GATE 2019
CHEMICAL
ENGINEERING Answer Keys
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Chemical Engineering Question Paper whish is solved by
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GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
1) Two cars start at the same time from the same location and go in the same direction.
The speed of the first car is 50 km/h and the speed of the second car is 60 km/h. the
number of hours it takes for the distance between the two cars to be 20 km is _______.
a) 1
b) 2
c) 3
d) 6
Ans. B
2) The search engine’s business model _________ around the fulcrum of trust.
a) Revolves
b) Plays
c) Sinks
d) Bursts
Ans. A
3) The expenditure on the project ___________ as follows: equipment Rs. 20 lakhs,
salaries Rs 12 lakhs, and contingency Rs. 3 lakhs.
a) Break down
b) Break
c) Breaks down
d) Breaks
Ans. C
4) A court is to a judge as _______ is to a teacher.
a) A student
b) a punishment
c) a syllabus
d) a school
Ans. D
5) ten friends planned to share equally the cost of buying a gift for their teacher. Where
two of them decided not to contribute, each of the other friends had to pay Rs 150 more.
The cost of the gift was Rs. _________.
a) 666 b) 3000 c) 6000 d) 12000
Ans. C
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
6) In a collage, there are three student clubs. Sixty students are only in the Drama club,
80 students are only in the dance club, 30 students are only in the Maths club, 40 students
are in both Drama and Dance clubs, 12 students are in both Dance and Maths clubs, 7
students are in both Drama and Maths clubs, and 2 students are in all the clubs. If 75% of
the students in the collage are not in any of these clubs, then the total number of students
in the college is _________.
a) 1000
b) 975
c) 900
d) 225
Ans. C
7) In the given diagram, teachers are represented in the triangle, researches in the
circle and administrators in the rectangle. Out of the total number of the people, the
percentage of administrators shall be in the range of ________.
a) 0 to 15
b) 16 to 30
c) 31 to 45
d) 46 to 60
Ans. C
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
8) The police arrested four criminals – P, Q, R and S. the criminals knew each other.
They made the following statements:
P says “Q committed the crime”
Q says “ committed the crime”
R says “I did not do it”
S says “What Q said about me is false”
Assume only one of the arrested four committed the crime and only one of the statements
made above is true. Who committed the crime?
a) P
b) R
c) S
d) Q
Ans. B
9) Three of the five students allocated to a hostel put in special requests to the warden.
Given the floor plan of the vacant rooms, select the allocation plan that will accommodate
all their requests.
Request by X : due to pollen allergy, I want to avoid a wing next to the garden.
Request by Y: I want to live as far from the washrooms as possible , since I am very
sensitive to smell.
Request by Z: I believe in Vaastu and so want to stay in the South- west wing.
The shaded rooms are already occupied. WR is washroom.
a)
b)
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
c)
d)
Ans. D
10) ‘ A recent High Court judgment has sough to dispel the idea of begging as a disease
which leads to its stigmatization and criminalization-and to regard it as a symptom. The
underlying disease is the failure of the state to protect citizen who fall through the social
security net’
a) Beggars are lazy people who beg because they are unwilling to work
b) Beggars are created because of the lack of social welfare schemes
c) Begging is an offence that that has to be deal with firmly
d) Begging has to be banned because if adversely affects the welfare of the state
Ans. B
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
TECHNICAL SECTION:
1) For a binary nonideal A-B mixture exhibiting a minimum boiling azeotrope, the
activity coefficients, yI (i= A,B), must satisfy
a) YA > 1, YE > 1
b) YA < 1, YE > 1
c) YA = 1, YE = 1
d) YA < 1, YE < 1
Ans. A
2) Producer gas is obtained by
a) Passing air through red hot coke
b) Thermal cracking of naphtha
c) Passing steam through red hot coke
d) Passing air and steam through red hot coke
Ans. D
3) The most common catalyst used for oxidation of o- xylene to phthalic anhydride is
a) V2O2
b) Pd
c) Pt
d) Ag
Ans. A
4) The correct expression for the colburn j-factor for mass transfer that relates
Sherwood number (Sh). Reynolds number (Re) and Schmidt number (Sc) is
a) 1/3
Sh
(Re)(Se)
b) 1/2
Sh
(Re) (Se)
c) 1/2 1/3
Sh
(Re) (Se)
d) Sh
(Re)(Se)
Ans. A
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
5) For a first order reaction in a porous spherical catalyst pellet, diffusional effects are
most likely to lower the observed rate of reaction for
a) Slow reaction in a pellet of smell diameter
b) Slow reaction in a peller of large diameter
c) Fast reaction in a pellet of small diameter
d) Fast reaction in a pellet of large diameter
Ans. D
6) In kraft process, the essential chemical reagents used in the digester are
a) Caustic soda, mercaptans and ethylene oxide
b) Caustic soda, sodium sulphide and soda ash
c) Quick lime, salt cake and dimethyl sulphide
d) Baking soda, sodium sulphode and mercaptans
Ans. B
7) A system of n homogeneous linear equations containing n unknowns will have non-
trivial solutions if and only if the coefficient matrix is
a) 1
b) -1
c) 0
d)
Ans. C
8) Consider a sealed rigid bottle containing CO2 and H2O at 10 bar ambient
temperature. Assume that the gas phase in the bottle is pure CO2 and follows the ideal gas
law. The liquid phase in the bottle contains CO2 dissolved in H2O and is an ideal solution.
The Henry’s constant at the system pressure and temperature is HCO2 = 1000 bar. The
equilibrium mole fraction of CO2 dissolved in H2O is _________ (rounded off to three decimal
places).
