CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 1 General Aptitude Questions Q.No-1-5 Carry One Mark Each 1.Choose the word most similar in meaning to the given word: Educe (A) Exert(B) Educate(C) Extract(D) Extend Answer:(C) 2.If logx (5/7) = -1/3, then the value of x is (A) 343/125(B) 12/343(C) -25/49(D) -49/25 Answer:(A) Exp: 31 3 1 35 7 7x x x 2.747 5 5| |= = = |\

3.Operators a b a b, and aredefined by : a b ; a b ; a b ab.a b a b + = = =+ Find the value ( ) ( ) 66 6 , 66 6 . (A)-2(B)-1(C)1(D) 2 Answer:(C) Exp: 66 6 60 566 666 6 72 6= = =+ 6666 6 72 6666 6 60 5+= = =

(666)(66 6) 5 616 5= = 4.Choose the most appropriate word from the options given below to complete the following sentence. The principal presented the chief guest with a ____________, as token of appreciation. (A)momento(B)memento(C)momentum(D)moment Answer:(B) CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 2 5.Choose the appropriate word/phrase, out of the four options given below, to complete the following sentence: Frogs ____________________. (A)Croak(B)Roar(C)Hiss(D)Patter Answer:(A) Exp:Frogs make croak sound. 6.A cube of side 3 units is formed using a set of smaller cubes of side 1 unit. Find the proportionof the number of faces of the smaller cubes visible to those which are NOT visible. (A)1 : 4(B)1 : 3(C)1 : 2(D)2 : 3 Answer:(C) Exp: Number of faces per cube = 6 Total number of cubes = 93 = 27 Total number of faces = 276 = 162 Total number of non visible faces = 162-54 = 108 Number of visible faces 54 1Number of non visible faces 108 2= = 7.Fill in the missing value 6 5 47 47 2 11 9 2 8 1 2 14 1 5 2 33 31 1 11111 1 11 1 11 11CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 3 Answer:(3) Exp:Middle number is the average of the numbers on both sides. Average of 6 and 4 is 5 Average of (7+4) and (2+1) is 7 Average of (1+9+2) and (1+2+1) is 8 Average of (4+1) and (2+3) is 5 Therefore, Average of (3) and (3) is 3 8.Humpty Dumpty sits on a wall every day while having lunch. The wall sometimes breaks. A person sitting on the wall falls if the wall breaks. Which one of the statements belowis logically valid and can be inferred from the above sentences? (A)Humpty Dumpty always falls having lunch (B)Humpty Dumpty does not fall sometimes while having lunch (C)Humpty Dumpty never falls during dinner (D)When Humpty Dumpty does not sit on the wall, the wall does not break Answer:(B) 9.The following question presents a sentence, part of which is underlined. Beneath the sentence you find four ways of phrasing the underlined part. Following the requirements of the standard written English, select the answer that produces the most effective sentence.Tuberculosis, together with its effects, ranks one of the leading causes of death in India. (A)ranks as one of the leading causes of death (B)rank as one of the leading causes of death (C)has the rank of one of the leading causes of death (D)are one of the leading causes of death Answer:(A) 10.Read the following paragraph and choose the correct statement. Climate change has reduced human security and threatened human well being. An ignored reality ofhumanprogressisthathumansecuritylargelydependsuponenvironmentalsecurity.Buton the contrary, human progress seems contradictory to environmental security. To keep up both at therequiredlevelisachallengetobeaddressedbyoneandall.Oneofthewaystocurbthe climatechangemaybesuitablescientificinnovations,whiletheothermaybetheGandhian perspective on small scale progress with focus on sustainability. (A)Human progress and security are positively associated with environmental security. (B)Human progress is contradictory to environmental security. (C)Human security is contradictory to environmental security. (D)Human