectorEngineers Institute of [email protected]

GATE Syllabus for Chemical Engineering (CH)ENGINEERING MATHEMATICS

Linear Algebra: Matrix algebra, Systems of linear equations, Eigen values and eigenvectors.

Calculus: Functions of single variable, Limit, continuity and differentiability, Taylor series,Mean value theorems, Evaluation of definite and improper integrals, Partial derivatives,Total derivative, Maxima and minima, Gradient, Divergence and Curl, Vector identities,Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’stheorems.

Differential equations: First order equations (linear and nonlinear), Higher order lineardifferential equations with constant coefficients, Cauchy’s and Euler’s equations, Initial andboundary value problems, Laplace transforms, Solutions of one dimensional heat and waveequations and Laplace equation.

Complex variables: Complex number, polar form of complex number, triangle inequality.

Probability and Statistics: Definitions of probability and sampling theorems, Conditionalprobability, Mean, median, mode and standard deviation, Random variables, Poisson,Normal and Binomial distributions, linear regression analysis.

Numerical Methods: Numerical solutions of linear and non-linear algebraic equations.Integration by trapezoidal and Simpson’s rule. Single and multi-step methods for numericalsolution of differential equations.

CHEMICAL ENGINEERING

Process Calculations and Thermodynamics: Steady and unsteady state mass and energybalances including multiphase, multicomponent, reacting and non-reacting systems. Use oftie components; recycle, bypass and purge calculations; Gibb’s phase rule and degree offreedom analysis.First and Second laws of thermodynamics. Applications of first law to close and opensystems. Second law and Entropy. Thermodynamic properties of pure substances: Equationof State and residual properties, properties of mixtures: partial molar properties, fugacity,excess properties and activity coefficients; phase equilibria: predicting VLE of systems;chemical reaction equilibrium.

Fluid Mechanics and Mechanical Operations:Fluid statics, Newtonian and non-Newtonian fluids, shell-balances including differential formof Bernoulli equation and energy balance, Macroscopic friction factors, dimensional analysisand similitude, flow through pipeline systems, flow meters, pumps and compressors,

elementary boundary layer theory, flow past immersed bodies including packed andfluidized beds, Turbulent flow: fluctuating velocity, universal velocity profile and pressuredrop.Particle size and shape, particle size distribution, size reduction and classification of solidparticles; free and hindered settling; centrifuge and cyclones; thickening and classification,filtration, agitation and mixing; conveying of solids.

Heat Transfer: Steady and unsteady heat conduction, convection and radiation, thermalboundary layer and heat transfer coefficients, boiling, condensation and evaporation; typesof heat exchangers and evaporators and their process calculations. Design of double pipe,shell and tube heat exchangers, and single and multiple effect evaporators.

Mass Transfer: Fick’s laws, molecular diffusion in fluids, mass transfer coefficients, film,penetration and surface renewal theories; momentum, heat and mass transfer analogies;stage-wise and continuous contacting and stage efficiencies; HTU & NTU concepts; designand operation of equipment for distillation, absorption, leaching, liquid-liquid extraction,drying, humidification, dehumidification and adsorption.

Chemical Reaction Engineering: Theories of reaction rates; kinetics of homogeneousreactions, interpretation of kinetic data, single and multiple reactions in ideal reactors, non-ideal reactors; residence time distribution, single parameter model; non-isothermal reactors;kinetics of heterogeneous catalytic reactions; diffusion effects in catalysis.

Instrumentation and Process Control: Measurement of process variables; sensors,transducers and their dynamics, process modelling and linearization, transfer functions anddynamic responses of various systems, systems with inverse response, process reactioncurve, controller modes (P, PI, and PID); control valves; analysis of closed loop systemsincluding stability, frequency response, controller tuning, cascade and feed forward control.

Plant Design and Economics: Principles of process economics and cost estimation includingdepreciation and total annualized cost, cost indices, rate of return, payback period,discounted cash flow, optimization in process design and sizing of chemical engineeringequipments such as compressors, heat exchangers, multistage contactors.

Chemical Technology: Inorganic chemical industries (sulphuric acid, phosphoric acid, chlor-alkali industry), fertilizers (Ammonia, Urea, SSP and TSP); natural products industries (Pulpand Paper, Sugar, Oil, and Fats); petroleum refining and petrochemicals; polymerizationindustries (polyethylene, polypropylene, PVC and polyester synthetic fibres).

Syllabus for General Aptitude (GA)

Verbal Ability: English grammar, sentence completion, verbal analogies, word groups,instructions, critical reasoning and verbal deduction.Numerical Ability: Numerical computation, numerical estimation, numerical reasoningand data interpretation.

Marking Scheme& Qualifying Marks*

General Aptitude: 15 Marks Engineering Mathematics: 15 Marks Technical: 70MarksTotal Marks: 100 Total Question: 65 Duration : 3:00 Hour

GATE 2016 GENERAL SC/ST/PD OBC(Non-Creamy) Total AppearedChemical Engineering 33.10 22.00 29.70 15495

GATE 2015 GENERAL SC/ST/PD OBC(Non-Creamy) Total AppearedChemical Engineering 27.52 18.34 24.77 15874

CONTENTS

1. Chemical Reaction Engineering ……………. 1-68

2. Heat Transfer …………………………………….... 69-122

3. Mass Transfer…….…………………………………. 123-182

4. Instrumentation and Process Control …… 183-234

5. Thermodynamics …………………………………. 235-278

6. Fluid Mechanics………………………………….… 279-332

7. Process calculations …………………………….. 333-358

8. Mechanical Operations…………………………. 359-374

9. Plant Design and Economics ………………... 375-392

10. Chemical Technology ………………………….. 393-418

11. Engineering Mathematics…………………….. 419-486

CHEMICAL ENGINEERING 1. CHEMICAL REACTION ENGINEERING [1]

Published by :Engineers Institute of India-E.I.I. © ALL RIGHT RESERVED28B/7, Jiasarai Near IIT Hauzkhas Newdelhi-110016 Ph. 011-26514888 www.engineersinstitute.com

1. CHEMICAL REACTION ENGINEERING(GATE Previous Papers)

GATE-2016

1. For a non-catalytic homogeneous reaction AB, the rate expression at 300K is

3 1 10(mol m )

1 5A

AA

Cr s

C

, where CA is the concentration of A (in mol/m3). Theoretically, the

upper limit for the magnitude of the reaction rate (-rA in mol m-3s–1, rounded off to the firstdecimal place) at 300K is ____________ (1-Mark)

2. The variations of the concentrations (CA, CR and CS) for three species (A, R and S) with time, inan isothermal homogeneous batch reactor are shown in the figure below. (1-Mark)

Select the reaction scheme that correctly represents the above plot. The numbers in the reactionschemes shown below, represent the first order rate constants in unit of s‒1.

(A) (B) (C) (D)

3. 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 _______ (1-Mark)



4. The liquid phase reversible reaction A B is carried out in an isothermal CSTR operating

under steady state conditions. The inlet stream does not contain B and the concentration of A inthe inlet stream is 10mol/lit. The concentrations of A at the reactor exit, for residence times of 1s and 5 s are 8 mol/lit and 5 mol/lit, respectively. Assume the forward and backward reactionsare elementary following the first order law. Also assume that the system has constant molardensity. The rate constant of the forward reaction (in s–1, rounded off to the third decimal place)is ________ (2-Marks)

5. A liquid phase irreversible reaction A→ B is carried out in an adiabatic CSTR operating understeady state conditions. The reaction is elementary and follows the first order rate law. For thisreaction, the figure below shows the conversion (XA) of A as a function of temperature (T) fordifferent values of the rate of reaction (–rA in mol m–3s–1) denoted by the numbers to the left ofeach curve. This figure can be used to determine the rate of the reaction at a particulartemperature, for a given conversion of A. (2-Marks)

The inlet stream does not contain B and the concentration of A in the inlet stream is 5 mol/m3.The molar feed rate of A is 100 mol/s. A steady state energy balance for this CSTR results inthe following relation: T = 350 + 25XA where T is the temperature (in K) of the exit stream andXA is the conversion of A in the CSTR. For an exit conversion of 80 % of A, the volume (in m3,rounded off to the first decimal place) of CSTR required is _______

6. A porous pellet with Pt dispersed in it is used to carry out a catalytic reaction. Following twoscenarios are possible. (2-Marks)

Scenario 1: Pt present throughout the pores of the pellet is used for catalyzing the reaction.Scenario 2: Pt present only in the immediate vicinity of the external surface of the pellet isused for catalyzing the reaction.



At a large value of Thiele modulus, which one of the following statements is TRUE?

(A) Since the reaction rate is much greater than the diffusion rate, Scenario 1 occurs(B) Since the reaction rate is much greater than the diffusion rate , Scenario 2 occurs(C) Since the reaction rate is much lower than the diffusion rate, Scenario 1 occurs(D) Since the reaction rate is much lower than the diffusion rate, Scenario 2 occurs

7. A CSTR has a long inlet pipe. A tracer is injected at the entrance of the pipe. The E-curveobtained at the exit of the CSTR is shown in the figure below. (2-Marks)

Assuming plug flow in the inlet pipe, the ratio (rounded off to the second decimal place) of thevolume of the pipe to that of the CSTR is _______

GATE-2015

8. An irreversible, homogeneous reaction A products , has the rate expression:

Rate22 0.1

,1 50

A A

A

C C

C

where CA is the concentration of A. CA varies in the range 0.5–50 mol/m3.

For very high concentrations of A, the reaction order tends to: (1-Mark)

(A) 0 (B) 1 (C) 1.5 (D) 2

9. For which reaction order, the half-life of the reactant is half of the full lifetime (time for 100%

conversion) of the reactant? (1-Mark)

(A) Zero order (B) Half order (C) First order (D) Second order

10. Consider two steady isothermal flow configurations shown schematically as Case-I and Case-II

below. In Case-I, a CSTR of volume V1 is followed by a PFR of volume V2, while in Case-II a

PFR of volume V2 is followed by a CSTR of volume V1. In each case, a volumetric flow rate Q

of liquid reactant is flowing through the two units in series. An irreversible reaction A

products (order n) takes place in both cases, with a reactant concentration CAO being fed into

the first unit. (2-Marks)



Choose the correct option:

(A) 1 1f

f

IA

IIA

Cfor n

C

(B) 1 1f

f

IA

IIA

Cfor n

C

(C) 1 1f

f

IA

IIA

Cfor n

C

(D) 1 1f

f

IA

IIA

Cfor n

C

11. An isothermal steady state mixed flow

reactor (CSTR) of 1m3 volume is used

to carry out the first order liquid-phase

reaction A products. Fresh feed at a

volumetric flow rate of Q containing

reactant A at a concentration CA0 mixes

with the recycle stream at a volumetric flow rate RQ as shown in the figure below.

