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GATE Solutions CHEMICAL ENGINEERING [1]
Published by: ENGINEERS INSTITUTE OF INDIA-E.I.I. ALL RIGHT RESERVED
28B/7 Jiasarai Near IIT Hauzkhas Newdelhi-110016 ph. 011-26514888. www.engineersinstitute.com
CHEMICAL CHEMICAL CHEMICAL CHEMICAL ENGINEERINGENGINEERINGENGINEERINGENGINEERING
GATE SOLUTION
Subject-wise descriptive solution.
Practice-Set for each topic with solution.
Register at www.gatechemical.com to get 3 Free GATE Mock Tests
Prepared by: K.Solanki & Eii Team
GATE Solutions CHEMICAL ENGINEERING [2]
Published by: ENGINEERS INSTITUTE OF INDIA-E.I.I. ALL RIGHT RESERVED
28B/7 Jiasarai Near IIT Hauzkhas Newdelhi-110016 ph. 011-26514888. www.engineersinstitute.com
2013 By Engineers Institute of India
ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be
reproduced, transmitted, stored or used in any form or by any means graphic, electronic, or
mechanical & chemical, including but not limited to photocopying, recording, scanning,
digitizing, taping, Web distribution, information networks, or information storage and
retrieval systems.
Engineers Institute of India
28-B/7, Jia Sarai, Near IIT Hauz Khas New Delhi-110016
Tel: 011-26514888
For publication information, visit www.engineersinstitute.com
ISBN: 978-93-5137-754-2
Price: Rs. 515
GATE Solutions CHEMICAL ENGINEERING [3]
Published by: ENGINEERS INSTITUTE OF INDIA-E.I.I. ALL RIGHT RESERVED
28B/7 Jiasarai Near IIT Hauzkhas Newdelhi-110016 ph. 011-26514888. www.engineersinstitute.com
This book is dedicated to all This book is dedicated to all This book is dedicated to all This book is dedicated to all
Chemical Engineers Chemical Engineers Chemical Engineers Chemical Engineers preparing for preparing for preparing for preparing for
GATE ExaminationGATE ExaminationGATE ExaminationGATE Examination....
GATE Solutions
Published by: ENGINEERS INSTITUTE OF INDIA
28B/7 Jiasarai Near IIT Hauzkhas Newdelhi
GATE examination is one of the most prestigious competitive examination
conducted for Graduate engineers. Over the past few years, it has become
more competitive as a number of aspirants are increasingly becoming
interested in M.Tech & government jobs due to d
options.
In my opinion, GATE exam test candidates’ basics understanding of
concepts, ability to apply numerical approach. A candidate is supposed to
smartly deal with the syllabus not just mugging up concepts. Thorough understanding
critical analysis of topics and ability to express clearly are some of the pre
this exam. The questioning & examination pattern has changed in few years, as numerical
answer type questions play a major role to score a good rank. K
difficulties of an average student, we have compose this booklet.
Established in 2006 by a team of IES and GATE toppers, we at
have consistently provided rigorous classes and proper guidance to engineering students over
the nation in successfully accomplishing their dreams. We believe
oriented teaching methodology
students stay ahead in the
professionals who have guided thousands to aspirants over the years. They are readily
available before and after classes to assist students and we maintain a heal
ratio. Many current and past years’ toppers associate with us for contributing towards our
goal of providing quality education and share their success with the future aspirants. Our
results speak for themselves. Past students of EII are
departments and PSU’s and pursuing higher specializations. We also give scholarships to
meritorious students.
R.K. Rajesh
Director
Engineers Institute of India
CHEMICAL ENGINEERING
ENGINEERS INSTITUTE OF INDIA-E.I.I. ALL RIGHT RESERVED
/7 Jiasarai Near IIT Hauzkhas Newdelhi-110016 ph. 011-26514888. www.engineersin
A word to the students
examination is one of the most prestigious competitive examination
conducted for Graduate engineers. Over the past few years, it has become
more competitive as a number of aspirants are increasingly becoming
interested in M.Tech & government jobs due to decline in other career
In my opinion, GATE exam test candidates’ basics understanding of
concepts, ability to apply numerical approach. A candidate is supposed to
smartly deal with the syllabus not just mugging up concepts. Thorough understanding
critical analysis of topics and ability to express clearly are some of the pre
this exam. The questioning & examination pattern has changed in few years, as numerical
answer type questions play a major role to score a good rank. Keeping in mind, the
difficulties of an average student, we have compose this booklet.
