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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




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

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ISBN: 978-93-5137-754-2

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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


[email protected]

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



interested in M.Tech & government jobs due to decline in other career



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


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


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


available before and after classes to assist students and we maintain a heal



results speak for themselves. Past students of EII are currently working in various


[4]

ALL RIGHT RESERVED

www.engineersinstitute.com

examination is one of the most prestigious competitive examination



ecline in other career



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


eeping in mind, the



in providing exam-

and test series so that our

The faculty at EII are a team of experienced


available before and after classes to assist students and we maintain a healthy student-faculty



currently working in various





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.


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.




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




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 & 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



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




PROCESS CALCULATIONS


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,


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


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

[9]

ALL RIGHT RESERVED


consists of a permeate

denote the fluoride concentrations in the feed,


= 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


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



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




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

[10]

ALL RIGHT RESERVED


GATE Solutions



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=




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

=

=

[11]

ALL RIGHT RESERVED





2. THERMODYNAMICS


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)




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

∆∆ = = =




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



3.


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




3. FLUID MECHANICS


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]

ALL RIGHT RESERVED


(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




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




4. MECHANICAL OPERATIONS


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=




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= =

∆ ∆




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= × =

sample file chemical - engineers institute · chemical engineering -e.i.i. all right...

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