GATE SOLUTIONS. Subject-wise descriptive Solution

Transcription

1 GTE SOLUTIONS Subject-wise descriptive Solution

2 This book is dedicated to all Chemical Engineers preparing for GTE & PSUs Entrance. 016 By Engineers Institute of India LL 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 8-B/7, Jia Sarai, Near IIT HauzKhas New Delhi Tel: For publication information, visit ISBN: Price: Rs

3 word to the students GTE examination is one of the most prestigious competitive examinations conducted for Graduate engineers. Over the past few years, it has become more competitive as a number of aspirants are increasingly becoming interested in M.Tech & government jobs due to decline in other career options. In my opinion, GTE exam test candidates basics understanding of concepts, ability to apply numerical approach. 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 answer type questions play a major role to score a good rank. Keeping in mind, the difficulties of an average student, we have composed this booklet. Established in 006 by a team of IES and GTE toppers, we at Engineers Institute 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 materialand test series so that our students stayahead in the competition. Many current and past year toppers associate with us for contributing towards our goal of providing quality education and share their success with the future aspirants. Past students of EII are currently working in various departments and PSU s and pursuing higher specializations. The basic objective of this book is to maintain the standard and uniformity in both teaching and learning. I sincerely express my appreciation and thanks to Mr. Pradeep Kumar Singh (M.Tech IITR) and team of EII who directly or indirectly contributed towards this book. R.K. Rajesh Director Engineers Institute of India eii.rkrajesh@gmail.com

4 GTE Syllabus for Chemical Engineering (CH) ENGINEERING MTHEMTICS Linear lgebra: Matrix algebra, Systems of linear equations, Eigen values and eigenvectors. Calculus: Functions of single variable, Limit, continuity and differentiability, Taylor series, Mean value theorems, Evaluation of definite and improper integrals, Partial derivatives, Total derivative, Maxima and minima, Gradient, Divergence and Curl, Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green 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: Complex number, polar form of complex number, triangle inequality. Probability and Statistics: Definitions of probability and sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Poisson, Normal and Binomial distributions, linear regression analysis. Numerical Methods: Numerical solutions of linear and non-linear algebraic equations. Integration by trapezoidal and Simpson s rule. Single and multi-step methods for numerical solution of differential equations. CHEMICL ENGINEERING Process Calculations and Thermodynamics: Steady and unsteady state mass and energy balances including multiphase, multicomponent, reacting and non-reacting systems. Use of tie components; recycle, bypass and purge calculations; Gibb s phase rule and degree of freedom analysis. First and Second laws of thermodynamics. pplications of first law to close and open systems. Second law and Entropy. Thermodynamic properties of pure substances: Equation of State and residual properties, properties of mixtures: partial molar properties, fugacity, excess properties and activity coefficients; phase equilibria: predicting VLE of systems; chemical reaction equilibrium. Fluid Mechanics and Mechanical Operations: Fluid statics, Newtonian and non-newtonian fluids, shell-balances including differential form of Bernoulli equation and energy balance, Macroscopic friction factors, dimensional analysis and similitude, flow through pipeline systems, flow meters, pumps and compressors,

5 elementary boundary layer theory, flow past immersed bodies including packed and fluidized beds, Turbulent flow: fluctuating velocity, universal velocity profile and pressure drop. Particle size and shape, particle size distribution, size reduction and classification of solid particles; free and hindered settling; centrifuge and cyclones; thickening and classification, filtration, agitation and mixing; conveying of solids. Heat Transfer: Steady and unsteady heat conduction, convection and radiation, thermal boundary layer and heat transfer coefficients, boiling, condensation and evaporation; types of heat exchangers and evaporators and their process calculations. Design of double pipe, shell and tube heat exchangers, and single and multiple effect evaporators. Mass Transfer: Fick s laws, molecular diffusion in fluids, mass transfer coefficients, film, penetration and surface renewal theories; momentum, heat and mass transfer analogies; stage-wise and continuous contacting and stage efficiencies; HTU & NTU concepts; 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, nonideal reactors; residence time distribution, single parameter model; non-isothermal reactors; kinetics of heterogeneous catalytic reactions; diffusion effects in catalysis. Instrumentation and Process Control: Measurement of process variables; sensors, transducers and their dynamics, process modelling and linearization, transfer functions and dynamic responses of various systems, systems with inverse response, process reaction curve, controller modes (P, PI, and PID); control valves; analysis of closed loop systems including stability, frequency response, controller tuning, cascade and feed forward control. Plant Design and Economics: Principles of process economics and cost estimation including depreciation and total annualized cost, cost indices, rate of return, payback period, discounted cash flow, optimization in process design and sizing of chemical engineering equipments such as compressors, heat exchangers, multistage contactors. Chemical Technology: Inorganic chemical industries ( sulphuric acid, phosphoric acid, chloralkali industry), fertilizers (mmonia, 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 fibres).

6 Syllabus for General ptitude (G) Verbal bility: English grammar, sentence completion, verbal analogies, word groups, instructions, critical reasoning and verbal deduction. Numerical bility: Numerical computation, numerical estimation, numerical reasoning and data interpretation. Marking Scheme& Qualifying Marks * General ptitude: 15 Marks Engineering Mathematics: 15 Marks Technical: 70 MarksTotal Marks: 100 Total Question: 65 Duration : 3:00 Hour GTE 016 GENERL SC/ST/PD OBC(Non-Creamy) Total ppeared Chemical Engineering GTE 015 GENERL SC/ST/PD OBC(Non-Creamy) Total ppeared Chemical Engineering

7

8 CONTENTS 1. Chemical Reaction Engineering Heat Transfer Mass Transfer Instrumentation and Process Control Thermodynamics Fluid Mechanics Process calculations Mechanical Operations Plant Design and Economics Chemical Technology Engineering Mathematics

9 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [1] 1. CHEMICL RECTION ENGINEERING (GTE Previous Papers) GTE For a non-catalytic homogeneous reaction B, the rate expression at 300K is C r (mol m) s, where C is the concentration of (in mol/m 3 ). Theoretically, the 1 5C upper limit for the magnitude of the reaction rate ( -r in mol m -3 s 1, rounded off to the first decimal place) at 300K is (1-Mark). The variations of the concentrations (C, C R and C S ) for three species (, R and S) with time, in an isothermal homogeneous batch reactor are shown in the figure below. (1-Mark) Select the reaction scheme that correctly represents the above plot. The numbers in the reaction schemes shown below, represent the first order rate constants in unit of s 1. () (B) (C) (D) 3. Hydrogen iodide decomposes through the reaction HI H + I. The value of the universal gas constant R is J mol 1 K 1. The activation energy for the forward reaction is J mol 1. The ratio (rounded off to the first decimal place) of the forward reaction rate at 600 K to that at 550 K is (1-Mark) Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