Established in 2006 by a team of IES and GATE toppers, we at Engineers Instit
have consistently provided rigorous classes and proper guidance to engineering students over
the nation in successfully accomplishing their dreams. We believe in
teaching methodology with updated study material and test series so
the competition. The faculty at EII are a team of experienced
professionals who have guided thousands to aspirants over the years. They are readily
available before and after classes to assist students and we maintain a heal
ratio. Many current and past years’ toppers associate with us for contributing towards our
goal of providing quality education and share their success with the future aspirants. Our
results speak for themselves. Past students of EII are currently working in various
departments and PSU’s and pursuing higher specializations. We also give scholarships to
[4]
ALL RIGHT RESERVED
www.engineersinstitute.com
examination is one of the most prestigious competitive examination
conducted for Graduate engineers. Over the past few years, it has become
more competitive as a number of aspirants are increasingly becoming
ecline in other career
In my opinion, GATE exam test candidates’ basics understanding of
concepts, ability to apply numerical approach. A candidate is supposed to
smartly deal with the syllabus not just mugging up concepts. Thorough understanding with
critical analysis of topics and ability to express clearly are some of the pre-requisites to crack
this exam. The questioning & examination pattern has changed in few years, as numerical
eeping in mind, the
Engineers Institute of India
have consistently provided rigorous classes and proper guidance to engineering students over
in providing exam-
and test series so that our
The faculty at EII are a team of experienced
professionals who have guided thousands to aspirants over the years. They are readily
available before and after classes to assist students and we maintain a healthy student-faculty
ratio. Many current and past years’ toppers associate with us for contributing towards our
goal of providing quality education and share their success with the future aspirants. Our
currently working in various
departments and PSU’s and pursuing higher specializations. We also give scholarships to
GATE Solutions CHEMICAL ENGINEERING [5]
Published by: ENGINEERS INSTITUTE OF INDIA-E.I.I. ALL RIGHT RESERVED
28B/7 Jiasarai Near IIT Hauzkhas Newdelhi-110016 ph. 011-26514888. www.engineersinstitute.com
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, 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’s theorems.
Differential equations: First order equations (linear and nonlinear), Higher order linear differential
equations with constant coefficients, Cauchy’s and Euler’s equations, Initial and boundary value
problems, Laplace transforms, Solutions of one dimensional heat and wave equations and Laplace
equation.
Complex variables: Analytic functions, Cauchy’s integral theorem, Taylor and Laurent series,
Residue theorem.
Probability and Statistics: Definitions of probability and sampling theorems, Conditional probability,
Mean, median, mode and standard deviation, Random variables, Poisson, Normal and Binomial
distributions.
Numerical Methods: Numerical solutions of linear and non-linear algebraic equations Integration by
trapezoidal and Simpson’s rule, single and multi-step methods for differential equations.
CHEMICAL ENGINEERING
Process Calculations and Thermodynamics: Laws of conservation of mass and energy; use of
tie components; recycle, bypass and purge calculations; degree of freedom analysis. First and
Second laws of thermodynamics. First law application to close and open systems. Second law
and Entropy. Thermodynamic properties of pure substances: equation of state and departure
function, properties of mixtures: partial molar properties, fugacity, excess properties and activity
coefficients; phase equilibria: predicting VLE of systems; chemical reaction equilibria.
Fluid Mechanics and Mechanical Operations: Fluid statics, Newtonian and non-Newtonian
fluids, Bernoulli equation, Macroscopic friction factors, energy balance, dimensional analysis,
shell balances, flow through pipeline systems, flow meters, pumps and compressors, packed and
fluidized beds, elementary boundary layer theory, size reduction and size separation; free and
hindered settling; centrifuge and cyclones; thickening and classification, filtration, mixing and
agitation; conveying of solids.