Ans. 0.010
9) A thermocouple senses temperature based on the
a) Nerst Effect
b) Maxwell Effect
c) Seebeck Effect
d) Peltier Effect
Ans. C
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
10) In the drying of non- dissolving solids at constant drying conditions, the internal
movement of moisture in the solid has a dominant effect on the drying rate during
a) The initial adjustment period only
b) The constant rate period only
c) The falling rate period only
d) Both the initial adjustment and constant rate periods
Ans. C
11) Consider the two countercurrent heat exchanger designs for heating a cold stream
from tin to tout . as shown in figure. The hot process stream is available at Tin . the inlet
stream conditions and overall heat transfer coefficients are identical in both the designs.
The heat transfer area in Design I and Design II are respectively I
HXA and II
HXA
If heat losses are neglected, and if both the designs are feasible, which of the following
statements holds true:
a) I I I II
HX HX out outA A T T
b) I I I II
HX HX out outA A T T
c) I I I II
HX HX out outA A T T
d) I I I II
HX HX out outA A T T
Ans. B
12) In petroleum refining operations, the process used for converting paraffins and
naphthes to aromatics is
a) Alkylation
b) Catalytic reforming
c) Hydrocracking
d) Isomerization
Ans. B
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
13) For a fully- developed turbulent hydrodunamic boundary layer for flow past a flat
plate, the thickness of the boundary layer increases with distance with distance x from the
leading edge of the plate, along the free- stream flow direction, as
a) x0.5
b) x1.5
c) x0.4
d) x0.8
Ans. D
14) the desired liquid phase reaction
1k2 0.3
F 1 D ED E F r k C C
Is accompanied by an undesired side reaction 2k
0.4 1.5
F 2 D ED E F r k C C
Four isothermal reactor schemes (CSTR: ideal Continuous- Stirred Tank Reactor; PFR :
ideal Plug Flow Reactor) for processing equal molar feed rates of D and E are shown in
figure Each scheme is designed for the same conversion. The scheme that gives the most
favorable product distribution is:
Ans. C
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
15) Consider a cylinder (diameter D and length D), a sphere (diameter D) and a cube
(side length D). Which of the following statements concerning the sphericity (φ) of the
above objects is true.
a) sphere cylinder cube
b) sphere cylinder cube
c) sphere cylinder cube
d) sphere cylinder cube
Ans. A
16) Prandtl number signifies the ratio of
a) MomentrmDiffusivity
ThermalDiffusivity
b) MassDiffusivity
ThermalDiffusivity
c) ThermalDiffusivity
MomentrmDiffusivity
d) ThermalDiffusivity
Mass Diffusivity
Ans. A
17) Three distillation schemes for separating an equimolar, constant relative volatility
ABC mixture into nearly pure components are shown. The usual simplifying assumptions
such as constant molal overflow, negligible heat loss, ideal trays are valids. All the schemes
are designed for minimum total reboiler duty.
Given that the relative volatilities are in the ratio A 5 C; ; 8: 2 :1, the correct option that
arranges the optimally- designed schemes in ascending order of total reboiler duty is
Scheme 1
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Scheme 2
A
Scheme 3
A
,
a) I, II, III
b) III, I, II
c) II, I, III
d) III, II, I
Ans. C
18) The value of the expression x
2
tan xlim is
x
a)
b) 0
c) 1
d) -1
Ans. A
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
19) Pool boiling equipment operating above ambient temperature is usually designed to
operate
a) Far above the critical heat flux
b) Near the critical heat flux
c) Far above the Leidenfrost point
d) Near the Leidenfrost point
Ans.
20) The product of the eigen values of the matrix 2 3
0 7
is ______(rounded off to one
decimal places)
Ans. 14.0
21) The liquid flow rate through an equal percentage control valve, when fully open, is 150
gal/min and the corresponding pressure drop is 50 psi. if the specific gravity of the liquid is
0.8, then the valve coefficient, vC in gal/(min psi 0.5 ___(rounded off to two decimal places)
Ans. 18.97
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
22) Consider a rigid. Perfectly insulated, container partitioned into two unequal parts by a
thin membrane ( see figure). One part contains one mole of an ideal gas at pressure Pi and
temperature Ti while the other part is evacuated. The membrane ruptures, the gas fils the
entire volume and the equilibrium pressure is / 4f iP P . If pC (molar specific heat capacity
at constant pressure). vC (molar specific heat capacity at constant volume) and R (universal
gas constant) have the same units as molar entropy, the change in molar entropy ( )f iS S
is
a) Cpln2 + Rln4
b) –Cpln2 + Rln4
c) Rln4
d) Cpln2
Ans. C
23) For a hydraulic lift with dimensions shown in figure, assuming g = 10 m/s2, the
maximum diameter Dleft ( in m) that lifts a vehicle of mass 1000 kg using a force of 100 N is
