gate 2012 chemical (ch) solution

7/29/2019 GATE 2012 Chemical (CH) Solution

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GATE 2012 Solutions

Detailed Solutions of the questions:

1. B because 2nd& 3rd equation have different RHS for same LHS which is not possiblefor any values of x1 and x2. Thus no soln as two equations are incompabitble.

2. Here normal auxillary approach solution is not in the option. So go for trial & error byfinding 2

ndorder derivative for the options. Hence D

3. Since t


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23. Steam reforming process has reactor (endo reaction) followed by shift reaction (high

T and low T), PSA for CO removal

24.

25.

26.27. Substitue for ax = t Then solve by gamma function to a2(1!/a2)

28. Apply Newtons formula to get x1 using f(x0) and f(x0)

29.

30. Using adiabatic process equation i.e. PV

= const and its variants get the relation for

T2/T1 & P2/P1

31 Here G = H-TS. Substitute the given expressions & then differentiate

G = 5 1 2 ( )( 1ln 1 2ln 2)x x RT x x x x wrt x1 with x2= 1-x1 Equate 1st order derivative to

zero to get x1 = 0.5

32.33. Shape factor F22 is to be found. For surface 1 F11 =0, F12 =1. So F21 get by

reciprocal rule Hence F22 =1-F21 = 1- (R1/R2)2

34. Get temperature on insulation surface by hA (Ts- 30) =100. Then use steady state HT

equation in case of cylindrical pipe for transfer from pipe to insulation So

T-Ts =

2ln(1)

2

RqR

k LGet T as ans.

35. Get T exit for cold fluid which is 60 C Then LMTD = 30 as difference same on bothsides avg would do. Equate heat duty with UATLMTD to get U

36. B because in the rate equation conc. Of vacant sites would come twice for A & B

hence the substitution of it would lead to B

37. Convert the conc to dry basis so 0.33 is the intial & 0.02 as final Then apply the

const. Drying rate & falling rate formula. Now t got is for water removal per kg of solid

whereas for solid loading of 50 kg/ m2 multiply by 50 to get ans. As C

38. Use relative volatility eqn1 ( 1)

xy

x

Subst x =0.4 to get y & then apply the mass

balance eqns to get V

39.

40.

41. For non isothermal, non adiabatic exoth rev rxn in PFR Graph C is preferred as for

rev exo with decrease in T eqlm conversion decreases & hence to min reactor volume

heat transfer to increase than evolved heat.

42.

43. Here Total cost =5

10340* 200

B B

B

P PP

Derivative wrt PB and equated to 0 gives

ans. C

44. For break even total investment = Energy cost that is saved in 3 yrsSo Investment + Interest (3 yr) = 20 * x*3

(1+0.15 *3)*(2*10^6) = 60 *x which gives x as 48300/- as ans. Option 43800 is near or itmay be error in the option

45. Here Nu = 0.2 * Re^0.6 Pr^0.33 correlation is helpful. Since Nu= hD/k where

pitch = D can be used for square arrangement.

46. Catalyst P & S are eay to identify then methanol conversion is oxidation so ironoxide as catalyst


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47. Polyester is fibre, polyethylene is thermoplastic, isoporene used in rubber i.e.

elastomer

48. Mass balance gives F*Y1* 0.99 = Smin(X1*) = Smin(Y1/2) gives Smin = 1485 Hence A

49. For Kresmer eqn i.e. equilibrium line is straight Substitutethe conc. Values to get n =5. Also conversion to mole compositions is required for conc.

Molar flow of feed.

50. Since Fb/Fa = 5 which is 2+0.3 FRG = 5 So FRG = 10

Using FRG & mole balance get the parent stream molar flow which splits into recycle &

purge . Also hint is C is pure & Recycle stream hence contains B & D only as A is

completely converted So parent stream flow = 12 mol/s Hence RG/PG = 10/2 =5

51. Now RG/PG = 4 Let PG =x so RG = 4x, total parent stream = 5x. Hence mole

balance across the reactor gives 3+4x = 5x+1 i.e. x = 2. Then use Fb/Fa = 5 where

2+y*8 = 5 gives y as 3/852.53.

54. Use the design equation of PFR in terms of conc to get k =0.2855. Using k =0.28 solve for the series of CSTR using the design rquation Remember intial

conc. Changes but space time remains same for all the CSTRs. Exit conc is 0.53 so

conversion is 73.7

gate 2012 chemical (ch) solution

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