Heat Transfer: Conduction, convection and radiation, heat transfer coefficients, steady and
unsteady heat conduction, boiling, condensation and evaporation; types of heat exchangers and
evaporators and their design.
Mass Transfer: Fick’s laws, molecular diffusion in fluids, mass transfer coefficients, film,
penetration and surface renewal theories; momentum, heat and mass transfer analogies;
stagewise and continuous contacting and stage efficiencies; HTU & NTU concepts design and
operation of equipment for distillation, absorption, leaching, liquid-liquid extraction, drying,
humidification, dehumidification and adsorption.
GATE Solutions CHEMICAL ENGINEERING [6]
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28B/7 Jiasarai Near IIT Hauzkhas Newdelhi-110016 ph. 011-26514888. www.engineersinstitute.com
Chemical Reaction Engineering: Theories of reaction rates; kinetics of homogeneous reactions,
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, transfer functions and dynamic responses of simple systems, process reaction curve,
controller modes (P, PI, and PID); control valves; analysis of closed loop systems including stability,
frequency response and controller tuning, cascade, feed forward control.
Plant Design and Economics: Process design and sizing of chemical engineering equipment such as
compressors, heat exchangers, multistage contactors; principles of process economics and cost
estimation including total annualized cost, cost indexes, rate of return, payback period, discounted
cash flow, optimization in design.
Chemical Technology: Inorganic chemical industries; sulfuric acid, NaOH, fertilizers (Ammonia,
Urea, SSP and TSP); natural products industries (Pulp and Paper, Sugar, Oil, and Fats); petroleum
refining and petrochemicals; polymerization industries; polyethylene, polypropylene, PVC and
polyester synthetic fibers.
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 reasoning and data
interpretation.
PATTERN OF GATE EXAMINATION
GATE exam would contain questions of four different types in engineering papers:
• Multiple choice questions carrying 1 or 2 marks each.
• Common data questions, where two successive questions use the same set of input data.
• Linked answer questions, where the answer to the first question of the pair is required in
order to answer its successor.
• Numerical answer questions, where the answer is a number, to be entered by the
candidate using the mouse and a virtual keypad that will be provided on the screen.
MARKING SCHEME
Engineering Mathematics : 15 Marks
General Aptitude Section : 15 Marks
Technical Section : 70 Marks
Total Marks : 100 Total Question : 65 Duration : 3:00 Hour
GATE 2013 Cut-off Marks GENERAL SC/ST/PD OBC(Non-Creamy) Total Appeared
Chemical Engineering 32.35 21.57 29.12 14,835
GATE Solutions CHEMICAL ENGINEERING [7]
Published by: ENGINEERS INSTITUTE OF INDIA-E.I.I. ALL RIGHT RESERVED
28B/7 Jiasarai Near IIT Hauzkhas Newdelhi-110016 ph. 011-26514888. www.engineersinstitute.com
CONTENTS
GATE Solution GATE Solution GATE Solution GATE Solution
1. Process Calculations …………………………… 1-19
2. Thermodynamics Engineering……………….. 20-55
3. Fluid Mechanics…………………………………… 56-110
4. Mechanical Operations………………………… 111-116
5. Heat Transfer …………………………………….. 117-161
6. Mass Transfer ……………………………………. 162-218
7. Chemical Reaction Engineering …………… 219-277
8. Instrumentation and Process Control …… 278-323
9. Plant Design and Economics ……………….. 324-339
10. Chemical Technology …………………………. 340-364
11. Engineering Mathematics…………………….. 365-429
12. Verbal Ability (General Aptitude)…………… 430-433
GATE Solutions CHEMICAL ENGINEERING [8]
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GATE & PSU’s Practice Set 435-427
1. Chemical Technology …………………………………………………… 437-439
2. Chemical Reaction Engineering ………………………………………… 440-443
3. Fluid Mechanics …………………………………………………………………… 444-446
4. Mechanical Operation ……………………………………………………….. 447-449
5. Heat Transfer ……………………………………………………………………… 450-452
6. Mass Transfer ……………………………………………………………………… 453-456