_____(rounded off to two decimal places).
Ans. 0.20
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
24) For a single component system, vapor (subscript g) and liquid (subscript f) coexist in
mechanical, thermal and phase equilibrium when
a) g fu u ( equality of specific internal energy)
b) g fh h ( equality of specific enthalpy)
c) g fs s ( equality of specific entropy)
d) g fg g ( equality of specific Gibbs free energy)
Ans. D
25) The combination that correctly matches the polymer in Group-I with the
polymerization reaction type in Group-II
Group –I Group-II
P) Nylon 6 I) Condensation polymerization
Q) Polypropylene II) Ring opening polymerization
R) Polyester III) Addition polymerization
a) P-II, Q-I, R-III
b) P-I, Q-III, R-II
c) P-III, Q-II, R-I
d) P-II, Q, III, R-I
Ans. D
26) A countercurrent absorption tower is designed to remove 95% of component a from an
incoming binary gas mixture using pure solvent B. the mole ratio of A in the inlet gas is
0.02. the carrier gas flow rate is 50 kmol/h. the equilibrium relations is given by Y=2Xm,
where Y and X are the mole ratios of A in the gas and liquid phases, respectively. if the
tower is operated at twice the minimum solvent flow rate, the mole ratio of A in the exit
liquid stream is_______
Ans. 0.005
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
27) The value of the complex number 12i (where i= 1 ) is
a) 1
(1 )2
i
b) 1
2i
c) 1
2i
d) 1
(1 )2
i
Ans. A
28) If x, y and z are directions in a Cartesian coordinate system and I, j and k are the
respective unit vectors, the directional derivative of the function 2( , , ) 3u x y z x yz at the
point (2, 0, - 4 ) in the direction ( 2 ) / 6i j k is______
Ans. 6.53
29) Stream A with heat capacity 2000 / ( )paC J kg K is cooled from 90 oC to 45 oC in a
concentric double pipe counter current heat exchanger having a heat transfer area of 8 m2
the cold stream B of specific heat capacity 1000 / ( )paC J kg K enters the exchanger at a
flow rate 1 kg/s and 45 oC. The overall heat transfer coefficient U=250 W/(m2 K). Assume
that the mean driving force is based on the arithmetic mean temperature difference, that is
, , , ,[ ]
2 2
A in A out B in B out
AMTD
T T T TT
where T1,in and T1,out refer to the temperature of
the ith stream (I = A, B) at the inlet and exit, respectively. the mass flow rate of stream A ( in
kg/s, is_____
Ans. 0.31
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
30) The elementary irreversible gas-phase reaction AB + C is carried out adiabatically
ion an ideal CSTR (Continuous Stirred Tank Reactor) operating at 10 atm. Pure A enters the
CSTR at a flow rate of 10 mol/s and a temperature of 450 K. assume A, B and C to be ideal
gases. The specific heat capacity at constant pressure ( )piC and heat of formation
0
iH , of component I ( I = (A, B C) are:
CPA = 30 J/(mol K) CPB = 10 J/ (mol K) CPC = 20 J/ (mol K)
0
AH = -90 kJ / mol 0
BH = - 54 kJ/ mol 0
CH = -45 kJ/mol/
The reaction rate constant k (per second) = 0.133 exp 1 1
{ ( )}450
E
R T , E = 31.4 Kj/MOL AND
UNIVERSAL GAS CONSTANT r = 0.082 L atm (mol K) = 8.314 (J/ (mol K). The shafts work
may be neglected in the analysis, and specific heat capacities do not vary with temperature.
All heats of formations are referenced to 273 K. The reactor volume ( in Liters) for 75%
conversion is _____
Ans. 89
31) A solid sphere of radius 1 cm and initial temperature of 25 oC is exposed to a gas
stream at 100 oC . For the solid sphere, the density is 104 kg/m3 and the specific heat
capacity is 500 J/(kg K.). The density of the gas is 0.6 kg/m3 and its specific heat capacity is
103 J/(kg K). the solid sphere is approximated as a lumped system (Biot number << 1) and
all specific heats are constant. If the heat transfer coefficient between the solid and gas is
50 W/(m2 K), the time (in seconds) needed for the sphere to reach 95 oC is ______
Ans. 903
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
32) A first-order irreversible liquid phase reaction AB (k= 0.1 min-1) is carried our under
isothermal, steady state conditions in the following reactor arrangement comprising an
ideal CSTR (Continuous-Stirred Tank Rector) and two ideal PFRs ( Plug Flow Reactors) fro
the information in the figure the volume of the CSTR (in Liters ) is______
Ans. 856
33) Consider a vessel containing steam at 180 oC. The initial steam quality is 0.5 and the
initial volume of the vessel is 1 m3. The vessel loses heat at a constant rate q under isobaric
conditions so that the quality of steam reduces to 0.1 after 10 hours. The thermodynamic
properties of water at 180 oC are (subscript g: vapor phase; subscript f: liquid phase).
Specific volume: 0.19405gv m3/kg, 0.001127fv m3/kg,
Specific internal energy: 2583.7 / ,gu kJ kg 762.08. /fu kJ kg
Specific enthalpy: 277.2. /gh kJ kg 277.2. /gh kJ kg
The rate of heat loss q ( in kJ/hour) is________
Ans. 747
34) A disk turbine is used to stir a liquid in a baffled tank. To design the agitator,
experiments are performed in a lab-scale model with a turbine diameter of 0.05 m and a
turbine impeller speed of 600 rpm. The liquid viscosity is 0.001 Pa s while the liquid
density 1000 kg/m3 the actual application has a turbine diameter of 0.5 m, and impeller
speed of 600 rpm, a negligible. If the power required in the lab-scale model is Pi and the
estimated power for the actual application is P2 then the ratoo P2/P1 is_____
a) 103 b) 104 c) 105 d) 106
Ans. C
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
35) Choose the option that correctly matches the step response curves on the left with the
appropriate transfer function on the right. The step input change occurs at time t = 0
a) P-III, Q-IV, R-II S-I
b) P-III, Q-I, R-IV, S-II
c) P-IV, Q-III, R- II, S-I
d) P-III, Q-II, R-IV, S-I
Ans. B
36) Carbon monoxide (CO) reacts with hydrogen sulphide (H2S) at a constant temperature
of 800 K and a constant pressure of 2 bar as:
CO + H2S 2COS H
The Gibbs free energy of the reaction 22972.3og rxn J/mol and universal gas constant R=
8.314 J/(mol K). both the reactants and products can be assumed to be ideal gases. If
initially only 4 mol of H2S and 1 mol of CO are present, the extent of the reaction ( in mol) at
equilibrium is________
Ans. 0.29
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
37) A centrifugal pump is used to pump water (density 1000 kg/m3)from an inlet pressure of 105 Pa to an exit pressure of 2 x 105 Pa. the exist is at an elevation of 10 m above the pump . the average velocity of the fluid is 10 m/s. the cross-sectional area of the pipes at the pump inlet and outlet is 10-3 m2 and acceleration due to gravity is g = 10 m/s2 Neglecting losses in the system , the power (in watts) delivered by the pump is ____________
Ans. 2000
38) A taxi – car is bought for Rs 10 lakhs . its salvage value is zero. The expected yearly
income after paying all expenses and applicable taxes is Rs 3 lakhs. The compound
interest rate is 9 % per annum the discount payback period (in years), is ______________
(rounded off to the next higher integer).
