GATE Solutions
CHEMICAL ENGINEERING
[1]
CHEMICAL ENGINEERING 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
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GATE Solutions
CHEMICAL ENGINEERING
[2]
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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GATE Solutions
CHEMICAL ENGINEERING
[3]
This book is dedicated to all Chemical Engineers preparing for GATE Examination. Examination.
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GATE Solutions
CHEMICAL ENGINEERING
[4]
A word to the students 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 decline decline in other career 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 with critical analysis of topics and ability to express clearly are some of the pre-requisites pre to crack 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 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 Institute Instit of India have consistently provided rigorous classes and proper guidance to engineering students over the nation in successfully accomplishing their dreams. We believe in providing examoriented teaching methodology with updated study material and test series so that our students stay ahead in 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 healthy heal 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 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 meritorious students.
R.K. Rajesh Director Engineers Institute of India
[email protected]
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GATE Solutions
CHEMICAL ENGINEERING
[5]
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. 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 Solutions
[6]
CHEMICAL ENGINEERING
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
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GATE Solutions
[7]
CHEMICAL ENGINEERING
CONTENTS 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
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GATE Solutions
[8]
CHEMICAL ENGINEERING
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
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GATE Solutions
[9]
CHEMICAL ENGINEERING
1. PROCESS CALCULATIONS (GATE Previous Papers) GATE-2013 Common Data for Questions 1 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 the reject is 60 % of the volumetric olumetric flow rate of the inlet stream, and CF = 2 mg/L and CP = 0.1 mg/L. Q.1 Q.2
The value of CR in mg/L, up to one digit after the decimal point, is ________ (2-Marks) 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 f , up to 2 digits after the decimal point, is __________ (2-Marks) GATE-2012
Common Data for Questions for 3 and 4 : The reaction A( liq ) + B( gas ) → C ( liq ) + D( gas ) , 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. Q. 3 If YRB = 0.3 , the ratio of recycle gas to, purge gas ( FRG / FPG ) is
(2-Marks)
(A) 2 (B) 5 (C) 7 (D) 10 Q. 4 If the ratio of recycle gas to purge gas ( FRG / FRB ) is 4 then FRB is (2-Marks) (A)
3 8
(B)
2 5
(C)
1 2
(D)
3 4
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GATE Solutions
CHEMICAL ENGINEERING
[10]
SOLUTIONS Explanations-2013 1. Feed (F)
RO
Reject (R)
CF = 2 mg/L
Permeate stream (P) CP = 0.1 mg/L
Mass balance F = P + R Given: R = 0.6 F …(i) F = P + 0.6 F ∴ P = 0.4F …(ii) Mass balance on fluoride content FCF = PCP + RC R From equation (i) and (ii) ii F × 2 = 0.4 × F × 0.1 + 0.6 × F × C R CR =
2 − 0.04 = 3.27 mg/ L 0.6
2.