It is observed that when the recycle ratio R = 0.5, the exit conversion 50%.AfX When the

recycle ratio is increased to R = 2, the new exit conversion (in percent) will be: (Marks-2)

(A) 50.0 (B) 54.3 (C) 58.7 (D) 63.2



ANSWER KEY

1 2 3 4 5 6 7 8 9 102.0 c 28.5 0.26671 8 b 0.25 B A B11 12 13 14 15 16 17 18 19 20A C 0.72 A D B C D 0.64 0.521 22 23 24 25 26 27 28 29 301 51.7 20 90% D D B B C B31 32 33 34 35 36 37 38 39 40C A B A B C D A A B41 42 43 44 45 46 47 48 49 50C C C C D B B A C C51 52 53 54 55 56 57 58 59 60C D B A C C B C B C61 62 63 64 65 66 67 68 69 70A D B B A A A C D D71 72 73 74 75 76 77 78 79 80B D C D B D A C A B81 82 83 84 85 86 87 88 89 90C A C A C A B A D C91 92 93 94 95 96 97 98 99 100C D C C A B B B B D

101 102 103 104 105 106 107 108 109 110A B C D C D A A B D

111 112 113 114 115 116 117 118 119 120B D B C B B B A A C

121 122 123 124 125 126 127 128A A C B C A C D



SOLUTIONS

GATE-20161. (2)

A B

AA

A

10Cr

1 5C

Upper limit of reaction rate will be at CA = CA0

So, A

A0

A0

10C 10r

11 5C 5C

As,A0

15 5

C

So,10

r 25

2. (C)For following parallel reaction the concentration of R (CR) and concentration of S (CS)will be same because k1 and k2 are same and order of reaction is also same.

A1 2 A A

dC(k k )C 2C

dt (k1 = k2 = 1 given)

On integrating the above equation, we getCA = CA0 e–2t …(1)

2tR1 A A A0

dCk C C C e

dt …(2)

On integrating the above equation, we get

2tA0R

CC 1 e

2

As, t

A0R

C 1C 0.5

2 2 [CA0 = 1 (given in figure)]

Similarly,

2tA0S

CC 1 e

2

As t

A0S

C 1C 0.5

2 2

Since, CR and CS are same and is equal to 0.5.Hence, option (c) is correct.



3. (28.5)

1 2 22H H IFrom Arrhenius law

2 2

1 1 1 2

k r E 1 1ln ln

k r R T T

2

1

r 184000 1 1ln

r 8.314 550 600

2

1

r28.5

r

2 1r 28.5r

4. (0.26671)for first reactor (CSTR)Given that,

0 1 110mol/lit , 1sec , 8mol/litA AC C

Hence, 0 11 1sec A A

A

C C

r

1 2Where, A A Br k C k C

1 2

10 81

A Bk C k C

1 2 2A Bk C k C

1 2 0 08 ( ) 2B A Ak k C C X

Since

10

0

81 1 0.2 and 0 (given)

10A

A BA

CX C

C

Hence,

1 28 2 2 ...(1)k k

For second reactor given

0 1 210 , 5 and 5secA AC C

Hence,

21 2

10 55sec

A Bk C k C

1 2 0 0

55

5 ( )B A Ak k C C X

Where, 1

0

51 1 0.5

10A

AA

CX

C and CB0 = 0 (given)

Hence, above equation reduce to

1 25 5 1k k …(2)

On solving equation (1) and (2), we get1

1 0.26671 seck



5. (8)For a chemical reaction conducting in MFR the conversion V/s Temperature data is given fordifferent value of (–rA)Given that

CA0 = 5 mol/m3

FA0 = 100 mol/secXA = 0.8

Since T = 350 + 25 XA (given)Hence at XA = 0.8T = 350 + 25 0.8

= 370K

From figure in question at T = 373K and XA = 0.8, we get ( ) 10Ar

Now for MFR

30 100 0.88m

( ) 10A A

A

F XV

r

6. (b)

Since (Thiele modulus) =surface reaction rate

.diffuison rate c

kL

D

Large value of implies that k>>D. It means the surface reaction rate is larger in compare todiffusion rate and scenario 2 occurs because it is not possible for Pt to dispersed to center ofcatalyst pellet because of diffusion rate is much lower than surface reaction rate. It means Ptpresent only in the immediate vicinity of the external surface of the pellet.

7. (0.25)Since for Ideal PFR

( ) ( ) ( 5)PE t t t

5min (from figure in question)p

Hence,

0

5minPV

…(1)

And for ideal CSTR the E(t) curve after p time is given as below

( )1( ) exp p

m m

tE t

At t = 5min

1 (5 5) 1(5) exp 0.05 (form figure in question)

m m m

E

Hence, m =0

120

0.05mV

v …(2)

Dividing equation (1) and (2)

equation 1 50.25

equation (2) 20P

m

V

V



GATE-2015

8. (B)22 0.1

1 50A A

AA

C Cr

C

22(50) 0.1 50 5000 5

1 50 50 1 2500Ar

[For very high CA put CA = 50]

2 25000 5 5000 2 0.1 2

& 1 2500 2500 1 50 50A A A

A A

C C C

C C

22 1

50 25A

A AA

Cr C

C So, at very high CA, order tends to 1.

9. (A) Prove that 1/2 2ft

t

Assume nth order reaction nAA

dCkC

dt

00

A

A

C tnA

A AnA C

dCkdt C dC k dt

C

1 1 1 10 0

1 (1 )

n n n nA A A AC C C C

kt tn k n

...(1)

For full life time CA = 0, now equation (1) becomes,1

0

(1 )

nA

fC

tk n

...(2)

For half life 0 ,2A

AC

C now equation (1) becomes,

1 11 1 100 0

1/2 1 1

2 12

2 (1 ) (1 )2

n nn n nAA A

n n

CC Ct

k n k n

...(3)

By equation (2), equation (3) becomes, 1

1/2 1

2 1

2

nf

n

tt

Put n = 0 in above equation, we get 1/2 2ft

t so, zero order.

10. (B) By taking 1st reaction (n = 1)

Case-I Consider 3 11 2 1 , 1secV V m k

3 301 sec and 1 /AQ m C mol m

For CSTR,

30 1 11

1 1

11 1 0.5 /A A A

AA A

C C CK C mol m

C C



For PFR:

11

3

1 ln

0.184 mol/m

Af

A

CAfA

A AC

Af

CdC

kC C

C

Case-2:

With same data1

0

1

0

31

For PFR 1 ln

0.3680 mol/m

A

A

CA A

A AC

A

dC C

kC C

C

And for CSTR

1 30.36801 1 0.184mol/mA Af Af

AfAf Af

C C CK C

C C

Hence for both cases, CAf is equal for 1st order reaction.

11. (A)

Apply mole A balance at steady state

0

0

0

( ) ( )A A A A

A A A

A A A

QC RQC RQ Q C kC V

QC QC kC V

C C kC

0

0 (1 ) 1A A

AA A

C C kX

kC X k

So conversion XA is independent of recycle ratio R. So, final conversion = 50%.

12. (C) From figure:

0 2 4

( ) 0 5 0

t

C t

0

( )( )

( )t

C tE t

C t dt

CHEMICAL ENGINEERING 2. HEAT TRANSFER [69]

Published by : Engineers Institute of India-E.I.I. © ALL RIGHT RESERVED28B/7, Jiasarai Near IIT Hauzkhas Newdelhi-110016 Ph. 011-26514888. www.engineersinstitute.com

2. HEAT TRANSFER(GATE Previous Papers)

GATE-2016

1. A composite wall is made of four different materials of construction in the fashion shownbelow. The resistance (in K/W) of each of the sections of the wall is indicated in the diagram.

The overall resistance (in K/W, rounded off to the first decimal place) of the composite wall, inthe direction of heat flow, is _______ (1-Mark)

2. Steam at 100oC is condensing on a vertical steel plate. The condensate flow is laminar. Theaverage 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 thetemperature range of interest. Using Nusselt equations for film-type condensation, what is the

value of the ratio 2

1

Nu

Nu? (1-Mark)

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

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

Group -1 Group-2 (1-Mark)

P Biot number Ibuoyancy force

viscous force

Q Schmidt number IIinternal thermal resistance of a solid

boundary layer thermal resistance

R Grashof number IIImomentum diffusivity

mass diffusivity

(A) P-II , Q-I , R-III (B) P-I , Q-III , R-II(C) P-III , Q-I , R-II (D) P-II , Q-III , R-I




GATE-2016





1

Nu

Nu? (1-Mark)

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




viscous force




mass diffusivity





GATE-2016





1

Nu

Nu? (1-Mark)

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




viscous force




mass diffusivity




4. In a 1- 1 pass shell and tube exchanger, steam is condensing in the shell side at a temperature(Ts) of 135oC and the cold fluid is heated from a temperature (T1) of 20oC to a temperature (T2)of 90oC. The energy balance equation for this heat exchanger is (2-Marks)

1

2

ln s

s p

T T UA

T T mc

Where U is the overall heat transfer coefficient, A is the heat transfer area, m is the mass flowrate of the cold fluid and cp is its specific heat. Tube side fluid is in a turbulent flow and theheat transfer coefficient can be estimated from the following equation:

Nu = 0.023 (Re)0.8 (Pr)1/3

Where Nu is the Nusselt number, Re is the Reynolds number and Pr is the Prandtl number. Thecondensing heat transfer coefficient in the shell side is significantly higher than the tube sideheat transfer coefficient. The resistance of the wall to heat transfer is negligible. If only themass flow rate of the cold fluid is doubled, what is the outlet temperature (in oC) of the coldfluid at steady state?

(A) 80.2 (B) 84.2 (C) 87.4 (D) 88.6

5. In an experimental setup, mineral oil is filled in between the narrow gap of two horizontalsmooth plates. The setup has arrangements to maintain the plates at desired uniformtemperatures. At these temperatures, ONLY the radiative heat flux is negligible. The thermalconductivity of the oil does not vary perceptibly in this temperature range. Consider fourexperiments at steady state under different experimental conditions, as shown in the figurebelow. The figure shows plate temperatures and the heat fluxes in the vertical direction.

(2-Marks)

What is the steady state heat flux (in W m‒2) with the top plate at 70oC and the bottom plate at40oC?