Ans. 5
39) For the closed loop system shown in figure, the phase margin(in degress) is ________
Ans. 45.4
40) 100 kg of a feed containing 50 wt% of a solute C is contracted with 80 kg of a solvent
containing 0.5 wt% of C in a mixer-settler unit. From this operations the resultant extract
and raffinate phase contains 40wt% and 20wt% of C, respectively .if E and R denote the
mass of the extract and raffinate phases respectively the ratio E/R is
a) 1/4
b) 1/2
c) 2/3
d) 1
Ans. C
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
41) Two spherical camphor particles of radii 20cm and 5cm, far away from each other, are
undergoing sublimation in a stream of air. The mass transfer coefficient is proportional to
1/ , where r(t) is the radius of the sphere at time t. Assume that the partial pressure of
camphor far away from the surface of the particle is zero. Also, assume quasi-steady state,
identical ambient conditions, and negligible heat effects. If t1 and t2 are the times required
for complete sublimation of the 20cm and 5cm camphor particles, respectively, the ratio
t1/t2 is______(rounded off to one decimal place).
Ans. 8.0
42) A fractionators recovers 95 mol n-propane as the distillate from an equimolar
mixture of n-propane and n-butane. The condensate is a saturated liquid at 55 C. The
Antoine equation is of the form, In (Psat [in bar])= A-
[ ] , and the constants are
provided below:
A B C n-propane 9.1058 1872.46 -25.16 n-butane 9.0580 2154.90 -34.42 Assuming Raoult’s law, the condenser pressure (in bar) is _______(rounded off to one
decimal place).
Ans. 17.9
43) Consider the reactor-separator-recycle process operating under steady state
conditions as shown in fig. The reactor is an ideal Continuous-Stirred Tank Reactor (CSTR),
where the reaction A+B C occurs. Assume that there is no impurity in the product and
recycle streams. Other relevant information are provided in the fig. The mole fraction of
B(xB)in the reactor that minimizes the recycle rate is__________(rounded off to two decimal
places).
Ans.
44) For a given binary system at constant at constant temperature and pressure, the molar
volume (in m3/mol) is given by 30 20 (15 7 )A B A B A Bv x x x x x x where xA and xB are the
mole fractions of components A and B respectively. the volume change of mixing maxv in
(in m3/mol) at x = 0.5 is__________
Ans. 1
45) A 20 cm diameter cylindrical solid peller of a nuclear fuel with density 6000 kg/m3 and
conductivity of 300 W/(m K) generates heat by nuclear fission at a spatially uniform rate of
104 W/kg. The heat from the fuel pellet is transferred to the surrounding coolant by
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
convection such that the pellet wall temperature remains constant at 300 0C . Neglecting
the axial and azimuthal dependence, the maximum temperature ( in 0C ) in the pellet at
steady state is___________
Ans. 800
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
46) A binary mixture with A and B is to be separated in a distillation column to obtain 95
mol% A as the top product. The binary mixture has a constant relative volatility 2AB .
The column feed is a saturated liquid containing 50 mol% A . under that usual simplifying
assumptions such as constant molar overflow, negligible heat loss ideal trays the minimum
reflux ratio for this separating is__________
Ans. 1.7
47) An incompressible Newtonian fluid flow in pipe diameter Di at volumetric flow rate Q.
Fluid with same properties flows in another pipe of diameter D2 = Di /2 at the same flow
rate Q. The transition length required for achieving fully- developed flow is iI for the tube
of diameter Di while it is l2 for the tube of diameter D2 Assuming steady laminar flow in
both case the ratio 1 2/l l is
a) 1/4
b) 1
c) 2
d) 4
Ans. B
48) Two unbiased dice are thrown. Each dice can show any number between 1 and 6. The
probability that the sum of the outcomes
Ans. 0.25
49) The Newton-Raphson method is used to determine the root of the equation f(x)=e-x – x.
If the initial guess for the root is 0, the estimate of the root after two iterations is
_______(rounded off to three decimal places).
Ans. 0.566
50) The solution of the ordinary differential equation
+ 3y=1, subject to the initial
condition y=1 at x=0, is
a)
(1+2e-x/3)
b)
(5-2e-x/3)
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
c)
(1+2e-3x)
d)
(1+2e-3x)
Ans. D
51) The combination that correctly matches the process in Group-I with the entries in
Group -2
Group -1 Group -2 P) Wulff process I) Sulfur mining Q) Sulfite Process II) Soda ash production R) Solvay process III) Acetylene production S) Frasch process IV) Pulp production
a) P-II,Q-IV,R-III,S-I
b) P-III,Q-IV,R-II,S-I
c) P-IV,Q-I,R-II,S-III
d) P-II,Q-I,R-IV,S-III
Ans. B
52) The elementary liquid –phase irreversible reactions 1 1
1 20.4min 0.1minK K
A B C
Take place in an isothermal ideal CSTR (continues- stirred tank Reactor). Pure A is
fed to the reactor at a concentration of 2 mol/Liter . for the residence time that
maximizes the exit concentration of B ,the percentage yield of B, defined as
100net formation rateof B
consumption rateof A
is __________________.
Ans. 67
53) Consider two competing equipment A and B for a compound interest rate of 10% per
annum in order for equipment B to be the economically cheaper option, its minimum
life (in years) is ___________
Equipment Capital Cost (Rs)
Yearly operating Cost(Rs)
Equipment life (years)
A 80,000 20,000 4 B 1,60,000 15,000 ?
Ans. 8
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
54) The elementary irreversible, liquid-phase parallel reactions 2A D and 2A U , take
place in an isothermal non-ideal reactor. The C-curve measured in a tracer experiment
is shown in the figure, where C(t) is the concentration of the tracer in g/m3 at the
reactor exit at time t(in min).
