1. business and industrial
2. energy
3. oil

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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
Answer:
2.
(B) Educate
(D) Extend
(C) -25/49
(D) -49/25
(C)
If logx (5/7) = -1/3, then the value of x is
(A) 343/125
(B) 12/343
Answer:
(C) Extract
(A)
3
Exp:
5
7
7
= x −1 3 ⇒ = x1 3 ⇒   ⇒ x = 2.74
7
5
5
3.
Operators , ◊ and → are defined by : a b =
( 66 6) , ( 66 ◊6) .
(A) -2
Answer:
Exp:
(C) 1
(D) 2
(C)
666 =
66
(B) -1
a−b
a+b
;a ◊ b =
;a → b = ab. Find the value
a+b
a−b
66 − 6 60 5
=
=
66 + 6 72 6
6=
66 + 6 72 6
=
=
66 − 6 60 5
5 6
(66 6)→(66 6) = × = 1
6 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
Answer:
(B) memento
(C) momentum
(D) moment
(B)
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5.
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Choose the appropriate word/phrase, out of the four options given below, to complete the
following sentence:
Frogs ____________________.
(A) Croak
Answer:
(B) Roar
(C) Hiss
(D) Patter
(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 proportion of
the number of faces of the smaller cubes visible to those which are NOT visible.
(A) 1 : 4
Answer:
(B) 1 : 3
(C) 1 : 2
(D) 2 : 3
(C)
Exp:
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Number of faces per cube = 6
Total number of cubes = 9×3 = 27
∴ Total number of faces = 27×6 = 162
∴ Total number of non visible faces = 162-54 = 108
∴
7.
Number of visible faces
54 1
=
=
Number of non visible faces 108 2
Fill in the missing value
1
6
5
4
7
4
7
2
9
2
8
1
4
1
5
2
3
1
2
1
3
3
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Answer:
Exp:
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(3)
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 below is 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
of human progress is that human security largely depends upon environmental security. But on
the contrary, human progress seems contradictory to environmental security. To keep up both at
the required level is a challenge to be addressed by one and all. One of the ways to curb the
climate change may be suitable scientific innovations, while the other may be the Gandhian
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)
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Section Name: Chemical Engineering
Q.No-1-25 Carry One Mark Each
1.
Benzene is removed from air by absorbing it in a non-volatile wash-oil at 100kPa in a countercurrent gas absorber. Gas flow rate is 100 mol/min, which includes 2 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.’
For this process, the value of the absorption factor (upto two decimal places) is _________.
Answer: 1.02
Exp: Absorption factor
A=
liquid
LS
Ls
mG S
(
50 mol
min
)
Given Raoult’s law is applicable
for gas phase
yP = xpsat
⇒ y × 100 = x × 50
⇒ y = 0.5 x
So, m = 0.5
A=
Gas ( 100mol min )
50
= 1.02
0.5 × 98
GS = 98mol min
Benzene = 2mol min
2.
The following set of three vectors
1
 
 2,
1
 
x
3
 
 
 6  and  4 
x
 
 
2
is linearly dependent when x is equal to
(A) 0
(B) 1
(C) 2
(D) 3
Answer: (D)
Exp:
For linearly dependent vectors A = 0
1 x 3
2 6 4 =0
1 x 2
⇒ (12 − 4x ) − x ( 4 − 4 ) + 3 ( 2x − 6 ) = 0
⇒ 2x = 6
⇒ x =3
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3.
For which reaction order, the half-life of the reactant is half of the full lifetime (time for 100%
conversion) of the reactant?
(A) Zero order
Answer:
Exp:
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(B) Half order
(C) First order
(D) Second order
(A)
For zero order reaction
−
dC A
= k C OA = k
dt
Where k = rate constant
CAO − C A = kt
For, full life time CA = 0
t = C AO K
and for half life
C A = C AO 2, So t1/ 2 =
C AO
2k
So, t = 2t1/2
4.
Match the output signals as obtained from four measuring devices in response to a unit step
I
change in the input signal.
Output signal time
II
P: Gas chromatograph, with a long capillary tube
III
Q : Venturi tube
IV
R : Thermocouple with first order dynamics
S : 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:
Exp:
(C)
If y(t) = output (response)
y(t)
( I ) Gas chromatograph
( II ) 1st order
( IV) Venturytube
( III ) 2nd order
Gas chromatograph – IV
time
Ventury tube - I
Thermocouple – II
Pressure transducer with second order dynamics - III
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5.
Match the chemicals written on the left with the raw materials required to produce them
mentioned on the right.
(I) Single Superphosphate (SSP)
(P)
(II) Triple Superphosphate (TSP)
(Q) Brine
(III) Diammonium phosphate (DAP)
(R) Rock Phosphate + Sulfuric Acid
(IV) Caustic soda
(S) Rock phosphate + phosphoric Acid
Rock phosphate + Sulfuric Acid +
Ammonia
(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:
Exp:
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(C)
Single super phosphate (SSP) → Phosphate rock + H 2SO4
Triple super phosphate ( TSP ) → phosphate rock + H3 PO4
Diammonium phosphate (DAD) → Rock phosphate + H 2SO 4 + NH 3
Caustic soda → Brine
6.
 4 3
For the matrix 
 , if
 3 4
Answer:
Exp:
 1
  is an eigenvector, the corresponding eigenvalue is _______.
 1
7
4 3
1
Let A = 
and x =  