(A) 26 (B) 39 (C) 42 (D) 63

6. The space between two hollow concentric spheres of radii 0.1 m and 0.2 m is under vacuum.Exchange of radiation (uniform in all directions) occurs only between the outer surface (S1) ofthe smaller sphere and the inner surface (S2) of the larger sphere. The fraction (rounded off tothe second decimal place) of the radiation energy leaving S2, which reaches S1 is _______



ANSWER KEY: HEAT TRANSFER

1 2 3 4 5 6 7 8 9 10

3.9 C D B A 0.25 A A 250.9 C11 12 13 14 15 16 17 18 19 20C A D 17.14 3.85 271.07ºC D D 5.67 A21 22 23 24 25 26 27 28 29 30D C B D A B B B A B31 32 33 34 35 36 37 38 39 40A A A B A A D B A B41 42 43 44 45 46 47 48 49 50B C B A D D B B D B51 52 53 54 55 56 57 58 59 60B D C B B A D B A C61 62 63 64 65 66 67 68 69 70D B A D D A D B B C71 72 73 74 75 76 77 78 79 80D B C A C A C B A A81 82 83 84 85 86 87 88 89 90D D A A B A B D A A91 92 93 94 95 96 97 98 99 100B B B D D A D C B B

101 102 103 104 105 106 107B C C D A D C



SOLUTIONS

GATE-2016

1. (3.9)

eqeq

1 1 1.251 5 R 0.2

R 0.25 0.25

totalR 3 0.2 0.7 3.9

2. (C)For film-type condensationNu (T)–1/4

Hence,1/4

2 2

1 1

Nu T

Nu T

Given that, 1 2T 100 10 90ºC and T 100 55 45ºC

Hence,1/4

2

1

Nu 451.189 1.19

Nu 90

3. (D)(1.) Biot number =

conductive resistance in solid internal thermal resistance of a solid

convective resistance in fluid boundary layer thermal resistance

(2.) Schmidt number =Momentum diffusivity

mass diffusivity

(3.) Grashof number =Buoyancy force

viscous force

4. (B)

1

2

s

s P

T T UAln

T T mC

…(1)



0

1 1 1

iU h h given that 0 ih h

Hence,0

10

h

1 1 i

i

or U hU h

Since,

0.8(Re) Re Re.

u d mdNu

A

0.8

.

mdNu

A

0.8 0.8Nu m h m

Hence, 0.8U m Put this result in equation (1)Hence,

1

2

s

s

T T Uln

T T m

0.2

2

135 20( )

135ln m

T

When, 2 90ºT C

2 2 2' ? , ' 2T m m Where ( ' ) indicates 2nd case when flow rate has increased twice to that of initial mass flowrate.Hence,

2

0.2 0.2

2

2

115145

' 2115135 '

ln m

mln

T

2

11511545

1.148 135 '

lnln

T

2

1150.817

135 'ln

T

2

1152.263

135 'T

135 – 2 'T = 50.81

2 'T = 84.18ºC



5. (A)In experiment 2 the hot plate is above the fluid and the cold plate is below the fluid. Hencethe heat transfer is possible only by conduction mode, convection is not possible because nofluid movement will be there and the same case is in experiment 4 where the hot plate isabove the fluid and cold plate is below the fluid again here, the heat transfer is only possibleby conduction hence, the heat flux in experiment 2 and 4 will be same is equal to 26 Watt/m2.

6. (0.25)

11 12 1.0F F

Here, F11 = 0 so, F12 = 1From Reciprocity theoremA1 F12 = A2 F21

121 12

2

AF F

A Where

221 1

22 2

4 0.1

0.24

A r

A r

221 (0.5) 0.25F

GATE-2015

7. (A)As we know that, for equall 1 2 3ε =ε =ε =ε (means all are same) then,

1

1with shield without any shield

q q

A n A

...(1)

Here 1 2 31 & ε =ε =ε =εn

So, by equation (1)

1

1 1with one shield without any shield

q q

A A

So,( / ) 1

0.5( / ) 2

with one shield

without any shield

q A

q A

8. (A)( ) ( ) T f time T f space

Now according to lumped parameter analysis,

( ) PdT

hA T T C Vdt

...(1)

CHEMICAL ENGINEERING 3. MASS TRANSFER [123]


3. MASS TRANSFER

(GATE Previous Papers)

GATE-20161. 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. Themixture follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off to thesecond decimal place) of benzene in contact with this liquid mixture at 350 K is _______

2. For what value of Lewis number, the wet-bulb temperature and adiabatic saturation temperatureare nearly equal?

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

3. A binary distillation column is to be designed using McCabe Thiele method. The distillatecontains 90 mol% of the more volatile component. The point of intersection of the q-line withthe equilibrium curve is (0.5, 0.7). The minimum reflux ratio (rounded off to the first decimalplace) for this operation is _______

4. Solute C is extracted in a batch process from itshomogenous solution of A and C, using solvent B. Thecombined composition of the feed and the extractingsolvent is shown in the figure below as point M, alongwith the tie line passing through it. The ends of the tieline are on the equilibrium curve.

What is the selectivity for C?

(A) 3.5 (B) 7 (C) 10.5 (D) 21

5. At 30ºC, the amounts of acetone adsorbed at partial pressure of 10 and 100 mmHg are 0.1 and0.4 kg acetone/kg activated carbon, respectively. Assume Langmuir isotherm describes theadsorption of acetone on activated carbon. What is the amount of acetone adsorbed (in kg perkg of activated carbon) at a partial pressure of 50 mmHg and 30ºC ?

(A) 0.23 (B) 0.25 (C) 0.30 (D) 0.35

6 Consider the following two cases for a binary mixture of ideal gases A and B under steady stateconditions. In case 1, the diffusion of A occurs through non-diffusing B. In Case 2, equimolalcounter diffusion A and B occurs. In both the cases, the total pressure is 100 kPa and the partialpressures of A at two points separated by a distance of 10mm are 10kPa and 5kPa. Assume thatthe Fick’s first law of diffusion is applicable. What is the ratio of molar flux of A in Case 1 tothat in Case 2?

(A) 0.58 (B) 1.08 (C) 1.58 (D) 2.18



GATE-2015

7. For a binary mixture of components A and B, NA and NB denote the total molar fluxes ofcomponents A and B, respectively. JA and JB are the corresponding molar diffusive fluxes.Which of the following is true for equimolar counter-diffusion in the binary mixture?(A) NA + NB = 0 and JA + JB 0 (B) NA + NB 0 and JA + JB = 0 (1-Mark)(C) NA + NB 0 and JA + JB 0 (D) NA + NB = 0 and JA + JB = 0

8. A spherical naphthalene ball of 2 mm diameter is subliming very slowly in stagnant air at 25ºC.The change in the size of the ball during the sublimation can be neglected. The diffusivity ofnaphthalene in air at 25ºC is 1.110–6 m2/s.

The value of mass transfer coefficient B10-3 m/s, where B (up to one decimal place) is________________ . (1-Marks)

9. Benzene is removed from air by absorbing it in a non-volatile wash-oil at 100 kPa in a counter-current gas absorber. Gas flow rate is 100mol/min, which includes 2 mol/min of benzene. Theflow rate of wash-oil is 50 mol/min. vapour pressure of benzene at the column conditions is 50kPa. Benzene forms an ideal solution with the wash-oil and the column is operating at steadystate. Gas phase can be assumed to follow ideal gas law. Neglect the change in molar flow ratesof liquid and gas phases inside the column. (1-Mark)For this process, the value of the absorption factor (up to two decimal places) is __________ .

10. Identify the WRONG statement amongst the following. (1-Mark)(A) Steam distillation is used for mixtures that are immiscible with water.(B) Vacuum distillation is used for mixtures that are miscible with water.(C) Steam distillation is used for mixtures that are miscible with water.(D) Vacuum distillation columns have larger diameters as compared to atmospheric columns

for the same throughput

11. A multi-stage, counter-current liquid-liquid extractor is used to separate solute C from a binarymixture (F) of A and C using solvent B. Pure solvent B is recovered from the raffinate R bydistillation, as shown in the schematic diagram below. (2-Marks)

Locations of different mixture for this process areindicated on the triangular diagram below. P is thesolvent-free raffinate, E is the extract, F is the feed and is the difference point from which the mass balance linesoriginate. The line PB intersects the bimodal curve at Uand T.

The lines P and FB intersect the bimodal at V and Wrespectively.

The raffinate coming out of the extractor is represented in the diagram by the point:(A) T (B) U (C) V (D) W



ANSWER KEY

1 2 3 4 5 6 7 8 9 100.396 C 1 C C B D 1.1 1.02 C

11 12 13 14 15 16 17 18 19 20B 0.22 1.4 6.4 D C A B B 1.2121 22 23 24 25 26 27 28 29 30

3.75 B D 0.5 66.85 A 1.14 A C B31 32 33 34 35 36 37 38 39 40B C C B B A D B D D41 42 43 44 45 46 47 48 49 50B D A A A D D A D C51 52 53 54 55 56 57 58 59 60C D B A A A B B D A61 62 63 64 65 66 67 68 69 70A A C C B B B B B C71 72 73 74 75 76 77 78 79 80C A A A C B D C C A81 82 83 84 85 86 87 88 89 90A A B C D A D C A D91 92 93 94 95 96 97 98 99 100D C B D C B C C D A

101 102 103 104 105 106 107 108 109 110C D B B A B C D D B

111 112 113 114 115 116 117 118 119 120D C A C C C A C B C

121 122 123 124 125 126 127 128B A B B A C D C



SOLUTIONS

GATE-20161. (0.396)

Apply Raoult’s lawsat

i i ip x p

And we also know i ii

T i

p py

P p

Mole fraction of Benzene (yB) =sat

B B Bsat sat

B B A AB A

p x p

x p x pp p

B

0.2 92y 0.396

0.2 92 0.8 35

2. (C)

For Lewis number 1 the wet-bulb temperature and adiabatic saturation temperature are nearly

equal to one because G

Y H

h1

K C and where G

Y H

h

K Cis called Lewis number

3. (1)Given that xD = 0.9 andThe intersection of q line with equilibrium curve is (0.5 , 0.7).Since, at minimum reflux condition the q line and upper operating line equation are intersectat equilibrium curve at same point (x , y).Hence, upper operating line equation

1 1DxR

y xR R

…(1)

At minimum reflux condition R Rm

Hence, equation (1) becomes

1 1m D

m m

R xy x

R R

Putting the value of (x , y) = ( 0.5 , 0.7) and xD = 0.9 in above equation, we get0.9

0.7 (0.5) 11 1

mm

m m

RR

R R

4. (C)

Selectivity =(wt fraction C in E)/(wt fraction A in E)

(wt fraction C in R)/(wt fraction A in R)

From diagramwt fraction C in E = 0.3 and

wt fraction A in E = 0.1

wt fraction C in R = 0.2

wt fraction A in R = 0.7

Putting the all values in formula, we get



=

0.30.1 10.50.20.7

5. (C)According to Langmuir Isotherm

'

max

...(1)1

A A AA

A A

q k pq

q k p

'

max

Amount of A adsorbedkg of adsorbent

Max amount of A adsorbedkg of absorbent

AA

qq

q

max 1a A

A A

k pq

q k p

For the 1st case

0.1 , 10mmHgAq p

max

100.1...(2)

