Chemical Sample Book PDF for Gate Exam (1)

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.

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Tel: 011-26514888

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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 doptions.

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 understandingcritical analysis of topics and ability to express clearly are some of the prethis exam. The questioning & examination pattern has changed in few years, as numerical answer type questions play a major role to score a good rank. Kdifficulties 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 believeoriented teaching methodologystudents 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 healratio. 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 aredepartments and PSUs and pursuing higher specializations. We also give scholarships to meritorious students.

R.K. Rajesh Director

Engineers Institute of India

[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

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


smartly deal with the syllabus not just mugging up concepts. Thorough understandingcritical analysis of topics and ability to express clearly are some of the prethis 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 Instithave 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 healratio. 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 PSUs 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


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 PSUs and pursuing higher specializations. We also give scholarships to



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 Greens theorems. Differential equations: First order equations (linear and nonlinear), Higher order linear differential equations with constant coefficients, Cauchys and Eulers equations, Initial and boundary value problems, Laplace transforms, Solutions of one dimensional heat and wave equations and Laplace equation. Complex variables: Analytic functions, Cauchys 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 Simpsons 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: Ficks 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 & PSUs Practice Set 435-427

1. Chemical Technology 437-439



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



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 Cpermeate, 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 afterQ.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 gasA 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 0.3RBY = , the ratio of recycle gas to, purge gas (A) 2 Q. 4 If the ratio of recycle gas to purge gas

(A) 38



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, 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 fthe 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 PGF F is (B) 5 (C) 7 (D) 10

If the ratio of recycle gas to purge gas ( )/RG RBF F is 4 then FRB is (B) 2

5

(C) 12

(D) 34

[9]

ALL RIGHT RESERVED


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



1.

Mass balance F = P + R Given: R = 0.6 F

F = P + 0.6 F P = 0.4F

Mass balance on fluoride contentF P RFC PC RC= +

From equation (i) and (ii F 2 = 0.4 F 0.1 + 0.6 F

R2 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 / secFBF mol=

1 / secAF 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

RGRB

F FFY

=

0.3 10 / secRB RGY given so F mol= =10 5.2

RG

PG

FF

= =

4. (A) (4RGPG

F givenF

=

2 50PGF from Question=

4RG PGF F= = = Material Balance at point (1) B FB RB RGF F Y F= +

-B FBRB

RG

F FYF

=

5- 28

=

38

=



Explanations - 2012

2 / secF mol

5 / secBA

Fmol

F=

1 / sec

A is completely converted Assume separator separatesall C 1 / secCF mol = Overall Material Balance across the dotted circle

A FB C PGF F F F+ = +

-PG A FB CF 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 RGY 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 / sec2 / 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, 0p pv v

< =

(B)

2

2, 0 0p pv v

>

(D)

2

20, 0p pv 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 FS S = =

(B) 2, 2E FS S = = (2-Marks) (C) 2, 2E FS S = < (D) 2, 2E FS 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 i denotes the liquid-phase activity coefficient of species i. If ( )1ln 0.09y = , then the value of ln ( )2 , 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 pv v

= =

4. (b) Isothermal compressibility

TT

1 VV P

= Unit of T is 1Pa .

5. (b) systemQ 600 J =

system surroundingQ Q 0 + =

surroundingQ 600 J =

We know revST

dq =

For surrounding T = constant

surrounding

surroundingsystem

Q 600S 2T 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 20T2

d+

+

= + = +

= 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

0.004 dVdy

dVdy

( ). 0Vt

+ =

( ). 0Vt

+ =

( R R R

2R

( )( )

11

2 1

r RR

R R+

+

0.20.046Re

16Re



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)

dVdy

( ) ( )V( ). 0V

t

+ =

( ). 0Vt

+ =

)2 1 1R R R



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

xyduKdy

=

1n

yxdu du duk ndy dy dy

= =

1ndun k

dy

=

0.3V0.004 dndy

=

1 0.3 1.3 1n n= = >

. ( ) 0ut

+ =

. 0u =

1A(R , 0) B( , V )r 2C(R . )r

2 1 1

V 00R R R

r

r

=

1

2 1

RVR Rr

r

=

2

2f v

=

16Re

f =

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

12

gRrpi

12

gRpi

12

grpi

12 -

gR rpi

1 Surface energy of the material

1 Surface energy of the material

12

2

dv kv cdt

= +1dv

dt kv c=

+

dv kvdt

=

2dv kvdt

=



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 Rittingers 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.

12 -c

gn

R rpi=

1 1 vesignr

sb sa

w Km D D

=

2

sm

dv PAx Vdt RA

=

+

2

1s mx RV

A P PA =

+

P Const for const press filtration =

1KV C

=

+ 2

s mx RK CA P PA

= =



Explanation - 2006

7. (C) Bonds Law

Where P = power required. T = feed rate Kg/hr. K = constant DP = Dia. Of product Df = Diameter of feed

1 1 -

P F

P KT D D

=

1 1 1 -

P FT D D

tan

P is sameK Cons t

=

2

1

1 1 -

1 101 1

-

0.5 10

TT

=

2

1

0.6841.098

TT

=2

1

0.623TT

=

2 1 10.623 1000 /T T Given T kg hr= = S1000 0.623 623 /kg hr= =

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