7. Process Calculation ……………………………………………………………… 457-459
8. Instrumentation & Process Control ………………………………… 460-463
9. Thermodynamics …………………………………………………………………… 464-466
10. Plant design & Economics …………………………………………………… 467-470
Solution with detailed explanations
1. Chemical Technology …………………………………………………………… 471-471
2. Chemical Reaction Engineering ………………………………………… 472-478
3. Fluid Mechanics …………………………………………………………………… 479-486
4. Mechanical Operation ………………………………………………………… 487-488
5. Heat Transfer ……………………………………………………………………… 489-495
6. Mass Transfer ……………………………………………………………………… 496-503
7. Process Calculation ……………………………………………………………… 504-509
8. Instrumentation & Process Control………………………………… 510-515
9. Thermodynamics ………………………………………………………………… 516-522
10. Plant design & Economics ………………………………………………… 523-526
GATE Solutions
Published by: ENGINEERS INSTITUTE OF INDIA
28B/7 Jiasarai Near IIT Hauzkhas Newdelhi
1. PROCESS CALCULATIONS
(GATE Previous Papers)
Common Data for Questions 1
A reverse osmosis unit treats feed water (F) containing fluoride and its output
stream (P) and a reject stream (R). Let C
permeate, and reject streams, respectively. Under steady state conditions, the volumetric flow rate of
the reject is 60 % of the volumetric flow rate of the inlet stream, and C
Q.1 The value of CR in mg/L, up to one digit after
Q.2 A fractionf of the feed is bypassed and mixed with the permeate to obtain treated water having
a fluoride concentration of 1 mg/L. Here also the flow rate of the reject stream is 60% of the
flow rate entering the reverse osmosis unit (after the bypass). The value of
the decimal point, is __________
Common Data for Questions for 3 and 4 :
The reaction ( ) ( ) (liq gas liq gas
A B C D+ → +
shown below.
Notation:
Molar flow rate of fresh B is F
Molar flow rate of A is F
Molar flow rate of recycle gas is F
Molar fraction of B in recycle gas is Y
Molar flow rate of purge gas is F
Molar flow rate of C is F
Here, FFB = 2 mol/s; FA = 1 mol/s, F
and A is completely converted.
Q. 3 If 0.3RB
Y = , the ratio of recycle gas to, purge gas
(A) 2
Q. 4 If the ratio of recycle gas to purge gas
(A) 3
8
CHEMICAL ENGINEERING
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PROCESS CALCULATIONS
(GATE Previous Papers)
GATE-2013
and 2:
A reverse osmosis unit treats feed water (F) containing fluoride and its output consists of a permeate
stream (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 of
olumetric flow rate of the inlet stream, and CF = 2 mg/L and C
in mg/L, up to one digit after the decimal point, is ________
of the feed is bypassed and mixed with the permeate to obtain treated water having
a fluoride concentration of 1 mg/L. Here also the flow rate of the reject stream is 60% of the
flow rate entering the reverse osmosis unit (after the bypass). The value of f
the decimal point, is __________
GATE-2012
for 3 and 4 :
) ( ) ,liq gas liq gasA B C D+ → + is carried out in a reactor followed by a separator as
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
= 1 mol/s, FB/FA = 5
and A is completely converted.
, the ratio of recycle gas to, purge gas ( )/RG PG
F F is
(B) 5 (C) 7 (D) 10
If the ratio of recycle gas to purge gas ( )/RG RB
F F is 4 then FRB is
(B) 2
5 (C)
1
2 (D)
3
4
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consists of a permeate
denote the fluoride concentrations in the feed,
permeate, and reject streams, respectively. Under steady state conditions, the volumetric flow rate of
= 2 mg/L and CP = 0.1 mg/L.
the decimal point, is ________ (2-Marks)
of the feed is bypassed and mixed with the permeate to obtain treated water having
a fluoride concentration of 1 mg/L. Here also the flow rate of the reject stream is 60% of the
flow rate entering the reverse osmosis unit (after the bypass). The value of f , up to 2 digits after
(2-Marks)
is carried out in a reactor followed by a separator as
(2-Marks)
(2-Marks)
GATE Solutions
Published by: ENGINEERS INSTITUTE OF INDIA
28B/7 Jiasarai Near IIT Hauzkhas Newdelhi
1.