10 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [] 4. The liquid phase reversible reaction B is carried out in an isothermal CSTR operating under steady state conditions. The inlet stream does not contain B and the concentration of in the inlet stream is 10mol/lit. The concentrations of at the reactor exit, for residence times of 1 s and 5 s are 8 mol/lit and 5 mol/lit, respectively. ssume the forward and backward reactions are elementary following the first order law. lso assume that the system has constant molar density. The rate constant of the forward reaction (in s 1, rounded off to the third decimal place) is (-Marks) 5. liquid phase irreversible reaction B is carried out in an adiabatic CSTR operating under steady state conditions. The reaction is elementary and follows the first order rate law. For this reaction, the figure below shows the conversion (X ) of as a function of temperature (T) for different values of the rate of reaction ( r in mol m 3 s 1 ) denoted by the numbers to the left of each curve. This figure can be used to determine the rate of the reaction at a particular temperature, for a given conversion of. (-Marks) The inlet stream does not contain B and the concentration of in the inlet stream is 5 mol/m 3. The molar feed rate of is 100 mol/s. steady state energy balance for this CSTR results in the following relation: T = X where T is the temperature (in K) of the exit stream and X is the conversion of in the CSTR. For an exit conversion of 80 % of, the volume (in m 3, rounded off to the first decimal place) of CSTR required is 6. porous pellet with Pt dispersed in it is used to carry out a catalytic reaction. Following two scenarios are possible. (-Marks) Scenario 1: Pt present throughout the pores of the pellet is used for catalyzing the reaction. Scenario : Pt present only in the immediate vicinity of the external surface of the pellet is used for catalyzing the reaction. Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

11 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [3] t a large value of Thiele modulus, which one of the following statements is TRUE? () Since the reaction rate is much greater than the diffusion rate, Scenario 1 occurs (B) Since the reaction rate is much greater than the diffusion rate, Scenario occurs (C) Since the reaction rate is much lower than the diffusion rate, Scenario 1 occurs (D) Since the reaction rate is much lower than the diffusion rate, Scenario occurs 7. CSTR has a long inlet pipe. tracer is injected at the entrance of the pipe. The E-curve obtained at the exit of the CSTR is shown in the figure below. (-Marks) ssuming plug flow in the inlet pipe, the ratio (rounded off to the second decimal place) of the volume of the pipe to that of the CSTR is GTE n irreversible, homogeneous reaction products, has the rate expression: Rate C 0.1 C 1 50C, where C is the concentration of. C varies in the range mol/m 3. For very high concentrations of, the reaction order tends to: () 0 (B) 1 (C) 1.5 (D) (1-Mark) 9. For which reaction order, the half-life of the reactant is half of the full lifetime (time for 100% conversion) of the reactant? (1-Mark) () Zero order (B) Half order (C) First order (D) Second order 10. Consider two steady isothermal flow configurations shown schematically as Case-I and Case-II below. In Case-I, a CSTR of volume V 1 is followed by a PFR of volume V, while in Case-II a PFR of volume V is followed by a CSTR of volume V 1. In each case, a volumetric flow rate Q of liquid reactant is flowing through the two units in series. n irreversible reaction products (order n) takes place in both cases, with a reactant concentration C O being fed into the first unit. (-Marks) Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

12 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [4] Choose the correct option: I Cf () 1 for n 1 II C f I Cf (B) 1 for n 1 II C f I Cf (C) 1 for n 1 II C f I Cf (D) 1 for n 1 II C f 11. n isothermal steady state mixed flow reactor (CSTR) of 1m 3 volume is used to carry out the first order liquid-phase reaction products. Fresh feed at a volumetric flow rate of Q containing reactant at a concentration C 0 mixes with the recycle stream at a volumetric flow rate RQ as shown in the figure below. It is observed that when the recycle ratio R = 0.5, the exit conversion X f 50%. When the recycle ratio is increased to R =, the new exit conversion (in percent) will be: (Marks-) () 50.0 (B) 54.3 (C) 58.7 (D) 63. Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

13 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [4] NSWER KEY c b 0.5 B B C 0.7 D B C D % D D B B C B C B B C D B C C C C D B B C C C D B C C B C B C D B B C D D B D C D B D C B C C C B D C C D C C B B B B D B C D C D B D B D B C B B B C C B C C D Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

14 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [5] SOLUTIONS 1. () B 10C r 1 5C GTE-016 Upper limit of reaction rate will be at C = C 0 10C 10 So, r 1 5C C s, C So, r 5. (C) For following parallel reaction the concentration of R (C R ) and concentration of S (C S ) will be same because k 1 and k are same and order of reaction is also same. dc (k1 k)c C (k 1 = k = 1 given) dt On integrating the above equation, we get C = C 0 e t (1) dcr t k1c C C0e () dt On integrating the above equation, we get C0 t CR 1 e s, t C0 1 CR 0.5 [C 0 = 1 (given in figure)] Similarly, C0 t CS 1 e s t C0 1 CS 0.5 Since, C R and C S are same and is equal to 0.5. Hence, option (c) is correct. Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

15 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [6] 3. (8.5) H H I 1 From rrhenius law k r E 1 1 ln ln k r R T T r ln r r r r 8.5 r 1 4. (0.6671) for first reactor (CSTR) Given that, C 10 mol/lit, 1sec, C 8mol/lit C0 C1 Hence, 1 1sec r Where, r k1c kcb k C k C 1 k 1 C k C B 8() k k CB C X 0 B Since C1 8 X and CB0 0(given) C 10 Hence, 8k k...(1) 1 For second reactor given C 10, C 5 and 5sec 0 1 Hence, sec k C k C () k1 k CB0 C0 X B C 1 5 Where, X and C B0 = 0 (given) C 10 0 Hence, above equation reduce to 5k 5k 1 () 1 On solving equation (1) and (), we get k sec 1 Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

16 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [7] 5. (8) For a chemical reaction conducting in MFR the conversion V/s Temperature data is given for different value of ( r ) Given that C 0 = 5 mol/m 3 F 0 = 100 mol/sec X = 0.8 Since T = X (given) Hence at X = 0.8 T = = 370K From figure in question at T = 373K and X = 0.8, we get () (b) Now for MFR F0 X V 8m () r 10 Since (Thiele modulus) = surface reaction rate. diffuison rate D 3 Large value of implies that k>>d. It means the surface reaction rate is larger in compare to diffusion rate and scenario occurs because it is not possible for Pt to dispersed to center of catalyst pellet because of diffusion rate is much lower than surface reaction rate. It means Pt present only in the immediate vicinity of the external surface of the pellet. 7. (0.5) Since for Ideal PFR E()()( t 5) t P t 5min(from figure in question) p Hence, V P 0 k Lc 5min (1) nd for ideal CSTR the E(t) curve after p time is given as below 1 E() t exp m t t = 5min () t m p 1(5 5) 1 E(5) exp 0.05(form figure in question) m m m Hence, m = V m Dividing equation (1) and () 0 () v 0 equation 1 5 VP 0.5 V equation() 0 m r Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

17 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [8] 8. (B) r C 0.1C 1 50C (50) r GTE-015 [For very high C put C = 50] C 0.1C C & C 50C C 1 r C So, at very high C, order tends to 1. 50C 5 t f 9. () Prove that t1/ ssume n th order reaction n C C 0 dc dt kc dc n kdt C dc k dt C C C C C kt t n 1(1) k n For full life time C = 0, now equation (1) becomes, n1 n1 1n 1n 0 0 1n 0 C t f k(1) n C For half life C 0, now equation (1) becomes, 1n 1n 1n 1n 1n C0 C C 0 0 1/ 1n 1n 1 t (1)(1) k n k n By equation (), equation (3) becomes, n t t 0 t f 1n 1/ 1n 1 Put n = 0 in above equation, we get t1/ so, zero order. 10. (B) By taking 1 st reaction (n = 1) 3 1 Case-I Consider V1 V 1 m, k 1sec Q 1m sec and C 1 mol / m t f...(1)...()...(3) For CSTR, C0 C1 1 C1 K 11 C1 0.5 mol / m C C Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