1 10A

A

k

q k

For 2nd case

0.4 , 100mmHgAq p

max

1000.4

1 100A

A

k

q k

…(3)

On dividing equation (2) and (3), wet get

max0.02 and 0.6Ak q

In 3rd caseGiven that pA = 50 mmHgHence from equation (1)

max

50 0.020.3

1 0.6 1 (50 0.02)a A

A A

k pq qq

q k p

6. (B)Case I (diffusion of A occurs through non-diffusing B)

1 21

( )AB T A AA

Blm

D P p pN

RTZ P

…(1)

Where, PBlm = 2 1

2

1

ln

B B

B

B

p p

p

p

Case II: (Equimolal counter diffusion of A and B )

2 1 2( )ABA A A

DN p p

RTZ …(2)

On dividing equation (1) and (2), we get



1

2

A T

A Blm

N P

N P …(3)

Given that PT = 100 kPa and pA1 = 10 kPa and pA2 = 5 kPa

Since PT = pA1 + pB1 = pA2 + pB2

Hence, pB1 = PT – pA1 = 100 – 10 = 90 kPa and , pB2 = PT – pA2 = 100 – 5 = 95 kPa

Hence PBlm = 2 1

2

1

95 9092.48

95lnln

90

B B

B

B

p p

p

p

Putting the value of PBlm and PT in equation (3)

1

2

1001.08

92.48A T

A Blm

N P

N P

GATE-2015

7. (D) As we know For component A,

...(1) A A AN Nx J

For component B,

...(2)B B BN Nx J

Add equation (1) and (2), we get

( ) ...(3)A B A B A BN N N x x J J

For equimolar counter diffusion, A BN N

0 A BN N N and 1A Bx x

Now equation (3) becomes 0 0 0A B A BJ J J J

8. (1.1)3 6 22 2 10 25º 1.1 10 /N airD mm m T C D m s

The relation between mass transfer coefficient and diffusivity can be given by SherwoodNumber.

...(1)c

AB

K DSh

D

Where D = diameter of naphthalene ballKc = mass transfer coefficientDAB = Diffusivity

Value of Sherwood number for spherical ball in stagnant medium is 2, so by equation (1)6

33

2 2 1.1 102 1.1 10 /

2 10c AB

AB

K D DK m s

D D

By comparison we get, B = 1.1

CHEMICAL ENGINEERING 4. INSTRUMENTATION & PROCESS CONTROL [183]

Published by :Engineers Institute of India-E.I.I. © ALL RIGHT RESERVED28B/7, Jiasarai Near IIT Hauzkhas Newdelhi-110016 Ph. 011-26514888. www.engineersinstitute.com

4. INSTRUMENTATION & PROCESS CONTROL


GATE-20161. Match the instruments in Group-1 with process variables in Group-2. (1-Mark)

Group-1 Group-2

P. Conductivity meter I Flow

Q. Turbine meter II Pressure

R. Piezoresistivity element III Composition

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

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

2. What is the order of response exhibited by a U-tube manometer? (1-Mark)

(A) Zero order (B) First order

(C) Second order (D) Third order

3. A system exhibits inverse response for a unit step change in the input. Which one of thefollowing statement must necessarily be satisfied? (1-Mark)(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

4. A liquid flows through an “equal percentage” valve at a rate of 2 m3/h when the valve is 10%open. When the valve opens to 20% the flowrate increases to 3 m3/h. Assume that the pressuredrop across the valve and the density of the liquid remain constant. When the valve opens to50%, the flowrate (in m3/h, rounded off to the second decimal place) is_______

(2-Marks)

5. A PI controller with integral time constant of 0.1 min is to be designed to control a process withtransfer function (2-Marks)

2

10( )

2 100pG s

s s

Assume the transfer functions of the measuring element and the final control element are bothunity (Gm = 1 , Gf = 1) . The gain (rounded off to the first decimal place) of the controller thatwill constitute the critical condition for stability of the PI feedback control system is _______

6. For a unit step input, the response of a second order system is (2-Marks)

2

2

11( ) 1 sin

1

t

py t k e t



Where, kp is the steady state, is the damping coefficient, is the natural period of oscillation

and is the phase lag. The overshoot of the system is exp21

. For a unit step input,

the response of the system from an initial steady state condition at t = 0 is shown in the figurebelow.

What is the natural period of oscillation (in seconds) of the system?(A) 15.9 (B) 50 (C) 63.2 (D) 100

GATE-2015

7. Match the output signals as obtained from four measuring devices in response to a unit stepchange in the input signal (1-Mark)

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

8. The transfer function for the disturbance response in an open-loop process is given ( ).opendG s

The corresponding transfer function for the disturbance response in a closed-loop feedback

control system with proportional controller is given by ( ).closeddG s Select the option that is

ALWAYS correct {O[G(s)] represents order of transfer function G(s)}: (1-Mark)

(A) ( ) ( )open closedd dO G s O G s (B) ( ) ( )open closed

d dO G s O G s (C) ( ) ( )open closed

d dO G s O G s (D) ( ) ( )open closedd dO G s O G s



ANSWER KEY

1 2 3 4 5 6 7 8 9 10D C D 10.124 2.5 MTA C D 3.6 0.8711 12 13 14 15 16 17 18 19 20A C D C 16.3 C D 0.50 A 1.478721 22 23 24 25 26 27 28 29 30C B D A C C C C B D31 32 33 34 35 36 37 38 39 40C B C C C B D B D D41 42 43 44 45 46 47 48 49 50A A A D A C C B D A51 52 53 54 55 56 57 58 59 60D B C D A A B D D B61 62 63 64 65 66 67 68 69 70A A A D C A D A A D71 72 73 74 75 76 77 78 79 80A B D B C B C D D D81 82 83 84 85 86 87 88 89 90C A C C A B B A D C91 92 93 94 95 96 97 98 99 100B C C D D A A D C A

101 102 103 104 105 106C B C B B A



SOLUTIONS

GATE-20161. (D)

Conductivity meter composition

Turbine meter flow

Piezoresistivity element Pressure

2. (C)U-tube Manometer second order response

3. (D)A system exhibits inverse response for unit step change in the input if transfer function of thesystem atleast one positive zero.

4. (10.124 m3/h)

For equal percentage valve1( )lf R

Where,max

andq x

f lq L

Given q1 = 2 m3/h and l1 = 0.1 , q2 = 3 m3/h and l2 = 0.2For 1st case

1 11 ( )lf R …(1)

And 2nd case2 1

2 ( )lf R …(2)On dividing equation (1) and (2), we get

1 2 1 21 ( 1)1

2

( )l l l lfR R

f

0.1 0.2 0.11

2

( ) ( )q

R Rq

0.1 0.12 3( ) 57.66

3 2R R R

Now, 3 0.5l

Then, 0.5 1 0.53 (57.66) (57.66)f …(3)

Dividing equation (3) and (1)

3 1 3 11 1 0.5 0.1 0.43

1

(57.66) (57.66)l l l lfR R

f

0.43

1

(57.66)q

q

0.4 3 33 12(57.66) 10.124m /h q 2m /h (given)q



5. (2.5)Since,

1 0 ...(1)OLG

Where, GOL = GP GC Gf Gm

2

10

2 100pGs s

1011

0.1c c

c c

K s KG K

s s

, Gm = Gf = 1 (given)

Putting all values in equation (1), we get

2

1 10( 10 )0

( 2 100)c cK s k

s s s

3 22 (100 10 ) 100 0c cs s s K K

From Routh-array3

2

0

1 (100 10 )

2 100

(100 40 ) 0

100 0

c

e

c

c

s K

s k

s K

s K

For system to be at critical condition of stability

100 40 0cK

1002.5

40cK

6. (MTA) Question is wrong by IISC

GATE-20157. (C)

P: Gas chromatograph, with a long capillary tube will have delayed output as compared to

others IV

Q: Venturi tube attains maximum velocity very quickly and remains constant IR : Thermocouple with Ist order dynamics will increase temperature, i.e. output signal with unittime. After unit time it will attain steady state and then remains constant II

S : Pressure transducer with second order dynamics shows S shape curve III8. (D)

Open loop process = without feed backClosed loop process = with feed backOrder of closed loop process with feedback is always equal to or greater than the order of open

loop process. So, ( ) ( )open closedd dO G s O G s

9. (5.03 )0.3

( ) ...(1)1.5 1

sc

OLK e

G ss

CHEMICAL ENGINEERING 5. THERMODYNAMICS [235]


5. THERMODYNAMICS(GATE Previous Papers)

GATE-20161. The partial molar enthalpy (in kJ/mol) of species 1 in a binary mixture is given

2 21 2 1 22 60 100h x x x , 1 2where x and x are the mole fractions of species 1 and 2,

respectively. the partial molar enthalpy (in kJ/mol, rounded off to the first decimal place) ofspecies 1 at infinite dilution is _________ (1-Mark)

2. An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a

closed system. The work required for the irreversible compression is 1.5 times the work that is

required for reversible compression from the same initial temperature and pressure to the same

final pressure. The molar heat capacity of the gas at constant volume is 30 J mol‒1 K‒1 (assumed

to be independent of temperature); universal gas constant, R is 8.314 J mol‒1 K‒1; ratio of molar

heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the

gas at the final state in the irreversible compression case is _______ (2-Marks)

3. A gas obeying the Clausius equation of state is isothermally compressed from 5 MPa to 15 MPa

in a closed system at 400 K. The Clausius equation of state is( )

RTP

v b T

where P is the

pressure, T is the temperature, v is the molar volume and R is the universal gas constant. Theparameter b in the above equation varies with temperature as 0 1( )b T b b T with

50 4 10b m3mol–1 and 7

1 1.35 10b m3mol–1K–1. The effect of pressure on the molar

enthalpy (h) at a constant temperature is given bypT

h vv T

p T

. Let hi and hf denote

the initial and final molar enthalpies, respectively. The change in the molar enthalpyhf–hi (in J mol–1, rounded off to the first decimal place) for this process is __ (2-Marks)

4. A binary system at a constant pressure with species ‘1’ and ‘2’ is described by the two-suffix

Margules equation, 1 23Eg

x xRT , where gE the molar excess Gibbs free energy, R is the

universal gas constant, T is the temperature and x1 , x2 are the moles fractions of species 1 and2, respectively.