Mass balance F = P + R
Given: R = 0.6 F
∴ F = P + 0.6 F
P = 0.4F
Mass balance on fluoride content
F P RFC PC RC= +
From equation (i) and (ii
F × 2 = 0.4 × F × 0.1 + 0.6 × F ×
R
2 0.04C 3.27 mg/
0.6
−= =
2.
Given: R = 60% of volumetric flow rate of the
Therefore, P = 0.4 × F (1
Fluoride content balance on by pass stream;
F 2 [P F] Cf f× × = + × ×
f × F × 2 = P + f
From equation (i)
f × F × 2 = 0.4 × F(1
2 × f = 0.4 × (1 –
0.4
0.2861.4
f = =
Feed (F)
C = 2 mg/LF
Permeate stream (P)C = 0.1 mg/LP
RO
CHEMICAL ENGINEERING
ENGINEERS INSTITUTE OF INDIA-E.I.I. ALL RIGHT RESERVED
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SOLUTIONS
Explanations-2013
…(i)
…(ii)
Mass balance on fluoride content
F P RFC PC RC
ii)
F × 2 = 0.4 × F × 0.1 + 0.6 × F × RC
C 3.27 mg/ L= =
Given: R = 60% of volumetric flow rate of the inlet stream
P = 0.4 × F (1 – f ) …(i)
Fluoride content balance on by pass stream;
PF 2 [P F] Cf f× × = + × ×
f × F ∵ PC 1= (Given)
× F × 2 = 0.4 × F(1 – f ) + f × F
– f ) + f
0.286
Reject (R)
Permeate stream (P)= 0.1 mg/L
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GATE Solutions
Published by: ENGINEERS INSTITUTE OF INDIA
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3. (B)
Given: 2 / secFB
F mol=
1 / secA
F mol=
A is completely converted
Assume separator separatesall C
Overall Material Balance across the dotted circle
A FB C PGF F F F+ = +
PG A FB CF F F F= +
Material Balance for component B at the point (1)
B FB RB RGF F Y F= +
-B FB
RG
RB
F FF
Y=
0.3 10 / secRB RG
Y given so F mol= =
105.
2
RG
PG
F
F= =
4. (A) (4RG
PG
Fgiven
F=
2 50PG
F from Question=∵
4
RG PGF F= × = × =
Material Balance at point (1)
B FB RB RGF F Y F= + ×
-
B FBRB
RG
F FY
F=
5- 2
8=
3
8=
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Explanations - 2012
2 / secF mol
5 / secB
A
Fmol
F=
1 / sec
A is completely converted
Assume separator separatesall C 1 / secC
F mol∴ =
Overall Material Balance across the dotted circle
A FB C PGF F F F+ = +
-PG A FB C
F F F F 1 2 -1= + 2 / secmol=
Material Balance for component B at the point (1)
B FB RB RGF F Y F
B FB
0.3 10 / secRB RG
Y given so F mol= =
5.
)given
( )2 50F from Question
4 2 8 / sec.mol= × =
Material Balance at point (1)
B FB RB RGF F Y F= + ×
5 / sec
2 / sec
B
FB
F molgiven
F mol
=
=
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GATE Solutions CHEMICAL ENGINEERING [12]
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2. THERMODYNAMICS
(GATE Previous Papers)
GATE-2013
Q.1 A gaseous system contains H2, I2, and HI, which participate in the gas-phase reaction
2 22HI H I+
At a state of reaction equilibrium, the number of thermodynamic degrees of freedom
is _________ (1-Mark)
Q.2 The thermodynamic state of a closed system containing a pure fluid changes from (T1, p1) to
(T2, p2), where T and p denote the temperature and pressure respectively. Let Q denote the heat
absorbed (> 0 if absorbed by the system) and W the work done (> 0 if done by the system).