3 4
1
Let eigen value is λ
So Ax = λx
1 0 
or [ A − Iλ ][ x ] = 0, where I = 

0 1 
λ 0 
So Ix = 

0 λ
3 
4 − λ
⇒ 
4 − λ 
 3
 4 − λ + 3
⇒ 
=0
3 + 4 − λ 
1
1 = 0

⇒ λ=7
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7.
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The transfer function for the disturbance response in an open-loop process is given by G open
(s) .
d
The corresponding transfer function for the disturbance response in closed-loop feedback control
system with proportional controller is given by G closed
( s ) . Select the option that is Always correct
d
{O G (S) represents order of transfer function G ( s )}:
(A) O G open
( s ) = O G dclosed ( s )
d
(B) O G open
( s ) ≠ O G dclosed ( s )
d
(C) O G open
( s ) ≥ O G dclosed ( s )
d
(D) O G open
( s ) ≤ O G dclosed ( s )
d
Answer: (A)
Exp: Open loop transfer function
G OL = GH = G open
(s)
d
x
and closed loop transfer function
G
G CL =
= G dclosed ( s )
1 + GH
Order of transfer function is order of denominator
Order of GH and 1 + GH are same
G
y
H
so, O  G open
( s ) = O G d closed ( s ) 
d
8.
If v,u, s and g represent 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.
(P)
− ( ∂u ∂v ) s
( Q ) ( ∂g
∂P ) T
( R ) − ( ∂g
( S ) ( ∂u
∂T ) P
∂s ) V
( I)
Temperature
( II )
Pr essure
( III )
V
( 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
and
dg = vdp − sdT
So,
∂u
∂u
= −P ⇒ −
=P
∂v S
∂v S
∂u
∂s
=T
V
∂g
∂g
= V and −
=S
∂p T
∂T P
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9.
Two different liquids are flowing through different pipes of the same diameter. In the first pipe,
the flow is laminar with centerline velocity, Vmax .1 , Whereas in the second pipe, the flow is
turbulent. For turbulent flow, the average velocity is 0.82 times the centerline velocity, Vmax .2 .
For equal volumetric flow rates in both the pipes, the ratio Vmax .1 Vmax .2 (up to two decimal
places) is ______.
Answer: 1.64
Exp: For laminar flow
Vavg.1 = 0.5Vmax.1
For Turbulent flow
Vavg.2 = 0.82 Vmax .2 ( given )
Since volumetric flow rate is some
Q1 = Q 2
A1Vavg.1 = A 2 Vavg.2
Given A 1 = A 2
So, Vavg.1 = Vavg.2
⇒ 0.5 Vmax .1 = 0.82 Vmax .2
⇒
Vmax .1
= 1.64
Vmax .2
dy
+ p ( x ) y = r ( x ) . Functions p ( x ) and r ( x ) are
dx
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
10.
Consider linear ordinary differential equation
y1 + p ( x ) y = r ( x )
Multiplying above equation by integrating factor e ∫
p ( x ) dx
makes the equation exact
A spherical naphthalene ball of 2mm diameter is subliming very slowly in stagnant air at 25OC.
The change in the size of the ball during the sublimation can be neglected. The diffusivity of
naphthalene in air at 25o C is 1.1×10-6 m2/s.
The value of mass transfer coefficient is B × 10-3 m/s, where B (up to one decimal place) is
_______.
Answer: 1.1
Exp: For spherical ball
Dimensionless sherwood number = 2
kCL
=2
D AB
11.
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Where L = characteristic length = diameter of
Spherical ball = d = 2 mm = 2 × 10−3 m
2 × D AB 2 × 1.1 × 10−6
=
d
2 × 10−3
K C = 1.1 × 10−3 m s
KC =
So, B = 1.1
12.
An irreversible, homogeneous reaction A → products, has the rate expression:
Rate =
2C2A + 0.1CA
, where CA is the concentration of A.
1 + 50CA
CA varies in the range 0.5 – 50 mol/m3.
For very high concentration of A, the reaction order tends to:
(A) 0
Answer:
Exp:
(B) 1
(C) 1.5
(D) 2
(B)
Rate =
2CA 2 + 0.1CA
1 + 50CA
0.5 < C A < 50 ( mol m3 )
For very high value of CA ( say 50 mol m 3 )
0.1CA << 2C2A
and 50C A >> 1
So, rate =
2C2A
1
=
CA
50CA 25
So reaction order is one
13.
A scalar function in the xy-plane is given by φ ( x, y ) = x 2 + y 2 . if ˆi amd ˆj are unit vectors in the x
and y directions, the direction of maximum increase in the value of φ at (1,1) is along:
(A) −2iˆ + 2jˆ
Answer:
Exp:
(B) 2iˆ + 2jˆ
(C) − 2iˆ − 2jˆ
(D) 2iˆ − 2jˆ
(B)
Direction of maximum increase
= ∇ φ at (1.1)
(
)
= ( 2iˆ + 2ˆj)
= 2x ˆi + 2yjˆ at (1.1)
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14.
For a gas phase cracking reaction A → B + C at 300OC, the Gibbs free energy of the reaction at
this temperature is ∆G O = − 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
Answer:
Exp:
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(B) 0.50
(C) 0.64
(D) 0.80
(D)
A →B+C
at t = 0, ⊥
0 0
at t = t 1 − α α α
where, α = fraction converted
Equilibrium constant
KP =
YB YC
.P
YA
 α  α 



α2
α2
1 + α  1 + α 
=
=
.P
=
1 − α2
1 − α2
1− α 


 1+ x 
Also
ln k =  −∆G O RT 
 −∆G O 
K = exp 
 : exp
 RT 
K = 1.78
 2750 
 8.314 × 573 