18 CHEMICL ENGINEERING 1. CHEMICL RECTION ENGINEERING [9] For PFR: C f dc C 1 ln kc C C f C 1 Case-: mol/m 3 f 1 With same data 1 dc C For PFR 1 ln kc C C 1 C C mol/m 3 nd for CSTR C Cf Cf K 1 1 Cf mol/m C C f Hence for both cases, C f is equal for 1 st order reaction. 11. () f pply mole balance at steady state QC RQC RQ Q C kc V ()() 0 QC QC kc V 0 C C kc 0 C0 C k 0 (1) X 1 kc X k So conversion X is independent of recycle ratio R. So, final conversion = 50%. 1. (C) From figure: t 0 4 C() t C() t E() t t C() t dt 0 Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

19 CHEMICL ENGINEERINGG. HET TRNSFER [69]. HET TRNSFER (GTE Previous Papers) GTE composite wall is made of four different materials of construction in the fashion shown below. The resistance (in K/W) of each of the sections of the wall is indicated in the diagram. The overall resistance (in K/W, rounded off to the first decimal place) of the composite wall, in the direction of heat flow, is (1-Mark). Steam at 100 o C is condensing on a vertical steel plate. The condensate flow is laminar. The average Nusselt numbers are Nu 1 and Nu, when the plate temperatures are 10 o C and 55 o C, respectively. ssume the physical properties of the fluid and steel to remain constant within the temperature range of interest. Using Nusselt equations for film-type condensation, what is the Nu value of the ratio Nu? (1-Mark) 1 () 0.5 (B) 0.84 (C) 1.19 (D) Match the dimensionless numbers in Group-1 with the ratios in Group-. Group -1 P Biot number Q Schmidt number R Grashof number () P-II, Q-I, R-III (C) P-III, Q-I, R-II I II III Group- buoyancy force viscous force internal thermal resistance of a solid boundary layer thermal resistance momentum diffusivity mass diffusivity (B) P-I, Q-III, R-II (D) P-II, Q-III, R-I (1-Mark) Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

20 CHEMICL ENGINEERING. HET TRNSFER [70] 4. In a 1-1 pass shell and tube exchanger, steam is condensing in the shell side at a temperature (T s ) of 135 o C and the cold fluid is heated from a temperature (T 1 ) of 0 o C to a temperature (T ) of 90 o C. The energy balance equation for this heat exchanger is (-Marks) T T1 U T T mc ln s s p Where U is the overall heat transfer coefficient, is the heat transfer area, m is the mass flow rate of the cold fluid and c p is its specific heat. Tube side fluid is in a turbulent flow and the heat transfer coefficient can be estimated from the following equation: Nu = 0.03 (Re) 0.8 (Pr) 1/3 Where Nu is the Nusselt number, Re is the Reynolds number and Pr is the Prandtl number. The condensing heat transfer coefficient in the shell side is significantly higher than the tube side heat transfer coefficient. The resistance of the wall to heat transfer is negligible. If only the mass flow rate of the cold fluid is doubled, what is the outlet temperature (in o C) of the cold fluid at steady state? () 80. (B) 84. (C) 87.4 (D) In an experimental setup, mineral oil is filled in between the narrow gap of two horizontal smooth plates. The setup has arrangements to maintain the plates at desired uniform temperatures. t these temperatures, ONLY the radiative heat flux is negligible. The thermal conductivity of the oil does not vary perceptibly in this temperature range. Consider four experiments at steady state under different experimental conditions, as shown in the figure below. The figure shows plate temperatures and the heat fluxes in the vertical direction. (-Marks) What is the steady state heat flux (in W m ) with the top plate at 70 o C and the bottom plate at 40 o C? () 6 (B) 39 (C) 4 (D) The space between two hollow concentric spheres of radii 0.1 m and 0. m is under vacuum. Exchange of radiation (uniform in all directions) occurs only between the outer surface (S 1 ) of the smaller sphere and the inner surface (S ) of the larger sphere. The fraction (rounded off to the second decimal place) of the radiation energy leaving S, which reaches S 1 is Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

21 CHEMICL ENGINEERING. HET TRNSFER [89] NSWER KEY: HET TRNSFER C D B C C D ºC D D D C B D B B B B B D B B B C B D D B B D B B D C B B D B C D B D D D B B C D B C C C B D D B B D B B B D D D C B B B C C D D C Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

22 CHEMICL ENGINEERING. HET TRNSFER [90] SOLUTIONS 1. (3.9) GTE Req 0. R eq R total (C) For film-type condensation Nu (T) 1/4 Hence, Nu Nu T T 1 1 1/4 Given that, T º C and T º C Hence, Nu 45 Nu / (D) (1.) Biot number = conductive resistance in solid internal thermal resistance of a solid convective resistance in fluid boundary layer thermal resistance 4. (B) (.) Schmidt number = (3.) Grashof number = Momentum diffusivity mass diffusivity Buoyancy force viscous force Ts T1 U ln T T mc s P (1) Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

23 CHEMICL ENGINEERING. HET TRNSFER [91] given that h0 hi U h h i Hence, 1 1 U h Since, i h 0 or U h i 0.8 u d md Nu (Re) Re Re. md Nu Nu m h m Hence, U m 0.8 Put this result in equation (1) Hence, T T U ln T T m s 1 s ln () m 135 T When, T 90º C T '?, m ' m 0. Where ( ') indicates nd case when flow rate has increased twice to that of initial mass flow rate. Hence, 115 ln 45 m m ' ln 135 T ' 115 ln ln T ' ln 135 T ' T ' 135 T ' = T ' = 84.18ºC Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

24 CHEMICL ENGINEERING. HET TRNSFER [9] 5. () In experiment the hot plate is above the fluid and the cold plate is below the fluid. Hence the heat transfer is possible only by conduction mode, convection is not possible because no fluid movement will be there and the same case is in experiment 4 where the hot plate is above the fluid and cold plate is below the fluid again here, the heat transfer is only possible by conduction hence, the heat flux in experiment and 4 will be same is equal to 6 Watt/m. 6. (0.5) F 11 F1 1.0 Here, F 11 = 0 so, F 1 = 1 From Reciprocity theorem 1 F 1 = F 1 F 1 1 F1 F1 (0.5) 0.5 Where 1 r1 4 r GTE () s we know that, for equall ε 1=ε =ε 3=ε (means all are same) then, q 1 q n 1 with shield Here n 1 & ε 1=ε =ε 3=ε So, by equation (1) q 1 q 11 So, 8. () with one shield without any shield ( q /) with one shield 1 ( q /) 0.5 without any shield without any shield T f ()() time T f space Now according to lumped parameter analysis, dt h() T T CPV...(1) dt...(1) Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