Neglect changes in kinetic and potential energies. Which one of the following is CORRECT
(A) Q is path-independent and W is path-dependent (1-Mark)
(B)Q is path-dependent and W is path-independent
(C) (Q − W) is path-independent
(D) (Q + W) is path-independent
Q.3 An equation of state is explicit in pressure p and cubic in the specific volume v. At the critical
point ‘c’, the isotherm passing through ‘c’ satisfies (1-Mark)
(A) 2
20, 0
p p
v v
∂ ∂< =
∂ ∂ (B)
2
2, 0 0
p p
v v
∂ ∂> <
∂ ∂
(C) 2
20, 0
p p
v v
∂ ∂= >
∂ ∂ (D)
2
20, 0
p p
v v
∂ ∂= =
∂ ∂
Q.4 The units of the isothermal compressibility are (1-Mark)
(A) m-3
(B) Pa-1
(C) m3 Pa
-1 (D) m
-3 Pa
-1
Q.5 In a process occurring in a closed system F, the heat transferred from F to the surroundings E
is600 J. If the temperature of E is 300 K and that of F is in the range 380 - 400 K, the entropy
Changes of the surroundings (∆SE) and system (∆SF), in J/K, are given by
(A) 2, 2E F
S S∆ = ∆ = −
(B) 2, 2E F
S S∆ = − ∆ = (2-Marks)
(C) 2, 2E F
S S∆ = ∆ < − (D) 2, 2E F
S S∆ = ∆ > −
Q.6 A binary; liquid mixture is in equilibrium with its vapor at a temperature T = 300 K. The liquid
mole fraction x1 of species 1 is 0.4 and the molar excess Gibbs free energy is 200 J/mol. The
value of the universal gas constant is 8.314 J/mol-K, and iγ denotes the liquid-phase activity
coefficient of species i. If ( )1ln 0.09y = , then the value of ln ( )2
γ , up to 2 digits after the
decimal point, is ________ (2-Marks)
Q.7 Calculate the heat required (in kJ, up to 1 digit after the decimal point) to raise the temperature
of1 mole of a solid material from 100 °C to 1000 °C. The specific heat (Cp) of the material (in
J/mol-K) is expressed as Cp = 20 + 0.005T, where T is in K. Assume no phase change.
_________ (2-Marks)
GATE Solutions CHEMICAL ENGINEERING [13]
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SOLUTIONS
Explanations - 2013
1. (c)
2 22HI( ) H ( ) I ( )g g g+
Degree of freedom = C – P + 2 = 2 – 1 + 2 = 3
2. (c) The difference of two path dependent functions ends up in a property which does not depend
on anything but the state of the system.
U Q W∆ = −
Q and W are path dependent. U depends only on the state of the system.
3. (d)
For pure substances, there is an inflection point in the critical isotherm (constant temperature
line) on a PV diagram. At the critical point
2
2T T
0p p
v v
∂ ∂ = =
∂ ∂
4. (b) Isothermal compressibility
T
T
1 V
V P
∂ β = −
∂
Unit of Tβ is 1Pa .−
5. (b) systemQ 600 J∆ = −
∵ system surroundingQ Q 0∆ + ∆ =
∴ surroundingQ 600 J∆ =
We know revST
dq∆ = ∫
For surrounding T = constant
∴ surrounding
surrounding
system
Q 600S 2
T 300
∆∆ = = =
GATE Solutions CHEMICAL ENGINEERING [14]
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6. 2 1 1 2G RT( ln ln )x v x v∆ = + and 2 1 1x x+ =
∴ 1 1 1 2G RT((1 ) ln ln )x v x v∆ = − +
Given, G 200 J/ mol∆ = R = 8.314 J/mol.K
f = 300 K 1 0.4x =
1ln 0.09v =
∴ 200 = 8.314 × 300((1 – 0.4) × 0.09 + 0.4 ln 2v )
After solving 2ln V 0.0654=
7 .