Given: R = 60% of volumetric flow rate of the inlet stream Therefore, P = 0.4 × F (1 – f ) …(i) Fluoride content balance on by pass stream; f × F × 2 = [P + f × F] × CP f×F×2=P+f×F
∵
CP = 1 (Given)
From equation (i) f × F × 2 = 0.4 × F(1 – f ) + f × F 2 × f = 0.4 × (1 – f ) + f 0.4 f = = 0.286 1.4
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GATE Solutions
CHEMICAL ENGINEERING
[11]
Explanations - 2012 3. (B)
FB = 5mol / sec FA
Given: FFB = 2mol / sec
FA = 1mol / sec A is completely converted Assume separator separatesall C ∴ FC = 1mol / sec Overall Material Balance across the dotted circle
FA + FFB = FC + FPG FPG = FA + FFB - FC = 1 + 2 -1 = 2 mol / sec Material Balance for component B at the point (1)
FB = FFB + YRB FRG FRG =
FB - FFB YRB
YRB = 0.3 given so FRG = 10mol / sec FRG 10 = = 5. FPG 2 4. (A)
FRG = 4 ( given ) FPG ∵ FPG = 2 from Question
( 50 )
FRG = 4 × FPG = 4 × 2 = 8 mol / sec. Material Balance at point
(1)
FB = FFB + YRB × FRG YRB =
=
FB - FFB FRG
FB = 5mol / sec given FFB = 2mol / sec
5-2 3 = 8 8
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GATE Solutions
[12]
CHEMICAL ENGINEERING
2. THERMODYNAMICS (GATE Previous Papers) GATE-2013 Q.1
Q.2
Q.3
A gaseous system contains H2, I2, and HI, which participate in the gas-phase reaction H 2 + I 2 2HI At a state of reaction equilibrium, the number of thermodynamic degrees of freedom is _________ (1-Mark) 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 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)
∂p ∂2 p < 0, 2 = 0 ∂v ∂v
(B)
∂p ∂2 p >, 0 2 < 0 ∂v ∂v
∂p ∂2 p ∂p ∂2 p (D) = 0, > 0 = 0, =0 ∂v ∂v 2 ∂v ∂v 2 The units of the isothermal compressibility are (1-Mark) -3 -1 3 -1 (A) m (B) Pa (C) m Pa (D) m-3 Pa-1 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) ∆S E = 2, ∆S F = − 2 (B) ∆S E = −2, ∆S F = 2 (2-Marks) (C)
Q.4 Q.5
(C) ∆S E = 2, ∆S F < − 2 Q.6
(D) ∆S E = 2, ∆S F > − 2
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
γi denotes the liquid-phase activity
coefficient of species i. If ln ( y1 ) = 0.09 , then the value of ln ( γ 2 ) , up to 2 digits after the Q.7
decimal point, is ________ (2-Marks) 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)
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GATE Solutions
CHEMICAL ENGINEERING
[13]
SOLUTIONS Explanations - 2013 1.
(c)
H2 ( g ) + I2 ( g ) 2HI( g ) 2.
3.
4.
5.
Degree of freedom = C – P + 2 = 2 – 1 + 2 = 3 (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. (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 p ∂p = =0 ∂v T ∂v 2 T
(b) Isothermal compressibility 1 ∂V βT = − V ∂ P T Unit of βT is Pa −1 . ∆ Q system = − 600 J (b) ∵
∆ Qsystem + ∆ Qsurrounding = 0
∴
∆ Qsurrounding = 600 J
We know
∆S = ∫
dqrev T
For surrounding T = constant
∴
∆ Ssurrounding =
∆ Q surrounding Tsystem
=
600 =2 300
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GATE Solutions
6.
∆ G = RT( x2 ln v1 + x1 ln v2 ) ∴
[14]
CHEMICAL ENGINEERING
and
x2 + x1 = 1
∆ G = RT((1 − x1 ) ln v1 + x1 ln v2 )
Given, ∆ G = 200 J/ mol
R = 8.314 J/mol.K x1 = 0.4
f = 300 K ln v1 = 0.09 ∴
200 = 8.314 × 300((1 – 0.4) × 0.09 + 0.4 ln v 2 )
After solving
ln V2 = 0.0654
7.