At a temperature T, 1 21 and 2g g

RT RT , where g1 and g2 are the molar Gibbs free energies of

pure species 1 and 2, respectively. At the same temperature, g represents the molar Gibbs freeenergy of the mixture. For a binary mixture with 40 mole % of species 1, the value (rounded off

to the second decimal place) ofg

RTis _______ (2-Marks)



GATE-20155. For a gas phase cracking reaction A B + C at 300ºC, the Gibbs free energy of the reaction at

this temperature is Gº = – 2750 J/mol. The pressure is 1 bar and the gas phase can be assumedto be ideal. The universal gas constant R = 8.314 J/mol K. The fractional molar conversion of Aat equilibrium is : (1-Mark)(A) 0.44 (B) 0.50 (C) 0.64 (D) 0.80

6. Which of the following can change if only the catalyst is changed for a reaction system?(A) Enthalpy of reaction (B)Activation energy (1-Mark)

(C) Free energy of reaction (D) Equilibrium constant

7. If v, u, s and g respectively the molar volume, molar internal energy, molar entropy and molarGibbs free energy, then match the entries in the left and right columns below and choose thecorrect option: (1-Mark)

P. ( / )su v I. Temperature

Q. ( / )Tu P II. Pressure

R. ( / )Pg T III. v

S. ( / )vu s IV. s

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

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

8. Three identical closed systems of a pure gas are taken from an initial temperature and pressure(T1, P1) to a final state (T2 P2), each by a different path. Which of the following is ALWAYSTRUE for the three systems? ( represents the change between the initial and final states; U. S,G, Q and W are Internal energy, entropy. Gibbs free energy, heat added and work donerespectively.) (1-Mark)(A) U, S, Q are same (B) W, U, G are same(C) S, W, Q are same (D) G, U, S are same

9. For a pure liquid, the rate of change of vapour pressure with temperature is 0.1 bar/K in thetemperature range of 300 to 350 K. If the boiling point of the liquid at 2 bar is 320K, thetemperature (in K) at which it will boil at 1 bar (up to one decimal place) is ____ (1-Mark)

10. An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2 m3. It is firstcompressed to 1 MPa by a reversible adiabatic process and then cooled at constant pressure to afinal volume of 0.2 m3. The total work done (in kJ) on the gas for the entire process (up to onedecimal place) is ________________ (2-Marks)

11. Given that molar residual Gibbs free energy, gR, and molar residual volume, vR, are related as

0,

R RPg v

dPRT RT

find gR at T= 27 ºC and P = 0.2 MPa. The gas may be assumed to follow the

Virial equation of state, z = 1 + BP / RT, where B = – 10–4 m3/mol at the given conditions(R = 8.314 J/mol.K). The value of gR in J/mol is……..... (2-Marks)(A) 0.008 (B) –2.4 (C) 20 (D) –20

12. A binary mixture of components (1) and (2) forms an azeotrope at 130ºC and x1 = 0.3. The

liquid phase non-ideality is described by ln 21 2Ax and ln 2

2 1 1 2, andAx are the activity

coefficients, and x1, x2 are the liquid phase mole fractions. For both components, the fugacitycoefficients are 0.9 at the azeotropic composition. Saturated vapour pressure at 130 ºC are

1 70satP bar and 2 30satP bar. The total pressure in bars for the above azeotropic system (up

to two decimal places) is _________ . (2-Marks)



ANSWERS KEY

1 2 3 4 5 6 7 8 9 10

–58 373 400 1.64 D B A D 310 750

11 12 13 14 15 16 17 18 19 20

D 27.54 B D C D 0.6965 3 C D

21 22 23 24 25 26 27 28 29 30

B B 0.0736 21.7 B A A A D C

31 32 33 34 35 36 37 38 39 40

B D B C D C C A B D

41 42 43 44 45 46 47 48 49 50

D D C D B B B C B A

51 52 53 54 55 56 57 58 59 60

D A B A C C D B C D

61 62 63 64 65 66 67 68 69 70

C B B B D C B B B A

71 72 73 74 75 76 77 78 79 80

C D B A B D D B C B

81 82 83 84 85 86 87 88 89 90

B D A D C A C A D C

91 92 93 94 95 96 97 98 99 100

B B C D A B C C B A

101 102 103 104 105 106 107 108 109 110

D B B C B D A A A D

111 112 113

C C A



SOLUTIONS

GATE-2015

1. (–58)2 2

1 2 1 2h 2 60x 100x x

For infinite dilution x1 = 0 and x2 = 1

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

1h 58

2. (373)Given

1 2

1

3 bar, 6 bar

T 300

P P

K

2 1

1 1.277 11.2772

2 11

( ) 1.5( )

J30

mole.KJ

8.314mole.K

1.277

Since ( )

2and 300 348.67

1

irr rev

V

P

V

rev V

W W

C

R

C

C

W C T T

PT T

P

Hence 30(348.67 300) 1460revJ

Wmole

1.5(1460) 2190

2190 [ 300]

irr

irr V f

JW

moleW C T

2190 30[ 300]

373

f

f

T

T K

3. (400)

...( )

RTP

v bRT

v b iP



0 1

50

71

and

where 4 10

1.35 10

b b b T

b

b

0 1

1 0 ( )

RTv b b T

PR

v T b b iiP

1

Since

P

T P

v Rb

T P

h vv T

P T

Putting value of v andP

v

T

in above equation

1 0 1 0T

h R RT b b T b b

P P P

50 4 10

T

hb

P

54 10 ( )h P

5 6 j4 10 [15 5] 10 400

moleh

4. (1.646)

Given

11

22

0.4 , 1

0.6 , 2

...( )E id

Gx

RTG

xRT

G G G i

From equation (i) and (ii)

,E idG G G given that 1 2(3 )EG x x RT

1 2 1 1 2 2 1 1 2 23 [ { ln ln }]G x x RT x G x G RT x x x x

1 21 2 1 2 1 1 2 23 { ln ln }

G GGx x x x x x x x

RT RT RT

Put values of 1 21 2,

G Gx x and

RT RTin above equation we get.

3 0.4 0.6 (0.4 1 0.6 2) (0.4ln 0.4 0.6ln 0.6) 1.646G

RT

CHEMICAL ENGINEERING 6. FLUID MECHANICS [279]


6. FLUID MECHANICS(GATE Previous Papers)

GATE-20161. For a flow through a smooth pipe, the Fanning friction factor (f) is given 0.2Ref m in the

turbulent flow regime, where Re is the Reynolds number and m is a constant. Water flowingthrough a section of this pipe with a velocity 1 m/s results in a frictional pressure drop of 10kPa. What will be the pressure drop across this section (in kPa), when the velocity of water is2m/s? (1-Mark)

(A) 11.5 (B) 20 (C) 34.8 (D) 40

2. 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. Theinterface between the two layers lies between the two points at which the L-tubes areconnected. The levels (in cm) to which the liquids rise in the respective tubes are shown in thefigure below. (1-Mark)

The distance (x in cm, rounded off to the first decimal place) of the interface from the pointat which the lower L-tube is connected is _______

3. Water (density=1000 kg m‒3) is pumped at a rate of 36 m3/h, from a tank 2 m below the pump,to an overhead pressurized vessel 10 m above the pump. The pressure values at the point ofsuction from the bottom tank and at the discharge point to the overhead vessel are 120 kPa and240 kPa, respectively. All pipes in the system have the same diameter. Take acceleration due togravity, g = 10 m s‒2. Neglecting frictional losses, what is the power (in kW) required to deliverthe fluid? (2-Marks)

(A) 1.2 (B) 2.4 (C) 3.6 (D) 4.8

4. The characteristics curve (Head – Capacity relationship) of a centrifugal pump is representedby the equation Hpump = 43.8 – 0.19Q, where Hpump is the head developed by the pump (in m)and Q is the flowrate (in m3/h) through the pump. This pump is to be used for pumping waterthrough a horizontal pipeline. the frictional head loss Hpiping = 0.0135QL

2 + 0.045QL. Theflowrate (in m3/h, rounded off to the first decimal place) of water pumped through the abovepipeline, is _________ (2-Marks)



5. Water flows through a smooth circular pipe under turbulent conditions. In the viscous sub-layer, the velocity varies linearly with the distance from the wall. The Fanning friction factor is

defined as,2 / 2wf

u where w is the shear stress at the wall of the pipe, is the density of

the fluid and u is the average velocity in the pipe. Water (density = 1000 kgm3,viscosity = 110–3 kg m–1 s–1) flows at an average velocity of 1 m s ‒1 through the pipe. For thisflow condition, the friction factor f is 0.005. At a distance of 0.05 mm from the wall of the pipe(in the viscous sub-layer), the velocity (in m s‒1, rounded off to the third decimal place), is_______

GATE-20156. For uniform laminar flow over a flat plate, the thickness of the boundary layer, , at a distance x

from the leading edge of the plate follows the relation: (1-Mark)

(A) 1( )x x (B) ( )x x (C) 1/2( )x x (D) 1/2( )x x

7. A cylindrical packed bed of height 1m is filled with equal sized spherical particles. Theparticles are nonporous and have a density of 1500 kg/m3. The void fraction of the bed is 0.45.The bed is fluidized using air (density 1 kg/m3). If the acceleration due to gravity is 9.8 m/s2,the pressure drop (in Pa) across the bed in incipient fluidization (up to one decimal place) is__________ . (1-Mark)

8. Two different liquids are flowing through different pipes of the same diameter. In the first pipe,the flow is laminar with centreline velocity, Vmax,1, whereas in the second pipe, the flow isturbulent. For turbulent flow, the average velocity is 0.82 times the centerline velocity, Vmax, 2.For equal volumetric flow rates in both the pipes, the ratio max,1 max, 2V / V (up to two decimal

places) is ________ . (1-Mark)

9. For Fanning friction factor f (for flow in pipes) and drag coefficient CD (for flow over immersedbodies), which of the following statements are true? (2-Mark)P. f accounts only for the skin frictionQ. CD accounts only for the skin frictionR. CD accounts for both skin friction and form frictionS. Both f and CD depend on the Reynolds numberT. For laminar flow through a pipe, f doubles on doubling the volumetric flow rate(A) R, S, T (B) P, Q, S (C) P, R, S (D) P, Q, S, T

10. A centrifugal pump delivers water at the rate of 0.22 m3/s from a reservoir at ground level toanother reservoir at a height H, through a vertical pipe of 0.2 m diameter. Both the reservoirsare open to atmosphere. The power input to the pump is 90 kW and it operates with anefficiency of 75%. (2-Mark)Data:Fanning friction factor for pipe flow is f = 0.004. Neglect other head losses.Take gravitational acceleration, g = 9.8m/s2 and density of water is 1000 kg/m3.The height H, in meters, to which the water can be delivered (up to one decimal place) is__________