PQ C Tn d∆ = ∫
12731000 273 2
273100 273
0.005 T(20 0.005T) T 20T
2d
+
+
= + = +
∫
= 29511.3225 – 7807.8225 = 21703.5 J = 21.7 kJ
GATE Solutions
Published by: ENGINEERS INSTITUTE OF INDIA
28B/7 Jiasarai Near IIT Hauzkhas Newdelhi
3.
(GATE Previous Papers)
Q.1 The apparent viscosity of a fluid is
where
(A) Bingham plastic
Q.2 The mass balance for a fluid with density
(A)
(C)
Q.3 An incompressible Newtonian fluid, filled in an annular gap between two concentric
cylinders of radii R1 and
The outer cylinder is rotating with an angular velocity of
stationary. Given that
can be approximated by,
(A)
(C)
Q.4 For a Newtonian fluid flowing in a circular pipe under steady state conditions in
developed laminar flow, the Fanning friction factor is
(A)
(C)
0.3
0.004dV
dy
dV
dy
( ). 0Vt
ρρ
∂+ ∇ =
∂
( ). 0Vt
ρρ
∂+ ∇ =
∂
( R R R
2R Ω
( )( )
1
1
2 1
r RR
R R
+Ω
+
0.20.046Re−
16
Re
CHEMICAL ENGINEERING
ENGINEERS INSTITUTE OF INDIA-E.I.I. ALL RIGHT RESERVED
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3. FLUID MECHANICS
(GATE Previous Papers)
GATE-2013
The apparent viscosity of a fluid is given by
is the velocity gradient. The fluid is
(B) dilatant (C) pseudo plastic
The mass balance for a fluid with density and velocity vector is
(B)
(D)
An incompressible Newtonian fluid, filled in an annular gap between two concentric
and R2 as shown in the figure, is flowing under steady state conditions.
The outer cylinder is rotating with an angular velocity of while the inner cylinder is
, the profile of the
can be approximated by,
(B)
(D)
For a Newtonian fluid flowing in a circular pipe under steady state conditions in
developed laminar flow, the Fanning friction factor is
(B)
(D)
dV
dy
( )ρ ( )V
( ). 0Vt
ρρ
∂+ ∇ =
∂
( ). 0Vt
ρρ
∂+ ∇ =
∂
Ω
)2 1 1R R R− << componentθ −
( )( )
2
2 1
r Rr
R R
−Ω
−
( )( )
1
2
2 1
r RR
R R
−Ω
−
0.32
0.1250.0014
Re+
24
Re
[15]
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(1-Mark)
(D) thixotropic
is (1-Mark)
An incompressible Newtonian fluid, filled in an annular gap between two concentric
as shown in the figure, is flowing under steady state conditions.
while the inner cylinder is
of the velocity
(1-Mark)
For a Newtonian fluid flowing in a circular pipe under steady state conditions in fully
(1-Mark)
component Vθ
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Solutions
Explanations– 2013
1. (b) In non-Newtonian fluid, the shear stress due to viscosity is
Where, K = Flow consistency index
n = Flow behavior index(dimensionless)
or
where,
For question
So,
n> 1 for dilatant
n< 1 for pseudo plastic
n = 1 for Newtonian fluid
2. (a) The continuity equation states that, in any steady state process, the rate at which mass enters
a system is equal to the rate at which mass leaves the system.