progress depends upon environmental security. Answer:(D) CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 4 Section Name: Chemical Engineering Q.No-1-25 Carry One Mark Each 1.Benzeneisremovedfromairbyabsorbingitinanon-volatilewash-oilat100kPainacounter-currentgasabsorber.Gasflowrateis100mol/min,whichincludes2mol/minofbenzene.The flowrateofwash-oilis50mol/min.Vapourpressureofbenzeneatthecolumnconditionsis50 kPa.Benzeneformsanidealsolutionwiththewash-oilandthecolumnisoperatingatsteady 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. For this process, the value of the absorption factor (upto two decimal places) is _________. Answer: 1.02 Exp:Absorption factor sSLAmG=Given Raoults law is applicablefor gas phase satyP xpy 100 x 50y 0.5xSo, m 0.5= = == 50A 1.020.5 98= = 2.The following set of three vectors 1 x 32 , 6 and 41 x 2||| | ||| | || | || | |\\ \ is linearly dependent when x is equal to (A)0(B)1(C)2(D)3 Answer:(D) Exp:For linearly dependent vectorsA 0 = 1 x 32 6 4 01 x 2= ( ) ( ) ( ) 12 4x x 4 4 3 2x 6 02x 6x 3 + = = = ( )SliquidLmol50min( )SGas 100mol minG 98mol minBenzene 2mol min==CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 5 3.Forwhichreactionorder,thehalf-lifeofthereactantishalfofthefulllifetime(timefor100% conversion) of the reactant? (A)Zero order(B)Half order(C)First order(D)Second order Answer: (A) Exp:For zero order reaction O AAdCkC kdt = =Where k = rate constant AO AC C kt =For, full life time CA = 0 AOt C K =and for halflife AOA AO 1/ 2CC C 2, So t2k= = 1/ 2So, t 2t = 4.Matchtheoutputsignalsasobtainedfromfourmeasuringdevicesinresponsetoaunitstep change in the input signal. Output signal time P: Gas chromatograph, with a long capillary tube Q : Venturi tube R : Thermocouple with first order dynamicsS : Pressure transducer with second order dynamics (A)P-IV, Q-III, R-II, S-I(B)P-III, Q-I, R-II, S-IV (C)P-IV, Q-I, R-II, S-III(D)P-II, Q-IV, R-III, S-I Answer: (C) Exp:If y(t) = output (response) Gas chromatograph IV Ventury tube- I Thermocouple II Pressure transducer with second order dynamics - III IIIIIIIV( ) I Gas chromatograph( ) IV Venturytube( ) II 1st order( ) III 2nd ordertime( ) y tCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 6 5.Matchthechemicalswrittenontheleftwiththerawmaterialsrequiredtoproducethem mentioned on the right. (I) Single Superphosphate (SSP)(P)Rockphosphate+SulfuricAcid+ Ammonia (II)Triple Superphosphate (TSP)(Q) Brine (III)Diammonium phosphate (DAP)(R)Rock Phosphate + Sulfuric Acid (IV) Caustic soda(S)Rock phosphate + phosphoric Acid (A)I Q, II R, III S, IV P (B)I S, II P, III Q, IV R (C)I R, II S, III P, IV Q (D)I S, II R, III P, IV Qa Answer:(C) Exp:Single super phosphate (SSP) Phosphate rock + 2 4H SOTriple super phosphate( )3 4TSP phosphate rock H PO +Diammonium phosphate (DAD) 2 4 3Rock phosphate H SO NH + +Caustic soda Brine 6.For the matrix 4 3 1, if3 4 1| | | | ||\ \ is an eigenvector, the corresponding eigenvalue is _______. Answer: 7 Exp: Let A = 4 3 1and x3 4 1 ((= (( Let eigen value isSoAx x = [ ][ ]1 0or A I x 0, where I0 1 ( = =( So 0Ix0(=( 4 3 103 4 14 303 47 (( = (( +( = (+ = CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 7 7.The transfer function for the disturbance response in an open-loop process is given by( )opendG s . The corresponding transfer function for the disturbance response in closed-loop feedback control system with proportional controller is given by( )closeddG s . Select the option that is Always correct( ) ( ) { }O G S represents order of transfer function G s :( (A)( ) ( )open closedd dO G s O G s = (B)( ) ( )open closedd dO G s O G s (C)( ) ( )open closedd dO G s O G s (D)( ) ( )open closedd dO G s O G s Answer:(A) Exp: Open loop transfer function ( )openOL