α2
= 1.78
1 − α2
α = 0.80
so,
15.
Two infinitely large parallel plates (I and II) are held at temperatures TI and TII ( T1 > TII )
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:
I
III
T1
(A) 0.5
II
TI1
(B) 0.75
(C) 0.25
(D) 0
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Answer:
Exp:
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(A)
4
4
Q12 σ ( T1 − T2 )
=
We have
1 1
A
+ −1
ε ε
Where ε = emissivity
4
4
Q12 σ ( T1 − T2 )
=
2
A
−1
ε
(i )
( iii )
T1
T3
( ii )
T2
In presence of shield
4
4
4
4
Q13 Q 23 σ ( T1 − T3 ) σ ( T3 − T2 )
=
=
=
2
2
A
A
−1
−1
ε
ε
T 4 + T24
or T34 = 1
2
(
4
4
T14 − ( T14 + T24 ) 2
Q13 σ ( T1 − T3 )
=
=σ
So,
2
2
A
−1
−1
ε
ε
1 Q12
=
2 A
Q13 1
=
So,
Q12 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:
Exp:
)
8079.61
For incipient fluidization
∆P
= g (1 − ε )( ρ P − ρ )
L
Where ε = 0.45
ρP =1500 kg m 3
ρ = 1kg m 3
And L = 1m (height of the bed)
∆P = 9.8 × (1 − 0.45 )(1500 − 1) × 1
= 8079.61Pa
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17.
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 2 bar is 320 K, the
temperature (in K) at which it will boil at 1 bar (up to one decimal place) is ___________.
Answer:
Exp:
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310
Given
dP
= 0.1⇒ dp = 0.1dT
dT
⇒ P = 0.1T + C........... (1)
Given at T = 320K. P = 2bar
2 = 0.1× 320 + C
⇒ C = −30
Equation (1) becomes
P = − 30 + 0.1T
Putting P = 1 bar
1 = −30 + 0.1T
⇒ T = 310 K
18.
For uniform laminar flow over a flat plate, the thickness of the boundary layer, δ, at a distance x
from the leading edge of the plate follows the relation:
(A) δ ( x ) α x −1
Answer:
Exp:
(B) δ ( x ) α x
(C) δ ( x ) α x1 2
(D)
δ ( x ) α x −1 2
(C)
For laminar flow
Boundary layer thickness
8( x ) =
4.99 x
Re
Or, δ ( x ) α x
19.
1
=
4.99x
ρVx
µ
2
Match the polymer mentioned on the left with the catalyst used for it manufacture given on the
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
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Answer:
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(C)
Exp:
20.
LDPE
Traces of oxygen
HDPE
Zeigler – Natta catalyst
PET
Antimony
PVC
Butyl Lithium
Three identical closed systems of a pure gas are taken from an initial temperature and pressure
( TL , P1 ) to a final state ( T2 , P2 ) , each by a different path. Which of the following in Always 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.)
(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.
( P1 . T1 )
P
1
2
( P2 . T2 )
T
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 for
the same throughout.
Answer: (C)
Exp: Steam distillation is used for the mixture immiscible with water.
Option (C) is the wrong statement
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22.
For a binary mixture of components A and B, NA and NB denote the total molar fluxes of
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) N A + N B = 0 and J A + J B ≠ 0
(B) N A + N B ≠ 0 and J A + J B = 0
(C) N A + N B ≠ 0 and J A + J B ≠ 0
(D) N A + N B = 0 and J A + J B = 0
Answer:
23.
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(A)
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 ∂u
=
+i
dz ∂y ∂y
(B)
df ∂u
∂v
=
+i
dz ∂x
∂x
(C)
df ∂v ∂u
=
−i
dz ∂y ∂y
(D)
df ∂v ∂v
=
+i
dz ∂y ∂x
Answer:
Exp:
(A)
f ( z ) = u ( x, y ) + iv ( x, y )
f '( z ) =
df ∂u
∂v ∂v ∂u
∂v
=
+i
=
=
+i
dz ∂x
∂x ∂x ∂y
∂y
- (i)
For analytic function
∂u ∂ υ
∂u −∂v
=
and
=
∂x ∂y
∂y ∂x
So, equation ( i ) can be written as
df ∂υ ∂v ∂v ∂u ∂u
∂v
=
+i
=
−i
=
+i
dz ∂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:
Exp:
(B)
Catalyst changes the activation energy of the reaction
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Match the technologies in Group 1 with the entries in Group 2:
Group – 1
Group 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) Gas – solid non-catalytic
reaction
(A)
(C)
Answer:
Exp:
P − I, Q − V, R − II, S − VI
(B)
P − IV, Q − I, R − III, S − II
P − IV, Q − VI, R − I, S − II
(D) P − V, Q − VI, R − IV, S − II
(B)
P : Urea manufacture → IV prilling tower
Q : coal gasification → VI Gas solid non − catalytic reaction
R : controlled release of chemical → I :micron capsulation.
S : Deep hydrodesulphurization → II ultra low sulphur diesel.
Q.No-26-55 Carry Two Marks Each
26.
Consider a solid block of unit thickness for which the thermal conductivity decreases with an
increase in temperature. The opposite faces of the block are maintained at constant but different
temperatures: T(x = 0) > T(x = 1). Heat transfer is by steady state conduction in x-direction only.
There is no source or sink of heat inside the block. In the figure below, identify the correct
temperature profile in the block.
T ( x = 0)
I
II
T
III
T ( x = 1)
0
(A) I
IV
x
(B) II
1
(C) III
(D) IV
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Exp:
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(C)
Let T ( x = 0 ) T1
and T ( x = 1) = T2
we have from Fourier’s law
q
dT
=− k
A
dx
At steady state conduction T1>T2
−k
dT
q
= const =
dx
A
⇒
q
dx = − kdT
A
____(1)
Here K is decreasing with temperature
Let K = K O (1 − αT )
⇒
T
q x
dx = − ∫ − k o (1 − αT ) dT
∫
T1
A 0
q
α
x = − k O ( T + T1 ) + ( T 2 − T12 )
A
2
____(11)
Similarly integrating equation (1) between x = 0 & x = 1
x =1
q
A
∫
⇒
q
α
(1) = − k o ( T2 − T1 ) + ( T22 − T12 ) − − − (iii)
A
2
x =0
dx = ∫ − k O (1 − αT ) dT
T2
T1
Dividing equation (ii) and (iii)
α ( T 2 − T12 ) − 2k o ( T − T1 ) = x {αT22 − T12 } − 2k 0 ( T2 − T1 )
From the above expression for temperature profile it is clear that the temp passes through a
minima
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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 is A → 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 ________.
0.3F
1
F
Answer:
Exp:
A
C
P
59.19
Basis: -
0.3F
100 mole of feed F
A in feed = 96 mol
Inert in feed = 4mol
Recycle = 0.3F = 30 mole
(1)
F
A→C
Product (P) Contains 50%C
( 2)
P
At steady state F = P = 100 mole
Product Contains 50 mole C
Product must also contain 4% inert
At point (2) total molar flow rate =100 + 30 = 130mole
 50 mole C