25 CHEMICL ENGINEERING 3. MSS TRNSFER [13] 3. MSS TRNSFER (GTE Previous Papers) GTE binary liquid mixture of benzene and toluene contains 0 mol% of benzene. t 350 K the vapour pressures of pure benzene and pure toluene are 9 kpa and 35 kpa, respectively. The mixture follows Raoult s law. The equilibrium vapour phase mole fraction (rounded off to the second decimal place) of benzene in contact with this liquid mixture at 350 K is. For what value of Lewis number, the wet-bulb temperature and adiabatic saturation temperature are nearly equal? () 0.33 (B) 0.5 (C) 1 (D) 3. binary distillation column is to be designed using McCabe Thiele method. The distillate contains 90 mol% of the more volatile component. The point of intersection of the q-line with the equilibrium curve is (0.5, 0.7). The minimum reflux ratio (rounded off to the first decimal place) for this operation is 4. Solute C is extracted in a batch process from its homogenous solution of and C, using solvent B. The combined composition of the feed and the extracting solvent is shown in the figure below as point M, along with the tie line passing through it. The ends of the tie line are on the equilibrium curve. What is the selectivity for C? () 3.5 (B) 7 (C) 10.5 (D) 1 5. t 30ºC, the amounts of acetone adsorbed at partial pressure of 10 and 100 mmhg are 0.1 and 0.4 kg acetone/kg activated carbon, respectively. ssume Langmuir isotherm describes the adsorption of acetone on activated carbon. What is the amount of acetone adsorbed (in kg per kg of activated carbon) at a partial pressure of 50 mmhg and 30ºC? () 0.3 (B) 0.5 (C) 0.30 (D) Consider the following two cases for a binary mixture of ideal gases and B under steady state conditions. In case 1, the diffusion of occurs through non-diffusing B. In Case, equimolal counter diffusion and B occurs. In both the cases, the total pressure is 100 kpa and the partial pressures of at two points separated by a distance of 10mm are 10kPa and 5kPa. ssume that the Fick s first law of diffusion is applicable. What is the ratio of molar flux of in Case 1 to that in Case? () 0.58 (B) 1.08 (C) 1.58 (D).18 Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

26 CHEMICL ENGINEERING 3. MSS TRNSFER [14] GTE For a binary mixture of components and B, N and N B denote the total molar fluxes of components and B, respectively. J and J B are the corresponding molar diffusive fluxes. Which of the following is true for equimolar counter-diffusion in the binary mixture? () N + N B = 0 and J + J B 0 (B) N + N B 0 and J + J B = 0 (1-Mark) (C) N + N B 0 and J + J B 0 (D) N + N B = 0 and J + J B = 0 8. spherical naphthalene ball of mm diameter is subliming very slowly in stagnant air at 5ºC. The change in the size of the ball during the sublimation can be neglected. The diffusivity of naphthalene in air at 5ºC is m /s. The value of mass transfer coefficient B10-3 m/s, where B (up to one decimal place) is. (1-Marks) 9. Benzene is removed from air by absorbing it in a non-volatile wash-oil at 100 kpa in a countercurrent gas absorber. Gas flow rate is 100mol/min, which includes mol/min of benzene. The flow rate of wash-oil is 50 mol/min. vapour pressure of benzene at the column conditions is 50 kpa. Benzene forms an ideal solution with the wash-oil and the column is operating at steady state. Gas phase can be assumed to follow ideal gas law. Neglect the change in molar flow rates of liquid and gas phases inside the column. (1-Mark) For this process, the value of the absorption factor (up to two decimal places) is. 10. Identify the WRONG statement amongst the following. (1-Mark) () Steam distillation is used for mixtures that are immiscible with water. (B) Vacuum distillation is used for mixtures that are miscible with water. (C) Steam distillation is used for mixtures that are miscible with water. (D) Vacuum distillation columns have larger diameters as compared to atmospheric columns for the same throughput 11. multi-stage, counter-current liquid-liquid extractor is used to separate solute C from a binary mixture (F) of and C using solvent B. Pure solvent B is recovered from the raffinate R by distillation, as shown in the schematic diagram below. (-Marks) Locations of different mixture for this process are indicated on the triangular diagram below. P is the solvent-free raffinate, E is the extract, F is the feed and is the difference point from which the mass balance lines originate. The line PB intersects the bimodal curve at U and T. The lines P and FB intersect the bimodal at V and W respectively. The raffinate coming out of the extractor is represented in the diagram by the point: () T (B) U (C) V (D) W Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

27 CHEMICL ENGINEERING 3. MSS TRNSFER [145] NSWER KEY C 1 C C B D C B D C B B B D C B B C C B B D B D D B D D D D C C D B B B D C C B B B B B C C C B D C C B C D D C D D C B D C B C C D C D B B B C D D B D C C C C C B C B B B C D C Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

28 CHEMICL ENGINEERING 3. MSS TRNSFER [146] SOLUTIONS 1. (0.396) pply Raoult s law p x p sat i i i nd we also know pi yi P T pi p Mole fraction of Benzene (y B ) = i B GTE-016 p x p p p sat B B B sat sat xbp B xp 0. 9 yb (C) For Lewis number 1 the wet-bulb temperature and adiabatic saturation temperature are nearly h G equal to one because 1 K C and where h G is called Lewis number K C Y H 3. (1) Given that x D = 0.9 and The intersection of q line with equilibrium curve is (0.5, 0.7). Since, at minimum reflux condition the q line and upper operating line equation are intersect at equilibrium curve at same point (x, y). Hence, upper operating line equation R x y x D (1) R 1 R 1 t minimum reflux condition R R m Hence, equation (1) becomes Rm xd y x R 1 R 1 4. (C) m m Putting the value of (x, y) = ( 0.5, 0.7) and x D = 0.9 in above equation, we get (0.5) Rm 1 Rm R 1 R 1 m Selectivity = From diagram wt fraction C in E = 0.3 and wt fraction in E = 0.1 wt fraction C in R = 0. wt fraction in R = 0.7 m (wt fraction C in E)/(wt fraction in E) (wt fraction C in R)/(wt fraction in R) Putting the all values in formula, we get Y H Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

29 CHEMICL ENGINEERING 3. MSS TRNSFER [147] 0.3 = (C) ccording to Langmuir Isotherm ' q k p q q 1 k p q q max max q max 1 ' q ka p q k p For the 1 st case q 0.1, p 10mmHg max max k q 1 10k mount of adsorbed kg of adsorbent Max amount of adsorbed kg of absorbent...(1)...() For nd case q 0.4, p 100 mmhg k (3) q k On dividing equation () and (3), wet get 0.0 and q 0.6 k max In 3 rd case Given that p = 50 mmhg Hence from equation (1) q ka p q q 1 k p 0.6 1(50 0.0) q max 6. (B) Case I (diffusion of occurs through non-diffusing B) DBPT () p 1 p N 1 (1) RTZ P Blm pb pb1 Where, P Blm = p B ln pb1 Case II: (Equimolal counter diffusion of and B ) DB N () p 1 p () RTZ On dividing equation (1) and (), we get 0.3 Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

30 CHEMICL ENGINEERING 3. MSS TRNSFER [148] N N 1 PT P Blm (3) Given that P T = 100 kpa and p 1 = 10 kpa and p = 5 kpa Since P T = p 1 + p B1 = p + p B Hence, p B1 = P T p 1 = = 90 kpa and, p B = P T p = = 95 kpa pb pb Hence P Blm = 9.48 p 95 B ln ln p 90 B1 Putting the value of P Blm and P T in equation (3) N N 1 PT P 9.48 Blm GTE (D) s we know For component, N Nx J...(1) For component B, N Nx J B B B...() dd equation (1) and (), we get N N N() x x J J...(3) B B B For equimolar counter diffusion, N NB N N N 0 and x x 1 B Now equation (3) becomes 0 0 J J J J 0 B B B 8. (1.1) 3 6 Nair D mm 10 m T 5º C D m / s The relation between mass transfer coefficient and diffusivity can be given by Sherwood Number. KcD Sh...(1) D B Where D = diameter of naphthalene ball K c = mass transfer coefficient D B = Diffusivity Value of Sherwood number for spherical ball in stagnant medium is, so by equation (1) KcD DB K m / s 3 D D 10 B By comparison we get, B = Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