dG GH G s = =and closed loop transfer function ( )closedCL dGG G s1 GH= =+ Order of transfer function is order of denominator Order of GH and 1 + GH are same ( ) ( )open closedd dso, O G s O G s(( = 8.If v,u, s and g represent respectively the molar volume, molar internal energy, molar entropy and molarGibbsfreeenergy,thenmatchtheentriesintheleftandrightcolumnsbelowandchoose the correct option. ( ) ( )sP u v ( ) I Temperature( ) ( )TQ g P ( ) II Pr essure( ) ( )PR g T ( ) III V ( )( )VS u s ( ) IV S (A)P II, Q III, R IV, S I (B)P II, Q IV, R III, S I (C)P I, Q IV, R II, S III (D)P III, Q II, R IV, S I Answer: (A) Exp: We have du Tds pdv = anddg =vdp sdT So, S Su uP Pv v = = VPTuTsg gV and Sp T= = = x yGHCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 8 9.Twodifferent liquidsare flowingthroughdifferentpipesof thesamediameter.Inthefirstpipe, theflowislaminarwithcenterlinevelocity, max.1V , Whereasinthesecondpipe,theflowis turbulent.Forturbulentflow,theaveragevelocityis0.82timesthecenterlinevelocity, max.2V .Forequalvolumetricflowratesinboththepipes,theratio max.1 max.2V V (uptotwodecimal places) is ______. Answer:1.64 Exp:For laminar flow avg.1 max.1V 0.5V =For Turbulent flow( )avg.2 max.2V 0.82V given =Since volumetric flow rate is some 1 2Q Q = 1 avg.1 2 avg.2A V A V =Given1 2A A =So, avg.1 avg.2V V = max.1 max.2max.1max.20.5 V 0.82VV1.64V = = 10.Considerlinearordinarydifferentialequation ( ) ( ) ( ) ( )dyp x y r x . Functions p x and r x aredx+ =defined and have a continuous first derivative. The integrating factor of this equation is non-zero. Multiplying this equation by its integrating factor converts this into a: (A)Homogeneous differential equation(B)Non-linear differential equation (C)Second order differential equation(D)Exact differential equation Answer:(D) Exp: Linear differential equation( ) ( )1y p x y r x + =Multiplying above equation by integrating factor ( ) p x dxe makes the equation exact 11.Asphericalnaphthaleneballof2mmdiameteris sublimingveryslowlyinstagnantair at25OC. Thechangeinthesizeoftheballduringthesublimationcanbeneglected.Thediffusivityof naphthalene in air at 25o C is 1.110-6 m2/s. ThevalueofmasstransfercoefficientisB10-3m/s,whereB(uptoonedecimalplace)is _______. Answer:1.1 Exp:For spherical ball Dimensionless sherwood number = 2 CABk L2D=CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 9 Where L = characteristic length = diameter ofSpherical ball = d = 2 mm = 32 10 m 6ABC 33C2 D 2 1.1 10Kd 2 10K 1.1 10 m sSo, B 1.1 = == = 12.An irreversible, homogeneous reactionA products, has the rate expression: 2A AAA2C 0.1CRate , where C is the concentration of A.1 50C+=+ ACvaries in the range 0.5 50 mol/m3. For very high concentration of A, the reaction order tends to: (A)0(B)1(C)1.5(D)2 Answer:(B) Exp:Rate = 2A AA2C 0.1C1 50C++ ( )3A0.5 C 50 mol m < = = 13.A scalar function in the xy-plane is given by( )2 2 x, y x y . if i amd j = +are unit vectors in the x and y directions, the direction of maximum increase in the value ofat (1,1) is along: (A) 2i 2j + (B) 2i 2j + (C) 2i 2j (D) 2i 2j Answer:(B) Exp:Direction of maximum increase ( )( )( )( )at 1.1 2x i 2yj at 1.1 2i 2j== += + CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 10 14.For a gas phase cracking reactionA B C +at 300OC, the Gibbs free energy of the reaction at this temperature is OG 2750J mol. = The pressure is 1 bar and the gas phase can be assumed to be ideal. The universal gas constant R = 8.314J/mol. K. The fractional molar conversion of A at equilibrium is: (A)0.44(B)0.50(C)0.64(D)0.80 Answer: (D) Exp:A B C + at t 0, 0 0at t t 1 where, fraction converted= = = Equilibrium constant B CPA2 22 2Y YK .PY1 1.P1 1 11 x= | || | | | + + \ \ = = = | | |+\ Also OO22ln k G