Product stream P is  46 mole A
 4 mole C

 15 mole C

Recycle stream R is 13.8 mole A
 1.2 mole I

Conversion of A in the reactor
=
=
A converted int o C at po int ( 2 )
A fed at point A
65
× 100 = 59.19%
( 96 + 13.8)
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The impulse response to a tracer pulse experiment for a flow reactor is given below:
t ( min )
6.0
C ( mol 1)
4.0
2.0
0.0
0.0
2.0
6.0
4.0
In the above figure, c is the exit tracer concentration. The corresponding E or E θ (normalized E)
curve is correctly represented by which of the following choices? Here, θ is dimensionless time.
(A)
E 1 ( min −1 )
0.30
(C)
2.0
4.0
t ( min )
θ
(− )
1.00
1.0
θ( −)
3.0
2.0
(D)
6.00
0.40
E 1 ( min
0.20
0.00
0.0
2.00
0.00
0.0
6.0
0.60
E 1 ( min −1 )
29.
E
0.10
0.00
0.0
Answer:
(B)
0.30
2.0
4.0
t ( min )
6.0
−1
)
4.00
2.00
0.00
0.0
6.0
2.0
4.0
t ( min )
(C)
Consider a steady state mass transfer process between well-mixed liquid and vapour phases of a
binary mixture comprising of components A and B. The mole fractions of component A in the
bulk liquid (xA) and bulk vapour ( y A ) 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. The vapour-liquid equilibrium can be approximated as y*A = 2x A for x A less than
0.4. The mole fraction of A in the liquid at the interface (up to two decimal places) is
Answer: 0.08
Exp: Given x A = 0.36
y A = 0.16
k L = 0.1mol m 2 sec
k g = 0.05mol m 2s
y*A = 2x A
for x A < 0.4
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So y Ai = 2x A ……….(i)
At steady state, mass flux is constant
k L ( x A − x Ai ) = k g ( y Ai − y A )
0.1( 0.36 − x Ai ) = 0.05 ( 2x A − 0.16 )
0.11( 0.36 − x Ai ) = 0.05 ( 2 × 0.36 − 0.16 )
⇒ x Ai = 0.08
30.
A heated solid copper sphere (of surface area A and volume V) is immersed in a large body of cold
fluid. Assume the resistance to heat transfer inside the sphere to be negligible and heat transfer
coefficient (h), density ( ρ ) , heat capacity (C), and thermal conductivity (k) to be constant. Then,
at time t, the temperature difference between the sphere and the fluid is proportional to:
 h