31 CHEMICL ENGINEERING 4. INSTRUMENTTION & PROCESS CONTROL [183] 4. INSTRUMENTTION & PROCESS CONTROL (GTE Previous Papers) GTE Match the instruments in Group-1 with process variables in Group-. (1-Mark) Group-1 Group- P. Conductivity meter I Flow Q. Turbine meter II Pressure R. Piezoresistivity element III Composition () P-II, Q-I, R-III (C) P-III, Q-II, R-I (B) P-II, Q-III, R-I (D) P-III, Q-I, R-II. What is the order of response exhibited by a U-tube manometer? (1-Mark) () Zero order (C) Second order (B) First order (D) Third order 3. system exhibits inverse response for a unit step change in the input. Which one of the following statement must necessarily be satisfied? (1-Mark) () The transfer function of the system has at least one negative pole (B) The transfer function of the system has at least one positive pole (C) The transfer function of the system has at least one negative zero (D) The transfer function of the system has at least one positive zero 4. liquid flows through an equal percentage valve at a rate of m 3 /h when the valve is 10% open. When the valve opens to 0% the flowrate increases to 3 m 3 /h. ssume that the pressure drop across the valve and the density of the liquid remain constant. When the valve opens to 50%, the flowrate (in m 3 /h, rounded off to the second decimal place) is (-Marks) 5. PI controller with integral time constant of 0.1 min is to be designed to control a process with transfer function (-Marks) Gp () s s 10 s 100 ssume the transfer functions of the measuring element and the final control element are both unity (G m = 1, G f = 1). The gain (rounded off to the first decimal place) of the controller that will constitute the critical condition for stability of the PI feedback control system is 6. For a unit step input, the response of a second order system is (-Marks) t 1 1 y() t k p 1 e sin t 1 Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

32 CHEMICL ENGINEERING 4. INSTRUMENTTION & PROCESS CONTROL [184] Where, k p is the steady state, is the damping coefficient, is the natural period of oscillation and is the phase lag. The overshoot of the system is exp. For a unit step input, 1 the response of the system from an initial steady state condition at t = 0 is shown in the figure below. What is the natural period of oscillation (in seconds) of the system? () 15.9 (B) 50 (C) 63. (D) 100 GTE Match the output signals as obtained from four measuring devices in response to a unit step change in the input signal (1-Mark) () P-IV, Q-III, R-II, S-I (C) P-IV, Q-I, R-II, S-III (B) P-III, Q-I, R-II, S-IV (D) P-II, Q-IV, R-III, S-I open 8. The transfer function for the disturbance response in an open-loop process is given G (). s The corresponding transfer function for the disturbance response in a closed-loop feedback closed control system with proportional controller is given by G (). s Select the option that is LWYS correct {O[G(s)] represents order of transfer function G(s)}: (1-Mark) open closed open closed ()()() O Gd s O Gd s (B)()() O Gd s O Gd s open closed open closed (C)()() O Gd s O Gd s (D)()() O Gd s O Gd s Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph d d

33 CHEMICL ENGINEERING 4. INSTRUMENTTION & PROCESS CONTROL [04] NSWER KEY D C D MT C D C D C 16.3 C D C B D C C C C B D C B C C C B D B D D D C C B D D B C D B D D B D C D D B D B C B C D D D C C C B B D C B C C D D D C C B C B B Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

34 CHEMICL ENGINEERING 4. INSTRUMENTTION & PROCESS CONTROL [05] SOLUTIONS 1. (D) Conductivity meter composition Turbine meter flow Piezoresistivity element Pressure GTE-016. (C) U-tube Manometer second order response 3. (D) system exhibits inverse response for unit step change in the input if transfer function of the system atleast one positive zero. 4. (10.14 m 3 /h) For equal percentage valve f () Where, l 1 R q f q max x and l L Given q 1 = m 3 /h and l 1 = 0.1, q = 3 m 3 /h and l = 0. For 1 st case l1 f 1 1 () R nd nd case () l f R 1 On dividing equation (1) and (), we get f1 l1 1( l1) l1 l () R R f q q 1 ()() R R () R R 3 R Now, l3 0.5 Then, f 3 (57.66)(57.66) Dividing equation (3) and (1) f3 l3 l1 l3 l1 R R (57.66)(57.66) f 1 q3 (57.66) q q (57.66) m /h q 1 m /h(given) (1) () (3) Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

35 CHEMICL ENGINEERING 4. INSTRUMENTTION & PROCESS CONTROL [06] 5. (.5) Since, (1) G OL Where, G OL = G P G C G f G m 10 G p s s Kcs 10Kc Gc Kc s s, G m = G f = 1 (given) Putting all values in equation (1), we get 110( Kcs 10) kc 0 s( s s 100) s s s 3 s (100 s s 10) 100K c 0 Kc From Routh-array 3 s 0 1(100 10) 100k (100 40) K 0 c 100 K 0 c For system to be at critical condition of stability K c 100 Kc (MT) Question is wrong by IISC e K c GTE (C) P: Gas chromatograph, with a long capillary tube will have delayed output as compared to others IV Q: Venturi tube attains maximum velocity very quickly and remains constant I R : Thermocouple with I st order dynamics will increase temperature, i.e. output signal with unit time. fter unit time it will attain steady state and then remains constant II S : Pressure transducer with second order dynamics shows S shape curve III 8. (D) Open loop process = without feed back Closed loop process = with feed back Order of closed loop process with feedback is always equal to or greater than the order of open open closed loop process. So, O Gd ()() s O Gd s 9. (5.03 ) G OL 0.3s Kce () s...(1) 1.5s 1 Published by :Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

36 CHEMICL ENGINEERING 5. THERMODYNMICS [35] 5. THERMODYNMICS (GTE Previous Papers) GTE The partial molar enthalpy (in kj/mol) of species 1 in a binary mixture is given h x x x, where x1 and x are the mole fractions of species 1 and, respectively. the partial molar enthalpy (in kj/mol, rounded off to the first decimal place) of species 1 at infinite dilution is (1-Mark). n ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol 1 K 1 (assumed to be independent of temperature); universal gas constant, R is J mol 1 K 1 ; ratio of molar heat capacities is The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is (-Marks) 3. gas obeying the Clausius equation of state is isothermally compressed from 5 MPa to 15 MPa in a closed system at 400 K. The Clausius equation of state is P RT where P is the v b () T pressure, T is the temperature, v is the molar volume and R is the universal gas constant. The parameter b in the above equation varies with temperature as b() T b0 b1t with 5 b m 3 mol 1 and 7 enthalpy (h) at a constant temperature is given by b m 3 mol 1 K 1. The effect of pressure on the molar h v v T p T T p. Let h i and h f denote the initial and final molar enthalpies, respectively. The change in the molar enthalpy h f h i (in J mol 1, rounded off to the first decimal place) for this process is (-Marks) 4. binary system at a constant pressure with species 1 and is described by the two-suffix E g Margules equation, 3x1 x RT, where ge the molar excess Gibbs free energy, R is the universal gas constant, T is the temperature and x 1, x are the moles fractions of species 1 and, respectively. g1 g t a temperature T, 1 and, where g 1 and g are the molar Gibbs free energies of RT RT pure species 1 and, respectively. t the same temperature, g represents the molar Gibbs free energy of the mixture. For a binary mixture with 40 mole % of species 1, the value (rounded off g to the second decimal place) of is RT (-Marks) Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