RTG 2750K exp : expRT 8.314 573K 1.78so, 1.7810.80 ( = ( (= ( ( == = 15.Two infinitely large parallel plates (I and II) are held at temperatures TI and TII( )1 IIT T >respectively, and placed at a distance 2d apart in vacuum. An infinitely large flat radiation shield (III) is placed in parallel in between I and II. The emissivities of all the plates are equal. The ratio of the steady state radiative heat fluxes with and without the shield is: (A)0.5(B)0.75(C)0.25(D)0 1ITIIII1IITCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 11 Answer:(A) Exp: We have( )4 41 212T TQ1 1A1 =+ Where= emissivity ( )4 41 212T TQ2A1 = In presence of shield ( ) ( )( ) ( ) ( )4 4 4 41 3 3 213 234 44 1 234 4 4 4 41 1 2 1 313121312T T T TQ Q2 2A A1 1T Tor T2T T T 2 T TQSo,2 2A1 1Q 12 AQ 1So,Q 2 = = = += + = = == 16.A cylindrical packed bed of height 1 m is filled with equal sized spherical particles. The particles are nonporous and have a density of 1500 kg/m3. The void fraction of the bed is 0.45. The bed is fluidized using air (density 1kg/m3). If the acceleration due to gravity is 9.8m/s2, the pressure drop (in pa) across the bed at incipient fluidization (up to one decimal place ) is _________. Answer:8079.61 Exp:For incipient fluidization ( )( )PPg 1L= Where0.45 = 3 3P1500kg m 1kg m = =And L = 1m (height of the bed) ( )( ) P 9.8 1 0.45 1500 1 18079.61Pa = = () i ( ) iii ( ) ii1T3T2TCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 12 17.Forapureliquid,therateofchangeofvapourpressurewithtemperatureis0.1bar/Kinthe temperaturerangeof300to350K.iftheboilingpointoftheliquidat2baris320K,the temperature (in K) at which it will boil at 1 bar (up to one decimal place) is ___________. Answer:310 Exp: Given dP0.1 dp 0.1dTdT= = ( ) P 0.1T C........... 1Given at T 320K. P 2bar2 0.1 320 CC 30 = += == + = Equation (1) becomes P 30 0.1T = +Putting P = 1 bar 1 30 0.1TT 310K= + = 18.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: (A)( )1x x (B)( ) x x (C)( )1 2x x (D)( )1 2x x Answer:(C) Exp:For laminar flow Boundary layer thickness ( )4.99x 4.99x8 xRe Vx= = Or,( )12x x 19.Matchthepolymermentionedontheleftwiththecatalystusedforitmanufacturegivenonthe right. (I)Low density Polyethylene(P)Ziegler-Natta catalyst (II)High density Polyethylene(Q)Traces of Oxygen (III)Polyethylene Terephthalate(R) Butyl Lithium (IV)Polyvinyl Chloride(S) Antimony (A)I Q, II R, III S, IV P (B)I S, II P, III Q, IV R (C)I Q, II P, III S, IV R (D)I S, II R, III P, IV Q CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 13 Answer:(C) Exp:LDPETraces of oxygen HDPEZeigler Natta catalyst PETAntimony PVCButyl Lithium 20.Threeidenticalclosedsystemsofapuregasaretakenfromaninitialtemperatureandpressure ( )L 1T , Pto a final state( )2 2T , P ,each by a different path. Which of the following in Always true forthethreesystems?( representsthechangebetweentheinitialandfinalstates:U,S,G,Q and W are internal energy, entropy, Gibbs free energy, heat added and work done, respectively.) (A)U, S, Q are same (B)W, U, G are same (C)S, W, Q are same (D)G, U, S are same Answer:(D) Exp:U,S and G are state variables (point function) so, U, S and G Remain same between two states 1 and 2. 21.Identify the WRONG statement amongst the following: (A)Steam distillation is used for mixtures that re 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 forthe same throughout. Answer: (C) Exp:Steam distillation is used for the mixture immiscible with water. Option (C) is the wrong statement 1( )1 1P . T( )2 2P . T2PTCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 14 22.ForabinarymixtureofcomponentsAandB,NAandNBdenotethetotalmolarfluxesof components A and B, respectively. JA and JB are the corresponding molar diffusive fluxes. Which of the