∆ Q = ∫ n CP d T 1000 + 273
0.005 T 2 = ∫ (20 + 0.005T)d T = 20T + 2 100 + 273
1273
273
= 29511.3225 – 7807.8225 = 21703.5 J = 21.7 kJ
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GATE Solutions
[15]
CHEMICAL ENGINEERING
3. FLUID MECHANICS (GATE Previous Papers) Q.1
GATE-2013 The apparent viscosity of a fluid is given by
dV 0.004 dy
0.3
dV is the velocity gradient. The fluid is dy
where
(A) Bingham plastic
(B) dilatant
(1-Mark)
(C) pseudo plastic
(D) thixotropic
Q.2
The mass balance for a fluid with density
∂ρ + ∇. ρ V = 0 ∂t ∂ρ (C) + ρ . ∇V = 0 ∂t
( )
(A)
( )
Q.3
( ρ ) and velocity vector (V ) is (1-Mark) ∂ρ + V . ( ∇ρ ) = 0 ∂t ∂ρ (D) + V . ( ∇ρ ) = 0 ∂t (B)
An incompressible Newtonian fluid, filled in an annular gap between two concentric cylinders of radii R1 and R2 as shown in the figure, is flowing under steady state conditions. The outer cylinder is rotating with an angular velocity of Ω while the inner cylinder is stationary. Given that
( R2 − R1 ) 1 n> 1 for dilatant n< 1 for pseudo plastic n = 1 for Newtonian fluid (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 ∂ρ + ∇ . (ρu ) = 0 ∂t Where, ρ – Fluid density u – Flow velocity vector field t – Time If, ρ = constantthen ∇ . u = 0 (d) Point: A(R 1 , 0) B( r , Vθ ) C(R 2 Ω . r ) Slope between A and C = Slope between B and A C Ωr − 0 Vθ − 0 B = A R 2 − R 1 r − R1 r − R1 ∴ Vθ = Ωr R 2 − R1
f ρv 2 τ= (c) 2 where, f – Fanning friction factor of the pipe τ – Shear stress at the wall ρ – Density of the fluid Fanning friction factor for laminar flow in round tubes f =
16 Re
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GATE Solutions
[17]
CHEMICAL ENGINEERING
4. MECHANICAL OPERATIONS (GATE Previous Papers) Q.1
GATE-2010 The critical speed (revolutions per unit time) of a ball mill of radius R, which uses balls of (1-Mark) radius r, is
1 (A) 2π Q.2
g Rr
1 (B) 2π
g R
1 (C) 2π
GATE-2008 The power required for size reduction in crushing is (A) Proportional to
(B) Proportional to
1 (D) 2π
g r
g R-r
(1-Mark)
1 Surface energy of the material
1 Surface energy of the material
(C) Proportional to Surface energy of the material (D) Independent of the Surface energy of the material Q.3
Q.4
GATE-2007 In Tyler series, the ratio of the aperture size of a screen to that of the next smaller screen is (1-Mark) 1 (A) (B) 2 (C) 1.5 (D) 2 2 Size reduction of coarse hard solids using a crusher is accomplished by
(1-Mark) Q.5
Q.6
(A) attrition (B) compression (C) cutting (D) impact In constant pressure filtration, the rate of filtration follows the relation (v: filtrate volume, t:time, k and c: constant), (1-Mark) dv dv 1 dv (A) (B) (C) (D) = kv + c = = kv dt dt kv + c dt dv = kv 2 dt 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 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 Solutions
[18]
CHEMICAL ENGINEERING
SOLUTIONS Explanations - 2010 1. (D)
nc =
1 2π
g R-r
Where
nc- Critical Speed r – radius of ball R – Radius of ball mill Mills run at 65 to 80% of the critical speed
2.
Explanations - 2008 (C) Energy required for size reduction is directly proportional to the change in surface area, According to Rittinger’s low
w 1 1 = Kr − ve sign m D sb D sa Surface energy is the work per unit area done by the forces that creates the new surface.
Explanations - 2007 Tyler mesh size is the number of openings per linear inch of mesh.
3.
(B)
4.
The ratio of the aperture size of a screen to that of the next smaller screen is 2 (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)
dv = dt
∆PA αx V µ s + Rm A
=
1
µα xs 2
V+
µ Rm
A ∆P ∆PA ∴∆P = Const for const press filtration = 6.
1 KV + C
Where
K =
µα xs 2
A ∆P
C =
µ Rm ∆PA
(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.
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GATE Solutions
[19]
CHEMICAL ENGINEERING
Explanation - 2006 7.
(C)
Bond’s Law
1 P = K T DP
1 DF
Where P = power required. K = constant
T = feed rate Kg/hr. DP = Dia. Of product
Df = Diameter of feed
∴
T2 T1 T2 T1
1 1 α T DP
1 DF
∴ P is same K = Cons tan t
1 1 1 10 = 1 1 10 0.5 0.684 T2 = = 0.623 1.098 T1
T2 = T1 × 0.623
Given T1 = 1000 kg / hrS
= 1000 × 0.623 = 623 kg / hr
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