37 CHEMICL ENGINEERING 5. THERMODYNMICS [36] GTE For a gas phase cracking reaction B + C at 300ºC, the Gibbs free energy of the reaction at this temperature is Gº = 750 J/mol. The pressure is 1 bar and the gas phase can be assumed to be ideal. The universal gas constant R = J/mol K. The fractional molar conversion of at equilibrium is : (1-Mark) () 0.44 (B) 0.50 (C) 0.64 (D) Which of the following can change if only the catalyst is changed for a reaction system? () Enthalpy of reaction (B)ctivation energy (1-Mark) (C) Free energy of reaction (D) Equilibrium constant 7. If v, u, s and g respectively the molar volume, molar internal energy, molar entropy and molar Gibbs free energy, then match the entries in the left and right columns below and choose the correct option: (1-Mark) P. ( u /) v s I. Temperature Q. ( u /) P T II. Pressure R. ( g /) T P III. v S. ( u /) s v IV. s () P-II, Q-III, R-IV, S-I (C) P-I, Q-IV, R-II, S-III (B) P-II, Q-IV, R-III, S-I (D) P-III, Q-II, R-IV, S-I 8. Three identical closed systems of a pure gas are taken from an initial temperature and pressure (T 1, P 1 ) to a final state (T P ), each by a different path. Which of the following is LWYS TRUE for the three systems? ( represents the change between the initial and final states; U. S, G, Q and W are Internal energy, entropy. Gibbs free energy, heat added and work done respectively.) (1-Mark) () U, S, Q are same (B) W, U, G are same (C) S, W, Q are same (D) G, U, S are same 9. For a pure liquid, the rate of change of vapour pressure with temperature is 0.1 bar/k in the temperature range of 300 to 350 K. If the boiling point of the liquid at bar is 30K, the temperature (in K) at which it will boil at 1 bar (up to one decimal place) is (1-Mark) 10. n ideal gas is initially at a pressure of 0.1 MPa and a total volume of m 3. It is first compressed to 1 MPa by a reversible adiabatic process and then cooled at constant pressure to a final volume of 0. m 3. The total work done (in kj) on the gas for the entire process (up to one decimal place) is (-Marks) 11. Given that molar residual Gibbs free energy, g R, and molar residual volume, v R, are related as R g RT R P v dp, 0 RT find g R at T= 7 ºC and P = 0. MPa. The gas may be assumed to follow the Virial equation of state, z = 1 + BP / RT, where B = 10 4 m 3 /mol at the given conditions (R = J/mol.K). The value of g R in J/mol is... (-Marks) () (B).4 (C) 0 (D) 0 1. binary mixture of components (1) and () forms an azeotrope at 130ºC and x 1 = 0.3. The liquid phase non-ideality is described by ln x 1 and ln x, and 1 1 are the activity coefficients, and x 1, x are the liquid phase mole fractions. For both components, the fugacity coefficients are 0.9 at the azeotropic composition. Saturated vapour pressure at 130 ºC are sat P bar and P 30 bar. The total pressure in bars for the above azeotropic system (up sat 1 70 to two decimal places) is. (-Marks) Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

38 CHEMICL ENGINEERING 5. THERMODYNMICS [51] NSWERS KEY D B D D 7.54 B D C D C D B B B D C B D B C D C C B D D D C D B B B C B D B C C D B C D C B B B D C B B B C D B B D D B C B B D D C C D C B B C D B C C B D B B C B D D C C Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

39 CHEMICL ENGINEERING 5. THERMODYNMICS [5] SOLUTIONS 1. ( 58) h 60x 100x x 1 1 For infinite dilution x 1 = 0 and x = 1 So, h1 60(1) 100(0)(1) h1 58. (373) Given P 3 bar, P 6 bar 1 T 300 K 1 Since() () W 1.5() irr rev J CV 30 mole.k J R mole.k C C rev P V 1.77 V W W C T T P 1.77 T T1 P 1 1 GTE-015 and Hence W 30( ) 1460 rev J mole J Wirr 1.5(1460) 190 mole W 190 C [ T 300] irr V f [ T 300] T f 373K 3. (400) RT P v b RT v b...() i P f Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

40 CHEMICL ENGINEERING 5. THERMODYNMICS [53] and b b b T where b b RT v b0 b1t P R v T b1 b0 () ii P v R b Since 1 T P P h v v T P T v Putting value of v and T T T P P in above equation h R R T b b T b b P P P h P T b h 4 10() P 5 h 410 [15 5] mole 4. (1.646) Given G1 x1 0.4, 1 RT G x 0.6, RT E id 5 6 j G G G...() i From equation (i) and (ii) E id G G G, given that G (3) x1 xrt G 3 x x RT [ x G x G RT{ x ln x x ln x }] G 1 3 G G x1 x x1 x { x1 ln x1 x ln x} RT RT RT E G1 G Put values of x1, x and RT RT in above equation we get. G ( )(0.4ln ln 0.6) RT Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

41 CHEMICL ENGINEERING 6. FLUID MECHNICS [79] 6. FLUID MECHNICS (GTE Previous Papers) GTE For a flow through a smooth pipe, the Fanning friction factor (f) is given f mre 0. in the turbulent flow regime, where Re is the Reynolds number and m is a constant. Water flowing through a section of this pipe with a velocity 1 m/s results in a frictional pressure drop of 10 kpa. What will be the pressure drop across this section (in kpa), when the velocity of water is m/s? (1-Mark) () 11.5 (B) 0 (C) 34.8 (D) 40. vertical cylindrical vessel has a layer of kerosene (of density 800 kg/m 3 ) over a layer of water (of density 1000 kg/m 3 ). L -shaped glass tubes are connected to the column 30 cm apart. The interface between the two layers lies between the two points at which the L-tubes are connected. The levels (in cm) to which the liquids rise in the respective tubes are shown in the figure below. (1-Mark) The distance (x in cm, rounded off to the first decimal place) of the interface from the point at which the lower L-tube is connected is 3. Water (density=1000 kg m 3 ) is pumped at a rate of 36 m 3 /h, from a tank m below the pump, to an overhead pressurized vessel 10 m above the pump. The pressure values at the point of suction from the bottom tank and at the discharge point to the overhead vessel are 10 kpa and 40 kpa, respectively. ll pipes in the system have the same diameter. Take acceleration due to gravity, g = 10 m s. Neglecting frictional losses, what is the power (in kw) required to deliver the fluid? (-Marks) () 1. (B).4 (C) 3.6 (D) The characteristics curve (Head Capacity relationship) of a centrifugal pump is represented by the equation H pump = Q, where H pump is the head developed by the pump (in m) and Q is the flowrate (in m 3 /h) through the pump. This pump is to be used for pumping water through a horizontal pipeline. the frictional head loss H piping = Q L Q L. The flowrate (in m 3 /h, rounded off to the first decimal place) of water pumped through the above pipeline, is (-Marks) Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