following is true for equimolar counter-diffusion in the binary mixture? (A) A B A BN N 0 and J J 0 + = + (B) A B A BN N 0 and J J 0 + + =(C) A B A BN N 0 and J J 0 + + (D) A B A BN N 0 and J J 0 + = + =Answer:(A) 23.A complex-valued function, f(z), given below is analytic domain D: ( ) ( ) ( ) f z u x, y iv x, y z x fy = + = +Which of the following is NOT correct? (A) df v uidz y y = + (B) df u vidz x x = + (C) df v uidz y y = (D) df v vidz y x = + Answer:(A) Exp:( ) ( ) ( ) f z u x, y iv x, y = +( )df u v v u vf ' z i idz x x x y y = = + = = + - (i) For analytic function u u vandx y y x = = So, equation() ican be written as df v v u u vi i idz y x y y x x = + = = + Only option (A) can not be deduced 24.Which of the following can change if only the catalyst is changed for a reaction system? (A) Enthalpy of reaction(B) Activation energy (C) Free energy of the reaction(D) Equilibrium constant Answer:(B) Exp:Catalyst changes the activation energy of the reaction CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 15 25.Match the technologies in Group 1 with the entries in Group 2: Group 1Group 2 (P)Urea manufacture(I) Microencapsulation (Q) Coal gasification(II)Ultra-low sulphur diesel (R) Controlled release of chemicals(III) Shale oil (S) Deep hydrodesulphurization(IV) Prilling tower (V) Gas hydrates (VI)Gassolidnon-catalytic reaction (A) P I, Q V, R II, S VI (B) P IV, Q VI, R I, S II (C) P IV, Q I, R III, S II (D)P V, Q VI, R IV, S II Answer:(B) Exp:P: Urea manufacture IV prilling tower Q: coal gasification VI Gas solid non catalytic reaction R : controlled release of chemical I :microncapsulation. S: Deep hydrodesulphurization II ultra low sulphur diesel. Q.No-26-55 Carry Two Marks Each 26.Considerasolidblockofunitthicknessforwhichthethermalconductivitydecreaseswithan increaseintemperature. Theopposite facesoftheblockaremaintainedatconstantbutdifferent temperatures: T(x = 0) > T(x = 1). Heat transfer is by steady state conduction in x-direction only. Thereisnosourceorsinkofheatinsidetheblock.Inthefigurebelow,identifythecorrect temperature profile in the block. (A)I(B)II(C)III(D)IV ( ) T x 0 =IIIIIIIV( ) T x 1 =0x1TCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 16 Answer: (C) Exp:Let( )1T x 0 T =and( )2T x 1 T = =we have from Fouriers law q dTkA dx=At steady state conduction T1>T2 dT qk constdx A = = qdx kdTA = ____(1) Here K is decreasing with temperatureLet( )OK K 1 T = ( )1x To0 Tqdx k 1 T dTA= ( ) ( )2 2O 1 1qx k T T T TA 2= + + ____(11) Similarly integrating equation (1) betweenx 0 & x 1 = = ( )21x 1 TOx 0 Tqdx k 1 T dTA=== ( ) ( ) ( )2 2o 2 1 2 1q1 k T T T T (iii)A 2 = + Dividing equation (ii) and (iii) ( ) ( ) { } ( )2 2 2 21 o 1 2 1 0 2 1T T 2k T T x T T 2k T T = From the above expression for temperature profile it is clear that the temp passes through aminima CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 17 27.The schematic diagram of a steady state process is shown below. The fresh feed (F) to the reactor consists of 96 mol% reactant A and 4 mol% inert I. The stoichiometry of the reaction isA C . A 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 A in the reactor based on A entering the reactor at point 1 in the figure (up to one decimal place) is ________. Answer:59.19 Exp: Basis: - 100 mole of feed F A in feed = 96 mol Inert in feed = 4mol Recycle= 0.3F = 30 mole Product (P) Contains50%C At steady state F = P = 100 mole ProductContains50 mole C Product must also contain 4% inert At point (2) total molar flow rate =100 + 30 = 130mole Product stream P is 50 mole C46 mole A4 mole C Recycle stream R is 15 mole C13.8 mole A1.2 mole I Conversion of A in the reactor ( )( )A converted int o C at point 2A fed at point A65100 59.19%96 13.8== =+ F1A CP0.3F0.3F( ) 1A C ( ) 2PFCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 18 28.The impulse response to a tracer pulse experiment for a flow reactor is given below: In the above figure, c is the exit tracer concentration. The corresponding E orE(normalized E) curve is correctly represented by which of the following choices? Here, is dimensionless time. (A)(B) (C)(D) Answer:(C) 29.Considerasteadystatemasstransferprocessbetweenwell-mixedliquidandvapourphasesofa binarymixturecomprisingofcomponentsAandB.ThemolefractionsofcomponentAinthe bulk liquid (xA) and bulk vapour( )Ay phases are 0.36 and 0.16, respectively. The mass transfer coefficients for component A in liquid and vapour phases are 0.1 mol/(m2.s) and 0.05 mol/(m2.s), respectively.Thevapour-liquidequilibriumcanbeapproximatedas *A Ay 2x = for Ax lessthan 0.4. The mole fraction of A in the liquid at the interface (up to two decimal places) is Answer:0.08 Exp:Given Ax 0.36 = Ay 0.16 = 2L2gk 0.1mol m seck 0.05mol m s==

*A A Ay 2x for x 0.4 = L) is used to conduct the first order reactionA B . At 450 K, the Thiele modulus for this system is 0.5. The activation energy for the first order rate constant is 100kJ/mol. The effective diffusivity of the reactant in the slab can be assumed to be independent of temperature, and external mass transfer resistance can be neglected. If the temperature of the reaction is increased to 470 K, then the effectiveness factor at 470 K (up to two decimal place) will be _________. Value of universal gas constant = 8.314 J/mol.K Answer:1.875 49.The diameter of sand particles in a sample range from 50 to 150 microns. The number of particles of diameter x in the sample is proportional to 150 x +. The average diameter, in microns, (up to one decimal place) is ________________ 50.ConsidertwosteadyisothermalflowconfigurationshownschematicallyasCaseIandCaseII below.IncaseI,aCSTRofvolumeV1isfollowedbyaPFRofvolumeV2,whileinCaseIIa PFR of volume V2 is followed by a CSTR of volume V1. In each case, a volumetric flow rate Q of liquid reactant is flowing through the two units in series. An irreversible reaction A products (order n) takes place in both cases, with a reactant concentration CA0 being fed into the first unit. Q1VA0CQ2VQAfCIQ2VQ1VQA0CfACIICH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 31 Choose the correct option: (A) AffIIIAC1 for n 1C> =(B) AffIIIAC1 for n 1C= =(C) AffIIIAC1 for n 1C< = (D) AffIIIAC1 for n 1C= > Answer:(B) Exp:We can solve problem by taking 1st order reaction. (n = 1) For Case I, Consider 3 11 2V V 1m , k 1.sec= = =

3 3AOQ 1m sec, and C 1mol m = =For CSTR, A0 A1A1C CkC = A1A11 C1 1C = 3A1C 0.5 mol m =For PFR. AfA1CACAdCkC= 3 AfAfA1C1 ln C 0.184mol mC= =Case II Q2VQAfCA1C1VA0CQ1VQ2VQA1CA0CQAfCCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 32 With same data, For PFR A1A0Cln 1C = = 3A1C 0.3680 mol m =and for CSTR, A1 Af AfAf AfC C 0.3680 CK 1C C = = = 3AfC 0.1840 mol m . =Clearly for both the case AfCis equal AfIIIAfCi, e 1 for n 1C= = 51.Select the wrong statement regarding water gas shift converters from the list below. (A)Inter-stage cooling is provided between the two stages of shift converters. (B)Usually high temperature shift (HTS) reactor has a iron-based catalyst and low temperature shift (LTS) reactor has a copper-based catalyst. (C)HTS reactor is followed by LTS reactor. (D)LTS reactor is followed by HTS reactor. Answer:(D) Exp:The correct sequence is Wrong statement is D 52.A vector u = -2yi 2xj, +where i and j are unit vector in x and y directions, respectively. Evaluate the line integral cI u.dr = Where C is a closed loop formed by connecting points (1,1) , (3,1), (3,2) and (1,2) in the order. The value of I is ___________. High TempReactorInterCoolerLow tempreactorCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 33 Answer:8 Exp:Given u 2y i 2xj = + dr dxi dy j = +