ANSWER KEY

1 2 3 4 5 6 7 8 9 10

C 10 B 48.9 0.125 C 8079.6 1.64 C 25

11 12 13 14 15 16 17 18 19 20

C D 8 8 A 80.8 B A D C

21 22 23 24 25 26 27 28 29 30

B 137500 C B A C B A D A

31 32 33 34 35 36 37 38 39 40

C D D C A C B A A D

41 42 43 44 45 46 47 48 49 50

A B C D B B B D D D

51 52 53 54 55 56 57 58 59 60

B B A C D B B C A D

61 62 63 64 65 66 67 68 69 70

A A B B C C A A D C

71 72 73 74 75 76 77 78 79 80

D B B C B B C B C D

81 82 83 84 85 86 87 88 89 90

C B C C D B D B C C

91 92 93 94 95 96 97 98 99 100

A D B D B A A B D C

101 102 103 104 105 106 107 108 109 110

A B D D D B C B D A

111 112 113 114 115 116 117 118 119 120

C A B A D B B C D B

121 122 123 124

B B A D



SOLUTIONS

GATE-2016

1. (C)

0.2

mf

(Re)

From Darcy’s law2

f

4fLvh

2gD , Since f

Ph

g

Hence,24 fLv

P2D

Putting the value of f in P , we get2

0.2

4 mLvP

2D(Re)

since Re =

vD

hence, 1.8P v

1.8

2 2

1 1

P v

P v

Given that 2 1 1v 2m/ sec , v 1m/ sec , P 10kPa

Hence, 1.82P 10(2) 34.8kPa

2. (10)

Pressure intensity along the line Y and Y’ will be same in each column. Hence(0.5 x)800g 1000xg 0.42(1000g)

On solving this equation, we getx = 0.1 m = 10 cm

3. (B)Given = 1000 kg/m3

Q = 36 m3/hr



P1 = 120 KPaP2 = 240 KPaSince diameter of pipes are same and volumetric flow rate are also same hence velocity insuction and discharge pipes will be same.V1 = V2

2 21 1 2 2

1 22 2p fP V P V

Z h Z hg g Pg g

Given that 1 20 ,fh V V

32 1

2 1 3

120 10( ) (12 0) 24

10 10

p

P Ph Z Z m

g

Now power (P) = ghQ36

1000 10 243600

= 2400Watt= 2.4 kW

4. (48.9 m3/h)

Given that,

Hpump = 43.8 – 0.19 Q …(1)

Hpiping = 0.0135QL2 + 0.045QL …(2)

On equating equation (1) and (2)

43.8 – 0.19 Q = 0.0135QL2 + 0.045QL

0.0135Q2 + 0.235Q – 43.8 = 0

On solving Q = 48.9 m3/h5. (0.125)

2

2wf u

=0.05 1000 1

2.52

Since wdu

dy

CHEMICAL ENGINEERING 7. PROCESS CALCULATIONS [333]


7. PROCESS CALCULATIONS(GATE Previous Papers)

GATE-2016

1. A liquid mixture of ethanol and water is flowing as inletstream P into a stream splitter. It is split into two streams, Qand R, as shown in the figure below.The flow rate of P, containing 30 mass% of ethanol, is 100kg/h. What is the least number of additional specification(s)required to determine the mass flowrates and compositions(mass%) of the two exit streams? (1-Mark)(A) 0 (B) 1 (C) 2 (D) 3

2. A jacketed stirred tank with a provision for heat removal is used to mix sulphuric acid andwater in a steady state flow process. H2SO4 (l) enters at a rate of 4 kg/h at 25oC and H2O (l)enters at a rate of 6 kg/h at 10oC. The following data are available: (2-Marks)Specific heat capacity of water = 4.2 kJ kg‒1K‒1.Specific heat capacity of aqueous solution of 40 mass% H2SO4 = 2.8 kJ (kg solution)‒1 K‒1.Assume the specific heat capacities to be independent of temperature.

Based on reference states of H2SO4 (l) and H2 O (l) at 25oC, the heat of mixing for aqueoussolution of 40 mass% H2SO4 = ‒ 650 kJ (kg H2SO4)

‒1.If the mixed stream leaves at 40oC, what is the rate of heat removal (in kJ/h)?(A) 1802 (B) 2558 (C) 5702 (D) 6458

3. A catalytic reforming plant produces hydrogen and benzene from cyclohexane by de-hydroaromatisation. In order to increase the production of hydrogen, the owner plans to change theprocess to steam reforming of the same feedstock that produces hydrogen and carbon dioxide.Stoichiometrically, what is the maximum ratio of pure hydrogen produced in the proposedprocess to that in the existing process? (2-Mark)(A) 1 (B) 2 (C) 5 (D) 6

GATE-2015

4. The schematic diagram of a steady state process is shown below. The fresh feed (F) to thereactor consists of 96 mol% reactant A and 4 mol% inert I. The stoichiometry of the reaction inA C. A part of the reactor effluent is recycled. The molar flow rate of the recycle stream is0.3F. The product stream P contains 50 mol% C. The percentage conversion of A in the reactorbased on A entering the reactor at point 1 in the figure (up to one decimal place) is_______________ (2-Marks)



GATE-2014

5. Two elemental gases (A and B) are reacting to form a liquid (C) in a steady state process as perthe reaction A + B C. The single-pass conversion of the reaction is only 20% and hencerecycle is used. The product is separated completely in pure form. The fresh feed has 49 mol%of A and B each along with 2 mol% impurities. The maximum allowable impurities in therecycle stream is 20 mol%. The amount of purge stream (in moles) per 100 moles of the freshfeed is ___________ (2-Marks)

6. Carbon monoxide (CO) is burnt in presence of 200% excess pure oxygen and the flametemperature achieved is 2298K. The inlet streams are at 25ºC. The standard heat of formation(at 25ºC) of CO and CO2 are – 110 kJ mol-1 and –390 kJ mol-1 respectively. The heat capacities(in J mol-1 K-1) of the components are (2-Marks)

3 3

2 225 14 10 25 42 10p po co

C T C T

Where, T is the temperature in K. The heat loss (in kJ) per mole of CO burnt is _________

GATE-2013

Common Data for Questions 7 and 8:A reverse osmosis unit treats feed water (F) containing fluoride and its output consists of a permeatestream (P) and a reject stream (R). Let CF, CP and CR denote the fluoride concentrations in the feed,permeate, and reject streams, respectively. Under steady state conditions, the volumetric flow rate ofthe reject is 60 % of the volumetric flow rate of the inlet stream, and CF = 2 mg/L and CP = 0.1 mg/L.

7. The value of CR in mg/L, up to one digit after the decimal point, is ________ (2-Marks)

8. A fraction f of the feed is bypassed and mixed with the permeate to obtain treated water havinga fluoride concentration of 1 mg/L. Here also the flow rate of the reject stream is 60% of theflow rate entering the reverse osmosis unit (after the bypass). The value of f , up to 2 digits afterthe decimal point, is __________ (2-Marks)

GATE-2012Common Data for Questions for 9 and 10:The reaction ,liq gas liq gasA B C D is carried out in a reactor followed by a separator as

shown below.

Notation:Molar flow rate of fresh B is FFB Molar flow rate of A is FA

Molar flow rate of recycle gas is FRG Molar fraction of B in recycle gas is YRB

Molar flow rate of purge gas is FPG Molar flow rate of C is FC

Here, FFB = 2 mol/s; FA = 1 mol/s, FB/FA = 5 and A is completely converted.

9. If 0.3RBY , the ratio of recycle gas to, purge gas /RG PGF F is (2-Marks)

(A) 2 (B) 5 (C) 7 (D) 10

10. If the ratio of recycle gas to purge gas /RG RBF F is 4 then YRB is (2-Marks)

(A)3

8(B)

2

5(C)

1

2(D)

3

4



GATE-2014



3 3

2 225 14 10 25 42 10p po co

C T C T


GATE-2013





shown below.






(A) 2 (B) 5 (C) 7 (D) 10


(A)3

8(B)

2

5(C)

1

2(D)

3

4



GATE-2014



3 3

2 225 14 10 25 42 10p po co

C T C T


GATE-2013





shown below.






(A) 2 (B) 5 (C) 7 (D) 10


(A)3

8(B)

2

5(C)

1

2(D)

3

4



ANSWER KEY

1 2 3 4 5 6 7 8 9 10B A D 45.4 10 34.62 3.27mg/L 0.264 B A11 12 13 14 15 16 17 18 19 20A B C D D C C B B C21 22 23 24 25 26 27 28 29 30D B C A C C D D C D31 32 33 34 35 36 37 38 39 40A B C B B B A D A C41 42 43 44 45 46 47 48 49 50D C D B B A D C B C51D



SOLUTIONS

GATE-20161. (B)

Given thatP = 100 kg/hTotal mass balanceP = 100 = Q + R …(1)Ethanol balanceP xP = QxQ + RxR

Given that, xP = 0.3P (0.2) = QxQ + RxR …(2)Where, xQ and xR compositions of ethanol in stream Q and R respectively,Because of splitting xQ and xR are same as xP = 0.2Hence, equation (2) reduced to 100 = Q + RHence, only one additional specification required to determine the mass flow rate andcomposition.

2. (A)Applying steady state energy balanceHeat rate in – Heat rate out + Heat for mixing of solution – Heat removed = 0(kJ/hr) (kJ/hr) (kJ/hr) (kJ/hr)

Heat rate in

2 4

3P H SO

kJ4 (C ) (25 25) 6 4.2 10 (10 25) 378

hr

Heat rate outkJ

10 2.8(40 25) 420hr.

Heat for mixingkJ

650 4 2600hr.