The rate constants are K1=0.2 Liter/(mol min) and K2=0.3 Liter/(mol min).pure A is fed to
the reactor at a concentration of 2 mol/liter. Using the segregated model, the percentage
conversation in the reactor is ________________-
Ans. 65
55) consider two non-interacting tank in series as shown in fig. water enters TANK 1at q
cm3/s and drains down to TANK 2 by gravity at a rate 3
1 ( / )k h cm s similarly water drains
from TANK 2 by gravity at a ratio of 3
2 ( / ).k h cm s where h1 and h2 represent levels of
TANK 1 and TANK 2, respectively (see figure). Drain value constant k =4 cm2.5/s and cross
sectional areas of the two tanks are A1=A2=28 cm2
At steady state operation, the water inlet flow rate is qxx=16 cm3/s. the transfer function
relating the deviation variable 2h
(cm) to flow rate 3( / )q cm s
is
a) 2
2
(56 1)s
b) 2
2
(62 1)s
c) 2
2
(36 1)s
d) 2
2
(49 1)s
Ans. A
GATE 2019 CHEMICAL ENGINEERING GATE 2019
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GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Detailed Solutions
Technical Section
Solution: 2)
Producer Gas is made when partial combustion of carbonaceous Substance usually coal or cake takes
place in an atmosphere of air and steam.
According to given options.
When air & steam both passes through red hot coke.
D) is the correct answer
Solution: 3)
Catalyst used in the production of Phthalic Anhydride from 0 – xylene is V2O5
A)
is the correct answer
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Solution: 4)
The correct expression used in the J-Factor analogy that relates the Sherwood number, Reynold’s
Number and Schmidt Number is______
1
3M
f SherwoodJ
2 Re ynolds.Schmidt
13
M
f SherwoodJ
2 Re ynolds.Pr andtl
A)
is the correct answer
Solution: 5)
Pore diffusion will control the process only and only when pore diffusion resistence will be high.
So for large pore diffusion resistance, Thiele modulus, 4
Thus, √
Hence, fast surface run & large pore size will be required.
D) is the correct answer
Solution: 6)
In the Kraft process, for the making of pulp, the chemical reagents used is white liquor, water and
woodchips.
50 Parts of water Caustic soda Sodium sulphide
25 Parts of white liquor (NaoH + Na2S)
25 Parts of woodchips (obtained from chipper bin)
White liquor essentially is NaOH but Na2S is added so that later on bleaching becomes easy.
B) is the correct answer
Solution: 8)
Answer:- According to Henry’s law
Pa = H.xA
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
For the pure Co2, particle pressure will be same as the pressure of Co2 in gas phase, which is 10
bar.
10=1000. XA
XA =
= 0.01
Solution: 9)
A Thermocouple senses temperature based on
a) Nernest Effect: It is observed when a sample allowing electrical conduction is subjected to a
magnetic field.
Hence it cannot be the answer
b) Peltier Effect: It is observed when electric current passes through or across a junction
between two dissimilar conductors.
Hence it cannot be the answer.
c) Maxwell Effect: It is a characteristic property of liquids related to birefringence
Hence It cannot be the answer.
d) Seeback Effect: Also known as thermoelectric effect that states an emf is generated
between two ends when they are at different temperatures.
C) is the Right Answer.
Solution: 11)
out in in out out in in out
1 X 1 Xout in out in
in out in out
T ' t T t T" t T tQ UA ' UA"
T ' t T" tln ln
T t T t
inT and outt are same
out out in in out out in in
1 X 1 Xout in out in
in out in out
T ' t T t T" t T tQ UA ' UA"
T ' t T" tln ln
T t T t
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
This is possible only if 1 X 1 XA" A" and out outT' T"
B) is the correct answer
Solution: 12)
Process for converting Napthenes and paraffins to Aromatics is Catalytic Reforming.
B) is the correct answer
Solution: 13)
for Turbulent boundary layer
0.2 0.2 0.2
1 0.2
0.8
0.385
Re Rex x
x x
x x
x
x
D) is the correct answer
Solution: 15)
Sphere=D
Cube=D
Cylinder= 2=D
sphere = 1
cube= 0.806
Cylinder =0.873
A) is the correct answer
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Solution: 17)
Relative volatilities of A.B. & C = 8, 2, 1 Arrange the given options into ascending order of
“Reboiler Duty”
Scheme 1
Scheme 2
A
Scheme 3
A
,
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
In Scheme 1
Reboiler Duty (Scheme) is being utilized for the vaporization of B & C in the first column and C
from the second colum.
In Scheme 2
Reboiler Duty is being utilized for the vaporization of B & C from the first and B from the second
column.
In Scheme 3
Reboiler Duty is being utilized for the partial valorization of B & C from the first and B and C
from the second and third column.
Ascending order for Increase in reboiler duty is Scheme 2 < Scheme 1 < Scheme 3
Answer - C
Solution: 18)
2
tan
x
xLim
x
R.H.L
0
tan2
2
h
h
Lim
h
0
coth
2
hLim
h
0
cot h
2
hLim
h
cot 0
2
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
L.H.L
2
0
0
0
tan
tan2
( )2
coth
2
cot
2
cot 0
2
x
h
h
h
xLim
x
h
Lim
h
Lim
h
hLim
h
R.L.S = L.H.S = Ans
Solution: 20)
7 6
0 2A
14 0 14A
Product of Eigen values = Determinant of Matrix =14
Ans. 14
Solution: 21)
Liquid Flow Rate |fully opened value= 150 gal/min
Pressure drop = 20 Psi
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Specific Gravity of the liquid = 0.8
Valve coefficient Cv= ?
We know
Q = Cv VP
G
Cv = ?