(A) exp  − A t 
 ρVC 
 ρVC 
t
(B) exp  −
 hA 
 4πk 
(C) exp  −
t
 ρCA 
 ρCA 
(D) exp  −
t
 4πk 
Answer:
Exp:
(A)
T = f(time)
Let apply first law of thermodynamics. If at any instant of time T. if T is body temperature then.
hA ( T∞ − T ) = ρ C P V
dT
dt
Where CP = heat capacity of the body
T
t
dT
hA
∫T T∞ − T = ∫0 ρVCP dt
0
⇒ ln
T − T∞
hA
=−
t
To − T∞
ρVC P
 −hA 
⇒ T − T∞ = e 
t
 ρVC 
An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressed
to 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volume
of 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is
____________
Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5R
Answer: 750
Exp: 1 → 2 : Re versible adiabatic process
31.
2 → 3 : Re versible isobaric process
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Given P1 = 0.1MPa
=105 Pa
V1 = 2m3
3
P
P2 = 10 Pa = P3
6
2
V3 = 0.2 m 3
CP = 2.5R So CV = 1.5R
2.5
=1.667
1.5
For process 1 – 2: P1V1γ = P2 V2γ
γ = CP CV =
1
P  γ
 105 
V2 =  1  .V1 = 2 ×  6 
 10 
 P2 
⇒ V2 = 0.50 m 2
V
1
1.667
Work done on gas for process 1 – 2
W1− 2 =
P1V1 − P2 V2 0.1 × 106 × 2 − 106 × 0.5
=
γ −1
0.667
⇒ W1− 2 = −450kJ
Work done for process 2 – 3
W2− 3 = P∆V
=106 × ( 0.2 − 0.5 )
= − 300kJ
Net work done = 450 + 350 = 750kJ
A centrifugal pump delivers water at the rate of 0.22 m3/s from a reservoir at ground level to
another reservoir at a height H, through a vertical pipe of 0.2m 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%.
Data:
Fanning friction factor for pipe flow is f = 0.004. Neglect other head losses
Take gravitational acceleration, g = 9.8 m/s2 and density of water is 1000kg/m3.
The height H, in meters, to which the water can be delivered (up to one decimal place) is _______.
Answer: 36
Exp: Given Q = 0.22 m 2 sec
32.
V2 =
0.22
π
2
× ( 0.2 )
4
W = 90 kW
= 7m sec
W in head unit =
90 × 103
ρQg
Applying Bernaulli’s equation between (1) and (2)
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ρ1
v2
z1 + 1 + hf + ηw
ρg
2g
=
ρ2
V2
+ z 2 + 2 − − − (1)
ρg
2g
and Z2 = H
hf =
4fHV 2 4 × 0.004 × H × 7 2
=
= 0.2 H m
2gD
2 × 9.8 × 0.2
90 × 103
49
=H+
0.22 × 1000 × 9.8
2 × 9.8
⇒ 0.2H + 31.308 = H + 2.5
(1) ⇒ 0.2 H + 0.75 ×
⇒ 0.8H = 28.808
H = 36.01m
So, H = 36m
33.
A multi-stage, counter-current liquid-liquid extractor is used to separate solute C from a binary
mixture (F) of A and C using solvent B. Pure Solvent B is recovered from the raffinate R by
distillation, as shown in the schematic diagram below:
F
R
P
extractor
distillation
E
B
B
Locations of different mixtures 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 interects the binodal curve at U and T. The lines P∆
and FB intersect the bimodal at V and W respectively.
C
F
E
P
V
U
∆
A
B
The raffinate coming out of extractor is represented in the diagram by the point:
(A) T
Answer:
(B) U
(C) V
(D) W
(B)
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The solution of the non-linear equation
x 3 − x = 0 is to be obtained using Newton-Raphson method. If the initial guess is x = 0.5, the
method converges to which one of the following values:
(A) -1
(B) 0
(C) 1
(D) 2
Answer: (A)
Exp:
f ( x ) = x 3 − x, f ' ( x ) = 3x 2 − 1
Initial guess = 0.5 = x0
First iteration
x1 = x 0 −
f ( x0 )
f '( x 0 )
( 0.5) − 0.5 = 0.5 − ( −0.375)
⇒ x1 = 0.5 −
2
( −0.25)
3 ( 0.5 ) − 1
3
⇒ x1 = −1
Solution to above equations are 0.1, - 1 but after first iteration it is converging to – 1
35.
An isothermal steady state mixed flow reactor (CSTR) of 1m3 volume is used to carry out the
first order liquid-phase reaction A → products. Fresh feed at a volumetric flow rate of Q
containing reactant A at a concentration CA0 mixes with the recycle steam at a volumetric flow
rate RQ as shown in the figure below.
Q = 0.5m 3 min
C A0 = 1mol m 3
Q = 0.5m 3 min
C
X
A
A
=
C
=
X
A f
A f
RQ
It is observed that when the recycle ratio R = 0.5, the exit conversion X Af = 50% When the
recycle ratio is increased to R = 2, the new exit conversion (in percent) will be:
(A) 50.0
(B) 54.3
(C) 58.7
(D) 63.2
Answer: (A)
Q
Exp: Mole balance
Input – Output
disappearance = Accum.
CA0 =
1mol
m3
At steady state,
Q
( QCA0 + RQCA ) − ( RQ + Q ) CA = kCA V
⇒ QC A 0 − QC A = kC A V
RQ
CA = CAf
x A = x Af
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⇒ CA 0 − C A = kC A τ
or τ =
C A0 − C A
kC A0 (1 − x )
or τ =
xA
x
⇒ τk = A
k (1 − x A )
1− xA
τk
1 + τk
So, conversion xA is independent of Recycle ratio R
So final conversion = 50%
Or , x A =
36.
A typical batch filtration cycle consists of filtration followed by washing, One such filtration unit
operating at constant pressure difference first filters a slurry during which 5 liters of filtrate is
collected in 100 s. This is followed by washing. Which is done for tw seconds and uses 1 liter of
wash water. Assume 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.
∆P
dv
= k1
; L = k 2 V, if L is changing
L
dt
k1 and k 2 can be taken to be constant during filtration and washing. The wash time tw, in seconds
(up to one decimal place) is __________
Answer: 40
∆P
dV
Exp: Given
= k1
and L = k 2 V
L
dt
so
dv 1 ∆P
=
dt k1 k 2 V
5
⇒ ∫ vdv =
0
∆P
k1k 2
∫
100
0
⇒
∆P
25
=
× 100
2 K1 K 2
or
∆P 1
=
k1k 2 8
dt
Also, Final rate of filtration
dV
∆P
1
=
=1
=
8
×
5
dt k1k 2 V
40
washing time t w =
tw =
volume of wash water used
finalrate of filtration
1
= 40sec
1
40
( )
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37.
Which one of the following transfer functions, upon a unit step change in disturbance at t = 0, will
show a stable time domain response with a negative initial slope (i.e., slope at t = 0):
1
2
1
2
(A) G ( s ) =
(B) G ( s ) =
−
+
s +1 s + 4
s +1 s + 4
1
2
1
2
(D) G ( s ) =
(C) G ( s ) =
+
+
s +1 s − 4
s −1 s − 4
Answer: (A)
1
2
Exp: From option (A) G ( s ) =
−
s +1 s + 4
so y ( s ) =
1 1
2 
−