42 CHEMICL ENGINEERING 6. FLUID MECHNICS [80] 5. Water flows through a smooth circular pipe under turbulent conditions. In the viscous sublayer, the velocity varies linearly with the distance from the wall. The Fanning friction factor is w defined as, f where w is the shear stress at the wall of the pipe, is the density of u / the fluid and u is the average velocity in the pipe. Water (density = 1000 kgm 3, viscosity = kg m 1 s 1 ) flows at an average velocity of 1 m s 1 through the pipe. For this flow condition, the friction factor f is t a distance of 0.05 mm from the wall of the pipe (in the viscous sub-layer), the velocity (in m s 1, rounded off to the third decimal place), is GTE For uniform laminar flow over a flat plate, the thickness of the boundary layer,, at a distance x from the leading edge of the plate follows the relation: (1-Mark) () ()x x 1 1/ (B) ()x x (C) ()x x (D) ()x x 1/ 7. cylindrical packed bed of height 1m is filled with equal sized spherical particles. The particles are nonporous and have a density of 1500 kg/m 3. The void fraction of the bed is The bed is fluidized using air (density 1 kg/m 3 ). If the acceleration due to gravity is 9.8 m/s, the pressure drop (in Pa) across the bed in incipient fluidization (up to one decimal place) is. (1-Mark) 8. Two different liquids are flowing through different pipes of the same diameter. In the first pipe, the flow is laminar with centreline velocity, V max, 1, whereas in the second pipe, the flow is turbulent. For turbulent flow, the average velocity is 0.8 times the centerline velocity, V max,. For equal volumetric flow rates in both the pipes, the ratio V max,1 / V max, (up to two decimal places) is. (1-Mark) 9. For Fanning friction factor f (for flow in pipes) and drag coefficient C D (for flow over immersed bodies), which of the following statements are true? (-Mark) P. f accounts only for the skin friction Q. C D accounts only for the skin friction R. C D accounts for both skin friction and form friction S. Both f and C D depend on the Reynolds number T. For laminar flow through a pipe, f doubles on doubling the volumetric flow rate () R, S, T (B) P, Q, S (C) P, R, S (D) P, Q, S, T 10. centrifugal pump delivers water at the rate of 0. m 3 /s from a reservoir at ground level to another reservoir at a height H, through a vertical pipe of 0. m diameter. Both the reservoirs are open to atmosphere. The power input to the pump is 90 kw and it operates with an efficiency of 75%. (-Mark) Data: Fanning friction factor for pipe flow is f = Neglect other head losses. Take gravitational acceleration, g = 9.8m/s and density of water is 1000 kg/m 3. The height H, in meters, to which the water can be delivered (up to one decimal place) is Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

43 CHEMICL ENGINEERING 6. FLUID MECHNICS [98] NSWER KEY C 10 B C C C D B D C B C B C B D C D D C C B D B C D B B B D D D B B C D B B C D B B C C D C D B B C B B C B C D C B C C D B D B C C D B D B B D C B D D D B C B D C B D B B C D B B B D Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

44 CHEMICL ENGINEERING 6. FLUID MECHNICS [99] SOLUTIONS 1. (C) m f (Re) 0. From Darcy s law h. (10) f 4fLv, Since hf gd P g Hence, Putting the value of f in P, we get 4mLv P since Re = vd 0. D(Re) 1.8 P v P1 v1 GTE-016 4fLv P D hence, 1.8 P v Given that v m/ sec, v1 1m/ sec, P1 10kPa 1.8 Hence, P 10() 34.8kPa Pressure intensity along the line Y and Y will be same in each column. Hence (0.5 x)800g 1000xg 0.4(1000g) On solving this equation, we get x = 0.1 m = 10 cm 3. (B) Given = 1000 kg/m 3 Q = 36 m 3 /hr Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

45 CHEMICL ENGINEERING 6. FLUID MECHNICS [300] P 1 = 10 KPa P = 40 KPa Since diameter of pipes are same and volumetric flow rate are also same hence velocity in suction and discharge pipes will be same. V 1 = V P1 V1 P V Z1 hp Z h f g g Pg g Given that hf 0, V1 V P P hp ()(1 Z Z0) 1 4 m 10 3 g 10 Now power (P) = ghq = 400Watt=.4 kw (48.9 m 3 /h) Given that, H pump = Q H piping = Q L Q L (1) () On equating equation (1) and () Q = Q L Q L Q Q 43.8 = 0 On solving Q = 48.9 m 3 /h 5. (0.15) f u w = du Since w dy Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

46 CHEMICL ENGINEERING 7. PROCESS CLCULTIONS [333] 7. PROCESS CLCULTIONS (GTE Previous Papers) GTE liquid mixture of ethanol and water is flowing as inlet stream P into a stream splitter. It is split into two streams, Q and R, as shown in the figure below. The flow rate of P, containing 30 mass% of ethanol, is 100 kg/h. What is the least number of additional specification(s) required to determine the mass flowrates and compositions (mass%) of the two exit streams? (1-Mark) () 0 (B) 1 (C) (D) 3. jacketed stirred tank with a provision for heat removal is used to mix sulphuric acid and water in a steady state flow process. H SO 4 (l) enters at a rate of 4 kg/h at 5 o C and H O (l) enters at a rate of 6 kg/h at 10 o C. The following data are available: (-Marks) Specific heat capacity of water = 4. kj kg 1 K 1. Specific heat capacity of aqueous solution of 40 mass% H SO 4 =.8 kj (kg solution) 1 K 1. ssume the specific heat capacities to be independent of temperature. Based on reference states of H SO 4 (l) and H O (l) at 5 o C, the heat of mixing for aqueous solution of 40 mass% H SO 4 = 650 kj (kg H SO 4 ) 1. If the mixed stream leaves at 40 o C, what is the rate of heat removal (in kj/h)? () 180 (B) 558 (C) 570 (D) catalytic reforming plant produces hydrogen and benzene from cyclohexane by de-hydro aromatisation. In order to increase the production of hydrogen, the owner plans to change the process to steam reforming of the same feedstock that produces hydrogen and carbon dioxide. Stoichiometrically, what is the maximum ratio of pure hydrogen produced in the proposed process to that in the existing process? (-Mark) () 1 (B) (C) 5 (D) 6 GTE The schematic diagram of a steady state process is shown below. The fresh feed (F) to the reactor consists of 96 mol% reactant and 4 mol% inert I. The stoichiometry of the reaction in C. part of the reactor effluent is recycled. The molar flow rate of the recycle stream is 0.3F. The product stream P contains 50 mol% C. The percentage conversion of in the reactor based on entering the reactor at point 1 in the figure (up to one de cimal place) is (-Marks) Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

47 CHEMICL ENGINEERINGG 7. PROCESS CLCULTIONS [334] GTE Two elemental gases ( and B) are reacting to form a liquid (C) in a steady state process as per the reaction + B C. The single-pass conversion of the reaction is only 0% and hence recycle is used. The product is separated completely in pure form. The fresh feed has 49 mol% of and B each along with mol% impurities. The maximum allowable impurities in the recycle stream is 0 mol% %. The amount of purge stream (in moles) per 100 moles of the fresh feed is (-Marks) 6. Carbon monoxide (CO) is burnt in presence of 00% excess pure oxygen and the flame temperature achieved is 98K. The inlet streams are at 5ºC. The standard heat of formation (at 5ºC) of CO and CO are 110 kj mol -1 and 390 kj mol -1 respectively. The heat capacities (in J mol -1 K -1 ) of the components are (-Marks) 3 3 C T C T p o Where, T is the temperature in K. The heat loss (in kj) per mole of CO burnt is Common Data for Questionss 7 and 8: reverse osmosis unit treats feed water (F) containing fluoride and its output consists of a permeate stream (P) and a reject stream (R). Let C F, C P and C R 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 flow rate of the inlet stream, and C F = mg/l and C P = 0.1 mg/l. 7. The value of C R in mg/l, up to one digit after the decimal point, is 8. fraction f 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 digits after the decimal point, is (-Marks) Common Data for Questions for 9 and 10: The reaction B C D is carried out in a reactor followed by a separator as shown below. p co liq gas, liq gas GTE-013 GTE-01 (-Marks) Notation: Molar flow rate of fresh B is F FB Molar flow rate of recycle gas is F RG Molar flow rate of purge gas is F PG Molar flow rate of is F Molar fraction of B in recycle gas is Y RB Molar flow rate of C is F C Here, F FB = mol/s; F = 1 mol/s, F B /F = 5 and is completely converted. 9. If Y 0.3, the ratio of recycle gas to, purge gas F / F is (-Marks) RB () (B) 5 (C) 7 (D) If the ratio of recycle gas to purge gas F / F is 4 then Y RB is (-Marks) RG RB RG PG () 3 8 (B) 5 (C) 1 (D) 3 4 Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