Cv = 150
18.9750
0.8V
q
P
G
Answer = 18.97
Solution: 22)
/ 4
f f
f c p n n
i i
in
i
T ps s c R
T p
pR
p
Tf = Ti, free expansion
= 4nR
C) is the correction answer
Solution: 23)
We have
1000 kg, g = 10 m/s100 N2
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
By pascal’s law
1 2
1 2
1 2
2 2
1
2
1 1
100 1000 10
(2)4 4
, 0.04 0.2
P P
F F
A A
d
Thus d d m
Solution: 24)
In Vapor-Liquid equilibrium system, where thermal, mechanical and chemical equilibrium exist, means
thermodynamic equilibrium exist, the criteria of thermodynamic equilibrium says,
(molar cuibbs energy)Liquid phase = (molar cuibbs energy)Vapor phase
D) is the correct answer.
Solution: 25)
Nylon -6 belongs to Ring Opening Polymerization, Polypropylene belongs to Addition and Polyester
belongs to Condensation
D) is the correct answer
Solution: 26)
For 95% removal of solute in absorption using solvent flow rate twice of the minimum solvent flow rate.
Exit composition is equal to______.
Equilibrium relation is y = 2X
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
2 1
2 1
y y 0.02 0.001m 1.9
x x 0.01 0
Also minLsm
Gs (slope of operating line)
minmin
Ls1.9 Ls 95
50 Gs 50 given
Ls operating = 2 Lsmin (Given) = 2 x 95 =190
Lsoperating
Gs=
1903.8
50
1
1
0.02 0.001 Lsoperatingm 3.8
X 0 Gs
1X 0.005
Ans= 0.005
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Solution: 27)
Find the value of 1
2i
We know
/2
1
2( 1/2) 42
cos sin
cos sin2 2
i
i
ii
e i
and
e i i
whichmeans
i e e
1
2 4
1
2
1
2
cos sin4 4
1 1 11
2 2 2
(1 )
2
i
i e i
i i i
ii
A)
Solution: 28)
2( , , ) 3u x y z x xy
u u uu i j k
x y z
= 2 ( 3 ) ( 3 )x i z j y k
(2,0, 4) 4 12 0u i j k
6
i j ka
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Directional Derivate = .u a
= 4 12 16
6.5326 6
Solution: 29)
A
45 0C
x
2
2000 /
1000 /
250 /
?
1 /
h
c
h
c
c J kg k
c J kg k
U w m k
m
m kg s
90 45 45 90( )
2 2 2
AMTD
x xT
( ) 2000 45 1 1000 ( 45) AMTD hQ UA T m x
90250 8 90000 1000( 45)
2
h
xm x
On solving above equations, we get:
0.31 / hm kg s
Solution: 30)
10AO
Pure A
F mol
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THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
To =450K
CPA = 30
10 20, ,
. .PB Pc
J J JC C
mol k mol mol K
= (-45) + (-54) – (-90)
9 /o
rxnH kj mol
K = 0.133 exp 1 1
450
E
R T
31.4 /
8.314 / .
E kJ mol
R J mol K
XA = 0.75 Vm ( ) = ?
FA= FAo ( 1-XA) = 2.5 mol
FB = FBO + . 7.5AO
bF XA mol
a
FC = FCO + . 7.5AO A
cF X mol
a
Energy in – Energy out = Energy due to reaction
1 2 2 0( ) ( )AO PA B PB C PC A PAF C T T F C F C F C T T 0.FAO XA H rxn
210 30 (450 273) [2.5 30 7.5 10 7.5 20] ( 273)T = 7.5 7.5 ( - 9000 )
253100 [300]( 273) 67500T
2
2
53100 67500 300( 273)
675
T
T K
Solution: 31)
We know that
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
4
3
ln
95 100 50ln
125 10010 500
3 100
2.708 3 10
902.68sec
i
T T ht
T T PLc
t
X XX
X t
t onds
Ans. 903 seconds rounded off to the nearest integer.
Solution: 32)
Here it is given for 1st order reaction,
K = 0.1 min-1
PFR in series can be replaced by single PFR of same equivalent volume
Thus, for PFR line
1
,2 ( )AO AO
CA CACA AOP A
nC C
O A CA A
d CV d C
V r k k C
100 1 1
,2 0.1 0.5
,2 14.427 / min
n
o
o
V
V
There for ,1 ,2 100 14.427Vo Vo Vo
,1 85.573Vo
For MFR, ,1
AO A AO A
A
C C C CVm
Vo rA k C
,1 85.573.( ) (1 0.5)
0.1 0.5
855.73
AO A
A
VoVm C C
k C
Vm
Solution: 33)
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
We have
(1 )
1 .5 .19405 .5 .001127
10.24711
g fV m xv x v
m
m kg of mixture
i.e 5.123555 kg of steam and
5.123555 kg of saturated water
Since total volume of 1 m3 remains constant
1 2 2 1 1 2
2 2 1 1
2 1 2 1
2 1
(1 ) u (1 )
( ) ( )
( )( )
(0.1 0.5)(2583.7 762.08) / 10
g f g f
g f
g f
q u u w
x u x x u x u
u x x x x u
x x u u
kJ kg in hrs
Total heat transfer = -728.648 kJ/kg , -ve sign indicates that heat is lost to the surroundings
728.648 10.24711
10
746.65 /
747 / .
Heat lost per hr
kJ hr
kJ hr rounded off to thenearest integer
Solution: 34)
After checking Reynolds no. it is coming out to be 22500 which is above 1000, hence the operating flow
regime is Turbulent. And For Turbulent flow regime,
Power Number = 3 5
P
cN D
180 oC
1 m3
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
1 2
3 5 3 5
1 1 2 2
5 3
2 2 2
1 1 1
2 2
1 1
5 3
2
1
52
1
0.5 , 6000
0.05 , 600
0.5 600
0.05 600
10
P P
N D N D
P D N
P D N
D m N rpm
D m N rpm
P
P
P
P
Solution: 35)
Match the given response functions with the transfer function.
P)
It looks like a first order system with a Transfer function of p1
p1
k
1 S
R)
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
It looks like an inverse response for which 1 1
2 2
k1
k
Q)
Response is going to be unbounded and correct option for this is
1
1
k S 1
S S 1
Also by applying final value theorem, it can be validated.