s  s +1 s + 4 
d {sy ( s )} =
dy
= e − t − 2e −4t
dt
dy
= −1
dt z = 0
So (A) has –ve initial slope
38.
Given that molar residual Gibbs free energy, g R , and molar residual volume VR, are related as
R
p v 
gR
R
O
=∫ 
 dP, find g at T = 27 C and P = 0.2 MPa. The gas may be assumed to follow the
0
RT
 RT 
viral equation of state, Z=1+BP/RT,where B= -10-4 m3/mol at the given conditions
(R = 8.314J/mol.K). The value of gR in J/mol is:
(A) 0.08
(B) −2.4
(C) 20
(D) −20
Answer:
Exp:
(D)
Given
R
PV
gR
=∫
dP
Rt 0 RT
We know, V R = V − V ig =
or
ZRT RT
RT
−
= ( Z − 1)
P
P
P
V R ( Z − 1)
=
RT
P
And Z = 1 +
So,
( Z − 1) = B
BP
⇒
RT
P
RT
P B
gR
=∫
dP
0
RT
RT
g R = B ( P − 0 ) = −10−4
m3
× 0.2 × 106 Pa
mol
g R = −20 J mol
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A spherical solid particle of 1mm diameter is falling with a downward velocity of 1.7mm/s
through a liquid (viscosity 0.04 Pa.s) at a low Reynolds number (Strokes regime). The liquid is
flowing upward at a velocity of 1 mm/s. All 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 _________
1.7 mm s
1mm s
40.
In the figure below, the temperature profiles of cold and hot fluids in counter current double pipe
heat exchanges (in different mode of operation) are shown on the left. For each case, match the
heat exchange process for the fluid represented by the bold curve with the options given on the
right.
( I)
(P)
Heating of sub − cooled feed
to sup er heated vapour
Temperature
Dis tan ce
( II )
(Q)
Condensation of sup er heated vapour
(R )
Boiling of sub − cooled liquid
Temperature
Dis tan ce
( III )
Temperature
Dis tan ce
( IV )
( S)
Condensation of Saturated vapour
followed by sub − cooling
Temperature
Dis tan ce
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(A) I − P, II − Q, III − R, IV − S
(B) I − P, II − Q, III − S, IV − R
(C) I − Q, II − P, III − S, IV − R
(D) I − Q, II − S, III − P, IV − R
Answer:
Exp:
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(C)
(P) Heating of sub cooled feed to super heated vapour
( 3)
(1)
( 2)
(1) Sub cooled to saturated
(2) Saturated up to boiling point
(3) Saturated to superheat
(Q) Condensation of superheated vapour.
(1)
( 2)
(1) Cooling of superheated vapour to saturated vapour
(2) Condensation of saturated vapour to liquid.
(R) Boiling of sub-cooled liquid
(1)
( 2)
( 2)
(1)
(1) Heating of sub cooled liquid to saturated liquid
(2) Boiling of saturated liquid
(S). Condensation of saturated vapour followed by sub-cooling
(1)
( 2)
(1) Condensation of saturated vapour to saturated liquid
(2) Sub cooling of saturated liquid.
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The block diagram for a process with feedback control for output deviation variable h is shown in
the figure below. All transfer functions are given with pre-factor of s in minutes. A unit step
change is made in the set-point at t = 0. The time required for h to reach 50% of its ultimate value,
in minutes (up to two decimal places), is: _____________.
1
Gd (s ) =
0.5s
h sp ( s )
G c ( s ) = 0.5s
+
Gf (s ) =
0.2
1.5s + 1
Gp (s) =
h (s)
1
0.5s
G m (s) = 1
Answer:
Exp:
0.8664
We have by making G d ( s ) = 0 (no load disturbance)
0.2
0.2
= 1.5s + 1 =
0.2
h sp ( s )
+ 1 1.2 + 1.5s
1.5s + 1
h (s)
Given h sp ( s ) = 1
s
1
0.2
A
B
So, h sp ( s ) = .
= +
s 1.5s + 1.2 S 1.5S + 1.2
1
−1
A = . and B =
6
4
1 1
1
So, h(s) = − .
6s 4 1.5s + 1.2
Taking inverse laplace transformation
h (t) =
1 1 −0.8t
− e
6 6
Ultimate value
= h (∞ ) =
1
6
Let t = time required to reach 50% of ultimate value
1 1 1 −0.8t
−1 −1 −0.8t
= − e
⇒
= e
12 6 6
12 6
0.693
⇒ e −0.8t = + 1 ⇒ t =
= 0.8664sec
2
0.8
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42.
Adsorption on activated carbon is to be used for reducing phenol concentration in wastewater
from 0.04 mol/1 to 0.008mol/1. The adsorption isotherm at the operating temperature can be
expressed as q = 0.025C1/3; where q is the phenol concentration in solid (mol/g solid) and C is the
phenol concentration in water (mol/1). The minimum amount of solid (in grams) required per liter
of wastewater (up to one decimal place) is ______________________
43.
Consider a control system with the open loop transfer function given by:
K e e −0.3c
1.5s + 1
In the above function, pre-factor of s is in minutes and KC is the gain of proportional controller.
G OL ( s ) =
The frequency for phase margin of 30O is 40.04rad/min. The value of KC for a gain margin of 1.7
(up to one decimal place) is _______