Put all values in energy balance equationHeat rate in – Heat rate out + Heat for mixing of solution – Heat removed = 0(kJ/hr) (kJ/hr) (kJ/hr) (kJ/hr)

– 378 – 420 + 2600 = Heat removed

Heat removed = 1802kJ

hr.3. (D)

Dehydrogenation of Cyclo-hexane

6 12 6 6 23C H C H H …(1)Steam reforming of Cyclo-hexane

6 12 2 2 212 6 18C H H O CO H …(2)

Hence, at complete conversion of 6 12C H , the ratio of H2 form in case (2) to that in case (1) is

186

3

CHEMICAL ENGINEERING 8.MECHANICAL OPERATIONS [359]


8. MECHANICAL OPERATIONS(GATE Previous Papers)

GATE-2016

1. In a cyclone separator used for separation of solid particles from a dust laden gas, theseparation factor is defined as the ratio of the centrifugal force to the gravitational force actingon the particle. Sr denotes the separation factor at a location (near the wall) that is at a radialdistance r from the centre of the cyclone. Which one of the following statements isINCORRECT? (1-Mark)(A) Sr depends on mass of the particle(B) Sr depends on the acceleration due to gravity(C) Sr depends on tangential velocity of the particle(D) Sr depends on the radial location ( r ) of the particle

2. An agitated cylindrical vessel is fitted with baffles and flat blade impellers. The power numberfor this system is given by where P is the power consumed for the mixing, ρ is the density ofthe fluid, n is the speed of the impeller and D is the diameter of the impeller. The diameter ofthe impeller is 1/3rd the diameter of the tank and the height of liquid level is equal to the tankdiameter. The impeller speed to achieve the desired degree of mixing is 4 rpm. In a scaled updesign, the linear dimensions of the equipment are to be doubled, holding the power input perunit volume constant. Assuming the liquid to be Newtonian and Np to be independent ofReynolds number, what is the impeller speed (in rpm) to achieve the same degree of mixing inthe scaled up vessel? (2-Marks)(A) 0.13 (B) 1.26 (C) 2.52 (D) 3.82

3. Consider a rigid solid sphere falling with a constant velocity in a fluid. The following data areknown at the conditions of interest: viscosity of the fluid = 0.1 Pas, acceleration due to gravity= 10 m s‒2, density of the particle = 1180 kg m‒3 and density of the fluid = 1000 kg m‒3.The diameter (in mm, rounded off to the second decimal place) of the largest sphere that settlesin the Stokes’ law regime (Reynolds number 0.1), is _______ (2-Marks)

GATE-2015

4. A typical batch filtration cycle consists of filtration followed by washing. One such filtrationunit operation at constant pressure difference first filters a slurry during which 5 liters of filtrateis collected in 100s. This is followed by washing which is done for tw seconds and uses 1 literof wash water. Assume the following relation to be applicable between the applied pressuredrop P, cake thickness L at time t, and volume of liquid V collected in time t: (2-Marks)

1 2; , if is changing.P dV

k L k V LL dt

k1 and k2 can be taken to be constant during filtration and washing. The wash time tw, in seconds(up to one decimal place), is _________________

5. The diameters of sand particles in a sample ranges from 50 to 150 microns. The number of

particles of diameter x in the sample is proportional to1

.50 x

The average diameter, in

microns, (up to one decimal place) is _______________ (2-Marks)



6. A spherical solid particle of 1 mm diameter is falling with a downwardvelocity of 1.7 mm/s through a liquid (viscosity 0.04 Pa.s) at a low Reynoldsnumber (Stokes regime). The liquid is flowing upward at a velocity of 1 mm/s.All velocities are with respect to a stationary reference frame. Neglecting thewall effects, the drag force per unit projected area of the particle, in Pa,(up totwo decimal places) is ____________ (2-Marks)

GATE-20147. In order to produce fine solid particles between 5 and 10 μm, the appropriate size reducing

equipment is (1-Mark)(A) Fluid energy mill (B) Hammer mill(C) Jaw crusher (D) Smooth roll crusher

8. Slurries are most conveniently pumped by a (1-Mark)(A) Syringe pump (B) Diaphragm pump(C) Vacuum pump (D) Gear pump

GATE-20139. In the Tyler standard screen scale series, when the mesh number increases from 3 mesh to 10

mesh, then (1-Mark)(A) the clear opening decreases (B) the clear opening increases(C) the clear opening is unchanged (D) the wire diameter increases

10. Taking the acceleration due to gravity to be 10 m/s2, the separation factor of a cyclone 0.5 m indiameter and having a tangential velocity of 20 m/s near the wall is _____ (1-Mark)

11. 100 ton/h of a rock feed, of which 80% passed through a mesh size of 2.54 mm, were reducedin size such that 80% of the crushed product passed through a mesh size of 1.27 mm. Thepower consumption was 100 kW. If 100 ton/h of the same material is similarly crushed from amesh size of 5.08 mm to a mesh size of 2.54 mm, the power consumption (in kW, to the nearestinteger) using Bond’s law, is _____________ (2-Marks)

GATE-201212. In a mixing tank operating at very high Reynolds number ( > 104), if the diameter of the

impeller is doubled (other conditions remaining constant), the power required increases by afactor of (1-Mark)(A) 1/32 (B) 1/4 (C) 4 (D)

GATE-201113. The particle size distributions of the feed and collected solids (sampled for same duration ) for

a gas cyclone are given below. (2-Marks)

What is the collection efficiency (in PERCENTAGE) of the gas cyclone?(A) 31 (B) 60 (C) 65 (D) 69

32

Size range Weight of feed in thesize range (g)

Weight of Collected solids in the sizerange (g)

1-55-1010-1515-2020-2525-30

2.03.05.06.03.01.0

0.10.73.65.52.91.0

μm



ANSWER KEY

1 2 3 4 5 6 7 8 9 10

A C 2.15 40 94.3 1.29 A B A 16011 12 13 14 15 16 17 18 19 20

70.71 D D D D D C B B B21 22 23 24 25 26 27 28 29 30B B B C A C C A B B31 32 33 34 35 36 37 38 39 40C D B C A D B D C A



SOLUTIONS

GATE-20161. (A)

Separation factor =

2tan

2centrifugal tan

gravity

muF urF mg rg

Separation factor2tanu

rg

Separation factor does not depend on mass of particle.

2. (C)

3 5pP

Nn D

Since (Re)pN f k constant

3 5P k n DGiven : Power per unit volume is constant

Volume of tank =2

4TD L

Given that D =1

3 TD

Hence, DT = 3DAnd L = DT = 3D

Volume = 29 34

D D

327

4D

3 5

3274

P k n D

V D

constant

3 2Pn D

V constant

Means3 2 3 2

1 1 2 2n D n D3 2

1 2

2 1

n D

n D

1 24rpm , ?n n

Given, 2 12D D

2 2.53n rpm

CHEMICAL ENGINEERING 9. PLANT DESIGN ECONOMICS [375]


9. PLANT DESIGN ECONOMICS


GATE-20161. Two design options for a distillation system are being compared based on the total annual cost.

Information available is as follows: (1-Mark)

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

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

2. Standard pipes of different schedule numbers and standard tubes of different BWG numbersare available in the market. For a pipe/tube of a given nominal diameter, which one of thefollowing statements is TRUE? (1-Mark)(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

3. Terms used in engineering economics have standard definitions and interpretations. Which oneof the following statements is INCORRECT? (1-Mark)1. The profitability measure ‘return on investment’ does not consider the time value of

money2. A cost index is an index value for a given time showing the cost at that time relative to a

certain base time3. 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

4. Payback period is calculated based on the payback time for the sum of the fixed and theworking capital investment

4. A vertical cylindrical tank with a flat roof and bottom is to be constructed for storing 150 m3 ofethylene glycol. The cost of material and fabrication for the tank wall is Rs 6000 per m2 andthe same for the roof and the tank bottom are Rs 2000 and Rs 4000 per m2, respectively. Thecost of accessories, piping and instruments can be taken as 10% of the cost of the wall. 10% ofthe volume of the tank needs to be kept free as vapour space above the liquid storage. What isthe optimum diameter (in m) for the tank? (2-Marks)

(A) 3.5 (B) 3.9 (C) 7.5 (D) 7.8



ANSWER KEY

1 2 3 4 5 6 7 8 9 10

A B D D 572.8 15.3 60393 2.62 1032386.23 216.5

11 12 13 14 15 16 17 18 19 20C B C B A B C B D C21 22 23 24 25 26 27 28 29 30C D A A B D C D C A31 32 33 34 35 36 37 38 39 40D C A B B A B C B C41 42 43 44 45 46 47 48 49 50D B B D C b B B D A



SOLUTIONS

GATE-2016

1. (A)For PTotal cost = (150 0.12) + 6 + 16

= 40 lakhs/year

For QTotal cost = (120 0.12) + 8 + 20

= 42.4 lakhs/yearSo, the better option is P having total cost of 40 lakhs/year.

2. (B)Correct statements is:-Wall thickness increases with increase in the schedule number and decreases with increase inBWG number

3. (D)Option “d” is INCORRECT because

P.B.P =Depreciable F.C.I

Average profit + Average depriciation

This is F.C.I, that does not include working capital investment

4. (D)90% V = 150 m3

3150 150 100166.7 m

0.90 90V

2 2 10( )6000 2000 4000 ( 6000)

4 4 100Td d

C dl dl

2

6600 (6000)4Td

C dl

2

2

6600 4(6000)

4Td V d

Cd

266004 6000

4Td

C Vd

26600 4 166.671500TC d

d

32

6600 4 166.673000 0 6600 4 166.61 3000

TCd

d dd d d

3 467.10 7 m 7.8m.75d d

CHEMICAL ENGINEERING 10. CHEMICAL TECHNOLOGY [393]


10. CHEMICAL TECHNOLOGY(GATE Previous Papers)

GATE-2016

1. India has no elemental sulphur deposits that can be economically exploited. In India, which oneof the following industries produces elemental sulphur as a by-product? (1-Mark)

(A) Coal carbonization plants (B) Petroleum refineries

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

2. Two paper pulp plants P and Q use the same quality of bamboo as a raw material. Thechemicals used in their digester are as follows: (1-Mark)

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

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

Group-1 Group-2 (1-Mark)

P Ethylene polymerization 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



GATE-2015

4. Match the chemicals written on the left with the raw materials required to produce themmentioned on the right. (1-Mark)(I) Single superphosphate (SSP) (P) Rock phosphate + Sulfuric Acid + Ammonia(II) Triple Superphosphate (TSP) (Q) Brine(III) Diammonium Phosphate (R) Rock phosphate + Sulfuric Acid(IV) Caustic sode (S) Rock phosphate + Phosphoric Acid(A) I-Q, II-R, III-S, IV-P (B) I-S, II-P, III-Q, IV-R(C) I-R, II-S, III-P, IV-Q (D) I-S, II-R, III-P, IV-Q

5. Match the technologies in Group 1 with the entries in Group 2: (1-Mark)Group 1 Group 2P. Urea manufacture I. MicroencapsulationQ. Coal gasification II. Ultra-low sulphur dieselR. Controlled release of chemicals III. Shale oilS. Deep hydrodesulphurization IV. Prilling towar

V. Gas hydratesVI. Gas-solid non-catalytic reaction

(A) P-I, Q-V, R-II, S-VI (B) P-IV, Q-VI, R-I, S-II(C) P-IV, Q-I, R-III, S-II (D) P-V, Q-VI, R-VI, S-II

6. Match the polymer mentioned on the left with the catalyst used for its manufacture given on theright. (2-Marks)(I) Low density Polyethylene (P) Ziegler-Natta catalyst(II) High density polyethylene (Q) Traces of Oxygen(III) Polyethylene Terephthalate (R) Butyl Lithium(IV) Polyvinyl Chloride (S) Antimony(A) I-Q, II-R, III-S, IV-P (B) I-S, II-P, III-Q, IV-R(C) I-Q, II-P, III-S, IV-R (D) I-S, II-R, III-P, IV-Q