Value = t s oLim f t Lim f s
d)
It matches with the last option remaining and can be matched by taking the inverse Laplace of the
transfer function.
B) is the correct answer
Solution: 36)
T=800 K
P=2 bar
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THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
CO +H2S COS + H2
0 22972.3 /
8.314 /
G J mol
R J mol
wkt, 0
n
GK
RT
thus, 22972.3 /
8.314 / 800n
J molK
J mol K
K = 0.03162
For ideal gas mixture & standard state to be at 1 bar,
K = Ka = Kp = 0.03612
We know that
Kp = Ky . Pv
Here,
Yco = 2 cos
1, ,
5 5 5Hy y
yH2S = 4
05
v
putting values, 0
55 0.03160 (2)
1 4
5 5
0.2889
Solution: 37)
We have
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THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
(2)
2 2
1 1 2 21 2
5 5
3 3
2 2
10 2 1010
10 10 10 10
pump
Pump
P V P Vy H y
g g g g
H
(Pump)
20 10 10 20
. .
Pump
Pump Pump
H m
Power WQH g Q H
3 2 3
2
10 10 10 / sec
P 1000 10 10 20
2000
Pump
Q AV m
W
Solution: 38)
payback time is defined as the time period at which the investment can be easily (fully) recovered with
the help of cash flow (profit)
1 2
3 3 310 ........
(1 ) (1 ) (1 )n
L L LLakhs
i i i
As interest Rate is given, time value of money concept is used.
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THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
(1 ) 1
(1 )3 (1 ) 1
10(1 )
Pr
3 1.12 110
0.12 1.12
n
n
n
n
n
n
R iP
i iL i
Lakhs wherei i
P esentValue
R uniform periodic payments
LLakhs
0.4(1.12 ) 1.12 1n n
1 0.6(1.12)n
1 1(1.12) ln(1.12)
0.6 0.6
n n lm
1
0.64.507 5
(1.12)
Ln
n yearsLn
Solution: 39)
Find the phase margin for the given block diagram.
Step 1. Find the open loop transfer function.
GOLTF =
Step 2. Identify the system involved in the diagram. One is dead time system with td = 0.1 other
is first order system with p =0.5
Step 3. Finding the phase margin.
180 PM Where is the value of phase at gain crossover frequency.
At Gain Cross over frequency
AR=1
AR of given system = 5 × 1 ×
√
Putting AR=1, w=9.7979 rad/time
= -td w ×
+ tan-1 (-w p )
Phase angle for phase angle for first
Dead time system order system
Putting w=wcog = 9.7979
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
= -56.16156 - 78.46297 = -134.6245
180 PM = 180 - 134.6245
PM = 45.4
Solution: 40)
Fxf + Sys = RxR + EyE
50+0.4=0.2+ 0.4E
Also
F + S = R + E
R + E = 180
R = 180-E
Equation (i) can be written as
50.4 = 0.2(180-E)+ 0.4E
50.4= 36-0.2E + 0.4E
14.4= 0.2E
E=72 and R= 180-E = 180
=
=
Solution: 41)
Corresponding radii of spherical camphor particles are 20 cm and 5 cm.
Mass transfer coefficient 1
( )r t
‘t1’ and ‘t2’ are the times required for the complete sublimation then t1/t2 = ?
We know that, in the case of spherical naphthalene ball in air, flux
(NA)|r = r1 = 1 2
1
AB A A
BLM
P D P P
RT r P
Also, NA = KG PA
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Where KG = 1
AB
BLM
P D
RT r P
Also, time taken for the complete Evaporation of naphthalene Ball
t = 2
1
2
BLM
A AB A
r RT P
M P D P
which means
t= 1
1
1
2 |A r r
r
M NA
t = 1
1
1
r
r
t r13/2
3/2
3/21 1
2 2
1 2
1
2
4 8
8
8
t r
t r
t t
t
t
Solution: 44)
We know that, volume change of mixing,
1 1 2 2
v
( v v )
i iV V x
V V x x
Molar volumes, will be
For component (1), 1
3
1 1 30m /molV x
For component (2), 3
2 2 1 20m /molV x
Thus, 21 2 1 1 2[30 20 (15 7 )]V x x x x x x 1 2(30 20 )x x
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
1 2 1 2(15 7 )V x x x x
1 20.5, 0.5at x x
1 0.5 0.5 0.5 (15 0.5 7 0.5)xV
3
1 0.5 1x
mV
mol
Solution: 45)
We know that
1 d dT Qr 0
r dr dr r
d dT Qr r 0
dr dr r
Integrating, 2dT Qr
r Gdr 2r
When r = 0, G = 0
dT Qr0
dr 2r _______(i)
Integrating again 2
2
QrT C
4r
When
21
2
Q Rr R,T Tw;Tw C
4R
1 2 21Q r R
T Tw Q4R 4R 4R
1 2 2
2
Q R rT Tw 1
4R R
_______(ii)
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
For max temp, dT
0dr
From (i) r = 0
In equation (ii)
1 2Q RT Tw
4R
Q is Heat generated/volume
2
4
3
.1wT 300 10 6000
m 4 300
= 300 + 500 = 0800 C
Solution: 46)
Find the value of minimum reflux ratio when the feed given is saturated liquid and equimolar.
Distillate Mole fraction is given as 0.95 Relative Volatility is 2.
In the condition of minimum reflux rectifying line, Equilibrium curve and feed line intersects at a
point.
Equilibrium Curve is given by y=
As on the feed line, x-coordinate is given as 0.5.
y-coordinate can be calculated from equilibrium curve as both intersects.
Y=
=
=
=
Now, Rectifying Line one end is (0.95, 0.95)and other point is (0.5,
).