Answer:
44.
5.019
For complex variable Z, the value of the contour integral
1
2πi
e −2z
∫c z ( z − 3) dz along the clockwise
contour C: z = 2 ( up to two decimal places ) is __________
Answer:
Exp:
1
2πi
-0.33
e −2z
∫c z ( z − 3)
C: z = 2 − 2 < z < 2
Poles are z = 0.3
Z = 3 is outside the countour C
R (o) =
e −2×0
1
= − = −0.33
( 0 − 3) 3
Value of integral = -0.33
45.
A proposed chemical plant is estimated to have a fixed capital (FC) of Rs. 24 crores. Assuming
other costs to be small, the total investment may be taken to be same as FC. After commissioning
(at t = 0 years), the annual profit before tax is Rs.10 crores/year (at the end of each year) and the
expected life of the plant is 10 years. The tax rate is 40% per year and a linear depreciation is
allowed at 10% per year. The salvage value is zero. If the annual interest rate is 12% the NPV
(net present value or worth) of the project in crores of rupees (up to one decimal place) is
_______.
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46.
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For fanning friction factor f (for flow in pipes) and drag coefficient CD(for flow over immersed
bodies), which of the following statements are true?
P: f accounts only for the skin friction
Q: CD accounts only for the form friction
R: CD accounts for both skin friction and form friction
S: Both f and CD depend on the Reynolds number
T: For laminar flow through a pipe, f doubles on doubling the volumetric flow rate.
(A) R,S,T
Answer:
Exp:
(B) P,Q,S
(C) P,R,S
(D) P,Q,S,T
(C)
Correct Statements are
P : f accounts only for the skin friction
R : CD accounts for both skin friction and form friction
S : Both f and CD depends on the Reynolds number
47.
A binary feed consisting of 25 mol% liquid and 75 mol% vapour is separated in a staged
distillation column. The mole fraction of the more volatile component in the distillate product is
0.95. The molar flow rate of distillate is 50% of the feed flow rate and McCabe-Thiele method
can be used to analyze the column . The q-line intersects the operating line of the enriching
section at (0.35,0.5) on the x-y diagram. The slope of the stripping section operating line (up to
one decimal place) is ________
Answer:
Exp:
1.4
Given q = 0.25
X D = 0.95
Equation of operating line for rectifying section is
y n +1 −
x
R
xn + D
R +1
R +1
V
Since this line passes through (0.35, 0.5).
0.5 =
R
0.95 0.35R + 0.95
× 0.35 +
=
R +1
R +1
R +1
⇒ 0.5R + 0.5 = 0.35R + 0.95
⇒ 0.15R = 0.45
⇒R = 3
L
Reflux ratio = O ⇒ L O = 3D
D
A (∞) F = D + w
n.mm
LO
D
F
m
m +1
W
⇒ 2D = D + W
W = D and F = 2D
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For stripping section
Wx w
L
y m +1 = m x m −
Vm +1
Vm +1
In stripping section Lm = Ln + qF
= 3D + 0.25 * 2D
Lm = 3.5D
and Vm +1 = Lm − W = 3D − 0.5D = 2.5D
Slope of operating line in stripping
Lm 3.5
Section =
=
= 1.4
Vm +1 2.5
48.
A catalyst slab of half-thickness L (the width and length of the slab>> L) is used to conduct the
first order reaction A → 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
1
of diameter x in the sample is proportional to
. The average diameter, in microns, (up to
50 + x
one decimal place) is ________________
50.
Consider two steady isothermal flow configuration shown schematically as Case I and Case II
below. In case I, a CSTR of volume V1 is followed by a PFR of volume V2, while in Case II a
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.
Q
C A0
Q
Q
V1
Q
Q
C A0
I
CAf
V2
Q
V2
V1
C Af II
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Choose the correct option:
(A)
(C)
Answer:
Exp:
CIA
C
II
Af
CIA
C
> 1 for n = 1
f
(B)
< 1 for n = 1
f
II
Af
(D)
CIA
C
CIA
C
f
II
Af
f
II
Af
= 1 for n = 1
= 1 for n > 1
(B)
We can solve problem by taking 1st order reaction. (n = 1)
For Case I,
Q
C A0
Q
V1
C A1
V2
Q
C Af
Consider V1 = V2 =1m , k = 1.sec−1
3
Q =1m 3 sec, and C AO = 1mol m 3
For CSTR, τk =
1×1 =
C A0 − C A1
C A1
1 − C A1
C A1
⇒ C A1 = 0.5 mol m3
For PFR.
τ= ∫
C Af
C A1
⇒1 = ln
dC A
kC A
C Af
⇒ C Af = 0.184 mol m 3
C A1
Case – II
Q
CA0
V2
Q
CA1
V1
Q
CAf
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With same data,
For PFR
τ = − ln
C A1
=1
C A0
⇒ C A1 = 0.3680 mol m 3
and for CSTR,
τK = 1 =
C A1 − C Af 0.3680 − C Af
=
C Af
C Af
CAf = 0.1840 mol m3 .
Clearly for both the case CAf is equal
i,e
51.
C IAf
= 1 for n = 1
CIIAf
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:
Exp:
(D)
The correct sequence is