GATE-20147. Decomposition efficiency (D) of an electrolytic cell used for producing NaOH is defined as

(A) D = (grams of NaOH produced /grams of NaCl decomposed) 100 (1-Mark)(B) D = (grams of a NaOH produced /grams of NaCl charged) 100

(C) D = (gram equivalents of NaOH produced/gram equivalents of NaCl charged) 100

(D) D = (theoretical current to produce one gram equivalent/actual current to produce one

gram equivalent) 100

8. Catalytic cracking is (1-Mark)(A) A hydrogen additions process (B) A carbon rejection process(C) An exothermic process (D) A coking process

9. Which ONE of the following statements is CORRECT? (1-Mark)(A) The major components of biodiesel are triglycerides(B) Biodiesel is essentially a mixture of ethyl esters(C) Biodiesel is highly aromatic(D) Biodiesel has a very low aniline point



87. Select the process form the right-hand column which matches with catalyst given on the left-hand column(I) Nickel (a) Hydrogenation of vegetable oils (2-Marks)(II) Zeolites (b) Ammonia synthesis

(c) Catalytic cracking of gas oil(d) Alkylation(e) Sulphuric acid manufacture(f) Nitric acid manufacture(g) Methanol synthesis(h) Ethylene oxide manufacture

(A) (B)I–a, b, fII–c, d, e, g, h

(C) (D)I–a, b,II–d, e, f, g, h

c

ANSWER KEY

1 2 3 4 5 6 7 8 9 10B D A C B C C B B B11 12 13 14 15 16 17 18 19 20A C D D A B C C A B21 22 23 24 25 26 27 28 29 30C C D D A D D D B A31 32 33 34 35 36 37 38 39 40A A A D D C D D B A41 42 43 44 45 46 47 48 49 50A D B D C C A B A B51 52 53 54 55 56 57 58 59 60D C D A C D A B B A61 62 63 64 65 66 67 68 69 70C D D B C B B B C A71 72 73 74 75 76 77 78 79 80C A D B D D A C A A81 82 83 84 85 86 87D B C B D A A

I –a, c, fII–b, d, e, g,h I –b, c, fII–a, d, e, g,h



SOLUTIONS

GATE-20161. (B)

Petroleum refineries

2. (D)In a Kraft process the chemicals used in the digester are Na2S.

3. (A)Ethylene Polymerization Aluminium triethyl with TiCl2 p

Petroleum feedstock cracking Zeolite

Oxidation of SO2 + SO3 Vanadium pentaoxide

Hydrogenation of oil Nickel

GATE-20154. (C)

SSP Reaction

( )3 4 2 2 2 4 4 4 2 43

( ) . 7 3 ( ) .7 2 SSPRP SA

aqCa PO CaF H SO CaH PO CaSO HF

TSP Reaction

3 4 2 2 3 4 4 4 23( ) . 14 10 ( ) 2

TSPPARP

Ca PO CaF H PO CaH PO HF

DAP Reaction(1) 3 3 4 4 2 4NH H PO NH H PO

(2) 3 4 2 4 4 2 4( ) ( )NH NH H PO NH HPO DAP

So, Ammonia PA RP DAP (Note: SA also used with PA)

Caustic soda: Electrolytic process for chlorine-caustic soda production by brine

5. (B)P : urea manufacture IV-Prilling tower

Q : Coal gasification VI. Gas solid non-catalytic reaction

R : Controlled release of chemical I: Microencapsulation

S : Deep hydrodesulphurisation II- Ultra-low sulphur diesel6. (C)

I. LDPE-Traces of oxygen (Q)II. HDPE-Ziegler-Natta Catalyst (P)III. PET-Antimony (S)IV. PVC-Butyl Lithium (R)

CHEMICAL ENGINEERING 11. ENGINEERING MATHEMATICS [419]


11. ENGINEERING MATHEMATICS(GATE Previous Papers)

GATE-20161. Which one of the following is an iterative technique for solving a system of simultaneous linear

algebraic equations? (1-Mark)(A) Gauss elimination (B) Gauss-Jordan(B) Gauss-Seidel (D) LU decomposition

2. The Laplace transform of e at sin (bt) is (1-Mark)

(A)2 2( )

b

s a b (B)

2 2( )

s a

s a b

(C)2 2

( )

( )

s a

s a b

(D)2 2( )

b

s a b

3. What are the modulus (r) and argument ( ) of the complex number 3+4i ? (1-Mark)

(A) 1 47 , tan

3r

(B) 1 3

7 , tan4

r

(C) 1 35 , tan

4r

(D) 1 4

5 , tan3

r

4. A set of simultaneous linear algebraic equations is represented in a matrix form as shownbelow. (2-Marks)

1

2

3

4

5

460 0 0 4 13

1612 5 5 2 10

610 0 2 5 3

300 0 0 4 5

812 3 2 1 5

x

x

x

x

x

The value (rounded off to the nearest integer) of x3 is _________

5. What is the solution for the second order differential equation2

20

d yy

dx , with the initial

conditions0

0

5 and 10 ?x

x

dyy

dx

(2-Marks)

(A) 5 10siny x (B) 5cos 5siny x x

(C) 5cos 10y x x (D) 5cos 10siny x x

6. The model y = mx2 is to be fit to the data given below. (2-Marks)

1 2 3

2 5 8

x

y

Using linear regression, the value (rounded off to the second decimal place) of m is _______



7. The Lagrange mean-value theorem is satisfied for f( x)= x3 + 5 , in the interval (1, 4) at a value(rounded off to the second decimal place) of x equal to _______ (2-Marks)

8. Values of f(x) in the interval [0, 4] are given below. (2-Marks)

0 1 2 3 4

( ) 3 10 21 36 55

x

f x

Using Simpson’s 1/3 rule with a step size of 1, the numerical approximation (rounded off to

the second decimal) of4

0

( )f x dx is

GATE-20159. A complex-valued function, f (z), given below is analytic in domain D: (1-Mark)

( ) ( , ) ( , )f z u x y iv x y z x iy

Which of the following is NOT correct?

(A)df v u

idz y y

(B)

df u vi

dz x x

(C)

df v ui

dz y y

(D)

df v vi

dz y x

10. A scalar function in the xy-plane is given by (x, y) = x2 + y2. If andi j are unit vectors in the

x and y directions, the direction of maximum increase in the value of at (1, 1) is along:

(1-Mark)

(A) 2 2i j (B) 2 2i j (C) 2 2i j (D) 2 2i j

11. The following set of three vectors (1-Mark)

1 3

2 , 6 4

1 2

x

and

x

is linearly dependent when x is equal to(A) 0 (B) 1 (C) 2 (D) 3

12. For the matrix4 3 1

, if3 4 1

is an eigenvector, the corresponding eigen value is _______ .

(1-Mark)

13. Consider a linear ordinary differential equation: ( ) ( ).dy

p x y r xdx Function ( )p x and ( )r x

are defined and have a continuous firs derivative. The integrating factor of this equation is non-zero. Multiplying this equation by its integrating factor converts this into a: (1-Mark)(A) Homogenous differential equation (B) Non-linear differential equation(C) Second order differential equation (D) Exact differential equation

14. The solution of the non-linear equation (2-Marks)3 0x x is to be obtained using Newton-Raphson’s method. If the initial guess is x = 0.5, the

method converges to which one of the following value:(A) –1 (B) 0 (C) 1 (D) 2



ANSWERS KEY

1 2 3 4 5 6 7 8 9 10

C A D 15 D 2.53 2.64 94.67 A B11 12 13 14 15 16 17 18 19 20D 7 D A 0.33 8 A D C A21 22 23 24 25 26 27 28 29 30

2.33 A B C 7 B 0.083 B D -0.2531 32 33 34 35 36 37 38 39 40B C 0.38 B D B D C D B41 42 43 44 45 46 47 48 49 50B C C D D D D C A C51 52 53 54 55 56 57 58 59 60C C C D A B D D B B61 62 63 64 65 66 67 68 69 70C C C C A B C D C A71 72 73 74 75 76 77 78 79 80A D D B B C B A A C81 82 83 84 85 86 87 88 89 90D C B C A B C B D C91 92 93 94 95 96 97 98 99 100D B D A C D A D B A

101 102 103 104 105 106 107 108 109 110C A A B B C D B C B

111 112 113 114 115 116 117 118 119 120B A D A C D B D D C

121 122 123 124 125 126 127 128 129 130D C B A C A D A A C

131 132 133 134 135 136 137 138 139 140C A B D A B D D C A

141 142 143 144 145 146 147 148 149A D C D B A B A A



SOLUTIONS

GATE-20161. (C)

Gauss-Seidel2. (A)

at2 2

bL e sin bt

(s a) b

3. (D)Z 3 4i (a ib)

2 2r a b 2 23 4

= 5

1 1b 4tan tan

a 3

4. (15)

4 54 13 46x x …(1)

4 54 5 30x x …(2)On solving this two equation, we get

4 5x & 5 2x

Now, 3 4 52 5 3 61x x x

Putting value of 4 5andx x , we get 3 15x 5. (D)

2

20

d yy

dx

2( 1) 0D y

D i

1 2cos siny C x C x Given that at 0 , 5x y

Hence,

15 0C So, 1 5C

And at 0 , 10dy

xdx

Hence, 1 2sin cosdy

C x C xdx

At 0x

20

10x

dyC

dx

Hence, C1 = 5 and C2 = 105cos 10siny x x

6. (2.53)



7. (2.64)

From Lagrange mean-value theorem( ) ( )

'( )f b f a

f xb a

… (1)

Given that,

b = 4 , a = 1 and 3( ) 5f x x

hence ( ) (4) 69 and ( ) (1) 6f b f f a f

putting the value of f(b) and f(a) in equation (1), we get69 6

'( ) 214 1

f x

and

2'( ) 3f x x

Hence,2 23 21 7 2.64x x x

8. (94.67)By Simpson’s 1/3 rule

4

0 2 1 3 40

( ) 2( ) 4( )3

hf x dx f f f f f 1

(3 2(21) 4(10 36) 553 = 94.67

GATE-20159. (A)

( ) ( , ) ( , )

'( ) ...(1)

f z u x y iv x y

df u i v u i vf z

dz x x y y

For analytic function

&

u v u v

x y y x

So equation (1) can be written as

df v v v u u vi i i

dz y x y y x x

Only option (A) can’t be deduced.

10. (B)For the largest increase of function

Grad f f ff f i j k

x y z

2 2

11

2 2

/ 2(1) 2(1) 2 2xy

f x y

f xi y j

Now f i j i j

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