48 CHEMICL ENGINEERING 7. PROCESS CLCULTIONS [34] NSWER KEY B D mg/L 0.64 B B C D D C C B B C D B C C C D D C D B C B B B D C D C D B B D C B C 51 D Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

49 CHEMICL ENGINEERING 7. PROCESS CLCULTIONS [343] SOLUTIONS GTE (B) Given that P = 100 kg/h Total mass balance P = 100 = Q + R (1) Ethanol balance P x P = Qx Q + Rx R Given that, x P = 0.3 P (0.) = Qx Q + Rx R () Where, x Q and x R compositions of ethanol in stream Q and R respectively, Because of splitting x Q and x R are same as x P = 0. Hence, equation () reduced to 100 = Q + R Hence, only one additional specification required to determine the mass flow rate and composition.. () pplying steady state energy balance Heat rate in Heat rate out + Heat for mixing of solution Heat removed = 0 (kj/hr) (kj/hr) (kj/hr) (kj/hr) Heat rate in 3 kj 4(C)(5 P HSO 5) (10 5) hr Heat rate out kj 10.8(40 5) 40 hr. Heat for mixing kj hr. Put all values in energy balance equation Heat rate in Heat rate out + Heat for mixing of solution Heat removed = 0 (kj/hr) (kj/hr) (kj/hr) (kj/hr) = Heat removed Heat removed = 180 kj hr. 3. (D) Dehydrogenation of Cyclo-hexane C6H1 C6 H6 3H Steam reforming of Cyclo-hexane C H 1H O 6CO 18H 6 1 (1) () Hence, at complete conversion of C6 H 1, the ratio of H form in case () to that in case (1) is Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

50 CHEMICL ENGINEERING 8.MECHNICL OPERTIONS [359] 8. MECHNICL OPERTIONS (GTE Previous Papers) GTE In a cyclone separator used for separation of solid particles from a dust laden gas, the separation factor is defined as the ratio of the centrifugal force to the gravitational force acting on the particle. S r denotes the separation factor at a location (near the wall) that is at a radial distance r from the centre of the cyclone. Which one of the following statements is INCORRECT? (1-Mark) () S r depends on mass of the particle (B) S r depends on the acceleration due to gravity (C) S r depends on tangential velocity of the particle (D) S r depends on the radial location ( r ) of the particle. n agitated cylindrical vessel is fitted with baffles and flat blade impellers. The power number for this system is given by where P is the power consumed for the mixing, ρ is the density of the fluid, n is the speed of the impeller and D is the diameter of the impeller. The diameter of the impeller is 1/3 rd the diameter of the tank and the height of liquid level is equal to the tank diameter. The impeller speed to achieve the desired degree of mixing is 4 rpm. In a scaled up design, the linear dimensions of the equipment are to be doubled, holding the power input per unit volume constant. ssuming the liquid to be Newtonian and N p to be independent of Reynolds number, what is the impeller speed (in rpm) to achieve the same degree of mixing in the scaled up vessel? (-Marks) () 0.13 (B) 1.6 (C).5 (D) Consider a rigid solid sphere falling with a constant velocity in a fluid. The following data are known at the conditions of interest: viscosity of the fluid = 0.1 Pas, acceleration due to gravity = 10 m s, density of the particle = 1180 kg m 3 and density of the fluid = 1000 kg m 3. The diameter (in mm, rounded off to the second decimal place) of the largest sphere that settles in the Stokes law regime (Reynolds number 0.1), is (-Marks) GTE typical batch filtration cycle consists of filtration followed by washing. One such filtration unit operation at constant pressure difference first filters a slurry during which 5 liters of filtrate is collected in 100s. This is followed by washing which is done for t w seconds and uses 1 liter of wash water. ssume the following relation to be applicable between the applied pressure drop P, cake thickness L at time t, and volume of liquid V collected in time t: (-Marks) P dv k1 ; L kv, if L is changing. L dt k 1 and k can be taken to be constant during filtration and washing. The wash time t w, in seconds (up to one decimal place), is 5. The diameters of sand particles in a sample ranges from 50 to 150 microns. The number of particles of diameter x in the sample is proportional to x The average diameter, in microns, (up to one decimal place) is (-Marks) Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph

51 CHEMICL ENGINEERING 8.MECHNICL OPERTIONS [360] 6. spherical solid particle of 1 mm diameter is falling with a downward velocity of 1.7 mm/s through a liquid (viscosity 0.04 Pa.s) at a low Reynolds number (Stokes regime). The liquid is flowing upward at a velocity of 1 mm/s. ll velocities are with respect to a stationary reference frame. Neglecting the wall effects, the drag force per unit projected area of the particle, in Pa,(up to two decimal places) is (-Marks) GTE In order to produce fine solid particles between 5 and 10 μm, the appropriate size reducing equipment is (1-Mark) () Fluid energy mill (B) Hammer mill (C) Jaw crusher (D) Smooth roll crusher 8. Slurries are most conveniently pumped by a (1-Mark) () Syringe pump (B) Diaphragm pump (C) Vacuum pump (D) Gear pump GTE In the Tyler standard screen scale series, when the mesh number increases from 3 mesh to 10 mesh, then (1-Mark) () the clear opening decreases (B) the clear opening increases (C) the clear opening is unchanged (D) the wire diameter increases 10. Taking the acceleration due to gravity to be 10 m/s, the separation factor of a cyclone 0.5 m in diameter and having a tangential velocity of 0 m/s near the wall is (1-Mark) ton/h of a rock feed, of which 80% passed through a mesh size of.54 mm, were reduced in size such that 80% of the crushed product passed through a mesh size of 1.7 mm. The power consumption was 100 kw. If 100 ton/h of the same material is similarly crushed from a mesh size of 5.08 mm to a mesh size of.54 mm, the power consumption (in kw, to the nearest integer) using Bond s law, is (-Marks) GTE In a mixing tank operating at very high Reynolds number ( > 10 4 ), if the diameter of the impeller is doubled (other conditions remaining constant), the power required increases by a factor of (1-Mark) () 1/3 (B) 1/4 (C) 4 (D) GTE The particle size distributions of the feed and collected solids (sampled for same duration ) for a gas cyclone are given below. (-Marks) Size range μm Weight of feed in the size range (g) What is the collection efficiency (in PERCENTGE) of the gas cyclone? () 31 (B) 60 (C) 65 (D) 69 3 Weight of Collected solids in the size range (g) Published by : Engineers Institute of India-E.I.I. LL RIGHT RESERVED 8B/7, Jiasarai Near IIT Hauzkhas Newdelhi Ph