High Temp
Reactor
Inter
Low temp
Cooler
reactor
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
I=
∫ u.dr
c
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 ___________.
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Answer:
Exp:
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8
Given u = −2y ˆi + 2xjˆ
dr = dxiˆ + dy ˆj
So,
∫ u.dx = − ∫ 2y.dx + ∫ 2x.dy − (1)
c
c
c
Now, for A → B y = 1, dy = 0
x changes from 1 to 3
(1) ⇒ ∫ u.dr = − ∫1 2 × dx = −4
3
c
For B → C, x = 3, dx = 0, y changes from 1 to 2
(1) ⇒ ∫ u.dr = ∫1 2 × 3dy = 6
2
c
For C → D, y = 2, dy = 0, x changes from 3 to 1
(1) ⇒ ∫ u.dr = − ∫3 2 × 2dx = 8
1
C
For D → A, x = 1, dx = 0, y changes from 2 to 1
(1) ⇒ ∫ u.dr = ∫ 2x1dy = −2
C
For whole loop
2
∫ u.dr = − 4 + 6 + 8 − 2 = 8
C
53.
A binary mixture of components (1) and (2) forms an azeotrope at 130OC and x1 = 0.3. The liquid
phase non-ideality is described by ln γ1 = Ax 22 and ln γ 2 = Ax12 , where γ1 , γ 2 are the activity
coefficients, and x1 , x 2 are the liquid phase mole fractions. For both components, the fugacity
coefficients are 0.9 at the azeotropic composition. Saturated vapor pressures at 130OC are
P1sat = 70 bar P2sat = 30 bar.
The total pressure in bars for the above azeotopic system (up to two decimal places) is
Answer:
Exp:
27.54
Given x1az = 0.3 = y1az ( At a zeotrope x az = y1az )
P1sat = 70 bar, P2sat = 30 bar
ln γ1 = Ax 22
and l n γ 2 = Ax12
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So ln
γ1
= A ( x 2 2 − x12 ) = A(x 2 − x1 ) ( x 2 + x1 )
γ2
γ1
= A (1 − 2x1az )
γ2
ln
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_____(1)
Also
y1az φ1P = x1az γ1P1sat
______(2)
y az2 φ 2 P = x az2 γ 2 P2sat
______(3)
We know x1az = y1az and yaz2 = x az2
So, (2)/(3)
⇒
φ1
γ Psat
= 1 = 1 1sat
φ2
γ 2s 2
⇒
γ1 P2sat 30 3
=
= =
γ 2 P1sat 70 7
From (1)
ln
3
= A (1 − 2 × 0.3)
7
A = − 2.1182
So ln γ1 = Ax 22 = −2.1182 × ( 0.7 )
2
γ1 = 0.3542
and ln γ 2 = Ax12 = −2.1182 × ( 0.3)
2
γ 2 = 0.8264
Adding equation (2) and (3) ( φ1 = φ2 ≠ φ ( say ) )
φ2 p + φ1p = γ1p1sat + γ 2 psat
2
2φp = 0.3542 × 70 + 0.8264 × 30
P = 27.54 bar
54.
Air is flowing at a velocity of 3m/s perpendicular to a long pipe as shown in the figure below. The
outer diameter of the pipe is d = 6 cm and temperature at the outside surface of the pipe is
maintained at 100o C. The temperature of the air far from the tube is 30OC.
Data for air Kinematic Viscosity, V = 18 × 10−6 m 2 s; Thermal conductivity, k = 0.3 W/(m.K)
Using the Nusselt number correlation : Nu =
hd
= 0.024 × Re 0.8 , the rate of heat loss per unit
k
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Length ( W m ) from the pipe to air (up to one decimal place) is ____________________
Surface temperature 100O C
6cm
Air velocity 3m s, Temperature 30O C
Answer: 250.945
Exp:
Given v = 3m s
d = 6 × 10−2 m
υ = 18 × 10−6 m 2 sec
k = 0.03 w mk
Given
Nu =
hd
= 0.024 R 0.8
e
k
Re =
ρVd Vd 3 × 6 × 10−2
=
=
= 10,000
µ
υ
18 × 10−6
hd
0.8
= 0.024 × (10,000 ) = 38.037
k
0.03 × 38.037
⇒h=
=19.0187 w m 2 k
6 × 10−2
So,
Rate of heat loss/length
= h.( πd )( TS − Ta )
= 19.0187 × π × 6 × 10−2 × (100 − 30 )
= 250.945 w m
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55.
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The cost of two independent process variable f1 and f2 affects the total cost CT (in lakhs of
rupees) of the process as per the following function:
CT = 100f1 +
1000
+ 20f 22 + 50
f1f 2
The lowest total cost CT, in lakhs of rupees (up to one decimal place), is ________.
Answer:
Exp:
572.8
Given CT =100f1 +
1000
20f 22 + 50
f1f 2
∂CT
1000
= P =100 =
∂f1
f 2 f12
and
∂CT
−1000
=q= 2
+ 40f 2
∂f 2
f 2 f1
For maxima of minima P = 0, q = 0
100 =
1000
1000
and
= 40f 2
2
f 2 f1
f1f 22
⇒ f12 f 2 = 10............. (1)
⇒ f1 f 23 = 25............(II)
Dividing equation (ii) by equation (1)
( ii ) ( i )
⇒
f1f 23
= 2.5
f12 f 2
⇒
f 22
f2
= 2.5 ⇒ f1 = 2
f1
2.5
From (1)
⇒
f 24
( 2.5)
2
f 2 = 10
⇒ f 25 = 62.5 ⇒ f 2 = 2.2865
so, f1 =
( 2.2865)
2
= 2.091
2.5
f1 = 2.091, f 2 = 2.2865
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Now,
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∂ 2 CT +1000 × 2 2000
=
= 3 =r
∂f12
f 2 f13
f1 f 2
S=
∂ 2 CT
1000
=+ 2 2
f1 f 2
∂f1∂f 2
t=
2000
∂ 2 CT
= 40 + 3
2
f 2 f1
∂f 2
rt − s 2 > 0 and r > 0
So, ( f1 , f 2 ) is point of minima
Lowest cost
C T = 100 × 2.091 +
1000
+ 20 × 2.8652 + 50
2.091 × 2.2865
CT = 572.82lakhs
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