Slope=
=
= Rmin = 1.7
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Solution: 47)
We have
12
1 2 1 2
1 2 1 2
2
22 1
1 1 1 1 1 1 1
222 2 2 2 2 2 2
2
1 2
1
2
,2
min
0.006Re
Re.
.
4.
4
1
Q Q Q P P P
DD D D
l Enhancelength
for fully developed la ar flow
l
d
l d
vdl d
l vd
dl V d V V A Q
l V d V V A Qd
Q Q Q
l
l
B) is the correct answer.
Solution: 48)
We know when two dice are thrown, minimum possible sum is 2 and Maximum possible sum is 12.
4, 8, 12 are divisible by 4.
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Number of favorable cases for getting the sum 4
(1, 3), (2, 2), (3, 1)
Number of favorable cases for getting the sum 8
(2,6), (3,5), (4,4), (5,3), (6,2)
Number of favorable cases for getting the sum 12.
(6, 6)
Total number of favorable cases = 9
Probability of getting the sum divisible by 4 = Number of favorablecases
Total number of cases
= 9 3 1
0.2536 12 4
Solution:49)
( ) xf x e x
0 0( )x Given
01 1 '
0
0
0
( )
( )
( 0 1 10
( 1) 1 1 1 2
f xx x Newton Rapson Method
f x
e
e
12 1 '
0
1/2
1/2
( )
( )
( 1/ 2 1 0.6065 0.51/ 2
( 1) 1 2 0.6065 1
f xx x
f x
e
e
0.10650.5 0.5 0.06629
1.6065
0.56629
After 2nd iteration, value of x is 0.5663
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Solution: 50)
(1 3 ) (0) 1
1 3
dyy y
dx
dydx
y
Integrating both sides
1
1
1 1
3
3
2 2
3
2
3
2
22 2
3
1 3
3
1 3 3 3
1 3 3 [ 3 ]
1 3 ,
1 3 , [ ]
1 3
10
3
1
1 (1)1 3 1 2
3
1 2
3
cx
cx
x
x
x
Ln yx c
Ln y x c
Ln y x c let c c
y e e
y e c let c e
c e y
c ey put x
y
cc c
ey
D) is the correct answer
Solution: 51)
Match the column
Sulfite process is used to make pulp which in turn is used to make white paper.
Solvay process is the oldest process to make Soda Ash commercially.
Frasch process is used to produce Sulphur from sulphur ore which in turn is used to make Sulphuric acid.
B) is the correct answer
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Solution: 52)
For First order Rxns, in MFR
For CB max, max
1 1
1. 2 0.4 0.1
k k
max 5min
Thus, for A ( ) ,
o oCA CA CA CA
rA K CA
2 25
0.4 3
CA molCA
CA
For ‘B’ 0
1 2A
CB CB CB
rB k C k CB
5 0.8892
0.4 0.13
CB molCB
CB
Hence, % yield of ‘B’ ' '
100' '
molesof B formeds
total molesof A reacted
= 100o
NB NBo
NA NA
100CB
CAo CA
YB/A = 0.889
100 66.67% 672 0.667
Solution: 53)
Capitalized cost factor = (1 )
[(1 ) 1]
n
n
i
i
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
The net present worth of yearly operating cost R is (1 ) 1
(1 )
n
n
iR
i i
If we convert it into capitalized cost by multiplying with capitalized cost factor, we get R
i
For equipment A: 80000 + 4
80000 20000452376.6
(1 0.1) 1 0.10
For equipment B: 160000 + 160000 15000 160000
310000(1 0.1) 1 0.10 (1.1) 1n n
For B. to be economical cheaper, n = 8 year
Solution: 54)
c (t)
5
0 6t
What will be the mean conversion using saggregation model?
Hence,
22 2
1 1 22 ( )A AA
rA k k k kCC C
Hence for 2nd order Rxn,
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
0
( 1 2)
1 ( 1 2)
oA
k k CA tX
k k CA t
0
0
0 0
1 )
. 1 ) 1 ): 1 )
1 ) 1 )
A A
A
A
X X E t dt
X c t dt c tX E t
c t c t dt
61 2 0
01 2 0
6
0
( )
1 (( )
1 )
AA
A
A
k k C tX
k k C t
C t dt
1 61 2 1 2
0 11 2 0 1 2 0
6
0
( ) 1 ) ( ) 1 )
1 ( ) 1 ( )
1 )
AA A
A
k k C t dt k k C t dtX
k k C t k k C t
C t dt
1 6
0 1
6
0
(0.5) 2 ( ) 0.5 2 ( )
1 0.5 2 1 0.5 2
1 )A
A
tc t dt t c t dt
t t
XC t dt
1 6
0 1
6
0
( ) (( )
1 1
( )A
A
tc t dt t c t dt
t tXC t dt
Here,
C 1t) = 5t for o < t <1
C1t) = (6 – t ) for 1 < t < 6
& 6
0
11 ) 15 ( : 6 5)
2c t dt Area
1 6
0 1
5 (6 )
1 10.6464
15
A
t t dt t t dt
t tX
Thus, % 64.64 65 AX
GATE 2019 CHEMICAL ENGINEERING GATE 2019
THE GATE COACH | 28, JIA SARAI, NEAR IIT, HAUZKHAS, NEW DELHI – 16, Ph. No. (+91) – 9873452122, 9818652587,
Solution: 55)
For Non-Interacting systems
According to the given data, A1 = A2 = 28 cm2
Steady state value of in let flow rate = 16 cm3/sec
Also, ssFo or ssFi =ss1ss 1k h 16 k h
k = 4 (Given) 1ssh 16cm
We also know
1R or
SS
ss1
1
sso
12 h1 2 16kR 2
k 42 hiF h
2 1 2
i ii
h s F s h s
F sF s F s
2
1 1 2 2
R1
A R S 1 A R S 1
Similarly the value of 2R is calculated.
As iss 1ss 2ssF F F 16
2
2i
h s 1 2 2
28 2 S 1 28 2 S 1F s 56s 1
A) Is the correct answer