Differential form of the continuity equation is given by
Where, – Fluid density
u – Flow velocity vector field
t – Time
If, = constant then
3. (d)
Point:
Slope between A and C = Slope between B and A
4. (c)
where, f – Fanning friction factor of the pipe
– Shear stress at the wall
– Density of the fluid
Fanning friction factor for laminar flow in round tubes
n
xy
duK
dy
τ =
1n
yx
du du duk n
dy dy dy
−
τ = =
1ndu
n kdy
−
=
0.3V
0.004d
ndy
=
– 1 0.3 1.3 1n n= ⇒ = >
. ( ) 0ut
∂ρ+ ∇ ρ =
∂ρ
ρ . 0u∇ =
1A(R , 0) B( , V )r θ
2C(R . )rΩ
2 1 1
V 00
R R R
r
r
θ −Ω −=
− −
∴ 1
2 1
RV
R R
rrθ
−= Ω
−2
2
f vρτ =
τ
ρ
16
Ref =
A B
C
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4. MECHANICAL OPERATIONS
(GATE Previous Papers)
GATE-2010
Q.1 The critical speed (revolutions per unit time) of a ball mill of radius R, which uses balls of
radius r, is (1-Mark)
(A) (B) (C) (D)
GATE-2008
Q.2 The power required for size reduction in crushing is (1-Mark)
(A) Proportional to
(B) Proportional to
(C) Proportional to Surface energy of the material
(D) Independent of the Surface energy of the material
GATE-2007
Q.3 In Tyler series, the ratio of the aperture size of a screen to that of the next smaller screen is
(1-Mark)
(A) (B) (C) 1.5 (D) 2
Q.4 Size reduction of coarse hard solids using a crusher is accomplished by
(1-Mark)
(A) attrition (B) compression (C) cutting (D) impact
Q.5 In constant pressure filtration, the rate of filtration follows the relation (v: filtrate volume,
t:time, k and c: constant), (1-Mark)
(A) (B) (C) (D)
Q.6 Sticky material are transported by (1-Mark)
(A) apron conveyor (B) screw conveyor
(C) belt conveyor (D) hydraulic conveyor
GATE-2006
Statement for Linked Answer Question 7&8:
A continuous grinder obeying the Bond crushing law grinds a solid at the rate of 1000 kg/hr from the
initial diameter of 10 mm to the final diameter of 1 mm.
Q.7 If the market now demands particles of size 0.5 mm, the output rate of the grinder (in kg/hr)
for the same power input would be reduced to (2-Marks)
(A) 227 (B) 474
(C) 623 (D) 856
1
2
g
Rrπ1
2
g
Rπ
1
2
g
rπ
1
2 -
g
R rπ
1
Surface energy of the material
1
Surface energy of the material
1
22
dvkv c
dt= +
1dv
dt kv c=
+
dvkv
dt=
2dvkv
dt=
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SOLUTIONS
Explanations - 2010
1. (D)
Where nc- Critical Speed
r – radius of ball
R – Radius of ball mill
Mills run at 65 to 80% of the critical speed
Explanations - 2008
2. (C) Energy required for size reduction is directly proportional to the change in surface
area, According to Rittinger’s low
Surface energy is the work per unit area done by the forces that creates the new
surface.
Explanations - 2007
3. (B) Tyler mesh size is the number of openings per linear inch of mesh.
The ratio of the aperture size of a screen to that of the next smaller screen is
4. (B) Size reduction of coarse hard solids crushers utilizes the compressive force between two solid
surface. Crushers are used for size reduction of minerals, gravel and ores.
5. (B)
Where
6. (B) Screw conveyor:Screw conveyor uses a rotating helical screw blade to move the liquid or
granular material. For sticy material we use shaftless screw conveyor.
1
2 -c
gn
R rπ=
1 1 vesignr
sb sa
wK
m D D
= −
2
s
m
dv PA
x VdtR
A
αµ
∆=
+
2
1
s mx RV
A P PA
µα µ=
+ ∆ ∆
P Const for const press filtration∴∆ =
1
KV C=
+ 2 s mx R
K CA P PA
µα µ= =
∆ ∆
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Explanation - 2006
7. (C) Bond’s Law
Where P = power required. T = feed rate Kg/hr.
K = constant DP = Dia. Of product
Df = Diameter of feed
1 1 -
P F
PK
T D D
=
1 1 1 -
P FT D D
α
∴ tan
P is same
K Cons t
∴
=
2
1
1 1 -
1 10
1 1 -
0.5 10
T
T
=
2
1
0.684
1.098
T
T= 2
1
0.623T
T=
2 1 10.623 1000 /T T Given T kg hr= × = S
1000 0.623 623 /kg hr= × =