So,( )c c cu.dx 2y.dx 2x.dy 1 = + Now, forA B y 1, dy 0 x changes from 1 to 3 = =( )31c1 u.dr 2 dx 4 = = ForB C, x 3, dx 0, ychanges from1 to 2 = =( )21c1 u.dr 2 3dy 6 = = For C D, y 2, dy 0, x changes from 3 to 1 = =( )13C1 u.dr 2 2dx 8 = = For D A, x 1, dx 0, ychanges from 2 to 1 = =( )C 21 u.dr 2x1dy 2 = = For whole loop Cu.dr 4 6 8 2 8 = + + = 53.A binary mixture of components (1) and (2) forms an azeotrope at 130OC and x1 = 0.3. The liquid phasenon-idealityisdescribedbyln21 2Ax = andln22 1Ax , = where 1 2, aretheactivity coefficients,and 1 2x , x aretheliquidphasemolefractions.Forbothcomponents,thefugacity coefficientsare0.9attheazeotropiccomposition.Saturatedvaporpressuresat130OCare sat1P 70 =bar sat2P 30 =bar. The total pressure in bars for the above azeotopic system (up to two decimal places) is Answer:27.54 Exp:Given ( )az az az az1 1 1x 0.3 y At a zeotrope x y = = = sat sat1 2P 70bar, P 30bar = = 21 2ln Ax =and 2n 2 1l Ax = CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 34 So ( ) ( )2 2 12 1 2 1 2 12ln A x x A(x x ) x x= = + ( )az 112ln A 1 2x= _____(1) Also az az sat1 1 1 1 1y P x P = ______(2) az az sat2 2 2 2 2y P x P = ______(3) We know az az az az1 1 2 2x y and y x = =So, (2)/(3) sat1 1 1sat2 2 2P1s = = sat1 2sat2 1P 30 3P 70 7 = = = From (1) ( )3ln A 1 2 0.37= A 2.1182 =So( )221 2ln Ax 2.1182 0.7 = = 10.3542 =and( )222 1ln Ax 2.1182 0.3 = = 20.8264 =Adding equation (2) and (3)( ) ( )1 2say = sat sat2 1 1 1 2 2p p p p2 p 0.3542 70 0.8264 30P 27.54bar + = + = + = 54.Air is flowing at a velocity of 3m/s perpendicular to a long pipe as shown in the figure below. The outerdiameterofthepipeisd=6cmandtemperatureattheoutsidesurfaceofthepipeis maintained at 100o C. The temperature of the air far from the tube is 30OC. Data for air Kinematic Viscosity, 6 2V 18 10 m s;= Thermal conductivity, k = 0.3 W/(m.K) Using the Nusselt number correlation : 0.8hdNu 0.024 Rek= = , the rate of heat loss per unit CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 35 Length( ) W mfrom the pipe to air (up to one decimal place) is ____________________ Answer:250.945 Exp:Givenv 3m s = 26 2d 6 10 m18 10 m seck 0.03w mk= = = Given 0.8ehdNu 0.024Rk= = 26Vd Vd 3 6 10Re 10, 00018 10 = = = = ( )0.822hdSo, 0.024 10, 000 38.037k0.03 38.037h 19.0187 w m k6 10= = = = Rate of heat loss/length ( )( )( )S a2h. d T T19.0187 6 10 100 30250.945w m= = = OSurface temperature 100 C6cmOAir velocity 3m s, Temperature 30 CCH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 36 55.Thecostoftwoindependentprocessvariablef1andf2affectsthetotalcostCT(inlakhsof rupees) of the process as per the following function: 2T 1 21 21000C 100f 20f 50f f= + + +The lowest total cost CT, in lakhs of rupees (up to one decimal place), is ________. Answer:572.8 Exp:Given 2T 1 21 21000C 100f 20f 50f f= + + T21 2 1C 1000P 100f f f= = = T2 22 2 1C 1000and q 40ff f f = = + For maxima of minima P = 0, q= 0 2 2 22 1 1 21000 1000100 and 40ff f f f= = 21 2f f 10............. = (1)

31 2f f 25............(II) = Dividing equation (ii) byequation (1) ( ) () ii i 31 221 22 22 211f f2.5f ff f2.5 ff 2.5 = = = From (1) ( )( )422 252 2211 2ff 102.5f 62.5 f 2.28652.2865so, f 2.0912.5f 2.091, f 2.2865 = = == == = CH-GATE-2015 PAPER|www.gateforum.com Indias No.1 institute for GATE Training 1 Lakh+ Students trained till date 65+ Centers across India 37 Now,2T2 3 31 2 1 1 2C 1000 2 2000rf f f f f + = = = 2T2 21 2 1 22T2 32 2 1C 1000Sf f f fC 2000t 40f f f= = + = = + 2rt s 0 and r 0 > >So,( )1 2f , fis point of minima Lowest cost 2T1000C 100 2.091 20 2.865 502.091 2.2865= + + + TC 572.82lakhs =